Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solve $\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $
Solve $$\left(\sqrt{\sqrt{x^2-5x+8}+\sqrt{x^2-5x+6}} \right)^x + \left(\sqrt{\sqrt{x^2-5x+8}-\sqrt{x^2-5x+6}} \right)^x = 2^{\frac{x+4}{4}} $$
Preface; I think there should be... | HINT:
Replace $\sqrt u-\sqrt{u-2}=\dfrac2{\sqrt u+\sqrt{u-2}}$ to form a Quadratic equation in $$\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}$$ which is
$$\left(\left(\dfrac{\sqrt2}{\sqrt u+\sqrt{u-2}}\right)^{x/2}-1\right)^2=0$$
| {
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"url": "https://math.stackexchange.com/questions/2068147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$ I have tried factoring it already, but it doesn't seem to evolve much:
First I multiply each side by $2$:
$ 2(a^2+b^2)>2ab$
Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:
$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$
and then:
$3(a^2+b^2)>(a+b)^2$
A... | We have that
$$0\le (a-b)^2=a^2+b^2-2ab.$$ Thus we have
$$2ab\le a^2+b^2.$$
Now, if $ab$ is positive then
$$ab< 2ab\le a^2+b^2.$$ And if $ab$ is negative then
$$ab< 0 <a^2+b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Special Orthogonal Group $SO(2)$ The special orthogonal group for $n=2$ is defined as:
$$SO(2)=\big\{A\in O(2):\det A=1\big\}$$
I am trying to prove that if $A\in SO(2)$ then:
$$A=\left(\begin{array}{cc}
\cos\theta& -\sin\theta\\
\sin\theta&\cos\theta
\end{array}\right)$$
My idea is show that $\Phi:S^1\to SO(2)$ define... | Let $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)\in \mathrm{SO}(2)$. Then,
$$\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
and
$$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc=1.$$
Th... | {
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Which of these quantity systems corresponds to sigma algebra? Given is the set $Ω = \left\{ 3,4,5,6,7\right\} $.
Determine which of the given quantity systems corresponds to an sigma algebra and justify it.
$1. \left\{\left\{\right\},\left\{3\right\},\left\{5\right\},\left\{3,4\right\},\left\{3,4,6,7\right\},\left\{... | *
*is not a $\sigma$-algebra because $\{3\}\cup \{5\}=\{3,5\}$ is not in 1.
*is a $\sigma$-algebra because $\{\}$, $\Omega$, and all the possible intersections and unions of the elements of 2. are in 2.
*works because all the $2^{\Omega}=32$ subsets (of $\Omega $) are in 3.
*would not work because $\Omega$ is not i... | {
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"timestamp": "2023-03-29T00:00:00",
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Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's T... | Your function is
$$\cos(x)-\frac{a}{b}-\frac{1-\frac{a}{b} }{ bx^2+1 }$$
$$=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{a}{b}-1+\frac{a}{b}+bx^2-ax^2-b(b-a)x^4+x^4\epsilon(x).$$
thus, we need
$b-a=-\frac{1}{2}$ and $b(b-a)=-\frac{1}{24}$.
which gives
$b=\frac{1}{12}$ and $a=\frac{7}{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate expected value for piecewise constant distribution function? The distribution function of a discrete random variable X is given
$F_X(x)=\begin{cases} 0, &x<1\\ \frac{5}{13},& 1\leq x< 2 \\ \frac{10}{13}, & 2\leq x<3 \\ \frac{11}{13}, & 3\leq x<4 \\ 1, & 4\leq x \end{cases} $
$A=(X=2)\cup (X=4)$
Cal... | the random variable $X$ can take four values, which are exactly the points of discontinuity of $F_X$:
$$
\mathbb P (X=1)= \frac 5 {13}, \quad \mathbb P (X=2)=\frac {10} {13}- \frac 5 {13}, \quad
\mathbb P (X=3)=\frac {11} {13} -\frac {10} {13}, \quad \mathbb P (X=4)=1- \frac {11} {13}. \quad
$$
Therefore
$$
\mathbb P(A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix determinant lemma derivation While reading this wikipedia article on the determinant lemma, I stumbled upon this expression (in a proof section):
\begin{equation}
\begin{pmatrix} \mathbf{I} & 0 \\ \mathbf{v}^\mathrm{T} & 1 \end{pmatrix}
\begin{pmatrix} \mathbf{I}+\mathbf{uv}^\mathrm{T} & \mathbf{u} \\ 0 & 1 \end... | I have no idea how this was invented and what was the original motivation, but let me outline a different proof for the formula $\det(I + uv^T) = 1 + v^T u$ which I think is much less magical and more natural.
Set
$$ B = uv^T = \begin{pmatrix} u_1 v_1 & \dots & u_1 v_n \\
u_2 v_1 & \dots & u_2 v_n \\
\vdots & \ddots &... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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The valid interval of the maclaurin series for $\frac{1}{1+x^2}$ The Maclaurin series for $\frac{1}{1-x}$ is $1 + x + x^2 + \ldots$ for $-1 < x < 1$.
To find the Maclaurin series for $\frac{1}{1+x^2}$, I replace $x$ by $-x^2$.
The Maclaurin series for $\frac{1}{1+x^2} = 1 - x^2 + x^4 - \ldots$.
This is valid for $-1 <... | Taking square roots in $a<x^2<b$ to get $\sqrt a < |x| < \sqrt b$ is valid if $a\ge0$ and $b\ge0$.
But notice that the solution of $-4< x^2$ is $-\infty<x<\infty$ since every square of a real number is $>-4.$
Since the square of a real number is never negative, the inequality $-1<x^2<1$ is equivalent to $x^2<1.$
That i... | {
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"timestamp": "2023-03-29T00:00:00",
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Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2... | Such coefficient is clearly
$$ 2n+\sum_{k=1}^{n-1} k(n-k) = \frac{n(n^2+11)}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$ proving $t^6-t^5+t^4-t^3+t^2-t+0.4>0$ for all real $t$
for $t\leq 1,$ left side expression is $>0$
for $t\geq 1,$ left side expression $t^5(t-1)+t^3(t-1)+t(t-1)+0.4$ is $>0$
i wan,t be able to prove for $0<t<1,$ could some help me with this
| If $0< t < 1$
$\frac {t^7 + 1}{t+1} \ge .6 \iff t^7 + 1 \ge .6t + .6 \iff t^7 - .6t \ge -.4$
$\frac {d(t^7 - .6t)}{dt} = 7t^6 - .6 = 0$ if $t = \sqrt[6] \frac 6{70}$
$\frac{d^2(t^7 - .6t)}{d^2t} = 42t^5 > 0 $ if $t > 0$ so $t = \sqrt[6] \frac 6{70}$ is a minimum value of $t^7 - .6t$. And so $t^7 - .6t \ge \sqrt[6] \f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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proving $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{n-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}$ proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdo... | Extended HINT: The result is incorrect as originally stated; it should read
$$\binom{n}0-\binom{n}1+\binom{n}2-\ldots+(-1)^{\color{crimson}m-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}\;,$$
or, more compactly,
$$\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^{m-1}\binom{n-1}{m-1}\;.\tag{1}$$
Fix $n\in\Bbb N$. For $m=1$ the desi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate Limit of $(1-\sin x)^{(\tan \frac{x}{2} -1)}$ when $x\to \frac{\pi}{2}$.
How to calculate Limit of $(1-\sin x)^{(\tan \frac{x}{2} -1)}$ when $x\to \frac{\pi}{2}$.
We can write our limit as $\lim_{x\to \frac{\pi}{2}}e^{(\tan \frac{x}{2} -1) \log(1-\sin x)}~ $ but I can not use L'Hopital rule.
Is ther... | Making the substitution $
x = \dfrac{\pi}{2} + y$ the required limit is
$\lim_{y \to 0} \exp h(y)$ where $h(y)= \ln(1-\cos y) \left( \tan(\pi/4 + y/2) - 1 \right) = \ln(1-\cos y) \times \dfrac{2\tan(y/2)}{1-\tan(y/2)}$.
Since $1-\cos y = 2 \sin^2(y/2)$ we have $$h(y) = (\sqrt{2}\sin(y/2)) \ln(2\sin^2(y/2)) \times \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$ Is it possible to know if $4-\sqrt{2}-\sqrt[3]{3}-\sqrt[5]{5} \gt 0$ without using decimal numbers?
| It is not hard to verify following inequalities (just power both sides and it should result into simple inequalities in natural numbers only):
\begin{align}
\frac{4}{3} &< \sqrt{2} < \frac{5}{3}\\
\frac{4}{3} &< \sqrt[3]{3} < \frac{5}{3}\\
\frac{4}{3} &< \sqrt[5]{5} < \frac{5}{3}\\
\end{align}
Summing these up will giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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Consider a function $f(x, y) = Ax^5 + Bx^4y + Cx^3y^2 + Dx^2y^3 + xy^4 − y^5$
Consider a function $f(x, y) = Ax^5 + Bx^4y + Cx^3y^2 + Dx^2y^3 + xy^4 − y^5$
where $A$, $B$, $C$, $D$ are unspecified real numbers. Determine the values of
$A$, $B$, $C$, $D$ such that $f(x,y)$ satisfies $f_{xx}(x,y) + f_{yy}(x,y) = 0$
Wha... | Add up and you get
$[20A+2C]x^3 + [12B+6D]x^2y + [6C+12]xy^2 + [2D-20]y^3=0$
So
$D = 10$
$C = -2$
$B = -5$
$A = \frac{1}{5}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.
| \begin{align}
I:&=\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dz\left(x^2-2xy\right)\exp\left(-3x^2 - 2y^2 - z^2 - 2xy\right)\\
&=\sqrt{\pi}\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}dy\left(x^2-2xy\right)\exp\left(-3x^2-2y^2-2xy\right) \tag1\\
\end{align}
We split the integral in $(1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$
Does this series converge? ($p$ is finite number greater than $1$)
| For $p>1$ the series converges.
Lets take the general geometric series;
$$|x|<1\\a_1(1+x+x^2+x^3+...)=\frac{a_1}{1-x}\\a_1(1+2x+3x^2+4x^3+...)=a_1(1+x+x^2+x^3+...)'=(\frac{a_1}{1-x})'=\frac{a_1}{(1-x)^2}$$
So in our case;
$$a_1=\frac{1}{p}\ \ ,\ \ x=\frac{1}{p}\\\frac{1}{p}+\frac{2}{p^2}+...=\frac{1}{p}\cdot\frac{1}{(1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute the $n$-th power of triangular $3\times3$ matrix I have the following matrix
$$
\begin{bmatrix}
1 & 2 & 3\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}
$$
and I am asked to compute its $n$-th power (to express each element as a function of $n$). I don't know at all what to do. I tried to compute some values manually to... | Here is another variation based upon walks in graphs.
We interpret the matrix $A=(a_{i,j})_{1\leq i,j\leq 3}$ with
\begin{align*}
A=
\begin{pmatrix}
1 & 2 & 3\\
\color{grey}{0} & 1 & 2\\
\color{grey}{0} & \color{grey}{0} & 1
\end{pmatrix}
\end{align*}
as adjacency matrix of a graph with three nodes $P_1,P_2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 6,
"answer_id": 5
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Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large n... | What you are asking to prove is incorrect, I believe.
By the substitution, we have that
$$\frac{x}{3}=\sec(\theta)\Leftrightarrow\frac{3}{x}=\cos(\theta).$$
By the Pythagorean identity,
$$\sin(\theta)=\sqrt{1-\frac{9}{x^2}}=\sqrt{\frac{x^2-9}{x^2}}.$$
Therefore,
$$ \tan(\theta)=\sqrt{\frac{x^2-9}{x^2}}\frac{x}{3}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90... | Claim 1 is vaccuously true, since given any $n$, $a_{n}\in\mathbb{Z}$.
It also looks like n=14 is a counterexample to claim 2.
Proof:
$3-14=-11$ and $2(-11)+5=-17$ has positive divisors $1$ and $17$ only.
| {
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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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Using the Arithmetic Mean-Geometric Mean Inequality Let a, b, c be positive real numbers. Prove that
$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq \sqrt{3(a^{2}+b^{2}+c^{2})}$
| Rewrite as $a^2b^2 + b^2c^2 + c^2a^2 \ge \sqrt{3(a^4b^2c^2 + a^2b^4c^2 + a^2b^2c^4)}$
Let $x = a^2b^2, y = b^2c^2, z = c^2a^2$.
Therefore, now we have to prove : $x + y + z \ge \sqrt{3(xy + yz + zx)}$
Squaring both sides, $(x + y + z)^2 \ge 3(xy + yz + zx)$
$x^2 + y^2 + z^2 \ge xy + yz + zx$.
Now, use A.M. $\ge$ G.M. $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using the Arithmetic Mean-Geometric Mean Inequality. Let a, b, c be real numbers. Prove that:
$(a+b-c)^{2}(b+c-a)^{2}(c+a-b)^{2}\geq(a^{2}+b^{2}-c^{2})(b^{2}+c^{2}-a^{2})(c^{2}+a^{2}-b^{2}) $
| We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$.
If $a^2+b^2-c^2<0$ and $a^2+c^2-b^2<0$ so $2a^2<0$, which is contradiction.
Thus, we can assume $a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $b^2+c^2-a^2\geq0$.
Since $(a+b-c)^2(a+c-b)^2-(a^2+b^2-c^2)(a^2+c^2-b^2)=2(b-c)^2(b^2+c^2-a^2)\geq0$,
we obtain:
$$\prod\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality with a+b+c=1 Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$
Progress: This is equivalent to show that
$$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$
I'm not sure how to proceed further. Also e... | Notice that the conditions imply that $0\leq a,b,c\leq 1$; in particular, we have
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{bc}{5}+\dfrac{ca}{5}+\dfrac{ab}{5} \leq \frac{b+c+a}{5} < \dfrac{1}{4}.$$
I don't think this works for the general case though, because for $n\geq 4$, we have $(n-1)^2\geq n+2$.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{... | First of all, you are right that there is trouble in multiplying by $\frac{4-x}{4-x}$ when $x=4$. But why bother with the domain $\langle-4,4]$ in the first place? You can change the integrand at one point without changing integral, so that one point is irrelevant. Thus, choose $D_g = \langle-4,4\rangle$.
The only oth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find all such $n$ that $a+b+c+d=0\implies a^7+b^7+c^7+d^7=0$ in $\mathbb{Z}/n\mathbb{Z}$
The problem is to specify all such $n>1$ that for any $a,b,c,d\in\mathbb{Z}/n\mathbb{Z}$ the following implication stands:
$$a+b+c+d=0\implies a^7+b^7+c^7+d^7=0.$$
One can note that when $n=7$ we have $(a+b+c+d)^7=a^7+b^7+c^7+d... | The examples $(2,-1,-1,0)$ and $(3,-1,-1,-1)$ show that you need $2^7=2$ and $3^7=3$. This means that $n$ must divide both $2^7-2$ and $3^7-3$, and their greatest common divisor is $42$.
It turns out that $k^7-k$ is divisible by $42$ for every natural $k$ (divisible by $7$ by Fermat's little theorem, by $2$ and $3$ by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Power series expansion of $\frac{z}{(z^3+1)^2}$ around $z=1$ I want to expand $f(z)=\frac{z}{(z^3+1)^2}$ around $z=1$. That is, I want to find the coefficients $c_n$ such that $f(z) = \sum_{n=0}^\infty c_n (z-1)^n$.
So far, my first strategy was using long division after I expanded both the numerator and the denominato... | Notice that
$$\frac z{(1+z^3)^2}=-z\frac{\partial}{\partial z^3}\frac1{1+z^3}=-z\frac\partial{\partial z^3}\sum_{n=0}^\infty(-1)^n(z^3)^n=\sum_{n=0}^\infty n(-1)^{n+1}z^{3n-2}$$
This expansion works for $|z|<1$, and since we know the expansion exists at $z=1$, we may apply the method described in this answer to get the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Find the numbers in A.P. those sum is $24$ and product is $440$
If the sum of three numbers in A.P. is $24$ and their product is $440$, find the numbers.
My Approach: Let the numbers be $a,a+d,a+2d$
So, according to question $$3a+3d=24$$ $$a+d=8$$ and $$a(a+d)(a+2d)=440$$ $$8a+ad=55$$ I can’t proceed from here.
Ple... | Another way to view it is let $m$ equal the middle term.
So they are $m -d, m, m+d$
$(m-d) + m + (m+d) = 3m = 24$ so $m = 8$.
So the numbers are $8-d, 8, 8+d$ and
$(8-d)8(8-d) = 8(64 -d^2)=440$
$64 -d^2 = 55$
$d^2 = 64 - 55 = 9$
$d = \pm 3$
so the numbers are $5,8,11$. Or $11, 8, 5$
Even if we did it your way.
$a + d =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Wrong solution - but why ?
Find all solutions to the ODE $$y'=\begin{pmatrix}0 & 1 \\ \frac{-2}{1-x^2} & \frac{2x}{1-x^2}\end{pmatrix}y$$
What I did :
Guessing $y_1=\begin{pmatrix}x \\1\end{pmatrix}$ and reduce the order:
complete $y_1$ to an invertible matrix such that $$H^{-1}=\begin{pmatrix}0 & 1 \\ 1 & -x\en... | In the wording of the question some symbols are undefined such as $B$, $C_1$ , $C_2$ , $z\quad$ This is confusing and makes difficult to answer with any certainty.
Nevertheless, the end of calculus might be :
$$z'=-\frac{2x}{x^2-1}z \quad\to\quad z=\frac{c_2}{x^2-1}$$
$$\int \frac{1}{(x^2-1)^2}dx=\frac{1}{2}\ln\left|\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity
$$\frac{1}{n(n+1)} > \frac{1}{(... | Another way:
$$\begin{align} \sum_{k=1}^{2^5} \frac{1}{k^2}
&\leq 1 + \sum_{k=2}^{2^5}\int_{k-1}^{k}\frac{1}{t^2} dt\\
&=1+\int_1^{2^5}\frac{1}{t^2} dt \\
&=2-\frac{1}{2^5}\\
&<2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
It's $(X+Y)^3- (X^2+Y^2)\in \mathbb{C}[X,Y]$ irreducible? It's $(X+Y)^3- (X^2+Y^2)\in \mathbb{C}[X,Y]$ irreducible? I can't apply Eisenstein Criterion.
| The brute force method :
See it as a polynomial of degree $3$ of $K[X]$ with coefficients in $K = \mathbb{C}(y)$. If it is not irreducible then $$X^3+y^3+3X^2y+3Xy^2-X^2-y^2=( X +a)(X^2+bX+c)=X^3+bX^2 +cX+aX^2+abX+ac$$
$c = \frac{y^3-y^2}{a}$, $b = 3y-1-a$
$$= X^3+(3y-1-a)X^2 +\frac{y^3-y^2}{a}X+aX^2+a(3y-1-a)X+1$$
I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this que... |
\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}
What is the value of $xyz$?
Surprisingly, the Ravi substitution
works in this case, despite that
not all the numbers $x,y,z$ are positive,
and hence, the corresponding t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Linear equation with different variable on the denominator? I have this given problem. That asking me to solve for $x$. Although this example has answered. I've had troubles on a certain part.
Here's the equation with answer
\begin{align*}
\frac{2x-a}b &= \frac{4x-b}a\\
a(2x-a) &= b(4x-b) \\
2ax-a^2 &= 4bx - b^2 \\
2a... | Maybe we need to write again the equation:
$$\frac{2x-a}{b}=\frac{4x-b}{a}.$$
Using the Multiplication Property of Equality, we get
$$ab\cdot\left(\frac{2x-a}{b}\right)=ab\cdot\left(\frac{4x-b}{a} \right).$$
Simplifying, we get
$$\frac{a\cdot b\cdot (2x-a)}{b}=\frac{a\cdot b\cdot (4x-b)}{a}.$$
Apply Cancellation Law in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Summation of $\arcsin $ series. What is $a $ if
$$\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?$$
Attempt: What I tried is to convert the series to $\arctan$ and then convert it telescoping series. So in terms of $\arctan $ it becomes
$$\arctan \left(\f... | Taking the principal branch of $\arcsin$ (with values in $\bigl[-\frac{\pi}{2}, \frac{\pi}{2}\bigr]$), we have
$$\tan\bigl(\arcsin s\bigr) = \frac{\sin \bigl(\arcsin s\bigr)}{\cos \bigl(\arcsin s\bigr)} = \frac{s}{\sqrt{1 - s^2}}.$$
With $s = \frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}$, we get
\begin{align}
1 - s^2 &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How prove $\left(1+\frac{4a}{b+c}\right)\left(1+\frac{4b}{c+a}\right)\left(1+\frac{4c}{a+b}\right)\ge 25$ let $a,b,c>0$ show that
$$\left(1+\dfrac{4a}{b+c}\right)\left(1+\dfrac{4b}{c+a}\right)\left(1+\dfrac{4c}{a+b}\right)\ge 25$$
It seem hard to prove AM-GM.Cauchy-Schwarz
| $$\Leftrightarrow a^{3}+b^3+c^3+7abc\geq ab(a+b)+bc(b+c)+ca(c+a) > 0$$
Right by Schur:
$$a^{3}+b^3+c^3+7abc> a^{3}+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | Note that
\begin{align*}
(a+b+c)^2 \ge 3ab+3bc+3ac.
\end{align*}
Therefore,
\begin{align*}
&\ \frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b}\\
=&\ \frac{a^2}{a^2+3ab+3ac}+\frac{b^2}{b^2+3ab+3bc}+\frac{c^2}{c^2+3ac+3bc}\\
\ge&\ \frac{(a+b+c)^2}{a^2+b^2+c^2 + 6ab + 6ac+6bc}\\
=&\ \frac{(a+b+c)^2}{(a+b+c)^2 + 4ab +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
Identifying left- and right-Riemann sums of $\int_9^{14}e^{-x^4}\ dx$
My attempt:
Relooking at it, I think $L_{20}$ would be the highest, so like
$R_{1200} < L_{1200} < L_{20}$, but I have no way to justify it, any help is appreciated.
| You're last thought is correct. One rigorous justification goes as follows:
$\displaystyle L_{20} = \sum_{i=0}^{19}\frac{5}{20}f\left(9+\frac{5i}{20}\right)$, and
$\displaystyle L_{1200} = \sum_{i=0}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.
Now, we are going to split up the $L_{1200}$ sum into groups of 60... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Possible projective duality between two determinantal formulas for triangle area The shoelace formula for the area of a polygon in terms of consecutive vertices is well-known. In the particular case of a triangle, this may be written using a 3-by-3 determinant as
$$\text{area of triangle}=\frac{1}{2}\begin{vmatrix} x_1... | I believe the connection between these two formulas is just a consequence of Cramer's rule.
Let $A=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{pmatrix}$ and $B=\begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1\end{pmatrix}$.
Let us assume that $a_jx_i+b_jy_i+c_j=0$, whenever... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find a polynomial of which this field is a splitting field? Consider the field $Q(\sqrt{2} + i \sqrt{5})$. I've proven that $\mathbb{Q}(\sqrt{2} + i \sqrt{5}) = \mathbb{Q}(\sqrt{2}, i \sqrt{5})$. Also, by the degree product rule for extension degrees, I have $$ [\mathbb{Q}(\sqrt{2}, i \sqrt{5}) : \mathbb{Q}] = ... | Let $\alpha = \sqrt 2 + \mathrm i \sqrt 5$ and consider successive powers:
\begin{eqnarray*}
\alpha &=& 0+1\sqrt 2 + \mathrm i \sqrt 5 + 0\sqrt{10}\\ \\
\alpha^2 &=& -3+0\sqrt 2 + 0\sqrt 5 + \mathrm i \sqrt{10} \\ \\
\alpha^3 &=& 0-13\sqrt 2+\mathrm i \sqrt 5 + 0\sqrt{10}\\ \\
\alpha^4 &=& -31+0\sqrt 2 + 0\sqrt 5 -12\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove $\sin2x+\sin4x+\sin6x=4\cos x\cos2x\sin3x$ Prove $\sin{2x}+\sin{4x}+\sin{6x}=4\cos{x}\cos{2x}\sin{3x}$
I have reached the point where the LHS equation has turned into $2\cos{x}\cos2x\sin{x}(2\sin2x+1)$
But I have no idea how to turn $\sin{x}(2\sin2x+1)$ into $2\sin3x$
A quicker method if it exists would be greatl... | To solve this problem, we can use the identities:
$$
\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A - B}{2},
$$
$$
\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A - B}{2},
$$
and
$$
\sin 2\phi = 2\sin \phi \cos \phi.
$$
Going back to the question,
$
\text{LHS} = \sin 2x + \sin 4x + \sin 6x \\
= 2\sin 3x \cos x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $n^5+n^4+1$ is not a prime I have to prove that for any $n>1$, the number $n^5+n^4+1$ is not a prime.With induction I have been able to show that it is true for base case $n=2$, since $n>1$.However, I cannot break down the given expression involving fifth and fourth power into simpler terms. Any help?
| $$n^5+n^4+1=n^5-n^2+n^4+n^2+1=n^2(n-1)(n^2+n+1)+(n^2+n+1)(n^2-n+1)=$$
$$=(n^2+n+1)(n^3-n^2+n^2-n+1)=(n^2+n+1)(n^3-n+1)$$
I think, the best way is the following:
$$n^5+n^4+1=n^5+n^4+n^3-(n^3-1)=(n^2+n+1)(n^3-n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
The notion of equality when considering composition of functions When going through some (very introductory) calculus homework I was given the function $f(x) = \frac{x}{1+x}$ and asked to find the composition $(f\circ f)(x)$ and then find its domain. Substituting, we find that $$(f\circ f)(x)= \frac{\frac{x}{1+x}}{1+\f... | Note that the notation $(f\circ f)(x)$ gives some intuitive notion that we are going to do $2$ different operations. The first will be to "feed" $x$ to $f$, and then "feed" $f(x)$ to $f$. Indeed we have $(f\circ f)(x) = f(f(x))$. We know that $f$ is not defined for $x = -1$ and therefore the inner $f$ in $f(f(x))$ cann... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Optimization and maximum geometry What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.
I created two equations:
Square: Area=$x^2$
Triangle: Area= $\sqrt{3 /4}$
However, how can I find the maximum??
I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which i... | Proof with some words
You can create a rectangle inside your triangle. Let $x_B$ and $x_G$ the $x$-coordinates of the blue and green points, respectively. It is clear that $y_B = y_G = 0$. Moreover, it is easy to see that:
$$x_B < 0, x_G >0 ~\text{and}~ x_B = -x_G.$$
The base of the rectangles is $$b = x_G - x_B = 2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve this limit $\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}$ I am trying to figure out how to make this limit, even with the hopital. I've tried using hopital two times, but the situation 0/0 is still there. I've tried to solve it using wolfram, but I don't the solution. Even with rationa... | \begin{align}
\lim_{x\to \frac{1}{2}^-}\frac{\arcsin{2x}-\frac{\pi}{2}}{\sqrt{x-2x^2}}&=\lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-4x^2}}\frac{2\sqrt{x-2x^2}}{1-4x}, \text{ L'hopital}\\
&= \lim_{x\to \frac{1}{2}^-} \frac{2}{\sqrt{1-2x}\sqrt{1+2x}}\frac{2\sqrt{x}\sqrt{1-2x}}{1-4x}\\
&=\lim_{x\to \frac{1}{2}^-} \frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Sum of series in GP $a+ ar+ar^2 + ar^3 +ar^4+ \cdots+ ar^{n-1}=S_n$
$a\left(1+r+r^2 +r^3+\cdots+r^{n-1} \right) = S_n$
Iam trying to get $S_n = \dfrac{a(r^n -1)}{r-1}$
I don't know how to get $1+r+r^2 +r^3+\cdots+r^{n-1} =\dfrac{r^n -1}{r-1}$
Any help will be appreciated $:)$
| Equation (1)
$S_n = a_1 + a_2 + a_3 + a_4 + …....... +a_n$
Putting value of each term,
Equation (2)
$S_n = a + ar + ar^2 + ar^3 + ar^4 + ... + ar^{n-1}$
Multiply equation (2) by r,
Equation (3)
$rS_n = ar + ar^2 + ar^3 + ar^4 + ........... + ar^{n}$
Subtract equation (2) from (3),
$rS_n - S_n = ar^n - a$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to show that $\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)?$ How do we show that?
$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)\tag1$$
$(1)$ it a bit difficult to start with
$$\int_{0}^{\pi}(1+2x)\cdot{\sin(x)[1-\sin^2(x)]\over 1+\cos^2(x)}\math... | $J=\displaystyle \int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx$
Perform the change of variable $y=\pi-x$,
$\displaystyle J=\int_0^{\pi} (1+2(\pi-x))\dfrac{\sin^3 x}{1+\cos^2 x}dx$
Therefore,
$\begin{align}\displaystyle 2J&=(2+2\pi)\int_0^{\pi}\dfrac{\sin^3 x}{1+\cos^2 x}dx\\
2J&=-(2+2\pi)\int_0^{\pi}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac... | You should have developed your result a bit more to obtain the assignment solution.
\begin{align}
-\frac{4x}{(x^2+1)^2\sqrt{1-\frac{(1-x^2)^2}{(x^2+1)^2}}}&=-\frac{4x}{(x^2+1)^2\sqrt{\frac{x^4+1+2x^2-1-x^4+2x^2}{(x^2+1)^2}}} \\
&=-\frac{4x}{(x^2+1)^2\sqrt{\frac{4x^2}{(x^2+1)^2}}}\\
&=-\frac{4x}{(x^2+1)^2\frac{2|x|}{(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding a strict Liapunov function I need to show that the equilibrium point $(0,0)$ is asymptotically stable using Liapunov function. That means I shall find some strict Liapunov function.
Its given me the non-linear system:
\begin{cases}
x_1' = -x_1 -\frac{1}{3}x_1^3 - x_1^2\sin(x_2) \\
x_2' = -x_2 -\frac{1}{3}x_2^... | As i can remember, we have to prove that $\left<\nabla V(x_1, x_2), (x_1', x_2')\right> < 0$ over some neighborhood $B$ of $(0,0)$.
We call $sg(x)$ the sign function defined by:$$sg(x)=1\quad if\ x>0$$
$$sg(x)=-1\quad if\ x<0$$
First, we have \begin{align*}
-1\leq\sin(x_2)\leq 1\\
-2a\leq-2a\sin(x_2)\leq 2a\\
-2a|x_1^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... |
which resulted in $a^2 - 2a(0) + 0^2 = 2a$
$a^2 - 2a(0) + 0^2 = a^2$, not $2a$.
or if I do not substitute the $b^2$ I end up with $a^2 + b^2$.
Why would you not substitute the $b^2$? If you're substituting $b=0$ then you need to do it in all occurrences of $b$. This includes $b^2$. When you do this you'll see th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
| Let $\epsilon>0$. Choose $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\epsilon}{4}$. If $n\geq N$, then we get
$$\begin{align}\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3 \right|&
=\left|\frac{n^3+3n}{2n^4-n+1}\right|\\&=\frac{n^3+3n}{2n^4-n+1}\\
&<\frac{n^3+3n}{2n^4-n}\\
&\leq\frac{n^3+3n}{2n^4-n^4}\\
&\leq\frac{n^3+3n^3}{n^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}... | $f(x) = \sqrt{1 + x}$
$f'(x) = \frac{1}{2\sqrt{1+x}}$
$f(0) = 1$
$f'(0) = 1/2$
$f(x + h) = f(x) + h f'(x) +.....$
$\sqrt{1.1} = f(.1) = f(0 + .1) \approx1 + 0.1 / 2 = 1.05$
you got confused by the fact they are using x + 1 in the function - i made a start for you, with two terms it is getting closer to the answer
do yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$ If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, prove that $a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$.
I guess that the given expression is true if and only if $a=b=c=0$. Is it true? Or,i... | We have $$a^{1/4} + b^{1/4} + c^{1/4} =0$$ $$\Rightarrow (a^{1/4} + b^{1/4} + c^{1/4})^2 = a^{1/2}+b^{1/2} + c^{1/2} + 2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}]=0$$ $$\Rightarrow a^{1/2}+b^{1/2}+c^{1/2}=-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}] $$ $$\Rightarrow (a^{1/2}+b^{1/2}+c^{1/2})^2 =(-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ w... | Can we just charge straight at this?
$y$ is odd. $x=2 \ (\Rightarrow y^2=7)$ is not a solution, so $x$ is an odd prime.
$x(5x-7) = (y-1)(y+1)$, so $x \mid (y-1) $ or $x \mid (y+1)$ ($x$ is prime)
so $kx=y\pm1$, $k$ even
$k\ge4$ is too large: $(kx\pm1)^2\ge (4x-1)^2 $ $= 16x^2-8x+1$ $>5x^2-7x+1$. So only $k=2$, that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
| $a^4-a^3+a^2+2=(a^2+a+1)(a^2-2a+2)=0$.
We can check also $a=1+i$ and $a=1-i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
| You have mistake.
$\binom{6}{4} . x^2 . \left(\frac 2x\right)^4$
= $15 . x^2 . \frac{16}{x^4}$
= $240 . \frac{1}{x^2}$
Here 240 is coefficient of $x^{-2}$ not $x^2$.
Correct term -
$\binom{6}{2} . x^4 . \left(\frac 2x\right)^2$
= $15 . x^4 . \frac{4}{x^2}$
= $60 . x^2$
So answer is 60.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all the solutions of the system $AX=B$ if $B$ is the difference between the first and the fourth column of $A$. Let
$$A \sim
\begin{pmatrix}
1 &2 &4&1\\
0&0&1&2\\
1&3&1&1\\
0&0&0&0\\
\end{pmatrix}$$
such that the equivalence is achieved by elementary row transformations. Find all the solutions of the system $AX=B... | Hint -
You have A and B.
Solve AX = B.
$$\begin{pmatrix}
1 &2 &4&1\\
0&0&1&2\\
1&3&1&1\\
0&0&0&0\\
\end{pmatrix} \cdot \begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix} = \begin{pmatrix}
0\\
-2\\
0\\
0
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given $G=\mathbb R$ and $x*y\equiv x+y+x^3 y^3$, is there an inverse? I am given a set $G = \mathbb{R}$ and $x*y = x+y+x^3y^3$. I need to find out whether there exists an inverse. Please note that this is not a group since * is not associative. The identity element $e$ is $0$. So I did the following:
For the inverse t... | I think you have already said the answer.
Since $x + y + x^3y^3 = 0$ is a cubic, it has at least one real root.
For any $x$ there exists a $y$ such that $x*y = y*x = 0$
However, we have not proven that for any $x$ there exists a unique $x^{-1}$
Rather than using implicit differentiation, continue with the assumption th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
synthetic division question. Synthetic division is possible when the Divisior is in the form of $x+a$ or $x-a$. but what if the divisor is in the form of $x^2+a$, $x^2-a$, $x^3-a$,... and higher powers. how can we perform synthetic division in such cases.
Thanks
| For the particular case of divisors in the form $x^n-a$ it is possible to replace the long division with $n$ synthetic divisions.
For a polynomial $P(x)$ being divided by $x^3-a$ for example, group the powers of $x$ in $P$ according to the remainder $\bmod 3$ and write it as:
$$
P(x) = P_0(x^3) + x P_1(x^3) + x^2P_2(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
| Hint:
$$(x+1)^5-13(x+1)^3+36(x+1)=0$$
$$\left[(x+1)^4-13(x+1)^2+36\right](x+1)=0$$
$$\left((x+1)^2-9\right)\left((x+1)^2-4\right)(x+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)
I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as... | Observe if $r<\sqrt{5}$, then there exists $\epsilon>0$ such that
\begin{align}
r+\epsilon<\sqrt{5} \ \ \Leftrightarrow& \ \ (r+\epsilon)^2 <5\\
\Leftrightarrow& \ \ r^2-5+2\epsilon r+ \epsilon^2=P(\epsilon)<0
\end{align}
which is true if we sketch $P(\epsilon)$ as a function of $\epsilon$. In particular, we see that ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope ... | $(a,b,c)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are opposite ordered.
Thus, by Rearrangement $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a\cdot\frac{1}{a}+b\cdot\frac{1}{b}+c\cdot\frac{1}{c}=3,$$ which gives
$$a^2c+b^2a+c^2b\geq3abc$$
Similarly we'll get $$a^2b+b^2c+c^2a\geq3abc$$ and after summing we a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$ I'm having trouble with the following derivation:
Q:
We can use Euler's Theorem ($e^{i\theta} = \cos\theta + i\sin\theta$), where $e$ is the base of the natural logarithms, and $i = \sqrt{-1}$, together with the binomial theorem as above, to derive a number of trigo... | I believe your claim is actually incorrect. You should have the identity
\begin{align}
\cos 3\theta = 4\cos^3\theta -3\cos\theta
\end{align}
not
\begin{align}
\cos 3\theta = (4\cos\theta)^3 -3\cos\theta.
\end{align}
Observe
\begin{align}
e^{i3\theta} =&\ (\cos\theta + i\sin\theta)^3 = \cos^3\theta+3(i\sin\theta)\cos^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the serie... | HINT
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$
Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$
Note that the second part ha... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there i... | Ok I was able to find the integral
$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
First note that
$$\int \frac{\log(1+x^2)}{x^2}\,dx = 2 \arctan(x) - \frac{\log(1 + x^2)}{x}+C$$
Using integration by parts
$$I = \frac{\pi^3}{12}+2\int^\infty_0\frac{\arctan(x)\log(1 + x^2)}{(1+x^2)x}\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}... | A worse bound, a slightly different approach:
$$ \frac{n\sqrt x}{1+n^2x^2} \le \frac{n\sqrt x}{\sqrt{1+n^2x^2}} = n \sqrt{ \frac{x}{1+n^2x^2}} = \frac{1}{\sqrt{x}} \sqrt{\frac{1}{1+\frac{1}{n^2x^2}}} \le \frac{1}{\sqrt{x}} $$
Where we used tha facts that $x \ge \sqrt{x} $ if $x \ge 1$ and $\sqrt{\frac{1}{1+x}} \le 1 $ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
| Let $$S_n= \sum_{n=1}^{\infty} \frac{n}{3^n}\tag1$$
$$\frac13S_n=\sum_{n=1}^{\infty} \frac{n}{3^{n+1}}=\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}-\sum_{n=1}^{\infty}\frac1{3^{n+1}}\tag2$$
$(1)-(2)$,
$$\begin{align}\frac23S_n&=\sum_{n=1}^{\infty} \frac{n}{3^n}-\sum_{n=1}^{\infty} \frac{n+1}{3^{n+1}}+\sum_{n=1}^{\infty}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Showing matrices in $SU(2)$ are of form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.
I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex n... | We have $tr^\ast=-us^\ast$ so $\left| r\right|^2 \left| t\right|^2 = \left| s\right|^2 \left| u\right|^2$ and $\left| r\right|^2 -\left| r\right|^2\left| u\right|^2 = \left| s\right|^2 \left| u\right|^2$ so $\left| r\right|^2 =\left| u\right|^2$. Hence $r,\,u$ have the same modulus, as do $s,\,t$.
If $tu\ne 0$ define ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | $\int \frac{x^2(x-2)}{(x-1)^2} dx$
$\int {x^2(x-1-1)\over(x-1)^2}dx$
$\int {x^2(x-1)-x^2\over(x-1)^2}dx$
$\int ({x^2\over(x-1)}-{x^2\over(x-1)^2})dx$
$\int {x^2\over(x-1)}dx-\int{x^2\over(x-1)^2}dx$
$t=x-1 => dt=dx$
$\int {(t+1)^2\over(t)}dt-\int{(t+1)^2\over(t)^2}dt$
$\int {(t^2+1+2t)\over(t)}dt-\int(1+{1\over t})^2dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Trick to this square root equations Okay, so this is a high school level assignment:
$$
\sqrt{x+14}-\sqrt{x+5}=\sqrt{x-2}-\sqrt{x-7}
$$
Here's a similar one:
$$
\sqrt{x}+\sqrt{x-5}=\sqrt{x+7}+\sqrt{x-8}
$$
When solving these traditionaly, I get a polynomial with exponent to the 4th which I cannot solve (I can guess the... | write like this
$$\sqrt{x+14}+\sqrt{x-7}=\sqrt{x-2}+\sqrt{x+5}$$
square both sides:
$$2x+7+2\sqrt{(x+14)(x-7)}=2x+3+2\sqrt{(x-2)(x+5)}\\
2+\sqrt{(x+14)(x-7)}=\sqrt{(x-2)(x+5)}$$
square again
$$4\sqrt{(x+14)(x-7)}+x^2+7x-94=x^2+3x-10\\
\sqrt{(x+14)(x-7)}=21-x \to (x+14)(x-7)=(21-x)^2$$
Can you finish?
Don't forget to te... | {
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"url": "https://math.stackexchange.com/questions/2136635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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$\lim_{z \to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}$ Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$
Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\... | First, note that:
$$z^3=\cos(\pi)+i\sin(\pi)=\color{red}{-1}$$
And also, you've evaluated $z$ correctly, but the substitution into your limit is wrong.
Therefore, you should be using:
$$\lim_{x\to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}=\frac{\color{red}{-1+8}}{\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)+4\color{red}{\... | {
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"url": "https://math.stackexchange.com/questions/2136937",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ then the sum of the first $mn$ terms of the sequence is $(mn+1)/2$
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ respectively then prove that the sum of the first $mn$ terms of the sequence is $(mn+1)/2$.
My Attempt ;
$$... | Let $u_t = u_0 + a \cdot t$
$$ u_m = \frac{1}{n} $$
$$ u_n = \frac{1}{m} $$
Therefore $$ u_n - u_m = a \cdot \left( n-m \right)= \frac{1}{m} - \frac{1}{n} $$
$$ a=\frac{1}{m \cdot n} $$
So the initial term is$$ u_n = u_0 + \frac{n}{m \cdot n} = u_0 + \frac{1}{m} = \frac {1}{m}$$
$$ u_0 = 0 $$
We have $$ u_t = \frac{t}{... | {
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"question_score": "3",
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Does there exist a triple of $distinct$ numbers $a,b,c$ such that $(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$?
Does there exist a triple of distinct numbers $a,b,c$ such that $$(a-b)^5 + (b-c)^5 + (c-a)^5 = 0$$ ?
SOURCE : Inequalities (PDF) (Page Number 4 ; Question Number 220.1)
I tried expanding the brackets and I ended up w... | Assume that $a,b,c$ are distinct.
Let $a-b=x \neq 0,\; b-c=y \neq 0$. Note that $a-c=x+y \neq 0$
Note that the equation becomes $$x^5+y^5=(x+y)^5$$
So $$(x+y)^5-x^5-y^5=5xy(x^3+2x^2y+2xy^2+y^3)=0$$
Note that this becomes $$xy(x+y)(x^2+y^2+xy)=0 \iff x^2+xy+y^2=0$$
Using the quadratic formula, we can find $x,y$. Note t... | {
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | Another way.
Let $c=\max\{a,b,c\}$.
Hence, $c\geq1$, $a+b\leq2$ and
$$(2a^2+3)(2b^2+3)\geq\left(\frac{(a+b)^2}{2}+3\right)^2$$
because it's just $$(a-b)^2(12-a^2-6ab-b^2)\geq0$$ and
$$12-a^2-6ab-b^2=12-(a+b)^2-4ab\geq12-2(a+b)^2\geq12-8>0.$$
Thus, it remains to prove that $$\left(\frac{(a+b)^2}{2}+3\right)^2(2c^2+3)\... | {
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"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sq... | $$\lim\limits_{x\to 2^{\pm}}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2} = \left[\frac{\sqrt{3-\sqrt{3}}}{0^{\pm}}\right]=\pm\infty$$
As you can see, the left- and right-side limits are different, so the limit $\lim\limits_{x\to 2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2}$ does not exists.
| {
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"question_score": "3",
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differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since p... | To put it into the form you requested:
$$
-(1+x^2) \,dy + (2xy + x^2 + x^4) \,dx = 0 \implies \frac{dy}{dx} - \frac{2xy + x^2 + x^4}{1+x^2} = 0 \\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = \frac{x^2 + x^4}{1+x^2}\\
\implies \frac{dy}{dx} + \left(-\frac{2x}{1+x^2}\right) y = x^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$
My Try (Edited from C... | Another way.
Since $ab+ac+bc\leq\frac{1}{3}(a+b+c)^2\leq1$, by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{1+a^2}}\leq\sum_{cyc}\frac{a}{\sqrt{ab+ac+bc+a^2}}=\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq$$
$$\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{b}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\left|z^3 + {1 \over z^3}\right| \le 2$ then $\left|z + {1 \over z}\right| \le 2$
$\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2$ prove that $\displaystyle \left|z + {1 \over z}\right| \le 2$
$$\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3... | It's all about identities.
Note that $\left( z + \frac 1z\right)^3 = z^3 + \frac 1{z^3} + 3\left(z + \frac 1z\right)$.
Apply the triangle inequality:
$$
\left|\left( z + \frac 1z\right)^3 \right| \leq \left|z^3 + \frac 1{z^3}\right| + 3\left|\left(z + \frac 1z\right)\right|
$$
Using what you know:
$$
\left|\left( z + \... | {
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"timestamp": "2023-03-29T00:00:00",
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List all the elements of $A = \{ n \in \mathbb{Z} \mid \frac{n^2-n+2}{2n+1} \in \mathbb{Z}\}$ I was given the following set $A = \{ n \in \mathbb{Z} \mid \frac{n^2-n+2}{2n+1} \in \mathbb{Z}\}$
I have to list all the elements of $A$.
I started using the Euclidean division:
$$n^2-n+2=(2n+1)(\frac{1}{2}n-\frac{3}{4}) + \f... | The answer based on the hint of @lhf
$$4\frac{n^2-n+2}{2n+1}= 2n-3+ \frac{11}{2n+1}$$
$$4\frac{n^2-n+2}{2n+1} \in \mathbb{Z} \Rightarrow \frac{11}{2n+1} \in \mathbb{Z}$$
$$\Leftrightarrow 2n+1 \in D_{11} \Leftrightarrow 2n+1 \in \{-11, -1, 1 , 11\}$$
$$\Leftrightarrow 2n \in \{-12, -2, 0 , 10\}$$
$$\Leftrightarrow n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to validate the following inequality? $\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$
| Assume that $z \neq 0$ and $x, y$ are not simultaneously zero. (For example, if $z$ is zero, $b$ can be any arbitrary negative number.)
Consider the equation $$b = -\sqrt{\frac{x + y}{z}}a + \sqrt{\frac{2(x + y + z)}{z}}.$$
Let $a^* = \sqrt{\frac{2(x + y + z)}{x + y}}$ and $b^* = \sqrt{\frac{2(x + y + z)}{z}}$, i.e., $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find value of $\sin 4x$ Given $\tan x = \frac { 1+ \sqrt{1+x}}{1+ \sqrt{1-x}}$. i have to find value of $\sin 4x$. i write $\sin 4x=4 \frac{ (1-\tan^2 x)(2 \tan x)}{(1+\tan^2 x)^2}$ but it seems very complicated to do this? Any other methods?
Thanks
| Write $\tan x=\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}$
As $-1\le y\le1$ for real $\tan x$
WLOG let $y=\cos2u$ where $0\le2u\le\pi$
$$\dfrac{1+\sqrt{1+y}}{1+\sqrt{1-y}}=\dfrac{1+\sqrt2\cos u}{1+\sqrt2\sin u}=\dfrac{\dfrac1{\sqrt2}+\cos u}{\dfrac1{\sqrt2}+\sin u}=\dfrac{2\cos\dfrac{45^\circ +u}2\cos\dfrac{45^\circ-u}2}{2\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $7^{13} \mod 40$ I wanted to compute $7^{13} \mod 40$. I showed that
$$7^{13} \equiv 2^{13} \equiv 2 \mod 5$$
and
$$7^{13} \equiv (-1)^{13} \equiv -1 \mod 8$$.
Therefore, I have that $7^{13} - 2$ is a multiple of $5$, whereas $7^{13} +1$ is a multiple of $8$. I wanted to make both equal, so I solved $-2 + 5k... | $$\phi(40)=16 \to 7^{16}\equiv 1$$mod 40 $$7^{16}\equiv 1 \\7^{13}7^3\equiv 1\\
343.7^{13}\equiv 1 \\
(320+23).7^{13}\equiv 1\\
23.7^{13}\equiv 1\\
23.7^{13}\equiv 1+40 \equiv 1+80\equiv 1+120\equiv 1+160\\
23.7^{13}\equiv 161 \\23.7^{13}\equiv 7(23)$$ divide by $23$ , $(23,40)=1
$
so
$$23.7^{13}\equiv 7(23) \to 7^{... | {
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"source": "stackexchange",
"question_score": "5",
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Can numbers in a sequence be equal after performing some operations I have a sequence of 21 numbers (-10, -9, -8, ..., -1, 0 , 1, 2, ... , 10). I can take any pair $(a, b)$ from the sequence and replace it with pair $((3a - 4b)/5 ), (4a + 3b)/5)$. Is it possible that all numbers in a sequence will be equal after perfor... | Note that after any replacement, all terms of the sequence are still rational numbers.
Suppose in a given step, you replace the pair $\small{\displaystyle{(a, b)}}$ by the pair
$\small{\displaystyle{\left(\frac{3a - 4b}{5}, \frac{4a + 3b}{5}\right)}}$.
Observing that
\begin{align*}
\left(\frac{3a - 4b}{5}\right)^2... | {
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"url": "https://math.stackexchange.com/questions/2150057",
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An integral of rational function with third power of cosine hyperbolic function Prove
$$\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx =
\frac{2}{\pi^3}\left(\pi \cosh\left(\frac{\pi}{4} \right)-4\sinh\left(
\frac{\pi}{4}\right) \right)$$
Attem... | $$I=\int_{-\infty}^{\infty}\frac{1}{(5 \pi^2 + 8 \pi x + 16x^2)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\int_{-\infty}^{\infty}\frac{1}{(4x+\pi+2i\pi)(4x+\pi-2i\pi)
}\frac{\cosh\left(x+\frac{\pi}{4} \right)}{\cosh^3(x)}dx$$
$$I=\frac{-i}{16\pi}\int_{-\infty}^{\infty}\left(\frac{1}{x+\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\fra... | Just in case you want to see another approach.
Consider $$A=\frac{\ln(\sqrt{x-2})}{x^2-9}=\frac 12\frac{\ln({x-2})}{x^2-9}$$ Now, as Simply Beautiful Art answered, let $x=u+3$ which makes $$A=\frac{\log (1+u)}{2 u (u+6)}$$ Now, use Taylor series around $u=0$ $$\log(1+u)=u-\frac{u^2}{2}+O\left(u^3\right)$$ which makes $... | {
"language": "en",
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What is the remainder when $15^{40}$ divided by $1309$? I know that $$ 15 \equiv 1\pmod{7}, N \equiv 1\pmod{7},$$ but cannot proceed further.
| Note that $1309=7 \times 11 \times 17$. So $$15^{40} \equiv 1^{40} \equiv 1 \pmod {7}$$
Using $15 \equiv 1 \pmod {7}$. Then, $$15^{40} \equiv \left(15^{10}\right)^4\equiv 1 \pmod {11}$$
And $$15^{40} \equiv (15+17 \times 2)^{40} \equiv 7^{80} \equiv \left(7^{16}\right)^5 \equiv 1 \pmod {17}$$
From Fermat's Little The... | {
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What is the sum of the solutions to $6x^3+7x^2-x-2=0$ What is the easy way to solve the problem?
The sum of the solutions to $6x^3+7x^2-x-2=0$ is:
$$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$
| Constant is -2.
Using remainder theorem
So you have 1,-1,2,-2.
By putting-1 in x you get to know that -1 satisfies the solution.
So -1is a solution and x+1 is a factor of above equation.
After dividing 6x^2+7x^2-x-2 by x+1 we get,
6x^2 +x-2 as quotient.
Solving this quadratic equation you get 1/2 and -2/3 as solution... | {
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Minimize $P=5\left(x^2+y^2\right)+2z^2$ For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$
I have no idea. Can you make a few suggestions?
| Let $x=y=2$ and $z=4$.
Hence, $P=72$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$5(x^2+y^2)+2z^2\geq72\left(\sqrt[3]{\frac{(x+y)(x+z)(y+z)}{144}}\right)^2.$$
It's enough to prove last inequality for non-negative variables.
Let $x+y=tz$.
Since $x^2+y^2\geq\frac{1}{2}(x+y)^2$ and $(x+z)(y+z)\l... | {
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"source": "stackexchange",
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Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| We have $7^4 = 49^2 < 50^2 = 4 \times 5^4 < 5^5$. Hence
$$7^{55} < 7^{56} = (7^4)^{14} < (5^5)^{14} = 5^{70} < 5^{72}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Smallest positive integral value of $a$ such that ${\sin}^2 x+a\cos x+{a}^2>1+\cos x$ holds for all real $x$
If the inequality $${\sin}^2 x+a\cos x+{a}^2>1+\cos x$$ holds for all $x \in \Bbb R$ then what's the smallest positive integral value of $a$?
Here's my approach to the problem $$\cos^2 x+(1-a)\cos x-a^2<0$$
Le... | The smallest positive integer is $1$. That doesn't satisfy the condition, because
$$ \sin^2 x + \cos x + 1 > 1 + \cos x $$
has equality when $\sin^2x = 0$, which happens, among other places, at $x=0$.
The next positive integer is $2$. Does that work?
$$ \sin^2 x + 2\cos x + 4 > 1 + \cos x $$
holds if and only if
$$ \si... | {
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Find the limit : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$ Find the limit: Without the use of the L'Hôspital's Rule
$$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$
My try:
$u=x-1$
Now:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-... | We can simplify the term of interest and rationalize terms to obtain
$$\begin{align}
\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}&=\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}\,)}\\\\
&=\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}\,\,\sqrt[n]{x-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}... | {
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For $a,b,c \in R$ and $a,b,c>0$. Minimize $A=a^3+b^3+c^3$ For $a,b,c \in R$ and $a,b,c>0$ satisfy $a^2+b^2+c^2=27$, minimize $$A=a^3+b^3+c^3$$
| By Power-Mean Inequality,
$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$
$a^3+b^3+c^3 \ge 81$
| {
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Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square. There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.
EDIT: I found solution by symmetric polynomials, which is posted below.
|
EDIT: I found solution by symmetric polynomials (in variables a, b, c)
The following more or less transcribes OP's solution in direct calculations, without explicitly using Newton's relations. From the assumption that $a+b+c=0\,$:
$$
0 = (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)
$$
$$
\implies 2(ab+bc+ca)=-(a^2+b^2+c^2)... | {
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Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to p... | By applying Caushy Schwartz inequality we get
( 1 + a + b + c ) ( 1 + 1/a + 1/b + 1/c ) >= 16
Equality holds when a^2 = b^2 = c^2 = 1
Therefore
a = b = c = 1
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this inequality $\sum\tan{\frac{A}{2}}-\frac{\sqrt{3}-1}{8}\prod\csc{\frac{A}{2}}\le 1$ In $\Delta ABC$,show that
$$\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}}-\frac{\sqrt{3}-1}{8}\csc{\frac{A}{2}}\csc{\frac{B}{2}}\csc{\frac{C}{2}}\le 1$$
I tried also $$\left(\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan... | Easy to show that $\sum\limits_{cyc}\tan\frac{\alpha}{2}=\frac{4R+r}{p}$ and $\prod\limits_{cyc}\sin\frac{\alpha}{2}=\frac{r}{4R}$.
Also, if $M$ is a center gravity of the triangle and $I$ is a center of the inscribed circle in the triangle,
so $MI^2=\frac{p^2+5r^2-16Rr}{9}$, which gives $p\geq\sqrt{16Rr-5r^2}$.
Let $... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
| Hint:
In general a double radical $\sqrt{a\pm \sqrt{b}}$ can be denested if $a^2-b$ is a perfect square and in this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
In Your case $a^2-b=1$ so....
| {
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"timestamp": "2023-03-29T00:00:00",
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How many combinations of pennies, dimes, nickels, and quarters create 0.32$? I need help solving this. I cannot find the complete number of combinations. I have already found $5$, but I can't find any more.
| Leaving out the pennies in each combo ...
There are 2 combos with a quarter:
25+5 (that is: 1 quarter + 1 nickel ...so 2 pennies...)
25 (So just a quarter ...so 7 pennies ... for combos below, you'll have to figure out how many pennies to add ...)
There is 1 combo with 3 dimes:
10+10+10
There are 3 combos with 2 dime... | {
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"url": "https://math.stackexchange.com/questions/2166054",
"timestamp": "2023-03-29T00:00:00",
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If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$... | I write it step by step:
With $\sin A+\sin^2 A=1$ we have
$\sin A=1-\sin^2 A=\cos^2A$
so
\begin{eqnarray}
&& a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0\\
&& a(\cos^2A)^6+b(\cos^2A)^4+c(\cos^2A)^3-1=0\\
&& a\sin^6A+b\sin^4A+c\sin^3A-1=0\\
&& a(1-\cos^2A)^3+b(1-\cos^2A)^2+c\sin A(1-\cos^2A)-1=0\\
&& a(1-\sin A)^3+b(1-\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the limit of this sequence created using Pascal's triangle? Let use the Pascal Triangle to create a sequence. We sum from the first term on the left to the term in the middle
and then we take the inverse root times two. We get then that the terms are
\begin{eqnarray}
&&a_0 := 2 \\
&&a_1 = 2 \frac{1}{1^{1/1}} ... | The sum of the elements in "half a row" of Pascal's triangle is either $2^{n-1}$ (if we are talking about the $n$-th row with $n$ being odd) or $2^{n-1}+\frac{1}{2}\binom{n}{n/2}$ if $n$ is even, due to $\binom{n}{k}=\binom{n}{n-k}$.
The central binomial coefficient $\binom{2n}{n}$ behaves in the following way
$$ \bin... | {
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"timestamp": "2023-03-29T00:00:00",
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How many integers between $10000$ and $99999$ ( inclusive) are divisible by $3$ or $5$ or $7$? How many integers between $10000$ and $99999$ (inclusive) are divisible by $3$ or $5$ or $7$ ?
My Try :
Total Integers between $10000$ and $99999$ are $89999$.
$\left\lfloor\frac{89999}{3}\right\rfloor+\left\lfloor\frac{899... | The number of integers between $3$ and $4$ inclusive divisible by $3$ is not $\lfloor\frac{(4-3+1)}{3}\rfloor=\lfloor\frac{2}{3}\rfloor=0$ but is instead $\lfloor \frac{4}{3}\rfloor - \lfloor\frac{2}{3}\rfloor = 1$.
In the same way, you should not be using $\lfloor \frac{90000}{7}\rfloor$ but instead you should be usin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of some natural numbers equal to $n$ In how many ways we can have some natural numbers that their sum is equal to $n$ and none of them is greater than $k$, for given $n$ and $k$?
NOTE: We don't know the number of the elements.
Can anyone help me with this problem? I can't find anything on the internet.
For example,... | How about using generating functions? The number of ways to add $m$ positive integers each of which is less than or equal to $k$ so that their sum is $n$ is the coefficient of $x^n$ in $(x+x^2+\ldots+x^k)^m$. So the coefficient of $x^5$ in $\sum_{m=1}^{5} (x+x^2)^m$ is the answer for your example, which is 8.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I add the terms in the binomial expansion of $(100+2)^6$? So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)$+$\binom {6} {1}$$100^5(2)$+....
I stumbled upon this step. How did they add... | $10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$
$= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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An integration $ \int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x.$
How can we evaluate
$$\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x?$$
Usually when having as the integrand a logarithmic function, ... | One may prove that
$$
\int_0^1 x\ln \left( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left( \sqrt {1+x} -\sqrt{1-x}\right)\:\mathrm{d}x=\frac{\ln^2 2}4-\frac{3\ln2}4+\frac1{16}. \tag1
$$
Hint. By using
$$
\begin{align}
4ab&=(a+b)^2-(a-b)^2
\end{align}
$$ giving
$$
\begin{align}
4\ln b \ln c&=\ln^2 (bc)-\ln^2 \left(\frac bc\rig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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