text stringlengths 83 79.5k |
|---|
H: Maximum Entropy Distribution When Mean and Variance are Not Fixed with Positive Support
I know when the mean and variance of $\ln x$ are both fixed, then the maximum entropy probability distribution is lognormal. When the mean of a random variable is fixed the MEPD is the exponential distribution. My question is, what is the MEPD in the continuous case when neither the mean or variance are fixed with support on $[0, \infty)$?
AI: Discrete case
In the discrete case you need to consider the functional
$$H[p]=-\sum_{i=1}^n p_i \ln(p_i)+\lambda(\sum_{i=1}^n p_i-1)$$
as we consider a single constraint.
Setting $\frac{\partial H[p]}{\partial p_i}=0$ for all $i=1,\dots,n$ we arrive at
$$-\ln(p_i)-1+\lambda=0\Leftrightarrow p_i=e^{\lambda-1}.$$
Imposing $\sum_{i=1}^n p_i-1=0$, one gets
$\lambda=1-\ln(n)$, or $p_i=e^{1-\ln(n)-1}=\frac{1}{n}$.
In summary, the wished distribution is the uniform probability distribution.
Continuous case
The continuous case needs more care, due to non trivial integration range.
We want to maximize the functional
$$H[p]=-\int_{0}^{\infty}p(x)\ln(p(x))dx+\lambda(\int_{0}^{\infty}p(x)dx-1), $$
where $p$ has support $[0,\infty]$ and $p(0)=p(\infty)=0$. We apply the calculus of variations by considering any distribution $\phi$ s.t. $p(0)=p(\infty)=\phi(0)=\phi(\infty)$. We compute the variation
$$\frac{\delta H}{\delta\phi}|_{p}=\lim_{\epsilon\rightarrow 0} \frac{H[p+\epsilon\phi]-H[p]}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left[\int_{0}^{\infty}\left(F(p+\epsilon\phi,x)-F(p,x)\right)dx+
\lambda(\int_{0}^{\infty}\epsilon\phi dx)\right],$$
where $F(p,x)=-p(x)\ln(p(x))$ and $F(p+\epsilon\phi,x)=-(p(x)+\epsilon\phi)\ln(p(x)+\epsilon\phi)$.
Using
$$F(p+\epsilon\phi,x)-F(p,x)=\epsilon\phi\frac{\partial F}{\partial p}(p,x)+O(\epsilon^2)$$
we have
$$\frac{dH}{d\phi}|_{p}=\int_{0}^{\infty}\left(\frac{\partial F}{\partial p}(p,x)+\lambda\right)\phi dx$$
where $\frac{\partial F}{\partial p}(p,x)=-\ln(p(x))-1$. In summary
$$-\ln(p(x))-1+\lambda=0 $$
or $p(x)=e^{\lambda-1}$, with $\int_0^{\infty}e^{\lambda-1}dx=1,$ which is not possible.
Roughly speaking, the absence of additional constraints like the fixed mean one
$$\int_{0}^{\infty} xp(x)dx=\mu$$
does not allow to arrive at "more interesting" differential equations for $p(x)$. Note that the $F$ does not depend on $p'(x)$: this leads to the simplified Euler Lagrange equation
$$\frac{\partial F}{\partial p}(p,x)+\lambda=0.$$ |
H: Mathematical news sources.
I'm studying my high school right now but I really like math and it would be great for me if I could find a place where I can find about what is going on in the math world nowadays. About a year ago I was subscribed to the scientific american magazine and one time they put an article in there about Terrence Tao http://adsabs.harvard.edu/abs/2012arXiv1201.6656T finding this proof to a weak version of the Goldbach conjecture.
Any good places where I can get a grasp of current mathematical research and what is being done right now? I know about papers and journals but I would like to get a wider picture without needing to be an expert in every field.
AI: There's an extensive mathblog circuit. While some blogs maintain a high level of rigor, I'd say that most intermix deep math, general thoughts and ramblings, and recent events of interest.
I'd like to recommend 4:
Firstly, the conglomerator mathblogging.org is a big compendium of bloggers that is somewhat differentiated based on the skill level and content.
Secondly, I recommend Terry Tao's blog, although much of the time, Terry talks about his own research which will probably fly above most audiences.
I also recommend Timothy Gower's blog, which tends to be almost entirely math news.
If you'll forgive the self-plugging, I also think my blog is pretty approachable, and I try to talk about things of interest to the greater math community some times.
Also, perhaps surprisingly, many mathematicians are very active of GooglePlus. Really, I'm serious. |
H: Conversion of $F(x) = {1 \choose 0} (2t^2 - 3t + 1) + {3 \choose 1} (4t - 4t^2) + {1 \choose 2} (2t^2 - t)$
I have a textbook with a calculation step that is pretty unclear to me:
$$F(x) = {1 \choose 0} (2t^2 - 3t + 1) + {3 \choose 1} (4t - 4t^2) + {1 \choose 2} (2t^2 - t)$$
$$= {-8 \choose 0} t^2 + {8 \choose 2} t + {1 \choose 0}$$
I guess there is some very simple math involved that I just don't know at all :.-(
AI: Coordinate-wise, you have
$$F_1(t) = 1(2t^2-3t+1)+3(4t-4t^2)+1(2t^2-t) = 2t^2-3t+1-12t^2+12t+2t^2-t$$
Group like terms:
$$F_1(t) = -8t^2+8t+1.$$
Perform a similar association with the second "row". |
H: Proving that the graph of a function is closed
Let $F$ be a metric spaces and $(x_n)$ a sequence in $F$ without limit points. Define $f:\{1/n:n\in\mathbb N\}\cup \{0\}\to F$ by letting $f(0)=t$ with $t\in F$ and $f(1/n)=x_n$.
How to prove that the graph of $f$ is closed?
AI: If $\{(1/n, x_n)\}$ had limit points, then so would $\{x_n\}$, hence it has no limit points. Then $\operatorname{Cl}(\{(1/n,x_n)\})=\{(1/n, x_n)\}$. Note that $\operatorname{Gr}(f)=\{(1/n, x_n)\}\cup\{(0, t)\}$. Since
$$
\operatorname{Cl}(A\cup B)=\operatorname{Cl}(A)\cup\operatorname{Cl}(B)
$$
then
$$
\operatorname{Cl}(\operatorname{Gr}(f))
=\operatorname{Cl}(\{(1/n,x_n)\}\cup \{(0,t)\})
=\operatorname{Cl}(\{(1/n,x_n)\})\cup \operatorname{Cl}(\{(0,t)\})
$$
$$
=\{(1/n,x_n)\}\cup \{(0,t)\}
=\operatorname{Gr}(f)
$$
So $\operatorname{Gr}(f)$ is closed. |
H: Prove that the null set is a subset of a set $A$.
Prove that $\; \emptyset \subseteq A$.
The statement seems obvious to me, but how do I prove it?
My instructor said to prove that the statement is vacuously true, but I'm not sure what that means.
AI: Recall that to prove any set $X$ is a subset of another set $Y$, we must prove the following implication:
$$\text{For all}\; x \in X,\,\; {\bf if}\;\, x \in X,\; {\bf then}\;\, x \in Y.$$
More formally, that's $$X\subseteq Y \;\;\text{if and only if}\;\;\forall x\in X,\; \left(x\in X\implies x\in Y\right)\tag{1}$$
Now, an implication (aka, a conditional statement) is true any time the antecedent of the conditional (which is the statement to the left of the implication sign $\implies$) is false. (When this happens, the implication is said to be vacuously true). Furthermore, an implication is true any time the statement to the right of the implication sign is true.
An implication is false if and only if it happens to be the case that the left-hand side of the implication is true, but (and) the right-hand side is false. (I'll include the truth-table for the conditional connective below (compliments of W|A) as a refresher:
$\qquad\qquad\qquad\qquad$
Now, the following assertion happens to be vacuously true for all sets $X$:
$$\text{Let X be any set, and $x$ be any element. Then}\;\;x\in \varnothing \implies x\in X\tag{2}$$
This implication is vacuously true because $x\in \varnothing$ never happens to be the case: there is no element $x \in \varnothing$, by the definition of $\varnothing$, so it cannot possibly be the case that there exists an $x \in \varnothing$ and such that $x \notin X$. So we know simply by virtue of this fact that the entire implication is thereby (vacuously) true.
Now, $(2)$ is exactly what we wanted to prove, and we've shown it is vacuously true. That is: For all sets $X$ and any element $x$, if $x\in \varnothing$, then $x \in X$. Therefore, $\varnothing \subseteq X$, for all sets $X$. |
H: Given the solution for a differential equation, find the corresponding differential equation
The solution of a differential equation is:
$$\left \{ \begin{array}{lcl}
x_1(t) & = & e^{-2t}-e^{-5t} \\
x_2(t) & = & e^{-2t}+e^{-3t}+e^{-5t} \\
x_3(t) & = & e^{-3t} + e^{-5t}
\end{array}\right .$$
with initial conditions
$$\left \{ \begin{array}{lcl}
x_1(0) & = & 0 \\
x_2(0) & = & 3 \\
x_3(0) & = & 2
\end{array}\right .$$
What is the corresponding differential equation?
I had worked another linear algebra question using back substitution and I know that is what I am going to need to do here. In the previous problem I was substituting $A=QR$ and finding $\mathcal A$. Just not sure which approach to use in this situation.
$Ax=b$
$A=x^{-1}b$
Any help is greatly appreciated. :)
AI: Writing
$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix},$$
we have
$$\begin{array}{ll} \color{DarkOrange}{\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}} & = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}-2e^{-2t} \\ -3e^{-3t} \\ -5e^{-5t}\end{pmatrix} \\ & = \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}} \begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & = \color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & \color{DarkOrange}{= \begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}.} \end{array}$$
One may solve for the unknown matrix via
$$\begin{array}{ll} & \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}}=\color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}} \\ \iff & \begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}^{-1}=\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}. \end{array}$$ |
H: Set $A$ has the cardinality of $\aleph_0$. Prove some properties of partial order $ \langle \mathcal P (A), \subseteq \rangle$.
I try to solve the following task:
Set $A$ has the cardinality of $\aleph_0$. The truth is that in the partial order
$ \langle \mathcal P (A), \subseteq \rangle$: (answer true or false)
a) every subset has upper bound
b) exists chain of the cardinality of the continuum
c) exists antichain of the cardinality of the continuum
What I have found so far:
a) The upper bound for every subset of set $A$ is simpy set $A$.
So here TRUE.
b) Set $A$ has the cardinality of $\aleph_0$, so every chain can be also of the cardinality of $\aleph_0$. Suppose that some chain that has the cardinality of continuum. Then, this chain must have two sets that have the same number of elements. But if two sets have the same number of elements, they cannot be compared (in the inclusion relation). So, they cannot be in the same chain simultaneously. So here is FALSE (some chain has cardinality of at most $\aleph_0$).
c) And here I suppose that there exists antichain of the cardinality of the continuum, but I cannot prove it. I have tried to prove it this way: In point b) I proved that all sets having the same number of elements cannot be compared. So they are in the antichain. But I cannot prove that this antichain has the cardinality of the continuum (maybe it has not such a cardinality).
I am grateful for any hint!
AI: There is an error in your argument for (b). It is quite possible for two sets to have the same infinite cardinality but have one be a proper subset of the other. For example, $\{0,1,2,\dots\}$ and $\{0,2,4,\dots,\}$ have the same cardinality, but they can be compared via inclusion. Indeed, one definition of being $X$ being infinite is that $X$ has a proper subset of the same cardinality.
Indeed, there is a definition of the real line which essentially gives for each real $x$ an $A_x\subset \mathbb Q$ such that if $x<y$ then $A_x\subset A_y$.
Since you were wrong in (b), your argument for (c) is incorrect, since it uses the same flawed argument.
For (c), for each $X\subseteq \mathbb N$, define $o(X)=\{2n: n\in X\}\cup \{2n+1:n\notin X\}$. So that if $X\neq Y$ then $o(X)$ and $o(Y)$ are incomparable. In particlar, $o(X)\neq O(Y)$, so we have an anti-chain in $1-1$ correspondence with $\mathcal P(\mathbb N)$. |
H: How to evaluate the derivative$\frac{d}{dx}\left(\ln\sqrt{\frac{4+x²}{4-x²}}\right)$?
How can I evaluate this derivative?:
$$\frac{d}{dx}\left(\ln\sqrt{\frac{4+x^2}{4-x^2}}\right)$$
Thank you.
AI: We'll exploit the properties of logarithms, recalling that $$\ln\left(\frac{a}{b}\right)^b = b \ln\left(\frac ab\right) = b (\ln a - \ln b)$$
$$\begin{align} {\bf f(x)} & = \ln\sqrt{\frac{4+x^2}{4-x^2}} \\ \\
&= \ln\left(\frac{4+x^2}{4-x^2}\right)^{1/2}\\ \\
& = \frac 12\ln\left(\frac{4 + x^2}{4 - x^2}\right)\tag{$\ln\left(\frac ab\right)^c = c\ln\left(\frac ab\right)$}\\ \\
& = \frac12 \left(\ln(4+x^2) - \ln(4-x^2)\right)\tag{$\ln\left(\frac ab\right) = \ln a - \ln b$}\\ \\
{\bf f'(x)} & = \frac 12\left( \frac{2x}{4 + x^2} + \frac{2x}{4 - x^2}\right)\tag{chain rule} \\ \\ & = \frac 12\left(\frac{2x[4-x^2 + 4 + x^2]}{16-x^4}\right) \\ \\ & =
\frac{8x}{16 - x^4}\end{align}$$ |
H: Using $\arctan(x)$ for integrating the following function
We want to find $\int \dfrac{x}{x^2 + 16} dx$
My method was as follows:
Rewrite it to: $\dfrac{\frac{1}{16x}}{{(\frac{1}{4x}})^2 +1}$
Take $u = \dfrac{1}{4}x$. We then have $ \int4u^2 \cdot \dfrac{1}{u^2+1} du = \dfrac{4}{3}u^3 \cdot \arctan(u) + c$.
However, this is not correct. What have I done wrong?
Also, my textbook uses the much easier method with $\ln$. I am however questioning how you could judge by 'looks' of the function whether to use the $\arctan$ or the $\ln$.
AI: You need to be systematic and use substitution correctly. As to how to know what method to use, after a while you will become efffective at it. At this time the priority is correct execution.
At the end there is the assertion
$$\int 4u^2 \cdot \frac{1}{1+u^2}=\frac{4}{3}u^3 \arctan u+C.$$
This is incorrect, the integral of a product is not the product of the integrals.
There are also errors earlier, though it is not easy to identify them clearly since no steps are given.
The first algebraic step is correct, although not helpful. Then there is the substitution $u=\frac{1}{4} x$. Perhaps that is a typo, and $\frac{1}{4x}$ was intended.
Whichever is the case, if we are going to make a substitution, we need to compute explicitly $dx$ as $g(u)\,du$ for the appropriate $g$. There is no indication that this process was followed. We are just told that the integral is equal to $\int 4u^2\cdot \frac{1}{u^2+1}\,du$. This is not true under the proposed substitution, or any variant of it.
Remark: For $\int \frac{x}{x^2+16}\,dx$, you will after a while see that the derivative of $x^2 +16$ is (almost) sitting on top. So soon the response try $u=x^2+16$ will become semi-automatic. |
H: Suppose A is a set of (x, y)', what is the name of the set that consists of all x in A?
Let $A$ be a set of a vector $(\mathbf{x}',\,\mathbf{y}')$. Here $\mathbf{x}'$ and $\mathbf{y}'$ could both be vectors. Is there a particular terminology for the set of all $\mathbf{x}'$ in the set $A$.
AI: Besides Git Gud's answer, we can also say that it is the image of $A$ under the projection map $\pi_1$, where $\pi_1$ is defined by:
$$\pi_1(\mathbf x',\mathbf y') = \mathbf x'$$
This can be denoted symbolically by $\pi_1(A)$ or $\pi_1[A]$; particularly set theorists often prefer the square brackets.
This is because in set theory, one frequently encounters that $A$ is both an element and a subset of a larger set, rendering the round brackets ambiguous. |
H: $\sum c_k^2<\infty$ then $A=\{\sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k \}$ is compact
Let $\{e_k\}_{k=1}^\infty$ be an orthonormal set in a Hilbert space $H$. If $\{c_k\}_{k=1}^\infty$ is a sequence of positive real numbers such that $\sum c_k^2<\infty$, then the set:
$$A=\left\{\sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k\right\}$$
is compact in $H$.
My effort: We need to prove that every sequence $\{f_n\}$ has a subsequence $\{{f_n}_k\}$ so that $\{{f_n}_k\}$ converges to a limit in $A$.I'm trying to build such a sequence:
First, for each $n$ look at $|\langle f_n,e_1\rangle|=|a_{n1}|\leq c_1$ then the sequence $\{a_{n1}\}$ is bounded thus there must be a subsequence that converges and we define the limit as $l_1$. So we continue for all the vectors in the given set and define the limit of these subsets as $f=\sum_{k=1}^{\infty} l_ke_k \in H$.
But something must be wrong with the proof since I havent used $\sum c_k^2<\infty$.
Thanks.
AI: What you did is more incomplete than wrong.
Firstly, we need to justify the sentence "we continue for all the vectors in the given set", writing explicitely the diagonal extraction.
Then we need to prove the limit element we get is indeed in $H$ (otherwise, we would have proved when $c_k=1$ that a set containing the unit ball is compact). That's where the conditions is used.
Also, there are criteria of compactness you can use here. |
H: Proof of closed form expression for a sum
For the following sum
$$
\sum_{i=0}^{n-r} \binom{n-r}{i}s^{r+i}(1-s)^{n-r-i}(1-t)^{i}
$$
Wolfram Alpha finds the following closed form
$$
s^{r}(1-st)^{n-r}.
$$
However, I have been unable to derive this myself. Without the binomial coefficient, it would be straightforward to simplify along these lines:
$$
S = 1 + x + x^2 + x^3 + ... + x^n\\
Sx = x + x^2 + x^3 + ... + x^{n} + x^{n+1}\\
Sx+1 = S + x^{n+1}\\
S=\frac{x^{n+1}-1}{x-1}
$$
When $x \neq 1$. But I have not been able to make this approach work since each term ends up with the binomial coefficient of the term that precedes it, thereby preventing the appearance of $S$ on the RHS. Any help is greatly appreciated.
AI: $$
\begin{align}
&\sum_{i=0}^{n-r} \binom{n-r}{i}s^{r+i}(1-s)^{n-r-i}(1-t)^{i}\\
&=s^r(1-s)^{n-r} \sum_{i=0}^{n-r}\binom{n-r}{i}\left(\frac{s(1-t)}{1-s}\right)^i\\
&=s^r(1-s)^{n-r}\left(1+\frac{s(1-t)}{1-s}\right)^{n-r}\\[6pt]
&=s^r(1-st)^{n-r}
\end{align}
$$ |
H: If $m^*(E)=\infty$, then $E=\bigcup_{k=1}^{\infty}E_k$, $E_k$ measurable and $m^*(E_k)<+\infty$
Reading Royden's fourth edition of Real Analysis. I'm working with outer measure defined as
$$m^*(E)=\inf\left\{\sum_{n=1}^\infty l(I_n):\,E\subset \bigcup_{n=1}^\infty I_n\right\},$$
where each $I_n$ is a bounded, open interval. Also, $E$ is measurable if and only if
$$m^*(A)=m^*(A\cap E)+m^*(A\cap E^C),$$
for every set $A$.
In reading the proof of Theorem 11 on page 40, I start with $E$ a measurable set. Then I suddenly read the statement: "Consider the case where $m^*(E)=\infty$. Then $E$ may be expressed as the disjoint union of a countable collection $\{E_k\}_{k=1}^\infty$ of measurable sets, each of which has finite outer measure.
I am stuck on this last sentence. How come this is true?
AI: This is false unless we assume that $E$ is measurable. We can construct nonmeasurable sets which have outer measure $\infty$ which contain no measurable set of positive measure (take a Bernstein set, where both the set $B$ and its complement have nonempty intersection with every uncountable closed set). Let $B$ be such a set.
Suppose $B = \bigcup_{i=1}^\infty E_i$ is the disjoint union of countably many measurable sets. Then the only possibility for $E_i$ are sets of measure $0$ (because $E_i \subset B$), which means that $B$ has measure $0$, which is clearly a contradiction.
If $E$ is measurable, then we can just take $E_i = E \cap ( i,i+1 ]$, which is the intersection of two measurable sets. $E = \bigsqcup_{i \in \mathbb Z} E_i$ ($\sqcup$ denotes disjoint union). |
H: Numbers with no finite representation on paper
It occurred to me that there must be a lot of numbers without any form of finite representation on paper. Is there a name for these numbers?
For example...
Integers and rationals have a very simple representation e.g. 3/4
Irrational numbers obviously can also have a finite representation:
1.41421356... can be written as "the solution to the equation x^2 = 2"
Transcendental numbers can also have a finite representation:
e can be written as "the limit of (1 + 1/n)^n as n approaches infinity"
In other words, with a finite amount of effort one can give the reader enough information to calculate the value of the specified number exactly (to any degree of accuracy the reader chooses)
However, there must be a lot of numbers where this simply is not possible.
Consider the number 1.2736358762987349862379358... where this is just a string of (genuinely) random digits. There is no way to provide a finite definition that will specify this number to an arbitrary degree of accuracy. Similarly, there is no equation to which this number is a solution (I think, although I don't know how one would prove this).
Does this mean there are "gaps" in the real numbers. The number above is definitely somewhere between 1.2 and 1.3 but there is no way I can specify the value of this number (without writing an infinite number of digits). The number exists on the number line but I will never be able to do anything with it.
Is there a name for these numbers? Can anyone point me to some interesting resources on this topic?
I'm only asking as an interested hobbyist so apologies if this question isn't very scientific.
AI: That depends on what you mean by "representation." One way to cash this out is to talk about the definable numbers. These are more general than the computable numbers, but they are still countable because there are still only countably many possible descriptions of a number in a language over a finite alphabet. |
H: Incompleteness theorem and $\mathbb{L}$.
Let $\alpha > \omega$ and $u = \{\ulcorner \sigma \urcorner : \sigma \in \mathrm{Th}(\mathbb{L}_\alpha, \in)\} \subseteq \omega$, where by $\mathbb{L}_\alpha$ we denote as usual the constructible sets up to level $\alpha$.
My question is "Can we have $u \in \mathbb{L}_\alpha$?".
Suppose we can. Let $\phi(x) := x \in u$. Then for every sentence $\sigma$ we have that $$\mathbb{L}_\alpha \models \phi(\ulcorner \sigma \urcorner) \Leftrightarrow \mathbb{L}_\alpha \models \sigma,$$
i.e. truth is definable in $(\mathbb{L}_\alpha, \in)$, which contradicts the incompleteness theorem.
So $\mathscr{P}(\omega) \nsubseteq \mathbb{L}_\alpha$ for all $\alpha \in \mathrm{ON}$.
AI: Note that the same can apply with the universe of set theory:
For example let $u=\{\ulcorner\sigma\urcorner : \sigma\in\mathrm{Th}(V_\alpha,\in)\}\subseteq\omega$. Then for $\alpha>\omega+1$ we have that $u\in V_\alpha$. Then the formula $x\in u$ yeilds $$V_\alpha\models\ulcorner\sigma\urcorner\in u\iff V_\alpha\models\sigma$$ i.e. there is nothing special about the constructible universe.
The problem with the argument is that the formula $x\in u$ is not definable from the language of the model (i.e. $\{\in\}$), but from the language with an extra constant term $u$. The model may not be able (in fact by Tarski's theorem it is not able) to define -using its language- the set $u$, creating no conflict with Tarski's theorem.
The subsets of $\omega$ that you describe are always in $\mathbb{L}$ since they are definable with parameters from $\mathbb{L}_{\alpha+2}$ for example. They are simply not definable in $\mathbb{L}_\alpha$.
As an interesting aside (that Jech notes), if $0^\sharp$ exists then the theory of $\mathbb{L}$ is the same as that of $\mathbb{L}_{\aleph_1}$, so the theory of $\mathbb{L}$ is an element of $\mathbb{L}$. This means that in that case the real $\aleph_1$ is not definable in $\mathbb{L}$, making it quite large. |
H: A peculiar binomial coefficient identity
While inventing exercises for a discrete math text I'm writing I came up with this
$$
\binom{\binom{n}{2}}{2}=3\binom{n+1}{4}
$$
It's an easy result to prove, but it got me wondering
Is this pure coincidence, or is it a special case of some more general result of which I'm unaware?
No matter what the answer to (1) is, is there a combinatorial proof?
AI: This is actually a well known identity. There are combinatorial ways to prove it.
Consider $n$ objects. Consider all ${n \choose 2}$ pairs. Consider all pairs of these pairs. We get the LHS.
Consider $n$ objects and 1 distinguished object. Consider all sets of 4 objects from these.
If the 4 objects do not include the distinguished object, they correspond to 3 possible pairs of pairs, whose 4 elements are distinct. I.E. $(A,B), (C,D)$ and $(A, C), (B, D)$ and $(A, D), (B,C)$.
If the 4 objects include the distinguished object, they correspond to 3 possible pairs of pairs, which have a common element, and whose union is the 3 objects. I.E. $(A, B) , (A, C)$ and $(B,A), (B,C)$ and $(C,A), (C,B)$.
This gives us the RHS.
I'm not sure if they are generalizations, though you can experiment with choosing triples and counting carefully. |
H: Homomorphism $f:S_{2009} \rightarrow S_{2009}$
Is there a non-trivial homomorphism $f:S_{2009} \rightarrow S_{2009}$ with $(1,2)$ in the kernel?
Are there more than $2009$ different homomorphisms $f:S_{2009} \rightarrow S_{2009}$ with $(1,2,3)$ in the kernel?
Prime factorization: $2009 = 7^2*41$
Tips and/or answers are welcome, thanks in advance!
AI: If the kernel contains $(1\;2)$, it also contains every conjugate of it, that is every transposition, hence also every product of transpositions, i.e. all of $S_{2009}$. Remark: Here, $2009$ can be replaced by any $n\ge 2$.
If the kernel contains $(1\;2\;3)$ it also contains every 3-cycle and products therof, including $(1\;2\;3)(1\;2\;4)=(1\;3)(2\;4)$ and again conjugates and products, hence all even permutations. So if the homomorphism is nontrivial the kernel must be $A_{2009}$, the image cyclic of order $2$. There are more than $2009\choose 2$ subgroups of order $2$ in $S_{2009}$, as there is already one for each two-cycle $(a\;b)$. Together with th etrivial homomorphism this gives us enough different homomorphisms. Remark: So here $2009$ can be replaced by any $n\ge 3$ (and in fact for $n=3$ there are exactly ${3\choose 2}=3$ subgroups of order $2$, but note that we still have the trivial homomorphism) |
H: On a proof of dimension of vector spaces
There is a proof in my notes showing that if $U$ is a subspace of a finite-dimensional vector space $V$ then $\dim U \le \dim V$. It proceeds by extending a basis of $U$ to a basis of $V$. Wouldn't it be shorter and neater to say if $\dim U > \dim V$ then there would be at least one vector $u$ not in the span of $v_1, \ldots, v_n$. But then $u$ would lie outside $V$ which is impossible?
Or is it not correct to argue like this?
AI: I'll limit the discussion to finite dimensional spaces, which seems what the question deals with.
The main point in the theorem is proving that also $U$ is finitely generated, which isn't obvious and in fact doesn't hold in more general situations like modules over non noetherian rings.
Without knowing the proof in your notes it's difficult to guess the order of arguments. Here's how I usually present it.
The exchange lemma guarantees that two bases of $V$ have the same number of elements (which is defined to be $\dim V$) and any linearly independent set in $V$ has at most $\dim V$ elements.
By induction, if no finite set of elements of $U$ spans $U$, we can build a linearly independent set in $U$ having $n$ elements, for all $n$.
The fact is clear for $n=0$, because the empty set is linearly independent.
If $\{u_1,\dots,u_n\}$ is linearly independent, since by hypothesis it doesn't span $U$, there exists $u_{n+1}\in U\setminus\operatorname{span}\{u_1,\dots,u_n\}$ and it's easy to see that $\{u_1,\dots,u_n,u_{n+1}\}$ is linearly independent.
Therefore the hypothesis that $U$ is not finitely generated leads to a contradiction.
From any finite set of generators of $U$ we can extract a basis which, by the exchange lemma mentioned above, has at most $\dim V$ elements. Thus $U$ is finite dimensional and $\dim U\le\dim V$.
Your argument seems to be as follows: $U$ cannot contain an infinite linearly independent set; therefore a maximal linearly independent set in $U$ is finite. Now, the number of its elements cannot exceed $\dim V$, because of the exchange lemma. A maximal linearly independent set spans $U$, otherwise (with the same argument as in the induction step above) it could be extended. In this form the argument is correct. |
H: Why does $AX=0$ have only the trivial solution when $A=\left(\int_a^b g_i(x)g_j(x)dx\right)$?
The system is $AX=0$, where
$$A_{m\times m}=\begin{pmatrix}
\int_a^bg_1(x)g_1(x)dx & \cdots & \int_a^bg_1(x)g_m(x)dx \\
\vdots & & \vdots \\
\int_a^bg_m(x)g_1(x)dx & \cdots & \int_a^bg_m(x)g_m(x)dx
\end{pmatrix},$$
$$X=\begin{pmatrix}
x_1\\
\vdots\\
x_m
\end{pmatrix},0=\begin{pmatrix}
0\\
\vdots\\
0
\end{pmatrix}\in \mathbb{R}^m,$$
and $g_1,...g_n:[a,b]\rightarrow\mathbb{R}$ are linearly independent continuous functions.
I've read that this system has only one solution, but I'm not able to prove it.
Thanks.
AI: Hint. (Presumably $g_1,g_2,\ldots,g_m$ are linearly independent on $[a,b]$ rather than on $\mathbb{R}$, otherwise $A$ can be singular.) The integrand is the matrix $\mathbf{g}(x)\mathbf{g}(x)^T$, where $\mathbf{g}(x)=(g_1(x),\ldots,g_m(x))^T$. If $A\mathbf{u}=0$ for some vector $\mathbf{u}$, then $\mathbf{u}^TA\mathbf{u}=0$ and in turn
$$
\int_a^b \mathbf{u}^T\mathbf{g}(x)\mathbf{g}(x)^T\mathbf{u}\,dx=\int_a^b \left(\mathbf{g}(x)^T\mathbf{u}\right)^2dx=0.\tag{1}
$$
Note that the integrand in $(1)$ is a square term, hence nonnegative. Using the continuity and linear independence of $g_1,g_2,\ldots,g_m$, argue that $\mathbf{u}$ must be zero.
Remark. $A$ is a Gramian matrix. In general, a Gramian matrix is always positive semidefinite. |
H: Notation to describe the adding of a constant to all terms of a sequence
I've been struggling to get down the proper mathematical notation for sequences. Suppose I have the following sequence:
$$A = (4, 3, 7, 3, 1)$$
How do I describe the addition of a constant to all terms of sequence $A$? For example, if I were to add $7$ to all the terms of $A$ to form a new sequence $B$:
$$B = (11,10,14,10,8)$$
How do I describe $B$ in terms of $A$? Would this be the proper notation:
$$B = A + 7$$
By the way, is even my notation for $A$ correct?
AI: It's unusual to see finite sequences written out in that way or any other. Given that you had a sequence (or net) $(a_i)_{i\in I}$, you could write $(a_i+1)_{i\in I}$ for the sequence obtained from the first by adding $1$ to each value. |
H: Does $a_n = \cos\left(n\ln \left(1+\frac{\pi}{n}\right)\right)$ converge?
I want to check if a sequence converges or diverges. The sequence is the following:
$$a_n = \cos\left(n\ln \left(1+\frac{\pi}{n}\right)\right)$$
I though of maybe using sandwich theorem, but can I use it, saying that the value will lie between $-1$ and $1$? If not, can anyone point me to where I should be looking at.
AI: $$\begin{align}a_n &= \cos \left(n \ln \left( 1+ \frac \pi n\right) \right)\\
&=\cos\left(\ln\left(1+\frac\pi n\right)^n\right)
\end{align}$$
You may recognize that $$\lim_{n\to \infty}\left(1+\frac x n\right)^n = e^x$$ for all $x$ (this being one way to define the exponential function).
Thus $\lim_{n\to\infty}\left(1+\frac \pi n\right)^n = e^\pi$.
Since $\ln$ is continuous throughout its domain, which includes $e^\pi$, and $\cos$ is continuous everywhere, your sequence will approach $\cos(\ln e^\pi)=\cos \pi = -1$.
Thanks for the tip on continuity from davin. |
H: Help with this definition of $(G:_M I)$
I didn't understand why in this definition $I$ has to be an ideal to make sense.
REMARK
This is from Steps in Commutative Algebra, page 107.
Thanks a lot
AI: The definition doesn't claim that $I$ has to be an ideal, and in fact it doesn't, but if $S$ is any subset of $R$ then $(G :_M S) = (G :_M I)$ where $I$ is the ideal generated by $S$, so $I$ might as well be an ideal. |
H: number theory equation involving GCD
Fix the natural number $b$. How can I solve ?
$$
x+\gcd(x,b) \equiv 0 \mod(b)
$$
Can anyone please give me a reference?
Best
AI: Let $x=cd$ where $d=\gcd(x,b)$, and let $a:=b/d$. Then we have $\gcd(a,c)=1$, and
$$cd+d\equiv 0 \pmod{ad}$$
so that $c+1\equiv 0\pmod a$.
It means, that for each divisor $d$ of $b$, we can choose $c:\equiv -1\pmod{b/d}$, then $x:=cd$ will be a solution, as $c\equiv -1\pmod{a}$ already implies $\gcd(a,c)=1$ hence $\gcd(b,x)=d$. |
H: Linear independence of functions: $x_1(t) = 3$, $x_2(t)=3\sin^2t$, $x_3(t)=4\cos^2t$
I want to determine whether 3 functions are linearly independent:
\begin{align*}
x_1(t) & = 3 \\
x_2(t) & = 3\sin^2(t) \\
x_3(t) & = 4\cos^2(t)
\end{align*}
Definition of Linear Independence: $c_1x_1 + c_2x_2 + c_3x_3 = 0 \implies c_1=c_2=c_3=0$ (only the trivial solution)
So we have:
\begin{align}
3c_1 + 3c_2\sin^2(t) + 4c_3\cos^2(t) = 0
\end{align}
My first idea is to differentiate both sides and get:
$6c_2\sin(t)\cos(t) - 8c_3\cos(t)\sin(t) = 0$
Then we can factor to get:
$\sin(t)\cos(t)(6c_2 - 8c_3) = 0$
So $c_3= \dfrac{6}{8}c_2$ gives the equation equals zero. Thus all $c$ are not $0$ and thus $x_1, x_2, x_3$ are linearly dependent.
Is this correct? Or is there a cleaner way to do this?
AI: Yes, indeed, your answer is fine. And it would have been a particularly fine determining the linear (in)dependence of a system of equations that doesn't readily admit of another observation about the relationship between $\cos^2 t$ and $\sin^2 t$ $(\dagger)$. Indeed, you're one step away from working with the Wronskian, which is a useful tool to prove linear independence.
$(\dagger)$ Now, to the observation previously noted: You could have also used the fact that $$x_1(t) - \left[(x_2(t) +\frac 34 x_3(t)\right] = 3 - (3 \sin^2 t + 3\cos^2 t)= 3 - 3\left(\underbrace{\sin^2(t) + \cos^2(t)}_{\large = 1}\right)=0$$
and saved yourself a little bit of work: you can read off the nonzero coefficients $c_i$ to demonstrate their existence: $c_1 = 1, c_2 = -1, c_3 = -\frac 34$, or you could simply express $x_1$ as a linear combination of $x_2, x_3$, to conclude the linear dependence of the vectors. (But don't count on just any random set of vectors turning out so nicely!) |
H: Interpretation of Linear Algebra and superposition
I have been told and have seen that knowing Linear Algebra is foundational in pursuing advanced mathematics. After dealing with it enough I have gotten "used to it" (I still need a lot more practice!) What I would like to know is the interpretation of Linear Algebra so that I may have more mathematical intuition when dealing with it.
Specifically, a transformation $L$ is linear if it satisfies the properties:
$aL(x) = L(ax)$ and $L(x+y)= L(x) + L(y)$ where $a$ is a scalar and $x,y$ are vectors.
How can I think about this intuitively? (Why does the term superposition accurately fit this?)
How else can one interpret results in Linear Algebra?
AI: Let $V$ and $W$ be vector spaces, and $L:V\to W$ a linear map. By either of the properties, we get $L(0)=0$.
Geometrically, a linear map assigns a vector in $W$ to each vector of $V$, so that
The origin goes to the origin. ($L(0)=0\,$.)
Lines go to lines. (Because $L(x_0+tv)=L(x_0)+t\,L(v)$ for any $t\in\Bbb R$.)
Triangles go to triangles (any two vector $x,y$ determines a triangle together with $x+y$, so this corresponds to $L(x+y)=L(x)+L(y)$).
It is invariant under zooming from the origin. ($L(ax)=a\,L(x)\,$.) |
H: Uniform Continuity and Cauchy Sequences
Let $(S,d)$ and $(S^*, d^*)$ be metric spaces. If $f:S \to S^*$ is uniformly continuous and if $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence in $S^*$.
$f:S \to S^*$ is uniformly continuous: $(\forall \varepsilon >0)( \exists > 0)(\forall s,t \in S)$ $d(s,t) < \delta \implies d(f(s), f(t)) < \varepsilon$.
$(s_n)$ Cauchy: $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(s_n,s_m)) < \varepsilon$.
Now I need to show that $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(f(s_n),f(s_m)) < \varepsilon$.
Since $f$ is uniformly continuous, $\delta$ will depend on $\varepsilon$ only. Then I separate the problem into two cases:
Suppose $\varepsilon < \delta$: Since I know $(s_n)$ is Cauchy, I can take the same $N$ to get $d(s_n,s_m) < \varepsilon < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$ by uniform continuity.
Suppose $\delta \leq \varepsilon$: Since $f$ is uniformly continuous, $d(s,t) < \delta \implies d(f(s),f(t)) < \varepsilon$ will hold for all $s,t \in S$. So let $N=1$, then $d(s_n,s_m) < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$.
Could someone give me feedback on my proof?
Edit For every $\varepsilon > 0$, since $f$ is uniformly continuous, there exists a $\delta >0$ such that for all $s, t \in S$, $d(s, t) < \delta \implies d(f(s),f(t)) < \varepsilon$. Then for any $\delta >0$ there exists an $N \in \mathbb{N}$ such that for all $n,m \geq N$, $d(s_n, s_m) < \delta$. By uniform continuity, $d(s_n, s_m) < \delta \implies d(f(s_n),f(s_m)) < \varepsilon$.
AI: The first part (case $\varepsilon<\delta$) is correct, but I can't follow the second part. The conclusion $N=1$ is clearly wrong for arbitrarily small $\varepsilon$'s.
In the definition of uniform continuity, the presence of $\delta$ is always around $S$ and $\varepsilon$ is around $S^*$. So, we want to prove $f(s_n)$ is Cauchy:
Let's assume an $\varepsilon>0$ is given. For this we can choose a $\delta$, and for this $\varepsilon':=\delta$ we can choose an $N$ for $s_n$ by the Cauchy property. |
H: Where does the theory of quadratic forms fail in characteristic 2?
Let $V$ be a finite-dimensional vector space over a field $k$, and $Q$ a nondegenerate quadratic form on $V$. If the characteristic of $k$ is not 2, then we can change coördinates on $V$ so that $Q(\vec{x})=\sum_ia_ix_i^2$ for some $a_i\in k.$ Thus we can decompose $V$ as $(V,Q)=\oplus(V_i,Q_i),$ where the $V_i$ are $1$-dimensional and $Q_i(x)=a_ix^2.$
My question is: can we do something similar in the case where $char(k)=2$? If not, why not, and what can we do instead?
AI: As to what can be done instead, I quite like this book, GROVE. The book you will see most on fields rather than rings is LAM. Also quite modern, but probably a bit easier to digest, is GERSTEIN.
Grove has the most detail, three chapters entirely on characteristic 2, one of those on Clifford Algebras but the other two on Orthogonal Groups/Geometry, meaning quadratic forms. |
H: Given a matrix $A$ and what it maps two vectors to, is $0$ an eigenvalue of it?
Studying for my Algebra exam, and this question popped out with no solution in a previous exam:
Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix}$.
(I) Is $0$ an eigenvalue of the matrix?
(II) Find a matrix like that, where the sum of its' eigenvalues is $0$.
So I (think) solved (I) but have no clue for (II).
Here's my solution for (I):
$A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = A \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 0$, and then, the vector $v = \begin{pmatrix} 2 \\ -1 \\0 \end{pmatrix}$ supplies that $Av = 0v = 0$ meaning that $0$ is an eigenvalue of $A$ with an eigenvector $v$.
AI: The answer for (I) looks good.
Good for (II) might be to note that
$$
A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\4\\-6\end{bmatrix}
$$
and
$$
A\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix}
$$
which gives the first two columns of $A$:
$$
A=\begin{bmatrix}1&2&x\\2&4&y\\-3&-6&z\end{bmatrix}
$$
The trace is the sum of the eigenvalues. Thus, if the sum of the eigenvalues is $0$, we need $z=-5$. $x$ and $y$ are arbitrary. |
H: definitions of the free abelian group
How one can use the universal mapping property of free abelian groups to prove that $\sum_{b\in S}<b>$ is a free abelian group generated by $S$?
AI: Let $A$ be an arbitrary Abelian group and $f:S\to A$ a function.
Then there is a unique way to extend $f$ to a homomorphism $\bar f:\sum_{b\in S}\langle b\rangle\,\to A$, namely, we have to set
$$\bar f(n_1b_1+n_2b_2+\dots+n_kb_k):=n_1\,f(b_1)+n_2\,f(b_2)+\dots +n_k\,f(b_k)\,.$$ |
H: The continuity of measure
Let $m$ be the Lebesgue Measure. If $\{A_k\}_{k=1}^{\infty}$ is an ascending collection of measurable sets, then
$$m\left(\cup_{k=1}^\infty A_k\right)=\lim_{k\to\infty}m(A_k).$$
Can someone share a story as to why this is called one of the "continuity" properties of measure?
AI: My understanding:
If a function $f$ is continuous and $\lim_{n\to \infty}x_n=x$, then $f(x)=f(\lim_{n\to \infty}x_n)=\lim_{n\to \infty}f(x_n)$. That is, you can take the limit out. Similarly, $\lim_{n\to \infty}\cup_{k=1}^nA_k=\cup_{k=1}^{\infty}A_k$, the "continuity" property of measure implies
$$
m(\cup_{k=1}^{\infty}A_k)=m(\lim_{n\to \infty}\cup_{k=1}^nA_k)=\lim_{n\to \infty}m(\cup_{k=1}^nA_k)=\lim_{n\to \infty}A_n
$$
Note $\{A_n\}_{n=1}^{\infty}$ is an ascending collection of measurable sets, so $A_n=\cup_{k=1}^nA_k$. |
H: Prove that $v_1^{T}v_2 = [0] $ (the 1 by 1 zero matrix)
This question is from a Linear Algebra past exam paper that I am reviewing as I study for finals. The full question is:
Suppose that $M$ is a square matrix. Recall that $M^T$ denotes the transpose of $M$, the result of interchanging columns and rows of $M$. Recall that
$$ (AB)^{T} = B^TA^T$$
for any square matrices $A$ and $B$ of the same size. (You do not need to prove this and may quote it freely).
Suppose that the remainder of this problem $M=M^T$ (we call this a symmetric matrix).
(i) Prove that if $v_1$ and $v_2$ are eigenvectors of $M$ with respect to eigenvalues $\lambda_1$ and $\lambda_2$ such that $\lambda_1 \neq \lambda_2$ then $v_1^{T}v_2 = [0] = 0 $ (the 1 by 1 matrix with entry zero).
The solution is given as follows:
We have that $Mv_1 = \lambda_1v_1$ and $Mv_2 = \lambda_2v_2$.
$$
v_1^{T}v_2 = (\frac{1}{\lambda_1}\lambda_1 v_1)^Tv_2 = \frac{1}{\lambda_1}(Mv_1)^Tv_2 = \frac{1}{\lambda_1} v_1^TM^Tv_2 = \frac{1}{\lambda_1}v_1^TMv_2 = \frac{1}{\lambda_1}v_1^T\lambda_2v_2 = \frac{\lambda_2}{\lambda_1}v_1^Tv_2
$$
so that $v_1^{T}v_2 - \frac{\lambda_2}{\lambda_1}v_1^Tv_2 = 0$. Then $(1 - \frac{\lambda_2}{\lambda_1})v_1^Tv_2=0$ and since $\lambda_1 \neq \lambda_2$, then $ v_1^Tv_2 = 0$.
What I am struggling to understand about this solution is how $ (AB)^{T} = B^TA^T$
can be applied to $(Mv_1)^T$, when it is clearly stated that this holds when $A$ and $B$ are square matrices of the same size? Could someone enlighten me?
Many thanks in advance!
AI: Well, I disagree with your teacher:
You should at least once prove for yourself, that for any matrices $A,B$ such that $\exists AB$, we do have
$$(AB)^T=B^TA^T\,.$$ |
H: Constructing a simple isomorphism from $\mathbb{R^{2n}}$ to $\mathbb{C^n}$
I need to construct an isomorphism from $\mathbb{R^{2n}}$ onto $\mathbb{C^n}$ (both over $\mathbb{R}$). I know couple of things:
This isomorphism will be a (bijective) linear map between two vector spaces that have equal dimensions.
The linear map is uniquely determined by action on the bases.
So I specify bases for both spaces, say, let $\{r_1,\dots,r_{2n}\}$ be a basis for $\mathbb{R^{2n}}$ and $\{c_1,\dots,c_{2n}\}$ be a basis for $\mathbb{C^n}$, where
\begin{align*}
r_1&=(1,0,0,\dots,0),\\
r_2&=(0,1,0,\dots,0),\ldots,\\
c_1&=(1,0,0,\dots,0),\\
c_2&=(i,0,0,\dots,0),\\
c_3&=(0,1,0,\dots,0),\\
c_4&=(0,i,0,\dots,0),\ldots.
\end{align*}
Then let the transformation be defined by $r_i\mapsto c_i$.
I could prove that it's linear, but I was wondering if that suffices to solve the problem, or not?
Thanks!
AI: Yes, that's all.
You don't even need to prove linearity, as any mapping defined on a basis uniquely extends to a linear mapping, and such is an isomorphism iff the image of the (or any) basis is again a basis in the target space.
However, if you want, you can write this up explicitly as
$$(a_1,b_1,a_2,b_2,\dots,a_n,b_n)\mapsto (a_1+ib_1,\,a_2+ib_2,\dots,\,a_n+ib_n)\,.$$ |
H: Conditional Probability P(A ∩ B)
So I found myself in a infinite loop while trying to do some probability. If A and B are independent, calculating P(A ∩ B) is as simple as P(A)P(B). However, how do I calculate P(A ∩ B) if they are dependent? I know P(A∣B)⋅P(B)=P(A∩B).
example:
Charlie and Doug take turns rolling two dice, Charlie goes first. If Charlie rolls a sum of 6 before Doug rolls a sum of 7, Charlie wins. What is the probability Charlie wins.
Solution attempt:
P(Charlie wins) = P(Charlie rolls 6, 1st turn) + P(Charlie rolls 6 | Charlie and Doug don't win previous rounds)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie and Doug don't win previous rounds)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)*P(Charlie doesn't win previous)P(Doug doesn't win previous)
= 5/36 + P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds)(31/36)*(30/36)
Now this is where I'm stuck, as I don't know how to calculate P(Charlie rolls 6 ∩ Charlie and Doug don't win previous rounds). Can anyone give me some pointers?
AI: Let $p$ be the required probability that Charlie wins the game. Charlie can win in two ways, fast or slow: (i) Fast: Charlie wins immediately, probability $\frac{5}{36}$; (ii) Slow: Charlie doesn't win immediately, neither does Doug on his first try, but Charlie ultimately wins, probability $\left(\frac{31}{36}\cdot \frac{30}{36}\right)p$. Thus
$$p=\frac{5}{36}+\left(\frac{31}{36}\cdot\frac{30}{36}\right)p.$$
Solve for this linear equation for $p$.
Remark: Without explicitly saying so, we have used conditional probability. The probability that Charlie (ultimately) wins given that neither player wins on his first try is just $p$.
We can instead use your general strategy. The probability that Charlie wins on his first try is $\frac{5}{36}$. The probability that Charlie wins on his second try is $\left(\frac{31}{36}\cdot\frac{30}{36}\right)\frac{5}{36}$. The probability Charlie wins on his third try is $\left(\frac{31}{36}\cdot\frac{30}{36}\right)^2\frac{5}{36}$. And so on. For the probability Charlie wins, we add up the infinite geometric series with first term $\frac{5}{36}$ and common ratio $\frac{31}{36}\cdot\frac{30}{36}$. We can find the sum using the ordinary formula for the sum of an infinite geometric series. |
H: How many 3 digit positive integers are divisible by 5?
Q1: How many 3 digit positive integers are divisible by 5?
Since it should be divisble by 5 it must end it 5?so
9 * 10 * 1
Is that right? But shouldn't I check if it ends in 0 or 5?
q2: how many odd non repeating 3 digit positive integers are there? It should end it either 1,3,5,7,9
So 9*9*5?
AI: Q1: Recall that a number is divisible by $5$ if and only if its last digit is either a $0$ or a $5$. So we have:
$$
9\cdot10\cdot2 = 180
$$
Q2: Observe that the only possibilities for the last digit that will make the number odd are $1,3,5,7,9$. Once you pick this digit, there will be $9-1=8$ remaining possibilities for the first digit. Once you pick these two digits, there will be $10-2=8$ remaining possibilities for the second digit. This yields:
$$
8 \cdot 8 \cdot 5 = 320
$$ |
H: Cauchy Schwarz Inequality for numbers
The CS inequality is given by
$$x_1y_1 + x_2y_2 \leq \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$$
I read that if $x_1 = cy_1$ and $x_2 = cy_2$, then equality holds.
But I reduced the above to $c \leq \sqrt{c^2} = |c|$. So isn't this only true if $c > 0$?
AI: Assume $x_1,\dots,y_2$ are real numbers. We have $$(x_1^2+x_2^2)(y_1^2+y_2^2)-(x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2.$$ This is the case $n=2$ of an identity due to Lagrange.
(See more on this, positive polynomials, sums of squares of polynomials, and Hilbert's 17th problem, in this old blog post of mine.)
This means that $|x_1y_1+x_2y_2|\le\sqrt{x_1^2+y_1^2}\sqrt{y_1^2+y_2^2}$, with equality iff $$\det\left(\begin{array}{cc}x_1&y_1\\ x_2&y_2\end{array}\right)=0,$$ that is, iff either there is a constant $c$ such that $x_1=cy_1$ and $x_2=cy_2$, or else $y_1=y_2=0$.
Since $a\le|a|$ for any real number $a$, it follows that $x_1y_1 + x_2y_2 \leq \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$, with equality as above, except that now we further need $c\ge 0$. |
H: Boundary Value Problem $y^{(4)} =-24$ a horizontal beam in theory of elasticity
The equation is $$y^{(4)} =-24$$
boundary values are $y(0) = y'(0) = y(4) = y'(4) = 0$. I integrated the equation four times and got $$y = -x^4 + ax^3 + bx^2 + cx + d$$ but if I used character equation $r^4 + 24 = 0$ I will get complex roots, the soln will be cos and sin?! The problem has a part 2, it wants me to prove maximum displacement occurs at the center of the beam $x=2$. I solved for a,b,c,d and use second derivative test, but only found y(2) is a minimum?! Anyone got an idea on this weird question? I am confused why the problem is named "boundary value problem" as well, didn't we assign initial value to the equation and solve it?
AI: The characteristic equation you quote does not apply to your equation, because the exponential $e^{r x}$ from which such an equation derives does not cancel. (Thik about the equations in which you make such an assumption about the solution.) The correct way is to integrate 4 times, as you did. The boundary values get applied as follows:
$$y(0)=0 \implies d=0$$
$$y(4) = 0 \implies 64 a+ 16 b+ 4 c + d = 256$$
etc. |
H: Area projection from cube to sphere
I have a regular cube of edge length = 2, and a sphere of radius = 1. Each face of the cube has been divided into N*N equal-area squares. How can I compute the projected area of every square on the sphere, so that the area of all the projections sum up to 4*pi.
I came to the following solution but the result does not sum to 4*pi:
Denote the area of each square on the cube as: A
Denote the length of the line segment connecting the center of the cube with the center of a square on the cube face as: R
Then I compute the cosine of the angle between the above line segment with a face normal (ex. if the square lies in +X face, then the angle is between the line segment and the normal of face +X).
so that the projection of the area from the square to the sphere can be computed as:
A * cosine(angle) / R^2
However, the above solution does not sum to 4*pi
AI: Your method is a (very) good approximation when N is (very) large.
If you want the exact formula, you will need to compute an integral (over the little square of area A of the function cos/R^2). |
H: What do I do in the case of this double series?
I am student who learns from examples, and I have yet to see what happens when two sums such as this one,
$$\displaystyle\sum\limits_{a=1}^2 \displaystyle\sum\limits_{b=a+1}^2 4\left[\left(\frac{1}{b-a}\right)^8 - \left(\frac{1}{b-a}\right)^4\right]$$
be solved. I am sure, when a = 1, the inner sum should solve itself like any "normal summation", but when a = 2, what happens? Do I skip the process of the inner summation; do I proceed with the inner sum, and if so, in what ways do I solve the inner sum; or do I solve this problem in a completely different way?
AI: Thinking inductively on the upper and lower indices, we can think of a summation as a difference of two functions on the integers:
$$
\sum_{k=a}^bf(k)=F(b)-F(a-1)
$$
Even in this generalization, if $b=a-1$, the sum is $0$ irregardless of $f$. |
H: About Baire spaces
I'm having difficult to solve this:
Determine whether or not $\mathbb{R}_l$ is a Baire space.
I tried to aplly the following lemma: "X is a Baire space iff given any countable collection $\mathbb{U}_n$ of open sets in X, each of which is dense in X, their intersection $\bigcap{U}_n$ is also dense in X."
Where $\mathbb{R}_l$ is the topology of lower limit, generated by the intervals [a,b).
Could you help to solve this?
Thank you.
AI: Suppose topologies $\sigma$ and $\tau$ on a set $X$ are such that every nonempty set from $\sigma$ contains a nonempty set from $\tau$, and the other way around. This is the case for the standard and lower-limit topology on the real line. You should be able to show that, for $Y \subset X$,
the $\sigma$-interior of $Y$ is empty if and only if the $\tau$-interior of $Y$ is empty.
$Y$ is $\sigma$-nowhere dense if and only if $Y$ is $\tau$-nowhere dense.
It follows that $\sigma$ is Baire if and only if $\tau$ is Baire. |
H: A Joint Density Problem Involving Change of Variable
A point $(X,Y)$ is picked at random uniformly in the unit circle.
Find the joint density of $R$ and $X$, where $R^2 = X^2 + Y^2$.
So the question is asking for $f_{R, X} (r, x) = \mathbb P(R=r, X=x)$.
Now, if I integrate$f_{R, X} (r, x)$ over all values of $r$, I will have
$$\int_{-\infty}^{\infty} f_{R, X}(r, x) dr = \mathbb P(R \le {\infty}, X=x) = f_X (x)$$
Of course, I can recover $f_X (x)$ from the following integral:
$$\int_{-\infty}^{\infty} f_{X, Y}(x, y) dy =
\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} {1 \over \pi} dy = {2 \over \pi} \sqrt{1-x^2}$$
The way the problem is set up suggests the following solution:
$$f_{R, X} (r, x) = {d \over dx} \left[ {2 \over \pi} \sqrt{1-x^2} \right]{dx \over dr}$$
However, I am not entirely sure of how to justify the above.
Let $x$ be a fixed value. Via the FOTC, I have the following:
$$f_{R, X} (r, x) = \mathbb P(R=r, X=x) = \lim_{h \rightarrow 0} {1 \over h}\left[ \int_{-\infty}^{r+h} f_{R, X}(r, x) dr \space - \int_{-\infty}^{r} f_{R, X}(r, x) dr \right]$$
I think I am on the right track. But after I'm given $f_{X, Y}$, a function of $x$ and $y$,
I am having a hard time imagining what $f_{R, X}$, a function of $x$ and $r$, looks like.
Specifically, what is the the region in the $XR$ plane over which $f_{R,X}$ is defined?
AI: The joint density function is defined everywhere. But in effect you are asking where anything interesting is happening. That is the part of the $X$-$R$ plane where $0\le r\le 1$ and $|x|\le r$. This is a triangle, really two triangles glued together. (There is symmetry across the $R$-axis.) The corners are $(0,0)$, $(1,1)$, and $(-1,1)$.
To see this, note that we are in the unit disk, so $0\le R\le 1$. And the only restriction on $X$ is that since $X^2+Y^2=R^2$, we have $|X|\le R$. Thus we must be above the "curve" $r=|x|$. |
H: Solving this system
I have been using substitution as the method to solve this. It doesn't work and I also do some trial-and-error but it takes a lot of time before I found the solution. Now, Is there a better way of solving this system of equation? Thank You.
$$\begin{cases} y =2000( 1 + 0.2x) \\ z= 2000(1.2)^x \\ z=2y\end{cases}$$
I am solving for $x$.
AI: Some simple substitutions lead to the equation:
$$(1.2)^x=2+0.4x$$
From the following picture, we see that there are $2$ solutions, which are approximately $-3.73$ and $9.73$.
There is no closed form for the solutions, but you can use Newton's method to achieve fairly accurate approximations. |
H: SOA Exam P Question: $P$ is a random point on the Cartesian Coordinate Plane. Find the variance of the area of a circle formed by $P$.
Caution: This problem was "passed down" to me and I think the wording was altered or lost along the way. I will post the problem as I have it and then make suggestions on what I think it should be. I am interested if anybody has seen this problem before and knows the correct wording or if anybody can "fill in the gaps" so that the problem makes sense.
Problem: $P$ is a random point on the Cartesian Coordinate plane. $P$ is uniformly distributed between 0 and 3. What is the variance of the circle formed by $P$.
Edit: What is the variance of the area of the circle formed by $P$.
Here are my thoughts: It doesn't make sense to say that $P$ is uniformly distributed between 0 and 3 since $P=(x,y)$ is an ordered pair. So this problem would only make sense if we have that $P=(x,y)$, where $X,Y$~ Uniform(0,3), or $||P||$~Uniform(0,3), where $||P||=\sqrt{x^2+y^2}$. In either case we get different variances.
I have tried this problem under both assumptions am I am beginning to believe that it's the latter case. If so, then I would write $A=\pi||P||^2$ and then do a random variable transformation to get the pdf of $A$. Then I would find the first and second moments of $A$ and thus have the variance.
What do you think? I am interested in knowing either a correct/appropriate wording of the problem or a solution (or both).
AI: To say $P$ is uniformly distributed between 0 and 3 means it's on a line, not a circle. If both $X$ and $Y$ were uniformly distributed on (0, 3), then you'd have a square. Also, to ask about the variance of a circle means no sense. Do you mean the radius or the area or what? You can't have a variance of a circle.
Next, I googled for 10 seconds and found this:
http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?p=2652448
I assume this is your problem. You can then google the first few words of that problem and get tons of results with that exact problem being asked.
UPDATE: Well, let me answer a question something like yours just for your benefit.
Question: Let $R$ be uniformly distributed on $(0, 3)$. After $R$ is chosen at random, form a circle $C$ of radius $R$. What is the variance of the area of $C$ throughout this process?
Answer: Let $A$ be the random variable representing the area of $C$. Then $A = \pi R^2$. So, we need to find
$$Var(A) = E(A^2) - E(A)^2 = E(\pi^2 R^4) - E(\pi R^2)^2 = \pi^2[ E(R^4) - E(R^2)^2].$$
Thus, the problem is now simply in terms of the uniform distribution and you can find the expected values with simple integrals. I will let you do the rest. |
H: Check solutions of vector Differential Equations
I have solved the vector ODE: $x\prime = \begin{pmatrix}1& 1 \\ -1 &1 \end{pmatrix}x$
I found an eigenvalue $\lambda=1+i$ and deduced the corresponding eigenvector:
\begin{align}
(A-\lambda I)x =& 0 \\
\begin{pmatrix}1-1-i & 1 \\-1& 1-1-i \end{pmatrix}x =& 0 \\
\begin{pmatrix} -i&1\\-1&-i\end{pmatrix}x =&0
\end{align}
Which is similar to:
$\begin{pmatrix}i&-1\\0&0 \end{pmatrix}x = 0$ By Row Reduction.
Take $x_2=1$ as $x_2$ is free. We then have the following equation:
\begin{align}
&ix_1 - x_2 = 0 \\
\iff& ix_1 = 1 \\
\iff& x_1 = \frac{1}{i}
\end{align}
Thus the corresponding eigenvector for $\lambda=1+i$ is: $\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix}$.
My solution should then be:
\begin{align}
x(t) =& e^{(1+i)t}\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\
=& e^t e^{it}\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\
=& e^t\left(\cos(t) + i\sin(t)\right)\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix} \\
\end{align}
By taking only the real parts we have the general solution:
$\left(c_1e^t\cos(t) + c_2e^t\sin(t)\right)\begin{pmatrix} \dfrac{1}{i} \\ 1\end{pmatrix}$
How can I quickly check this is correct? Idealy I would like to use Sage to verify. I think this would be faster than differentiating my solution and checking whether I get the original equation.
AI: Let me work through the other eigenvalue, and see if you can follow the approach.
For $\lambda_2 = 1-i$, we have:
$[A - \lambda_2 I]v_2 = \begin{bmatrix}1 -(1-i) & 1\\-1 & 1-(1-i)\end{bmatrix}v_2 = 0$
The RREF of this is:
$\begin{bmatrix}1 & -i\\0 &0\end{bmatrix}v_2 = 0 \rightarrow v_2 = (i, 1)$
To write the solution, we have:
$\displaystyle x[t] = \begin{bmatrix}x_1[t]\\ x_2[t]\end{bmatrix} = e^{\lambda_2 t}v_2 = e^{(1-i)t}\begin{bmatrix}i\\1\end{bmatrix} = e^te^{-it}\begin{bmatrix}i\\1\end{bmatrix} = e^t(\cos t - i \sin t) \begin{bmatrix}i\\1\end{bmatrix} = e^t\begin{bmatrix} \sin t + i \cos t\\ \cos t -i \sin t \end{bmatrix} = e^t\begin{bmatrix}c_1 \cos t + c_2 \sin t\\ -c_1 \sin t + c_2 \cos t\end{bmatrix}$
Note, I put $c_1$ with the imaginary terms, and $c_2$ with the other terms, but this is totally arbitrary since these are just some constants.
For the validation:
take $x'[t]$ of that solution we just derived.
then, take the product $Ax$ and verify that it matches the $x'[t]$ expressions from the previous calculation.
I would recommend emulating this with the other eigenvalue/eigenvector and see if you can get a similar result. Lastly, note that $1/i = -i$ (just multiply by $i/i$). |
H: Why should $|2^\mathbb{N}|>|\mathbb{N}^2|$ be true?
I've been thinking a bit about infinite things lately, and this question I had wondered about came back to me.
One of the classic expository demonstrations of Cantor's work is the two equally surprising facts that there are as many rationals as natural numbers, but there are more reals than natural numbers. This can be reduced to a statement in cardinal arithmetic, namely that
$$|2^\mathbb{N}|>|\mathbb{N}^2|$$
Now, we can think of these sets as two collections of functions, $2^\mathbb{N}=\{f:\mathbb{N}\to \{0,1\}\}$ and $\mathbb{N}^2=\{f:\{0,1\}\to\mathbb{N}\}$. So it seems that what this statement (and others like it) is saying, is that if you want more functions, you're better off having a large domain than a large codomain.
Is there an intuitive explanation for why this should be true?
AI: The inequality is made somewhat plausible by the observation that $|\mathbb N^2|$ is almost counting the number of $1$-element and $2$-element sets of natural numbers, while $|2^{\mathbb N}|$ counts all the sets of natural numbers. ("Almost" refers to the facts that each $2$-element set $\{a,b\}$ is counted twice, as $(a,b)$ and as $(b,a)$.) |
H: (1)Determination of local and global maximum/minimum points
Be $I=[-2,2]\subset \mathbb{R}$ and a function $f:I \Leftarrow \mathbb{R} $ (continuous and differentiable on $I$)
$$ f(x) =
\begin{cases}
xe^{x-1} & \text{if $x \leq 0$}\\
xe^{1-x} & \text{if $x>0$ }
\end{cases}$$
Determine the local & global maximum and minimum points of $f$ on $I$.
1) For $x\leq 0:$
$f'(x)=(1+x)e^{x-1}=0$
$\Leftrightarrow x=-1$
$f''(x)=(2+x)e^{x-1}$
$f''(-1)=e^{-2}>0$
So that there is a local minimum point for x=-1.
$f(-2)=-2e^{-3}$
$f(-1)=-e^{-2}<-2e^{-3}=f(-2) \Leftrightarrow f(-1)$is also the global minimum point.
2) For $x>0:$
$f'(x)=(1-x)e^{1-x}=0$
$\Leftrightarrow x=1$
$f''(x)=(x-2)e^{1-x}$
$f''(1)=-e^{0}=-1<0$
So that there is a local maximum point for x=1.
$f(2)=2e^{-1}$
$f(1)>f(2) \Leftrightarrow f(1)$is also the global maximum point.
My questions:
1) Is this correct?
2) and is it enough or do I also have to regard the limits for -2 and 2 (or is this unnecessary by saying the function is continuous & differentiable on I)?
AI: The solution is not entirely correct. For example, it contains the assertion that $f'(x)=(1+x)e^{x-1}$ for $x\le 0$.
This is not quite true. The derivative is correct for $x\lt 0$. But $f$ is not differentiable at $x=0$. You will be able to quickly check that
$$\lim_{x\to 0^-}\frac{f(h)-f(0)}{h} \ne \lim_{x\to 0^+}\frac{f(h)-f(0)}{h},$$
so the two-sided limit of the difference quotient does not exist.
Now in fact there is no real "trouble" at $0$, we do not have a local max or min at $0$. But one has to say that, with some explanation, since we cannot use the derivative criterion.
Remark: Perhaps one could get away with the mistake about the derivative, if the (incorrect) assertion that the function is differentiable was part of the problem statement. |
H: $f''$ bounded implies $f'$ is bounded
Suppose that $f''$ exists on $[0,1]$ and that $f(0)=0=f(1)$. Suppose also that $|f''(x)|\le K$ for $x\in(0,1)$. Prove that $|f'(1/2)|\le K/4$ and that $|f'(x)|\le K/2$ for $x\in [0,1]$.
The mean-value theorem might help, but I can't see how.
AI: Suppose $f'\left(\dfrac12\right)>K/4$ and $f\left(\dfrac12\right)>0$ (other cases are handled similarly.)
For $\dfrac12<t<1$ we have $f'(t)=f'\left(\dfrac12\right)+\int_{\frac12}^{t}f''(x)dx>\dfrac{K}{4}+\int_{\frac12}^{t}-Kdx = \dfrac{3K}{4}-Kt$.
Then $f(1)=f\left(\dfrac12\right)+\int_{\frac12}^{1}f'(x)dx > 0 + \int_{\frac12}^1\left(\dfrac{K}{4}-K\left(t-\dfrac12\right)\right)dt=0$, a contradiction. |
H: If $\mathcal A$ is diagonalizable, determine $\mathcal S$ and $\Lambda$ such that $\mathcal A=\mathcal S\Lambda\mathcal S^{-1}$
$$\mathcal A=\pmatrix{-5.25&-4.125&8.25\\-15&-0.5&15\\-4.75&-0.375&7.75}$$
If $\mathcal A$ is diagonalizable, determine $\mathcal S$ and $\Lambda$ such that $\mathcal A=\mathcal S\Lambda\mathcal S^{-1}$
So far I have calculated $det(\lambda I-\mathcal A)=0$
and ended up with:
$$\lambda^3-2\lambda^2-59.005\lambda+170.813=0$$
Now I'm stuck.....how do I find the eigenvalues from this?? Any help would be great appreciated! :)
The ultimate goal of this question is to find $\mathcal S$ and $\Lambda$.
This is what I have found using $\lambda_1=3$, $\lambda_2=7$ and $\lambda_3=-8$:
$$\mathcal S=\pmatrix{1&0&0\\0&2&0\\1&1&0}$$
$$\Lambda=\pmatrix{3&0&0\\0&7&0\\0&0&-8}$$
But when I try to produce $\mathcal S^{-1}$ I get a matrix of infinities!! Is it possible for someone to explain this?? It is greatly appreciated. :)
AI: ${\lambda}^{3}- 2{\lambda}^{2}- 59\lambda+168=0$
I did it with maple. |
H: Simple algebraic logic question
$2x^2 + 3y^2=0$. This is possible only when both the value of $x$ and $y$ are zero. But the thing is I fail to understand the significance of this equation. Why would such an equation exist or why we create them?
AI: As you note, the only solution to $\;2x^2 + 3y^2 = 0$, is indeed $(x, y) = (0, 0)$. It is not a very interesting question, in itself. No one with the least bit of creativity would simply create the equation to pose as an exercise: few teachers or texts would do so, I imagine.
But it may happen that something of the sort "crops up" in the process of solving a more complex equation, after simplifying/canceling, or when searching for "zeros."
If you did mean to write $2x^2 + 3x^2 = 5x^2 = 0 \implies x = 0$, that might be even less interesting of an equation, but may crop up when finding the zero's of, say, $f(x) = 5(x^4 - x^2) = 5x^2(x^2 - 1) = 5x^2(x-1)(x+1),\;$ which is not too terribly interesting, but the point is, many exercises are created simply to test one particular concept or technique, as this slightly more complicated function might be used. |
H: The minimum possible value of $\,\,|w|^2+|w-3|^2+|w-6i|^2$
I am stuck on the following problem :
What is the minimum possible value of $\,\,|w|^2+|w-3|^2+|w-6i|^2\,\,,w \in \Bbb C,i=\sqrt{-1}\,\,$?
The options are $\,\,15,45,20,30.$
I have no idea how to tackle it effectively. Some detailed explanation will be of great help.
AI: Let $w=x+iy$, where $x$ and $y$ are real. Then the expression we are looking at is
$$x^2+y^2 +(x-3)^2+y^2 +x^2+(y-6)^2.$$
Expand, and complete the squares. I expect that part will give you no trouble. If it does, I can show in detail how to finish. We will get lucky and be able to divide by $3$, making the arithmetic simpler. |
H: Is "have the same cardinality" a equivalence relation?
A relation is a subset of the Cartesian product of two sets, if "have the same cardinality", denoted as $R$, is a relation, then there must exist set $A, B$, such that $R \subset A \times B$. What are $A, B$ then? They cannot be "set of all sets", because there is no such set according to axiom of regularity (ZFC set theory).
Did I miss something?
Thanks.
AI: You are right, this is not an equivalence relation if you require that the domain be "all sets". However, if you restrict yourself to, say, all subsets of a given set $X$, then this is an equivalence relation now. (And, by picking an appropriate $X$, we can always ensure to include in the domain of the relation whatever sets we may actually be interested in.)
It is common, however, not to worry about this when arguing about cardinalities. After all, it is for sure true that for any sets $A,B,C$, we have that $A$ has the same size as $A$, if $A$ has the same size as $B$ then $B$ has the same size as $A$, and if $A$ has the same size as $B$ and $B$ has the same size as $C$ then $A$ has the same size as $C$. This is what matters when actually working with cardinalities, rather than whether something or other satisfies the formal definition of equivalence relation.
That being said, there is one reason why we may want this to be an equivalence relation, namely so that we can take equivalence classes and work with them. If we just work with the class of all sets, then the equivalence classes would be proper classes (that is, they would be "too big" to be sets), and this may lead to issues when formalizing what one is doing. In axiomatic set theory, the standard solution to this problem is provided by the axiom of foundation. This assigns to each set a rank (formally, an ordinal), which allows us, via a little detour, to define equivalence classes as sets. Another solution is provided by the axiom of choice, that allows us to pick canonical representatives from each equivalence class (certain ordinals). |
H: $f(x)=x$ if $x$ irrational and $f(x)=p\sin\frac1q$ if $x$ rational
Define the real-valued function $f$ on $\mathbb{R}$ by setting $f(x)=x$ if $x$ is irrational, and $f(x)=p\sin\frac1q$ if $x=\frac{p}q$ is written in lowest terms. At what points is $f$ continuous?
I'm pretty sure $f$ is continuous at no point. For any point $x$, we can find $\frac{p}q$ that's really close to $x$, but such that the value $p\sin\frac1q$ is nowhere near $x$. For example, if $x=0$, we have $f(x)=0$, and we can choose $y$ as something like $\frac{34567}{10000000000}$ so that its value is really close to $x$, but $\frac{p}{q}$ is unlikely to be close to $x$. But since the value of $\sin\frac{1}{q}$ fluctuates, I'm unable to turn this into a rigorous proof.
AI: Let $x \in \mathbb{Q}^+$. One can show that $f(x) < x$. Let $\epsilon = x - f(x)$. For any $\delta > 0$, we can choose an irrational number $y$ such that $x < y < x + \delta$. $f(y) - f(x) = y - f(x) > x - f(x) = \epsilon$. Thus $f$ is not continuous at $x$.
For $x \in \mathbb{Q}^-$, note that $f(-x) = -f(x)$, and negating a function doesn't affect continuity.
For $x = 0$, we want to find $\delta$ such that, if $|y - 0| < \delta$, then $|f(y) - f(0)| < \epsilon$. Let $\delta = \epsilon$. Since $|f(y)| \le |y|$, this is a good choice of $\delta$. Our argument for rationals does not work, because $f(x) \not< x$.
$f$ should be continuous at irrationals, but I can't nail down something rigorous. Informally: a small $\delta$ means that all $y = p/q$ that are less than $\delta$ away from $x$ will have a large $q$. This makes $f(y)$ very close to $y$, and so we only need to pick a $\delta$ slightly smaller than our $\epsilon$. I think one would have to bound $f(y) - y$ somewhere to show that. (Taylor series?)
So the argument would go something like this: choose an arbitrary $\epsilon$, determine a large enough $q$, determine a $\delta$ such that $(x - \delta, x + \delta)$ doesn't contain any smaller $q$s. The "large enough $q$" I have trouble with.
EDIT: Think I have something rigorous.
First, let's find a bound on $|f(y) - y|$. By Taylor series,
$$\sin{\frac{1}{q}} = \frac{1}{q} - \frac{1}{3! \cdot q^3} + \frac{1}{5! \cdot q^5} - \frac{1}{7! \cdot q^7} + \cdots$$
We can truncate this and put a bound on the error:
$$\sin{\frac{1}{q}} = \frac{1}{q} + R(x), \quad |R(x)| \le 1 \frac{(1/q)^2}{2!} = \frac{1}{2q^2}$$
Thus, $|\sin{\frac{1}{q}} - \frac{1}{q}| \le \frac{1}{2q^2} \implies |p\sin{\frac{1}{q}} - \frac{p}{q}| \le \frac{p}{2q^2} \implies |f(\frac{p}{q}) - \frac{p}{q}| \le \frac{p}{2q^2}$
Choose an arbitrary $\epsilon > 0$. We want to find a "large enough" $q$. If we restrict $\delta < 1$, then we know that $\frac{p}{q} < x + 1$. Then $|f(\frac{p}{q}) - \frac{p}{q}| \le \frac{p}{2q^2} < \frac{x + 1}{2q}$. If $q > \frac{x + 1}{2\epsilon}$, then $|f(\frac{p}{q}) - \frac{p}{q}| < \epsilon$. So we take the distance from $x$ to the nearest rational with denominator $\lceil \frac{x + 1}{2\epsilon} \rceil$, and let that be $\delta$ (or $1$, whichever is smaller). |
H: If the product of two integers is odd, then both are odd
Prove $ \forall x, y \in \mathbb{Z}$ if $x\times y$ is odd then $x$ and $y$ are odd.
Is this valid proof?
Proof by contrapositive. The contrapositive of the implication is: $ \forall x, y \in \mathbb{Z} $if $x$ and $y$ are not both odd then $x \times y$ is even.
To prove the contrapositive of the implication, let $x = 2k$ for some arbitrary $k \in \mathbb{Z}$. We cannot assume anything about $y$ to keep generality. Then $x \times y = 2k \times y = 2ky = 2(ky)$. Because $x \times y$ is a multiple of $2$ the implication holds.
AI: You have a good idea, but that's not quite the contrapositive. The contrapositive would be: "If $x$ and $y$ (integers) are not both odd, then $xy$ is also not odd." Another way of phrasing this would be "Let $x,y\in\Bbb{Z}$. If $x$ or $y$ is even, then $xy$ is even." I think you can finish it from here! |
H: Show that $7 \mid( 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47})$
I am solving this one using the fermat's little theorem but I got stuck up with some manipulations and there is no way I could tell that the residue of the sum of each term is still divisible by $7$. what could be a better approach or am I on the right track? Thanks
AI: $6^{47} \equiv (-1)^{47} = -1^{47}\mod 7$
$5^{47} \equiv (-2)^{47} = -2^{47}\mod 7$
$4^{47} \equiv (-3)^{47} = -3^{47}\mod 7$
Hence $ 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47} \equiv 0 \mod 7$. |
H: Are these combinations and permutations correct?
I wanted to know if I did these 3 questions correctly:
There are 100 distinct people in line. How many arrangements are there?
Ans: Combination - 100!
There are 30 distinct objects. How many (unordered) selections of 6 objects can be obtained?
Ans: Combination - (30, 6) = $$\frac{30!}{6!*24!}$$
How many different strings can be formed by reordering the letters of the phrase
’TYRANNOSAURUS REX’, counting the space?
Ans: Permutation - $$\frac{17!}{3!2!2!2!2!}$$ because there are 3 R's and 2 A, N, S, and U's and the rest are 1's.
Please correct me if I'm wrong. It's for a sample exam and the professor did not post solutions.
AI: Your resulting expressions are all "spot on," but for problem $(1)$, we have a permutation of 100 folks, (and not a combination, though you do indeed have a combination in your second problem). Otherwise, it seems clear that you know what you're doing in each of these problems. |
H: Logical Implications Quiz Help
These are fill in the blank questions. I scored an 18 out of 25, but I don't see where I am making mistakes. Please help me understand these conditional logic statements!
The answer options are:
IMPLIES (-->)
IS IMPLIED BY (<--)
IFF (<-->, both "IMPLIES" and "IS IMPLIED BY")
NONE OF THE ABOVE
I'm viewing these problems by saying "if P, then Q" in my head. If that statement is true, then I mark IMPLIES. I also check the converse: "if Q, then P." If the converse is true, then I also mark IS IMPLIED BY. Marking both is equivalent to IFF. If neither "if P, then Q" or its converse are true, I mark NONE OF THE ABOVE.
x > 1 ________ x $\ge$ 1 Answer: NONE
x > 1 ________ x > 0 Answer: IMPLIES
x > 1 ________ x + 1 > 0 Answer: IMPLIES
x > 1 ________ x - 1 > 0 Answer: IFF
x > 1 ________ x$^2$ > 1 Answer: IMPLIES
x > 1 ________ x + y > y Answer: IMPLIES
x > 1 ________ xy>y Answer: IFF
x>1 ________ x>-x Answer: IMPLIES
x>1 ________ x>y AND y>1 Answer: IMPLIED BY
x >1 ________ x>y OR y>1 Answer: NONE
x>1 ________ xy>y AND y>0 Answer: IMPLIED BY
x>1 ________ xy>y OR y>0 Answer: NONE
x<1 ________ x$\le$1 Answer: NONE
x<1 ________ x<0 Answer: IMPLIED BY
x<1 ________ x+1 < 0 Answer: IMPLIED BY
x<1 ________ x-1 < 0 Answer: IFF
x<1 ________ x$^2$ < 1 Answer: NONE
x<1 ________ x+y $\le$ y Answer: IMPLIED BY
x<1 ________ xy < y Answer: NONE
x<1 ________ x < z-x Answer: IFF
x<1 ________ x < y OR y < 1 Answer: NONE
x<1 ________ x < y AND y < 1 Answer: IMPLIED BY
x<1 ________ xy < y OR y > 0 Answer: NONE
x<1 ________ xy < y AND y > 0 Answer: IMPLIED BY
x<1 ________ x < 0 OR x$^2$ < 1 Answer: IMPLIED BY
AI: 1: If $x>1$, then it is certainly true that $x\geq 1$. The converse is false ($1\geq1$, but it is not the case that $1>1$). So, the answer should be "IMPLIES".
7: You have ignored the possibility that $y$ is negative. Take $x=2$ and $y=-1$. Then $x>1$, but $xy=-2<-1$. So, the forward implication is false. The backward implication is false as well: take $x=-1$ and $y=-2$. Then $xy=2>-2$, but $x\leq 1$. So, the answer should be "NONE".
13: If $x< 1$, then certainly $x\leq1$; the converse doesn't need to hold. So the answer should be "IMPLIES".
17: It is true that $x<1$ does not imply $x^2<1$; take $x=-2$, for instance. However, the converse holds: $x^2<1$ is equivalent to $-1<x<1$, which implies $x<1$. So, this should be "IMPLIED BY".
20: If $x=0$ and $z=-2$, then $x<1$ but $0=x>-2=z-x$; so, the forward implication doesn't hold. On the other hand, if $x=2$ and $z=5$, then $x<z-x$ but it is not true that $x<1$. So, this should be "NONE".
21: The forward implication is true: if $x<1$, there are two cases: either $y<1$ or $y\geq 1$. If $y<1$, then the RHS is true; if $y\geq 1$ and $x<1$, then certainly $x<y$, making the RHS true. The backward implication is false: if $y=3$ and $x=2$, then RHS is satisfied, but it is not true that $x<1$. So, this should be "IMPLIES".
25: If $x<1$, there are two cases: $x<0$ or $0\leq x<1$. In the first case, RHS is satisfied; in the second, $0\leq x^2<1$, and so RHS is still satisfied. So, the forward implication is true. For the backward: if $x<0$, then $x<1$; if $x^2<1$, then $-1<x<1$, and so $x<1$. So, 25 should be "IFF". |
H: The order of a group with two generators
Let $G$ be an abelian group generated by $x$ and $y$ such that the order of $x$ is $16$, the order of $y$ is $24$ and $x^2=y^3$. What is the order of $G$?
The elements of $G$ are of the form $x^ny^m$ with $n=1,3,5,7,\cdots,15$ and $y=1,2,3,4,\cdots,23$. But I don't know how to quickly eliminate elements further, such as $xy^3=x\cdot x^2=x^3$. Can you help me with this? Thank you.
AI: Consider the map ${\bf Z}/16{\bf Z}\times{\bf Z}/24{\bf Z}\to G$ extending $(1,0)\mapsto x$, $(0,1)\mapsto y$.
More specifically, invoke the first isomorphism theorem. What is the order of $(2,-3)$?
(Technically this yields that maximal possible order of $G$. Other possible orders are divisors of it.) |
H: Is it possible for a function(f) to be $O(f)$ but not $o(f)$?
Is it possible for a function(f) to be $O(f)$ but not $o(f)$? or $o(f)$ but not $O(f)$?
I guess it might be possible for a function that is not monotonically increasing.
Is there an example of this case?
Added: Is it correct if I say subtracting $\theta(f)$ from $O(f)$ equals $o(f)$?
AI: Certainly: the function $f(x)=x$ is such a function. In fact $f$ is always $O(f)$ and never $o(f)$ (when these both make sense).
Added: As Did points out in the comments, I am assuming here that $f$ is not identically $0$ in a neighborhood of the target; in particular, if we’re looking at the behavior of $f$ as $x\to\infty$, I’m assuming that $f$ is not eventually $0$. |
H: Can one show me how to plot this graph by hand (composition).
Can one show me how to plot this graph by hand (composition).
$$\frac{1}{1-x^2}$$
Plot[1/(1 - x^2), {x, -5, 5}]
Output result see picture in the end
Fine, thanks, I got it by plot 1/(1 + x) and 1/(1 - x)
Plot[{1/(1 - x), 1/(1 + x), 1/(1 - x) 1/(1 + x)}, {x, -5, 5}]
Output result see picture in the end
AI: $$\frac{1}{1-x^2}$$ has the domain $\Bbb{R}-\{-1,1\}$
If $x\to\{-1^{+},1^{-}\}$, then $\frac{1}{1-x^2}$ approaches $+\infty$, and if $x\to\{-1^{-},1^{+}\}$, then $\frac{1}{1-x^2}$ approaches $-\infty$
For $x\in(-\infty,-1)\cup (1,\infty)$, $\frac{1}{1-x^2}$ is negative, and tends to $0$ as $x\to\pm\infty$
This should give you some hints as to the general shape of the graph.
For greater accuracy, you may calculate the derivative of the graph at various points. |
H: Need helping solving this reflexive, symmetric, and transitive closure question
I'm working on a sample exam with no solutions and I've googled examples of similar problems but can't understand them either. Can someone show me a clear cut way to solve this problem:
Let R be a relation on { a,b,c,d,e } defined by R = { ( a,c ) , ( b,d ) , ( c,a ) , ( d,b ) , ( e,d ) } .Find S= the reflexive closure of R, T = the symmetric closure of R, and U = the transitive closure of R.
I know what it means to be reflexive, symmetrical, and transitive but don't understand the closure part. Thank you.
AI: The reflexive closure of $R$ is the smallest relation $S$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq S$ and (2) $S$ is reflexive. To form this, just add to $R$ the ordered pairs required by reflexivity:
$$S=R\cup\{\langle a,a\rangle,\langle b,b\rangle,\langle c,c\rangle,\langle d,d\rangle,\langle e,e\rangle\}\;.$$
Similarly, the symmetric closure $R$ is the smallest relation $T$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq T$ and (2) $T$ is symmetric. Here again you need only add to $R$ the missing ordered pairs needed for symmetric. There is only one; what is it?
Finally, the transitive closure of $R$ is the smallest relation $U$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq U$ and (2) $U$ is transitive. Finding the transitive closure can be little trickier than finding the reflexive or symmetric closure, because you may have to go through several stages of adding ordered pairs. $R$ has the ordered pairs $\langle a,c\rangle$ and $\langle c,a\rangle$, so $U$ will have those pairs as well. In order to be transitive, $U$ will therefore also have to include the pair $\langle a,a\rangle$. Similarly reasoning, reversing the rôles of $a$ and $c$, shows that $U$ must contain the ordered pair $\langle c,c\rangle$. And we can apply the same reasoning to the pairs $\langle b,d\rangle,\langle d,b\rangle\in R$ to conclude that $U$ must contain $\langle b,b\rangle$ and $\langle d,d\rangle$. Is there anything else? Yes: $R$ includes both $\langle e,d\rangle$ and $\langle d,b\rangle$, so $U$ does as well and must therefore contain $\langle e,b\rangle$ in order to be transitive. At this point we know that we must have at least
$$R\cup\{\langle a,a\rangle,\langle b,b\rangle,\langle c,c\rangle,\langle d,d\rangle,\langle e,b\rangle\}\;.\tag{1}$$
Is the relation $(1)$ transitive? If so, it’s the transitive closure of $R$. If not, what do you need to add to it to make it transitive? |
H: How to show that $\log_{10} n$ is not a rational number if $n$ is any integer not a power of $10.$
How to show that $\log_{10} n$ is not a rational number if $n$ is any integer not a power of $10.$
If not, let $\log_{10}n=\dfrac{p}{q}$ for some $p,q(\ne0)\in\mathbb Z$ where $(p,q)=1.$
Then $q\log_{10}n=p\implies\log_{10}n^q=p\implies\log_{10}n^q=\log_{10}10^p\implies n^q=10^p$.
I don't know what to do next.
Added: I can see $5,2$ are the only prime factors of $n.$ But I cant get why the same number of 5 and 2 would occur in $n?$
AI: What you had written brought you close to the end. We show how to finish.
From the assumption $\log_{10} n=p/q$ we obtain $10^{p/q}=n$ and therefore $10^p=n^q$.
The only primes that divide $n$ are $2$ and $5$. Suppose that $2^a$ is the highest power of $2$ that divides $n$. Then $aq=p$. Similarly, if $5^b$ is the highest power of $5$ that divides $n$, then $bq=p$. It follows that $a=b$.
We have implicitly used the Fundamental Theorem of Arithmetic: Every integer $\gt 1$ has a unique (apart from order) factorization as a product of prime powers. |
H: Piecewise linear function close to continuous function
A continuous function $\phi$ on $[a,b]$ is called piecewise linear provided there is a partition $a=x_0<x_1<\ldots<x_n=b$ of $[a,b]$ for which $\phi$ is linear on each interval $[x_i,x_{i+1}]$. Let $f$ be a continuous function on $[a,b]$ and $\epsilon$ a positive number. Show that there is a piecewise linear function $\phi$ on $[a,b]$ with $|f(x)-\phi(x)|<\epsilon$ for all $x\in[a,b]$.
I want to construct $\phi$ bit-by-bit starting from the left at $a$. I consider the set $C=\{c\mid$the linear function $\phi$ from connecting $f(a)$ and $f(c)$ satisfies $|f(x)-\phi(x)|<\epsilon$ for all $x\in[a,c]\}$. This set contains $a$ and also some interval close to it (since $f$ is continuous we can find $\delta$ such that $|f(a+y)-f(a)|<\epsilon/2$ for all $0<y<\delta$. Consider its least upper bound, and choose something slightly less than it. But I have no way to know why this process will eventually cover the entire interval $[a,b]$.
AI: As julien hinted, the function is uniformly continuous by compactness of $[a,b]$. This means there exists $\delta$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon/2$. Define $\phi(x)=f(x)$ for $x=a,a+\frac{\delta}{2},a+2\frac{\delta}{2},\ldots$. Then "connect the dots" to make $\phi$ piecewise linear. |
H: How do I understand the meaning of the phrase "up to~" in mathematics?
I am reading a book that explains elementary number theory: Number Theory: A Lively Introduction with Proofs, Applications, and Stories by James Pommersheim, Tim Marks and Erica Flapan.
The authors say,
"We express this idea in the statement of the Fundamental of Arithmetic by saying that prime factorization are unique up to order. ... for example, 40 and -40 are equal up to sign, the functions $12x^2$ and $x^2$ are equal up to a constant factor, and so forth."
Because I am not an English native speaker, the phrase "up to" is ambiguous for me a little bit. I would like to figure out the meaning of the phrase and use it properly. How do I do?
AI: When one says "$X$ is true up to $Y$", then one means that it is not strictly speaking correct that $X$ is true, and the $Y$ which occurs after the up to clarifies the sense in which $X$ is not quite true.
Thus it really means "Roughly speaking $X$ is true, except for $Y$".
As you can see, this is a very informal construction. It could be used very abusively. In order to be acceptable, the sense in which $Y$ "corrects" $X$ needs to be very familiar to the reader, or the underlying logic should be made more explicit.
So when one says that factorizations in $\mathbb{Z}$ are unique up to order, then it is understood that of course we could have $a*b*c= b*a*c = c*a*b = \ldots$ and we don't want to count these as being essentially different factorizations. In this case this sloppy language is probably a good expository choice, since most readers have an intuitive idea of what the "exception" is, whereas spelling it out explicitly would probably involve something like the following:
Uniqueness of Prime Factorizations: If $r$ and $s$ are positive integers, $p_1,\ldots,p_r,q_1,\ldots,q_s$ are prime numbers and $p_1 \cdots p_r = q_1 \cdots q_s$, then $r = s$ and there is a bijection $\sigma: \{1,\ldots,r\} \rightarrow \{1,\ldots,s\}$ such that $p_i = q_{\sigma(i)}$ for all $1 \leq i \leq r$.
The point is that this more precise statement will be gibberish to someone who does not have a certain amount of mathematical sophistication. (On the other hand, at some point a student of mathematics should be able to supply the statement above or something equally explicit and logically equivalent. In fact I happen to recall one colleague of mine, a veteran research mathematician, who told me that he was always uncomfortable with the statement of uniqueness of factorization because of the vague phrase "up to". He has very high standards for clear exposition and thinking.) The book that you speak of is written for a much more general audience, so their expository choice is a good one.
(As an aside, I was first introduced to number theory in a course taught by Marks and Pommersheim, a course they taught to talented high school students during the summer over a period of many years. I haven't read the book that you mention, coauthored with Flapan, but I'd have to think that it is largely based on these courses. This course was one of the great influences on my intellectual life -- not coincidentally I am now a number theorist! -- so I expect the book is pretty good.) |
H: linear algebra (matrix representation of a linear mapping) problem
Find the matrix representation $A$ of a linear mapping $T$ : $\mathbb R^2 \to \mathbb R^2$ that rotates points $\pi$ radians clockwise, then reflects points through the line $x_2 = - x_1$. Determine the range of $T$. Determine if $(1,1)$ is in the range of $T$. Determine if this linear mapping $T$ is invertible by calculating the determinant of $A$.
Alright, I've been trying to solve the following problem for nearly 3 hours now, but no avail. I don't even fully understand the question. I didn't find any tutorials online that met my needs. I'd appreciate if someone could go through the solution with (brief if possible) explanation of the problem.
AI: It is sufficient to see how $T$ behaves on a basis of $\mathbb{R}^2$.
Without doing any computation, you can guess that $T$ is invertible, since both operations of reflecting through the line, and rotating by $\pi$ are invertible.
$T$ takes the point $e_1=(1,0)^T$, rotates by $\pi$ to get $(-1,0)^T$ and reflects in the line $L$ to get $e_2=(0,1)^T$. ($L$ is the line $y=-x$.)
$T$ takes the point $e_2=(0,1)^T$, rotates by $\pi$ to get $(0,-1)^T$ and reflects in the line $L$ to get $e_1=(1,0)^T$.
Hence the representation of $T$ in terms of the basis $e_1,e_2$ is given by: $A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.
To see if $(1,1)^T$ is in the range, solve $A(x,y)^T = (1,1)^T$, which quickly gives $x=y=1$, so yes, the point is in the range.
In fact, it is clear from this that the equation $A(x,y)^T = (a,b)^T$ has the solution $(x,y)^T = (b,a)^T$, for any $a,b$, hence $T$ (equivalently, $A$) is invertible.
The determinant is straightforward to compute as $\det A = 0 \cdot 0 - 1 \cdot 1 = -1$. |
H: Trigonometry: Law of Cosines
How to solve using rule of cosines? I can solve using law of sines but trying to check using rule of cosines is tripping me up, can anyone help clear things up?
AI: By writing down the law of cosines for this situation, we have
$$ 5^2=a^2+4^2-2\cdot a\cdot 4\cdot \cos 65^\circ$$
i.e.
$$ a^2-3.3809 a-9\approx 0.$$
That is a quadratic in $a$, hence you'll find two solutions, but one will be negative. |
H: If $f$ is twice differentiable in $(-1,1)$ and $f(0) = f'(0) =0$, does $\sum\limits_{n=1}^\infty f(1/n) $ converge?
In calculus, I'm trying to understand if the following forces the convergence of the sum. I think it does, but I have no clue as how to prove it.
Similar to subject line, here is the question:
If $f$ is twice differentiable in $(-1,1)$ and $f(0) = f'(0) =0$, does $\sum\limits_{n=1}^\infty f(1/n)$ converge?
AI: Using Taylor's theorem, we can write $f(x)=cx^2+h(x)x^2$, where $c$ is a constant and $h$ tends to $0$ as $x$ tends to $0$. This means we have the bound
$$|f(x)| < Ax^2$$
where $A$ is some positive real constant. Then your assertion follows immediately from comparison with the series $$\sum_1^\infty \frac{1}{n^2}.$$ |
H: Numerical Analysis and Big O
How can I show that $e^x -1$ is not $O(x^2)$ as $x\to0$
I'm not sure where to start. We can use Taylor's Theorem with remainder:
\begin{equation}
e^x = \sum\limits_{k=0}^n\dfrac{x^n}{n!} +\dfrac{f^{n+1}(\xi)}{(n+1)!}x^{n+1}
\end{equation}
Where $\infty < x < \infty$ and $\xi$ is between $0$ and $x$. However, when and why would you cut off at $n=1$? Also, what is the significance that $x\to0$?
AI: Saying $e^x-1=_0O(x^2)$ is similar to say that the function $x\mapsto\frac{e^x-1}{x^2}$ is bounded as $x$ tend to $0$ which isn't the case since
$$\lim_{x\to0}\frac{e^x-1}{x^2}=\infty$$ |
H: Linear algebra - Coordinate Systems
I'm preparing for an upcoming Linear Algebra exam, and I have come across a question that goes as follows: Let U = {(s, s − t, 2s + 3t)}, where s and t are any real numbers. Find the coordinates of x = (3, 4, 3) relative to the basis B if x is in U . Sketch the set U in the xyz-coordinate system.
It seems that in order to solve this problem, I'll have to find the basis B first! how do I find that as well?
The teacher hae barely covered the coordinate systems and said she will less likely include anything from the section on the exam, but I still want to be safe. The book isn't of much help. It explains the topic but doesn't give any examples.
Another part of the question ask about proving that U is a subspace of R3, but I was able to figure that one out on my own. I'd appreciate if someone could show me how to go about solving the question above.
AI: This might be an answer, depending on how one interprets the phrase "basis $B$", which is undefined in the question as stated:
Note that $(s, s - t, 2s + 3t) = s(1, 1, 2) + t(0, -1, 3)$. Taking $s = 1$, $t = 0$ shows that $(1, 1, 2) \in U$. Likewise, taking $s = 0$, $t = 1$ shows $(0, -1, 3) \in U$ as well. Incidentally, the vectors $(1, 1, 2)$ and $(0, -1, 3)$ are clearly linearly independent; to see this in detail, note that if $(s, s - t, 2s + 3t) = s(1, 1, 2) + t(0, -1, 3) = (0,0,0)$, then we must obviously have $s = 0$, whence $s - t = -t = 0$ as well. Assuming $B$ refers to the basis $(1, 1, 2)$, $(0, -1, 3)$ of $U$, it is easy to work out the values of $s$ and $t$ corresponding to $x$: setting $(s, s - t, 2s + 3t) = (3, 4, 3)$, we see that we must have
$s = 3$ whence $t = -1$ follows from $s - t = 4$. These check against $2s + 3t = 3$, as the reader may easily verify. The desired coordinates for $x$ in the basis $B$ are thus $(3, -1)$.
Think that about covers it, if my assumption about $B$ is correct.
Can't provide a graphic, but one is easily constructed noting that the vectors $(1, 1, 2)$, $(0, -1, 3)$ span $U$ in $R^3$ (the "$xyz$" coordinate system). |
H: Factor $x^4 - 11x^2y^2 + y^4$
This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (examples of my working below):
U-Substitution.
Guess and Check.
Reversing the question (multiplying the answer out).
Here is my working for each case so far.
(1) U-Substitution.
I tried a simpler case via u-substitution.
Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.
(2) Guess and Check.
To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:
$(x^2 \pm y^2)(x^2 \pm y^2)$.
But I'm at a loss as to how to get the final solution's middle term from guessing and checking.
(3) Reversing the question
Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.
The original answer:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$
$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$
$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$
$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$
The $3y^3x$ terms cancel out, and we are left with:
$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.
The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).
AI: \begin{align*}
x^4 - 11x^2y^2 + y^4
&= x^4 - 2x^2y^2 + y^4-9x^2y^2 \\
&= (x^2-y^2)^2-(3xy)^2 \\
&= (x^2-y^2-3xy)(x^2-y^2+3xy).
\end{align*} |
H: If $|f(z)|\le1$ if $|z|\le 1$ then $|a_k|\le1$
Given the polynomial $f(z)=a_nz^n+a_{n-1}z^{n-1}+ \dots+a_0$ is bounded by $1$ on a unit disc, which means $|f(z)|\le1$ if $|z|\le 1$. Prove that $|a_k|\le1$ for all $k$.
I haven't found any idea for this problem.
AI: Hint: Notice that $a_k= \frac{1}{k!}f^{(k)}(0)$ and use Cauchy's integral formula. |
H: How can an oblique asymptote be $y = x$ , as $x\to \infty$?
In my Calculus book, an oblique asymptote defined as:
Oblique Asymptote:
the function $y = f(x)$ has an oblique asymptote $y = mx + n$, if:
$$\lim_{x\to \infty} {f(x) \over x} = m$$ where $m$ is a finite number.
$$\lim_{x\to \infty} [{f(x) - mx}] = n$$
where $n$ is a finite number as well.
My problem:
In the exercises book, there's an exercise:
$$f(x) = x + {\sin(x) \over x} $$
I found there's no vertical asymptote, as in the solutions.
But, I also found that there's no oblique asymptote,
but the solutions tell the opposite.
My solution:
$$m = \lim_{x \to \infty} x + {\sin x \over x} = \lim_{x \to \infty} x + {0} = \lim_{x \to \infty} {x} = {\infty}. $$
As we can see, by his own definition, the asymptote couldn't be - because $m$ is infinity,
as the definition defines that $m$ and $n$ must be finite.
So how the oblique asymptote is $y = x$?
Thank you!
AI: We have
$$m=\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}1+\frac{\sin x}{x^2}=1$$
and
$$n=\lim_{x\to\infty}f(x)-x=\lim_{x\to\infty}\frac{\sin x}{x}=0$$
so
$$y=x$$
is the oblique asymptote of the function at $\infty$ |
H: Calculating 7^7^7^7^7^7^7 mod 100
What is
$$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$
I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.
AI: $7^4 = 2401 \equiv 1 \pmod{100}$, so you only need to calculate $7^{7^{7^{7^{7^7}}}} \pmod{4}$. We know that $7 \equiv -1 \pmod 4$ and $7^{7^{7^{7^7}}}$ is odd, so $7^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3 \pmod 4$, and then $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$ |
H: Is my understanding of limit points wrong?
From Munkres p. 163:
"The notion of compactness is not nearly so natural as that of connectedness. From the beginnings of topology, it was clear that the closed interval $[a, b]$ of the real line had a certain property that was crucial for proving such theorems as the maximum value theorem and the uniform continuity theorem. But for a long time, it was not clear how this property should be formulated for an arbitrary topological space. It used to be thought that the crucial property of $[a, b]$ was the fact that every infinite subset of $[a, b]$ has a limit point."
Isn't that true for $(a,b)$ also? Isn't $a$ a limit point of $(a,b)$?
AI: The boldface sentence should be understood as follows:
It used to be thought that the crucial property of $[a,b]$ was the fact that every infinite subset of $[a,b]$ has a limit point in $[a,b]$.
Yes, $a$ is a limit point of $(a,b)$ in the space $\Bbb R$, but the set $\left\{a+\frac{b-a}n:n\ge 2\right\}$ is a subset of $(a,b)$ with no limit point in $(a,b)$. That difference turns out to be crucial. |
H: Prove that $(a-b) \mid (a^n-b^n)$
I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. But I found that:
\begin{align*}
(a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &= a^n-b^n.
\end{align*}
Doesn't this imply that $(a-b) \mid (a^n-b^n)$ as $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked?
Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated!
AI: The induction step can be handled by observing that $a^{(k+1)} - b^{(k+1)} = aa^k - ab^k + ab^k - bb^k = a(a^k - b^k) + (a - b)b^k$. Then by the inductive hypothesis, $a - b$ divides each summand.
Would say more but I must crash. Perhaps I can add the details of the full inductive proof manana. But it should be easy from here . . .
G'night, fellow math-heads! |
H: Air Strike Game
This is an Air Strike Game with the solution, I have added some questions regarding the solution and I would appreciate if someone could answer them.
Army $A$ has a single plane with which it can strike one of three possible targets. Army $B$ has one anti-aircraft gun that can be assigned to one of the targets. The value of target $k$ is $v_k$, with $v_1 > v_2 > v_3 > 0$. Army $A$ can destroy a target only if the target is undefended and $A$ attacks it. Army $A$ wishes to maximize the expected value of the damage and army $B$ wishes to minimize it. Formulate the situation as a (strictly competitive) strategic game and find its mixed strategy Nash equilibria.
The solution is take from Solution Manual for a course of Game Theory by Martin Osborne.
Let $(p^∗ , q^∗ )$ be a mixed strategy equilibrium.
Step 1. If $p^∗_i = 0$ then $q^*_i = 0$ (otherwise $q^*$ is not a best response to $p^*$ ); but if $q^*_i = 0$ and $i≤2$ then $p_{i+1} = 0$ (since player $i$ can achieve $v_i$ by choosing $i$).
Q: may be there is a typo here and instead of $p_{i+1} = 0$ should be written $p^*_{i+1} = 0$, I am not sure, for me $p_{i+1} = 0$ doesn't make sense. On the other hand, defender knowning that $p^*_{i+1}=0$ has incentive to increase $q^*_i$ because chances to attack $i$ target with $p^*_{i+1} =0$ actually increased.
Thus if for $i ≤ 2$ target $i$ is not attacked then target $i + 1$ is not attacked either.
Step 2. $p^∗ \neq (1, 0, 0)$ it is not the case that only target 1 is attacked.
Q: why it is actually so, in general I think there is no pure strategies at all, because, knowing that $p^∗ = (1, 0, 0)$ second player will play $q^∗ = (1, 0, 0)$, so not much sense in pure strategies. On the other hand, the decisions are made simultaneously, therefore it might be possible to play pure strategy. Why it so?
Step 3. The remaining possibilities are that only targets 1 and 2 are attacked or all three targets are attacked.
• If only targets $1$ and $2$ are attacked the requirement that the players be indifferent between the strategies that they use with positive probability implies that $p^∗ =(v_2/(v_1+v_2), v_1/(v_1+v_2), 0)$ and $q^∗=(v_1/(v_1+v_2),v_2/(v1 + v2 ),0)$. Thus the expected payoff of Army A is $v_1 v_2 /(v_1 + v_2 )$. Hence this is an equilibrium if $v_3 ≤ v_1 v_2 /(v_1 + v_2 )$.
• If all three targets are attacked the indifference conditions imply that the probabilities of attack are in the proportions $v_2 v_3 : v_1 v_3 : v_1 v_2$ and the probabilities of defense are in the proportions $z − 2v_2 v_3 : z − 2v_3 v_1 : z − 2v_1 v_2$ where $z = v_1 v_2 + v_2 v_3 + v_3 v_1$ . For an equilibrium we need these three proportions to be nonnegative, which is equivalent to $z − 2v_1 v_2 ≥ 0$, or $v_3 ≥ v_1 v_2 /(v_1 + v_2 ).$
Q: I cannot derive proportions from the Step 2.
AI: Regarding Step 1 you are correct that it should be $p_{i+1}^*=0$. The reasoning is that if $q_i=0$ then whenever $A$ is tempted to play $i+1$ he should play $i$ instead, since whenever he plays $i$ he will certainly gain the value $v_i$, which is better than gaining $v_{i+1}$.
Re Step 2: If $p^*=(1,0,0)$ then the best reply for $B$ is $q^*=(1,0,0)$. But with these strategies player $A$ never destroys anything and could do better by changing strategy to $(0,1,0)$. Hence we can not have that $A$ plays $(1,0,0)$ in equilibrium.
Re the last step: Let us consider if there is an equilibrium where $A$ attacks target 1 and 2 only with positive probability. Clearly $q_3^*=0$, in that case. A standard fact from game theory tells us that $A$ in equilibrium must be getting the same expected payoff from playing 1 and 2. (If you are not familiar with this I recommend you reread the chapter in your textbook where this is discussed.) The payoff from playing 1 for $A$ is
$ v_1q(1-q_1^*) = v_1q_2^*$. The payoff from playing 2 is $v_2 q_1^*$. Thus we must have $$ v_1 (1-q_1^*) = v_2 q_1^*.$$ This equation can be solved to give $$q_1^*=\frac{v_1}{v_1+v_2}.$$ Hence $$q_2^* = 1-q_1^*= \frac{v_2}{v_1+v_2}.$$
The values of $p_1^*$ and $p_2^*$ can be found in a similar way, and the case where all three targets are attacked can also be handled in a similar way. |
H: A question regarding propositional logic
Good day, I'm currently studying for an exam and need to learn about propositional logic. Well, since I'm not good at English I'll just write what I've done so far:
$(A \land (B \rightarrow \neg A)) \rightarrow \neg B$
I started like this:
$(A \land (\neg B \lor \neg A)) \rightarrow \neg B$
$\neg (A \land (\neg B \lor \neg A)) \lor \neg B$
$\neg A \lor \neg (\neg B \lor \neg A)) \lor \neg B$
$\neg A \lor (B \land A) \lor \neg B$
$\neg A \lor \neg B \lor (B \land A) $
And I think it is the same as:
$( \neg A \lor A) \land (\neg B \lor B) $
Am I on the right road? Because I'm really not sure anymore. However, I think this is a tautology.
AI: Well, in this style of argument, why not proceed from your fourth line
$$\neg A \lor (B \land A) \lor \neg B$$
by rearranging
$$(B \land A) \lor \neg B \lor \neg A$$
whence
$$(B \land A) \lor \neg(B \land A)$$
which is of the form
$$\varphi \lor \neg\varphi$$
and hence a tautology. So the original wff is a tautology as you wanted to show. |
H: How to show that the index $[G: H]$ is invertible?
I am reading the book Elements of representation theory of associative algebras, volume 1. I have a question on page 176, the proof of Corollary 5.2. In order to applied (5.1)(b), we have to show that the index $[G: H]$ is is invertible as an element of $A=K$. In the proof of Corollary 5.2, $[G: H]=|G|/|H|=|G|$. We know that $p$ does not divide $|G|$. But how could we show that $|G|$ is invertible as an element in $K$? Thank you very much.
AI: Every integer not divisible by $p$ is invertible when viewed as element of $\mathbb F_p$ because not being divisible by $p$ in $\mathbb Z$ corresponds to not being $0$ in the field $\mathbb F_p$. |
H: Negative binomial with a rational power?
I didn't quite understand the expansion of, for instance $1 \over (1-x)^\alpha$, for $\alpha \in \mathbb Q$, for instance for $\alpha = {1\over 2}$ using the binomial coefficients. I know that for $\alpha \in \mathbb N$, ${1 \over (1-x)^\alpha}=\sum_{n=0}^∞ {n+\alpha-1 \choose \alpha-1}x^n$, but I don't know how to apply it when $\alpha$ isn't whole.
Thanks in advance for any explanation!
AI: Consider the definition of $\displaystyle {m\choose k}\color{grey}{=\frac{m!}{(m-k!)k!}}$.
Now note that $\dfrac{m!}{(m-k!)k!}=\dfrac{m\cdot (m-1)\cdots(m-k+1)\cdot(m-k)!}{(m-k)!k!}=\dfrac{m\cdot (m-1)\cdots(m-k+1)}{k!}$.
Consider $\dfrac{m\cdot (m-1)\cdots(m-k+1)}{k!}$. What do you need $m$ to be for it to make sense? |
H: How to show that $K[t]/(t^d)$ is indecomposable as a $K[t]$-module and $\operatorname{End}_{K[t]} (K[t]/(t^d)) \cong K[t]/(t^d)$?
How to show that
(1) $K[t]/(t^d)$ is indecomposable as a $K[t]$-module?
(2)$\operatorname{End}_{K[t]} (K[t]/(t^d)) \cong K[t]/(t^d)$?
I think that if (2) is true, then $\operatorname{End}_{K[t]} (K[t]/(t^d))$ is local since $K[t]/(t^d)$ has only two idempotents $0, 1$. Therefore $\operatorname{End}_{K[t]} (K[t]/(t^d))$ is local and hence $K[t]/(t^d)$ is indecomposable as a $K[t]$-module. Is this true? How to show (2)? Thank you very much.
AI: In general if $\varphi: R \rightarrow S$ is a surjective ring homomorphism and $M$ is a (left) $S$-module, $M$ is also an $R$-module via $\varphi$ and $\operatorname{End}_R(M) = \operatorname{End}_S(M)$. (Note that we always have $\operatorname{End}_S(M) \subseteq \operatorname{End}_R(M)$ and for the reverse inclusion, actually $\varphi$ being an epimorhism in the category of rings is enough: see these notes)
Thus if $I$ is a two-sided ideal of a ring $R$, considering $R/I$ naturally as a left $R/I$-module, we have $\operatorname{End}_R(R/I) = \operatorname{End}_{R/I}(R/I) \cong (R/I)^{\text{op}}$, where op denotes the opposite ring - you can ignore it in the commutative case. The last ring isomorphism above comes from a general fact: if $A$ is a ring and we consider $A$ as a left $A$-module and write $_{A}A$ for this module, we have $\operatorname{End}_A(_AA) \cong A^{\text{op}}$ via the maps $f \mapsto f(1)$ and $ \text{"right multiplication by $a$"}\leftarrow a$.
The assignments $R = K[t]$ and $I = (t^d)$ yields (2) , noting that $R$ is commutative here. |
H: Find area of the surface obtained by rotating curve around x-axis?
I got a curve $y=a \cdot cosh \frac{x}{a}$ where $|x|\le b$. The task is to find area of the surface obtained by rotating curve around x-axis.
Here is my solution. Unfortunately the result is not identical with the result of the textbook. Would you please look at this and tell me where is my mistake? Maybe there is a mistake in my textbook. Thanks
My attempt at a solution:
Formula for area of the surface obtained by rotating curve:
$$S=2\pi \int_{a}^{b}f(x)\sqrt{1+(f'(x))^2}\,dx$$
I calculated $(f'(x))$:
$$f'(x)=\frac{d}{dx}(a\cdot \cosh\frac{x}{a})=\ldots =\sinh\frac{x}{a}$$
so: $(f'(x))^2=\sinh^2 \frac{x}{a}$
Then I put it into the formula for area of the surface obtained by rotating curve:
$$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a \int_{a}^{b} \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}dx$$
I made 1st substitution $u=\frac{x}{a} $ and $du=\frac{1}{a}dx $ then I obtained:
$$2\pi a^2 \int_{a}^{b} \cosh(u)\sqrt{1+\sinh^2 (u)}\,du $$
I made 2nd substitution $v=\sinh(u)$ and $dv=\cosh(u)du$ then I got:
$$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}dv$$
To solve previous integral I need to calculate this integral $\int_{}^{}\frac{1}{\cos^3\varphi }=\int_{}^{}\sec^3\varphi $:
$$\int_{}^{}\sec^3\varphi =\frac{1}{2} \left( \sec\varphi \cdot \tan\varphi +\ln\right|sec\varphi +\tan\varphi \left|\right)$$
Then I returned back to $2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv$ and I made 3rd substitution: $v=\tan\varphi $ so $dv=\sec^2\varphi d\varphi$ then I got:
$$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sqrt{\sec^2\varphi }\sec^2\varphi d\varphi=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi$$
Now we can see that we just calculated the value of $\int_{...}^{...}\sec^3 \varphi d\varphi$ so:
$$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|sec\varphi +\tan\varphi \right|\right]_{...}^{...}$$
Then I made a substitution back from $\varphi $ to $x$ , I obtained:
$$S=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|\sec\varphi +\tan\varphi \right|\right]_{...}^{...}=2\pi a^2\frac{1}{2}\left[ v\sqrt{1+v^2}+\ln\left|\sqrt{1+v^2}+v\right|\right]_{...}^{...}=$$
$$=\pi a^2 \left[ \sinh\frac{x}{a}\sqrt{1+\sinh^2\frac{x}{a}}+\ln \left| \sqrt{1+\sinh^2\frac{x}{a}}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$
$$=\pi a^2 \left[ \sinh\frac{x}{a}\cosh\frac{x}{a}+\ln \left| \cosh\frac{x}{a}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$
$$=\pi a^2 \left[ \frac{1}{2}\sinh\frac{2x}{a}+\ln \left| e^{\frac{x}{a}}\right|\right]_{a}^{b}=\pi a^2 \left( \frac{1}{2}\sinh\frac{2b}{a}+\ln \left| e^{\frac{b}{a}}\right| - \frac{1}{2}\sinh 2 -\ln \left| e \right| \right)$$
AI: Hint: Use whe well-known hyperbolic identity
$$\cosh^2 x-\sinh^2x=1\iff 1+\sinh^2x=\cosh^2x$$
Use the above and you'll save about 80% of the work you did...:) |
H: How can I know the time difference between two cities almost at the same latitude?
Well I know that's the earth rotation speed is:
$v=1669.756481\frac{km}{h}$
I have two cities New York, Madrid almost at the same latitude and the distance between them is:
$d=5774.39$ $km$
I know that's :
$\Delta t=\frac{d}{v}=3.4582$ $hours$ --> +3:27
But after I calculated it I found it wrong (according to http://www.timeanddate.com/worldclock/) and (http://www.happyzebra.com/timezones-worldclock/difference-between-New%20York-and-Madrid.php) time difference = +6:00 ???!!!
AI: Speed of the earth at equator is equal to $1669.756481$ km/h. Speed of the earth at latitude of $40.400$ degrees equals
$$ 1669.756481 \cos\left(40.400\right )\,km/h $$ |
H: $w_1,w_2$ are distinct complex numbers such that $|w_1|=|w_2|=1$ and $w_1+w_2=1$
I am stuck on the following problem:
Let $w_1,w_2$ are distinct complex numbers such that $|w_1|=|w_2|=1$ and $w_1+w_2=1$.Then the triangle in the complex plane with $w_1,w_2,-1$ as vertices
must be isosceles ,but not necessarily equilateral
must be equilateral
I have to decide which of the aforementioned options is correct.
I tried with $\,\,w_1=e^{i\theta},w_2=e^{i \phi},w_3=e^{i \pi}$ but messed it up . Can someone explain how to tackle it?
AI: $$w_k=e^{i\phi_k}\;,\;\;k=1,2\;,\;\;\phi_k\in\Bbb R\implies 1=w_1+w_2=\cos\phi_1+i\sin\phi_1+\cos\phi_2+i\sin\phi_2\implies$$
$$\sin\phi_1=\sin(-\phi_2)\iff \begin{cases}\phi_1=-\phi_2\\{}\\\phi_1=\pi+\phi_2\end{cases}$$
In the first case we have
$$2\cos\phi_1=1\implies \phi_1=\pm\frac\pi3$$
In the second one we get
$$1=\cos\phi_1+\cos(\phi_1-\pi)=0\implies\;\;\text{contradiction}$$
Well, finish the argument... |
H: Cardinality of Hausdorff Space
Here is a theorem which proof I don't understand (taken from R. Engelking, General Topology).
Theorem: For every Hausdorff space $X$ we have $|X| \le \operatorname{exp}\operatorname{exp} d(X)$. ($d(X)$ denotes the cardinality of the smallest dense subset of $X$)
Proof: Let $A$ be a dense subset of $X$ such that $|A| = d(X)$ and $\{ \mathcal B(x) \}_{x\in X}$ a neighbourhood system of $X$.
To prove the first inequality assign to every $x \in X$ the family $\mathcal A(x) = \{ U \cap A : U \in \mathcal B(x) \}$ of subsets of $A$. From the equality $\overline{U \cap A} = \overline{U}$ it follows that the intersection of the closures of all members of $\mathcal A(x)$ equals $\{ x \}$; hence $\mathcal A(x) \ne \mathcal A(y)$ for $x \ne y$. Since the number of all distinct families $\mathcal A(x)$ is not larger than $\operatorname{exp}\operatorname{exp}|A|$, we have $|X| \le \operatorname{exp}\operatorname{exp} d(X)$.
I don't understand why "the number of all distinct families $\mathcal A(x)$ is not larger than $\operatorname{exp}\operatorname{exp}|A|$", in my opinion, because of $\mathcal A(x) \subseteq A$, there could be no more distinct $\mathcal A(x)$ than subsets of $A$, and so the cardinality could be at most $\operatorname{exp} A$ (that of the powerset), and not twice exponential?
AI: $\mathcal{A}(x)$ is not a subset of $A$. It is a set of subsets of $A$. |
H: Contour integration: $\int_0^{\infty} x^p /(x^4+1) dx$ where $-1 < p < 3$
I want to calculate $\int_0^{\infty} x^p /(x^4+1) dx$ where $-1 < p < 3$. My first guess is to let $f(z) := \frac {z^p}{z^4+1}$ and integrate this over $\gamma_R$ where
$$
\gamma_r = [-R,R] \cup \{Re^{i\theta} : \theta \in [0,\pi] \}
$$
The simple poles of $f$ are in $e^{k \pi i /4}$ where $k\in \{1,3,5,7\}$ so two of the poles lie in $\gamma_R$ for $R > 1$.
Is this a good approach ?
AI: Your approach will work just fine. However, you can reduce this to a problem where only one singularity needs to be considered.
In this answer, it is shown that
$$
\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)
$$
Setting $m=4$ yields
$$
\int_0^\infty\frac{x^p}{1+x^4}\,\mathrm{d}x=\frac{\pi}{4}\csc\left(\pi\frac{p+1}{4}\right)
$$ |
H: An Exercise in Kunen (A Model for Foundation, Pairing,...)
This is exercise I.4.18 in Kunen's Set Theory.
Derive $\forall y (y \notin y)$ from the Axioms of Comprehension and Foundation. Don't use the Pairing or Extensionality Axioms. Then find a 2 element model for Foundation, Extensionality, Pairing, and Union, plus $\exists y \forall x (x \in y)$ (so of course, $y \in y$).
Hint. If $y \in y$ use Comprehension to form $x := \{y\}$; the fact that there may be more than one $x$ doesn't affect the proof.
My attempt:
Suppose for contradiction $\exists y (y \in y)$. Fix such a $y$. Let $x = \{z \in y \mid z = y\}$ so by Comprehension we have that $x = \{y\}$. But then we see that $y \in x$ and $y \in y$ contradicting the Foundation Axiom which says: $\exists y (y \in x) \implies \exists y (y \in x \wedge \neg\exists z (z \in x \wedge z \in y)$.
That part I didn't have too much trouble with, but I can't for the life of me figure out the second part of the question. Every configuration I make of two elements hasn't worked. I'm sure there is a very simple and obvious answer, so please let me know! This has stumped me for the last couple days!
Thanks!
AI: Take the universe to be $\{x,y\}$. Let $\in=\{(x,y),(y,y)\}$. That is $x$ is the "empty set" and $y$ contains everything. Extensionality follows trivially. Foundation follows trivially as well because the only non-empty set contains the empty set. Union is also true since $\bigcup y = y$ and $\bigcup x=x$. Pairing is trivially true because Kunen defines pairing as $\forall x\forall y\exists z(x\in z\land y\in z)$, so the universal set is there to provide the pair of any sets.
Notice however that if the pairing axiom was $\forall x\forall y\exists z(\forall w(w\in z\leftrightarrow w=x\lor w=y))$, then there is no 2 element model that satisfies all these axioms. To see this let $\{x,y\}$ to be the universe and let the universal set be $y$. Then by pairing $\{x\}$ exists and by extensionality it is different from $y$ (since one contains one element while the other contains two). Hence $\{x\}=x$. From this foundation fails, exactly as you showed it in your proof. |
H: Question on step in the proof of Itō's formula (along the book of Revuz and Yor)
I am working through the proof of Itō's formula contained in the book "Continuous Martingales and Brownian Motion" by Revuz and Yor and am stuck at a point in the proof.
Theorem (Itō's formula). Let $X = (X^1, \ldots, X^d)$ be a continuous vector semimartingale and $F \in C^2(\mathbb{R}^d, \mathbb{R})$; then, $F(X)$ is a continuous semimartingale and
$$
F(X_t) = F(X_0) + \sum_i \int_0^t \frac{\partial F}{\partial x_i}(X_s)dX_s^i + \frac{1}{2}\sum_{i,j}\int_0^t\frac{\partial^2 F}{\partial x_i\partial x_j}(X_s)d\langle X^i, X^j\rangle_s.
$$
The equation immediately shows that $F(X)$ is a continuous semimartingale. Through stopping and approximation it suffices to prove the equation in the case where $F(\cdot)$ is a polynomial. Therefore it is enough to show the following claim, since the equation is true for constant functions.
Claim. If the equality holds for the polynomial $F$, then it holds for all polynomials
$$G(x_1, \ldots, x_d) := x_{i_0}F(x_1, \ldots, x_d), \;\; 1 \leq i_0 \leq d$$
Using the integration by parts formula, one obtains
$$
G(Y_t) = G(Y_0) + \int_0^t Y_s^{i_0}dF(Y)_s +\int_0^t F(Y_s)dY_s^{i_0} + \langle Y^{i_0}, F(Y)\rangle_t$$
This is now the point where I am not sure how to proceed exactly. Clearly, I want to "replace" the $dF(Y)_s$ in the first integral using Itō's formula for $F$ (which we assumed to hold) to obtain
\begin{align}
G(Y_t) = G(Y_0) & + \sum_{i=1}^d \int_0^t Y_s^{i_0}\partial_i F(Y_s) dY_s^i + \int_0^t F(Y_s) dY_s^{i_0} \\
& + \frac{1}{2}\sum_{i,j}^d \int_0^t Y_s^{i_0}\partial_{i,j}^2F(Y_s)d\langle Y^i, Y^j\rangle_s + \langle Y^{i_0}, F(Y) \rangle_t
\end{align}
My question now is, how to justify this in a rigourous manner or which intermediate steps I am missing, since it is not obvious to me why one can just make this formal replacement.
Thanks for any input.
AI: I think what you are missing is the "composition property" (I do not remember the name exactly) of stochastic integrals (this is proposition IV.2.4 in Revuz-Yor, this is the case $L^2$ but it can be later be generalized), if
$$ V_t = \int_0^t X_s dY_s $$
where
$$ Y_t = \int_0^t W_s dZ_s $$
then
$$ V_t = \int_0^t X_s W_s dZ_s$$
This can be reinterpreted in the differential notation as
$$ dV_s = X_s W_s dZ_s$$
In your case $Y = F(X) = \text{Ito's formula}$ and $G = V$ |
H: $q$ be a real polynomial of real variable $x$ of the form $q(x)=x^n+a_{n-1}x^{n-1}+....+a_1x-1 .\,\,$
I am stuck on the following problem:
Let $q$ be a real polynomial of real variable $x$ of the form $q(x)=x^n+a_{n-1}x^{n-1}+....+a_1x-1 .\,\,$ Suppose $q$ has no roots in the open unit disc and $q(-1)=0.$
Then which of the following options are correct?
$\lim_{x \to \infty}q(x)=\infty$
$q(3)=0$
$q(2) >0$
$q(1)=0.$
I have also found the following result interesting but I am not sure it has anything to do with the current problem.
I need some detailed clarification to get the desired result.
AI: (1) is correct because the leading coefficient is positive. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $q$. We have $\prod\alpha_i=(-1)^{n+1}$. We see that if one of the roots had modulus $>1$ then another would have modulus $<1$, which contradicts the condition on the roots of $q$. Therefore all the roots lie on the unit circle so the roots split into real roots $\alpha_i=\pm 1$ and complex roots $\beta_j$ on the unit circle and since complex roots come in pairs it is of the form
$$
q(X)=(X-1)^k(X+1)^l\prod_{j\in J}|X-\beta_j|^2
$$
So $q(2)=3^l\prod_j|2-\beta_j|^2>0$. Also $q(3)$ can't be zero, which answers $(2)$. The formula $\prod\alpha_i=(-1)^{n+1}$ becomes $q(0)=(-1)^k=-1$, so $k>0$ and odd, which implies $q(1)=0$.
Summing up: (1), (3) and (4) are true and the opposite of (2) is true, i.e. $q(3)\neq 0$. Note also that every polynomial of the above form with $l>0$ satisfies the conditions, so this also gives all possible $q$'s.
Remark: This is not in contrast to Tomas answer. The difference here is that Tomas did not take into consideration the complex roots of $q$, in this answer I do. |
H: Limit-preserving point-wise maximum
Let $F$ denote the set of monotonic functions in $N^{\infty} \rightarrow N^{\infty}$ (where $N^{\infty}$ includes 0 and $\infty$) ordered by the point-wise application of $\le$ i.e., for $f,g \in F$
$$
f \le g \iff \forall n \in N^{\infty} :~ f(n) \le g(n)
$$
then I'd like to show that the point-wise maximum ($\vee$), defined such that $(f \vee g)(n) = \max(f(n), g(n))$ is limit-preserving i.e.
$$
\inf_{(f,g) \in C} (f \vee g) = (\inf_{(f,g) \in C} f) \vee (\inf_{(f,g) \in C} g)
$$
for a chain (totally ordered subset) $C \subseteq F \times F$ ordered by the element-wise comparison.
AI: In short: since these functions take values in $\mathbb{N}^\infty$, pointwise, these inf are attained min by the well-ordering principle. The point is that over a chain $C$, the inf on the right are attained jointly for the same element $(f^*,g^*)\in C$.
I assume your definition of monotonic is $n\leq m\Rightarrow f(n)\leq f(m)$. But it could be strictly monotonic with the same conclusions.
First, it is clear that $f\vee g\in F$ if $(f,g)\in F\times F$. Now for any non-empty set $S\subseteq F$, and for every $n$, $\{f(n)\,;\,f\in S\}$ is nonempty in $\mathbb{N}^\infty$ whence $\inf_S f(n)$ exists and is a least element $\min_S f(n)\in \mathbb{N}^\infty$. So the function $\inf_S f=\min_S f$ is well-defined. Then for $n\leq m$, and for every fixed $g\in S$ we have
$$
\min_S f(n)\leq g(n)\leq g(m).
$$
Taking the inf over $g$ shows that $\min_S f$ is motonic and belongs to $F$.
So what you want to prove is a min-max theorem: for every $n\in\mathbb{N}^\infty$
$$
\min_{(f,g)\in C} \max (f(n),g(n))=\max \left(\min_{(f,g)\in C} f(n), \min_{(f,g)\in C} g(n) \right).
$$
Proof: note that $n$ is fixed. For every $(f_0,g_0)\in C$ fixed, it is clear that
$$
\max (f_0(n),g_0(n))\geq \max \left(\min_{(f,g)\in C} f(n), \min_{(f,g)\in C} g(n) \right).
$$
Hence the inequality $\geq $ holds by taking the inf over all $(f_0,g_0)\in C$.
The other direction is where the fact that $C$ is a chain is useful. Take $(f_0,g_0)\in C$ such that $\min_{(f,g)\in C} f(n)=f_0(n)$ and $(f_1,g_1)\in C$ such that $\min_{(f,g)\in C} g(n)=g_1(n)$. Then either $(f_0,g_0)\leq (f_1,g_1)$, or $(f_0,g_0)\geq (f_1,g_1)$ by chain condition. In the former case, we have
$$
g_0(n)\leq g_1(n)=\min_{(f,g)\in C} g(n)\leq g_0(n)\quad\Rightarrow \quad g_0(n)=\min_{(f,g)\in C}g(n).
$$
In the latter case, we would conclude that $f_1(n)=\min_{(f,g)\in C} f(n)$. So in any case, there exists $(f^*,g^*)\in C$ such that
$$
f^*(n)=\min_{(f,g)\in C} f(n)\quad\mbox{ and}\quad g^*(n)=\min_{(f,g)\in C} g(n).
$$
Finally
$$
\min_{(f,g)\in C}\max(f(n),g(n))\leq \max(f^*(n),g^*(n))
$$
which proves the desired reverse inequality. $\Box$ |
H: How to understand the limit in the generalized real system?
As stated on the title:
How to understand a sequence limit in the generalized real system $[-\infty,\infty]$?
AI: Exactly as in the typical case of $\mathbb{R}$. The only difference is that now, having a limit of $\infty$ or $-\infty$ counts as converging, rather than diverging. |
H: What does $K^{1/p}$ for a field $K$ mean?
In the proof of the finite generation of the invariant ring of a finite group acting on $k[x_1,\dots,x_n]$, at one time there is a symbol I don't understand. The situation is as follows.
$k$ is a field of characteristic $p<\infty$, and $P$ its prime field. Suppose that $k$ arose from $P$ by adjoining finitely many (algebraic or transcendental) elements. Then she writes:
Because the prime field $P$ is perfect, it follows that $k^{1/p}$ is finite with respect to $k$.
I would have suspected that $k^{1/p}=\{x\in\bar k:x^p\in k\}$. This was called "Wurzelkörper" earlier in the paper. What is it, and why does the above follow from $P$ being perfect?
Thanks!
AI: $P$ being perfect means that exponentiation by $p$ is an isomorphism of $P$, so we can take $p$-th roots of any element in $P$ . You are right with the definition of $k^{1/p}$. To prove the statement suppose $k$ is generated by $x_1,\ldots,x_n$ over $P$. I claim that $k^{1/p}$ is (finitely) generated by $x_1^{1/p},\ldots, x_n^{1/p}$. Indeed let $y\in k$ be of the form
$$
y=a_1x_1+\ldots+a_nx_n,\quad a_i\in P.
$$
If $b_i\in P$ is the $p$-th root of $a_i$ then the $p$-th root of $y$ in $\overline{k}$ is
$$
y^{1/p}=b_1x_1^{1/p}+\ldots+b_nx_n^{1/p}
$$
Remark: $p$-th roots of elements are unique in characteristic $p$. If $a^p=b^p$ then $a^p-b^p=(a-b)^p=0$, so $a=b$. |
H: How to describe a n-tuple of sequences
When I write computer programs, I often use something called a multidimensional array. I think the concept would be equivalent to an n-tuple of finite sequences.
Suppose I have the following four sequences and 4-tuple called $t$:
$$A = (0)$$
$$B = (3,4,2)$$
$$C = (6,7,2,5)$$
$$D = (3,2)$$
$$t = (A, B, C, D)$$
I'm trying to determine how to describe the $Nth$ term of a sequence within the tuple. For example, in the C programming language -- where arrays start counting at the $0th$ term -- I would write arr[3][1] to describe the $2nd$ term of the $4th$ sequence in the multidimensional array arr.
What is the mathematical notation to describe the same term in a sequence in an n-tuple? Would it be $t_{(3)(1)}$?
AI: The mathematical notation is, frankly, whatever you decide to make it! It is all about defining the notation you use, and making sure it doesn't collide with other notation.
For instance: you could have the elements (sets) of your tuple be $A^1,\ldots,A^n$, and let $A^i_j$ denote the $j$th element of $A_i$.
There are many other possibilities. The most important thing is that, unless you are using a very standard notation, you explain what you mean. |
H: Vector bundle definition
Is the condition
$$ \pi \circ \varphi (x,v) = x $$
in the definition of a vector bundle needed? In Milnor/Stasheff "Characteristic classes" the definition is given without it.
AI: It's not strictly necessary, because it is a consequence of the following
the map $v \mapsto \varphi (x, v)$ is an isomorphism between the vector spaces $\mathbb{R}^k$ and $\pi^{−1}(\{x\})$.
However, for a site like wikipedia, it is preferable to state the fact plainly instead of relying on all readers to deduce it. |
H: Plotting in maple/MATLAB
How do you plot the following parametric equation (equation of an ellipse) in the same graph
$$ x = a\cos{t} $$
$$ y = b\sin{t} $$
with varying value of $a$ and $b$ in either maple or MATLAB ?
Many thanks.
AI: And in MATLAB:
min_a = 1; max_a = 10;
min_b = .5; max_b = 5;
t = linspace(0,2*pi,101);
cost = cos(t);
sint = sin(t);
N = 21; % Number of distinct values of a, b to plot
a = linspace(min_a,max_a,N);
b = linspace(min_b,max_b,N);
x = kron(a',cost);
y = kron(b',sint);
figure; hold on
for i=1:size(x,1)
plot(x(i,:),y(:,:))
end
Code comparison for loop versus vectorization techniques:
clear all
clc
time = zeros(5,1);
for k=1:5
close all
tic
min_a = 1; max_a = 10;
min_b = .5; max_b = 5;
t = linspace(0,2*pi,101);
cost = cos(t);
sint = sin(t);
N = 21; % Number of distinct values of a, b to plot
a = linspace(min_a,max_a,N);
b = linspace(min_b,max_b,N);
x = kron(a',cost);
y = kron(b',sint);
figure(1); hold on
for i=1:size(x,1)
plot(x(i,:),y(:,:))
end
time(k) = toc;
end
mean(time(2:end))
clear all
time = zeros(5,1);
for k=1:5
close all
tic
min_a = 1; max_a = 10;
min_b = .5; max_b = 5;
t = linspace(0,2*pi,101);
N = 21; % Number of distinct values of a, b to plot
a = linspace(min_a,max_a,N);
b = linspace(min_b,max_b,N);
figure(1); hold on
for i=1:length(a)
for j=1:length(b)
for l=1:length(t)
x = a(i)*cos(t(l));
y = b(j)*sin(t(l));
plot(x,y)
end
end
end
time(k) = toc;
end
mean(time(2:end)) |
H: Doubling a point on an elliptic curve
I've a programming background and am just about to get into a project where Elliptic Curve Cryptography (ECC) is used. Although our libraries deal with the details I still like to do background reading so started with the ECC chapter of Understanding Cryptography. Everything was fine until I came upon this example of point doubling over $Z_{17}$:
I can't figure out how he gets $s$. For example the $(2\cdot1)^{-1}(3\cdot5^2+2)$ why does that evaluate to $2^{-1} \cdot 9$? From looking at it I would have thought $(3\cdot5^2+2)$ evaluates to $77$ so giving $2^{-1}\cdot77$ or $77\div2$ for the whole expression. Obviously the math doesn't work in the way I expect, is it something to do with the dot product not being normal multiplication? Or something else?
p.s. Sorry about formatting, I'm looking up how to do the Tex for the site now.
AI: All arithmetic is done modulo $17$, so
$$(2\cdot 1)^{-1}(3\cdot 5^2+2)\equiv 2^{-1}\cdot 9\bmod 17$$
is a congruence, not an equality in $\mathbb{Q}$.
Moreover, $9\cdot 9\equiv 13 \bmod 17$ because
$$9\cdot 9 = 81 = 13 + 68 = 13 + 4\cdot 17 \equiv 13 \bmod 17,$$
or, equivalently,
$$9\cdot 9\equiv (-8)\cdot (-8)\equiv 64\equiv 16\cdot 4\equiv -4\equiv 13\bmod 17.$$ |
H: Fourier Transform of Derivative
Consider a function $f(t)$ with Fourier Transform $F(s)$. So $$F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt$$
What is the Fourier Transform of $f'(t)$? Call it $G(s)$.So $$G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt$$
Would we consider $\frac{d}{ds} F(s)$ and try and write $G(s)$ in terms of $F(s)$?
AI: The Fourier transform of the derivative is (see, for instance, Wikipedia)
$$
\mathcal{F}(f')(\xi)=2\pi i\xi\cdot\mathcal{F}(f)(\xi).
$$
Why?
Use integration by parts:
$$
\begin{align*}
u&=e^{-2\pi i\xi t} & dv&=f'(t)\,dt\\
du&=-2\pi i\xi e^{-2\pi i\xi t}\,dt & v&=f(t)
\end{align*}
$$
This yields
$$
\begin{align*}
\mathcal{F}(f')(\xi)&=\int_{-\infty}^{\infty}e^{-2\pi i\xi t}f'(t)\,dt\\
&=e^{-2\pi i\xi t}f(t)\bigr\vert_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}-2\pi i\xi e^{-2\pi i \xi t}f(t)\,dt\\
&=2\pi i\xi\cdot\mathcal{F}(f)(\xi)
\end{align*}
$$
(The first term must vanish, as we assume $f$ is absolutely integrable on $\mathbb{R}$.) |
H: A sufficient and necessary condition of a continuous map
Prove that a map $f:X\to Y$ is a continuous iff $$\text{cl}f^{-1}(A)\subseteq f^{-1}(\text{cl}A）$$ for each $A\subset Y$, where $X$ and $Y$ are both topological spaces and $\text{cl}$ denotes the closure.
In fact,for any $A\subset Y$,we have $A\subset \text{cl}A$ where $\text{cl}A$ is closed in $Y$, thus $\text{cl} f^{-1}(A)\subseteq f^{-1}(\text{cl}A）$. But how about the converse?
AI: Let $A$ be closed in $Y$. This can be expressed by $A=\text{cl}A$. By hypothesis $f^{-1}(A)=f^{-1}(\text{cl}A)\supseteq \text{cl}f^{-1}(A)$. Can you go on from here? |
H: Prove that: the center of any group is characteristic subgroup .
Let $G$ be any group , $Z(G)$ is the center of the group $G$ , prove that :
$\forall \tau \in Aut(G) , \tau [(Z(G)] = Z(G)$
My first trial was to prove that the center of any group is the unique subgroup of its order hence is a characteristic . but i found a counterexample easily which is $D_8$ the Dihedrial group of order $8$ ( i use the notation $D_{2n}$ )
So , any hints ?
AI: Can you show that for any $x\in Z(G)$ and $y\in G$, $\tau(x)\tau(y)=\tau(y)\tau(x)$? This will say $\tau(x)$ commutes with every element of $G$ (since $y$ was arbitrary), thus $\tau(x)\in Z(G)$. This shows containment in one direction, can you show containment in the other direction now? |
H: Showing that the mean of translations of a function approaches 0 in $L_p$
Given $p \in (1,\infty)$, $f \in L^p(\Bbb R)$ and $T: \Bbb R \to \Bbb R,x \mapsto x+1$. How do I show that for $n \to \infty$ $$\frac{1}{n}\sum_{k=0}^n f \circ T^k \to 0$$ in $L^p$? I see that for very large $N$ $f$ and $f\circ T^N$ have their masses concentrated in different areas and hence one has $||f+T^N\circ f||_p^p \approx ||f||_p^p+||T^N\circ f||_p^p$, but I don't see how to make this explicit.
AI: In other words, you want to prove that the sequence of ergodic mean operators
$$
U_n:L^p(\mathbb{R})\longrightarrow L^p(\mathbb{R}) \qquad U_nf:=\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k
$$
converges pointwise to $0$.
First note that $U_T:f\longmapsto f\circ T$ is a surjective isometry from $L^p$ to itself. As a consequence, $\|U_n\|\leq 1$ for every $n$. So it suffices to prove the result for simple functions, by density. And if the property is true for $f_1,\ldots,f_k$, it is true for every linear combination of them. So it suffices to prove the result for characteristic functions $1_A$ for bounded measurable sets $A$. Finally, up to splitting $A$, we can assume that $A\subseteq (j,j+1)$. It is easy in this case. See below if needed.
The property is false for $p=1$, as shown by the example $f=1_{(0,1)}$. And false as well for $p=\infty$, as shown by $f=1$.
Now for every $k$, $1_A\circ T^k=1_{T^{-k}A}\subseteq (j-k,j-k+1)$. Since the latter are disjoint, it follows that $\|U_n1_A\|_p\leq \frac{1}{n}n^\frac{1}{p}\longrightarrow 0$ for $p>1$. |
H: Making a function in $W^{1,2}$ continuous
Let $\Omega$ be an open domain in $\mathbb{R}^n$, $u\in W^{1,2}(\Omega)$ and assume that for any $y$ in $\Omega$
$$\lim_{\varrho \to 0} \operatorname{osc}(u,B(y,\varrho)) \rightarrow 0 , \varrho \rightarrow 0$$ where $$\operatorname{osc}(u,B(y,\varrho))= \sup_{x\in B(y,\varrho)} u(x)-\inf_{x\in B(y,\varrho)} u(x)$$
how can I prove that $${\bar{u}}(x)=\lim_{\varrho \rightarrow 0} \frac{\int_{B(x,\varrho)} u(y) dy }{\operatorname{meas} B(x,\varrho)}$$
is continuous in $\Omega$ ?
EDIT: I am always referring to essential suprema, infima and oscillations, i.e. ignoring sets of measure zero.
AI: OK, I will try myself, please let me know what you think of this...
Essentially we have $$\bar u (x) =\lim_{\varrho\rightarrow 0 } \sup_{t\in B(x,\varrho)} u(t) = \lim_{\varrho\rightarrow 0 } \inf_{t\in B(x,\varrho)} u(t)$$
then let $\epsilon > 0$ and $|x-y|< \delta$
$\sup_{B(y,\delta)} u(t) \geq \bar u(x)$
But also:
$$-\bar u (y) =-\lim_{\varrho \rightarrow 0}\inf_{B(y,\varrho)} u(t)=\lim_{\varrho \rightarrow 0}\sup_{B(y,\varrho)} -u(t)\leq\sup_{B(x,\delta)}-u(t)= -\inf_{B(x,\delta)} u(t) $$
Hence $$\bar u (x) -\bar u (y) \leq \sup_{B(y,\delta)} u(t) -\inf_{B(x,\delta)} u(t)\leq osc(u,B(x,2\delta)) $$
and I can make the RHS as small as I want by assumption, in particular choosing small $\delta$ we can make it small than epsilon to give continuity. |
H: Limit on the expression containing sides of a triangle
To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$ and in case of equilateral triangle, the limit is 3.
Now how to proceed further?
AI: Hint: The issue is that the condition "sides of a triangle," though it poses an easily stated set of constraints ("Triangle Inequality") is not easy to work with algebraically.
There is, however, a standard trick. The positive numbers $a$, $b$, and $c$ are sides of a triangle if and only if there exist positive numbers $p$, $q$, and $r$ such that $a=q+r$, $b=p+r$, $c=p+q$.
(The numbers $p,q,r$ have a simple geometric interpretation. Draw the incircle of the triangle. This meets the sides of the triangle at $3$ points, which divide each side of the triangle into $2$ segments. These segments are equal in pairs, and $p$, $q$, and $r$ are the three lengths of these segments.)
If we substitute for $a$, $b$, and $c$ in our expression, we get fairly quickly the desired bounds. |
H: if f(x,y) = 1 for all (x,y) where R has a nice shape like a rectangle or triangle what is another geometric interpretation of....?
If $f(x,y)$ is greater than or equal to $0$ on region $R$ in the plane, then $\iint_R f(x,y)dA$ can be interpreted geometrically as the volume of the solid under the surface $z=f(x,y)$ and above $R$. If $f(x,y)=1$ for all $(x,y)\in R$, where $R$ has a nice shape such as a rectangle, triangle, etc what is another geometric interpretation of $\iint_R f(x,y) dA$ ?
Could the answer simply be a circle, since it seems to be asking for just another "nice shape" for which z always equals 1? I am a bit confused as to how to prove that it could also be a circle though.
This is for my Multivariable Calculus class.
Thank you!
AI: If $f(x,y)=1$ for all $x,y\in R$ then $$\iint_R f(x,y) \mathrm{d}A=\iint_R 1 \mathrm{d}A=A$$
i.e. the integration just returns the area of $R$ |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.