Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-... | $n+1$ certainly divides $3n+3$. If it divides $3n+11$, it must also
divide $(3n+11)-(3n+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How is this Taylor expansion computed? I am reading a paper, where we consider the following function,
$$F(r) = \frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)$$
where $b>0$ is a constant such that $r>2/b.$ After stating this definition the author writes that the taylor expression of $F$ is,
$$F(r) = ... | Here you need the following expansions centered at $x=0$:
$$\sqrt{1+x}=1+\frac{x}{2}+O(x^2)\quad,\quad\ln(1+x)=x+O(x^2).$$
Hence, for $r$ large enough and $x=-\frac{4}{b^2r^2}$,
$$\begin{align*}F(r)&=\frac{r}{4}\sqrt{b^2r^2-4} - \frac{1}{b}\ln(\sqrt{b^2r^2-4} + br)\\
&=\frac{br^2}{4}\sqrt{1-\frac{4}{b^2r^2}}-\frac{1}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz$, where $\gamma$ is the circle with center 1/4 and radius 1/2 I need to calculate the following complex integral:
$$
\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz
$$
where $\gamma$ is the circle with center $1/4$ and radius $1/2$. It s... | Without residues you may proceed as follows:
The function $f(z) = \frac{e^{iz}}{z^2+2}$ is holomorph on your disc and contains $z= 0$.
So, Cauchy integral formula gives
$$f^{(5)}(0) = \frac{5!}{2\pi i }\oint_{\gamma}\frac{\frac{e^{iz}}{z^2+2}}{z^6}dz$$
Now, you know that $\frac{f^{(5)}(0)}{5!}$ is the coefficient of $z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Real Analysis Inf and Sup question I am hung up on this question for real analysis ( intro to anaylsis ).
Find $\inf D$ and $\sup D$
$$\mathrm{D}=\left\{\frac{m+n\sqrt{2}}{m+n\sqrt{3}} :m,n\in\Bbb{N}\right\}$$
I have spent enough time staring at this thing that I know the $\sup D=1$ and $\inf D=\frac{\sqrt{2}}{\sqrt{... | Well, $\frac {\sqrt 2}{\sqrt 3} < \frac{m+n\sqrt 2}{m+n\sqrt3} \iff$
$m\sqrt 2 + n\sqrt 6 < m\sqrt 3 + n\sqrt 6 \iff$
$m\sqrt 2 < m\sqrt 3$ which is always the case if $m > 0$.
.... so $\frac {\sqrt 2}{\sqrt 3}$ is a lower bound of D....
And $\frac{m+n\sqrt 2}{m+n\sqrt3} < \frac {\sqrt 2}{\sqrt 3} + \epsilon\iff$
$m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Existence of integers s.t. $Y^2 = X^3 + n^3 - 4m^2$ Given $m, n \in \mathbb Z$ with $n = -1 (mod 4)$. Let's say that $m$ has no prime divisors that are congruent to $3 (mod 4)$.
Why are there no integers $X, Y$ such that $Y^2 = X^3 + n^3 -4m^2$ ?
| Assume for the sake of contradiction that a solution $(X,Y)$ exists. Write $$Y^2+(2m)^2=X^3+n^3=(X+n)(X^2-nX+n^2).$$ We first claim that any prime factor $p$ of $Y^2+(2m)^2$ satisfies $p=2$ or $p\equiv 1\pmod{4}$.
To show this, let $p\equiv 3\pmod{4}$ be a prime factor of $Y^2+(2m)^2$. Because $\left(\frac{-1}{p}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{s... | As I have written, the idea of proof taking $c=mid(a,b,c)$ is just a matter of notation. Without loss of generality (WLOG), You can take the other mid one if $c$ is not the one. Note that there is always something in the middle. Even if $a=b=c$ you can take any of them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Solving $3\sin(2x+45^\circ)=2\cos(x+135^\circ)$ for $x$ between $0^\circ$ and $360^\circ$ Please find the value of $x$ in degree from this equation, with explanation
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
For $x$ between $0^\circ$ and $360^\circ$.
| Starting from S. Dolan's answer
$$6\sin x\cos x+6\cos^2 x-3=-2\cos x-2\sin x$$ let $c=\cos x$ ans $s=\sin x$ to make
$$6sc+6c^2-3=-2c-2s$$
Solving for $s$ gives $$s=\frac{-6 c^2-2 c+3}{2 (3 c+1)}$$, Square both sides and remember that $c^2+s^2=1$ to get
$$1-c^2=\left(\frac{-6 c^2-2 c+3}{2 (3 c+1)} \right)^2$$ Assuming ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equilateral triangle ABC has side points M, D and E. Given AM = MB and $\angle$ DME $=60^o$, prove AD + BE = DE + $\frac{1}{2}$AB I would like to ask if someone could help me with solving the following probelm.
The triangle $ABC$ is equilateral. $M$ is the midpoint of $AB$. The points $D$ and $E$ are on the sides $CA$ ... | To simplify, assume all lengths relative to $|AM|$. In other words, $AM=1$
Let $x = CD, \quad y = CE,\quad z=DE$
Assume the statement to be proved is true, this follows:
$AD + BE = (2-x)+(2-y) = z + 1 \quad → x+y+z=3$
$ΔCDE:\; z^2 = x^2 + y^2 - x y$
$$(3-x-y)^2 = x^2+y^2-xy$$
$$x^2+y^2+2xy-6(x+y)+9 = x^2+y^2-xy$$
$$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\fra... | Here is my take on the second and a remark on the first limit:
$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$:
So, consider
\begin{eqnarray*} \tan(2x)\ln(\tan x)
& \stackrel{\tan(2x)= \frac{2\tan x}{1-\tan^2x}}{=} & 2\frac{\tan x\cdot \ln(\tan x)}{1-\tan^2x}\\
& \stackrel{t = \tan x, t\to1}{=} & 2\frac{t \ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM... | After full expanding we need to prove that:
$$\sum_{cyc}(4a^{5}b+4a^{5}c-12a^{4}b^{2}+12a^{4}c^{2}+5a^{3}b^{3}+8a^{4}bc-19a^{3}b^{2}c+5a^{3}c^{2}b-7a^{2}b^{2}c^{2})\geq0$$ or
$$6\sum_{cyc}ab(a^{2}-b^{2}-2ab+2ac)^{2}+
\sum_{sym}(2a^{5}b-a^{3}b^{3}-4a^{4}bc+10a^{3}b^{2}c-7a^{2}b^{2}c^{2})\geq0.$$
Thus, it's enough to pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and the... | Some naive bounding gives
$(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{\frac 13} < (\sqrt 5 + 2)^{\frac 13} < (3+2)^{\frac 13} < 8^\frac 13 = 2$.
It is also easy to show that this quantity is positive, so if it is an integer then it must be $1$.
Some Galois theory shows that if $(\sqrt 5 + 2)^{\frac 13} - (\sqrt 5 - 2)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Solve Euler Project #9 only mathematically - Pythagorean triplet The "Euler Project" problem 9 (https://projecteuler.net/problem=9) asks to solve:
$a^2$ + $b^2$ = $c^2$
a + b + c = 1000
I find answers solving it with brute-force and programmatically, but is there a way to solve the problem ONLY mathematically? Can some... | Hint Euclid's parameterization of the Pythagorean triples (Elements, Book X, Proposition XXIX) is:
$$a = k (m^2 - n^2), \qquad b = 2 k m n, \qquad c = k (m^2 + n^2),$$
where $m > n > 0$ and $m, n$ coprime and not both odd.
Substituting in our condition gives
$$1000 = a + b + c = 2 k m (m + n),$$ and clearing the consta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How how do I find the measure of the sides of an equilateral triangle inscribed in a 30-60-90 triangle?
I am completely lost, I have no idea where to even start. I am sure someone here would be able to do this easily which is why I'm posting this here. Basically the point of the problem is to find the measure of line ... |
Let the side length of the equilateral triangle be $x$, and let the tilt angle $\theta$ be as shown. Then
$ x \cos \theta = 1 $
And the top right vertex of the equilateral triangle has coordinates
(Assuming the right angle vertex is the origin)
$P = (1 - x \cos(\theta + 60^\circ) , x \sin(\theta + 60^\circ) ) $
And t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integral $\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx$ The following problem was posted earlier this year by Cornel Ioan Valean:
Prove that
$$I=\int_0^1 \frac{2x-1}{1+x-x^2}\left(4\ln x\ln(1+x)-\ln^2(1+x)\right)dx=\frac{127}{20}\zeta(3)-\frac{8\pi^2}{5}\ln(\varphi)$$
My idea was to conside... | We start by writing
\begin{align}
I &= \int \limits_0^1 \frac{1-2x}{1+x -x^2} \log(1+x) [\log(1+x)-4\log(x)] \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{1-2x}{1+x-x^2} \left[\log^2\left(\frac{x^2}{1+x}\right) - 4 \log^2(x)\right] \mathrm{d} x \equiv J -4K \, .
\end{align}
Then
\begin{align}
J &= \int \limits_0^1 \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
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prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
| We have
$$
3^0 \equiv +1 \bmod 4
\\
3^1 \equiv -1 \bmod 4
\\
3^2 \equiv +1 \bmod 4
\\
3^3 \equiv -1 \bmod 4
\\
$$
Therefore, $3^{3n}+3^{2n}+3^{n} + 1 \equiv 1^n +(-1)^n + 1^n + (-1)^n \equiv 0 \bmod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Logarithmic equation, some variables in bases and in arguments This is the exercise, there are no clues in the book about it.
$$
40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2
$$
Solutions given by the book: $x=1; x=4; x=\frac{\sqrt{2}}{2}.$
And this is what I did so far:
*
*conditions for existenc... | We can use that
$$\log_a^b=\frac{\log a}{\log b}$$
therefore
$$40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2$$
$$40\frac{\log x^\frac{1}{2}}{\log {4x}}-14\frac{\log x^3}{\log {16x}}=-\frac{\log x^2}{\log {\frac{1}{2}x}}$$
$$20\frac{\log x}{\log {x}+\log 4}-42\frac{\log x}{\log {x}+\log 16}=-2\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Limit of $a_n := \frac{5^n}{2^{n^2}}$ Consider the sequence $(a_n)$ defined by $a_n := \frac{5^n}{2^{n^2}}$.
1. Prove that the sequence $(a_n)$ is bounded below by $0$.
We note that $a_n > 0$ for $n\geq 0$. Thus, the sequence is bounded from below.
2. Prove that the sequence $(a_n)$ is strictly decreasing by showing th... | For the first question: The are many ways to show this but we will use induction. Well $a_1 = \frac{5}{2} > 0$. Now suppose that $a_n > 0$ for some $n$. Then
\begin{equation*}
a_{n+1} = \frac{5^{n+1}}{2^{(n+1)^2}} = \frac{5^{n+1}}{2^{n^2+2n+1}} = \frac{5}{2^{2n+1}}\cdot \frac{5^n}{2^{n^2}} = \frac{5}{2^{2n+1}}\cdot a_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Markov Chain: Calculating Expectation Reach a Certain Set of States Suppose I have a Markov chain $Z_k$ with $6$ states, as depicted below:
The probability of moving from one node to a neighboring node is $1/2$. For example, the probability of moving from node $1$ to node $2$ is $1/2$ and the probability of moving fr... | Taking two steps in the Markov chain can lead to one of two things, with equal probability:
*
*$1 \to 2 \to 3$ or $1 \to 6 \to 5$ and we're done.
*$1 \to 2 \to 1$ or $1 \to 6 \to 1$ and we're back where we started.
We took $2$ steps. In the first case, we have $0$ steps left, and in the second case, we have $\mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove this formula for the $\sin\left(\frac{x}{2^n}\right), x \in [0,\frac{\pi}{2}[, n \in \Bbb{N}$ The formula in question:
$$\sin\left(\frac{x}{2^n}\right) = \sqrt{a_1-\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\dots+\sqrt{a_{n-1}+\sqrt{\frac{a_{n-1}}{2}\left(1-\sin^2(x)\right)}}}}}}$$
where
$$a_k = \frac{1}{2^{2^k-1}} \quad \for... | It's a bit tedious, but nonetheless doable:
Since we have
\begin{align}\sin\left(\frac{x}{{2^{n+1}}}\right)&= \sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{\sin^2\left(\frac{x}{{2^{n}}}\right)}{4}}}
\end{align}
We can use the induction hypothesis to calculate
\begin{align*}\frac{1}{4}\cdot\sin^2\left(\frac{x}{{2^{n}}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Evaluate $\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}$ How to prove
$$\sum _{k=0}^{\infty } \frac{L_{2 k+1}}{(2 k+1)^2 \binom{2 k}{k}}=\frac{8}{5} \left(C-\frac{1}{8} \pi \log \left(\frac{\sqrt{50-22 \sqrt{5}}+10}{10-\sqrt{50-22 \sqrt{5}}}\right)\right)$$
Where $L_k$ denotes Lucas number and $C$ C... | This is more of an extended comment at this point until I link to the final result, but numerically the above relation is plausible.
There are two observations to make to take the main step forward:
$$L_n = \phi^n + (1-\phi)^n $$
where $\phi = (\sqrt{5}+1)/2$ is the golden ratio, and
$$\sum_{k=0}^{\infty} \frac{x^{2 k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Isocline in dynamical system I solved differential equation and now should draw graph. So, we have:$$y'=\sqrt{3+y^2}$$ $$x'=x^2+x$$
I get isocline $x=-1$ (from $x'=0$). Is it correct or maybe here is more?
| Solving both equations we get the following:
$$y = \sqrt{3}\sinh(t+C)$$
$$ t + C = \int \frac{1}{x^2+x}dx = \int \frac{\frac{1}{x^2}}{1+\frac{1}{x}}dx = -\log\left(1+\frac{1}{x}\right)$$
Then plugging in, we get these nice graphs which we can plot:
$$y = \frac{\sqrt{3}}{2}\left(\frac{C}{1+\frac{1}{x}} - C^{-1}\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @... | I asked Cornel for a solution to the nice key result from Ahmad Bow's solution. Here is a solution in large steps.
We need two known results, that is $\displaystyle \int_0^1\frac{x^n}{1+x}\textrm{d}x=H_{n/2}-H_n+\log(2)$ and $\displaystyle \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Eigenvalues of almost symmetric matrix after discretizing a PDE are negative
Let $h > 0$. Define, for every positive integer $n$, $a_n = a + n \cdot h,$
and $a_{n+1/2} = a + \left( n + \frac{1}{2} \right) \cdot h, $where $a \in (0,1)$. How can we prove that the eigenvalues of the matrix $$A_{n+1} = \begin{pmatrix} ... | Let $\{e_1,e_2,\ldots,e_{n+1}\}$ be the standard basis of $\mathbb R^{n+1}$. Then $A=DP$ where $D=-\operatorname{diag}\left(\frac{1}{a_1},\ldots,\frac{1}{a_{n+1}}\right)$ is a negative diagonal matrix and
$$
P=a_{0+\frac12}e_1e_1^T+\sum_{i=1}^na_{i+\frac12}(e_i-e_{i+1})(e_i-e_{i+1})^T+a_{n+1+\frac12}e_{n+1}e_{n+1}^T.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$
Show that if $x_1,x_2,...,x_n$ are positive numbers then $\left(\dfrac{1}{n}\sum_{i=1}^n x_i^5\right)^{1/5}\geq \dfrac{1}{n}\sum_{i=1}^n x_i$.
I'm a little stuck on this question. I think I might need to use convex functio... | One may also use Tchebecheff's inequality repeatedly. If $a \ge b \ge c $ and $p \ge q \ge r$, then $$(ap+bq+cr) \ge \frac{(a+b+c)(p+q+r)}{3}~~~~(*)$$ We have
$$(a^2+b^2+c^2) \ge \frac{(a+b+c)(a+b+c)}{3} \ge \frac{(a+b+c)^2}{3} ~~~(1)$$ Next $$(a^3+b^3+c^3) \ge \frac{(a+b+c)(a^2+b^2+c^2)}{3} \ge \frac{(a+b+c)^3}{9} ~~~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laplace Inverse of the problem. What is the Laplace Inverse of the given question?
$$\frac{\sqrt{4+s^3}}{s^3}$$
I tried solving it by expanding $\sqrt{1+\frac{s^3}{4}}$ but the terms will not have Laplace Inverse.
If I expand it like $\sqrt{1+\frac{4}{s^3}}$i.e.$$s^{-3/2}* \sqrt{1+\frac{4}{s^3}}$$
It will still have m... | With CAS help I have:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{4+s^3}}{s^3}\right](t)=\frac{2 \sqrt{t} \,
_1F_3\left(-\frac{1}{2};\frac{1}{2},\frac{5}{6},\frac{7}{6};-\frac{4 t^3}{27}\right)}{\sqrt{\pi }}$$
For general a:
$$\mathcal{L}_s^{-1}\left[\frac{\sqrt{a+s^3}}{s^3}\right](t)=\\\mathcal{L}_s^{-1}\left[\sum _{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. Prove it!
$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{r... | In general, the period of $2$ in $\Bbb Z/(3^{N+1})^\times$ is $2\cdot3^N$. And we expect that the $2$-adic expansion of $-3^{N+1}$ should be purely periodic, period $2\cdot3^N$.
Indeed, since $3^{N+1}|(2^{2\cdot3^N}-1)$, say with quotient $Q_N$, we get the results
\begin{align}
Q_N&=\frac{2^{2\cdot3^N}-1}{3^{N+1}}\\
-\... | {
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"source": "stackexchange",
"question_score": "3",
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How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumpe... | Your condition isn't quite right. When $x = 1.1$, $x^3−x−1 = 1.331 - 1.1 - 1 < 0$.
From $\log(x^3 - x - 1) < \log(x^3 + x - 1)$, subtract one side from the other, then use a logarithm property to get one logarithm. \begin{align*}
0 &< \log(x^3 + x - 1) - \log(x^3 - x - 1) \\
0 &< \log \left( \frac{x^3 + x - 1}{x^3 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$
Evaluate
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\lef... | Using the facts that $(e^{\pi i/5})^5=e^{\pi i}=-1$ and $i^5=i$, we have
$$\begin{align}
(1+\sin\pi/5+i\cos\pi/5)^5+i(1+\sin\pi/5-i\cos\pi/5)
&=(1+ie^{-\pi i/5})^5+i(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5})^5(1+ie^{-\pi i/5})^5+i^5(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5}+i)^5+(i+e^{\pi i/5})^5\\
&=0
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinations company A company has $4$ workers who can build a wall, $5$ workers who can paint a wall and 2 workers who can do both. A job requires $ 3$ builders and $3$ painters. In how many ways can the company choose the workers for the job?
I found this resolution, is it correct?
We will have three disjoint cases:
... | Your solution is correct. I have the same result by making another approach. I´ve made a table which shows the number of builder (B), painter (P) and mixed workers (M)
B P M
1. 3 3 0
2. 2 3 1
3. 3 2 1
4. 2 2 2
5. 1 3 2
6. 3 1 2
For 1. we calculate the number of ways to select $3$ builder out o... | {
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A question about modulus for polynomials The other day my friend was asked to find $A$ and $B$ in the equation
$$(x^3+2x+1)^{17} \equiv Ax+B \pmod {x^2+1}$$
A method was proposed by our teacher to use complex numbers and especially to let $x=i$ where $i$ is the imaginary unit. We obtain from that substitution
$$(i+1)^{... | *
*It comes from the exponential form of complex numbers: $1+i$ has modulus $\sqrt 2$ and argument $\frac\pi 4$, so it writes as
$$1+i=\sqrt2\,\mathrm e^{\tfrac\pi 4},\quad\text{and similarly}\quad 1-+i=\sqrt2\,\mathrm e^{-\tfrac\pi 4} $$
The substitution is valid because of the meaning of the congruence:
\begin{alig... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that sequence is convergent. if $S_1$ and $S_2$ are two positive real number and $S_{n+2}=\sqrt{S_{n+1}S_n}$
If $s_1$ and $s_2$ are two positive real number and $$s_{n+2}=\sqrt{s_{n+1}.s_n}$$ then prove that sequence is convergent and find its limit.
I know that its solve by using concept of monotonic and boun... | Since $s_{n+2}=\sqrt{s_{n+1}.s_n}$, $\frac{s_{n+2}}{s_{n+1}}=\sqrt{\frac{s_n}{s_{n+1}}}$. Let $T_n=\frac{s_{n+1}}{s_n}$. Then $T_{n+1}=T_n^{-1/2}$, or $log \left(T_{n+1}\right) = -\frac{1}{2}log\left(T_{n}\right)$. Hence $log\left(T_n\right)={\left(-\frac{1}{2}\right)}^{n-1}log(T_1)$, and $T_n=T_1^{{\left(-\frac{1}{2}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matrix for reflection about the line $y = \tan (\theta) \, x$ How would I show that a reflection about the line $y = \tan (\theta) \, x$ is the following?
\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
| The transformation\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
maps any arbitrary vector $P=\begin{pmatrix}\alpha\\\beta\end{pmatrix}\in\mathbb R^2$ to the vector
$P'=\begin{pmatrix}\alpha\cos2\theta+\beta\sin2\theta\\\alpha\sin2\theta-\beta\cos2\theta\end{pmatrix}$.
It is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integers m,n and primes $p\geq5$ such that $m(4m^2+m+12)=3(p^n-1)$ I did it like this. We can manipulate the equation to come to
$\frac{(m^2+3)(4m+1)}{3}=p^n$
Now 3 divides either $(m^2+3) or (4m+1) $
If we assume 3 divides $(m^2+3)$ and it is equal to $(4m+1)$ (my intuition) .we get
$\frac{m^2+3}{3... | You have that
$$\frac{(m^2+3)(4m+1)}{3}=p^n \tag{1}\label{eq1A}$$
I don't see any particular way to directly prove your assumption that $\frac{m^2 + 3}{3} = 4m + 1$. Instead, here is a proof it's the only solution, so your assumption is true in that manner.
Since both $m^2 + 3$ and $4m + 1$ are $\gt 3$ for positive int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sum \frac{x}{x^2+7}\le \frac{3}{8}$ Let $x,y,z>0$ such that $xy+yz+xz=3$. Show that $$\frac{x}{x^2+7}+\frac{y}{y^2+7}+\frac{z}{z^2+7}\le \frac{3}{8}$$
We have: $$x+y+z\ge \sqrt{3\left(xy+yz+xz\right)}=3\rightarrow \frac{3}{8\left(x+y+z\right)}\le \frac{3}{8\cdot 3}$$
Then i will prove $$\sum \frac{x}{3x^... | Hint: Use the substitution $x = \sqrt 3 \tan A$ and so on for some acute $\triangle ABC$, and then Jensen’s inequality as $t \mapsto \dfrac{\sqrt 3 \tan t}{3\tan^2t + 7}$ is concave for $t \in (0, \frac\pi2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is the following series $\sum_{n=0}^\infty \frac{n^3}{n!}$ convergent and if so what is the sum? I started by converting the following $\sum_{n=0}^\infty \frac{n^3}{n!}$ into
$$\sum_{n=1}^\infty \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!}.$$ Did i get it right and where should I go from here ?
| Note that by limit comparison test with $\sum \frac 1{n^2}$
$$\frac{\frac{n^3}{n!}}{\frac 1{n^2}}=\frac{n^5}{n!}\to 0$$
therefore the series converges.
For the sum we can use that
$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}=\sum_{n=0}^\infty \frac{n^2-1+1}{(n-1)!}=\sum_{n=2}^\infty \frac{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation of a hyperbola given its asymptotes
Find the equation of the hyperbola whose asymptotes are $3x-4y+7$ and $4x+3y+1=0$ and which pass through the origin.
The equation of the hyperbola is obtained in my reference as
$$
(3x-4y+7)(4x+3y+1)=K=7
$$
So it make use of the statement, the equation of the hyperbola = e... | $$
\frac{4x+3y+1}{5}=\pm\frac{3x-4y+7}{5}\\
\implies x+7y-6=0\;;\; 7x-y+8=0\text{ which are the axis of the hyperbola with centre }(-1,1)\\
$$
Since $m_1m_2=-1\implies$ asymptotes are perpendicular $\implies$ rectangular hyperbola
$$
\frac{(x+7y-6)^2}{50a^2}-\frac{(7x-y+8)^2}{50a^2}=\pm1\\
\text{At }(0,0): \frac{18}{25... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the ... | A more thorough version of the answer given by fleablood, avoiding assumptions and unexamined cases.
The stated question, find all solutions for $n$ such that $n^2=(n-1)!+1$
Since for all non-negative integers, $k!\ge 1$, we conclude $n^2\ge 2 \Rightarrow n>1 \Rightarrow (n-1)>0$
Rearranging, $(n-1)!=n^2-1=(n-1)(n+1)$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of ways to distribute 10 pieces of candy using this generating function The question asks:
a) Find a generating function for the number of ways to distribute identical pieces of candy to 3 children so that no child gets more than 4 pieces. Write this generating function in closed form, as a quotient of... |
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain
\begin{align*}
\color{blue}{a_{10}}&\color{blue}{=[x^{10}]\left( \frac{1-x^5}{1-x} \right)^3}\\
&=[x^{10}]\frac{1-3x^5+3x^{10}-x^{15}}{(1-x)^3}\tag{1}\\
&=[x^{10}]\left(1-3x^5+3x^{10}\right)\sum_{j=0}^\infty \binom{-3}{j}(-x)^j\tag{2}\\
&=\left([x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \... | You want $2^{42} \mod 10$. You found that $2^4\equiv6\mod 10$. Now $6^n\equiv 6$ for all $n\in\mathbb N$ (you could prove that by induction), so $2^{42}\equiv2^{40}2^2\equiv(2^4)^{10}2^2\equiv6^{10}\times4\equiv6\times4\equiv4\mod10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the integral $\int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}$
I'm trying to solve the following indefinite integral:
$$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}$$
The integral is a general case which comes from a physics problem of p... | Start by using the substitution $x=\frac{2z}{1+z^2}$ then:$$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}=-4\int \frac{\ln z}{1-z^2}dz=2\int \frac{\ln z}{z-1}dz-2\int \frac{\ln z}{z+1}dz$$
$$=-2\operatorname{Li}_2(1-z)-2\operatorname{Li}_2(-z)-2\ln z \ln(1+z)+C,\quad z=\frac{1-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430785",
"timestamp": "2023-03-29T00:00:00",
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g... | $$f(x)=(x^2+5x+4)(x^2+5x+6)+5=(x^2+5x+5)^2-1+5\geq4.$$
The equality occurs for $x^2+5x+5=0,$ which happens on $[-6,6],$ which says that $4$ is a minimal value.
Now, $f(6)=5045$ and since for any $-6\leq x\leq6$
$$5045-f(x)=(6-x)(x+11)(x^2+5x+76)\geq0,$$ we see that $5045$ is a maximal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need ... | For a more accurate bound consider the following:
$$
\left|a_{n} - \frac{1}{2} \right| = \left|\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2} \right| =
\left|\frac{2-(\sqrt{1+\frac{1}{n}}+1)}{2\left( \sqrt{1+\frac{1}{n}}+1 \right)} \right| \leq \left| \frac{2-(\sqrt{1+\frac{1}{n}}+1)}{4} \right| \implies \left| 2-(\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a closed form expression for $\sum_{k=0}^n(k^2+3k+2)$ using generating functions First of all, i know that there is already a similar question (Closed form expression for $\sum_{k=0}^{n}(k^2 + 3k + 2)$) regarding my equation in this forum, but my question is only about verification of my thought process because... | Yes, this approach of splitting into three sums is valid, but you forgot the $\sum_{k=0}^n$. Here's a correct solution. Let $a_n=\sum_{k=0}^n(k^2+3k+2)$. Then
\begin{align}
\sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty \left(\sum_{k=0}^n(k^2+3k+2)\right) x^n\\
&= \sum_{k=0}^\infty (k^2+3k+2) \sum_{n=k}^\infty x^n\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the interpretation of the difference? We have the linear maps \begin{equation*}f:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+x_2 \\ x_3 \\ x_1+x_2\end{pmatrix} \ \text{ and } \ h:\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} \mapsto \begin{pmatrix}x_1+ x_2 \\ x_3 \\ x_1-x_2\end{pmat... | The matrices corresponding to the linear transformations $f,h$ are $$\begin{bmatrix}1&1&0\\0&0&1\\1&1&0\end{bmatrix},\begin{bmatrix}1&1&0\\0&0&1\\1&-1&0\end{bmatrix}$$
respectively. Obviously the determinant of the former is $0$ since there are two identical (and hence linearly dependent) rows, while you can show that ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$
Domain of $f(x)=\sec^{-1}\dfrac{x}{\sqrt{x-[x]}}$ is:
$$
x-[x]\neq0\implies x\notin \mathcal{Z}\implies x\in\mathcal{R}-\mathcal{Z}\\
\sec^{-1}:\mathcal{R}-(-1,1)\to(0,\pi)\implies \dfrac{x}{\sqrt{x-[x]}}\in\mathcal{R}-(-1,1)\\
\dfrac{x}{\sqrt{x-[x]}}\notin(-... | In order to take the square root (and have a real result) we need $x - \lfloor x \rfloor \ge 0$ , but that is not a problem.
We cannot divide by $0$
$x - \lfloor x \rfloor \ne 0 \implies x\notin \mathbb Z$
For inverse secant to be defined we need
$\left|\frac {x}{\sqrt{x - \lfloor x \rfloor}}\right|\ge 1$
$|x| \ge \sqr... | {
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"url": "https://math.stackexchange.com/questions/3438460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove that $\sum_{i=3}^{n} \frac{i-2}{\binom{i}{2}} < \frac n 4$ for all natural $n > 3$? How do I prove:
$$\frac{3-2}{\binom 3 2} + \frac {4 - 2}{\binom 4 2} + \dots + \frac{n-2}{\binom n 2} < \frac n 4$$
I tested this sum on a variety of $n$ from $2$ to $100$ and they all seem to be less than $\frac n 4$, ho... | The inequality is equivalent to:
$$\sum_{i=3}^{n} \frac{i-2}{i(i-1)} < \frac n 8$$
Using partial fractions, $\frac{i-2}{i(i-1)} = \frac 2 n - \frac 1 {n-1} = \frac 1 n - \left(\frac 1 {n-1} - \frac 1 n\right)$.
The given sum becomes:
$$\sum_{i=3}^{n} \frac 1 i - \sum_{i=3}^{n} \left(\frac 1{i-1} - \frac 1 i\right)$$
$$... | {
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"answer_id": 2
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prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$
1) For $n\ge 2$, prove that $\frac{n!}{n^n}\le(\frac{1}{2})^k$ where $k=[\frac{n}{2}]$, the greatest integer $\le\frac{n}{2}$.
2) Deduce the value of $\underset{n\rightarrow\infty}{\lim}\frac{n!}{n^n}$.
3) ... | If $n$ is even, the first $\frac n2$ factors of $$\frac {1\cdot 2 \cdot 3 \ldots n}{n^n}$$ are less than or equal to $\frac 12$ and the rest are less than or equal to $1$, so the fraction is less than or equal to $\left(\frac 12\right)^{n/2}$. If $n$ is odd we just need to account for one more factor $\frac 12$. Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow ... | Instead of representing $\frac{a}{b}$ as a fraction, represent it as the vector $\left( \begin{array}{c} a \\ b \end{array} \right)$.
Then, all you are doing to generate your sequence is repeatedly multiplying by the matrix $\left( \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right)$. One of the eigenvectors of this m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 1
} |
Strategy for the Limit: $\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n} $ I do not understand how to properly solve this limit:
$$
\lim_{n\to\infty} \frac{2^{n+1}+3^{n+1}}{2^n+3^n}
$$
I thought of breaking it up:
$$
\lim_{n\to\infty} \frac{2^{n+1}}{2^n+3^n} +\lim_{n\to\infty} \frac{3^{n+1}}{2^n+3^n}
$$
But I do no... | Since the 3s dominate,
I would expect the limit
to be 3.
Check:
$\begin{array}\\
\dfrac{2^{n+1}+3^{n+1}}{2^n+3^n}-3
&=\dfrac{2^{n+1}+3^{n+1}-3(2^n+3^n)}{2^n+3^n}\\
&=\dfrac{2^{n+1}+3^{n+1}-3\cdot 2^n-3^{n+1}}{2^n+3^n}\\
&=\dfrac{2^{n+1}-3\cdot 2^n}{2^n+3^n}\\
&=\dfrac{-2^n}{2^n+3^n}\\
&\to 0
\qquad\text{since } 2^n/3^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1... | Another way:
From $\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac\pi3$
Let $\cos^{-1}\dfrac{1-x^2}{1+x^2}=2y\implies0\le2y\le\pi$
$\cos2y=?,x^2=\tan^2y$
Case$\#1:$
If $x=\tan y\ge0$
$$\tan^{-1}(\tan2y)=\dfrac\pi3-2y\implies\tan(2y)=\tan\left(\dfrac\pi3-2y\right)$$
$\implies2y=n\pi+\dfrac\pi3-2y\iff y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt... | Divide all the terms by $n^{2}$ and take the limit. you will see that the limit is $\frac 2 {1+1+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statem... | Hint
Like Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$
If $f(m)=4^{2m+1}+5^{2m+1}+6^{2m+1},$
$$f(n+1)-4^2f(n)=5^{2n+1}(5^2-4^2)+6^{2n+1}(6^2-4^2)$$ will be clearly divisible by $15$ if $n\ge0$
So, if $15$ divides $f(n),15$ will divide $f(n+1)$
Now establish the base case i.e., $m=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then Find $6xyz$. If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then
Find $6xyz$.
How can I approach this problem?
I need some hints. Thanks.
| $x + y + xy = 3 \implies ( 1 + x ) ( 1 + y ) = 4\tag1$
$y + z + yz = 8 \implies ( 1 + y ) ( 1 + z ) = 9\tag2$
$z + x + zx = 15 \implies ( 1 + z ) ( 1 + x ) = 16\tag3$
Multiplying $(1)$ and $(3)$, we have
$$( 1 + x )^2 ( 1 + y )( 1 + z )=4\cdot 16\implies ( 1 + x )^2= \dfrac{64}{9}\qquad \text{{using $(2)$}}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Stuck on a probability law problem I'm currently trying to solve a problem, I completed the first question but I am stuck at the second one, here is the problem:
One person roll a die until the result is a $1$, a second person toss a coin until he gets $3$ tails.
*
*How many tries are they going to make on average.
... | We have $$\begin{align}
\Pr(X=Y)&=\sum_{k=3}^\infty\frac16\left(\frac56\right)^{k-1}\left(\frac12\right)^k\binom{k-1}{2}\\
&=\frac15\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k\binom{k-1}{2}\\
&=\frac{1}{10}\sum_{k=3}^\infty\left(\frac{5}{12}\right)^k(k-1)(k-2)\\
&=\frac{1}{24}\sum_{k=3}^\infty\left(\frac{5}{12}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to 0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$
If $$f(x)= \frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}$$ is continuous at $x=0$ then find $f(0)$
$$
f(0)=\lim_{x\to0}\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)}\\
=\lim_{x\to0}\big[\frac{a^x-1}{x}\big]^3.\frac{x\log a}{\sin(x\log a)}... | Using series expansions and $o$-notation you get for $x\to 0$:
$$\frac{(a^x-1)^3}{\sin(x\log a)\log(1+x^2\log a^2)} = \frac{(e^{x\log a}-1)^3}{(x\log a +o(x))(2x^2\log a+ o(x^3))}$$
$$= \frac{(x\log a+o(x))^3}{2x^3\log^2 a+ o(x^3)} = \frac{x^3\log^3 a+o(x^3)}{2x^3\log^2 a+ o(x^3)}$$
$$=\frac{\log^3 a+o(1)}{2\log^2 a+ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin... | Other way is used identities double angle and sum-product
\begin{eqnarray*}
2\cos^2x+\sin x& = & 1 \\
2\cos^2x-1+\sin x& = & 0\\
\cos(2x)+\sin x& = & 0\\
\cos(2x)+\cos\left(\frac{\pi}{2}-x\right) & = & 0\\
2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)\cos\left(\frac{3x}{2}-\frac{\pi}{4}\right) & = & 0\\
\end{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
If $\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$ where $x\in [0,2\pi]$ If we solve the above expression, it’s very clear that the equation doesn’t depend upon x. Then all that is left to is to get
$$\sin x -\cos x \not =0$$
$$x\not = \pi /4, 5\pi/4$$
But the answer give... | Since
$$\frac{\sin^3 x -\cos^3 x}{\sin x - \cos x}-\frac{\cos x}{\sqrt{1+\cot^2x}}-2\tan x \cot x=-1$$
$$\sin^2x+\sin x\cos x+\cos^2x-\frac{\cos x}{\sqrt{\frac{\sin^2 x+\cos^2x}{\sin^2x}}}-2=-1$$
$$\sin x\cos x-|\sin x|\cos x=0$$
which is always true for $\sin x\ge 0$ otherwise it is true for $x=0,\frac \pi 2, \frac{3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is there an integer-sided right triangle with square perimeter and square hypotenuse? Is there an integer-sided right triangle with square perimeter and square hypotenuse?
Frenicle[89] noted (pp. 71-8) that if the hypotenuse and perimeter of a right triangle both are squares, the perimeter has at least 13 digits.
"His... | An elementary proof for primitive triangles
A primitive triangle has sides $2mn,m^2-n^2,m^2+n^2$ where $m,n$ are coprime and of opposite parity. We require $m^2+n^2$ and $2m(m+n)$ to be squares.
Then $m=2u^2, n=v^2-2u^2$ and $$v^4-4v^2u^2+8u^4=z^2,$$ where $u,v,z$ are coprime in pairs and $v,z$ are odd.
Then $(v^2-2u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is ... | To conclude your proof we can use that by AM-GM
$$x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2$$
$$\big{|}\log(1+x^2y^2)\big{|} \le x^2y^2 \le \left(\frac{x^2+y^2}2\right)^2=\frac{\left(\sqrt{x^2+y^2}\right)^4}{4}<\epsilon $$
as $\sqrt{x^2+y^2} < \sqrt[4]{4\epsilon}$.
As an alternative, we have that
$$\dfrac{y^2\log(1+x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to prove $\int^{\infty}_0 e^{-c^2/a^2}c^4\,dc=\frac{3}{8}a^5\sqrt\pi$? How do i prove this integral can someone give me proof i've been trying this too long i tried direct integration by parts, differentiation all to no avail.Please support us :)
$$\int^{\infty}_0 e^{\frac{-c^2}{a^2}}c^4\,dc=\frac{3}{8}a^5\sqrt\pi... | Two times by parts:
$\int\limits_0^\infty e^\frac{-x^2}{a^2}dx=
\left.xe^\frac{-x^2}{a^2}\right|_0^\infty+
\int\limits_0^\infty \frac{2x^2}{a^2}e^\frac{-x^2}{a^2}dx=
\frac{2}{a^2}\int\limits_0^\infty e^\frac{-x^2}{a^2}d\left(\frac{x^3}{3}\right)=$
$
\frac{2}{a^2}\left(\left.\frac{x^3}{3}e^\frac{-x^2}{a^2}\right|_0^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\... | $${\sqrt{\cosh(3x^2)}e^{4x^3}-1\over x^2\tan(2x)}={1\over\sqrt{\cosh(3x^2)}e^{4x^3}+1}\cdot{x\over\tan(2x)}\cdot{\cosh(3x^2)e^{8x^3}-1\over x^3}$$
and
$${\cosh(3x^2)e^{8x^3}-1\over x^3}={e^{8x^3}-1\over x^3}+e^{8x^3}{\cosh(3x^2)-1\over x^3}={e^{8x^3}-1\over x^3}+{e^{8x^3}\over\cosh(3x^2)+1}\left(\sinh(3x^2)\over x^2\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$4x≡2\mod5$ can you divide both sides by $2$ to get $2x≡1\mod5\,?$ Since gcd$(2,5)=1$ , could you treat $4x$ as $2(2x)$ and cancel the $2$ on both sides?
i.e. $$2(2x)≡2\mod5\implies 2x≡1\mod5$$
Thanks!
| Not really.
$4x \equiv 2\pmod 6$ (so $x$ could be $2\pmod 6$ or $x$ could be $5\pmod 6$) does not mean $2x \equiv 1\pmod 6$ (which is impossible).
But you can say
$ka \equiv kb \pmod n \implies a \equiv b \pmod {\frac n{\gcd(k, n)}}$
And so
$4x \equiv 2 \pmod 6$ meams $\frac {4x}2\equiv \frac 2{2}\pmod {\frac {6}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Checking if series converge: $\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$ and $\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$ etc. I want to see if these series converge:
$$\text{1. }\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$$
$$\text{2. }\sum_{k=1}^\infty \frac{(1+i)^k}{k!}$$
$$\text{3. }\sum_{k=1}^\infty \frac{k^2+2}{k^4+1}... | *
*What you did is fine.
*The absolute value of a complex number is always real and non-negative. So you have
$$
\left|\frac{a_{k+1}}{a_k}\right|= \left|\frac{(1+i)^{k+1}}{(k+1)(1+i)^k} \right|
=\left|\frac{(1+i)}{k+1}\right|=\frac{|1+i|}{k+1}=\frac{\sqrt2}{k+1}.
$$
*Your limit is wrong. It's $1$, which gives you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $(x^2+1)y''-2xy'+2y=0$
Solve $$(x^2+1)y''-2xy'+2y=0$$
Seems I can't use Euler Differential method. I tried it
\begin{align}
(x^2+1)y''-2xy'+2y&=0\\
\text{Let }y&=xv\\
(x^2+1)(xv''+2v')-2x(xv'+v)+2(xv)&=0\\
x(x^2+1)v''+2v'&=0\\
\frac{v''}{v'}&=-\frac{2}{x(x^2+1)}\\
\frac{v''}{v'}&=-\frac{2}{x}+\frac{2x}{x^2+1}
... | Hint: Put $x^2+1 = t$ and differentiate and put back in your ODE
Think backward, let $y=x^n$ maybe one of solution of your ODE. Then it must satisfy the ODE,
\begin{align}
y'&=nx^{n-1}\\
y''&=n(n-1)x^{n-2}
\end{align}
\begin{align}
(x^2+1)n(n-1)x^{n-2}-2xnx^{n-1}+2x^n&=0\\
(n^2-3n+2)x^n+n(n-1)x^{n-2}&=0\\
(n-1)(n-2)x^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\pri... | If $f = x^2 + y^2$ then we have $f' = 2\sqrt{f}$. That can be solved easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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how to prove the identity that $2^{ab}-1=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+.....+2^a+1)$ I am really confused.
The identity is given but I really want to know why$?$
Show that if $2^n -1$ is a prime , then $n$ is a prime . [Hint : Use the identity]
$$2^{ab} -1 = (2^a-1)\cdot(2^{a(b-1)} + 2^{a(b-2)} + \cdots + 2^a +1 )$$
... | Consider the following G.P :
$$1,2^a , 2^{2a} ,\cdots ,2^{a(b-2)} , 2^{a(b-1)} $$
We use the Formula to find it's sum :
$$S = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}$$
Or
$${\color{#d05}{1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)} = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}}}$$
Rearranging this gives us the deisred result :
$${\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Can any number of squares sum to a square? Suppose
$$a^2 = \sum_{i=1}^k b_i^2$$
where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).
Can any positive integer be the value of $k$?
The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we hav... | Solutions exist for every $k>0$. The simplest forms use most of the $b_n$ values as $1$. I will list them as "$b_*$". I suspect there are infinitely many distinct answers for each $k\ge2$, but I can't prove it.
if $k = 2n$ (k is even) and large enough ($k\ge4$ or $n>1$):
*
*$ a = n $
*$ b_1 = n-1 $
*$ b_* = 1 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 8,
"answer_id": 5
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | I will be giving a $\varepsilon -\mathcal{N}$ proof
notice
$$\begin{align}\frac{n}{1+n^2} + \frac{n}{2+n^2} + \cdots + \frac{n}{n+n^2} - 1=
n\Big( \frac{1}{1+n^2} - \frac{1}{n^2} + \cdots + \frac{1}{n+n^2} - \frac{1}{n^2} \Big) = h_n\end{align}$$
Claim $\dfrac{n}{n+n^2} > \dfrac{r}{r+n^2}$
So
$|h_n| < \Big|n\cd... | {
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Limit of sequence-term not always positive Let the sequence $a_n=\frac{3^n+2(-4)^n}{(-4)^n-2^n}$.
We have $$\frac{3^n+2(-4)^n}{(-4)^n-2^n}=\frac{3^n+2(-1)^n 4^n}{(-1)^n 4^n-2^n}=\frac{(-1)^n 4^n \left( 2+ \frac{3^n}{(-1)^n 4^n}\right)}{(-1)^n 4^n \left( {1-\frac{2^n}{(-1)^n 4^n}}\right)}$$
In this case the ineqaulity $... | You have
\begin{align}
\frac{3^n+2(-4)^n}{(-4)^n-2^n}
&=\frac{3^n}{(-4)^n-2^n}+\frac{2(-4)^n}{(-4)^n-2^n}\\ \ \\
&=\frac{3^n}{4^n((-1)^n-(1/2))^n}+\frac{2}{1-2^n/(-4^n)}\\ \ \\
&\to 0+2=2
\end{align}
| {
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Finding the eigenvalues of a $3 \times 3$ matrix without the determinant I am trying to find the eigenvalues of
$$A = \begin{pmatrix} 0 & -2 & -3 \\ -1 & 1 & -1 \\ 2 & 2 & 5\end{pmatrix}$$
Without using the determinant. I first tried doing something like I did here Finding the eigenvalues of $\begin{pmatrix} a & b \\ ... | \begin{cases}
-\lambda x_1 & -2x_2 & -3x_3 &= 0 \\
-x_1 &+ \ (1-\lambda)x_2 &- x_3 &= 0 \\
2x_1 &+ \ 2x_2 &+ \ (5 - \lambda)x_3 &= 0
\end{cases}
What you are doing above is finding the null space of the matrix $A-\lambda I$. This consists of all vectors $v$ such that $Av=\lambda v$. This gives you equations that will ... | {
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Approximating Integrals within an error with Maclaurin Series "Use Maclaurin's Series to approximate the integral to 3dp accuracy"
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}} $$
I was wondering if it is possible to solve this question by identifying out the number of terms needed to obtain a 3dp accuracy. I thought of us... | Using the binomial expansion, we have
$$ \frac{1}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{2 n}$$ which, integrated gives
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \frac{2^{-(2 n+1)}}{2 n+1} \binom{-\frac{1}{4}}{n}$$
So, summing $p$ terms, you look for $p+1$ such that
$$R_p=\Big| ... | {
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Differentiation by Substitution Find $\frac{dy}{dx}$ if $$y=sin^{-1}\frac{1-x^2}{1+x^2}$$
Let us put $t=\frac{1-x^2}{1+x^2}$ then $y=sin^{-1}t$ So $$\frac{dy}{dx}=\frac{dy}{dt}×\frac{dt}{dx}$$ so we get $$\frac{dy}{dx}=\frac{d}{dt}(sin^{-1}t)×\frac{d}{dx}(\frac{1-x^2}{1+x^2})=\frac{1}{\sqrt{1-(\frac{1-x^2}{1+x^2})^2}}×... | Both your questions can be explained by looking at the graph of $y$.
*
*You'll notice that $y$ is not differentiable at $x = 0$ so your concern about $x/x = 1$ doesn't matter because there is no derivative there.
*You'll notice that for $x > 0$ the derivative is negative and for $x < 0$ the derivative is positive. ... | {
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Showing that for $p_k\ \ge 31, p_{k-1}\# > (p_{k+1})^2$ I am trying to show that for $p_k \ge 31, p_{k-1}\# > (p_{k+1})^2$
Does the following work? Is there a stronger or more straight forward result that can be shown?
Let:
*
*$p_k$ be the $k$th prime
*$p\#$ be the primorial for $p$.
Base Case: $7\# = 210 > 13... | I think your solution if fine. But you might find this a bit tighter ?
We have
\begin{eqnarray*}
p_{k-1} < p_{k+1} < p_{k+2} < p_{k+3} < 2p_{k-1}.
\end{eqnarray*}
So
\begin{eqnarray*}
p_{k} \#= p_k p_{k-1} \# >p_k p_{k+1}^2 > 4p_k^2 > p_{k+2}^2.
\end{eqnarray*}
| {
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Would this be a legal move in a Matrix? Here I have a matrix:
\begin{pmatrix}
b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{pmatrix}
I was wondering can I do the following operation on this:
$C2 \to C2 \times C2$
So it would give me the following:
\begin{pmatrix}
b^2c^2 & b^2c^... | You need to think in terms of homogeneous symmetric polynomials' degrees. Subtract $ab+bc+ca$ times the second column from the first, thus changing $b^2c^2$ to $-abc(b+c)$. Now add $abc$ times the third column to the first; it should vanish then. Explicitly$$\left(\begin{array}{c}
b^{2}c^{2}\\
c^{2}a^{2}\\
a^{2}b^{2}
\... | {
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If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$ I am studying general equation of the second degree. While studying that chapter I came across
$$\tan2\theta=\frac{b}{a-c} \tag{1}$$
Now from (1), the author computed
$$\cos2\theta=\frac... | From
$$\sin^2\alpha+\cos^2\alpha=1,$$
draw
$$\tan^2\alpha+1=\frac1{\cos^2\alpha}$$
and
$$\cos\alpha=\pm\frac1{\sqrt{\tan^2\alpha+1}}.$$
Then
$$\sin\alpha=\tan\alpha\cos\alpha=\pm\frac{\tan\alpha}{\sqrt{\tan^2\alpha+1}}.$$
| {
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n... | And if you want to generate
$(a_i)_{i=1}^m$ repeatedly,
the i-th term is generated by
$a_ix^i$,
and for this to repeat every $m$
this needs
$\dfrac{a_ix^i}{1-x^m}
$
so the final result is
$\sum_{i=1}^m\dfrac{a_ix^i}{1-x^m}
$.
In your case,
with $(1, 2)$
repeated,
this is
$\dfrac{x}{1-x^2}+\dfrac{2x^2}{1-x^2}
=\dfrac{x+... | {
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There is no continuous surjective multiplicative map from $M_n(\mathbb H)$ to $\mathbb H$ Let $\mathbb H$ denote the field of quaternions. I would like to prove that there does not exist any function $f:M_n(\mathbb H)\rightarrow \mathbb H$ for $n\geq 2$ that is continous surjective and multiplicative.
I have been thi... | I will work out the case $n = 2$ in detail. The same proof works for general $n$, I just want to save the labor of typing $n$ by $n$ matrices...
Thus assume that $f:\operatorname M_2 = \operatorname M_2(\Bbb H) \rightarrow \Bbb H$ is a surjective multiplicative map.
Lemma 1. Whenever $A\in \operatorname M_2$ is invert... | {
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Prove that matrix is invertible I have a symmetric real-valued matrix with the following structure:
\begin{bmatrix}
1 + (n-1)\alpha_{1}^2 & 1-\alpha_{1}\alpha_{2} & ... & 1-\alpha_{1}\alpha_{n} \\
1-\alpha_{1}\alpha_{2} & 1+(n-1)\alpha_{2}^2 & ... & 1-\alpha_{2}\alpha_{n} \\
\vdots & \vdots & \ddots & \vdots\\
1-\alpha... | Let us assume that $\alpha_i\neq 0$ for $i\in\{1,\ldots,n\}.$
Let $X$ be the original matrix and
$$
A = n\cdot \begin{pmatrix}
\alpha_1^2 & & & \\
& \alpha_2^2 & & \\
& & \ddots & \\
& & & \alpha_n^2
\end{pmatrix}
$$
and
$$
U = \begin{pmatrix}
1 & \alpha_1 \\
1 & \alpha_2 \\
\vdots & \vdots \\
1 & \alpha_n
\end{pmatri... | {
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$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
| Note,
$$\frac17(x^2-y^2)=\frac1{37}( x^2+xy+y^2)$$
or
$$44y^2+7xy-30x^2=0$$
which leads to $x=\frac43y$ and $x=-\frac{11}{10}y$. Plug them into $x^2-y^2=7$ to obtain the solutions
$$(4,3),\>(-4,-3), \>(-\frac{11}{\sqrt3},\frac{10}{\sqrt3}), \>(\frac{11}{\sqrt3},-\frac{10}{\sqrt3})$$
| {
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What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$?
What is the angle between the asymptotes of this hyperbola?
$$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$
I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?
| Rearranging lab's answer ...
Omit the lower-degree terms of $5x^2-2\sqrt 7 xy-y^2-2x+1=0$ to get $5x^2-2\sqrt 7 xy-y^2=0$. This is the equation of two lines which are parallel to your asymptotes. So find the angle between these two lines.
Factor:
$$
5x^2-2\sqrt 7 xy-y^2 =
-\frac{\left( 2\,y\sqrt {3}-\sqrt {7}y+5\... | {
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$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it r... | If you want more than the limit itself
$$y_n= \left(1+ \frac{3}{n^2+n^4}\right)^n\implies \log(y_n)=n \log\left(1+ \frac{3}{n^2+n^4}\right)$$
Using Taylor expansion
$$ \log\left(1+ \frac{3}{n^2+n^4}\right)=\frac{3}{n^4}-\frac{3}{n^6}+O\left(\frac{1}{n^8}\right)$$
$$\log(y_n)=\frac{3}{n^3}-\frac{3}{n^5}+O\left(\frac{1}{... | {
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Number of ordered pairs of $A,B$ in Probability
Let $|X|$ denote the number of elements in a set $X,$
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur.
If $A$ and $B$ are independent events associated with
$S,$
Then the number of ordered pairs $(A,B)$
such that $1 \leq |... | The equation
$$z=\frac{xy}{6}$$
shows that the events $A$ and $B$ can be independent only if
*
*$A=S$, or
*$|A|=3,|B|=2$
*$|A|=4,|B|=3$
In the 2nd case $A$ and $B$ have 1 common element; in the 3rd case $A$ and $B$ have 2 common elements.
In the 1st case, we have $2^6-2$ solutions ($B$ can be anything except $S$ a... | {
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Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$. Problem: Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.
My efforts: $(y-a)^2(x^2-a^2)=x^4+a^4\implies (y-a)^2=\dfrac{x^4+a^4}{x^2-a^2}\implies x=\pm a$ are vertical asymptotes. What are the remaining asymptotes?
| Rewrite the equation as
$$y=a \pm \sqrt{\frac{x^4+a^4}{x^2-a^2}}$$
The vertical asymptotes $x=\pm a $ are directly identified. Since
$$\lim_{x^2\to\infty} y(x) = a\pm x$$
$y=a\pm x $ are the slant asymptotes.
| {
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Minimal polynomial with $f(\alpha)=0$ I'm learning this stuff for the first time so please bear with me.
I'm given $\alpha=\sqrt{3}+\sqrt{2}i$ and asked to find the minimal polynomial in $\mathbb{Q}[x]$ which has $\alpha$ as a root.
I'm ok with this part, quite sure I can find such a polynomial like so:
$\alpha = \sqrt... | It suffices to prove that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$. Writing them in the basis $1,\sqrt{3},\sqrt{2}i,\sqrt{6}i$ of $\mathbb Q[\sqrt{3},\sqrt{2}i]$ we get:
$$
\begin{pmatrix}
1 \\ \alpha \\ \alpha^2 \\ \alpha^3
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\... | {
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"answer_count": 2,
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Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}... | $$P_n=\prod_{k=1}^n\cos(kx)\implies \log(P_n)=\sum_{k=1}^n\log(\cos(kx))$$ Now, by Taylor expansion
$$\log(\cos(kx))=-\frac{k^2}{2}x^2-\frac{k^4 }{12}x^4+O\left(x^6\right)$$
$$\log(P_n)=-\frac{x^2}{2}\sum_{k=1}^n k^2-\frac{x^4 }{12}\sum_{k=1}^n k^4+\cdots$$ $$\log(P_n)=-\frac{n (n+1) (2 n+1)}{12} x^2-\frac{n (n+1) (2 ... | {
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"answer_count": 5,
"answer_id": 4
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Normalising Beal's conjecture Beal's conjecture
Can it be shown that
If
$$ \sum_{q=0}^{u}(n+qd)^{m_{q}} =a^b $$
,where $n,u,d,a,m_q$ and $b$ are positive integers with $m_q,b> 3$, then $n,n+d,n+2d,...,n+ud$ and $a$ have a common prime factor.
Example
*
*$(n,u,d,a)=(98,2,98,98)$ and $(m_0,m_1,m_2,b)=(4,4,4,5)$
... | No. For example, $3^3 + 4^3 + 5^3 = 6^3$ and $\gcd(3,4,5,6) = 1$. You can find other examples from the identity
$$ 1^3 + 2^3 + \cdots + n^{3} = \left(\frac{n(n+1)}{2}\right)^{2}. $$
Since there are infinitely many triangular numbers that are also squares, there are infinitely many integers $n$ for which the right hand ... | {
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$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have ... | HINT
From the trigonometric point of view, one can substitute $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get
\begin{align*}
\begin{cases}
r^{2}\cos(2\theta) = 5\\\\
r^{2}\sin(2\theta) = 12
\end{cases} \Longrightarrow \frac{25}{r^{4}} + \frac{144}{r^{4}} = 1 \Longrightarrow r^{4} = 169 \Longrightarrow r... | {
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How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{... | I will show that
$$\int_0^1 \frac{1 - x}{1 + x} \frac{x^n}{\sqrt{x^4 - 2x^2 + 1}} \, dx = -\frac{1}{2} + n (-1)^{n + 1} \left (\ln 2 + \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right ), n \geqslant 1.$$
Note here we interpret the empty sum as being equal to zero (the empty sum is the case when $n = 1$ in the finite sum) a... | {
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Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)... | Your approach is sound and the arguments besides the last case $|a|=1$ are clear. Here is a slightly more stringent variant, which does not bring anything new, besides maybe the modified argument for the case $|a|=1$.
We are looking for all polynomials $P\in\mathbb{R}[x]$ such that
\begin{align*}
P(x)P(x-3)=P(... | {
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How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$ Prove or disprove:
if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfra... | Note that
\begin{align*}
\int_{-\pi}^\pi f(x) \sin x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) + f(\pi - x) - f(-x)) \sin x \,dx \\
&= \int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx
\end{align*}
and similarly
\begin{align*}
\int_{-\pi}^\pi f(x) \cos x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) - f(\pi - x) + f(-x)) \cos x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Proof that $n^3 \leq 2^n$ for $n \geq 10$ by induction I'm trying to prove this statement and it got a bit clumsy but at least it seems to make sense. I'd like to know if this proof is valid. Here's what I managed to do:
First the base case:
$n = 10$
$$n^3 \leq 2^n$$
$$1000 \leq 1024$$
I assumed that it's true for $k$:... | Your solution looks good, but you might want to add more details as to why
$$(k+1)^3\leq k^3+k^3$$
This can be seen explicitly as
$$(k+1)^3=1 + 3 k + 3 k^2 + k^3=k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)$$
Since $k\geq 4$ we have
$$k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)\leq k^3+k^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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How to do partial fractions with a denominator to the power of a variable? How can I take the following sum and simplify it with partial fractions? $$\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}$$
I know the denominator can be rewritten as $(2^k)(2)$, but how do I deal with the $2^k$ when doing partial fractions?
Usually,... | TO MAKE A SENCE $$a=\sum^{\infty}_{k=1}\frac{k-1}{2^{k+1}}=\frac{0}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...$$multiply by $\frac 12$
$$\frac12a=\frac{0}{4}+\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...$$ now notice to $a-\frac 12 a$
$$a-\frac 12 a=\frac{0}{2}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Why do I get a different power series? My question revolves around the power series of the following function:
$$ f(x) = \frac{1}{1+x}$$
Now, it is almost immediate to have the power series of this function by substituting $-x$ into the geometric series to get:
$$\frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (... | Note that the center of the $$\frac{1}{1+x} = \frac{1}{2}\sum_{k=0}^\infty \left(-\frac{1}{2}(x-1)\right)^k = \sum_{k=0}^\infty (-1)^k\frac{1}{2^{k+1}}(x-1)^k$$
is the point $x=1$ while the center of $$ \frac{1}{1 + x} = \frac{1}{1-(-x)} = \sum_{k=0}^\infty (-x)^k = \sum_{k=0}^\infty (-1)^kx^k = 1 - x + x^2 - x^3 + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | $$(-1)^3=(x^4+x^2)^3=(x^3)^4+(x^3)^2+3(x^3)^2(-1)$$
Set $x^3=y$
$$y^4-2y^2+1=0$$ whose roots are $\alpha^3$ etc.
$$(y^4-2y^2+1)^2=0$$
Set $y^2=z$
$$z^4-4z^3+\cdots+1=0$$ whose roots are $(\alpha^3)^2$ etc.
Now apply Vieta's formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/... | $x^{2(n-1)} \not= x^2 x^{n-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to prove that a statement is false using principle of mathematical induction? I came across this question while solving some problems based on principle of mathematical induction ,
$P(n) : 1+2+3+....+n < \frac{(n+2)^2}{8}, n\in\mathbb{N}$, is true for
$(A) \, n\geq1\\
(B) \, n\geq2\\
(C) \text{ all } n\\
(D) \tex... | Just prove by induction that $\sum_{j=1}^nj\ge\frac18(n+2)^2$ for all $n\ge2$, using $k+1\ge(2k+5)/8$ (which is equivalent to $6k\ge3$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Cost Revenue Profit Question involving First Order Conditions A Cobb-Douglas production function is given by:
$Q(x,y) = x^{1/2}y^{1/4}$
where x and y are the input variables. The price of each product is p, and the cost of production is $C(x, y) = ax + by$.
I need to write down the profit, Π and find the stationary poi... | As Henry has already commented set the derivatives equal to zero
$$\frac{1}{2}px^{-1/2}y^{1/4} - a=0\Rightarrow \frac{1}{2}px^{-1/2}y^{1/4} =a \quad (1)$$
$$\frac{1}{4}px^{1/2}y^{-3/4} - b=0\Rightarrow \frac{1}{4}px^{1/2}y^{-3/4} = b \quad (2)$$
Divide the first equation by the second equation. The laws of exponents ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find minimum of $\frac{n}{S(n)}$ For every Natural Number like $n$ consider:$\frac{n}{S(n)}$ so that $S(n)$ is sum
of the digits of the number $n$ in base-10. find minimum of $\frac{n}{S(n)}$ when:
a)$9<n<100$
b) $99<n<1000$
c)$999<n<10000$
d)$9999<n<100000$
for $9<n<100$ I tried:
$n=10a+b$ and $Min(\frac{10a+b}{a+b})... | For a mathematical proof of (b)
$$\frac{10a+b}{a+b}=2+\frac{8a-b}{a+b}$$
The only way this can be less than $2$ is if $a=1,b=9$. So the minimum is $2-\frac{1}{10}=1.9$.
Part (d)
$$\frac{10^4a+10^3b+10^2c+10d+e}{a+b+c+d+e}-100=\frac{9900a+900b-90d-99e}{a+b+c+d+e}$$
The numerator of the RHS is clearly positive and so the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3499173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
ways in which $A,B$ refuse to be the member of same team
Consider a class of $5$ Girls and $7$ boys . The number of different teams consisting of $2$ girls and $3$ boys that can be formed from this class , If there are two specific boys $A$ and $B,$ who refuse to be the member of same team, is
what i try
Method $(1... | As John Omielan has pointed out, your second answer is incorrect since you omitted those cases in which exactly one of the boys $A$ or $B$ is a member of the team. Let's correct your count.
$A$ is a member, but $B$ is not: Since $A$ is on the team and $B$ is not, we must choose two of the five girls and two of the fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do we get $(b-a)^2/12$ from factorisation? I am in the final steps of calculating the variance of the uniform distribution for $(a,b)$. I'd like to see the steps involved in getting $$\frac{1}{12}(b-a)^2$$ from $$\frac{4(b^3-a^3)-3(b-a)(a+b)^2}{12(b-a)}$$ which is the result after integrating the variable $X^2$ bet... | $4(b^3-a^3)-3(b-a)(b+a)^2=4b^3-4a^3-3(b^2-a^2)(b+a)$
[using $(b-a)(b+a)=b^2-a^2$]
$=4b^3-4a^3-3b^3-3b^2a+3a^2b+3a^3$
$=b^3-3b^2a+3ba^2-a^3$
$=(b-a)^3\quad(*)$.
Dividing by $12(b-a)$ gives $\frac{1}{12}(b-a)^2$
So how did we do $(*)$? Well I did it because I recognize a few well-known factorisations like:
$(a+b)^2=a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.