Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the area of the surface generated when the given curve is revolved about the x-axis Find the area of the surface generated when the given curve is revolved about the x-axis
The part of the curve $y=12x-2$ between the points $(\frac{5}{12},3)$ and $(\frac{13}{12},11)$
I understand how to solve this when I am dealin... | Since $\displaystyle S=\int_{\frac{5}{12}}^{\frac{13}{12}} 2π \left(12x-2\right)\sqrt{145} $, letting $u=12x-2$ and $du=12\;dx$ gives
$\displaystyle\frac{2\pi\sqrt{145}}{12}\int_3^{11} u\;du=\frac{\pi\sqrt{145}}{6}\left[\frac{u^2}{2}\right]_3^{11}=\frac{\pi\sqrt{145}}{12}(121-9)=\frac{28\pi\sqrt{145}}{3}$.
Alternative... | {
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How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know... | Hint: The numerator can be factorized as $$x^4-3x^3+2=x^4-1-3(x^3-1)=(x-1)((x^2+1)(x+1)-3(x^2+x+1))$$ and the denominator as $$x^4-5x^2+3x+1\\=x^4-1-(5x^2-3x-2)=(x-1)((x^2+1)(x+1)-(5x+2))$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Surface area of ellipsoid given by rotating $\frac{x^2}{2}+y^2=1$ around the x-axis Calculate the surface area of the ellipsoid that is given by rotating
$\frac{x^2}{2}+y^2=1$ around the x-axis.
My idea is that if $f(x)=\sqrt{1-\frac{x^2}{2}}$ rotates around the x-axis we will end up with the same figure. The formula ... | Looks like one of those times you'll just have to tangle with the algebra monster. $$\begin{align} \sqrt{1-\frac{x^2}{2}}\sqrt{1+\frac{x^2}{4-2x^2}} = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{4-2x^2}} \\ = \sqrt{\frac{2-x^2}{2}}\sqrt{\frac{4-x^2}{2(2-x^2)}} \\ = \frac{1}{2} \sqrt{\frac{(2-x^2)(4-x^2)}{2-x^2}}\end{align... | {
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Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the ini... | The nested radical $\sqrt{16 - 4 \sqrt{7}}$ is equal to $\sqrt{14}-\sqrt{2}$. Simply square both sides to prove that. Using $(a-b)^2=a^2+b^2-2ab$, we get$$(\sqrt{14}-\sqrt{2})^2=14+2-2\sqrt{2}\sqrt{14}=16-2\sqrt{2}\sqrt{2}\sqrt{7}=16-4 \sqrt{7}$$
You can read more about denesting on Wikipedia.
EDIT: Lucian's answer is ... | {
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove... | \begin{align} x^3-y^3 &=x^3\color{red}{-x^2y+x^2y}-y^3\\
&= x^2(x-y)+y(x^2-y^2)\\
&= x^2(x-y)+y(x-y)(x+y)\\
&=(x-y)\Big( x^2+y(x+y)\Big)\\
&=(x-y)( x^2+xy+y^2)
\end{align}
| {
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Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$
How prove this inequality. Thanks
a suggestion please to prove this inequality
| The inequality $a^ab^bc^c\le(abc)^{\frac{a+b+c}{3}}$ is incorrect. Let $a=b=1$, $c=4$
Then
$$a^ab^bc^c=4^4$$
while
$$(abc)^{\frac{a+b+c}{3}}=4^2$$.
However, the opposite inequality $a^ab^bc^c\ge(abc)^{\frac{a+b+c}{3}}$ is correct as demonstrated in Angel's answer.
| {
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How do I solve the recurrence relation without manually counting? Given the recurrence relation : $a_{n+1} - a_n = 2n + 3$ , how would I solve this?
I have attempted this question, but I did not get the answer given in the answer key.
First I found the general homogenous solution which is $C(r)^n$ where the root is 1 s... | Let $$f(x) = \sum_{n=0}^\infty a_n x^n.$$
Multiplying both sides of the given recurrence equation by $x^n$ and summing over $n\geqslant0$, the left-hand side becomes
$$\sum_{n=0}^\infty a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n = \frac1x\sum_{n=0}^\infty a_{n+1}x^{n+1} - f(x) = \frac1x\left(f(x)-1\right)+f(x).$$
The right-ha... | {
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Find all primes of the form Find all primes of the form $3\cdot2^{2^t-1}+1$.
For $t=1$ it works.And I was thinking that it should have some logic because the other numbers go really large and is meaningless to take them one by one.
| For $t>1$, the expression can never be prime.
Modulo 5 we get
$3 \cdot 2^{2^t - 1} + 1 \equiv (-2) \cdot 2^{2^t - 1} + 1 \mod 5$
$\equiv - 2^{2^t} + 1 \mod 5$
Since $t>1$, write $2^t = 4\cdot 2^{t-2}$. So $2^{2^t} = (2^4)^{2^{t-2}}$ and
$- 2^{2^t} + 1 \equiv -(2^4)^{2^{t-2}} + 1 \mod 5$
$\equiv -(1^{2^{t-2}}) + 1 \equi... | {
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with this inequality condition $xyz(x+y+z)=3$ Let $x,y,z$ be postive real numbers such that $xyz(x+y+z)=3$,show that
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{54}{(x+y+z)^2}\ge 9$$
I found this problem on math magazine and the solution does not hold,see:
and I want to use Holder's inequality
$$\dfrac{1}{x^... | We'll prove a stronger inequality.
Let $a$, $b$ and $c$ be positives such that $ abc(a+b+c)=3$. Prove that:
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}\ge 11$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$.
In another hand, $\frac{1}{a^2}+\frac{1}{b^2}+\... | {
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2-adic expansion of (2/3) I have been asked in an assignment to compute the 2-adic expansion of (2/3). It just doesn't seem to work for me though. In our definition of a p-adic expansion we have $x= \sum_{n=0}^{\infty}a_np^n$ with $0 \leq a_0<p$.
So I use the same method as for similar questions which seems to work fin... | If I am not mistaken you can also do the following:
$\frac{2}{3}=1+\frac{1}{-3}$ where
$\frac{1}{-3}=\frac{1}{1-4}=\frac{1}{1-2^2}=\sum_{n=0}^{\infty}2^{2n}$ (geometric series expansion).
This yields the whole expansion of $\frac{2}{3}$.
| {
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Problem with a simple convergent power series Create a power series around $x=0$ and find its radius. What is $f(1)$?
$$f(x)=\frac{1}{2+x}$$
Well according to me it is easy to see that the radius is $\left|x\right|<2$ and $f(1)=\frac{1}{3}$...right?
If I try to manipulate it to the geometric serie $f(x)=\frac{1}{1-x}$ ... | I start with
$$\frac{1}{t+1}=\sum _{k=0}^{\infty } (-1)^k t^k$$
Now I substitude
$$t=\frac{x}{2}$$
It gives me
$$\frac{1}{\frac{x}{2}+1}=\sum _{k=0}^{\infty } (-1)^k \left(\frac{x}{2}\right)^k$$
Multiply with $\frac{1}{2}$ I get
$$\frac{1}{2 \left(\frac{x}{2}+1\right)}=\frac{1}{2} \sum _{k=0}^{\infty } (-1)^k \left(\fr... | {
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Solve the following in vector form:
So i did a substitution to solve the system normally, and got
$x=17.67$
$y=9.67$
$z=10.67$
Where I am stuck is how to represent something like this in a vector form, maybe my solution was wrong in the first place.
| For example
\begin{align}
(1,2,-3)^T \cdot x &= 5 \\
(2,1,-3)^T \cdot x &= 13 \\
(-1,1,0)^T \cdot x &= -8
\end{align}
where $x = (x_1,x_2,x_3)^T$.
But more usual is using a matrix equation
$$
A x = b
$$
with $A$ using the above first argument vectors as row vectors and
$b = (5, 13, -8)^T$.
The solution would be $x = A... | {
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order of operation For the following expressions,why we can get the right answer even we do addition /subtraction first ?
*
*$3+4\times 11-5 = (3+4)\times(11-5) = 42$
*$6+4\times7-4$
*$5+2\times13-10$
*$4+7\times16-6$
*$3-2\times1+5$
| Expressed algebraically, you have:
$$\begin{align}
a+bc+d&=(a+b)(c+d)\\
a+bc+d&=ac+ad+bc+bd\\
a+d-ac-ad-bd&=0\\
a(1-c-d)+d(1-b)&=0\\
\end{align}$$
This is a single simultaneous equation in 4 unknowns, therefore it has an infinite number of solutions - you have just found some of them.
Rewriting to address OP comment
$$... | {
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The $n$’th Harmonic number The $n$’th Harmonic number is defined as follows: $$H_n = 1 + \frac12 + \frac13 +\ldots + \frac1n$$ for $n\geq 1$.
a) Show that for all $n≥0$: $H_{2^n} \geq 1+\frac n2$.
b) Show that for all $n≥0$: $H_{2^n} \leq 1+n$.
| (a). Base case:
for $n = 0,$ we have $H_{2^n}= H_1 = 1,$ and $1 + 0/2 = 1.$
Induction Step:
Let's assume $k \geq 0$ is an arbitrary number, and it holds for $k$:
$H_{2^k} \geq 1+ k/2.$
Then we prove it for $k+1$:
$$\begin{eqnarray}
H_{2^{k+1}} & = & H_{2^k} + \underbrace{\frac{1}{2^k+1} + \ldots +
... | {
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Prove that there exists a choice of $\pm$ signs Let $x$, $y$ be two real numbers. Prove that there exists a choice of $\pm$ signs so that:
$$ \pm \cos x \pm \cos y \pm \cos (y - x) \leq -1 $$
| Without loss of generality, we can assume $-\pi\le y\le x\le \pi$.
(If $x<y$, swap $x$ and $y$ and note that $\cos (x-y)=\cos(y-x)$.) Then
$-\frac{\pi}2\le \frac{x}2 \le \frac{\pi}2,\
-\frac{\pi}2\le \frac{y}2 \le \frac{\pi}2,\ \text{and}\ 0\le \frac{x-y}2\le \pi.$
\begin{align*}
A&=\cos x+\cos y+\cos (x-y)+1\\
&=... | {
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From the set $\{1,2,\ldots,n\}$ two numbers are chosen uniformly, with replacement. Find the probability that the product of the numbers is even. The answer is given, just the reasoning is unclear. Result: $${2\over n}\left[{n\over 2}\right]-\left({\left\lfloor{n \over 2}\right\rfloor\over n}\right)^2,\text{ where $\lf... | First Problem
There are $\left\lfloor\frac{n+1}2\right\rfloor$ odd numbers in $1\dots n$, so the probability that each number chosen is odd is
$$
\frac1n\left\lfloor\frac{n+1}2\right\rfloor
$$
Therefore, the probability that the product is even is
$$
\bbox[5px,border:2px solid #F0A000]{1-\frac1{n^2}\left\lfloor\frac{n+... | {
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Why does $\binom{n}{2} = \frac{n^2 - n }{2}$? In a proof in Introduction to Algorithms, the book says $\binom{n}{2} \cdot \frac{1}{n^{2}} = \frac{n^2 - n }{2}\cdot \frac{1}{n^{2}}$, which implies $\binom{n}{2} = \frac{n^2 - n }{2}$.
Why are these equivalent? Is it a special case of a more general rule?
| $\binom{n}{2}$ is the number of subsets of size $2$ from a set of size $n$. There are $n$ ways to choose the first element, and $n-1$ ways to choose the secon element. but sets are not "ordered", so we must divide by two and we have $\frac{n\cdot(n-1)}{2}$ such subsets. Thus $\binom{n}{2}=\frac{n(n-1)}{2}=\frac{n^2-n}{... | {
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Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a ... | You should prove one more thing: $k^3\neq3k^2+9k+7$. Since the equation $k^3-3k^2-9k-7=0$ has no integer solutions, the conclusion follows.
| {
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Infinite sequence series. Limit If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\ldots+\frac{x^{2^n}}{1-x^{2^{n+1}}},$$
then $\lim_{n\to\infty} A_n$ is
$$\text{a) }\ \dfrac{x}{1-x} \qquad\qquad \text{b) }\ \frac{1}{1-x} \qquad\qquad \text{c) }\ \frac{1}{1+x} \qquad\qquad \text{d) }\ \frac{x}{1+x}$$
How to do thi... | If you expand each of the geometric series in the $A_n$, and combine each of the series together as one sum (you can since these are each absolutely convergent), then you can demonstrate that this is tending to $x(1-x)^{-1}$.
What you need to make sure of though, is that $x^k$ can appear only once between each of the t... | {
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$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
| $$
\begin{align}
\int_0^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,\mathrm{d}x
&=\int_0^1\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1a}\\
&=\int_1^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1b}\\
&=\frac12\int_0^\infty\frac{(1+x^6)(1+x^2)}{1+x^{10}}\,\mathrm{d}x\tag{1c}
\end{align}
$$
Explanation:
$\mathrm{(1a... | {
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1... | First use partial fractions to get rid of the sine in the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$
We have the identity
$$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right... | {
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"timestamp": "2023-03-29T00:00:00",
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Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac... | Here is another way, and it combines integration and probability.
First off, consider the triple integral:
$$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$
Convert this integrand into a geometric series:
$$\frac{1}{1+x^2y^2z^2}=\sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}.$$
Replace the integrand with th... | {
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"timestamp": "2023-03-29T00:00:00",
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Fake induction proofs Question: Can you provide an example of a claim where the base case holds but there is a subtle flaw in the inductive step that leads to a fake proof of a clearly erroneous result? [Note: Please do not answer with the very common all horses are the same color example.]
Comment: Sometimes inductive... | Claim:
If $a$ is an odd square modulo $m$, then $a$ is a square modulo $2m$.
Proof:
Let $m=2^v k$ where $k$ is $m$'s largest odd factor. Proceed by induction on $v$.
Base case:
$v=0$, so $m=k$ is odd.
$a$ is a square modulo $k$, so, for some $x$, $k\mid x^2-a$. Modulo $2k$,
\begin{align*}
(k+x)^2 &= k^2+2kx+x^2\\
&= k+... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the following limit without L'Hopital I tried to evaluate the following limits but I just couldn't succeed, basically I can't use L'Hopital to solve this...
for the second limit I tried to transform it into $e^{\frac{2n\sqrt{n+3}ln(\frac{3n-1}{2n+3})}{(n+4)\sqrt{n+1}}}$ but still with no success...
$$\lim_{n... | For the first limit, it breaks into 2 factors with finite limits.
$$\lim{n \to \infty} \frac{2n^2-3}{7-n^2} = \frac{2n^2}{-n^2} =-2\\
\lim{n \to \infty} \frac{3^n-2^{n-1}}{3^{n+2}+2^n2} = \frac{3^n}{3^{n+2}} = \frac{1}{9}
$$
so the answer is $-\frac{2}{9}$.
For the second, rewrite it as
$$
\left(\frac{(3n-1)(2n-3)}{4n^... | {
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} |
Jacobi Triple product cases Hello I want some guidance how can I prove these identities of Jacobi Tripple Product
$$\sum_{n=-\infty}^\infty q^{2n^2+n}=\prod_{n=1}^\infty\frac{(1-q^{2n})^2}{(1-q^n)}$$
$$\sum_{n=-\infty}^\infty q^{n^2}=\prod_{n=1}^\infty\frac{(1-(-q)^n)}{(1+(-q)^n)}$$
$$\sum_{n=-\infty}^\infty q^{n(n+1)... | Start with the triple product identity
$$\sum_{n = -\infty}^\infty z^n q^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1}).$$
*
*Replacing $q$ by $q^2$, then setting $z = q$, we obtain
\begin{align}
\sum_{n = -\infty}^\infty q^{2n^2+n} &= \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n-1})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y... | Hint: Multiply by conjugate:
$$\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|=\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|\frac{|\sqrt{x^2+x}+\sqrt{y^2+y}|}{|\sqrt{x^2+x}+\sqrt{y^2+y}|}$$
Note that this part of the proof only works with $[A,\infty)$, where $A > 0$. For the interval $[0,A]$, observe that $[0,A]$ is closed, and $f$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Product of repeated cosec.
$$P = \prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$$
I realize that there must be some sort of trick in this.
$$P = \csc^2(1)\csc^2(3).....\csc^2(89) = \frac{1}{\sin^2(1)\sin^2(3)....\sin^2(89)}$$
I noticed that: $\sin(90 + x) = \cos(x)$ hence,
$$\sin(89) = \cos(-1) = \cos(359)$$
$$\sin(1) = \co... | I'm using a non-standard notation inspired by various programming languages in this evaluation because I think it's a bit easier to follow than the traditional big pi product notation. Hopefully, it's self-explanatory. Also, all angles are given in degrees.
This derivation uses the following identities:
$$
\cos x = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$.
My (wrong) solution I thought that I would get the s... | First,I donnot think the orthogonal projection is proper here. The orthogonal projections are often used to make infinity series to approach function,because the coefficients there is more simple,you can try to use error square or the maximun of error
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?
I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{... | Assume $$\sqrt{3}+ \sqrt{5}+ \sqrt{7}= \frac{a}{b}$$ for some integers $a,b$. Multiblying both sides by $\sqrt{3}+ \sqrt{5}- \sqrt{7}$ yields $$ (\sqrt{3}+\sqrt{5})^2 - 7 = \frac{a(\sqrt{3}+ \sqrt{5}- \sqrt{7})}{b}=\frac{a\left(a/b-2\sqrt{7}\right)}{b} \\ 2\sqrt{15}+8-7=\frac{a^2/b-2a\sqrt{7}}{b} \\ 2b\sqrt{15} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
How to show that $zw=1\implies w = z^{-1}, z = w^{-1}$? I need to show that
$$zw=1\implies w = z^{-1}, z = w^{-1}$$
But I don't know how to start. I've tried to consider $z=a+bi, w = c+di$ then:
$$zw = ac+adi+bci-bd\implies zw = ac-bd + (ad+bc)i = 1\implies \\ac-bd = 1\\ad+bc = 0$$
$$ac = 1+bd\\ad = -bc$$
$$c = \frac{1... | If you know in advance that the complex numbers form a field then $$zw = 1 \implies z^{-1}zw = z^{-1}1 \implies w = z^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring derivatives Well I have $$r\frac{d^2T}{dr^2} + \frac{dT}{dr} = -\frac{qr}{k}$$
I wonder how to do algebraic manipulation to go from this step to
$$\frac{d}{dr}(r\frac{dT}{dr}) = \frac{-qr}{k}$$
Is it even possible ?
I wonder how to write a second derivative in terms of a first derivative ?
| Use
\begin{align}
\frac{d}{dx}(f(x) g(x)) = f(x) \, \frac{d g(x)}{dx} + \frac{d f(x)}{dx} \, g(x)
\end{align}
to put
\begin{align}
x \frac{d^{2} f}{dx^{2}} + \frac{d f}{dx} = - \frac{q x}{k}
\end{align}
into the form
\begin{align}
\frac{d}{dx} \left( x \frac{df}{dx} \right) = - \frac{q x}{k}.
\end{align}
For the sake o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\int\log(\sqrt{1-x}+\sqrt{1+x}) \, dx$ Integration of
$$\int\log\left(\sqrt{1-x}+\sqrt{1+x}\right) \, dx$$
Please help to go through this problem as i have started with putting $x$= $cos2y$.
| Consider the integral
\begin{align}
I = \int\log(\sqrt{1-x}+\sqrt{1+x})dx
\end{align}
It is seen that
\begin{align}
I &= \int \ln\left[ \sqrt{1 -x} \, \left(1 + \sqrt{\frac{1+x}{1-x}} \right) \right] dx \\
&= \frac{1}{2} \int \ln(1-x) \, dx + \int \ln\left(1 + \sqrt{\frac{1+x}{1-x}} \right) dx
\end{align}
Now make the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Number of solution of Frobenius equation Oke I am trying to find all presentable of a number $n$ as sum $ax+(a+1)y$ where $a=0,1,\ldots$ and $x,y\geq0$ are integers.
I find that
$5=1+1+1+1+1=1+1+1+2=1+2+2=2+3=5$ so we have $5$ ways.
$7=1+1+1+1+1+1+1=1+1+1+1+1+2=1+1+1+2+2=1+2+2+2=2+2+3=3+4$ so we have $7$ ways.
I suppo... | The problem is slightly misstated; what you're actually counting is the number of different ways to write $n=ax+(a+1)y$ with $a,x\ge1$ and $y\ge0$, not $a,x,y\ge0$. Changing the lower limit on $a$ and $x$ rules out what would otherwise be an infinite number of solutions, $(a,x,y)=(0,x,n)$ with $x=0,1,2,\ldots$, as wel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simultaneusly solving $2x \equiv 11 \pmod{15}$ and $3x \equiv 6 \pmod 8$ Find the smallest positive integer $x$ that solves the following simultaneously.
Note: I haven't been taught the Chinese Remainder Theorem, and have had trouble trying to apply it.
$$
\begin{cases}
2x \equiv 11 \pmod{15}\\
3x \equiv 6 \pmod{8}
\en... | You must start from Bézout's identity between the moduli $15$ and $8$: $\,2\cdot 8-1\cdot 15=1$. From the solutions to the individual congruences $\color{red} {13} \pmod{15}$ and $\color{red}2 \pmod{8}$, you deduce the solutions to the system:
$$2\cdot 8\cdot\color{red} {13}-1\cdot 15\cdot\color{red}2=178\equiv 58\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove that the map $f(z)=\frac{1}{z}$ sends any line onto either a line or a circle. Show the cases in which the image is a line and the case in which the image is a circle.
I understand that representing the equation of line, ($ax+by+c=0$ $a,b,c\in\Bbb R$ $a,b\neq0$ at the same time), as $pz+\overline {pz} +2c=0$ w... | First of all, I think that the rquation of the line $ax + by + c = 0$, when written in terms of $z = x + iy$, should be
$\bar p z + p \bar z + 2c = 0, \tag{1}$
and not
$pz + \bar p z + 2c = 0; \tag{2}$
here $p = a + bi$. To see the problem with (2), note that it may be written
$(p + \bar p)z + 2c= 0, \tag{3}$
and that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Can flooring and ceiling thirds sum up to more than a whole? I'm working on a web layout which needs to divide an area into three columns, but do so using whole pixel values (due to subpixel rendering issues on some mobile devices). For this purpose I've decided to go with the following approach:
*
*The first two co... | The answer is no:
*
*$x\equiv0\pmod3\implies\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lceil\frac{x}{3}\right\rceil=\frac{x}{3}+\frac{x}{3}+\frac{x}{3}=\frac{3x}{3}=x$
*$x\equiv1\pmod3\implies\left\lfloor\frac{x}{3}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\left\lceil\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $e_0 = \frac{1}{2} $and $\forall n\in\Bbb{N}:e_n=\frac{(2n-1)^2}{2n(2n+1)}e_{n-1}$, find $\sum_{n\geq 0} e_n$ Consider the following optimal stopping game: The controller is presented with steps in a fair random walk (fair coin flips, $P(h)=P(t) = \frac{1}{2}$) and at each stage of the game, the controller can stop... | $$\begin{align}e_n &= \frac{(2 n-1)^2}{(2 n+1) (2 n)} e_{n-1} \\ &=\frac{(2 n-1)^2}{(2 n+1) (2 n)} \frac{(2 n-3)^2}{(2 n-1) (2 n-2)}\cdots \frac{1^2}{3 \cdot 2} \cdot\frac12 \\ &= \frac{(2 n)!}{(2 n+1) n!^2} \frac1{2^{2 n+1}} \end{align} $$
Thus,
$$ \sum_{n=0}^{\infty} e_n = \sum_{n=0}^{\infty} \frac{(2 n)!}{(2 n+1) n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}$ How to evaluate $$\left(\cos{\frac{5\pi}{9}}\right)^{11}+\left(\cos{\frac{7\pi}{9}}\right)^{11}+\left(\cos{\frac{11\pi}{9}}\right)^{11}?$$
I found the problem on this page.
| Using for an odd $n$
$${\cos ^n}(x) = \frac{1}{{{2^{n - 1}}}}\sum\limits_{k = 0}^{\frac{{n - 1}}{2}} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)} \cos ((n - 2k)x)$$
gives me:
$${\cos ^{11}}(t) = \frac{{462\cos (t) + 330\cos (3t) + 165\cos (5t) + 55\cos (7t) + 11\cos (9t) + \cos (11t)}}{{1024}}$$
It... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Residue Integration I am attempting to calculate the integral of $\frac{(1+sin(\theta))}{(3+cos(\theta))}$ from $0$ to $2\pi$. I have already changed $sin$ and $cos$ into $\frac{1}{2i(z-z^{-1})}$ and $\frac{1}{2(z+z^{-1})}$. I am really stuck now. Can anyone please guide me?
| Since the integrand is periodic, then
$$\begin{align}
\int_0^{2\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta&=\int_{-\pi}^{\pi} \frac{1+\sin \theta}{3 + \cos \theta} d\theta\\
&=\int_{-\pi}^{\pi} \frac{1}{3 + \cos \theta} d\theta
\end{align}$$
where we exploited the fact that $\frac{\sin \theta}{3+\cos \theta}$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Extension of the additive Cauchy functional equation Let $f\colon (0,\alpha)\to \def\R{\mathbf R}\R$ satisfy $f(x + y)=f(x)+f(y)$
for all $x,y,x + y \in (0,\alpha)$, where $\alpha$ is a positive real number. Show that there exists an additive function $A \colon \R \to \R$ such that $A(x) = f(x)$ for all $x \in (0, \alp... | Let $x > 0$. Choose $n \in \def\N{\mathbf N}\N$ with $\frac xn < \alpha$. Define $A(x) := nf(\frac xn)$. Note that this is well-defnined: If $m\in \N$ is another natural number such that $\frac xm < \alpha$, we have
\begin{align*}
mf\left(\frac xm\right) &= mf\left(\sum_{k=1}^n \frac x{mn}\right)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to solve $2(2y-1)^{\frac{1}{3}}=y^3+1$ how can I solve $2(2y-1)^{\frac{1}{3}}=y^3+1$
I got by $x=\frac{y^3+1}{2}$ that $y=\frac{x^3+1}{2}$ but I was told earlier that can't say that $x=y$.
So I can I solve this equation?
Thanks.
| By cubing you get
$$ y^9+3 y^6+3 y^3 - 16 y + 9 = 0$$
It factors (using Wolfram alpha) into
$$ (y-1) (y^2+y-1) (y^6+2 y^4+2 y^3+4 y^2+2 y+9) = 0$$
According to WA, all the roots of the last polynomial are complex. The second one has $ \frac{(-1-\sqrt{5})}{2}$ and $ \frac{(\sqrt{5}-1)}{2}$ as roots. I don't know if you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is there a closed-form of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^4}$ Is there a closed-form summation result for Fourier series:
$$\sum_{n=1}^{\infty}\frac{\sin(n)}{n^4}?\tag{1}$$
I tried using available result of the following (odd) function :
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}.\tag{2}$$
However I coul... | Consider the series
\begin{align}
f(x) = \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}}
\end{align}
for which upon differentiation the following is obtained.
\begin{align}
f'(x) &= \sum_{n=1}^{\infty} \frac{\cos(n x)}{n^{3}} \\
f''(x) &= - \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{2}} \\
f'''(x) &= - \sum_{n=1}^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
| $x^3+x^2=2^y+16$. The RHS is positive, so $x^2(x+1)>0\iff x\ge 1$. Since $2^y$ is an integer, we have $y$ is a positive integer too ($y=0$ won't give a solution).
$x,y$ are positive integers.
$x^3+x^2-16=2^y$. You see a cubic polynomial on the LHS that could easily be strictly bounded between two consecutive cubes (n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$... | for part (a), $$|x+ y| \ge |y|-|x| = 11-7 = 4$$ this is achieved when $x = -y|x|/|y|.$
for part (b), you can find the angle $t$ between the $x,y$ by using the cosine rule in the triangle made up of $x, y, x+y.$ that is $$\cos t = \frac{5^2 + 4^2 - 7^2}{2 \times 5 \times 4} = -\frac 15 \to t =\pi-\cos^{-1}\left(\frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the 3-volume of the 3-parallelepiped defined by $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}$? We have $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\3\\4\end{bmatrix}\right\}$
QR-factorization gives $\text... | $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Since volume of a parallelipiped spanned by a set of vectors is invariant under the operation of adding a scalar multiple of one vector to another, it suffices to compute the volume of the parallelipiped spanned by
$$
\left[\begin{array}{@{}c@{}}
1 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...... | The combinatorial identity:
$$ \sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}\tag{1}$$
can be easily proved by induction. It gives:
$$\sum_{j=1}^{n-1}j(j+1) = 2\sum_{j=1}^{n-1}\binom{j+1}{2} = 2\binom{n+1}{3} = \frac{(n+1)n(n-1)}{2}\tag{2}$$
as well as:
$$\begin{eqnarray*} \sum_{k=1}^{n}k(1+2+\ldots+k) &=& 2\sum_{k=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the integral $\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx$ Help me find the integral. I think we have to somehow replace apply.
$$\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx =\int \frac{1}{\sqrt{x} (1+x^{\frac{1}{4}})^{10}} dx $$
| You may just perform the change of variable $x=u^4$, to get
$$
\begin{align}
\int \:x^{-\frac{1}{2}}\cdot \left(1+x^{\frac{1}{4}}\right)^{-10} dx&=4\int \:u^{-2}\cdot \left(1+u\right)^{-10} u^3du\\\\
&=4\int \frac{1}{(1+u)^9} du-4\int \frac{1}{(1+u)^{10}}du\\\\
&=-\frac{1}{2 (1+u)^8}+\frac{4}{9 (1+u)^9}+C\\\\
&=-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the sum of all odd numbers between two polynomials I was asked this question by someone I tutor and was stumped.
Find the sum of all odd numbers between $n^2 - 5n + 6$ and $n^2 + n$ for $n \ge 4.$
I wrote a few cases out and tried to find a pattern, but was unsuccessful.
Call polynomial 1, $p(n) = n^2 - 5n + 6,$ t... | Because $n^2+n$ is even we can write the odd numbers between the two polynomials to sum as:
$$n^2+n-1,n^2+n-3,\cdots,n^2+n-(6n-7)$$
and because the sum of odd numbers between $1$ and $2k+1$ is $(k+1)^2$ we have :
$$\sum_{i=0}^{3n-4}(n^2+n-(2i+1))=(3n-3)(n^2+n)-(3n-3)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|... | Hint: $\delta$ should not be depending of $x$.
Consider $c > 0$ then
$$ c - 1 < x < c+ 1 \implies |x| < c + 1 $$
Then $$\begin{align}|x^3 - c^3| &= |x - c| |x^2 + xc +c^2| \\&\leq |x-c|\bigg(|c+1|^2 + |c+1||c| + + |c|^2\bigg) \\&=|x-c|\bigg[|c+1|(|c + 1| + |c|) + |c|^2\bigg]\end{align} $$
Take $$\delta = \min \Bigg\{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove that if $k \in \mathbb{N}$, then $k^4+2k^3+k^2$ is divisble by $4$ I am trying to solve by induction and have established the base case (that the statement holds for $k=1$).
For the inductive step, I tried showing that the statement holds for $k+1$ by expanding $(k+1)^4+2\cdot(k+1)^3+(k+1)^2$, but this equals $16... | First note that $n^4+2n^3+k^2= n^2(n^2+2n+1)=n^2(n+1)^2$. Hence, the problem is to show that for every $n\geq 1$,
$$
S_n : 4\mid n^2(n+1)^2
$$
is true.
Without induction, the direct proof follows from the fact that among $n$ and $n+1$, one is even, and so one of $n^2$ and $(n+1)^2$ has a factor of $4$. But we will d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
| $x^2=s,\ y^2=t \geq 0$ then $$ (2s)^2+ (3t)^2=8^2 \Rightarrow
2s=4\cos\ \theta,\ 3t=4\sin\ \theta,\ 0\leq\theta \leq
\frac{\pi}{2} $$
Then $$ s+t=2\cos\ \theta + \frac{4}{3}\sin\ \theta=\sqrt{2^2+
\bigg(\frac{4}{3}\bigg)^2} \sin\ (\theta+\alpha),\ 0< \alpha <
\frac{\pi}{2} $$
Let $\theta =\frac{\pi}{2}-\alpha$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the value of $\sum_{m=1}^\infty \tan ^ {-1}\frac{2m}{m^4+m^2+2}$ How to find value of this sum?
$$\sum\limits_{m=1}^\infty \tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$$ I can't understand how to simplify this. Should I use any trigonometric substitution to simplify the fraction? Hints and help needed!
| \begin{eqnarray}
\frac{2m}{m^4+m^2+2}&=&\frac{2m}{m^4+2m^2-m^2+1+1}\\
&=&\frac{2m}{(m^2+1)^2-m^2+1}\\
&=&\frac{2m}{(m^2+m+1)(m^2-m+1)+1}\\
&=&\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1}
\end{eqnarray}
so almost done!
$$
\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)
$$
\begin{eqnarray}
\arctan(m^2+m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Induction for Vandermonde Matrix
Given real numbers $x_1<x_2<\cdots<x_n$, define the Vandermonde
matrix by $V=(V_{ij}) = (x^j_i)$. That is, $$V =
\left(\begin{array}{cccccc} 1 & x_1 & x^2_1 & \cdots &
x^{n-1}_1 & x^n_1 \\
1 & x_2 & x^2_2 & \cdots & x^{n-1... | Add the last row multiplied by -1 to all other rows, and we get,
det$|V|$=det$\left|\begin{array}{}
0 & x_1-x_n & x_1^2-x_n^2 & \cdots & x_1^n-x_n^n \\
0 & x_2-x_n & x_2^2-x_n^2 & \cdots & x_2^n-x_n^n \\
\vdots & \vdots & \vdots & & \vdots \\
0 & x_{n-1}-x_n & x_{n-1}^2-x_n^2 & \cdots & x_{n-1}^n-x_n^n \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove... | Here is an alternate solution. We want to prove
$$
\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1
$$
Or, equivalently, that
$$
\frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1}>\frac{1}{n^2 - n + 1}
$$
where the left-hand side is the arithmetic mean of all the fractions. We know that the arithmetic mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
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Inverse function. A function $h$ is defined by $h:x\rightarrow 2-\frac{a}{x}$, where $x\neq 0$ and $a$ is a constant. Given $\frac{1}{2}h^2(2)+h^{-1}(-1)=-1$, find the possible values of $a$.
Can someone give me some hints? Thanks
| Firstly, find the inverse function $h^{-1}$. I.e. let $y=2-a/x$ and hence $h^{-1}$ is defined as
\begin{align}x=2-\frac{a}{y} \ \ & \Longrightarrow \ \ \frac{a}{y}=2-x\\
&\Longrightarrow y=\frac{a}{2-x}.
\end{align}
Now, $h(2)=2-a/2$ and $h^{-1}(-1)=a/3$. You should be able to continue from here and get a quadratic equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find basis so matrix is in Jordan Canonical Form $M = \left(\begin{array}{ccc}0 & -3 & -2 \\1 & 3 & 1 \\1 & 2 & 3\end{array}\right)$
I want to find a basis $B$ such that matrix for $M$ w.r.t $B$ has the form:
$\left(\begin{array}{ccc}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{array}\right)$
The eigenvalues for $M = 2,2,2$.
... | let us call $$A- 2I = B = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right).$$ row reducing we find that $B \to \pmatrix{1&0&1\\0&1&0\\0&0&0}$ so that the null of $B$ has dimension one and $$u = \pmatrix{1\\0\\-1}, Bu= 0$$ is a basis.
you can also find that null of $B^2$ has dimension $2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of an integral. $$I_1=\int \frac{a^2\sin^2(x)+b^2\cos^2(x)}{a^4\sin^2(x)+b^4\cos^2(x)}dx$$
I tried writing it as $$\int \frac{a^2+\cos^2(x)(b^2-a^2)}{a^4+\cos^2(x)(b^4-a^4)}dx$$
But I don't know how to proceed. What's the procedure to evaluate this integral?
| $$\frac{a^2u^2+b^2}{a^4u^2+b^4}=\frac{(a^2+b^2)(a^2u^2+b^2)}{(a^2+b^2)(a^4u^2+b^4)}=\cdots =\frac{1}{a^2+b^2}+\frac{a^2(u^2+1)}{(a^2+b^2)(\frac{a^4}{b^2}u^2+b^2)}.$$
So if $u=\tan{x}$ and integrate the previous you get
$$\frac{x}{a^2+b^2}+\frac{a^2}{a^2+b^2}\int \frac{\tan^2{x}+1}{\frac{a^4}{b^2}\tan^2{x}+b^2} dx=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1... | First of all, in general
$$x(x+1)^{n-1} \neq \binom{n-1}{0} x^0+\cdots+\binom{n-1}{k - 1}x^{k-1}+\cdots
+ \binom{n+1}{n-1}x^{n-1}.$$
Notice that as $x\to 0$ the left side goes to zero but the right side does not.
Also, in general
$$1(x+1)^{n-1} \neq \binom{n-1}{0}x^0+\cdots+\binom{n-1}{k - 1}x^{k}+\cdots+
\binom{n-1}{n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that a polynomial over $\mathbb{Z}_{2}$ is irreducible Given the polynomial: $p(x)=x^4+x^3+x^2+x+1$ over $\mathbb{Z}_{2}$, to show that it is irreducable, is it enough to show that $p(0)=p(1)=1$?
| On factorising into quadratics, just try by hand. The coefficient of $x^2$ has to be $1$ in both factors to give $x^4$ and the constant in each factor has to be $1$ too to give a constant term $1$ in the product.
So $$(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+abx^2+(a+b)x+1=x^4+x^3+x^2+x+1$$
We have $a+b=1$ so $a$ and $b$ have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Sum of polynomial coefficient
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x)=2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"
After some calculations, Jon says, "There is more than o... | We have
$$2(r_1r_2+r_2r_3+r_3r_1) = a^2 - 81 = (r_1+r_2+r_3)^2 - 81$$
This gives us
$$r_1^2 + r_2^2 + r_3^2 = 81$$
Since $81 \equiv 1\pmod8$, we obtain that one of them is odd, say $r_1$ and the other two ($r_2 = 2k_2$ and $r_3 = 2k_3$) are even. Further, $1 \leq r_1,r_2,r_3 \leq 8$, i.e., $1 \leq k_2,k_3 \leq 4$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $\sqrt{x^2} = (\sqrt x)^2$? Take $x=4$ for example:
$ \sqrt{(4)^2} = \sqrt{16} = \pm4 $
However:
$ (\sqrt{4})^2 = \sqrt{\pm2}$
Case 1: $ (-2)^2 = 4$
Case 2: $ (2)^2 = 4$
Solution : $+4$
How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ?
What is missing?
| Disclaimer: In the following we restrict ourselves to real numbers, not taking complex numbers and the like into consideration.
By convention, the square root of a positive number $t$, written as $\sqrt t$, has been defined to be the positive solution to the equation $x^2=t$. This gives meaning to the following way of... | {
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"source": "stackexchange",
"question_score": "4",
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Write a function as $\sum _{n=0} ^{\infty} a_n x^n$ We have $f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4$.
Now I want to write this as $\sum _{n=0} ^{\infty} a_n x^n$.
What I got:
$f(x) = (x+ x^2 + x^3 + x^4 + x^5 + x^6)^4 = x^4 (1+ x + x^2 + x^3 + x^4 + x^5)^4$
We know: $\frac{1}{1-x} = \sum _{n=0} ^{\infty} x^n$.
Note ... | We know that for any Taylor series representation:
$$ \sum_{n=0}^{\infty} \frac{ f^{(n)}(a) } {n!} (x-a)^n $$
It looks like you want the series centered at a=0.
So to get the coefficient for terms 1,2,3,4,.... evaluate:
$$b_n=\frac{ f^{(n)}(a)}{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Number of times $2^k$ appears in factorial
For what $n$ does: $2^n | 19!18!...1!$?
I checked how many times $2^1$ appears:
It appears in, $2!, 3!, 4!... 19!$ meaning, $2^{18}$
I checked how many times $2^2 = 4$ appears:
It appears in, $4!, 5!, 6!, ..., 19!$ meaning, $4^{16} = 2^{32}$
I checked how many times $2^3 = ... | How many times $2$ divides the product $\prod_{i=1}^{19}i!$ ?
Let's call each term inside a factorial $i$. That way, $i = 1$ occurs in 19 factorials, $i = 2$ occurs in 18 factorials, and $i = 3$ occurs in 17 factorials etc.
$i = 2$ occurs 18 times. $1 \times 18 = 18$
$i = 4$ occurs 16 times. $2 \times 16 = 32$
$i = 6$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fractions - different results when done in steps than not We have: $\frac 1 {(1-x)(1+x)(1-2x)}$
If I do the partial fractions straight: $\frac 1 {(1-x)(1+x)(1-2x)}= \frac a {1-x} + \frac b {1+x} + \frac c {1-2x}$
I get: $a=-\frac 12, b = \frac 1 6, c=\frac 4 3$.
But when I do it in steps, i.e: $\frac 1 {(1-x^2)... | $\frac 1 {(1-x^2)(1-2x)}\ne\frac a {1-x^2} + \frac b {1-2x}$
$\frac 1 {(1-x^2)(1-2x)}=\frac {a+bx} {1-x^2} + \frac c {1-2x}$
One finds that $a=-1/3$, $b=-2/3$ and $c=4/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do i evaluate the following integral? Hi I was wondering if someone can help me evaluate the following integral.
Show that if $-1 < x < 1$, then
$$\int_{0}^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}}dy= \pi \arcsin{x} $$
thank you in advance.
| Setting
$$
f(x)=\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy, \quad x\in (-1,1),
$$
we have $f(0)=0$, and
$$
f'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}\,dy=\int_0^\pi\frac{1}{1+x\cos y}\,dy\quad \forall x\in (-1,1)
$$
Setting
$$
t=\tan\frac{y}{2},
$$
we have
$$
y=2\arctan t,\, \frac{1}{1+x\cos y}=\frac{1}{1+x\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Real Canonical Form Question: Let us consider the quadratic form $q: R^3 -> R$, $$q(x,y,z) = x^2+25y^2+10xy+2xz$$Find the corresponding symmetric bilinear form $f$ and a basis $B$ such that $[f]_B$ has the real canonical form. State the signature and rank of $[f]_B$.
I thought that the best way to start this question w... | A way to proceed to get the signature and the rank is to reduce $q$ thanks to the Gauss' algorithm :
$$q\left(x,y,z\right)=x^2+25y^2+10xy+2xz
=\left(x+5y+z\right)^2-10yz-z^2
=\left(x+5y+z\right)^2-\left(z+5y\right)^2+25y^2.$$
Let then $X=\left(x+5y+z\right)$, $Y=\left(z+5y\right)$ and $Z=5y$. Then the reduced equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| In case $n$ is a multiple of $5$ the $n^2$ on the rhs is also.
If $n=2,3\pmod 5$ then $n^2\equiv4\pmod5$ , hence $n^2+1$ would be a multiple of $5$.
In the remaining cases, i.e.,$n=1,4\pmod 5$ then in $n^2-1= (n+1)(n-1)$ one of the term on the right will be a multiple of $5$.
So $n^2(n^2+1)(n^2-1)$ is always a multip... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 7
} |
Solving a cubic equation involving trigonometric functions Question:
If $$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
Then $$\sin^3(x) - \cos^3(x) = ?$$
I have turned first equation into a quadratic so I got $$\sin(x) = \frac{\sqrt{3}\mp\sqrt{5}}4$$ and $$\cos(x) = \frac{-\sqrt{3}\mp\sqrt{5}}4$$ But stuck here. I don't know... | $$\sin(x) - \cos(x) = \frac{\sqrt{3}}2$$
$$\implies \sin^2(x) + \cos^2(x)-2\sin x \cos x = \dfrac 34$$
$$\implies \sin x\cos x=\dfrac 18$$ (using $\sin^2x+\cos^2x=1$)
Now, using $a^3-b^3=(a-b)(a^2+ab+b^2)$,$$\sin^3(x)-\cos^3x = (\sin x-\cos x)(\sin^2x+\cos^2x+\sin x\cos x)$$
$$=\dfrac {\sqrt 3}{2}\times \left (1+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I simplify $\sqrt{3^2 + 3^2\tan^2\theta}$? $$\sqrt{3^2 + 3^2\tan^2\theta}$$
$$ = (3)(3\tan\theta) = 9\tan\theta $$
I've simplified it like this but I'm not sure if that's correct.
| Hint
$$5x+5y=5(x+y) \text{ distributive property}$$
$$\sqrt{9\cdot 16} = \sqrt{9}\cdot\sqrt{16}\text{ root of a product property}$$
$$\begin{align}
opposite^2 + adjacent^2 &= hypotenuse^2\\
\frac{opposite^2 + adjacent^2}{adjacent^2}&=\frac{hypotenuse^2}{adjacent^2}\\
\bigg(\frac{opposite}{adjacent}\bigg)^2 + \bigg(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find conditions for $\left(\frac{-3}{p}\right)=1$ $\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)^{\frac{p-1}{2}}$
$=\begin{cases}1,\:p\equiv 1\pmod{4}\text{ or }\begin{cases}p\equiv 3\pmod{4}\\p\equiv 2\pmod {3}\end{cases}\\-1,\:\text{ot... | $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{\frac{p-1}{2}}$$
The problem is in your application of the quadratic-reciprocity law.
| {
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"timestamp": "2023-03-29T00:00:00",
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Algorithm to generate an arbitrary matrix of special linear group $SL(2,\mathbb{Z})$ I have a given $2\times 2$ special linear matrix, for example
\begin{equation}
m=\begin{pmatrix} 55 & 8469 \\ 1 & 154 \end{pmatrix}
\end{equation}
and I would like to get the generating form of it from the s and t matrices which are th... | This rather small example can also be done by hand without any algorithms.
Note that
$$
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
a & a+b \\
c & c+d
\end{pmatrix}
$$
While $S$ switches the columns and puts a minus sign in.
Then by guesstimating powers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13... | $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}=n^6(n^2-1)(n^2+1)^2+n^{13}(n^8-1)=$$
$$=n^6\left((n^2-1)(n^2+1)^2+n^7(n^2-1)(n^2+1)(n^4+1)\right)=$$
$$=n^6(n^2-1)(n^2+1)\left[n^2+1+n^7(n^4+1)\right]=$$
$$=\color{red}{n(n-1)(n+1)}\cdot n^5(n^2+1)\left[n^2+1+n^7(n^4+1)\right]$$
Can you see why the red part is already a multiple of $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the sum of values of $x$ such that $|x+2| +|x-3| +|x+4| + |x+5| = 18$ I tried it by finding the different values at the $4$ inflection points of the graph. Then didn't know how to proceed. Am I correct till here?
| *
*$x \le -5 $ We need to solve $-(x+2)-(x-3)-(x+4)-(x+5)=18$
*$-5 \le x \le -4 $ We need to solve $-(x+2)-(x-3)-(x+4)+(x+5)=18$
*$-4 \le x \le -2$ We need to solve $-(x+2)-(x-3)+(x+4)+(x+5)=18$
*$-2 \le x \le 3$ We need to solve $(x+2)-(x-3)+(x+4)+(x+5)=18$
*$x \ge 3$ We need to solve $(x+2)+(x-3)+(x+4)+(x+5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Is Vieta the only way out? Let $a,b,c$ are the three roots of the equation $x^3-x-1=0$. Then find the equation whose roots are $\frac{1+a}{1-a}$,$\frac{1+b}{1-b}$,$\frac{1+c}{1-c}$.
The only solution I could think of is by using Vieta's formula repeatedly which is no doubt a very messy solution. Is there any easier and... | Since the polynomial $x^3-x-1$ is irreducible over$\def\Q{\Bbb Q}~\Q$ by the rational root test, one approach would be to identify the element $\frac{1+a}{1-a}$ where $a$ is the image of $x$ in the field $K=\Q[x]/(x^3-x-1)$, and to compute its minimum polynomial over$~\Q$; since the Galois group of the splitting field ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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How to determine if $\sum_{n=1}^{\infty}\left (\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$ converges or diverges $$\sum_{n=1}^{\infty} \left(\frac{n^2-5n+1}{n^2-4n+2}\right)^{n^2}$$
Using root test seems not a efficient way since I got stuck without knowing what to do next
$$\lim_{n\to\infty}\left(\frac{n^2-5n+1}{n^2-4n+2... | To complete your root test:
$$=\lim_n\left[\left(1+\frac{-5n-1}{n^2-4n+2}\right)^{\frac{n^2-4n+2}{-5n-1}}\right]^{\frac{n(-5n-1)}{n^2-4n+2}}=e^{\lim_n\frac{n(-5n-1)}{n^2-4n+2}}=e^{-5}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Calculate sides of right triangle with hypotenuse and area or perimeter I'm trying to find if it is possible to find the lengths of the base and height of a right triangle with only the hypotenuse and the area (or the perimeter) of the triangle. I would have just figured that it was impossible, but I found an online ca... | For hypotenuse $h$ and area $A$, let the legs be $x$ and $y$. Then $x^2+y^2=h^2$ and $xy=2A$. It follows that
$$(x+y)^2=x^2+y^2+2xy=h^2+4A$$
and
$$(x-y)^2=x^2+y^2-2xy=h^2-4A.$$
Thus $x+y=\sqrt{h^2+4A}$ and $x-y=\pm\sqrt{h^2-4A}$ and now we can solve for $x$ and $y$.
For hypotenuse $h$ and perimeter $p$, we have $x^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Cyclic Equation. Prove that: $\small\frac { a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3 }{ (a-b)(b-c)(c-a) } = ab + bc + ca$? This is how far I got without using polynomial division:
\begin{align}
\tiny
\frac { a^{ 2 }(b-c)^{ 3 }+b^{ 2 }(c-a)^{ 3 }+c^{ 2 }(a-b)^{ 3 } }{ (a-b)(b-c)(c-a) }
&\tiny=\frac { { a }^{ 2 }\{ { b }^{ 3... | Set $f(a,b,c) = a^2(b-c)^3+b^2(c-a)^3+c^2(a-b)^3$. Note that $f(a,b,c)$ is cyclic, i.e., $f(a,b,c) = f(c,a,b) = f(b,c,a)$.
Further, we also have $$f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$$
This means $f(a,b,c) = (a-b)(b-c)(c-a)g(a,b,c)$, where $g(a,b,c)$ is also cyclic polynomial of degree $2$.
The most general cyclic polyn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Factor $z^4 +1$ into linear factors $z$ is a complex number, how do I factor $z^4 +1$ into linear factors?
Do I write z in terms of $x+yi$ so that $z^4+1=(x+yi)^4+1?$
| If $z^4+1=0$ then $z^4=-1$ so $z^2=\pm i$. If $z^2=i=\cos90^\circ+i\sin90^\circ$ then $z=\pm(\cos45^\circ+i\sin45^\circ) = \pm\left(\frac1{\sqrt2} + i\frac 1 {\sqrt2}\right)$, and similarly for $-i=\cos(-90^\circ)+i\sin(-90^\circ)$.
So
$$
z^4+1 = \left(z-\frac{1+i}{\sqrt2}\right)\left(z-\frac{-1-i}{\sqrt2}\right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
T... | We can simplify and rigorize your proof by scaling by $\,4\,$ to clear denominators, yielding
$$2a\equiv -3b\,\overset{\rm square}\Rightarrow\,\color{#c00}{4a^2\equiv 9b^2}\,\Rightarrow\,\color{#c00}4(\color{#c00}{a^2}-5b^2) \equiv \color{#c00}{9b^2}-20b^2 \equiv -11b^2\equiv 0\!\pmod{\!11}\quad $$
Remark $\ $ In fact... | {
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"url": "https://math.stackexchange.com/questions/1271972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? This isn't a homework question but one I found online.
Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?
We just started doing field theory in my class and I want extra practice, but I have no idea how to even start this problem. It's no, right? I'm... | On the other hand, $-1$ is the sum of two squares here. What Lam calls Siegel's Theorem is that, in an algebraic number field, the stufe is one of $1,2,4,\infty.$ Either $-1$ is not the sum of any number of squares in the field, or it is itself a square, or it is the sum of two or four squares.
I will see if I can fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
I have a ball that intersects a cylinder and I need the volume. How do I do it? I have an exam coming up and I am stressing out about it really hard. I don't even know how to actually do this. Is it a triple integral?
I have a ball $\{(x,y,z)|x^2 + y^2 + z^2 \leq 9\}$ and a cylinder $\{(x,y,z) | x^2 + y^2 \leq 5, 0 \l... | Let's start with picking convenient coordinate system. Since the $Oz$ axis is an axis of symmetry for both of the bodies, let's consider cylindrical coordinate system $(r, \theta, z)$ with $r = \sqrt{x^2+y^2}$ so $x = r \cos \theta, y = r \sin \theta$.
Now the ball $B$ is described as $r^2 + z^2 \leq 3^2$ and the cylin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Numbers that can be expressed as the sum of two cubes in exactly two different ways It seems known that there are infinitely many numbers that can be expressed as a sum of two positive cubes in at least two different ways (per the answer to this post: Number Theory Taxicab Number).
We know that
$$1729 = 10^3+9^3 = 12^3... | In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
*
*There exists a divisor $m|N$ with $N^{1/3}\leq m \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation.
What was shown in class: $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz=\large \int_{|z-2|=2} \frac {5z+7}{(z+3)(z-1)}=\cdots= \int_{|... | Cauchy's Integral Formula states $$ \oint_C \frac{g(z)}{z-a}dz =g(a)2\pi i.$$ for every $a$ in the interior of $C$.
Here $g(z) = \dfrac {5z+7}{(z+3)}$, and $a= 1$ hence $$ \oint_C \frac{g(z)}{z-1}dz = \dfrac {5(1)+7}{(1+3)}2\pi i = 6\pi i.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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confusion regarding 'o' function . could one explain me the following steps ?
my books have written ,
$$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$
the the expression for 'sec' from Taylor series expansion I understand. but how can w... | $\frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$
$[1-(x^2/2-o(x^2))]^{-1}=1+x^2/2-o(x^2)+(x^2/2-o(x^2))^2+O(f(x))$
Here for sufficiently small "$x$" we can ignore $2nd$ terms and merge $O(f(x))$ with $o(x^2)$ because it will contain terms at least larger than $x^2$.
Gives $1+x^2/2-o(x^2)+O(f(x))=1+x^2/2+o'(x^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding real coefficients of equation given that $a+ib$ is a root Below is the question present in a past examination paper. I'll be giving my attempts and how I thought it through. Do feel free to point out any mistakes I make throughout my working even if unrelated to the question itself.
*(a) Find $a+ ib$ suc... | First $px^2 + qx + r = 0$ and $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ have the same set of solutions. Note that we are sure that $p \neq 0$ (Why??) Think.
We have $\frac{q}{p} = 1$ and $\frac{r}{p} = \frac{1}{2}.$ We get then the equation $x^2 + \frac{q}{p}x + \frac{r}{p} = 0$ as $x^2 + x + \frac{1}{2} = 0,$ so one of s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is the value of $ \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +....+ \frac{1}{n \sin^2 x}\right]^{\sin^2x} $ I took out $\sin^2x$ out of the brackets . Inside the brackets , I think I should use the formula $\frac{ n(n-1)}{2}$ . Am I doing right ? If yes, then what should I do next ? Thanks ! ... | we have $$ \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} = e^{\sin^2x \ln \frac{1}{\sin^2x}}$$knowing that ${\sin^2x} \rightarrow 0$ as $x \rightarrow 0$, and $ \lim_{x\rightarrow 0} x\ln \frac{1}{x} = 0 $ then the limit is
$$\lim_{x \rightarrow 0} \Big( \frac{1}{\sin^2x} \Big)^{\sin^2x} =1. $$
Similarly, $$\lim_{x \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of sequence and Riemann sum problem work verification I have to calculate $$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}}$$ My attempt:
$$\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{n}{k^2-4n^2}}} = \lim_{n \to \infty}{\frac{1}{n^2}\frac{1}{n}\sum_{k=1}^{n}{\frac{1}{\frac{k^2}{n^2}-4}}} = \left(\lim_{n \t... | $$\sum_{k=1}^{n} \frac{n}{k^2-4n^2}=\sum_{k=1}^{n} \frac{n}{n^2\left(\frac{k^2}{n^2}-4\right)}=\frac{1}{n}\sum_{k=1}^{n} \frac{1}{\left(\frac{k^2}{n^2}-4\right)}$$
So
$$\lim_{n \to\infty} \sum_{k=1}^{n} \frac{n}{k^2-4n^2}= \int_0^1 \frac{1}{x^2-4}\mathrm{d}x=\frac{-\ln(3)}{4} $$
I hope I make no mistake !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^2y''+(2x^2+x)y'+(2x^2+x)y=0$ A Bessel equation $$x^2y''+(2x^2+x)y'+(2x^2+x)y=0$$
The solution is $$e^{-x}J_o(x)+e^{-x}Y_o(x)$$
How does one approach a problem like this?
| One method is to recognize that $x^2 y''$ is frequently seen in many differential equations whereas the other terms are not. With this then one can consider a function of the type $y(x) = f(x) g(x)$.
\begin{align}
y(x) &= f g \\
y' &= f g' + f' g \\
y'' &= f g'' + 2 f' g' + f''
\end{align}
Now,
\begin{align}
0 &= x^2y'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Determining the sign of a term I have a problem in proving the sign of a term. It is as follows:
$$x=\dfrac{1-a}{b_1b_2-a}+1,\qquad y=\dfrac{1-a}{b_1-a}+\dfrac{1-a}{b_2-a},\qquad z=x-y$$
with $0<b_1<1,\quad 0<b_2<1,\quad 0<a<b_1b_2$.
Based on some simulations I have noticed that $z>0$ for several instances of $a$, $b_1... | We can write
\begin{align}
x-y &= \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a} - \frac{1-a}{b_1-a} - \frac{1-a}{b_2-a} + 1 + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_2-a}\\
&= \biggl(\frac{1-a}{b_1-a} - 1\biggr)\biggl(\frac{1-a}{b_2-a}-1\biggr) + \frac{1-a}{b_1b_2-a} - \frac{1-a}{b_1-a}\cdot \frac{1-a}{b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove... | You can also do a direct proof following your first computations. You can write
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \sum_{i = 0}^{n-1} \binom{k}{i}x^i$$
using the fact that $\binom{k}{i}=0$ when $i$ is greater that $k+1$. Then, if you invert the sums, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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For which values does the Matrix system have a unique solution, infinitely many solutions and no solution? Given the system:
$$\begin{align}
& x+3y-3z=4 \\
& y+2z=a \\
& 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?... | Simply by Gauss-Jordan elimination you get:
$$\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
2 & 5 & a^2-9 & 9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 &-1 & a^2-3 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 3 & -3 & 4\\
0 & 1 & 2 & a\\
0 & 0 & a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Probability distribution of number of columns that has two even numbers in a chart
We distribute numbers $\{1,2,...,10\}$ in random to the following chart:
Let $X$ be the number of columns that has two even numbers.
What is the distribution of $X$?
My attempt:
$|\Omega|=10!$
$P(x=0)=\frac {(5!)^2\cdot 2 ^5}{10!}=\fr... | You can first simplify the problem:
Let 1 denote an odd number and 0 and even number.
Case 1: P(X=0)
$11111$
$00000$
The even number can be arranged in 5! ways. The odd numbers can be arranged in 5! ways, too. Thus you have the factor $(5!)^2$ in all cases. In each coloumn 0 can be in the first row or in the second row... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2x... | The correct inequality is
$$(a+b)^2=(c+h)^2-xy<(c+h)^2$$
which implies $$a+b<c+h$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to find differential equation $$\frac{dy}{dx}-8x=2xy^2\quad y=0\,x=1$$
I separated $x$ and $y$.
\begin{align*}
\color{red}{\frac{dy}{y^2}}&=\color{red}{2x+8x dx}\\
\frac{dy}{y^2}&=\color{red}{10x dx}\\
\color{red}{\ln y^2} &= 5x^2\\
y^2&=Ae^{5x^2}
\end{align*}
When I plug $y$ and $x$ in, i get $A=0$.
I think I did... | You've made a couple of mistakes.
$\frac{dy}{dx}-8x = 2xy^2 \Rightarrow \frac{dy}{dx} = 2x(y^2+4) $
Therefore
$\frac{dy}{(y^2+4)} = 2xdx$
And try to continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding out $S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$ I was willing to determine the sum of following
$$S:=1+\frac12-\frac13-\frac14+\frac15+\frac16-\frac17-\frac18+\cdots$$
I tried the following
\begin{align*}
S=&\sum\limits_{n=0}^\infty (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right)\\
=... | Let $$S = \sum_{n = 1}^{\infty} (-1)^n\left(\frac{1}{2n-1}+\frac{1}{2n}\right).$$ Then
$\begin {eqnarray}
S & = & \sum_{k = 1}^{\infty} (-1)^{2k}\left(\frac{1}{4k-1}+\frac{1}{4k}\right) + \sum_{k = 1}^{\infty} (-1)^{2k+1}\left(\frac{1}{4k+1}+\frac{1}{4k+2}\right) \nonumber \\ & = & \sum_{k = 1}^{\infty} \left(\frac {1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Determinant of a real skew-symmetric matrix What will be the value of the determinant of a skew-symmetric matrix of even order when a single element is interchanged between first row and first column?
For,
$\left| \begin{array}{cccc}
0 & 1 & 2 & -1 \\
-1 & 0 & 1 & 2 \\
-2 & -1 & 0 & 1 \\
1 & -2 & -1 & 0 \end{ar... | I suspect the answer to "when does the value of the determinant change?" is "nearly always". Consider what happens when you interchange the $(1,3)$ and $(3,1)$ elements. Doing the algebra (by computer or by hand) gives
$$\eqalign{\det\pmatrix{0&a&b&c\cr -a&0&d&e\cr-b&-d&0&f\cr -c&-e&-f&0\cr}
-&\det\pmatrix{0&-a&b&c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Uniform convergence: $\sum \frac{x^2}{(1 + x^2)^n}$ Consider the series of functions:
$$\sum_{n \ge 1} \frac{x^2}{(1 + x^2)^n}$$
Q: Where does this series converge uniformly?
We have if $x \neq 0$:
$$\lim_{n \to \infty} \left| \frac{x^2}{(1 + x^2)^{n+1}} / \frac{x^2}{(1 + x^2)^n} \right|= \frac{1}{1 + x^2} < 1$$
And... | Perhaps I am missing something, but I believe we can write your sequence as:
$$
\frac{x^2}{(1 + x^2)^n} = \frac{x^2 + 1 - 1}{(1 + x^2)^n} = \frac{1 + x^2}{(1 + x^2)^n} - \frac{1}{(1 + x^2)^n} = \frac{1}{(1 + x^2)^{n - 1}} - \frac{1}{(1 + x^2)^n}
$$
This gives two infinite sums:
\begin{align}
\sum_1^\infty \frac{x^2}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if som... | Let us look at the RHS only: (It is fine to manipulate both sides but it is usually preferred if you manage to make one into the other.)
$$
\frac{\cot{\theta}-1}{\cot\theta +1} = \frac{\frac{\cos\theta}{\sin\theta}-1}{\frac{\cos\theta}{\sin\theta} +1}
$$
Now we multiply the numerator and denominator by $\sin\theta$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:
Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)
I know one of solutions is $5776=(76)^2$.
| $ABCD=(CD)^2$ can be translated as $100n + x = x^2$ which implies
$x^2 - x \equiv 0 \pmod{100}$.
$$ \text{$x^2-x \equiv 0 \pmod{100} \iff x(x-1) \equiv 0 \pmod{4}$
and $x(x-1) \equiv 0 \pmod{25}$}$$
$$x^2-x \equiv 0 \pmod{4} \iff x \in \{\bar 0_{4}, \bar1_{4} \}$$
$$x^2-x \equiv 0 \pmod{25} \iff x \in \{\bar 0_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sq... | If you need a parametrization, it is best to just complete the square, rather then exploiting the full power of the spectral theorem.
$$ x^2+xy+3y^2 = 1 $$
is equivalent to:
$$ \left(2x+y\right)^2 + 11 y^2 = 4$$
hence $2x+y=2\cos\theta,y=\frac{2}{\sqrt{11}}\sin\theta$ is a valid parametrization, that leads to:
$$ x = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.