Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Dimension of the spanned subspace Let $a,b\in \mathbb{R}.$ Determine the dimension of the subspace of $\mathbb{R}^5$ generated by the following vectors.
$$\begin{pmatrix} 1\\2\\1\\2\\0 \end{pmatrix}, \begin{pmatrix} 2\\5\\1\\3\\1 \end{pmatrix}, \begin{pmatrix} 5\\8\\3\\6\\2 \end{pmatrix}, \begin{pmatrix} 2\\1\\1\\1\\1 ... | Yes, it is, that is the quickest/most common method to compute the dimension of the subspace generated by those vectors :)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that numerator of fraction $\sum \limits_{k=1}^{p-2}\frac{k}{(k+1)^2}$ is divisible by $p$ for all prime $p>3$ Let $p>3$ be a prime number. Let $\sum \limits_{k=1}^{p-2}\frac{k}{(k+1)^2}=\frac{r}{s}$, where $r,s$ are some integers numbers. Prove that $r$ is divisible by $p$.
My work. Let $\sum \limits_{k=1}^{p-1}... | Note that $$\sum_{k=1}^{p-2} \frac{k}{(k+1)^2} = \sum_{i=2}^{p-1} \frac{1}{i} - \frac{1}{i^2} = \sum_{i=1}^{p-1} \frac{1}{i} - \frac{ 1}{i^2}. $$
We work mod $p$.
*
*$ \{ \frac{1}{i} | i = 1\text{ to }p-1\} = \{ j | j = 1\text{ to }p-1\}$.
Hence $ \sum \frac{1}{i} = \sum j = \frac{p(p-1)}{2} \equiv 0\pmod{p}$... | {
"language": "en",
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If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{uv^2}{w^3}-\sqrt{\frac{3u^2-2v^2}{v^2}}.$$
We see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ ( https://artofproblemsolving.com/communit... | {
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"timestamp": "2023-03-29T00:00:00",
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Roots of equation form infinite sequence
The sequence $a_n$ has the property that $a_n$ and $a_{n+1}$ are the roots of the equation
$$x^2-c_nx+\frac{1}{3^n}=0$$ and $a_1=2$. What is $\sum_{n=1}^{\infty}c_n?$
By Vieta's,
$a_{n+1}=\frac{1}{3^na_n}$ and $c_n=a_n+a_{n+1}$. Additionally listing the first numbers in $a_n... | Instead of simplifying all the terms, try to find a relation between every term in terms of $a_1$.
Now, your sequence of $a$ shall look like this $a_1 ,\frac{1}{3a_1},\frac{a_1}{3},\frac{1}{9a_1}\cdots$
Notice that, every even terms of the sequence are in a $g.p$ with common ratio $\frac{1}{3}$
And every odd terms are ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the Maclaurin series for $f(x)=x\ln(x+1)$
Find the Maclaurin series for the function
$$f(x)=x\ln(x+1)$$
So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.
$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0... | Note that $$\ln(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}x ^k}{k}$$
So the Mclaurin series for $x\ln(1+x)$
is $$ x\ln(1+x)= \sum_{k=1}^{n} \frac{(-1)^{k+1} x^{k+1}}{k}=x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{4}+....$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$ This question was asked in the Crux Mathematicorum , October edition , Pg -$5$ , which can found be Here.
The question states that :
Both $4$ and $52$ can be expressed as the sum of two squares as well as
exceeding another square by $3$... | There are infinitely many even numbers. Let us assume that $b$ is even, as doing so does not limit the number of possible solutions. We can rearrange the original equation to $$c^2-a^2=b^2-3$$ where $b^2-3$ must be odd.
It is well know that the sum of the first $n$ odd numbers equals $n^2$. Consequently, every odd numb... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\ti... | We need to find the radius of the large circle and the rest is straight forward:
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\f... | Given differential equation is$$(D^2+4)y=4x^2\cos 2x$$
We have to find the particular integral (P.I.).
\begin{equation}
\text{P.I.}~=4\dfrac{1}{D^2+4}~x^2\cos 2x\\
=\text{R.P. of }~\left(4\dfrac{1}{D^2+4}~x^2~e^{2ix}\right)\\
=4~\cdot~\left[\text{R.P. of }~\left\{e^{2ix}~\dfrac{1}{(D+2i)^2+4}~x^2\right\}\right]\\
=4~\c... | {
"language": "en",
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What is the expected length of the hypotenuse formed by bending a unit length randomly at a right angle? This is easy enough to simulate and find the answer is somewhere around .812. However I am not finding it so easy to solve the integral involved which I believe is...
$$\int_0^1 \sqrt{x^2+(1-x)^2} dx$$
I don't seem ... | Use the substitution $u=x-\frac{1}{2}$
$$\int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{\left(u+\frac{1}{2}\right)^2+\left(u-\frac{1}{2}\right)^2}du = \int_{-\frac{1}{2}}^{\frac{1}{2}} \sqrt{2u^2 + \frac{1}{2}}du = 2\sqrt{2}\int_0^{\frac{1}{2}} \sqrt{u^2+\frac{1}{4}}du$$
Then there are two ways to approach this problem, eith... | {
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If point $P$ varies on a circle about the center of rectangle $\square ABCD$, then $PA^2+PB^2+PC^2+PD^2$ remains constant
Let $\square ABCD$ be a rectangle with center $O$. Prove that if a $P$ is a point that varies over a circle about $O$, then $PA^2+PB^2+PC^2+PD^2$ remains constant
Attempt:
| Place $O$ at $(0,0)$ on the Cartesian Plane, and place $P$ at a point $(x,y)$ at a distance $r$ from $O$. Furthermore, place the vertices of the rectangle at the points of the form $(\pm a,\pm b)$, for the corresponding $a$, $b$. Then, $$PA^2+PB^2+PC^2+PD^2=$$ $$2\left(\left(x-a\right)^2+\left(x+a\right)^2+\left(y-b\ri... | {
"language": "en",
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evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos... | $$\lim_{n\rightarrow \infty}\dfrac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}=\lim_{n\rightarrow \infty}\dfrac{1}{n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n\pi}{2}}$$
As $n\to\infty$, since $\sin^2u$ and $\cos^2u$ both oscillate in $[0,1]$ and $n,n^2\to\infty$, So, $$(n\cos^2\frac{n\pi}{2}+n^2\sin^2\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don... | A standard trick to solve such problems is to use the substitution $a= y + z, b =x+z, c = x+y$, where $x, y, z>0$ (to see why this is so, draw the inscribed circle and pairs of equal tangents: $x, y, z$ are the lengths of the tangents)
So, we have $$x+y+z = \frac{1}{2}$$
and the inequality is $$ (x+y)^2 + (x+z)^2 + (y... | {
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"url": "https://math.stackexchange.com/questions/3468328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did... | $$ 2 \cos \left( \frac{2 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{4 \pi}{7} \right) \; , \; \; 2 \cos \left( \frac{8 \pi}{7} \right) \; , \; \; $$
are the roots of
$$ x^3 + x^2 - 2x - 1 $$
so
$$ \cos \left( \frac{2 \pi}{7} \right) \; , \; \; \cos \left( \frac{4 \pi}{7} \right) \; , \; \; \cos \left( \frac{... | {
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Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other way... | $$a<b\iff a^2<b^2\iff 7+10+2\sqrt {70}<3+19+2\sqrt {57}$$ $$\iff 2\sqrt {70}<5+2\sqrt {57}$$ $$\iff (2\sqrt {70})^2<(5+2\sqrt {57})^2$$ $$\iff 280<25+228+20\sqrt {57}$$
$$\iff 27<20\sqrt {57}$$ and we have $20\sqrt {57}>20\sqrt 4=40>27.$
Another way, from one of the intermediate steps above, is $$a<b\iff 2\sqrt {70}<5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471412",
"timestamp": "2023-03-29T00:00:00",
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Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$ Wikipedia's "List of formulae involving $\pi$" entry states that
$$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$
Why is this true?
If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we ... | $$F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}=2 \left[\sin ^{-1}\left(\frac{x}{2}\right)\right]^2$$ makes
$$F(\varphi)=\frac{9 \pi ^2}{50}\qquad \text{and}\qquad F(\psi)=\frac{\pi ^2}{50}$$
since
$$\varphi=1-2 \cos \left(\frac{3 \pi }{5}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the fol... | The original inequality is equivalent to
$$
(5 m n+1)^2\leq 75 n^4
$$
or
$$
75 n^4\leq (5 m n-1)^2
$$
This reduces to
$$
\left(n\leq -1\land \left(m\leq \frac{1}{5 n}-\sqrt{3} \sqrt{n^2}\lor m\geq -\sqrt{3} \sqrt{n^2}-\frac{1}{5 n}\right)\right)
$$
or
$$
\left(n\geq 1\land \left(m\leq \sqrt{3} \sqrt{n^2}-\frac{1}{5 n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the value of the double integral: $\int\limits_0^{2a}\int\limits_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$ Evaluate $$\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$$
Here I changed the order of integration but after that I am not able... | Changing the order and taking $x^2+y^2=t,dt=2xdx$ gives,$$\frac12\int_0^a\phi'(y)\int_{2a^2-2a\sqrt{a^2-y^2}}^{2a^2+2a\sqrt{a^2-y^2}}\frac{t~dt}{\sqrt{4a^2(t-y^2)-t^2}}~dy$$Now$$\frac t{\sqrt{4a^2(t-y^2)-t^2}}=\frac{-1}2\left[\frac{4a^2-2t}{\sqrt{4a^2(t-y^2)-t^2}}\right]+\frac{2a^2}{\sqrt{4a^2(a^2-y^2)-(t-2a^2)^2}}$$an... | {
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Solve system of equations $\frac{x}{y}+\frac{y}{x}=\frac{10}{3}$, $x^2-y^2=8$
Solve the system:
$$\begin{array}{l}\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{10}{3} \\
x^2-y^2=8\end{array}$$
First, we have $x \ne 0$ and $y \ne 0$. We can rewrite the first equation as $$\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}$$
What should I do nex... | Multiplying the first equation by $xy$ gives us
$$3x^2-10xy+3y^2=0$$
Let $a=x+y$ and $b=x-y$ to get
$$4b^2-a^2=3x^2-10xy+3y^2=0$$
$$ab=8$$
Solving for $a$ in the second equation we get $a=8/b$, and substituting into the first equation gives
$$b^2-\frac{64}{b^2}=0$$
Solving for $b$ we get $b^4=16$, which has the four so... | {
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"timestamp": "2023-03-29T00:00:00",
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If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$. Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.
My effort:
If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+... | As question is closed let me bring short answer here:
Having $I_n=\int\frac{dx}{(a+b\cos x)^n}$ partial integration gives
$$I_2(a^2-b^2) = aI_1-\frac{b\sin x}{a+b\cos x}$$
$$I_3 2(a^2-b^2) = 3aI_2 - I_1 -\frac{b\sin x}{a+b\cos x}$$
So $I_3$ can be represented as sum of constant member with $I_1$ plus function members.
... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $ f [f (x)] $ Where $ f (x) = \frac {1} {1 + x ^ 2} $, $ f [f (x)] $ is equal to:
a
I was factoring and found $f(f(x))= \frac{1}{1+(\frac{1}{1+x^2})^2} = \frac{1}{\frac{x^4+2x^2+2}{x^4+2x^2+1}}= \frac{x^4+2x^2+1}{x^4+2x^2+2}$, but the answer is x. I am wrong?
| The answer is not $x$, as Desmos (and the rest of math!) suggest(s):
| {
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Find the total number of 7-digit Find the total number of 7-digit natural numbers whose digits sum is 20.
Solution: This is not a exactly answer, as well we can use a way to subtracts the cases when exists a digit greater than $9$.
Let $x_1,\cdots, x_7$ the digits.
As well there are ${25\choose 6}$ ways to give us the ... | We count the number of integers $x$ with $1\,000\,000\leq x\leq 9\,999\,999$ which have a digit sum equal to $20$ with the help of generating functions. It is convenient to use the coefficient of operator $[x^{n}]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{20}]}&\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478419",
"timestamp": "2023-03-29T00:00:00",
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Proof by mathematical induction proof verification If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.
The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.
Now to show that $S_k\Rightarrow S_{k+1}$, we assume that ... | A couple of your steps could possibly be arranged in a more readable way. However, I do believe your proof is core.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots... | $$L=\lim_{x \rightarrow 0} \frac{x^n \sin^{n} x}{x^n-\sin^n x}$$ $$= \lim_{x \rightarrow 0} \frac{x^{2n}(1-x^2/6)^n}{x^{n}-x^n(1-x^2/6)^n}$$
$$=\lim_{x \rightarrow 0} \frac{x^n (1-nx^2/6)}{nx^2/6}=\lim_{x \rightarrow 0}\frac {6x^{n-2}}{n}= 3,~ if~ n =2,$$
$$= 0, ~if~ n > 2, $$ $$=\infty ~~if~~n= n <2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
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Integer triangle $ABC$ such that $IHO$ is also an integer triangle. An infinite number of such non-similar triangles $ABC$.
Let $I, H, O$ be the incenter, orthocenter and circumcenter of non-isosceles triangle $ABC$ respectively. Prove that there are infinitely many integer triangles $ABC$, none of which are similar, ... | (Some small examples that I found satisfy $IO=IH$. So...)
Let us consider $\triangle{ABC}$ satisfying $IO=IH$. We have
$$IO=IH\iff (a^2 - a b + b^2 - c^2) (a^2 - b^2 + b c - c^2) (a^2 - a c - b^2 + c^2)=0$$
So, in the following, let us consider $\triangle{ABC}$ such that $$a^2=b^2-bc+c^2\tag1$$
(which means that $\ang... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluate $\int \frac1{\left(1+x^4\right)\left(\sqrt{\sqrt{1+x^4}}-x^2\right)}dx$ Evaluate $\int \dfrac{1}{\left(1+x^4\right)\left(\sqrt{\sqrt{1+x^4}}-x^2\right)}dx$
My attempt is as follows:-
$$x^2=\tan\theta$$
$$2xdx=\sec^2\theta d\theta$$
$$dx=\dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$$
$$\int \dfrac{1}{\left(... | In the same spirit as Mariusz Iwaniuk's comment, a CAS gives
$$\int \dfrac{\sqrt{1-y^4}}{\sqrt{1-y^4}-y^2}dy=\frac{1}{3} y^3 F_1\left(\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};y^4,2
y^4\right)+\frac{1}{8} \left(4 y+2^{3/4} \left(\tan ^{-1}\left(\sqrt[4]{2}
y\right)+\tanh ^{-1}\left(\sqrt[4]{2} y\right)\right)\right... | {
"language": "en",
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"source": "stackexchange",
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Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\i... | I agree with J. W. Tanner's question comment that what you've done looks all right.
I have just one small suggestion. With your $x^2−y^2=4a$ and $y^2−z^2=4b$ equations, you don't need to do any rearranging. Instead, you can just add these $2$ equations, as the $y^2$ terms cancel, to more directly get your result of $x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the det... | Hint:
$$\triangle=\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \dfrac1a\begin{vmatrix}a(b^2+c^2)&ab&ac\\a(ba)&c^2+a^2&bc\\a(ca)&cb&a^2+b^2\end{vmatrix} $$
$$C_1'=C_1-bC_2-cC_3$$
$$\triangle=\dfrac1a\begin{vmatrix}0&ab&ac\\-2bc^2&c^2+a^2&bc\\-2bc^2&cb&a^2+b^2\end{vmatrix}=-2bc\begin{vmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matrix with 2 unknown variables, how to solve? Given that
$$Ax=b$$
where $$x = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},$$
$$A= \begin{bmatrix}
1 & 1 & 3 \\
1 & 2 & 4 \\
k_1 & 3 & 5
\end{bmatrix},$$
and
$$ b= \begin{bmatrix}
2\\
3\\
k_2
\end{bmatrix}$$
and $k_1 ,k_2\in\mathbb R$, I need to ... | I'll call $k_1=x$, $k_2=y$ for simplicity. I don't think you did your row reduction right. you have
$
\begin{bmatrix}
1&1&3&2\\
1&2&4&3\\
x&3&5&y
\end{bmatrix}
\xrightarrow{R_2-R_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
x&3&5&y
\end{bmatrix}
\xrightarrow{R_3-xR_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
0&3-x&5-3x&y-2x
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confu... | Square both sides, multiply both sides by $5$, then move all terms to the left side. Note that your original equation is really two equations, but after squaring they are the same:
\begin{align}x-\tfrac45=\pm\tfrac{\sqrt{31}}{5}
&\implies \left(x-\tfrac45\right)^2=\left(\pm\tfrac{\sqrt{31}}{5}\right)^2 \\
&\implies x^2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Turning $\frac{2x-y}{2x+y}\frac1{\left(1-\frac{y}{2x}\right)^2}$ into $\frac1{\frac{y}{x}(1-\frac{y}{2x})+(1-\frac{y}{2x})^2}$ I stopped at this equality in the solution of a question in my book
$$\left(\frac{2x-y}{2x+y}\right)\left(\frac{1}{\left(1-\frac{y}{2x}\right)^2}\right) = \frac{1}{\big( \frac{y}{x}\big) \Big(1... | Divide top and bottom by $2x-y$. The numerator becomes 1 and the denominator becomes
$$\frac{2x+y}{2x-y}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{1-\frac{y}{2x}+\frac{y}{x}}{1-\frac{y}{2x}}\left(1-\frac{y}{2x}\right)^2$$
$$=\frac{y}{x}\left(1-\frac{y}{2x}\right)+\left(1-\frac{y}{2x}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Given sequence $(a_n)$ such that $a_{n + 2} = 4a_{n + 1} - a_n$. Prove that $\exists \frac{a_i^2 + 2}{a_j}, \frac{a_j^2 + 2}{a_i} \in \mathbb N$.
Given sequence $(a_n)$ such that $a_1 = a_2 = 1$ and $$\large a_{n + 2} = 4a_{n + 1} - a_n, \forall n \ge 2$$
Prove that $$\large \exists i,j \in \mathbb N \colon \frac{a_i^... | The condition holds true for all pairs $a_i, a_{i+1}$.
In fact, the relation $a_i^2+2=a_{i-1}a_{i+1}$ holds for all $i$, from which the above result follows.
This can be proven by induction. Assume $a_i^2+2=a_{i-1}a_{i+1}$ is true for $1<i<n$. That makes
$$
\begin{split}
a_{n-1}a_{n+1}
& = a_{n-1}\cdot(4a_n-a_{n-1})
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2... | Factoring multivariable polynomials usually needs a bit of creativity. However, when looking for an approach, you can always think of multivariable polynomials as polynomials over a single variable whose coefficients are another polynomials!
With a bit of creativity: $$\begin{align}
1 - y-x^2-y^2-yx^2+y^3 &= \underbrac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$ Inspired by a problem of Vasile Cirtoaje I propose this :
Let $a,b>0$ such that $a+b=1$ then we have :
$$a^{4b}+b^{4a}\geq \frac{1}{2}$$
I compute the derivative of $f(x)=x^{4(1-x)}+(1-x)^{4x}$ on $]0,1]$ we get :
$$ f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + ... | Alternative proof:
I give a proof following @Malper's nice idea.
Fact 1: Let $x\in (0, 1/2]$ and $t \in (0, 1/2]$. Then $x^{4 - 4t} + (1 - x)^{4t} \ge \frac12$.
The proof is given at the end.
By Fact 1, we have
$x^{4 - 4x} + (1 - x)^{4x} \ge \frac12$
for all $x$ in $(0, 1/2]$.
We are done.
Proof of Fact 1: Let $f(x, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Sum of matrices value Let $\mspace{10mu}A,B\in M^{4\times 4}(\mathbb{R})\mspace{10mu}, $
$A=\begin{pmatrix}
1 &0 &2 &-1 \\
0 &1 &-1 &1 \\
1 &1 &1 &0 \\
1 &-1 &3 &-2
\end{pmatrix} \mspace{10mu} \mathrm{and} \mspace{10mu} \mathrm{rk}(B)=1$
What values may have $\mspace{10mu}\mathrm{rk}(A+B)\mspace{10mu}?$
... | It cannot be $A = -B$ since $\operatorname{rank}(A) = 2$ and $\operatorname{rank}(B) = 1$ so $A+B \ne 0$. Therefore by the subadditivity of rank we have
$$1 \le \operatorname{rank}(A+B) \le \operatorname{rank}(A) + \operatorname{rank}(B) = 3$$
so $\operatorname{rank}(A+B) \in \{1,2,3\}$.
All possibilities are indeed at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\left( \frac{2}{7} \right) \left(\log_2 \frac{y}{x} \right)$ if- If $$x= (\cos1°) (\cos2°) (\cos3°) .............(\cos89°)$$
and
$$y=(\cos 2°)(\cos 6°)(\cos 10°).............(\cos 86°)$$
Then what is the integer nearest to $$\left( \frac{2}{7} \right) \left( \log_2 \frac{y}{x} \right) ?$$
I tried putting it in $... | $$
x = \Bigl(\cos(1^{\circ})\cos(2^{\circ})\cdots\cos(44^{\circ})\Bigr) \cdot \cos(45^{\circ}) \cdot \Bigl(\sin(44^{\circ})\sin(43^{\circ})\cdots\sin(1^{\circ})\Bigr)\tag{1}
$$
$$
=\dfrac{1}{\sqrt{2}} \cdot 2^{-44} \sin(2^{\circ})\sin(4^{\circ})\cdots\sin(88^{\circ}).
$$
Next step:
denote $$z=\sin(2^{\circ})\sin(4^{\ci... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of the given function- Find the minimum value of-
$$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$
I tried opening the brackets and trying to cancel the terms and using AM-GM.
| Write $z=x+\frac1x$. Then the function is equal to
$$\frac{z^6-(z^6-6z^4+9z^2-2)-2}{z^3+z^3-3z}=\frac{6z^4-9z^2}{2z^3-3z}=\frac{3z^2(2z^2-3)}{z(2z^2-3)}$$
Now $z$ can never assume a value in $(-2,2)$ and $2z^2-3=0\implies z=\sqrt{3/2}\in(-2,2)$, so we can cancel and get $3z$. Since $z$ has no minimum over the reals, so... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich
Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$
What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \... | With the polar coordinates $\frac xa = r\cos t$ and $\frac yb =r \sin t$,
$$z=\frac12ab r^2\sin 2t \sqrt{1-r^2}$$
For any given $r\le 1$,
$$|z(r)|\le \frac12abr^2 \sqrt{1-r^2}$$
Then, set $(r^2 \sqrt{1-r^2})'=0$ to locate the extreme at $r^2=\frac23$. Thus,
$$-\frac{ab}{3\sqrt3}\le z \le \frac{ab}{3\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502217",
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"source": "stackexchange",
"question_score": "1",
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Summation to n terms of series \begin{gather}
In\ a\ problem\ given\ to\ us\ by\ our\ teacher,\ one\ part\ of\ it\ requires\ the\ summation\ of\ the\ \notag\\
following\ sequence\ to\ a\ finite\ number\ of\ terms\ ( n) , \notag\\
t_{r\ } \ =\ \frac{1}{r\times 2^{r}} \notag\\
\notag\\
S\ =\ \frac{1}{1\times 2} +\frac{... | Let $$f(x):=\sum_{r=1}^n\frac{x^r}r$$ and
$$f'(x)=\sum_{r=1}^n x^r=\frac{1-x^r}{1-x}.$$
Then by integration,
$$f(\tfrac12)=-\log(\tfrac12)-\int_0^{1/2}\frac{x^r}{1-x}dx.$$
The last integral is known as an incomplete Beta, and has no closed-form expression (other than the explicit summation for integer $r$).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Solve in $\mathbb{R^{3}}$ : $\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$ Solve the system in $\mathbb{R^{3}}$ :
$$\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$$
My try :
By imaging I see $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ is a soluti... | We see that all are nonnegative.
If one of them is $0$ then all are $x=y=z=0$. Now suppose all are $>0$.
Now $1+4x^2\geq 4x$ so $$4z^2=y(1+4x^2)\geq 4xy\implies z^2\geq xy$$
In the same manner we get $y^2\geq xz$ and $x^2\geq yz$ so, if say $x^2>yz$ we get $$x^2y^2z^2>xyxzzy$$
a contradiction. So $x^2=yz$ and similary... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Every natural number is covered by consecutive numbers that sum to a prime power. Conjecture. For every natural number $n \in \Bbb{N}$, there exists a finite set of consecutive numbers $C\subset \Bbb{N}$ containing $n$ such that $\sum\limits_{c\in C} c$ is a prime power.
A list of the first few numbers in $\Bbb{N}$ has... | While nickgard's answer shows how to solve the problem using sums being squares of increasing primes, this answer shows how to do it using the sums being just odd powers of $3$.
As suggested in joriki's question comment, for any integers $1 \le j \le k$, you have
$$\begin{equation}\begin{aligned}
\sum_{i=j}^{k}i & = \s... | {
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"url": "https://math.stackexchange.com/questions/3506020",
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"source": "stackexchange",
"question_score": "24",
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Prove that : $\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$ Prove that : ($n\in\mathbb{N}$)
$$\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$$
I know that from little theorem :
$$10^{18}\equiv 1\pmod{19}$$
So :
$$10^{18k}\equiv 1\pmod{19}$$
Now i will go to this step if $\operatorname{correct}$
$$10^{12}=100^{6}\equiv 5^{6}... | Let $a=10^{18k+12}-7$.
If $19|a$ and $3|a$, then $19|(\frac a3)3$, and $19\nmid3$,
so, by Euclid's lemma, $19|\frac a3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove the inequality $\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}$? I want to prove that for all $a,b>0$: $$\sqrt{\frac{a^2+b^2}2}+\frac{a+b}2\geq\frac{a^2+b^2}{a+b}+\sqrt{ab}.$$
My attempts:
Failed attempt. We know by QM-AM inequality that $\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2$ so it w... | Let
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2} \ge \frac{a^2+b^2}{a+b}+\sqrt{ab}$$
$$\implies \sqrt{\frac{a^2+b^2}{2}}\ge \frac{a^2+b^2}{a+b}-\frac{a+b}{2}+\sqrt{ab}$$
Let $a=1, b=t^2$, then
$$\implies \sqrt{\frac{1+t^4}{2}} \ge \frac{1+t^4}{1+t^2}-\frac{1+t^2}{2}+t=\frac{(1-t^2)^2+2t(1+t^2)}{2(1+t^2)}$$ So now we need t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2... | By Hölder's inequality:
$$ ..... \geq \frac{(a+b+c)^3}{4(ab+bc+ac)^2} \geq \frac{9(a+b+c)^3}{4(a+b+c)^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Find the coordinates of $3x^3-2x^2+x+4$ in the base $(3,-1+x,-2+x^2,-3+x^3)$ I did:
$$\alpha_1(3)+\alpha_2(-1+x)+\alpha_3(-2+x^2)+\alpha_4(-3+x^3) = \\
x^3(\alpha_4)+x^2(\alpha_3)+x(\alpha_2)+(3\alpha_1-\alpha_2-2\alpha_3-3\alpha_4)$$
$$\left\{
\begin{array}{c}
\alpha_4 = 3 \\
\alpha_3=-2\\
\alpha_2=1 \\
3\alpha_1-\al... | Looks like you did the first part right, barring a careless mistake.
For the second part, $x^3-2x^2+3x-3=3a+(-1+x)b+(-2+x^2)c+(-3+x^3)d$. So solve as above.
$d=1, c=-2, b=3$ and $a=-1/3$.
So $(-1/3, 3, -2, 1)$.
This problem can also be done by inverting the change of basis matrix: $\begin{pmatrix}3&-1&-2&-3\\0&1&0&0\\... | {
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"source": "stackexchange",
"question_score": "1",
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How to deal with negative area when evaluating a definite integral
Find the area bounded by the curves $y=2-x^2$ and $x+y=0$
$$
-x=2-x^2\implies x^2-x-2=(x-2)(x+1)=0
$$
My Attempt
$A_1:$ Area above the x-axis and $A_2:$ Area below the x-axis
$$
A_1=\int_{-1}^\sqrt{2}(2-x^2)dx-\int_{-1}^0(-x)dx=\Big[2x-\frac{x^3}{3}\... | Example 1:
When trying to find the area between curves $f(x)$ and $g(x)$ you can achieve this by integrating the function $H(x) = |f(x) - g(x)|$. However integration of an absolute value function is piecewise. In this case though $f(x) \ge g(x)$ for $-1\le x \le 2$ so integrating $\int_{-1}^2 (f(x) - g(x)) dx$ is va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divide the rational expressions below and simplify as much as possible: The question is :
$$\frac{x^2 + 8}{2x^2 + x - 3}\div\frac{x + 2}{x - 1}$$
the answer I get is
$$\frac{x^2 + 8}{(2x + 3)(x + 2)}$$
but a further factor of $x+2$ is factored out somehow from the $x^2+8$.
I'm having doubts about this. please help... | In the answer given by the prof $(x+2)(x^2-2x+4)=x^3+8$. It seems the problem statement has the wrong exponent on the $x$ at the start. Your answer is correct for the problem as given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the differential equation $(1-x^2)y' - 2xy^2 = xy$ I solved this equation as a Bernoulli equation:
$y' - \frac{x}{1-x^2}y = \frac{2x}{1-x^2}y^2$
I get expression for $y$:
$y = -C(x)\sqrt{|1-x^2|}$
And then I don’t know how to differentiate the module
| $$y' = \frac{x}{1-x^2}y + \frac{2x}{1-x^2}y^2=\frac{x}{1-x^2}(y+2y^2)$$
This is a separable ODE.
$$\frac{2dy}{y+2y^2}=\frac{2x\:dx}{1-x^2}$$
Integrate :
$$2\ln\left|\frac{y}{2y+1}\right|=\ln\left|\frac{1}{1-x^2}\right|+\text{constant}$$
$$\left(\frac{y}{2y+1}\right)^2=C\frac{1}{1-x^2}\quad\begin{cases}
C>0 \quad \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there som... | $\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$
$\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}=\dfrac{5}{20}$
$\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{2}{8}=\dfrac{5}{20}$
$\dfrac{1}{9}+\dfrac{1}{11}=\dfrac{20}{99}>\dfrac{4}{20}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=... | The polynomial factors fully over the complex numbers,
$$ (a+b+c)(a+b \omega + c \omega^2)(a + b \omega^2 + c \omega) \; , \; $$
where $\omega$ is a primitive cube root of unity, either solution of $x^2 + x + 1 =0$
There is actually a concrete calculation that tells us whether a homogeneous cubic factors completely ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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JBMO TST macedonia Number Theory
Find all $x,y$ positive integers $$x + y^2 + (\gcd(x, y))^2 = xy \cdot
\gcd(x, y)$$
I tried supposing $\gcd(x,y)=d$ and letting $x=ad$, $y=bd$ but didn't get anything the closest thing I think is usefull that $(b+1)^2+4a^2b$ is a perfect square.
| Your start is very good. Now you have $$ad+b^2d^2+d^2=abd^3$$ thus $$a+db^2+d=abd^2\implies d\mid a$$ so $a=dc$ for some integer $c$ and now we have: $$c+b^2+1=cbd^2$$
From here we have $$c={b^2+1\over bd^2-1}\implies \boxed{bd^2-1\mid b^2+1}$$
Now we have $$\color{red}{bd^2-1}\mid (b^2+1)d^2-(bd^2-1)b =\color{red}{ d^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagra... | To get rid of some square roots and simplify, substitute $(a,b)\to (2x, 2y)$. The inequality can be written as:
Let $x,y >0$ and $\frac{1}{x}+\frac{1}{y} = 2$. Prove that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq (x+y)^2$$
In this case, from AM-GM we can see that:
$$\sqrt{2(x^2+1)}+\sqrt{2(y^2+1)} \leq \frac{x^2+1+2}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0... | $\int_{0}^{1}\dfrac{b^3}{1-\theta(1-b)^2}db$. Integrating by parts, taking u=b and dv as the integral of the rest we have:
$(b\int\dfrac{b^2}{1-\theta(1-b)^2}db)\Big|_0^1 -\int_{0}^{1}(\int\dfrac{b^2}{1-\theta(1-b)^2}db)db$.
Lets solve the first integral.
$\int\dfrac{b^2}{1-\theta(1-b)^2}db=\dfrac{-1}{\theta}\int\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = ... | $N = 3^55^77^9$ gives $(7+1)((3)(5)+(3)(5)) =240$ the first $(3)(5)$ deals with $\{0,2,4\}$ and $\{0,2,4,6,8\}$ the second with $\{1,3,5\}$and $\{1,3,5,7,9\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore... | Why not a Taylor series expansion at $x=-2$? That would be
$$f(x) = \sum_{n\geq 0} \frac{f^{(n)}(-2)}{n!}(x+2)^n, $$
with radius of convergence of $2$. You may calculate the $n$-th derivative of $f(x)=1/x^2$ to find
$$\frac{d^nf}{dx^n}(x)= \frac{(-2)(-3) \cdots (-2-n+1)}{x^{n+2}} = (-1)^n \frac{(n+1)!}{x^{n+2}} $$
and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Integral of $\left(\frac{x^4}{1-x^4}\right)^k$ I'm stumped. How does one solve the following expression:
$$\int\left(\frac{x^4}{1-x^4}\right)^kdx, k\in\mathbb{N}$$
I thought it would be a fun challenge, but it beats me.
| You can obtain a recurrence relation and then find a formula using sums and products for a closed form solution. $k\geq 1$, let
$$ I_k(x) = \int \left( \frac{x^4}{1-x^4} \right) ^k {\rm d}x$$
Using by parts with $u = (1-x^4)^{-k}$ and ${\rm d}v = x^{4k} {\rm d}x$, we will get
$$
\begin {align}
I_k(x) &= \frac{x^{4k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$. So i had to prove:
Numbers $x,y,z$ are such that $x+y+z=0$ and $x^2+y^2+z^2=1$. Show that at least one of $xy,xz,yz$ is not bigger than $-\frac{1}{3}$.
Here is my proof:
So we have
$(x... | Regarding my notations here, I refer to my comments under the question. The idea of the parametrization in my comments arises from the following observation.
For $(x,y,z)\in S$, the maximum value of $|x|$ is $\sqrt{2/3}$. To show this, we note that $x=-y-z$ and
$$1=x^2+y^2+z^2=(-y-z)^2+y^2+z^2=2(y^2+yz+z^2) \geq \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $S_{100} - k + k² = 7500$ and $S_{100}$ it is the sum of $100$ consecutive positive integers, then what it is the value o k? $k$ must be one of the 100 consecutive integers.
I know the answers are $50$ and $26$.
But I got stuck into getting these numbers.
Here is what I tried:
$S_{100} = \frac{(n+n+99)\times100}{2} ... | Let's start with $$k^2-k = 50(51-2n)$$
Then we must have $k^2-k \equiv 0 \pmod{50}$.
We split this up into $k^2-k \equiv 0 \pmod 2$ and $k^2-k \equiv 0 \pmod{25}$.
\begin{align}
k^2-k &\equiv 0 \pmod{25} \\
k(k-1) &\equiv 0 \pmod{25} \\
k &\equiv 0, 1 \pmod {25}
\end{align}
Similarly, $k \equiv 0,1 \pmod 2$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving that the limit is bounded. I am trying the following exercise that's actually stated as a lemma on an Econometrics textbook:
Let $\{a_t\}_{t = 1,2.,..}$ be a sequence of nonnegative scalars such that $\sum\limits_{t = 1}^{T}\frac{a_t}{T} < M < \infty$ for every $T$ for some finite $M$. Prove that $\lim\limits_... | We have that
$$
\forall T \in \mathbb{N}^+ \qquad \sum_{i=1}^{T}a_i < MT.
$$
We can define $s_i$ as the cumulative sum, i.e.,
$$
s_i = \sum_{j=1}^{i} a_j
$$
and we notice that
$$
A
\left(
\begin{array}{c}
s_1\\
s_2\\
\vdots\\
s_T
\end{array}
\right )
=
\left(
\begin{array}{c}
a_1\\
a_2\\
\vdots\\
a_T
\end{array}
\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) ... | There is a subtle way to get rid of the extra solutions that uses the work you've done.
When you render
$\sin 2x=2\sin x\cos x =2-2\sqrt{2}$
you then have with $u=\sin x, v=\cos x$:
$uv=1-\sqrt{2}$
$\color{blue}{u+v=uv=1-\sqrt{2}}$
where the blue equation reimposes the original requirement and it's goodbye extraneous r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Solve for the real scalar value of $k$, if $z =\frac{2}{1+ki} -\frac{i}{k-i}$ and $z$ lies on the line, $y= 2x$ The complex number $z$ is given by $\frac{2}{1+ki} - \frac{i}{k-i}$. If it is given that $z$ lies on the line, $y = 2x$, find the value of the real scalar $k$.
So far this is what I have come up with : (See ... | The equations derived in the picture for $(x,y)$ are wrong from line 4 onward. The corrected ones are
$$z=\frac{3k-3i}{2k-i(1-k^2)}\times\frac{2k+i(1-k^2)}{2k+i(1-k^2)}=\frac{3k^2+3+i(-3k^3-3k)}{4k^2+(1-k^2)^2}$$
We have $y=2x$ and $x = \frac{3k^2+3}{4k^2+(1-k^2)^2}$ and $y = \frac{-3k^3-3k}{4k^2+(1-k^2)^2}$ therefore ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What's wrong with my calculation of a limit here? So, I have to find the following limit:
$\lim _{x\to \:0+}\left(\frac{\left(1-\cos \left(2x\right)\right)^{14}\left(1-\cos \left(7x\right)\right)^2\sin ^{14}\left(9x\right)}{\tan ^{14}\left(x\right)\left(\ln \left(8x+1\right)\right)^{30}}\right)$
I solved it by splittin... | Sanity check:
For small $x$,
$$1-\cos x=\Theta(x^2),\sin x=\Theta(x),\tan x=\Theta(x),\log(x+1)=\Theta(x).$$
Then the expression is of order
$$\frac{x^{28}x^4x^{14}}{x^{14}x^{30}}=x^2.$$
(Which tends to $0$, of course.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.... | It can be geometrically seen that once you draw the diameter of the circle that passes through $(4,2)$, contributes the maximum and minimum distances. That diameter must have an equation like $y={x\over2}$ which together with $x^2+y^2=a$, leads to two points $(2\sqrt{a\over 5},\sqrt{a\over 5})$ and $(-2\sqrt{a\over 5},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
| \begin{align}
I&=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\\
&=\int \frac{1}{\sqrt{\cot x}+1}dx
=\frac12x -\frac12 \int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx\tag1 \\
\end{align}
where, with $t=\sqrt{\cot x}$
\begin{align}
\int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx
&=\int \frac{2t(1-t)}{(1+t)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Is $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\dots$ convergent?
Let $a_n= (-1)^k$ ,when $2^k \leq n < 2^{k+1} $.
Then determine if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converge or not.
[my attempt]
I try to apply Dirichlet tset but I fail.
So I tried following method
$$1-\... | Bunching terms with the same sign together already gives you enough information. Each such bunched term has a magnitude of at least $\frac12$, which means that the series diverges by Cauchy's test (all sums of sufficiently far sequences of arbitrary length must be less than any $\varepsilon$ for the series to converge)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding $\sqrt{625}$ without a calculator We were factoring $25n^2+15n-4$ in class and my professor used the quadratic formula. This root came up and he wrote the answer directly. He doesn't want us to use calculators.
Is there any trick for this?
| I guess your professor has been using the following trick: the square of a number of the form $10n + 5$ is $100n(n+1) + 25$. Indeed $(10n + 5)^2 = 100n^2 + 100n + 25 = 100n(n+1) + 25$.
For instance:
\begin{align}
n &= 0: && 5^2 = 25 &&\text{since $0 \times 1 = 0$} \\
n &= 1: && 15^2 = 225 &&\text{since $1 \times 2 = 2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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About the sequence: $1, 3, 8, 24, 29, 87, 92, ?...$ find $a_n$ This is a classic and curious recurrence sequence used in logic tests. Your rule can be determined as follows:
for $n$ even: $a_n = 3a_{n-1}$
for $n$ odd: $a_n = a_{n-1} + 5$
That is, the ratio alternates with each term.
Will there be a single formula for $... | A bit of a sneaky answer. The subsequence of only odd terms, $1,8,29,92,\ldots$, is listed in the OEIS as sequence A116952, $b_n=3b_{n-1} + 5$ with $b_0=1$ and has the explicit formula $b_n=\frac723^n-\frac52$. So then, the even terms satisfy $c_n=3b_n=3\left(\frac723^n-\frac52\right)$. Simple indicator functions for o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Deriving $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ using Fourier Series for $f(x)=|x|$ One can find that $$f(x) = \frac{\pi}{2} + \sum_{n=1}^{\infty}\frac{2(-1+(-1)^n)}{\pi n^2}\cos(nx)$$
Now look at the case for $x=0$. We can find that $\sum_{n \text{ odd}, n \geq 1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}$.... | \begin{align}
& \text{sum of even terms} = \sum_{n=1}^\infty \frac 1 {(2n)^2} \\[8pt]
= {} & \frac 1 4 \sum_{n=1}^\infty \frac 1 {n^2} = \left( \frac 1 4 \times\big(\text{sum of all terms} \big)\right).
\end{align}
If the sum of the even terms is $1/4$ times the sum of all of the terms, then the sum of the odd terms is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Consider the linear function$ f : \mathbb{R}^{2\times2}\to \mathbb{R}^{3\times2}$ defined as follows: Consider the linear function $f : \mathbb{R}^{2\times2} → \mathbb{R}^{3\times2}$ defined as follows:
$$\begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \\ \end{pmatrix} \in \mathbb{R}^{2\times2}\mapsto f\begin{pmatrix} r_1 & r... | The rank of a matrix is the number of non empty rows once it is une the reduced echelon form. Your matrix could be reduced by doing the following operations $3L_2+2L_1$ then $7L_3-4L_2$.
$$\begin{pmatrix}3&2\\ -2&1\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&4\end{pmatrix}\sim\begin{pmatrix}3&2\\ 0&7\\ 0&0\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Proof that both eigenvalues lie inside the unit circle iff $\det < 1$ and $\operatorname{tr }< 2$ Let $A = \begin{pmatrix}a&b \\c &d \end{pmatrix}$, prove that both eigenvalues lie within the unit circle if and only if $\det(A) < 1$ and $|\operatorname{tr}(A)| < 1 +\det(A)$.
This is what I have done so far:
$|\operato... | Hint:
Assume $x,y \in \mathbb{R}$.
We have $0 \le |x+y| < 1+xy$ which implies that actually $ -1 \le xy < 1$, or $x^2y^2 \le 1$.
Squaring the relation $|x+y| < 1+xy$ gives
$$x^2+y^2 < 1+x^2y^2 \implies (1-x^2)(1-y^2) > 0$$
so $x^2, y^2 > 1$ or $x^2, y^2 < 1$. However, the first possibility contradicts $x^2y^2 \le 1$.
| {
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"source": "stackexchange",
"question_score": "2",
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Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
... | what you need to show after applying the inductive method is that:
$\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}+(k+1)^2=\frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{k+1}{6}$
this expands to:
$2k^3+3k^2+k+6k^2+12k+6=2k^3+6k^2+6k+2+3k^2+6k+3+k+1$
after simplifying:
$2k^3+9k^2+13k+6=2k^3+9k^2+13k+6$
left side = right side.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\... | You have some mistakes in $f(t)$. Correct is, it should be:
$$f(t) = \frac{(t-1)^3}{t^3-3t\color{red}{-1}}$$
and it's important to notice that $|t| \geq 2$ for any real $x$. So, we have to minimize $f:(-\infty,-2] \cup [2,\infty) \to \mathbb{R}$ with
$$f(t) = \frac{(t-1)^3}{t^3-3t-1}$$
We have:
$$f'(t) = \frac{3 (t-1)^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
... | You did right, but gave up soon. For
$$\int\frac{x-1}{x^2+x+1}$$
write
$$\frac{x-1}{x^2+x+1}=\frac{1}{2}\frac{2x-2}{x^2+x+1}=\frac{1}{2}\frac{2x+1}{x^2+x+1}
+\frac{1}{2}\frac{-3}{x^2+x+1}$$
then the integral becomes
$$\int\frac{x-1}{x^2+x+1}dx=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{dx}{x^2+x+1}$$
and then...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(... | I'm not sure how you got
Then,
6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]
So, also
6 | (a+b+c)^3
6 | (a+b+c)(a+b+c)(a+b+c)
As Iris's question comment says, because $6$ divides the first expression of $(a+b+c)^3 - 3(a+b)(b+c)(c+a)$ doesn't mean it divides the first term, i.e., $(a+b+c)^3$ and then a different expression of $(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Direct formula for sequence I want to find out the function to generate the sequence with the following pattern.
1 2 3 1 1 2 2 3 3 1 1 1 1 2 2 2 2 3 3 3 3 ....
Where the lower bound number is 1 upper number bound number is 3. Each time numbers start from 1 and each number repeats 2 ^ n times, with n starting with 0.
| The following PARI/GP code will do what you want
a(n) = (n-3*2^(k=exponent((n-1)\3+1))+2)\2^k+1;
Note that exponent(n) is the binary exponent of $n$. In other words, the floor of the base $2$ logarithm of $n$. Also, n\m is the floor of $n/m$. Here is an example of use of the code:
? vector(45,n,a(n))
[1,2,3,1,1,2,2,3,... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limit... | HINT:
Use the inequality
$$\left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1}$$
so
$$e^{\frac{1}{n+1}}< \left(1+\frac{1}{n}\right)< e^{\frac{1}{n}}$$
We get lower and upper bounds:
$$e^{\frac{n^4}{n^3+1}}< \left(1+\frac{1}{n^3}\right)^{n^4}< e^n $$
$$e^{\frac{(n+1)^4}{(n+1)^3 +1}}< \left(1 + \frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 0
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Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63}$. Why is this assumption necessary? Question:
Show that if $\gcd(a,3)=1$ then $a^7 \equiv a\pmod{63} $. Why is this assumption necessary?
Proof:
Since $\gcd(a,3)=1$ $\Leftrightarrow a\equiv 1\pmod 3$ $\Leftrightarrow a^7\equiv 1\pmod3\equiv a\pmod3$
Then using Fer... | $a^7-a=a\left(a^6-1\right)=a(a^2-1)(a^4+a^2+1)=\underbrace{(a-1)}_{\equiv 0\pmod{3}}a(a+1)(a^4+a^2+1)$ is clearly always divisible by $3$ because it's a product of $3$ consecutive integers.
$(\forall a\in\mathbb Z)$
$a^7\equiv a\pmod{7}$
$$a^7-a\equiv 0\pmod{3}\;\land\;a^7-a\equiv 0\pmod{7}\implies a^7-a\equiv 0\pmod{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{1}{(p+q)^3}\left(\frac{1}{p^3}+\frac{1}{q^3}\right)+...$
Simplify $\dfrac{1}{(p+q)^3}\big(\dfrac{1}{p^3}+\dfrac{1}{q^3}\big)+\dfrac{3}{(p+q)^4}\big(\dfrac{1}{p^2}+\dfrac{1}{q^2}\big)+\dfrac{6}{(p+q)^5}\big(\dfrac{1}{p}+\dfrac{1}{q}\big)$ if $p\ne -q, p\ne0$ and $q\ne 0$.
$\dfrac{1}{(p+q)^3}\big(\dfrac... | $$
=\dfrac{(p+q)^2}{(p+q)^5} \times \dfrac{p^3 + q^3}{p^3 q^3} + \dfrac{3(p+q)}{(p+q)^5} \times \dfrac{(p^2+q^2)pq}{p^3q^3} + \dfrac{6}{(p+q)^5} \times \dfrac{(p+q)p^2q^2}{p^3q^3}
$$
$$
=\dfrac{(p+q)^2(p^3+q^3)+3(p+q)(p^2+q^2)pq+6(p+q)p^2q^2}{(p+q)^5p^3q^3}
$$
$$
=\dfrac{(p+q)(p^3+q^3)+3(p^2+q^2)pq+6p^2q^2}{(p+q)^4p^3q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to know the most efficient way of finding the rank of matrix and how do I apply it to this example For the matrix $
\begin{bmatrix} 1 & 2 & -3 \\ 1 & -2 & 3 \\ 4 & 8 & -12 \\ 1 & -1 & 5 \end{bmatrix}
$
The reduced row echelon I obtained is $
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{b... | Doing Gaussian Elimination, you get
$$
\begin{split}
\begin{bmatrix}
1 & 2 & -3 \\
1 & -2 & 3 \\
4 & 8 & -12 \\
1 & -1 & 5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & -3 \\
0 & 4 & -6 \\
0 & 0 & 0 \\
0 & -3 & 8
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & -3 \\
0 & 1 & -3/2 \\
0 & 0 & 0 \\
0 & -3 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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finding a formula for the generating function for a recurring sequence I have the sequence
$a_0=0$, $a_1=3$, $a_2=0$, $a_3=23$,
and
$a_n=6a_{n-2} + 8a_{n-3} + 3a_{n-4}$ for $n\ge 4$
and I have to find the formula for the generating function $A(t)=\sum_{n=0}^\infty a_n t^n$ and find a formula for $a_n$.
So far I h... | The recurrence relation and initial conditions imply
$$A(t)-0 t^0-3 t^1-0 t^2-23 t^3=6t^2 (A(t)-3t^1) + 8t^3 A(t) + 3t^4 A(t).$$
Solving for $A(t)$ yields
$$A(t)=\frac{3t+5t^3}{1-6t^2-8t^3-3t^4}=\frac{1/2}{1-3 t} - \frac{3/2}{1+t} + \frac{3}{(1+t)^2} - \frac{2}{(1+t)^3},$$
which immediately implies that
$$a_n=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x... | You got all the derivatives right, with $f(0) = 3, $$f'(0) = -2$, $f''(0) = \dfrac{3}{2}$, $f'''(0) = -\dfrac{5}{4}$, and so on. As a side note, keep in mind you're evaluating $f^{(n)}$ at $x = 0$, so you saying $f'(x) = -2$, $f''(x) = \dfrac{3}{2}$, and so on isn't correct. (Note that by "derivatives," I'm referring t... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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can the number $x^2 +y^2$ when with x and y positive integers
can the number $x^2 +y^2$ when with $x$ and $y$ positive integers,
end in $03$?
I know that $x^2+y^2$ can never end with unit digit 03 but am not sure how would I show the proof of that.
| Let $x = 10a \pm b$ and $y = 10c \pm d$ where $b,d = 0,1,2,3,4$ or $5$.
So $x^2 + y^2 = 100(a^2 + c^2) + 20(\pm ab \pm cd) + (b^2 + d^2)$.
The only way for $b^2 +d^2$ to end in $3$ is if $b=2$ and $d=3$ (or vice versa)
So $x^2 + y^2 =100(a^2 + c^2) + 20(\pm 2a \pm 3c) + 13$
And the only we for that to end in $03$ is fo... | {
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"source": "stackexchange",
"question_score": "2",
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Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$... | To prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 \ge \frac{2(a^2+ab+b^2)}{a+b}$$
we can use AM-GM:
$$
\begin{aligned}
a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 &\geq 2\sqrt{a^2+b^2+\frac{a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+b^2)^2+2(a^2+b^2)ab+a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+ab+b^2)^2}{(a+b)^2}}\\
&=2\left|\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Taylor approximation is different from the actual value The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$.
I found that the $3$rd Taylor polynomial with remainder term is \begin{align*}
x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos... | This just means that teh third order is not sufficient in particular because $-1.09$ is very far away from $-2$.
To make the story short, what you have is
$$x^2 \cos(x)=\sum_{n=0}^\infty \frac{\left(-n^2+n+4\right) \cos \left(2-\frac{\pi n}{2}\right)+4 n \sin \left(2-\frac{\pi n}{2}\right)}{n!}(x+2)^n$$ So, if you ... | {
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"source": "stackexchange",
"question_score": "2",
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Find the value of C for which the following integral converges $$I=\int_0^\infty \frac {x} {x^2+1} - \frac {C} {3x+1} dx$$
Find the value of C for which the integral converges.
I've gotten as far as solving the integral
$\frac 1 2 \ln|x^2+1| - \frac 1 3 c\ln|3x+1|$ but not sure where to go from here as evaluating it ... | You can do this without finding the actual integral. The basic question is what is the behaviour of the integrand as $x \to \infty$. We have $$ \frac{x}{x^2+1} = \frac{1}{x (1+1/x^2)} = \frac{1}{x} \left(1 - \frac{1}{x^2} + \ldots \right) = \frac{1}{x} + O\left(\frac{1}{x^3}\right)\ \text{as}\ x \to \infty$$
while
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Prove that if $a,b,c > 0$ and $a + b + c = 1$, we have: $\frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$ Prove that if $a,b,c > 0$ such that $a + b + c = 1$, then the following inequality holds:
$$S = \frac{a^2}{a^3 + 5} + \frac{b^2}{b^3 + 5} + \frac{c^2}{c^3 + 5} \leq \frac{1}{4}$$
Wh... | Based on Michael Rozenberg's idea, we also have:
$$\frac{a^2}{a^3+5}=\frac{a}{6}-\frac{a(1-a)(5-a^2-a)}{a^3+5}\leq \frac{a}{6}$$
so that:
$$\frac{a^2}{a^3+5}+\frac{b^2}{b^3+5}+\frac{c^2}{c^3+5}\leq \frac{1}{6}(a+b+c)=\frac{1}{6}$$
Equality is attained when $(a,b,c)=(1,0,0)$ up to any cyclic permutation.
We can use the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$ If $a,b,c>0$, prove that:
$$\sqrt[3]{\frac{(a^4+b^4)(a^4+c^4)(b^4+c^4)}{abc}}\ge \sqrt{\frac{(a^3+bc)(b^3+ca)(c^3+ab)}{1+abc}}$$
My try: I used $a^4+b^4 \geq ab(a^2+b^2)$ to get:
$$(a^4+b^4)(a^4+c^4)(b^4+c^4) \geq... | It's wrong.
Let $a=b=c.$
Thus, we have $$2a^2\geq\sqrt{\frac{(a^3+a^2)^3}{1+a^3}}$$ or
$$2\geq a\sqrt{\frac{(a+1)^2}{a^2-a+1}},$$ which is wrong for $a\rightarrow+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569404",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\... | By AM-GM
$$0=\sum_{cyc}\log_{yz}\frac{x^2+4}{4\sqrt{yz}}\geq\sum_{cyc}\log_{yz}\frac{2\sqrt{x^2\cdot4}}{4\sqrt{yz}}=\sum_{cyc}\left(\log_{yz}x-\frac{1}{2}\right)=$$
$$=\sum_{cyc}\frac{2\ln{x}-\ln{y}-\ln{z}}{2(\ln{y}+\ln{z})}=\frac{1}{2}\sum_{cyc}\frac{\ln{x}-\ln{y}-(\ln{z}-\ln{x})}{\ln{y}+\ln{z}}=$$
$$=\frac{1}{2}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the relationship between and (complex numbers) $ z = x + \text{j} \cdot y $ and $ \arg(z) = \frac{\pi}{4} $. Find the relationship between $x$ any $y$, then find $z$ for
$|z - 3 + 2 \cdot \text{j}| = |z +3 \cdot \text{j}|$, where $\text{j}^2= -1$,
In the above question do I have to compare like this:
$Z= x+\t... | Well, when we know that the argument of a complex number is $\frac{\pi}{4}$ we can write:
$$\arg\left(\text{a}+\text{b}i\right)=\arctan\left(\frac{\text{b}}{\text{a}}\right)=\frac{\pi}{4}\space\Longleftrightarrow\space\frac{\text{b}}{\text{a}}=\tan\left(\frac{\pi}{4}\right)=1\space\Longleftrightarrow\space\text{a}=\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
| Let $m$ and $n$ be integers such that $m^2=n^6+n^4+1$. Without loss of generality, we may assume that $m$ and $n$ are nonnegative. Clearly, $(m,n)=(1,0)$ is the only solution when $n\in\{0,1\}$.
If $n\ge 2$, then $n^2\geq 4$, so that $$(2n^3+n)^2=4n^6+4n^4+n^2\geq 4n^6+4n^4+4=4m^2\,.$$ On the other hand, $$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$ Problem: Prove $\sum_{n=1}^{\infty}\frac{n-1}{2^{n+1}}=\frac{1}{2}$, without using any $\sum\frac{n}{2^{n}}$.
I manage to make progress, but then end up always getting a $\sum\frac{n}{2^{n}}$. For example, $\sum_{n=1}^{\... | Here is a fairly general approach. Note that $n-1=\sum_{m=1}^{n-1}(1)$ so that
$$\begin{align}
\sum_{n=1}^\infty (n-1)a_n &=\sum_{n=2}^\infty (n-1)a_n\\\\
&=\sum_{n=2}^\infty \sum_{m=1}^{n-1} a_n\\\\
&=\sum_{m=1}^\infty \sum_{n=m+1}^\infty a_n\tag1
\end{align}$$
Now, take $a_n=x^n$, $|x|<1$ in $(1)$ and using $\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For non-negative reals $a$, $b$, $c$, show that $3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$ A 11th grade inequality problem:
Let $a,b,c$ be non-negative real numbers. Prove that
$$3(1-a+a^2)(1-b+b^2)(1-c+c^2)\ge(1+abc+a^2b^2c^2)$$
Do you have any hints to solve this inequality? Any hints would be fine.
I t... | Another way.
Since $$3(1-a+a^2)^3-1-a^3-a^6=(a-1)^4(2a^2-a+2),$$ by Holder we obtain:
$$3\prod_{cyc}(1-a+a^2)\geq\sqrt[3]{\prod_{cyc}(1+a^3+a^6)}\geq1+abc+a^2b^2c^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Looking for a short proof of a harmless looking binomial identity I managed to prove for this MSE post the rather harmless looking binomial identity for natural $1\leq k\leq n$:
\begin{align*}
\color{blue}{\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}=\binom{n+k}{n-k}\frac{4^kn}{n+k}}\tag{1}
\end{align*}
using the coe... | Here is an alternate solution, where the number of steps is about
the same as what OP provided. Could use additional streamlining by
removing some of the simpler proceedings. Start as follows:
$$\sum_{j=0}^k {2n\choose 2j} {n-j\choose k-j}
= \sum_{j=0}^k {2n\choose 2k-2j} {n-k+j\choose j}
\\ = [z^{2k}] (1+z)^{2n} \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this... | Amazing : the tangent half angle substitution works. Let $x=2\tan^{-1}(t)$ to get
$$I=-4\int\frac{ t \left(t^2-1\right)^5 \left(t^2+1\right)}{\left(t^8-4 t^6+22 t^4-4
t^2+1\right)^2}\,dt$$ Now, a non-trivial substitution
$$t=\sqrt{1+\frac{2 \left(\sqrt{z+1}\right)}{z}}\implies dt=-\frac{1}{2} \sqrt{\frac{z+ 2 \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My... | Let's denote $f(x)=ax^2+3x−4$, $g(x)=-4x^2+3x+a$, and $h(x)=f(x)/g(x)$.
Since $a>0>-9/16>-4$, $f(x)$ and $g(x)$ cannot have the same roots, and each of them has two distinct roots. However, they may have a common root. If $a=1,7$ then there exists $y_0\in \mathbb{\mathbb{R}}$ for which only one $x$ may give $h(x)=y_0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum) I'm looking for a specific clarification for a part of the solution in proving the following identity.
$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$
Here I'm taking, $\theta=\tan^{-1}m;$ $-\pi/2<\theta<\p... | Using this
We need $\tan^{-1}m+\tan^{-1}n\ge0$ to admit the equality
$$\tan^{-1}m+\tan^{-1}n\ge0\iff\tan^{-1}m>-\tan^{-1}n=\tan^{-1}(-n)$$
$$\iff m\ge-n\iff m+n\ge0$$
Now use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the following system of equations by Cross Multiplication Method Solve the following System of equations by Cross Multiplication Method
$$2x-2y+3z=20$$
$$x+3y-z=12$$
$$3x-y+4z=22$$
My Attempt:
Let us solve first two equations:
$$\dfrac {x}{2-9}=\dfrac {y}{3-(-2)}=\dfrac {z}{6-(-2)} = k(let)$$
$$\dfrac {x}{-7}=\df... | We want to use the Cross Multiplication Method (CMM) to solve
$$\begin{align} 2x-2y+3z&=20 \tag 1 \\ x+3y-z&=12 \tag 2 \\ 3x-y+4z&=22 \tag 3 \end{align}$$
We can write $(1)$ and $(2)$ as
$$\begin{align} 2x-2y+ (3z - 20)&=0 \\ x+3y-(z+12)&=0 \end{align}$$
By the CMM
$$\dfrac{x}{-2(-z-12)-3(3z - 20)} = \dfrac{y}{(3z-20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\sqrt{4x^2+8x}dx$ using $\int\sqrt{u^2-a^2}dx=\frac u2\sqrt{u^2-a^2}-\frac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+C.$ I completed the square to get $$2 \sqrt{(x+1)^2 - 1}$$
set $u=x+1$ and $a=1$, and you end up with
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
This is not the ... | Your answer is equivalent to the answer provided by Symoblab
$$x\sqrt{x^2+2x}-\ln\left\lvert x + \sqrt{x^2+2x}+1\right\rvert+\sqrt{x^2+2x}+C$$
since starting from
$$(x+1) \sqrt{(x+1)^2 - 1} - \ln\left\lvert x+1 + \sqrt{(x+1)^2-1}\right\rvert+C$$
we have that
$$\sqrt{(x+1)^2 - 1}=\sqrt{x^2+2x}$$
therefore
$$(x+1) \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$ What should be the proper way to answer this question?
My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?
$x+y=q \tag{1}$
$x^2-2x+2y^2=3 \... | Your calculations are correct. Here you can find also for what values of $q$ the line $x+y=q$ and the ellpise $x^2-2x+2y^2=3$ have two points in common.
First, you can rewrite the equation of the ellipse in the form:
$$(x-1)^2+2y^2=4$$
Now, substituing $y=q-x$, you have:
$$(x-1)^2+2(q-x)^2=4\leftrightarrow 3x^2-2x(2q+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$ Solve for $x$: $\log_{\frac {3}{4}}(\log_{8} (x^2+7))+\log_{\frac {1}{2}} (\log_{\frac {1}{4}} (x^2+7)^{-1})=-2$
My Attempt:
$$\log_{\frac {3}{4}}(\frac {1}{3} \log_2 (x^2+7))+\log_{\frac {1}{2}} (\frac {1}... | Let's use
\begin{eqnarray*}
\log_b(a) = \frac{\ln(a)}{\ln(b)}
\end{eqnarray*}
to change everything into natural logarithms.
Your equation becomes
\begin{eqnarray*}
\frac{\ln \left( \frac{\ln(x^2+7)}{\ln(8)} \right) }{\ln(3/4)} + \frac{\ln \left( - \frac{\ln(x^2+7)}{\ln(1/4)} \right) }{\ln(1/2)} =-2 \\
\end{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(... | The hint:
Prove that $$-(|x|+|y|)\leq\frac{x^2-y^2}{\sqrt{x^2+y^2}}\leq|x|+|y|.$$
The right inequality.
We need to prove that:
$$\frac{(|x|-|y|)(|x|+|y|)}{\sqrt{x^2+y^2}}\leq|x|+|y|$$ or $$ |x|-|y|\leq\sqrt{x^2+y^2}.$$
If $|x|-|y|\leq0$ it's obvious.
But for $|x|-|y|\geq0$ it's enough to prove that
$$(|x|-|y|)^2\leq x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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Fractions in Questions and Answers
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