Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the domain of $f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$ Find the domain of definition of $$f(x)=\log_{10}(1-\log_7(x^2-5x+13))+\arccos\left(\frac{3}{2+\sin\frac{9\pi x}{2}}\right)$$
I found the domain of $\log_{10}(1-\log_7(x^2-5x+13))$ to be $x\in(2,3)$ and the do... | Since $\sin(\frac{9\pi x}{2})=1$,
$$\frac{9\pi x}{2}\in(\frac{\pi}{2}, \frac{5\pi}{2}...\frac{21\pi}{2}...) \quad or \quad x\in(\frac{1}{9}, \frac{5}{9},..., \frac{21}{9}, \frac{25}{9}...)$$
We have seen that the domain of the first part of equation should be in $(2,3)$, So the only feasible solution is $\frac{21}{9}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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let $a,b,c \in \mathbb{R^+} \ \ a+b+c =1$ Then prove that : $a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$
Let $\{a,b,c\}\subset\mathbb{R^+}$ such that $a+b+c =1$. Prove that :
$$a^3+b^3+c^3 \geq \dfrac{1}{3}(a^2+b^2+c^2)$$
$$a^3+b^3+c^3-3abc=(x+b+c)(a^2+b^2+c^2-(ab +ac+bc))$$
$$a^2+b^2+c^2=(a+b+c)^2-2(ac+bc+ab)$$
N... | Let us consider the vectors $u=(\sqrt{a},\sqrt{b},\sqrt{c})$, $v=(a\sqrt{a},b\sqrt{b},c\sqrt{c})$.
By the Cauchy-Schwarz inequality
$$ a^2+b^2+c^2 = \left|u\cdot v\right|\leq \|u\|\cdot\|v\|=\sqrt{a+b+c}\sqrt{a^3+b^3+c^3} $$
but we also have
$$ 3(a^2+b^2+c^2)=(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2 = 1 $$
hence
$$ a^3+b^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ I have the following problem:
How to show $\exp(ix)=\cos x+i\sin x$ using $\exp(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$
My attempt:
\begin{align*}
\exp(ix)=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}
\end{align*}
Now split the sum in 4 su... | You split the sum like that:
\begin{align*}
\exp(ix) = \sum_{n=0}^{\infty}\frac{(ix)^n}{n!}&=\sum_{n=0}^{\infty}\frac{(ix)^{4n}}{(4n)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+1}}{(4n+1)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+2}}{(4n+2)!}+\sum_{n=0}^{\infty}\frac{(ix)^{4n+3}}{(4n+3)!}\\
&=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $ then prove that :
Let $a_1,a_2,. . . ,a_n \in \mathbb{R^+} $. Prove that :
$$\frac{ a_{1} }{ a_{2} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\frac{ a_{2} }{ a_{1} ^{2}+ a_{3} ^{2}+\cdots+ a_{n} ^{2}}
+\cdots+\frac{ a_{n} }{ a_{1} ^{2}+ a_{2} ^{2}+\cdots+ a_{n-1} ^{2}}
\geq \fr... | I think we can start by C-S:
$$\sum_{k=1}^n\frac{a_k}{\sum\limits_{i\neq k}a_i^2}=\sum_{k=1}^n\frac{a_k^2}{a_k\sum\limits_{i\neq k}a_i^2}\geq\frac{\left(\sum\limits_{k=1}a_k\right)^2}{\sum\limits_{k=1}^na_k\sum\limits_{i\neq k}a_i^2}.$$
Thus, it remains to prove that
$$\left(\sum\limits_{k=1}a_k\right)^3\geq4\sum\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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System of ellipses with same foci Consider a system of ellipses $E_u$ with same foci $(\pm u,0)$, here $u>0$ is a constant. In other words, we consider a system of equations
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a^2-b^2=u^2.$$
Q: How to show that any ellipse in $E_u$ satisfies
$$xyy'^2+(x^2-y^2-u^2)y'-xy=0,$$
here ... | We have $$\frac{2x}{a^2}+\frac{2yy'}{b^2}=0$$ or
$$y'=-\frac{b^2x}{a^2y}.$$
Thus, we need to prove that
$$\frac{xy\cdot b^4x^2}{a^4y^2}-\frac{(x^2-y^2-a^2+b^2)b^2x}{a^2y}-xy=0$$ or
$$x(b^2-a^2)(b^2x^2-a^2y^2-a^2b^2)=0$$ or
$$(b^2-a^2)x\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)=0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $\sum_{cyc} \frac{a}{c+a^3+b^2} \leq \frac{1}{2}$
Let $a$, $b$ and $c$ be positive real numbers such that $abc \geq 10+6\sqrt{3}$. Prove that: $$ \displaystyle\sum_{cyc} \frac{a}{c+a^3+b^2} \leq \frac{1}{2}.$$
I've tried to bound the LHS of Michael Rozenberg's last inequality,
$\frac{3\left(7.5^{7.5}\cdot... | Let $a=(1+\sqrt3)kx^2$, $b=(1+\sqrt3)ky^2$ and $c=(1+\sqrt3)kz^2$, where $k>0$ and $xyz=1$.
Hence, $10+6\sqrt3\leq abc=(10+6\sqrt3)k^3xyz$, which gives $k\geq1$ and
$$\sum_{cyc}\frac{a}{a^3+b^2+c}=\sum_{cyc}\frac{x}{k^2(1+\sqrt3)^2x^3+(1+\sqrt3)ky^2+z}\leq$$
$$\leq\sum_{cyc}\frac{x}{(1+\sqrt3)k(2\sqrt{x^3z}+y^2)}\leq\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Possibilities for the side and angle of the triangle We have the triangle ABC with $c=18$, $\alpha=\frac{\pi}{6}$ and the median through $C$ has the length $5$. I want to determine all the possibilities for $a$ and $\gamma$.
I have done the following:
From the median we have the following:
$$m_c=\sqrt{\frac{2a^2+2b^2... | Let $CM$ is a median of our triangle.
Thus, $$\frac{5}{\sin30^{\circ}}=\frac{9}{\sin\measuredangle ACM}$$
Hence, we have to cases.
*
*$\measuredangle ACM=\arcsin\frac{9}{10}$.
Thus, $$AC=\sqrt{9^2+5^2-2\cdot9\cdot5\cdot\cos\left(150^{\circ}-\arcsin\frac{9}{10}\right)}=$$
$$=\sqrt{106-90\left(-\frac{\sqrt{3}}{2}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\t... | If you wish to exploit the residue theorem, then first exploit the fact that the integral is even. In addition, use Euler's formula to write $2\cos(2\theta)+\cos(3\theta)=\text{Re}(2e^{i2\theta}+e^{i3\theta})$.
Then, we have
$$\begin{align}
\int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Categorizing the conic section given by $x^2-4xy+y^2+8x+2y-5=0$
Find the conic section of $x^2-4xy+y^2+8x+2y-5=0$
I came to the following using diagonalization and bi-linear form
$$-\left(x-\frac{5}{\sqrt{2}}\right)^2+\left(\sqrt{3}y-\frac{\sqrt{3}}{\sqrt{2}}\right)^2-19=0$$
So I can conclude (using eigenvalues)that ... | First thing is to remove the offset and center the curve on the origin. The easiest way I do this is to take $f(x,y) = x^2-4xy+y^2+8x+2y-5$ and solve the system $$\left. \begin{align} \frac{\partial f}{\partial x} & = 0 \\ \frac{\partial f}{\partial y} & = 0 \end{align} \right\} \begin{aligned} x &= 2 \\
y &= 3 \end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Computing $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}$ I don't understand why I can't get the telescopic sum after the partial fraction decomposition:
$$\frac{1}{n^3-n}= \frac{1}{n(n-1)(n+1)}=\frac{-1}{n}+\frac{1}{2(n-1)}+\frac{1}{2(n+1)}=\frac{-1}{n}+\frac{n}{(n-1)(n+1)}.$$
I have $\sum\limits_{n=2}^{\infty }\frac{1}{... | $$\sum _{ n=2 }^{ \infty } \frac { 1 }{ n^{ 3 }-n } =\sum _{ n=2 }^{ \infty } \frac { 1 }{ \left( n-1 \right) n\left( n+1 \right) } =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty } \left[ \frac { 1 }{ \left( n-1 \right) } +\frac { 1 }{ \left( n+1 \right) } -\frac { 2 }{ n } \right] =\\ =\frac { 1 }{ 2 } \sum _{ n=2 }^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Probability that at least $2$ people will not receive any ace.
I've a deck with 52 french cards ($13$ values for each of $4$ suits) and $4$ players. Randomly dealing out all cards, what's the probability that at least
$2$ people will not receive any ace?
My try:
$$p=\frac{\frac{4!}{2!}\binom{48}{13,13,12,10}+\binom... | With reference with my last comment against your question.
Imagine $4$ labelled rooms, each with $13$ labelled beds, to be occupied by $4$ travellers (aces)
As you have found, either $4$ are in one room, or $3-1 \;or\; 2-2$ in two rooms to satisfy the question's constraints
$Pr = \dfrac{4\binom{13}4 + (4\cdot3)\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to solve the following Cauchy problem. Solve the following Cauchy problem for a first order PDE:
$$(2x_1 + x_2)u_{x_1} + (x_2 + 1)u_{x_2} = u^2, \ \ u(x_1, 1) = x_1^2 + 1, \ \ x_1 \ge 0, x_2 \ge 1$$
and find an implicit conldition over $x_1$ and $x_2$ under which this Cauchy problem has a bounded solution.
Attempt... | Consider writing your system as follows
$$\left( 2x+t \right){{u}_{x}}+\left( t+1 \right){{u}_{t}}={{u}^{2}}$$ where $$u\left( x,1 \right)={{x}^{2}}+1,\,\,t\ge 1$$
Let $u\left( x,t \right)=u\left( x\left( s \right),t\left( s \right) \right)\Rightarrow \frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$
If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then
$$S\gt 1,$$
but how?
I understood that there are $451$ pair of terms. So clubbed two terms together.
$\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$
But ... | numbers from $\frac{1}{100}$ to $\frac{1}{200}$ are larger than $\frac{1}{200}$
and there are $100$ of them (101 actually, but doesn't matter) so their sum is larger than $100\times \frac{1}{200}=\frac12$
You can repeat a similar argument to show that from $\frac{1}{200}$ to $\frac{1}{300}$ the sum is larger than $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.
| Lemma(1): Let $a$ & $b$ to be integers such that $ab \mid a^2+b^2$.
If $\gcd(a,b)=1$, then prove that $a=\pm b$.
Proof: We claim that $ab=\pm 1$.
*
*Proof of the claim: Suppose on contrary; that $1 < |ab|$.
So there exist a prime number $p$, which divides $ab$; i.e. $p \mid ab$.
Without loss of generality we can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$?
Given that
$$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
What is the value of $a+b$
My try
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(... | As $x(1+a\cos x)-b\sin x\sim (1+a-b)x$ as $x\to0$ then the given limit
can only exist when $b=1+a$. A series expansion then gives
$$x(1+a\cos x)-(1+a)\sin x\sim-\frac{a}2x^3+(1+a)\frac{x^3}6.$$
One needs then $(1+a)/6-a/2=1$ for the given limit to equal $1$.
Now one can obtain $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $a+b+c=0$ find the value of $\big(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\big)\big(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\big)$ I already have a solution, which is correct, it is $9$. I'm just wondering if there is a simpler method.
First we just expand and find
$3+\frac{b-c}{a}\big(\frac{b}{c-a}+\frac{c}{a-... | Observe that $(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)=a^3+b^3+c^3-3abc$ so $a^3+b^3+c^3=3abc$. Now
\begin{eqnarray*}
&&\left( \frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c} \right) \left( \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \right)\\
&=&3+\frac{a}{b-c}\left(\frac{c-a}{b}+\frac{a-b}{c}\right)+
\frac{b}{c-a}\left(\frac{a-b}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What significance do squares of trigonometric ratios have algebraically and geometrically? I recently came upon a problem which asked me to:
"Show that $(\cos x + 1)^2 = \frac{\cos2x}{2} + 2\cos x + \frac32$ and hence determine $\int(\cos x + 1)^2 dx$.";
from Mathematics for the International Student-IB Diploma: SL. B... | Expanding the expression we have $$(\cos x + 1)^2 = \cos ^2 x + 2 \cos x + 1$$
However, $\cos^2 x = \frac {\cos 2x + 1}{2}$ so we have $$\frac {\cos 2x + 1}{2} + 2 \cos x + 1$$ Factoring $1/2$ from the first term $\frac {\cos 2x + 1}{2}$, we then have $$\frac {\cos 2x}{2} + \frac {1}{2} + 2 \cos x + 1 $$ or $$\frac {\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ... | The indefinite integral is
$$\int \frac{1}{x^2-2 x-3} \, dx=\frac{1}{4} \left[\log |3-x|-\log (x+1)\right]+C$$
Indeed the fraction can be split in the sum of two fractions
$\dfrac{1}{(x-3) (x+1)}=\dfrac{a}{x-3}+\dfrac{b}{x+1}=\dfrac{(a+b) x+a-3 b}{(x-3) (x+1)}$
which is equal to the given fraction if
$\left\{ \begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the value of expression $^2+^2+^2$ given $++=1$ and $\frac 1x +\frac 1y +\frac 1z=0$ Find the value of expression $$^2+^2+^2$$ given $$++=1$$ and $$\frac 1x +\frac 1y +\frac 1z=0$$
I am sorry I don't really know how to tag this question.
| Assume that $x,y,z$ are the roots of a monic, cubic polynomial $p(t)$
$$ p(t) = (t-x)(t-y)(t-z) = t^3-e_1 t^2+e_2 t-e_3. $$
Since $x+y+z=1$ we have $e_1=1$. Similarly, from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ we get $e_2=0$.
$$x^2+y^2+z^2 = (x+y+z)^2-2(xy+xz+yz) = e_1^2-e_2$$
then leads to $x^2+y^2+z^2=\color{blue}... | {
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"url": "https://math.stackexchange.com/questions/2389486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $a_n = \sum\limits_{r=0}^n \binom{n}{r} b_r$, prove $(-1)^n b_n = \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s$ Suppose that sequences of real numbers satisfy:
\begin{align*}
a_n &= \sum\limits_{r=0}^n \binom{n}{r} b_r \\
\end{align*}
Prove that:
\begin{align*}
(-1)^n b_n &= \sum\limits_{s=0}^n \binom{n}{s} (-1)^... | Here is a direct proof
that uses only
$\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
=1$
when $m=0$
and
$=0$ when
$m \ge 1$,
This last is
from the expansion
$0
=(1-1)^m
=\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
$.
$\begin{array}\\
\sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s
&=\sum\limits_{s=0}^n \binom{n}{s} (-1)^s
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Maximum of sum of fractions with tans Let $ABC$ be an acute triangle. What is the maximum value of
$$\frac{\tan^2A+\tan^2B}{\tan^4A+\tan^4B}+\frac{\tan^2B+\tan^2C}{\tan^4B+\tan^4C}+\frac{\tan^2C+\tan^2A}{\tan^4C+\tan^4A}?$$
For the equilateral triangle, $\tan A=\tan B=\tan C=\sqrt{3}$, and the sum is $1$. For the isosc... | Say $x= \tan A$, $y= \tan B$ and $z = \tan C$. It is
well-known that $x+y+z=xyz$. Then we have $$E = {x^2+y^2\over
x^4+y^4} +{y^2+z^2\over y^4+z^4} + {z^2+x^2\over
z^4+x^4}$$
By Cauchy inequality and then by $AM-GM$ we have
$${x^2+y^2\over x^4+y^4} \leq {2\over x^2+y^2}\leq {1\over xy}$$
So $$ E \leq {1\over xy} +{1\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Formulation of the Wallis product for $\pi/2$ using the $\Gamma$ function I am trying to write the Wallis Product (WP) using the $\Gamma$ function, thereby hoping for either a new identity or a simple derivation of the formula, but I get a result I am not sure its correct.
This is the WP
$$ \frac{\pi}{2} = \frac{2}{1}\... | Let's say $\rho(n) = \Gamma(n)/\Gamma(n-1/2).$ Then, yes, $\rho(n) \approx \sqrt n$ for large $n$. More precisely, $\rho(n)/\sqrt n \to 1$ as $n \to \infty$.
You should expect this. After all, $$\rho(n)\cdot \rho(n-1/2) = \frac{\Gamma(n)}{\Gamma(n-1/2)} \cdot \frac{\Gamma(n-1/2)}{\Gamma(n-1)} = \frac{\Gamma(n)}{\Gamma(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve this problem on complex numbers? Let $ Z $ be a complex number with nonzero imaginary part such that
$$ (2Z + 1)(3Z + 1)(5Z + 1)(30 Z + 1) = 10 $$
Then compute $$ \frac{\text {sum of all values of Z}}{\text {product of all values of Z} } $$
| Hint: $1/2+1/30=1/3+1/5 = 2 \cdot 4/15\,$, which suggests the substitution $z = x - 4/15\,$, which gives a biquadratic in $x\,$:
$$
2\cdot3\cdot5\cdot30\left(z+\frac{1}{2}\right)\left(z+\frac{1}{3}\right)\left(z+\frac{1}{5}\right)\left(z+\frac{1}{30}\right) \\ = 900\left(x+\frac{7}{30}\right)\left(x+\frac{1}{15}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Reference (database) for polynomials
Suppose I want to know what is known about the following polynomial
$$x^4 - 2x^2y^2 - 2x^2z^2 + y^4 - 2y^2z^2 + z^4.$$
Where can one find such information?
| It's just
$$(x+y+z)(x-y-z)(x+y-z)(x-y+z).$$
We can get it by the following way.
$$x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2=$$
$$=x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2-4y^2z^2=$$
$$=(x^2-y^2-z^2)^2-(2yz)^2=$$
$$=(x^2-y^2-z^2-2yz)(x^2-y^2-z^2+2yz)=$$
$$=(x^2-(y+z)^2)(x^2-(y-z)^2),$$
which gives the final factorization.
Also, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$ for $A$, $B$, $C$ the angles of a triangle Given that $A$,$B$ and $C$ are angles of a triangle, show that
$$1 + \cos 2C - \cos 2A - \cos 2B = 4\sin A\sin B\cos C$$
| We have: $1+\cos(2C) = 2\cos^2C, \cos(2A)+\cos(2B) = 2\cos(A+B)\cos(A-B)=-2\cos C\cos(A-B)\implies LHS = 2\cos C(\cos C+ \cos(A-B))= 2\cos C(\cos(A-B) - \cos(A+B)) = 2\cos C(-2\sin A\sin (-B)) = RHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Computing determinant without expansion $$\begin{align}\mathrm D &= \left|\begin{matrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)\left|\begin{matrix} b+c - a & a^2 & a^2 \\ b - a -c & (a+c)^2 & b^2 \\ 0 & c^2 & (a+b)^2 \end{matrix}\right| \\ &= (a+b+c)^2\left|\b... | We already reduced the problem to calculate
$$D’=\left|\begin{matrix}
b+c - a & 0 & a^2 \\
0 & a+c - b & b^2 \\
b & a & -ab
\end{matrix}\right|$$
If $a=0$ then
$$D’=\left|\begin{matrix}
b+c & 0 & 0\\
0 & c - b & b^2 \\
b & 0 & 0
\end{matrix}\right|=0.$$
If $b=0$ then
$$D’=\left|\begin{matrix}
c - a & 0 & a^2 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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prove that $(x^2 + y^2 - z^2)^2 = 4x^2y^2$. We have
$$x\cos \theta+y\cos \phi = -z\cos \psi \tag 1$$
$$x\sin \theta+y\sin \phi = -z\sin \psi \tag 2$$
$$x\sec \theta+y\sec \phi = -z\sec \psi \tag 3$$
and we have to prove that
$$(x^2 + y^2 - z^2)^2 = 4x^2y^2$$
squaring (1) & (2), and adding them we have
$$x^2 + y^2 - z^2... | multiplying (2) & (3), we have
$$x^2\tan\theta+y^2\tan\phi+xy\left(\sin\theta \sec\phi+\sin\phi\sec\theta \right)=z^2\tan\psi$$
Therefore in simplest terms $\tan\theta=1$, $\tan\phi=1$ and $\tan\psi=1$ whilst at the same time
$\cos(\theta - \phi) =-1$ from squaring (1) and (2).
One solution that meets these constraint... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of the constraints and... | (pqr method) Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
We need to prove that $p^2 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$
or $9 - 2q \le (27 - 15\sqrt{3})\frac{q}{r}$.
Since $(a-b)^2(b-c)^2(c-a)^2 = -4p^3r+p^2q^2+18pqr-4q^3-27r^2 \ge 0$,
we have
$r\le q - 2 + \frac{2}{9}\sqrt{3(3-q)^3}$.
It suffices to prove t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show that $f(x)$ is an Odd Function
Show that $$f(x) = \ln \left(x+\sqrt{x^2+1}\right)$$ is an odd function.
My attempt:
$$f(-x)=\ln\left(-x+\sqrt{(-x)^2+1}\right)=\ln\left(-x+\sqrt{x^2+1}\right).$$
How should I proceed? I know that if $f(-x)=-f(x)$, the function is odd.
| One more:
$f(x) = \ln (x + \sqrt{x^2+1})$, $x \in \mathbb{R}.$
$\star)$ $f(-x) = \ln(-x + \sqrt{x^2 +1}) =$
$\ln (\frac{1}{x+\sqrt{x^2+1}} )=$
$- \ln (x + \sqrt{x^2+1})$.
Combining:
$f(-x) = - f(x)$.
Used :
$ \ln [( -x +\sqrt{x^2+1}) \frac{x + \sqrt{x^2 +1}}{x+\sqrt{x^2+1}}] =$
$\ln ( \frac{1}{x + \sqrt{x^2+1}} )=$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Proving the sequence is increasing Suppose that a sequence $a_n$ of real numbers satisfies $7a_{n+1 }= a_n^3+6$
for $n ≥ 1$. If $a_1 =\frac{1}{2}$
, prove
that the sequence increases and find its limit.
Now I calculated the difference of two successive terms and factorized it. I got $$a_{n+1}-a_{n}=\frac{(a_n-1)(a_n-2)... | Use induction you can show $a_n$ is always less than $1$ so the product is positive.
$a_{n+1} = \frac{a_n^3+6}{7}$, so when $a_n<1$, $a_n^3<1$, which implies $a_{n+1} = \frac{a_n^3+6}{7}<1$. Base case is $a_1=\frac{1}{2}$.
So $(a_n-1)(a_n-2)(a_n+3)>0$. Thus $a_{n+1}-a_n>0$.
Now the limit can be found by investigating ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $y^3 + 3a^2x + x^3 = 0,$ then prove that $y'' + \frac{2a^2x^2}{y^5} = 0$ I am comfortable with second derivatives but I am just unable to set all the variables up in such a format that I get the latter (the part which needs to be proven). A hint would be a lot of help.
| You can also differentiate explicitely:
$$y=(-3a^2x-x^3)^{1/3}.$$
$$y'=\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-3a^2-3x^2).$$
$$y''=-\frac{2}{3^2}(-3a^2x-x^3)^{-5/3}\cdot (-3a^2-3x^2)^2+\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-6x)=$$
$$-\frac{2(a^2+x^2)^2}{y^5}-\frac{2x}{y^2}=\frac{-2(a^2+x^2)^2-2x(-3a^2x-x^3)}{y^5}=\frac{2a^2x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\... | \begin{align*}\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}&=\lim_{x\to1}\frac{-\pi\sin\pi x}{2\pi\sin\pi x\cos\pi x}\\&=-\lim_{x\to1}\frac1{2\cos\pi x}\\&=\frac12\end{align*}and therefore$$\lim_{x\to1}\frac{1+\cos\pi x}{\tan^2\pi x}=\lim_{x\to1}\frac{1+\cos\pi x}{\sin^2\pi x}\times\lim_{x\to1}\cos^2\pi x=\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Correct logic of permuting 5 men and 5 women to find probability of different highest women rank The problem reads like this:
Problem
Five men and $5$ women are ranked according to their scores on an examination. Assume that no two scores are alike and all $10!$ possible rankings are equally likely. Let $X$ denote t... | In your calculations,
*
*$5$ is the number of possible top-ranked women
*$\binom{5}{k}$ is the number of ways $k$ of the five men can have a lower rank than the top-ranked woman
*$(9 - k)!$ is the number of ways of arranging the $9 - k$ people whose rankings are lower than that of the top-ranked woman
*$10!$ is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving an infinite product of consecutive square roots Given $a$ and $b$ calculate $ab$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$
I simplified the terms and further obtained that $ab$ is equal to:
$$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7... | First answer:
Notice that:
$$\color{Blue}{a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}} \ \ \ ;$$
$$\color{Red}{b=}\sqrt{\color{Red}{2}\color{Blue}{\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}} \ \ \ ;$$
which implies that:
$$
\color{Red}{b}=\sqrt{\color{Red}{2}\color{Blue}{a}}
\ \ \ ;
$$
similarly we have:
$$a=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
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How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is......
... | Using combinatorics we find that the answer is coefficient of $x^{20} $ in $(x+x^2+x^3+x^4+x^5)(x^2+x^3+..x^6)(x^3..+x^7)(x^4+..+x^8)(x^5+..+x^9)=x^{15}(1+x+x^2+x^3+x^4)^5$ which is $7$ ie ways are $(1,1,1,x,x^4), (1,1,x,x,x^3), (1,1,1,x^2,x^3), (x,x,x,x,x), (1,1,x,x^2,x^2),(1,x,x,x,x^2), (1,1,x,x,x^3) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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Find $\lim_{n\to\infty}x_n\sqrt{\log n}$.
Let $0<x_0<\pi$ and consider the sequence defined by $x_{n+1}=x_n-\frac{x_n-\sin(x_n)}{n+1}$. Calculate $\lim_{n\to\infty}x_n\sqrt{\log n}$.
I did a little code in C to guess an answer and it seems like this limit is $\sqrt{3}$.
I proved that $0<x_n<\pi$ for all $n$. However ... | By writing $x_{n+1} = \frac{n}{n+1} x_n + \frac{1}{n}\sin x_n$, we can prove that $(x_n)$ is monotone decreasing. So $(x_n)$ converges to some value $\ell \in [0, x_0)$.
Now introduce the function $f(x) = \frac{x - \sin x}{x^3}$ and notice that (1) this can be continuated at $x = 0$ with value $f(0) = \frac{1}{6}$, and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not su... |
$\iff 0= a^2+b^2+ab$
Multiplying by $a-b$ gives $a^3-b^3=0 \iff a^3 = b^3\,$, then taking the (real) cube roots $a=b\,$. But it is readily verified that $\frac{1}{a+a} \ne \frac{1}{a} + \frac{1}{a}\,$ for any $\forall a \in \mathbb{R}\,$, so there are no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$
What is the remainder of $\frac{P(X)}{x^3+x}$?
I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^... | $x^3+x=x(x+i)(x-i)$
Remainder of $P(x)/(x^3+x)$ is a (at most) second degree polynomial
$R(x)=a x^2+bx+c$
We have $P(i)=-1+2i$ and $R(i)=-a + i b + c$ so $-a + i b + c=-1+2i$
then $P(-i)=-1-2i$ and $R(-i)=-a - i b + c$ so $-a - i b + c=-1-2i$
and $P(0)=1$ and $R(0)=c$
So $c=1$. Substitute in the first two equations
$-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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How does this proof that $xyz \le\left (\frac{x+y+z}{3}\right)^3$ work? I have shown that the function $f(x)=x\left(\frac{na-x}{n-1}\right)^{n-1}$, $n \ge 2, a \ > 0$ with domain $0 \le x \le na$ achieves a global maximum at $x=a$. I am then supposed to show, using this function when $n=2$ and $n=3$ that if $x + y + z ... | First, show that $xy\le\left(\frac{x+y}{2}\right)^2$.
This can be done by applying your result with $n=2$ and $a = \frac{x+y}{2}$. Indeed, we get
$$ x\left(\frac{2a-x}{2-1}\right)^{2-1} \le a\left(\frac{2a-a}{2-1}\right)^{2-1} = a^2 $$
since the function $x\mapsto x\left(\frac{2a-x}{2-1}\right)^{2-1}$ attains a maximum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{... | Let $\,P_n(x)=1+x+\cdots+x^{n-1}=(x^n-1)/(x-1)\,$, then the problem is equivalent to finding the remainder of the division $\,P_{12}(x^2) / P_{12}(x)\,$.
The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
$$
\begin{align}
P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solve the system $x^2-3xy+2y^2+x-y=0$, $2x^2-2xy-3y^2-2x+5y=0$
Solve the following system of equations:($x,y \in \Bbb R)$
$$\begin{cases} x^2-3xy+2y^2+x-y=0 \\ 2x^2-2xy-3y^2-2x+5y=0
\end{cases}$$
It's easily seen that $(0,0)$ and $(1,1)$ are two set of answers,but I can't analyze it to find all answers...
A... | $$\left\{\begin{array}{l} x^2-3xy+2y^2+x-y=0 \\
2x^2-2xy-3y^2-2x+5y=0
\end{array}\right.$$
substitute $x=X+1;\;y=Y+1$ in the first equation
$(X+1)^2+-3 (X+1) (Y+1)+2 (Y+1)^2+X+1-Y-1=0$
becomes $X^2 - 3 X Y + 2 Y^2=0$
$X=x-1;\;Y=y-1$
Solve $X=Y;\;X=2Y$ and then $x=y;\;x-1=2y-2\to x=2y-1$
Plug in the second equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | Ok, here is a different approach: Using graphs. In the $a,b$ coordinate plane, $a+b=1$ represents a line and $a^2+b^2=\frac{1}{2}$ represents a circle. Now when you graph (You can do that) you notice that the line falls completely outside the circle, except for one point where the line is tangent. Conclusion? Note:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent?
If we drop out the $n=0$ term and fold the remaining sum, the qu... | Your two expressions are not equivalent. The sum is not uniformly convergent, so the order you sum the terms in can matter. The second has an explicit order of summing while the first does not and will cancel out the constant terms. Even the second one does not converge, however.
We have $\sqrt {(n+a)^2+b^2}-n \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $... | Using Macaulay2 (version 1.16) with package SumsOfSquares:
i1 : needsPackage( "SumsOfSquares" );
i2 : R = QQ[x,y,z];
i3 : f = (x^2 + y^2 + z^2)^2 - 3*( x^3 * y + y^3 * z + z^3 * x);
i4 : sosPoly solveSOS f
1 2 1 1 1 2 2 9 1 2 2 2 1 2 2
o4 = (3)(- -x + x*y - -x*z -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Partial Fractions Decomposition $\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$ explanation repeated factors I am trying to solve the fraction $$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2}$$
into partial fractions.
Now, I thought it could be solved into the following
$$\frac{6x^2 - 29x - 29}{(x+1)(x-3)^2} = \frac{A}{x+1} + \frac{B}{(... | Both decompositions used by your friend and the text are analogous since
$$\frac{Bx+C}{(x-3)^2}=\frac{Bx}{(x-3)^2}+\frac{C}{(x-3)^2}=\frac{B(x-3)+3B}{(x-3)^2}+\frac{C}{(x-3)^2}$$
$$=\frac{B}{x-3}+\frac{3B+C}{(x-3)^2}=\frac{B'}{x-3}+\frac{C'}{(x-3)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the sum $\sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$ Let $a$ and $b$ be natural numbers such that $a \geq b \geq 1$. How can we evaluate the following sum?
$$ \sum_{n=1}^{b}n\binom{a}{n}\binom{b}{n}$$
| For any polynomial $p(z) = a_0 + a_1 z + \cdots a_n z^n$, let $[z^k]p(z)$ be the
coefficient $a_k$ for the monomial $z^k$ in $p(z)$. We have
$$\begin{align}\sum_{n=1}^b n \binom{a}{n}\binom{b}{n}
&= \sum_{n=0}^b \binom{a}{a-n}\times n\binom{b}{n}
= \sum_{n=0}^b [z^{a-n}]\{(1+z)^a\} \times [z^n]\{z\frac{d}{dz}(1+z)^b\}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the value of $x$, for which the function $h(x) = \frac 1{(x-5)^2+4(x-5)+4}$ is undefined Picture of problem
I'm having an issue with a simple SAT question. If you can't see from the picture, I'm being asked to find the value of $x$, for which the function:
$$h(x) = \frac 1{(x-5)^2+4(x-5)+4}$$
is undefined.
I figu... | The denominator has to be zero for the chosen $x$ values of course;
$(x-5)^2+4(x-5)+4=0$
We can name $(x-5)=a$ to make things easier.
$a^2+4a+4=0$
That is a perfect square;
$(a+2)^2=0$
$a=-2$ so $x-5=-2 \rightarrow x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
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Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}... | \begin{align}
Hint:(3\cos x-4\sin x)^2&=25-(3\sin x+4\cos x)^2\\
\int\frac{(3\cos x-4\sin x)^2}{3\sin x+4\cos x}dx&=\int\frac{25-(3\sin x+4\cos x)^2}{3\sin x+4\cos x}dx\\
&=25\int\frac{dx}{3\sin x+4\cos x}-\int(3\sin x+4\cos x)dx\\
&=25\int\frac{dx}{3\sin x-4+4(1+\cos x)}+3\cos x-4\sin x\\
&=25\int\frac{\frac{1}{1+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is al... | Basic things to note about $(2^x + y)(2^x - y) = ab = 60; a = 2^x + y; b=2^x-y)$.
1) $2^x + y > 0$ so $2^x - y > 0$.
2) $a= 2^x + y > 2^x -y=b > 0$ so $a > \sqrt{60} > 7$ and $b < \sqrt{60} < 8$.
3) $a$ and $b$ can't both be odd so they are both even.
So $b = 2,\color{red}4,6$ and $a=30,\color{red}{15},10$
4) $a+b = 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Prove that $\left|\begin{smallmatrix}a&-b&-c&-d\\b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{smallmatrix}\right|=(a^2+b^2+c^2+d^2)^2$
Let $a, b, c, d \in \mathbb K$ where $\mathbb K$ is a field. Prove that
$$\det \begin{bmatrix}
a & -b & -c & -d\\
b & a & -d & c\\
c & d & a & -b\\
d & -c & b & a
\end{bmatrix} = (a^2+b^2+c^2+d... | Calculate $$ P P^T $$ then think about it.
Or
$$ P^T P $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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What special functions do we need to get a closed form for $\sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}$? This question generalizes the considerations in Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ .
Let $p\ge 3... | We provide a partial answer to question number one. It is not hard to see that one can construct anti-derivatives of the integrand by using poly-logarithms.
As a matter of fact we have:
\begin{eqnarray}
\int \log(z) \log(z+1) \frac{1}{z+1} dz&=& \frac{1}{2} \imath \pi [\log(1+z)]^2- \log(1+z) Li_2(1+z)+Li_3(1+z)\\
\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2... | You made a mistake when you wrote $$a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
In fact,
$$a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2\color{red}{-}2b^2)=2.$$
From here, you can conclude that $a^2=2b^2$ as the right-hand side is not a function of $t$. Consequently, $a^2=2$, and $b^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:
Prove by induction that $$ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$$
My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and $\ 2^{n+1} > n^{2} + 3$
WTS $ \ 2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 =... | $n^2-2n+2=(n-1)^2+1>0$.
By the way, this reasoning gives also the proof without an induction:
$$2^{n+1}-n^2-3=(1+1)^{n+1}-n^2-3\geq$$
$$\geq1+(n+1)+\frac{(n+1)n}{2}+\frac{(n+1)n(n-1)}{6}-n^2-3=$$
$$=\frac{(n-1)(n^2-2n+6)}{6}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Polynomials : $P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$ Find all polynomials $P(x) \in \mathbb{R}[x]$ such that $$P(xy)+P(yz)+P(zx) =P(xy+yz+zx)$$
$\forall x, y, z \in \mathbb{R}$ satisfying the equation $x+y+z=0$
My attempt :
Consider $P(0,0,0)$, $3P(0)=P(0)$, so $P(0)=0$
Since $x+y+z=0$, let $x=a-b, \;y=b-c, \;z=c-a, \;\fo... | Working first with $y=x$ and $z=-2x$, we obtain
$$P(x^2)+ 2 P(-2 x^2) = P(-3 x^2)$$
Let $a_n x^n$ be the highest order term in $P(x)$, this relation implies
$$a_n x^{2n} + 2 (-2)^n a_n x^{2n} = (-3)^n a_n x^{2n} \quad\Longrightarrow\quad 1 + 2 (-2)^n = (-3)^n$$
The only solutions are $n = 1$ or $n = 2$. So that $P$ has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\... | After squaring of the both sides you'll get
$$(2k+1)^2(3k+4)\leq4(k+1)^2(3k+1)$$ or
$$19k\leq20k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to show that the deltoid is a plane algebraic curve of degree $4$?
How can I show that a deltoid is a plane algebraic curve of degree 4?
I have searched that the parametric equation for deltoid is given by
$$\begin{align}
x&= 2 \cos t + \cos 2t \\
y&=2 \sin t - \sin 2t
\end{align}$$
So, by using some trigonometri... | In M2 setting $c=\cos(t), s=\sin(t)$ using $c^2+s^2-1=0$:
R=QQ[c,s,x,y,MonomialOrder=>Lex]
I=ideal(c^2+s^2-1,x-(2*c+c^2-s^2),y-(2*s-2*s*c))
gens gb I
yields $x^4-8x^3+2x^2y^2+18x^2+24xy^2+y^4+18y^2-27$ or as wikipedia puts it (for $b=3, a=1$): $$(x^2+y^2)^2+18(x^2+y^2)-27=8(x^3-3xy^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Computing derivatives with fractional exponents I'm trying to compute the following derivative:
$$
\text{Using first principles, differentiate}: f'(x) = (x)^\frac{1}{4}\\\\
$$
I'm used to the functions being whole numbers or some simple algebra, i'm a little confused with what exactly to do when we're working with $(... | Note that
\begin{align*}
f(x + h) - f (x)&=\left(\sqrt[4]{x+h}-\sqrt[4]{x}\right)\cdot
\frac{\sqrt[4]{x+h}+\sqrt[4]{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt[4]{x+h}+\sqrt[4]{x}}\cdot
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2428132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^... | Assuming that the manipulations up till (4) are correct,
$$y^2 - x^2y=x^2$$
$$(y-\frac{x^2}{2})^2-\frac{x^4}{4}=x^2$$
$$(y-\frac{x^2}{2})^2=x^2+\frac{x^4}{4}$$
$y=\frac{x^2}{2}+\sqrt{x^2+\frac{x^4}{4}}$ or $y=\frac{x^2}{2}-\sqrt{x^2+\frac{x^4}{4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solve the equation $ z^2 + \left\vert z \right\vert = 0 $, where $z$ is a complex number. I've tried solving this, but I'm stuck at one point.
Here's what I did:
Let $ z = x + yi $, where $x, y \in \mathbf R$
Then , $ (x + yi)^2 + \sqrt{x^2 + y^2} = 0 $
Or, $x^2 + {(yi)}^2 + 2xyi + \sqrt{x^2 + y^2} = 0 $
Or, $ x^2 - y^... | Let $x,y\in\mathbb{R};\;z=x+iy$ the given equation $z^2+|z|=0$ becomes
$(x+iy)^2+\sqrt{x^2+y^2}=0$
$x^2-y^2+\sqrt{x^2+y^2}+ixy=0$
which translates in the system
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
y-2 x y=0 \\
\end{array}
\right.
$
$\left\{
\begin{array}{l}
x^2-y^2+\sqrt{x^2+y^2}=0 \\
2xy=0 \\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determine all real $x$ such that $\arccos{\frac{1-x^2}{1+x^2}}=-2\arctan{x}.$ I just need you guys to check this solution and tell me where I can improve. The key questions to be answered are:
*
*Are things clear?
*Is there any unnecessary redundancy?
*Any logical fallacies?
*Quality of stringency and mathematic... | $\cos (-2\arctan x) = \frac {1-x^2}{1+x^2}$
However
the range of $\arccos \theta$ is $[0,\pi]$
the range of $-2\arctan x$ is $(-\pi,\pi)$
These two functions can equal one another when $x$ is such that $-2\arctan x$ is in $[0,\pi)$
or $x\in(-\infty, 0]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Multiplying the first three equations with $x+y$, we get
\begin{eqnarray*}
(ax+by)(x+y) &=& ax^2+by^2 +(a+b)xy \\
(ax^2+by^2)(x+y) &=& ax^3+by^3 +(ax+by)xy \\
(ax^3+by^3)(x+y) &=& ax^4+by^4 +(ax^2+by^2)xy
\end{eqnarray*}
or
\begin{eqnarray*}
7(x+y) &= & \;\;49 +(a+b)xy \\
49(x+y) &= &133 +\;\;7xy \\
133(x+y) &=& 406 +4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Evaluating $\int_0^{2\pi}\frac{1}{1-\rho \sin{2 \phi }}\,\mathrm{d}\phi$ in verifying density function Let $(X,Y)$ have a bivariate normal density centered at the origin with $\mathbb{E}(X^2)=\mathbb{E}(Y^2)=1$ and $\mathbb{E}(XY)=\rho$. In polar coordinates, $(X,Y)$ becomes $(R,\Phi)$ where $R^2=X^2+Y^2$. Then it can ... | \begin{align}
t & = \tan\varphi \\[10pt]
dt & = \sec^2\varphi\,d\varphi = (1+\tan^2\varphi)\,d\varphi = (1+t^2)\,d\varphi, \\[10pt]
\text{so } \frac{dt}{1+t^2} & = d\varphi \\[10pt]
\text{and } \sin(2\varphi) & = 2\sin\varphi\cos\varphi = 2\sin(\arctan t)\cos(\arctan t) \\[10pt]
& \phantom{{}= 2\sin\varphi\cos\varphi} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| A different viewpoint, in case you don't like the (smart) change of variables $y:=x+1$:
First a bit of calculus: Observe that $f(x):=(x+2)^4+x^4-82$ is a twice-differentiable function, with second derivative $f''(x)=12(x+2)^2+12x^2$, which is positive everywhere. Hence $f$ is convex, so it has at most two real roots. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Given the problem : find the coefficient of $x^{100}$ in $\frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}$, why this solution works? We will use partial fractions so :
$\\ \frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}=\frac{1}{\left(1-x^5\right)(1-x^5)\left(1+x^5\right)}=\:\frac{A}{1-x^5}+\frac{B}{\left(1-x^5\rig... | example -
$(1-x^5)^2$ will be zero when $x=1$
if you multiply by $(1-x^5)^2$ on both sides of the equation (considering your second and third terms), you then have
$\frac{1}{1+x^5} = A(1-x^5) + B + \frac{C(1-x^5)^2}{1 + x^5}$
if you then go back to your case where x=1, the first and third terms on the right vanish (s... | {
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"url": "https://math.stackexchange.com/questions/2444393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the PDF of $Y = (X - \frac{1}{\theta})^2$ Let $X$ be a random variable with probability density function
$f(x) = \theta e^{(-\theta x)}$ for $x \geq 0$, otherwise $0$, when $\theta > 0$.
Let $Y = (X - \frac{1}{\theta})^2$. Find pdf of $Y$.
I'm just not sure how to incorporate $\theta$.
| Let $Y = (X -\frac{1}{\theta})^{2}$ we want the PDF of $Y$,
So Lets find the CDF of Y that is $F_{Y}(y)$ and to find the PDF we will differentiate $F_{Y}(y)$ so that $f(y) = F'_{Y}(y)$.
$F_{Y}(y) = P(Y \leq y) = P((X -\frac{1}{\theta})^{2} \leq y) = P(-\sqrt{y}+\frac{1}{\theta} \leq X \leq \frac{1}{\theta}+\sqrt{y})$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Is this a Taylor series? $\ln(x) +1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln(x))^n}{x}$ Can you provide a proof of this identity using only calculus?
$$\ln x + 1 = \sum_{n=0}^{\infty} \frac{n+1}{n!} \cdot \frac{(\ln x)^n}{x}$$
By the way, here is how I arrived at it:
There is string of length $x$ units. Se... | In more steps than maybe necessary:
$$\sum_{n=0}^\infty{n+1\over n!}{(\ln x)^n\over x}={1\over x}\left(\sum_{n=1}^\infty{(\ln x)^n\over(n-1)!}+\sum_{n=0}^\infty{(\ln x)^n\over n!} \right)={1\over x}\left(\ln x\sum_{n=1}^\infty{(\ln x)^{n-1}\over(n-1)!}+e^{\ln x} \right)={1\over x}\left(\ln x\sum_{k=0}^\infty{(\ln x)^k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to solve :$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$ $$\operatorname{arcsin}x-\operatorname{arcsin}\left(x/2\right)=\operatorname{arcsin}\left(\frac{x\sqrt3}2\right)$$
Why can't I figure this one out??
Is it possible to cancel out the arcsins?
I ... | this makes sense ony if $|x|<1$
\begin{split}
&&\arcsin x - \arcsin \dfrac x 2 = \arcsin \dfrac{x\sqrt3}2 \\
&&\sin\left(\arcsin x - \arcsin \dfrac x 2\right) =\sin\arcsin \dfrac{x\sqrt3}2 \\
&&\sin(\arcsin x )\cos(\arcsin \frac x 2)- \cos(\arcsin x )\sin (\arcsin \dfrac x 2) = \frac{x\sqrt3}{2} \\
&&x\sqrt{1-\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convergence and limit of $\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$ as $n \to \infty$? I was calculating in WA some values of some sequences and constructed this sequence:
$$a_n=\left((n+3)^{1/(n+3)}-(n+4)^{1/(n+4)}\right)^{1/(n+5)}$$
It is immediate that expression in parentheses tends to $0$ since both... | Define $f(x) := (1/x)^x$. Then we have
$$
(n + 3)^\frac{1}{n + 3} - (n + 4)^\frac{1}{n + 4} = f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right).
$$
By the mean value theorem
$$
f\left(\frac{1}{n + 3}\right) - f\left(\frac{1}{n + 4}\right) = f'(c_n) \left(\frac{1}{n + 3} - \frac{1}{n + 4}\right)
$$
for some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Matrix whose characteristic polynomial does not splits over field. I need an example of matrix which is non-zero and not nilpotent whose characteristic polynomial does not splits over field $\mathbb{R}$
| $$
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
Thursday morning: if we try to diagonalize over the reals, we have real numbers $a,b,c,d$ with $ad-bc \neq 0,$ with
$$
\frac{1}{ad-bc} \; \;
\left(
\begin{array}{rr}
d & -b \\
-c & a
\end{array}
\right)
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Equation of a plane by given 3 points I've read of solutions, with let's say points $P(1,2,3)$, $Q(3,1,2)$, $R(2,3,1)$, where one subtracts point $P-Q$ and $P-R$ to get two vectors:
$\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\\1\end{pmatrix}$
$\begin{pmatrix}1\\2\\3\end... | That's because “cross-product” doesn't mean to multiply each entry be the corresponding entry of the other vector. In this case$$\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}3\\3\\3\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer? For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer?
I managed to figure out:
$5p +49 = n^2$
$5p = (n+7)(n-7)$
But can't think of anything more than that, anyone able to solve this without bruteforcing?
For primenumbers under $10^7$, the solutions are $3$... | Then $n-7=5$ and $n+7=p$, that is $n=12$ and $p=19$.
Or $n-7=-5$ and $n+7=-p$, that is $n=2$ and $p=-9$ which is negative.
Or $n+7=5$ and $n-7=p$, that is $n=-2$ and $p=-9$ which is negative.
Or $n+7=-5$ and $n-7=-p$, that is $n=-12$ and $p=-19$, which is negative.
Or $n-7=1$ and $n+7=5p$, that is $n=8$ and $p=3$.
Or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$ I have to show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$.
I've already proven that if $f_n$ is the nth Fibonnaci number, then,
$$f_{n+1} > \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} $$
But I really don't get how to go from this statement to what I need to prove.
A... | Binomially expand
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 &=& \frac{1+5 \sqrt{5} +10 \times 5 +10 \times 5 \sqrt{5}+5 \times 25 +25 \sqrt{5}}{32} \\
&=&\frac{11+5 \sqrt{5}}{2}.
\end{eqnarray*}
Now $125>121 $ square root this and we have the stronger result that
\begin{eqnarray*}
\left( \frac{1+ \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving this Bernoulli ODE I have the following nonlinear ODE which I am trying to solve.
$$
\begin{align}
\frac{dr}{dt} &= ar(1-r) + b (1-r)^2\\
&= (b-2a) r^2 + (a+2b)r + b
\end{align}
$$
We then let $r = y + z$.
$$
\begin{align}
y '
&= (b-2a) (y-z)^2 + (a+2b)(y-z) + b\\
&= (b-2a) y^2 + y(2z(b-2a)+(a+2b))+((b-2a)z... | if $$r=y+z$$ it must be
$$y'=(b-2a)(y+z)^2+(a+2b)(y+z)+b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int \frac{1}{\sqrt{25y^2-30y-7}} \,dy$ $$\int \frac{1}{\sqrt{25y^2-30y-7}} \, dy$$
I start by completing the square: so that $(5y-3)^2-16=25y^2-30y-7$.
Then I let $u=5y-3$, and then let $ u=4 \sec \theta$.
This leads to the equation becoming $\frac{1}{4\tan\theta}$ which is the same as $\frac{1}{4}\cot \the... | I realized my mistake! I forgot to put in the du in my substitutions. I ended up with $\frac{1}{5}\int sec \theta$ which ends up being $\frac{1}{5}ln|sec\theta+tan \theta|$ which is $\frac{1}{5}ln|\frac{5y-3+\sqrt(25y^2-30y-7)}{4}$|+C. Hooray!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Area enclosed by quarter circle arcs in a square
The problem I have here asks to find the area $A$ enclosed by the two quarter circle arcs in terms of $a$. The first circle has radius $a/2$, and the second has radius $a$. This is what I did.
Firstly I determined that $$A=\frac{a^2}{8}(2\theta_1-\sin(2\theta_1))+\frac... | $\sin\theta_1=2\sin\theta_2$
$\frac a 2\cos\theta_1+a\cos\theta_2=a\sqrt 2 \implies\cos\theta_1+2\cos\theta_2=2\sqrt 2\implies\cos\theta_1=2\sqrt 2-2\cos\theta_2$
Then you can square both equations and add to each other
$$\sin^2\theta_1+\cos^2\theta_1=4\sin^2\theta_2+(2\sqrt2-2\cos\theta_2)^2$$
$$8\sqrt2 \cos\theta_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty o... | All Pythagorean triples can be generated by the formula:
$$A=(m^2-n^2)k\qquad B=2mnk\qquad C=(m^2+n^2)k$$
For a triple to be primitive, it is necessary but not sufficient that $k=1$. This leaves $B=2mn$. The smallest $m$ is $2$; $m,n$ must be of opposite parity; $m>n$, meaning 2mn must be a multiple of $4$
Every side-B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Napier analogy and algebra in triangle.
If in a $\triangle{ABC}$, we define $x=\tan\frac{B-C}{2}\tan\frac{A}{2}$, $y=\tan\frac{C-A}{2}\tan\frac{B}{2}$ and $z=\tan\frac{A-B}{2}\tan\frac{C}{2}$, then show that $x+y+z=-xyz$.
My attempts:
By Napier analogy,
$x=\frac{b-c}{b+c},\ y=\frac{c-a}{c+a},\ z=\frac{a-b}{a+b}$
Then... | Let $u = \tan \frac{B}{2}\tan \frac{C}{2}$, $v = \tan \frac{C}{2}\tan \frac{A}{2}$, $t = \tan \frac{A}{2}\tan \frac{B}{2}$. One has $u+v+t = 1$.
$x = \tan \frac{B-C}{2}\tan \frac{A}{2} = \frac{\tan\frac{B}{2}-\tan\frac{C}{2}}{1+\tan\frac{B}{2}\tan\frac{C}{2}}\tan \frac{A}{2} = \frac{t-v}{1+u}$.
$y = \frac{u-t}{1+v}$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluating limit using taylor expansions How can I evaluate this limit? Trying with Taylor doesn't seem to give me the right result, why ?
$$\lim_{x\to 0} \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$
with Taylor I can approximate $\sin(2x)$ to $ 2x $
and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits ... | $$\log(x^2+1)\sim x^2$$
$$\sin(2x)\sim 2x -\frac{8}{6}x^3$$
$$ \sqrt{1+4x} -1\sim 2x-2x^2 $$
since $$(\sqrt{1+4x} -1)'|_{x=0} = 2~~~and ~~~(\sqrt{1+4x} -1)"|_{x=0} = -4$$
$$\lim_{x\to0} \frac{\sqrt{1+4x} -1 -\sin{2x}}{\log{1+x^2}}\sim \lim_{x\to0}\frac{ -2x^2+\frac{8}{6}x^3}{ x^2} = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Prove by Induction that $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{2^n}\leq n$ The question asks to prove by induction that for every integer, $n\geq3$.
We have the sequence: $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\leq n$$
So from here, I started with a base case scenario where ... | Ok, so you want to show that
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \frac{1}{5} + \dots+\frac{1}{2^n}\leq n
$$
for all $n\geq 3$. I assume that the $3$ is there because the sum actually has $2^n$ terms. If
$$
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^k}\leq k
$$
for some $k$, you then want to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Arithmetic of complex numbers
If $z = \cos x + i\sin x$ , show that
$$\frac{2}{1+z} =1-i\tan\left(\frac{x}{2}\right),$$
I get up to:
$$\frac{1+\cos x -i\sin x}{1+\cos x}.$$
| You can apply the Weierstrass substitutions (a bit of a misnomer, but that's how they're commonly known).
Note that if $\displaystyle t = \tan \frac x2$, then:
$$\sin x = \frac{2t}{1+t^2}$$
and
$$\cos x = \frac{1-t^2}{1+t^2}$$
So:
$$z = \frac{1}{1+t^2}(1-t^2 + i2t)$$
$$1 + z = \frac{1}{1+t^2}(1-t^2 + i2t) + \frac{1+t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve $a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$ I am looking for a characterization of the solutions of
$$a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$$ in positive integers.
| I suppose that you mean
$$
a^2 + b^2+c^2+d^2 + 2ab - 2ac - 2ad - 2bc - 2bd + 2cd =N^2.
$$
Since the left hand side is $(a+b-c-d)^2$, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the limit of $2c^{n-1}-c^{n-2}-...-c-1$ when $n\to\infty$ I have a sequence in which the value of the $n^{th}$, $x_n$ is given by
$$x_n=2c^{n-1}-c^{n-2}-...-c-1$$
Where $c>0,$
then in the limit when $n\to\infty$ what is $x_n$?
| First simplify the expression. When $c \ne 1$, we have
$$\begin{align*}
2c^{n-1} - c^{n-2} - \cdots - c - 1
&= 2c^{n-1} - \sum_{i=0}^{n-2} c^i \\
&= 2c^{n-1} - \dfrac{c^{n-1} - 1}{c-1} \\
&= \frac{2c^n-3c^{n-1}+1}{c-1}
\end{align*}$$
Now consider what happens if $0<c<1$ and if $c>1$ separately.
When $c = 1$, the expres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y... | There's a mistake in the first equality: it's false that
$$
|x^2y-xy^2|=x^2|y|-|x|y^2
$$
However, by the triangle inequality,
$$
|x^2y-xy^2|\le x^2|y|+|x|y^2<x^2\sqrt{x^2+y^2}+y^2\sqrt{x^2+y^2}
$$
so
$$
\left|\frac{x^2y-xy^2}{x^2+y^2}\right|
<\frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}=\sqrt{x^2+y^2}
$$
Now the squeeze the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the constant and limit Here is the limit
$$\lim_{x\rightarrow-2} \frac{4x^2 + ax + a + 12}{x^2 + x - 2}.$$
a) Find the constant $a$
b) find the limit
I dont think it is solvable because it didn't tell me as $x \rightarrow -2$, $y \rightarrow $?
| I think it's saying that if the limit exists, find $a$ and find the limit.
If $x \to -2$ then the denominator $\to 0$, so in order for the limit to exist, then the numerator must also $\to 0$.
$(4x^2 +ax +a+12) \to (16 -2a +a+12) = (28-a)$ therefore $a = 28$
So that expression becomes:
$$\lim_{x \to -2} \frac{4x^2 + 28... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forg... | The first approach is good, except for its last step. The correct argument is
$$f(x,y,z)\ge -4x^2y^2\ge -(x^2+y^2)^2\ge -(x^2+y^2+z^2)^2 = -1.$$
Minimum attains at $x=y=\pm\sqrt2,z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a... | Substituting $$c=-a-b$$ in $$\frac{(a^2+b^2+c^2)(a^3+b^3+c^3)}{6}$$ we obtain $$-(a^2+ab+b^2)ab(a+b)$$ and so is the right-hand side
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Show that $\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $ I wonder why this is true
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$
Where the sum omits the case $n = m = 0$ ofcourse.
| (Too long for a comment.)
We have,
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 10n^2} = - \frac{2\pi \ln( \sqrt2\; U_{5})}{\sqrt {10}} $$
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{2\pi \ln( \sqrt2\; U_{29})}{\sqrt {58}} $$
with fundamental units $U_5 = \frac{1+\sqrt5}2$ and $U_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Euclidean Algorithm for polynomials I know how to use the extended euclidean algorithm for finding the GCD of integers but not polynomials. I can't really find any good explanations of it online. The question here is to find the GCD of
m(x) = $\ x^3+6x+7 $ and n(x) = $\ x^2+3x+2 $.
I try to use it the same way as for... | For $m(x) = ^3 + 6 + 7$ and $n(x) = ^2 + 3 + 2$,
Set
$m(x) = ^3 + 6 + 7 = p(x)n(x) + r_1(x) = p(x)(^2 + 3 + 2) + r_1(x)$
Where $r_1(x)$ is your remainder term. You want to find some p(x) that multiplies with n(x) to give you the first term of m(x). You also want $deg r_1(x) < deg n(x)$.
In this case, p(x) should be l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Prove $\frac {2 - \csc^2 A}{\csc^2 A\space + \space2\cot A} \equiv \frac {\sin A \space -\space \cos A} {\sin A \space+\space \cos A}$ So I have this rather simple trigonometric identity that, for the life of me, I cannot solve. I have worked on it for about 2 hours and still can't get it.
Show that
$$\frac {2 - \c... | You are nearly there! Let us go on with the identity $\sin^2 + \cos^2 = 1$ and some clever factorizations:
$$\frac{2 \sin^2 A - 1}{2 \sin A \cos A + 1} = \frac{\sin^2 A - \cos^2 A}{2\sin A \cos A + \sin^2 A + \cos^2 A} = \frac{(\sin A + \cos A)(\sin A - \cos A)}{(\sin A + \cos A)^2} = \frac{\sin A - \cos A}{\sin A + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
a prime divisor of $1+q+q^2+q^3+q^4$ Suppose $q$ is a prime power. Can we say anything about a prime divisor of $1+q+q^2+q^3+q^4$? Is any prime divisor congruent to 1 modulo 10?
| Let $p$ be a prime divisor of $1+q+q^2+q^3+q^4$, then
$$1+q+q^2+q^3+q^4 \equiv 0 \pmod{p} \implies q^5-1 \equiv 0 \pmod{p}$$
So $q$ has order $1$ or $5$ in the group $(\mathbb{Z}/p\mathbb{Z})^{\times}$. If it has order $1$, then $q\equiv 1 \pmod{p}$, hence $$5 \equiv 1+q+q^2+q^3+q^4 \equiv 0 \pmod{p} \implies p=5$$
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Understanding $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ by means of determinants. The identity comes from expanding the determinant
$$\begin{vmatrix}
a & b & c \\
c & a & b \\
b & c & a \\
\end{vmatrix}$$
in two ways.
The LHS comes from expanding the determinant by Sarrus' rule. The RHS comes from adding up all ... | Add $\omega$ times the second row and $\omega^2$ times the third row to the first row to get
$\begin{align}
\begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c\omega+b\omega^2 & b+a\omega+c\omega^2 & c+b\omega + a\omega^2 \\ c & a & b \\ b & c & a\end{vmatrix} \\
&=\begin{vmatrix}a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to solve the equations? $\begin{cases}x_1+x_2+x_3+x_4 = 6,\\
x_1^2+x_2^2+x_3^2+x_4^2 = 10,\\
x_1^3+x_2^3+x_3^3+x_4^3 = 18,\\
x_1^4+x_2^4+x_3^4+x_4^4 = 34.\end{cases}$
I have found a solution of $(1,1,2,2)$. And I tried to solve it by MAPLE.
All solutions are in the form of $(1,1,2,2)$ (and any possible arrangem... | As Rene comments, generally speaking, to solve equations of this kind, you should first transform it to
\begin{cases}x_1+x_2+x_3+x_4 = c_1,\\
x_1 x_2 + x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = c_2,\\
x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = c_3,\\
x_1x_2x_3x_4=c_4.\end{cases}
Then, by Vieta's formulas, you immediately ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2477938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Conjugate of Quaternion
The conjugate of $$\alpha=\left[ {\begin{array}{cc}
a+bi & c+di \\
-c+di & a-bi \\
\end{array} } \right]$$ is$$\overline{\alpha}=\left[ {\begin{array}{cc}
a-bi & -c-di \\
c-di & a+bi \\
\end{array} } \right]$$
The norm of $\alpha$ is $a^2+b^2+c^2+d^2$ and is written $\lvert\lver... | You did everything right except you need to go one more step:
$a,b,c,d$ are ordinary numbers and they commute multiplicatively, and they commute with $i$ as well. So $bibi = b^2i^2 = -b^2$ and
$$
a^2 - bibi + c^2 -didi = a^2+b^2+c^2+d^2 = t
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n... | Very short Answer
From this What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression
we have that,
$$ \sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}.$$
Therefore, By polynomial limit we have,
$$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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What is the probability that player A and B play against each other?
Players with same skill take part in a competition. The probability of
winning each game is 0.5. At first, we divide a group of $2^n$ people
to random pairs that play against each other. Then we will do the same
for $2^{n-1}$ winners and this c... | Hmm...I don't see any real difference between this and an ordinary single-elimination tournament where the seeding is set from the start. Maybe I'm missing something.
Anyway, assuming that there isn't any difference, we observe that Player A has $2^n-1$ different possible opponents in the first round. Of those, $2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{1}{xy}\ge4$ given that $x+y=1$ and conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge2$
Let $x,y\in\mathbb R^+$ and $x+y=1$
1- Prove that $\frac{1}{xy}\ge4$
2- Conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge25$
I have tried to start from $x+y=1$ or $x\ge0\land y\ge0$ and reach $\frac{1}{xy}\ge4$ but ... | \begin{eqnarray*}
x+y &=& 1 \\
x^2+2xy+y^2 &=& 1 \\
x^2-2xy+y^2 &=& 1-4xy \\
0 \leq (x-y)^2 &=& 1 -4xy.
\end{eqnarray*}
Now rearrange and we have the desired inequality.
For the second part
\begin{eqnarray*}
\left(\frac{1}{x} -\frac{1}{y}\right)^2 \geq 0 \\
\frac{1}{x^2} +\frac{1}{y^2} \geq \frac{2}{xy}
\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ Find Number of solutions of $\sin (2x)+\cos (2x)+\sin x+\cos x=1$ in $\left [0 \:\: 2\pi\right]$
The equation can be written as:
$$\sin (2x)+1-2 \sin^2x+\sin x+\cos x=1$$ $\implies$
$$\sin x+\cos x=2\sin^2 x-2 \sin x\cos x$$ $\implies$
$$\sin x+\cos x=2\sin ... | Another way would be to let $t=\tan(\frac x2)$ and arrive to $$t^4+2 t^3+8 t^2-6 t-1=0$$ Now, using the formulae for the quartic equation, the discriminant is $\Delta=-309248$ which shows that the equation has two distinct real roots and two complex conjugate non-real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the surface area of a cube given a diagonal? I encountered this problem while practicing for a mathematics competition.
A cube has a diagonal length of 10. What is the surface area of the cube? No Calculators Allowed.
(Emphasis mine)
I'm not even sure where to start with this, so I scribbled down so... | This problem has a simple solution: $s=2d^2$ where $s$ is the surface area and $d$ is the spacial diagonal.
Explanation:
I discovered through Google (while writing the question) that the side length of a cube can be calculated with $d = a\sqrt{3}$ where $d$ is the diagonal of the cube and $a$ is the length of a single ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Maximizing a polynomial expression involving the sidelengths of an inscribed pentagon
Given that $ABCDE$ is a convex pentagon and is inscribed in a circle of radius $1$ unit with $AE$ as diameter. If $AB=a,BC=b ,CD=c$ and $DE=d$, then maximum possible integral value of $a^2+b^2+c^2+d^2+abc+bcd$ is ?
What I tried
$AE ... | In quadrilateral $ABCE$ By Ptolemy's theorem
$$2b+av=(\sqrt {4-a^2})u$$
And similarly in quadrilateral $CDEA$
$$2c+du=(\sqrt {4-d^2})v$$
On squaring these equations and adding them we get $$4b^2+a^2v^2+4abv+4c^2+d^2u^2+4cdu$$
$$=4u^2-a^2u^2+4v^2-v^2d^2$$
Which simplifies to
$$4(b^2+c^2)+(a^2+d^2)(u^2+v^2)+4(abv+cdu)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Showing $ \left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq 1/3 $ I'm trying to prove the following inequality using Holder's Inequality:
$$ \left(\frac{a}{a + 2b}\right)^2 + \left(\frac{b}{b + 2c}\right)^2 + \left(\frac{c}{c + 2a}\right)^2 \geq \frac{1}{3}, $$
w... | By C-S twice we obtain:
$$\sum_{cyc}\left(\frac{a}{a+2b}\right)^2=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}\left(\frac{a}{a+2b}\right)^2\geq\frac{1}{3}\left(\sum_{cyc}\frac{a}{a+2b}\right)^2=$$
$$=\frac{1}{3}\left(\sum_{cyc}\frac{a^2}{a^2+2ab}\right)^2\geq\frac{1}{3}\left(\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}\right)^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.