Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
point on a line closest to another point Question:
Find the point on the line $2x+y+3=0$ that is closest to the point $(2,-6)$
This question really made me think for a while as to how to approach it.
I calculated distance as $\frac{1}{\sqrt{14}}$ using $D = \frac{ax_0+by_0+C}{\sqrt{a^2+b^2+c^2}}$
So would that mean t... | The equation for the line perpendicular to $y = -2x - 3$ passing through $(2, -6)$ is $y = 0.5x - 7$.
($0.5$ is the negative reciprocal of $-2$ and the $y$ intercept is $-6$ minus $2\cdot 0.5$)
Subtracting one from the other, their intersection is where $0 = 2.5x - 4$, which is at $x = 1.6$ and therefore $y = -6.2$.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\c... | Avoid squaring whenever possible as it immediately introduces
extraneous root(s).
$$\sin x-\sqrt3\cos x=1$$
Method$\#1:$
Use Prosthaphaeresis Formulas
Method$\#2:$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
Method$\#3:$
Use Double Angle formula,
$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Evaluate $\lim\limits_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$ Problem
Evaluate
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$$
Attempt
First, we may obtain
$$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right]=\lim_{x ... | The key here is obtaining the expansions for $x^{\sin x} $ and $\sin^xx$ but it is simpler to deal with their logarithms. Consider
\begin{align}
f(x) &=\sin x \log x-x\log\sin x\notag\\
&=(x\log x)\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots \right)-x\log x-x\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots\right)\notag\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Prove $\tan\frac\pi{16}+2\tan\frac\pi8+4=\cot\frac{\pi}{16}$
Prove that $\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$
My Attempt
\begin{align}
&\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\dfrac{1}{\cot\dfrac{\pi}{16}}+\dfrac{2}{\cot\dfrac{\pi}{8}}+4\\
&=\dfrac{1}{\cot\dfrac{\pi}{16}}+2\dfrac{2\cot\df... | Given $$\tan\dfrac{\pi}{16}+2\tan\dfrac{\pi}{8}+4=\cot\dfrac{\pi}{16}$$
Now $$\cot \theta-\tan\theta=\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}=\dfrac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{\cos2\theta}{\dfrac12\sin2\theta}=2\cot2\theta$$
Since $\cot\theta-\tan\theta=2\cot2\theta\ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\;\lim\limits_{x\to 4}\left (\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right) $ without L'Hôpital? $$
\lim_{x\to 4}\left(\frac{\frac{\pi}{6} - \arcsin\left(\frac{\sqrt{x}}{4}\right)}{\sqrt[3]{2x-7}-1}\right)
$$
Hello! I need to solve this limit. I had solved it with the ... | HINT
We have that by definition of derivative by $f(x)=\arcsin(\frac{\sqrt{x}}{4})$ and $g(x)=\sqrt[3]{2x-7}$ we have
$$\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{\sqrt[3]{2x-7}-1}=\lim_{x\to 4} \frac{\frac{\pi}{6} - \arcsin(\frac{\sqrt{x}}{4})}{x-4}\frac{x-4}{\sqrt[3]{2x-7}-1}=-\frac{f'(4)}{g'(4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$
Evaluate $$\lim_{x \to \infty}x^4\left(\arctan\frac{2x^2+5}{x^2+1}-\arctan\frac{2x^2+7}{x^2+2}\right)$$
My Solution
Denote
$$f(t):=\arctan t.$$
By Lagrange's Mean Value Theorem，we have
$$f\left(\frac{2x^2+5}{x^2+1}\r... | Recall that by Arctangent addition formula
$$\arctan\dfrac{2x^2+5}{x^2+1}-\arctan\dfrac{2x^2+7}{x^2+2}=\arctan \left(\frac{\dfrac{2x^2+5}{x^2+1}-\dfrac{2x^2+7}{x^2+2}}{1+\dfrac{2x^2+5}{x^2+1}\dfrac{2x^2+7}{x^2+2}}\right)=$$
$$=\arctan \left(\frac{3}{5x^4+27x^2+37}\right)$$
therefore
$$x^4\left(\arctan\dfrac{2x^2+5}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998279",
"timestamp": "2023-03-29T00:00:00",
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Prove that $C_{3 \over 2}^n$ is bounded given $C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}$
Let:
$$
\begin{cases}
C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}\\
C_{a}^0 = 1
\end{cases}
$$
Prove $C_{3 \over 2}^n$ is bounded.
I've started with finding a reduced formula:
$$
C_{3\over 2}^n = \frac{{3\over 2}\left({... | We should start from using the known expression of Gamma function:
$\Gamma(z)= \frac{\Gamma(z+n+1)}{z(z+1)....(z+n)}$ following that $z(z+1)....(z+n-1) =\frac{\Gamma(z+n+1)}{\Gamma(z) (n+z)}$
On the other hand:
$C_{a}^n = \frac{a(a-1)(a-2)\dots(a-n+1)}{n!}=(-1)^n\frac{(-a)(-a+1)(-a+2)\dots(-a+n-1)}{n!}$
If $z=-a=-\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Characteristic polynomial modulo 12
Consider the vector space $V =\left\{a_0+a_1x+\cdots+a_{11}x^{11},\;a_i\in\mathbb{R}\right\}$. Define a linear operator $A$ on $V$ by $A(x^i) = x^{i+4}$ where $i + 4$ is taken modulo $12$.
Find $(a)$ the minimal polynomial of $A$ and $(b)$ the characteristic polynomial of $A$.
My t... | It usually helps to find some nontrivial relation that the given operator satisfies. Clearly
$$A^3(x^i)=x^{i+12}=x^i,$$
for all $i$, so $A$ is a zero of $X^3-I$. This factors as
$$X^3-I=(X-I)(X^2+X+I),$$
which has no repeated factors, so this is the minimal polynomial of $A$. The characteristic polynomial has the same ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum... | $$\sum_{k=1}^{n}k\binom{n}{k}^{2}=n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k}=n\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-k-1}=n\binom{2n-1}{n-1}$$
Applying Vandermonde's identity in third equality.
| {
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Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only Q:Determine whether the solution space of the system $Ax=0$ is a line through the origin, a plane through the origin, or the origin only. If it is a plane, find an equation for it; if i... | Assuming that the reduction has been done correctly, (i) has a two-dimensional image space, so its null space is one-dimensional, hence a line.
To find a parametrisation, note that $y$ can be chosen freely, $z=0$ is forced by the second equation in reduced form, and the first one forces $x=2y$, so
$$\begin{pmatrix}2t\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving trigonometric identities I’ve had a bit of difficulty of this question:
(1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA
I tried to do:
(SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA
But then I’m kind of lost. Any help will be appreciated! Additionally, I am not allowed to move one side to anoth... | $$
\frac{1+\sin A +\cos A}{1-\sin A + \cos A} = \frac{1+\sin A}{\cos A}\\
\cos A + \sin A \cos A + \cos^2 A = (1- \sin A + \cos A) (1+\sin A)\\
\cos A + \sin A \cos A + \cos^2 A = 1- \sin A + \cos A + \sin A - \sin^2 A + \sin A \cos A\\
\cos A + \sin A \cos A + \cos^2 A + \sin^2 A = 1- \sin A + \cos A + \sin A + \sin A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003758",
"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikip... | Let $P_n:=\prod\limits_{k=1}^n\,\dfrac{2k-1}{2k}$ for each integer $n\geq 1$. We shall prove that $$\frac{2}{3\sqrt{2n}}<P_n<\frac{1}{\sqrt{2n}}\tag{*}$$
for every positive integer $n$, as suggested by Kemono Chen (I think there should be a factor $\dfrac{1}{\sqrt{2}}$ there so that (*) implies the OP's inequality for... | {
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"url": "https://math.stackexchange.com/questions/3004689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What am I doing wrong finding the derivative of $\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$? $$y=\frac{3-x}{2}\sqrt{1-2x-x^2}+2\arcsin{\frac{1+x}{\sqrt{2}}}$$
For convenience, let
$$A=\frac{3-x}{2}\sqrt{1-2x-x^2},$$
$$B=2\arcsin{\frac{1+x}{\sqrt{2}}}.$$
$$y'=A'+B'$$
$$A'=(-\frac{1}{2})(\sqrt{1-2x-x^2}... | In the first line of your derivation of $A'$: after the first summand, your second summand should be $$\color{blue}{(-2 -2x)}\left(\frac{1}{2}\cdot \frac{1}{{\sqrt{1-2x-x^2}}}\right)\left(\frac{3-x}{2}\right)$$
So in fact, $$A'=\left(-\frac{1}{2}\right)\left(\sqrt{1-2x-x^2}\right)+ \color{blue}{(-2 -2x)}\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009739",
"timestamp": "2023-03-29T00:00:00",
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How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{... | Compute
$$\left(\frac{3+\sqrt{13}}2\right)^3=18+5\sqrt{13}.$$
Therefore
$$\sqrt[3]{18+5\sqrt{13}}=\frac{3+\sqrt{13}}2.$$
Similarly
$$\sqrt[3]{\pm 18\pm 5\sqrt{13}}=\frac{\pm 3\pm\sqrt{13}}2$$
where the signs on both sides correspond. Then
$$\sqrt[3]{-18+ 5\sqrt{13}}-\sqrt[3]{-18-5\sqrt{13}}
=\frac{-3+\sqrt{13}}2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?
| HINT
By tangent half-angle substitution with $t=\tan \frac x 2$ we have
$$3\sin x +\cos x =2 \iff 3\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}=2 \iff 3t^2-6t+1=0$$
then
$$\dfrac{3\sin x}{4\sin x+3\cos x}=\frac{6t}{-3t^2+8t+3}=\frac{6t}{(-3t^2-1)+8t+4}=\frac{6t}{2t+4}=\frac{3t}{t+2}$$
As an alternative since $\cos x=\pm \frac1... | {
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"url": "https://math.stackexchange.com/questions/3020530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The commutative law of multiplication. The commutative law of multiplication states that:
If '$a$' and '$b$' are any whole numbers,
then
'$a · b = b · a$'.
I've been looking for why multiplication has this property and how is saying that 3 times 4 the same thing as saying 4 times 3. I've found two interpretatio... | a * b = $\sum_{n=1}^b a$ = (a / b) (b / a) $\sum_{n=1}^b a$ = (a / b) $\sum_{n=1}^b b$ = $\sum_{n=1}^a b$ = b * a
For example, 3 * 4 = 3 + 3 + 3 + 3 = (3 / 4) (4 / 3) (3 + 3 + 3 + 3) = (3 / 4) (4 + 4 + 4 + 4) = 4 + 4 + 4 = 4 * 3
In words, 3 * 4 is 3 added 4 times (3 + 3 + 3 + 3). If you take 4/3 of that, all the 4s bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021523",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domai... | For $x \in (0,\pi/4)$,
the integral $$\frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t$$ can also be written as $$\frac12 \int_0^1 \frac{1+1/t^2}{(t-1/t)^2+1} \text{d}t$$
Substituting $u=t-\frac1t$, the integral becomes $$\frac12 \int_{-\infty}^0 \frac{1}{u^2+1} \text{d}u$$ which is equal to $$\frac12[\arctan(u)]_{-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $ \frac{1}{5}+ \frac{1}{9}+ \frac{1}{13}+\ldots+ \frac{ 1}{2005}< \frac{ 7}{4}$. Prove that
$ \dfrac{1}{5}+ \dfrac{1}{9}+ \dfrac{1}{13}+\ldots+ \dfrac{ 1}{2005}< \dfrac{ 7}{4}$.
Solution
$\sum_{k=1}^{501}\frac 1{4k+1}$ $=\frac 15+\sum_{k=2}^{501}\frac 1{4k+1}$ $<\frac 15+\int_1^{501}\frac{dx}{4x+1}$ $=\frac ... | It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=\frac{1}{4x+1}$ is convex over $\mathbb{R}^+$,
$$ \int_{1/2}^{501+1/2}f(x)\,dx \geq f(1)+f(2)+\ldots+f(501) $$
hence
$$ \sum_{k=1}^{501}\frac{1}{4k+1}\leq \frac{\log(669)}{4} $$
and it is enough to show that $e^7>669.$ On the other hand $e>2+\frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$$
My try:
The limit can be written as follows:
$$\lim_{n\to\infty}\... | Thanks to user xbh for the hint:
Using the Stolz-Cesàro theorem, we have:
$a_n=\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}$
$b_n=n^2$
Two monotone and increasing sequences.
Apply the theorem to get:
$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=\frac{(n+2)(n+2)^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1... | Use the properties of determinant to show that the given relationship does not hold.
$$\det ((A^2)^{-1})=-\frac1{64}=\det (A^{-1})\cdot(-4)$$
from the given relationship and properties of determinant. So $\det (A)=64\cdot 4$ and this cannot be true.
| {
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"url": "https://math.stackexchange.com/questions/3033706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$
Find the simplified form of $\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]$, where $x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]$
My reference gives the ... | Your book is wrong. $x\in\Big[-3\pi/4, 3\pi/4\Big]$, which is an interval of length $3\pi/2$. Whatever be the value of $\alpha, \alpha-x$ will belong to an interval of length $3\pi/2$, which means $\alpha-x$ is not confined to $[0, \pi].$
So the answer is $\begin{cases}2\pi-\alpha+x,&x\in[-3\pi/4,\alpha-\pi)\\\alpha-x,... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $n$ be the product of all positive factors of $10^4$. Find $log_{10} n$ I can't see any relationship between the products of the positive factors of powers of 10.
The answer given here is 50, by the way, but I simply don't know where to start.
| As @NL1992 pointed out, the key is write $10^4 = 2^4 \times 5^4$.
Consider the factors which have no contribution from the "2". These are $5^0, 5^1, 5^2, 5^3$, and $5^4$. Their product is $5^{10}$.
Now, consider the factors which exactly one "2". These are $2 \times 5^0$, $2 \times 5^1$, $2 \times 5^2$, $2 \times 5^3$... | {
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"url": "https://math.stackexchange.com/questions/3036038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the followin... | Recall that
$$3^2+4^2=5^2 \implies (3\cdot 7)^2+(4\cdot 7)^2=(5\cdot 7)^2$$
and note that
$$(21, 220, 221)$$
is a primitive triple.
Your criterion doesn't works because the remainder of squares $\pmod 4$ are $0,1$ therefore we can't comclude that
$$z^2-y^2\equiv 0 \pmod 4$$
What we need to solve is
$$21^2=441=3^2\cdot... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $f(x) = x^2\sin\left(\frac{1}{x}\right)$ is Lipschitz (no use of derivative)
Prove that $f:\mathbb{R}\to \mathbb{R}$ such that
$$ f(x) = \left\{
\begin{array}{c l}
x^2\, \sin\left(\frac{1}{x}\right) & ,\quad x\neq 0\\
0 & ,\quad x=0
\end{array} \right.$$
is Lipschitz (without use of derivatives)... | Writing
$$
f(x) - f(y) = (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )
$$
is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| \le |x|$:
$$
\left| (x^2- y^2) \sin \frac 1x \right| \le |x-y| |x+y|| \frac 1x| \le 2|x-y|
$$
and
$$
\left|y^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum\frac1{2-A_k}$ for $(A_k)$ the $n$th roots of unity
If $1,A_1,A_2,A_3....A_{n-1}$ are the $n^{th}$ roots of unity then prove that
$$\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}$$
What I did: I tried to use some of the following formulas:
$$1+ A_1 ... | Two ingredients for this proof:
*
*The numbers $A_k$ for $1\leqslant k<n$ and $A_n=1$ are the roots of the polynomial $X^n-1$.
*If $Y=\frac1{2-X}$ then $X=\frac{2Y-1}Y$.
Thus, the numbers $B_k=\frac1{2-A_k}$ for $1\leqslant k<n$ and $B_n=\frac1{2-A_n}=1$ are the solutions of the equation $$\left(\frac{2Y-1}Y\rig... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Isomorphism between two versions of $GF(2^3)$ I have $GF(2^3)$ generated by $\Pi_1(\alpha)=x^3+x+1$ and $GF(2^3)$ generated by $\Pi_2(\alpha)=x^3+x^2+1$.
$\bullet$ $\Pi_1(\alpha)=x^3+x+1$ $000=0, 100=1,010=\alpha,001=\alpha^2,110=\alpha^3, 011=\alpha^4,111=\alpha^5,101=\alpha^6$
$\bullet$ $\Pi_2(\alpha)=x^3+x^2+1$ $000... | $\Pi_1(\lambda^3) = (\lambda^3)^3 + \lambda^3 + 1 = \lambda^9 + \lambda^3 + 1$
But you know that $\lambda^7=1$. So $\lambda^9=\lambda^2$ and
$\Pi_1(\lambda^3) = \lambda^3 + \lambda^2 + 1 = \Pi_2(\lambda) = 0$
The complete isomorphism is as follows:
$0 \leftrightarrow 0 \\ 1 \leftrightarrow 1 \\ \alpha \leftrightarrow \... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:
Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.
The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod ... | $u^3=(r+s)(r-s)$ and $\gcd(r+s,r-s)=1$, so $r+s$ and $r-s$ are odd, coprime perfect cubes.
So let $r+s=a^3$, $r-s=b^3$. Then
$$r=\frac{a^3+b^3}2$$
$$s=\frac{a^3-b^3}2$$
where $a$ and $b$ are odd and coprime.
Conversely, if $a$ and $b$ are odd and coprime, let $r=(a^3+b^3)/2$ and $s=(a^3-b^3)/2$, which are coprime and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12... | $$ ( n^2 + n + 2 )^2 = n^4 + 2n^3 + 5n^2 + 4n + 4 $$
$$ ( n^2 + n + 3 )^2 = n^4 + 2n^3 + 7n^2 + 6n + 9$$
The second one is larger than yours, meaning yours cannot be square, when
$$ 7n^2 + 6n+9 - 6n^2 - 12 n - 25 > 0 \; , \; $$
$$ n^2 - 6n - 16 > 0 \; , \; $$
$$ (n-8)(n+2) > 0 \; . $$
You need check only $0 \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, ... | $$\frac{a+b-c}{c}=\frac{a-b+c}{b}$$
$$ab+b^2-bc=ac-bc+c^2\implies a(b-c)=(b+c)(c-b)$$
$$\implies b=c \quad OR\quad a+b+c=0$$
If we take $b=c$, $$\frac{a}{c}=\frac{-a+2c}{a}\quad using (1)and (3).$$
$$\implies ({\frac{a}{c}})^2=1\quad OR\quad -2$$
Obviosuly $-2$ isn't possible, $a=c$ will give $x=1$ and $a=-c$ will giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Diagonal entries are zero, others are $1$. Find the determinant. $\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=?$
Attempt:
First I tried to use linearity property of the determinants such that $$\det\binom{ v+ku }{ w
}=\det\binom{v }{ ... | For $n = 1$, we have
$$ \det \left[ \begin{matrix} 0 \end{matrix} \right] = 0.$$
For $n=2$, we have
$$ \det \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] = -1. $$
For $n = 3$, we have
$$ \det \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] = 0 \det \left[ \begin{matrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inequality with $(x+y)(y+z)(z+w)(w+x)=1$ Let $x,y,z,w>0$ and such that
$$(x+y)(y+z)(z+w)(w+x)=1.$$
Show that
$$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le 2.$$
I'm trying to use Holder's inequality
$$(\sqrt[3]{xyz}+\sqrt[3]{yzw})^3
\le (x+y)(y+z)(z+w)$$
$$(\sqrt[3]{zwx}+\sqrt[3]{wxy})^3\le (z+w)(w+x)(x+y... | Since $$\prod_{cyc}(x+y)-\sum_{cyc}x\sum_{cyc}xyz=(xz-yw)^2\geq0,$$ by Holder we obtain:
$$2=2\left(\prod_{cyc}(x+y)\right)^{\frac{1}{4}}\geq2\left(\sum_{cyc}x\sum_{cyc}xyz\right)^{\frac{1}{4}}=$$
$$=\sqrt[4]{\left(4^2\sum_{cyc}xyz\right)\sum_{cyc}x}\geq\sqrt[4]{\left(\sum_{cyc}\sqrt[3]{xyz}\right)^3\sum_{cyc}x}.$$
Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to c... | Starting at $ x - x^2 = 1/4 $, subtract $ 1/4 $ to each side:
$$\mathrm{(1)} \qquad x - x^2 - 1/4 = 0 $$
Arrange the left hand side of the equation $ (1) $ in descending degree of $ x $:
$$\mathrm{(2)} \qquad -x^2 + x - 1/4 = 0 $$
I would recommend having the coefficient of the term with the highest degree of $ x $ not... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 8
} |
Integrate $\int_{0}^{\infty}e^{-pt}\sin\left(\sqrt{t}\right)\mathrm dt$ I need the following Laplace transform to solve the Differential Equation
$$\int_{0}^{\infty}e^{-pt}\sin\sqrt{t}\, dt, \quad \text{where} \ \ \ p>0$$
I tried Integration by parts after substituting $t=x^2$, but didn't work.
\begin{align}
\int_{0}... | Hint:
Use the expansion of $\sin$
$$\sin\sqrt{t}=\sum_{n=0}^{\infty}(-1)^n\dfrac{t^{n+1/2}}{\Gamma(2n+2)}$$
Edit:
$${\cal L}(\sin\sqrt{t})=\sum_{n=0}^{\infty}(-1)^n\dfrac{\Gamma(n+3/2)}{\Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${\cal L}(\sin\sqrt{t})=\dfrac{1}{p^{3/2}}\sum_{n=0}^{\infty}(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$ I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$
Although I could solve it, I am not quite comfortable with the way I did it.
But first I will show the ... | To show that $$(-1)^{n-1} \int_{0}^{\infty}\frac{ \mathrm dx}{ (\pi^{2}+\ln^{2} x)(1+x)^{n}} $$ is an integral representation of the Gregory coefficients, some textbooks integrate the function $\frac{1}{(\ln z - i\pi)(1+z)^{n}} $ around a keyhole contour. We can do something similar here.
Let's integrate the function ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 2
} |
How do I find out that the following two matrices are similar? How do I find out that the following two matrices are similar?
$N =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$
and $M=
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
... | The two matrices are made of Jordan blocks; in $2\times2$ block format, they are
$$
N=\begin{bmatrix} J & 0 \\ 0 & 0 \end{bmatrix}
\qquad
M=\begin{bmatrix} 0 & 0 \\ 0 & J \end{bmatrix}
$$
You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.
My attempt:
$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\r... | For a constrained problem
\begin{align}
\max{} & f(x,y) \\
& g(x,y)\le 0
\end{align}
you must write the Lagrangian:
$$
L(x,y,\lambda)=f(x,y)+\lambda g(x,y)
$$
Then you must find stationary points of your Lagrangian (attention this is only necessary conditions, see wiki)
\begin{align}
\partial_x L &= 0 = \partial_x f+ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Algebraic Closed Form for $\sum_{n=1}^{k}\left( n- 3 \lfloor \frac{n-1}{3} \rfloor\right)$ Let's look at the following sequence:
$a_n=\left\{1,2,3,1,2,3,1,2,3,1,2,3,...\right\}$
I'm trying to calculate:
$$\sum_{n=1}^{k} a_n$$
Attempts:
I have a Closed Form for this sequence.
$$a_n=n- 3 \bigg\lfloor \frac{n-1}{3} ... | Sorry can't comment on @5xum's answer. Building on the quadratic function we can derive:
$S_k=\frac{1}{2}((k-3\lfloor\frac{k}{3}\rfloor)^2+k+9\lfloor\frac{k}{3}\rfloor)$, which is pretty neat.
To derive this, note that from @5xum's answer, $S_k=2k-1+\frac{1}{2}(x^2 -3x+2)$ where $x=k-3\lfloor\frac{k}{3}\rfloor$. We ca... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Not getting the right answer with alternate completing the square method on $\int\frac{x^2}{\sqrt{3+4x-4x^2}^3}dx$ So I've looked up how to do this problem and when they complete the square it's using the $(\frac{-b}{a})^2$ method where, in the denominator, they factor out the $4$, and then make:
$$x^2-x=x^2-x+{1\over4... | Except some minor mistakes, your solution is correct but you led to wrong answer.
Mistake 1
In the first integral after substituting $u=\sin \theta$ you used $dx=du$ while $dx={1\over 2}du$.
Mistake 2
From $2$nd to $3$rd integral, you used $|\cos \theta|=\cos \theta$ while you should have dealt with $|\cos \theta|$ ti... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The exponential of a skew-symmetric matrix in any dimension. The Rodrigues rotation formula gives the exponential of a skew-symmetric matrix in three dimensions, and the exponential of a skew-symmetric matrix in two dimensions is given by Euler's formula. Is there a general formula (or set of formulas) for the exponent... | The spectral decomposition of any skew-symmetric matrix $A$ is given by $A=U Q U^\dagger$ where $U$ is unitary and
\begin{align*}
Q=\begin{bmatrix}
0 & \lambda_1 & \\
-\lambda_1 & 0 & \\
& & 0 & \lambda_2\\
& & -\lambda_2 &0\\
& & & & \ddots\\
& & & & & 0 & \lambda_r\\
& & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we... | From “harmonic mean $\le$ arithmetic mean” one gets
$$
\frac{2}{f(x, y)} \le \frac12 \left(\frac{4-x^2}{4}+ \frac{9-y^2}{9} \right)
= \frac 12 \left(2- \frac{x^2}{4} - \frac{y^2}{9} \right)
$$
and from “arithmetic mean $\ge$ geometric mean”
$$
\frac{x^2}{4} + \frac{y^2}{9}\ge 2 \frac{|xy|}{6} = \frac 13
$$
so that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Expressing $1-\exp{(\lambda_1p+\lambda_3q)}$ as $x+y$, where $x$ is in terms of $\lambda_1$ and $y$ is in terms of $\lambda_3$ I have this simple equation
$$c = 1 - \exp\left(\lambda_1 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right) + \lambda_3 R^2 \left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)\right)$$
I will like ... | Write $$c(\lambda_1,\lambda_3)=1 - e^{(\lambda_1 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}) + \lambda_3 R^2 (\frac{2\pi}{3}-\frac{\sqrt{3}}{2}))}$$
Note first that $c(0,0)=1-e^0=0$.
Suppose we had the desired expression $c(\lambda_1,\lambda_3)=x(\lambda_1)+y(\lambda_3)$. Then:
$$x(\lambda_1)+y(0)=c(\lambda_1,0)=1 - e^{\l... | {
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"source": "stackexchange",
"question_score": "2",
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Proof of $ x^2 + y = y^2 + x$ when $ x+ y =1$ and $x$ is larger than $y$ I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 ... | Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Counting, Probability and Binomial Coefficients If $$P_{2n+2}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^n-{2n \choose {n+1}}p^{n+1}q^{n+1}$$
$\mathbf {Inspiration:}$ A and... | We have for the first sum
$$S = \sum_{k=0}^n {2n+2\choose k+n+2} p^{k+n+2} q^{n-k}$$
and for the second one
$$T = \sum_{k=0}^{n-1} {2n\choose k+n+1} p^{k+n+1} q^{n-1-k}$$
and seek to show
$$S-T =
{2n\choose n} p^{n+2} q^n - {2n\choose n+1} p^{n+1} q^{n+1}$$
where $p+q=1.$ We see that with
$$Q_n = \sum_{k=0}^n {2n+2\cho... | {
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"url": "https://math.stackexchange.com/questions/3071705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving for $x$ in $y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$ I want to confirm my solution of $x$ from
$$y=N\cdot\left(\frac{10}{x}\right)^{-2.6}$$
My answer is:
$$x=\frac{N^{2.6}}{10\cdot y^{2.6}}$$
Is this right? How would you solve?
| First, notice that
$$
y = N \cdot \left( \dfrac{10}{x} \right)^{-2.6} = N \cdot \left( \dfrac{x}{10} \right)^{2.6}
$$
Next, divide both sides by $N$:
$$
\dfrac{y}{N} = \left( \dfrac{x}{10} \right)^{2.6}
$$
Now exponentiate with $\frac{1}{2.6}$ to get
$$
\left( \dfrac{y}{N} \right)^{\frac{1}{2.6}} = \dfrac{x}{10}
$$
Fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative? The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
| Does not exist.
Try $x\rightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$\frac{x^2+1}{x+1}=\frac{x^2-1+2}{x+1}=x-1+\frac{2}{x+1}=$$
$$=x+1+\frac{2}{x+1}-2\geq2\sqrt{(x+1)\cdot\frac{2}{x+1}}-2=2\sqrt2-2.$$
The equality occurs for $x+1=\frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine all $a$ and $b$ natural numbers such that $\frac {a^2+2b} {b^2-2a}$ and $\frac {b^2+2a} {a^2-2b}$ are whole numbers. I proceeded in the following way:
It is clear that $a \ne 0$ and $b \ne 0$.
Let $\frac {a^2+2b} {b^2-2a} = k, k \in \mathbb{Z} \tag 1$
and $\frac {b^2+2a} {a^2-2b} = m, m \in \mathbb{Z} \ta... | So we have $b\in \{a-2,a-1,a,a+1,a+2\}$
Case 1: $b= a-2$, then $$(a-2)^2-2a\mid a^2+2(a-2)$$
so $$a^2-6a+4\mid a^2+2a-4$$
and thus $$a^2-6a+4\mid (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4\leq 8a-8\implies a^2-14a+12\leq 0$$
so $$ (a-7)^2\leq 37\implies |... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $a^{n} - 1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3} + \cdots + a + 1)$ for all $n \ge 1$ using induction Problem 1.1.3 in Burton's Elementary Number Theory (6th ed.) is stated as follows:
Use the Second Principle of Finite Induction to establish that for all $n \ge 1$,
\begin{align} a^{n} - 1 = (a-1)(a^{n-1}+a^{n-2}+... | The only problem with your proof is that you failed to check and confirm that for the base case, when $n=1$, the given proposition holds. Add that to your proof, and you're good to go.
It may seem utterly obvious and not worth mentioning in a proof, but without proving the base case $n=1$, your proof is meaningless in... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrals depending on a parameter: $\int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx $ I'm trying to calculate this integral:
$$ \int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx $$
For $a > 0$.
This is what I did:
$$ I(a) = \int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x})dx \\\
I'(a) = \int_{0}^{\pi/2} \frac{2a\sin^2{x}... | you can also use the series of $\ln x$ to find it
$$\begin{aligned}
\int_{0}^{\pi/2} \ln(a^2\sin^2{x} + \cos^2{x}) \mathrm{d}x
& = \int_{0}^{\pi/2} \ln(1+(a^2-1)\sin^2{x}) \mathrm{d}x\\
& = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{\pi/2}(a^2-1)^n\sin^{2n}{x} \mathrm{d}x\\
& = \pi \sum_{n=1}^{\infty} \frac{(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Finding the tenth derivative of $f(x) = e^x\sin x$ at $x=0$ I came across this Question where I have to find $$f^{(10)}$$ for the following function at $x = 0$
$$f(x) = e^x\sin x$$
I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.
| Using power series: it is well-known that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ and $\sin(x)=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}$ for any real number $x$, so
$$ e^x\sin(x)=(1+x+\frac{x^2}{2!}+\dots+\frac{x^{10}}{10!}+\dots)(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\dots) $$
By expand... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 10,
"answer_id": 8
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Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}... | Avoiding hyperbolic functions, I believe you can do a trigonometric substitution: consider a right triangle with angle $\theta$ for which the opposite side is $x$, the adjacent side is $1$, and (thus) the hypotenuse is $\sqrt{x^2 + 1}$.
Then $x = \tan{\theta}$, so $dx = \sec^2{\theta}d\theta$; note that $\sec{\theta} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the coefficient of $x$ in the given polynomial We have polynomial
$$(x-2^0)(x-2^1)(x-2^2)···(x-2^n)$$
I know that the coefficient of $x$ equals to:
$$(-1)^{n-1}((2^0×2^1×···×2^{n-1})+(2^0×2^1×···×2^{n})+···+(2^1×2^2×···×2^{n}))$$
But it's hard to calculate this sum.
Any help is appreciated.
| The coefficient of $x$ in $(x-a_1)\cdots(x-a_n)$ is
$$
(-1)^n \sum_{k=1}^n \frac{a_1 \cdots a_n}{a_k}
$$
If $a_k=2^{k-1}$, then $a_1 \cdots a_n = 2^0 2^1 \cdots 2^{n-1} = 2^{0+1+\cdots +(n-1)}=2^{n(n-1)/2}$.
Thus, the coefficient is
$$
(-1)^n \sum_{k=1}^n \frac{2^{n(n-1)/2}}{2^{k-1}}
=
(-1)^n 2^{n(n-1)/2} \sum_{k=1}^n ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$ $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.
| $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
$$ = \lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-9-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \frac{x^{2}-9+\sqrt{x+6}-3}{x^{2}-9}$$
$$ = \lim_{x\to 3} \Big(\frac{x^{2}-9}{x^{2}-9}+\frac{\sqrt{x+6}-3}{x^{2}-9}\Big) $$
$$ = \lim_{x\to 3} \frac{x^{2}-9}{x^{2}-9} + \lim_{x\to 3} \frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trouble showing $a_n=\frac{\sin \frac{n\pi}{3n+1}}{\sqrt{n+3}}$ is monotonically decreasing I've tried reducing the expression to something obviously true, but I couldn't get anywhere. Also tried induction but I got stuck at the inductive step and failed to obtain anything useful. I've also tried to prove it by contrad... | The fraction $\frac{n\pi}{3n+1}$ increases strictly with $n$. Therefore, by Aristarchus's inequality,
$$
\frac{\sin\left(\frac{(n+1)\pi}{3n+4}\right)}{\sin\left(\frac{n\pi}{3n+1}\right)} < \frac{(n+1)(3n+1)}{n(3n+4)} = 1 + \frac1{n(3n+4)}.
$$
Write $t = \frac1{n(3n+4)} < 1$. For $n \geqslant 2$,
$$
\left(1 + \frac1{n(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically ... | Multiplying both sides by $\sqrt {49-x^2} + \sqrt {25-x^2}$, yields:
$$(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$
$$(49-x^2 - (25-x^2)=3(\sqrt {49-x^2} + \sqrt {25-x^2})$$
Thus,
$$\sqrt {49-x^2} + \sqrt {25-x^2}=8.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ 0=1 $ ? Where is the mistake? I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=\sqrt{0-0+\sqrt{0-0+....}}$$
$$0=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equ... | The problem arises because you didn't really defined the radical expression
$$\sqrt{n^2-n+\sqrt{n^2-n+\sqrt{n^2-n+\cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=\sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=\sqrt{a_k}=\sqrt[4]{a_{k-1}}=\cdots\sqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does the CRT formula yield a solution of a congruence system? I understand there is a method for solving simultaneous modular equations. For example;
$$x = 2 \mod{3}$$
$$x = 3 \mod{5}$$
$$x = 2 \mod{7}$$
We find numbers equal to the product of every given modulo except one of them - giving $5 \cdot 7$, $3 \cdot 7$ ... | This is a generalisation of the formula for the solutions of a system of two congruences modulo two coprime numbers $a$ and $b$?. This formula uses a Bézout's relation: $\;ua+vb=1$ and it is:
$$\begin{cases}
x\equiv \alpha\mod a,\\
x\equiv \beta\mod b,
\end{cases}
\quad\text{which is }\qquad x\equiv \beta ua+\alpha vb\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Evaluate $\int \frac{x^2}{x-1} \,dx$ Evaluate $\int \frac{x^2}{x-1} \,dx$
(A) $2x^2+x+\ln|x-1|+C$
(B) $\frac{x^2}{2}+x+\ln|x+1|+C$
(C) $\frac{x^2}{2}+x+\ln|x-1|+C$
(D) $x^2+x+\ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$\int \frac{(u+1)^2}{u}\,du\\
=\int u +2+\frac{1}{u}\,du\\
=\frac{u^2}{2}+2u+\ln|u|+... | After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:
$$
\begin{align}
\int \frac{x^2}{x-1} \,dx
&=\int \frac{x^2-1+1}{x-1} \,dx\\
&=\int \left(\frac{x^2-1}{x-1}+\frac{1}{x-1}\right) \,dx\\
&=\int \frac{(x-1)(x+1)}{x-1}\,dx+\int\frac{1}{x-1} \,dx\\
&=\int \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$. The following was a question in the final of the Flanders Mathematics Olympiad 2018:
In the triangle $\triangle ABC$, $|AB|^3 = |AC|^3 + |BC|^3$. Prove that $\angle ACB > 60^\circ$.
In this competition, points are assigne... | In the standard notation we need to prove that
$$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$\sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/3096866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convergence of $\sum_{n=1}^\infty\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)$ where $a_1=1$ and $a_n=2-\frac 1n$ for $n\geq 2$ Let $a_1=1\ $ and $\ a_n=2-\frac 1n$ for $n\geq 2$. Then $$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)$$
converges to?
I started by expanding the sum:
$\sum_{n=1}^\in... | Let $$S_N=\sum_{n=1}^N\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)=\frac{1}{a_1^2}-\frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_n\to 2$, we have
$$\lim_{N\to\infty}S_N=1-\frac{1}{2^2}=\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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GCSE Probability question relating to trees, help please. There are only red counters, yellow counters and blue counters in a bag.
Kevin takes at random a counter from the bag.
He puts the counter back in the bag.
Lethna takes at random a counter from the bag.
She puts the counter back in the bag.
The probability that ... | Let $R$, $Y$ and $B$ be the events in which a red, yellow and blue counter is drawn from the bag, respectively. We have:
$$P(R) \cdot (1 - P(R)) = \frac{1}{4} \iff P(R)^2 - P(R) + \frac{1}{4} = 0 \iff P(R) = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1}{2}$$
$$P(R) \cdot P(R) + P(Y) \cdot P(Y) = \frac{13}{36} \iff P(Y)^2 = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can you solve $\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$? $$\frac{dy}{dx}=(1-x)y^2+(2x-1)y-x$$
This is a form of riccati differential equation, which can be reduced to Bernoulli's equation if one particular solution is given. Here $y=1$ is a given solution.
Hence the general solution must be in the form $y=1+z$ for some $z(x)$... | Bernoulli's equation has form,
$$
\frac{dy}{dx}+p(x)y=q(x)y^n
$$
Now, consider this,
$$
\frac{dz}{dx}+z^2x=z^2+z
$$
This easily simplifies to,
$$
\frac{dz}{dx}-z=(1-x^2)z^2
$$
where $p(x)=-1$ and $q(x)=1-x^2$ .
And similarly other one simplifies to,
$$
2x^2\frac{dy}{dx}=(x-1)(y^2-x^2)+2xy
$$
$$
2x^2\frac{dy}{dx}=xy^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Equality in triangle obtuse
$m_1, m_2, m_3 $ are sides-lengths of a triangle such that $m_1\sqrt{m_1}+m_2\sqrt{m_2}=m_3\sqrt{m_3}$.
Prove that this triangle is an obtuse-angled triangle.
I don't have idea make run this example
when I raised to square argument it makes a little bit complicated like this
$m_1^3+2m_1\sq... | The hint.
Let $m_1=a$, $m_2=b$ and $m_3=c$.
Thus, $$c^2=\left(\sqrt{c^3}\right)^{\frac{4}{3}}=\left(\sqrt{a^3}+\sqrt{b^3}\right)^{\frac{4}{3}}>a^2+b^2.$$
The last inequality it's
$$(\sqrt{a^3}+\sqrt{b^3})^4>(a^2+b^2)^3,$$ which is true because
$$(\sqrt{a^3}+\sqrt{b^3})^4=a^6+4\sqrt{a^9b^3}+6a^3b^3+4\sqrt{a^3b^9}+b^6>$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove by induction: $C(2n, 2) = 2C(n, 2) + n^2$
Show that if $n$ is a positive integer, then $C(2n, 2) = 2C(n, 2) + n^2$. Here, $C(a, b)$ means the binomial coefficient $\dbinom{a}{b}$.
Prove this by induction.
Here is my calculation:
$n$ cannot be $1$ because $n$ should be equal or larger than $2$ for $C(n, 2)$.
If ... | Hint:
Show and then use that
$$\binom nk=\binom{n-1}{k-1}+\binom{n-1}k$$
| {
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"url": "https://math.stackexchange.com/questions/3105685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a generating function for which $A(n)={n \choose 2}$ In the book I'm using, $A(x)$ denotes the formal power series (generating function), $A(x) = \sum a_ix^i$.
I'm really stuck on this problem. Thanks for any help.
My attempt after the given hint:
$$\begin{align}
A(x)&=\sum_{n\geq 0} \binom{n}{2}x^n\\
&=\sum_{n\ge... | Do you mean $a_n=\binom{n}{2}$ and $A(x) =\sum_{n\geq 0}a_nx^n= \sum_{n\geq 0} \binom{n}{2}x^n$?
Then note that
$$\binom{n}{2}x^n=\frac{x^2}{2}\cdot n(n-1) x^{n-2}=\frac{x^2}{2}\cdot\frac{d^2}{dx^2}(x^{n})$$
and recall the basic generating function $\sum_{n\geq 0}x^n=\frac{1}{1-x}$.
What is $A(x)$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the values of $x$ in $\sqrt{2x-5} = \sqrt[3]{6x-15}$
$$\sqrt{2x-5} = \sqrt[3]{6x-15}$$
*
*Evaluate the values of $x$
I believe that there would be an easier approach to this problem because I will have to expand 3th degree binomial as shown below
$$(2x-5)^3 = (6x-15)^2$$
What am I missing?
Regards
|
because I will have to expand 3th degree binomial as shown below: $(2x-5)^3 = (6x-15)^2$
Which is not a difficult thing to do:
$8x^3 - 60x^2 + 150x - 125 = 36x^2 - 180x +225$ so
$8x^3 -96x^2 + 330x - 350=0$
which is not an unreasonable thing to expect a student to be able to factor and solve (although I pity the stu... | {
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"url": "https://math.stackexchange.com/questions/3107800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\alpha$ denote the image of $x$ in $\mathbb Q[x]/(x^3 + 3x + 3)$
Let $\alpha$ denote the image of $x$ in $\mathbb Q[x]/(x^3 + 3x + 3)$. Express each of $1/α,1/(1+ \alpha),1/(1+ \alpha^2)$ in the form $c_2\alpha^2+c_1\alpha+c_0$ with $c_0, c_1,c_2 \in \mathbb Q$.
I can prove $x^3 + 3x + 3$ is irreducible in $\mat... | Let $P(x)=x^3+3x+3$, $Q_1(x)=x$, $Q_2(x)=1+x$ and $Q_3(x)=1+x^2.$
By Euclid's algorithm, we have that $$P(x)=(x^2+3)Q_1(x) + 3.$$
If we view this relation in $\mathbb{Q}[x]/(P(x))$, we get that $$0=P(\alpha)=(\alpha^2+3)Q_1(\alpha)+3,$$ hence $$3 = -(\alpha^2+3)\alpha,$$ and is $$\frac{1}{\alpha} = \frac{-(\alpha^2+3)... | {
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"timestamp": "2023-03-29T00:00:00",
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Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$ Provide a counterexample: If $n^2-1$ is divisible by $5$, then $n$ is divisible by $2$ or $3$
My book doesn't have an answer to this question, but I think it's $n=6$.
Since $6^2-1=35$, which is divisible by $5$ but not by $2$... | $n^2 - 1= (n-1)(n+1)$ so if $5|n^2 -1$, as $5$ is prime we know either $5|n-1$ or $5|n+1$.
Now if we're clever we can try to take counter examples $n=6$ or $4$. nope. $n = 11$ or $9$. Oh, wait there's a counter example. $n=11$. (Also $n=1$ and $5|0$ is a counter example.)
But plugging in numbers in the hope of a co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing an inequality using Cauchy-Schwarz I managed to solve the following inequality using AM-GM:
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} \geq \frac{3}{4}
$$
provided that $a,b,c >0$ and $abc=1$.
However it was hinted to me that this could also be solved with Cauchy-Schwarz inequality but I ... | Let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S we obtain:
$$\sum_{cyc}\frac{a}{(a+1)(b+1)}=\sum_{cyc}\frac{\frac{y}{x}}{\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)}=\sum_{cyc}\frac{y^2}{(x+y)(y+z)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x+y)(y+z)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\... | The limit of a sequence, if it exists, is equal to its lim inf and lim sup. Accordingly, if the lim inf and lim sup of a sequence are different, then its limit cannot exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
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Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$ I want to compute the product
$$
(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,
$$
for a natural number $n$. For $n$ equal to 0 or 1, the computation is very simple but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I canno... | For convenience let us rewrite $x,\partial_x$ with $a,b$ so $[a,b]=x\partial_x-\partial_x x=-1$ (in the operator sense on the Schwartz space).
Lemma. For all $n\in\mathbb N$
$$
[(a+b)^n,a-b]=2n(a+b)^{n-1}
$$
Proof. As darij pointed out, one has $[a+b,a-b]=2$ (i.e. the case $n=1$). The trick then is
$$
\begin{split}
[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x... | No, what you did was wrong.
this is because :
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ does not imply $(x+1)-(x+2)=x+3$.
You need to square both sides, which means:
$(\sqrt{x+1} - \sqrt{x+2})^2=x+3$
$(x+1)+(x+2)-2\sqrt{(x+1)(x+2)}=x+3$.
As a sidenote, you can plug x=-4 into the original equation to check whether x=-4 i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all primes $p$ for which there are positive integers $x, y$ such that $p+1=2x^2$ and $p^2+1=2y^2$ Find all primes $p$ for which there exist positive integers $x, y$ such that $p+1=2x^2$ and $p^2+1=2y^2$.
I have tried coming up with an equation for $p$ or $p^2$ and this is what I've got
$p=2x^2-1$;
$p^2=(2x^2-1)^2$... | So $$p(p-1)=p^2-p=2y^2-2x^2=2(y+x)(y-x)$$
One of the three factors on the right must be $p$ or a multiple thereof, and it must be the largest among the three. We conclude $p\mid x+y$. But clearly $x<p$ and $y<p$, hence we can only have $p=x+y$. Then $p-1=2(y-x)$, hence $y=3x-1$.
Now by simple algebraic manipulations,$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this a Riemann sum (if so, I can't figure out which one)? This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$\lim_{n\to\infty}{\frac{1}{n} {\sum_{k=3}^{n}{\frac{3}{k^2-k-2}}}}$$
Well, even the fact that $\frac{3}{k^2-k-2} = \frac{1}{... | $$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=3}^n \frac{1}{k-2}- \sum_{k=3}^n \frac{1}{k+1}$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$\sum_{k=3}^n \frac{1}{k-2}=\sum_{k=0}^{n-3} \frac{1}{k+1}$$
Insert above you get
$$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=0}^{n-3} \frac{1}{k+1} - \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \fra... | Note that
$$\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}= \frac{|x|}{x}\stackrel{x<0}{=}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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If $\sec\theta=-\frac{13}{12}$, then find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$. The official answer differs from mine.
Given $\sec\theta=-\frac{13}{12}$ find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$.
If the $\sec\theta$ is $-\frac{13}{12}$ then, the $\cos \theta$ is $-\frac{12}{13}$, ... | The answer they gave $\left(\frac {5 \sqrt{26}}{26}\right)$ is the value for $$\sin \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1-\cos \theta}{2}}$$ however they're looking for $$\cos \dfrac {\theta}{2} = \pm \sqrt {\dfrac {1+\cos \theta}{2}}$$
EDIT (thanks, DanielWainfleet!): For the range $\pi/2 \lt \theta \lt \pi$, $\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Limit involving inverse functions When I am given the limit
$$\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2 +1}}{\sqrt{x^2+1}}$$
would it be possible to evaluate it giving some substitution?
L'Hospital's rule seemed an option but I ended up going in circles.
| When $x>0$, $|x|=x$ and obviously if $x\rightarrow\infty$, then $\sqrt{x^2+1}\rightarrow\infty$. And the last thing that you will need is the fact that the inverse tangent function approaches a value of $\pi/2$ as its argument goes to infinity:
$$
\lim\limits_{x \rightarrow \infty}\frac{x\arctan\sqrt{x^2+1}}{\sqrt{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3118910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum number of steps in water measure problem I've been struggling with this problem for a while and gone through questions about the "Water Jug Problem/Puzzle".
A person wants to have $2$ separate $1$ L measures of water at the same time. However the only measures she has are for $6, 10$ and $15$ L. Show how this ... | Solutions in $9$ steps (without proof of optimality yet):
denote jugs as $a$($6$L), $b$($10$L), $c$($15$L); then:
\begin{array}{|c|c|c|c|c|}
\hline
\# & move & a & b & c \\
\hline
1) & \bullet \rightarrow a & 6 & - & - \\
2) & \bullet \rightarrow c & 6 & - & 15 \\
3) & c \rightarrow b & 6 & 10 & 5 \\
4) & b \rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding the ratio of a side of $\triangle ABC$ and its segment where one cevian line from the opposite vertex intersect the side in any point
In $\triangle ABC$, $L$ and $M$ are two points on $AB$ and $AC$ such that $AL = \frac{2AB}{5}$ and $AM = \frac{3AC}{4}$. $BM$ and $CL$ intersect at the point $P$ and the extensi... | $$\frac{BN}{NC}\cdot\frac{CM}{MA}\cdot\frac{AL}{LB}=\frac{S_{\Delta PBN}}{S_{\Delta PCN}}\cdot\frac{S_{\Delta PCM}}{S_{\Delta PAM}}\cdot\frac{S_{\Delta PAL}}{S_{\Delta PBL}}=$$
$$=\frac{\frac{1}{2}PB\cdot PN\sin\measuredangle BPN}{\frac{1}{2}PC\cdot PN\sin\measuredangle CPN}\cdot\frac{\frac{1}{2}PC\cdot PM\sin\measured... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding $\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$
Find
$$\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$$
What I tried:
$$\sin^4(x)+\cos^4(x)=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$$
and $$\sin^3 x+\cos^3 x=(\sin x+\cos x)(1-\sin x\cos x)$$
so
$$\int\frac{1-\sin^2 x\cos^2 x}{(\sin ... | A brutal way is just to enforce the substitution $x=2\arctan t$, leading to
$$ \int\frac{\sin x\cos x}{\sin^3 x+\cos^3 x}\,dx=2\int\frac{\frac{2t(1-t^2)}{(1+t^2)^3}}{\frac{(2t)^3}{(1+t^2)^3}+\frac{(1-t^2)^3}{(1+t^2)^3}}\,dt=\int\frac{4t(1-t^2)}{(2t)^3+(1-t^2)^3}\,dt $$
an ugly integral, but perfectly solvable by partia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Integral problem. Unsure of the approach. I have this integral:
$$\int_0^1 \frac{1 + 12t}{1 + 3t}dt$$
I can split this up into:
$$\int_0^1 \frac{1}{1+3t} dt + \int_0^1\frac{12t}{1+3t}dt$$
The left side:
$u = 1+3t$ and $du = 3dt$ and $\frac{du}{3} = dt$
so $$\frac{1}{3} \int_0^1 \frac{1}{u} du = \frac{1}{3} \ln |u| + C... | $u=1+3t$:
$$
\begin{align}
\int\frac{1+12t}{1+3t}\,dt
&=\int\frac{1}{1+3t}\,dt+12\int\frac{t}{1+3t}\,dt\\
&=\frac{1}{3}\int\frac{1}{1+3t}\frac{d}{dt}(1+3t)\,dt+\frac{12}{3}\int\frac{3t}{1+3t}\,dt\\
&=\frac{1}{3}\int\frac{1}{u}\,du+4\int\frac{1+3t-1}{1+3t}\,dt\\
&=\frac{1}{3}\ln{|u|}+4\left(\int\frac{1+3t}{1+3t}\,dt-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Solve $\log_2(3^x-1)=\log_3(2^x+1)$
Solve the following equation over the real number(preferably without calculus):
$$\log_2(3^x-1)=\log_3(2^x+1).$$
This problem is from a math contest held where I learn; I was unable to do much at all tinkering with it; I have observed the solution $x=1$ but haven't been able to ... | We'll use the following: $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
So: $ \log_2(3^x - 1) = \log_3(3^x-1) \iff \frac{\ln(3^x-1)}{\ln(2)} = \frac{\ln(2^x+1)}{\ln(3)}$
That is:
$ \ln(3)\ln(3^x-1) = \ln(2)\ln(2^x+1) $
So we're left with a function $f:(0,+\infty) -> \Bbb R$ , $f(x) = \ln(3)\ln(3^x-1) -\ln(2)\ln(2^x+1) $
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find matrix $B$ from kernel of $A$ Here's the question:
For the matrix
$$
A=\begin{bmatrix}
1 & 1 & 0 & 1 & 1\\
1 & 0 & 1 & 1 & 0\\
0 & 1 & 1 & 1 & 1\\
\end{bmatrix}
$$
find a matrix B such that $\text{Im}(\phi_B)=\text{Ker}(\phi_A)$ and $\text{Ker}(\phi_B)=0$
I have found the null space of $A$ which must be the imag... | I like using the language of column spaces and null spaces when discussing matrices.
Presumably, the notation $\phi_M$ refers to the linear map $\phi_M:\Bbb R^n\to \Bbb R^m$ given by $\phi_M(\vec{x})=M\vec{x}$ where $M$ is an $m\times n$ matrix. Then, in our notation, for any matrix $M$ we have
\begin{align*}
\operato... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Question from the 2011 IMC (International Mathematics Competition) Key Stage III paper, about the evaluation of a quadratic equation
When $a=1, 2, 3, ..., 2010, 2011$, the roots of the equation $x^2-2x-a^2-a=0$ are $(a_1, b_1), (a_2, b_2), (a_3, b_3),\cdots, (a_{2010}, b_{2010}), (a_{2011}, b_{2011})$ respectively. Ev... | $$\sum_{k=1}^{2011}\left(\frac{1}{a_k}+\frac{1}{b_k}\right)=\sum_{k=1}^{2011}\frac{a_k+b_k}{a_kb_k}=\sum_{k=1}^{2011}\frac{2}{-k(k+1)}=$$
$$=-2\sum_{k=1}^{2011}\left(\frac{1}{k}-\frac{1}{k+1}\right)=-2\left(1-\frac{1}{2012}\right)=-\frac{2011}{1006}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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prove that the triangle is isosceles
In a $\triangle ABC$, If
$\begin{vmatrix}
1 & \;\;1\;\;&\;\; 1\;\;\\\\
\displaystyle \cot \frac{A}{2} & \displaystyle \cot \frac{B}{2} & \displaystyle \cot \frac{C}{2} \\\\
\displaystyle \tan\frac{B}{2}+\tan \frac{C}{2} &\;\;\displaystyle \tan \frac{C}{2}+\tan\frac{A}{2} & \... | Hint:
Applying $C_2'=C_2-C_1,C_3=C_3-C_1$
$$\begin{vmatrix}1& 1& 1\\
\displaystyle \frac{1}{p}& \displaystyle \frac{1}{q}& \displaystyle \frac{1}{r}\\
q+r & r+p& p+q\end{vmatrix} =\begin{vmatrix}1& 1-1& 1-1\\
\displaystyle \dfrac1p & \displaystyle \dfrac1q-\dfrac1p & \displaystyle \dfrac1r-\dfrac1p\\
q+r & r+p-(q+r)& p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find $\mathbf a$ for which the following equation has triple roots Find $\mathbf a \in\mathbb{R}$ for which the following equation has triple roots:
$$ x^4-5x^3+\mathbf ax^2-7x+2=0$$
| Suppose $b$ is the triple root. Then the polynomial can be factored as $(x-b)^3(x-c)$ and we get
$$
x^4 - (3b+c)x^3 + (3b^2 + 3bc)x^2 - (b^3+3b^2c)x + b^3c
$$
so we need
\begin{cases}
3b+c=5 \\
3b^2+3bc=a\\
b^3+3b^2c=7\\
b^3c=2
\end{cases}
Multiplying the third equation by $b$ and taking into account the fourth equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\ x^{12}\equiv 87\pmod{101}$ Can someone help me to solve the following congruent equation. I have tried to use Fermat's little theorem but failed to solve this:
$x^{12}=87(\mod 101)$
| Numbers congruent to $87 \pmod {101}$ are $87,\; 87+101=188,\; 87+2\times 101=\color{red}{289}, ...$
and $87-101=-14,\;87-2\times 101=-115,\;87-3\times 101=\color{blue}{-216},...$
Note that therefore $17^2=289 \equiv87\pmod{101}$
and $(-6)^3=-216 \equiv 87 \pmod{101}.$
Furthermore $17\times6=102 \equiv 1 \pmod{101}.$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Determinant of block matrix with equal diagonals
Let $$Q=\begin{bmatrix}A&B\\-B&A\end{bmatrix}$$ where $A,B\in \mathbb{R}^{n\times n}$. Prove that $$\det(Q)=\det(A^2+B^2)$$
Since $A$ and $B$ do not commute, I cannot use Schur's formula. Also, $A$ and $B$ may not be invertible. Any comment or response is appreciated.... | You need to require that the matrices $A$ and $B$ commute (i.e., that
$AB=BA$). Otherwise, for example, $A=\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & 0\\
1 & 1
\end{pmatrix}$ yield a counterexample (since $\det\begin{pmatrix}
A & B\\
-B & A
\end{pmatrix}=\det\begin{pmatrix}
1 & 1 & 1 & 0\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | If $h$ denotes getting a heads and $p$ denotes getting tails, let's write $\frac{1}{2}h+\frac{1}{2}p$ for the notion of a fair coin: half the time it turns up heads, and half the time tails.
If we expand out the following product while keeping track of multiplication order,
$$\left(\frac{1}{2}h+\frac{1}{2}p\right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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"answer_id": 2
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Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \f... | Let
$$2^{b}\le n<2^{b+1}$$
For all $1\le k\le b$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=\frac{2^b}{2^k}+\left\lfloor\frac{n-2^b}{2^k}+\frac12\right\rfloor.$$
For $k=b+1$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=1.$$
And for $k>b+1$,
$$\left\lfloor\frac{n}{2^k}+\frac12\right\rfloor=0.$$
This allows u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
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Prove that $1+\frac{1}{8n}\sqrt{\pi_1n}<\frac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<(1+\frac{1}{8n}+\frac{1}{128n^{2}})\sqrt{\pi_2n}$ Within the confines of the O-level syllabus, prove that:
$$1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{1... | $$
\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)} = \dfrac{2n!}{(2n-1)!} = 2n \\ \therefore \;\; 1+\dfrac{1}{8n}\sqrt{\pi_1n}<\dfrac{2.4.6\ldots(2n-2)(2n)}{1.3.5\ldots(2n-3)(2n-1)}<\left(1+\dfrac{1}{8n}+\dfrac{1}{128n^{2}}\right)\sqrt{\pi_2n}
\\ = 1+\dfrac{1}{8n}\sqrt{\pi_1n}<2n<\left(1+\dfrac{1}{8n}+\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the least possible value of perimeter of $\triangle ABC$ with given ranges of angle
In $\triangle ABC$,$\angle A >2\angle B$ and $\angle C > 90^\circ$. If the length of all side of triangle $\triangle ABC$ are positive integers, then what is the least possible value of perimeter of $\triangle ABC$?
However, I... | So, how about a really simple approach: run through positive integer triples $b<a<c$. If they're sides of a triangle, the angles will be in order $B<A<C$.
*
*$(1,2,3)$: Not a triangle. In fact, we can't have a triangle if the small side is $1$.
*$(2,3,4)$: Since $4^2=16>13=2^2+3^2$, it's obtuse. Now, the smaller ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tr... | From the second equation we get
$$\sqrt{x+y}=x+3-\sqrt{x}$$ y squaring this equation we obtain
$$y=(x+3-\sqrt{x})^2-x$$ plugging this in the first equation we get
$$\sqrt{(x+3-\sqrt{x})^2}+\sqrt{x+3}=\frac{1}{x}((x+3-\sqrt{x})^2-x-3)$$
Can you proceed?
Hint: $$x=1,y=8$$
Ok, we can also eliminate the square root:
$$\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of attempts to see all faces $1, 2, 3, ..., 6$ when rolling a dice? Recently I have been thinking about the following random experiment: we repeatedly roll a dice until we see all the faces $1, 2, 3, 4, 5, 6$ of the dice at least once.
Let $X$ = number of attempts necessary to see all the faces.
Obviously $X(\O... | Set up the problem as a Markov chain with states being the number of faces that have shown up to present, thus from 1 to 6. The transition matrix (with columns adding to 1) is $$P=\frac{1}{6}\begin{pmatrix}1&0&0&0&0&0\\5&2&0&0&0&0\\0&4&3&0&0&0\\0&0&3&4&0&0\\0&0&0&2&5&0\\0&0&0&0&1&6\end{pmatrix}$$ The 'absorbing' state ... | {
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"url": "https://math.stackexchange.com/questions/3140954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime numbers $pFind the prime numbers $p<q<r$ such that $r^2-q^2-p^2$ is a perfect square.
I think the only solution is (2,3,7) but i cannot prove it.
The equation would be $r^2-q^2-p^2=k^2$ equivalently
$q^2+p^2+k^2=r^2$ which is somehow a classical diophantine equation (of the pythagorean quadruples) which have a ... | Hint
All prime numbers $p\geq5$ satisfy $p\equiv \pm 1\mod 6$
Therefore, if $p,q,r\geq5$
$$r^2-q^2-p^2\equiv (\pm1)^2-(\pm1)^2-(\pm1)^2\equiv 1-1-1\equiv5\mod 6$$
Observe now, that there's no perfect square with the residue $5$ modulo 6.
This follows from the simple fact that (taking the equations modulo $6$)
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did ... | From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$\frac{dy}{dx}= 3x^2+18x +36 $$
$$ \frac{d^2y}{dx^2}=6x+18 $$
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $n^2+3n+6$ divisible by 25, where $n$ is a integer? If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.
$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not div... | Hint: We have
$$
n^2+3n+6=(n+4)^2 \bmod 5
$$
Now look at Bill's answer at this duplicate:
$n^2 + 3n +5$ is not divisible by $121$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Last digit of large powers Define the sequence $a_1, a_2,...$ by $a_1=7$ and where $a_{n+1} = a_n^7$.
1) Find the last digit of $a_{1876}$, ie $7(^7)(^7)(^7)(^7)(^...)$
Now, I understand that the last digit of powers of 7 repeat every 4 powers (i think), so the answer has something to do with integers modulo 4. While I... | The order of $7$ mod $100$ is $4$, i.e. $7^4 = 2401 \equiv 1 \mod 100$, so $7^{4i+j} \equiv 7^j (\bmod 100)$ for any integers $i$ and $j$.
Since $4$ divides $100$, this is also true mod $4$.
Now $a_n = 7^{7^{n-1}}$, and $2008$ is divisible by $4$ so $7^{2008} \equiv 7^0 = 1 (\bmod 4)$, and $a_{2009} \equiv 7^{1} = 7 (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If rank of a given matrix of order $3 \times 4$ is $2$ then the value of $b$ is Q) Suppose the rank of the matrix
$\begin{pmatrix} 1 &1 &2 &2 \\ 1&1 &1 &3 \\ a&b &b &1 \end{pmatrix}$
is $2$ for some real numbers $a$ and $b$. Then $b$ equals
$(A)$ $1\;\;\;$ $(B)$ $3\;\;\;$ $(C)$ $1/2\;\;\;$ $(D)$ $1/3\;\;\;$
My Approach... | Consider the submatrix $M$ obtained from the first two rows
$$
M = \left[\begin{array}{rrrr}
1 & 1 & 2 & 2 \\
1 & 1 & 1 & 3
\end{array}\right]
$$
The two vectors
\begin{align*}
\vec{n}_1 &= \left\langle1,\,3,\,-1,\,-1\right\rangle & \vec{n}_2 &= \left\langle0,\,4,\,-1,\,-1\right\rangle
\end{align*}
form a basis of $\o... | {
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"url": "https://math.stackexchange.com/questions/3144398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Weierstrass Zeta function Series Let $\Gamma := \mathbb{Z} + \mathbb{Z} \tau$ be a lattice.
Then \begin{align*} \zeta_\Gamma(z+1) - \zeta_\Gamma(z) \end{align*} is a function independent from $z$. Why is that so?
The function is defined as follows
\begin{align}
\zeta_\Gamma(z) = \frac{1}{z} + \sum_{(m,n)\neq (0,0)} \f... | Its derivative is $-\wp_\Gamma(z+1)+\wp_\Gamma(z)=0$, as $1\in\Gamma$
and $\Gamma$ is the
period lattice of $\wp_\Gamma$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving an inequality without an integral: $\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}$ I would like to prove the following inequality without integration; could you help?
$$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x > 0. $$
I can however differentiate this.
Thanks in advance.
| Lemma: For all real $x, e^x \ge 1 + x$.
Proof of lemma: Consider $f(x)=e^x-x-1.$ Its first derivative ($e^x-1$) is $0$ when and only when $x=0$, and its second derivative ($e^x$) is positive for all real $x, $ so $e^x-x-1$ has a global minimum when $x=0\; (f(0)=0$) and $e^x-x-1\ge 0$ for all real $x.$
From the lemma... | {
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"timestamp": "2023-03-29T00:00:00",
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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.