Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integral of $\frac{\ln(1+x^2)}{1+x^2}$ Compute $\int\frac{\ln(1+x^2)}{1+x^2}dx$
Attempt:
I tried to do Integration by Parts
$$u=\ln(1+x^2)\Rightarrow du=\frac{2xdx}{1+x^2}\\
dv=\frac{dx}{1+x^2}\Rightarrow v=\arctan x\\
\int\frac{\ln(1+x^2)}{1+x^2}dx=\ln(1+x^2)\arctan x-\int\frac{2x\arctan x dx}{1+x^2}$$
Since it seems ... | $$\text{Let }I=\int\dfrac{\ln(1+x^2)}{1+x^2}\mathrm dx=\int\left(\underbrace{\dfrac{i\ln(1+x^2)}{2(x+i)}}_{I_1}-\underbrace{\dfrac{i\ln(1+x^2)}{2(x-i)}}_{I_2}\right)\mathrm dx$$
Solving $2/i\cdot I_1$:
$$\begin{bmatrix}u \\ \mathrm du\end{bmatrix}=\begin{bmatrix}x+i\\ \mathrm dx\end{bmatrix}$$ $$\begin{align}\int\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions to the diophantine equation $a^3+b^3+c^3=6$ I once thought that this equation $x^3+y^3+z^3=6$ has only a few smaller integer solutions. Until somebody told me this $6=192722201207819^3+162765491944499^3+(-225522344776678)^3$ existed. I don't know how it came about.
Question: How to find more integer s... | Nothing more than brute force computation I'm afraid.
The method used to crack $33$ was to rearrange to $$(a^2-ab+b^2)=\frac{33-c^3}{a+b}$$
The computer can then assign a random $c$ and test all values of $a,b$ such that $(a+b)|(33-c^3)$ which requires much less testing, but still very large amounts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}... | I assume:
$$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x\color{red}-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot \sqrt{x+1}+(\sqrt{x-1})^2}{ \sqrt{4x-4}+(\sqrt{x-1})^2(x+1)} =\\
\lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot (\sqrt{x+1}+\sqrt{x-1})}{\sqrt{x-1}\cdot (2+(\sqrt{x-1})(x+1))} =\frac{\sqrt{2}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$ Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$
My try: Since $y=x^2+ax+a^2+6a$ is an open upward Parabola, the roots $\alpha,\beta$ should be distinct and satisfy $1 \lt \alpha \lt 2$ and $1 \lt \beta \lt 2$ implies b... |
Find values of $a$ such that $x^2+ax+a^2+6a<0, ∀x\in (1,2)$.
As commented by Marty Cohen, the roots must be outside the interval:
$$x_1<1 \ \text{and} \ x_2>2 \iff \\
x_1=\frac{-a-\sqrt{-3a^2-24a}}{2}<1 \ \text{and} \ x_2=\frac{-a+\sqrt{-3a^2-24a}}{2}>2 \iff \\
1) \ \sqrt{-3a^2-24a}>-a-2 \ \text{and} \ 2) \ \sqrt{-3a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
proving complex Binomial Identity Proving the result $\displaystyle \sum^{\infty}_{n=0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$
what i try
$\displaystyle (1+x)^{n}=\sum^{n}_{r=0}\binom{n}{r}x^r$
$\displaystyle (x+1)^n=\sum^{n}_{r=0}\binom{n}{n-r}x^{n-r}$
Campare coefficient of $x^n$ on left and right
$\displaystyle \bin... | Note that
$$
\begin{align}
\binom{2n}{n}
&=\binom{2n-2}{n-1}\frac{2n(2n-1)}{n^2}\\
&=\binom{2n-2}{n-1}\frac{4n-2}{n}\tag1
\end{align}
$$
Multiply $(1)$ by $4^{-n}$ and set $a_n=4^{-n}\binom{2n}{n}$:
$$
\begin{align}
\overbrace{4^{-n}\binom{2n}{n}}^{\large a_n}
&=4^{-n}\binom{2n-2}{n-1}\frac{4n-2}{n}\\
&=\underbrace{4^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equality of $2^x + 2^{4-x} \geq 8$
Want to find the value(s) of $x$ for which equality holds in $$2^x + 2^{4-x} \geq 8$$
I've found it by solving $2^x + 2^{4-x} = 8$:
$$2^x + 2^{4-x} = 8 \Rightarrow 2^{2x} - 8 \cdot 2^x + 16 = 0 \Rightarrow (2^x - 4)^2 = 0$$
so clearly $x = 2$.
However, the notes I've been reading ... | You could argue it by the arithmetic-geometric mean inequality:
$$\frac{2^x + 2^{4 - x}}{2} \ge \sqrt{2^x \cdot 2^{4 - x}} = \sqrt{2^4} = 4,$$
with equality if and only if $2^x = 2^{4 - x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equival... | Let $f(x)=\frac{1}{x(k-x)}-\frac{x^2+(k-x)^2}{2},$ where $0<x<k$.
Thus, $$f'(x)=-\frac{k-2x}{(kx-x^2)^2}-x-x+k=(2x-k)\left(\frac{1}{(kx-x^2)^2}-1\right)=$$
$$=\frac{(2x-k)(1-kx+x^2)(1+kx-x^2)}{(kx-x^2)^2}.$$
We see that $$1+kx-x^2=1+x(k-x)>0.$$
Consider two cases.
*
*$0<k\leq2.$
Thus, $$1-kx+x^2=\left(x-\frac{k}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$.
What is the sum of all real numbers $x$ such that
$(x^2-5x+5)^{(x^2-7x+12)}=1$?
So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up.
Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.
Solv... | Three cases:
$$x^0=1$$
$$1^x=1$$
$$(-1)^{\text{even #}}=1$$
Substituting into the original equation gives $$x=3, 4$$ $$x=1, 4$$ $$x=2, 3$$
We don't add in the solutions already accounted for, namely, 3 and 4.
$\sum=3+4+1+2=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expected length of time until the mouse reaches the compost heap. I'm assuming it's a Markov Chain question but I have no idea how to do this. Thanks in advance!
A mouse lives in a mousehole with three exits.
*
*The first exit leads
to a compost heap after 1 minute of scurrying.
*The second exit leads
... | More difficult way! Refer to the table, where $X$ is the number of minutes:
$$\begin{array}{c|c|c}
\text{Excursions ($n$)}&X&P(X)&XP(X)\\
\hline
1&1&1/3&1\cdot 1/3\\
\hline
2&3+1&1/3^2&(3+1)\cdot 1/3^2\\
2&4+1&1/3^2&(4+1)\cdot 1/3^2\\
\hline
3&3+3+1&1/3^3&(3+3+1)\cdot 1/3^3\\
3&3+4+1&1/3^3&(3+4+1)\cdot 1/3^3\\
3&4+3+1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken? A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red o... | You can use the Hypergeometric distribution.
a) The probability that the first person pick up 4 red ones is $\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}$
Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$
b) The probability that the first person does not pick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Convergence of $\sum_{n=1}^\infty\frac{k^2}{k^2-2k+5}$, $\sum_{n=1}^\infty\frac{6\cdot 2^{2n-1}}{3^n}$, and $\sum_{n=1}^\infty\frac{2^n+4^n}{e^n}$ I have three series questions:
1) $$\sum_{n=1}^{\infty} \frac{k^2}{k^2 - 2k + 5}$$
I'm going to use this theorem:
So $$\lim_{x\to \infty} \frac{x^2}{x^2-2x+5}$$
So $$\li... | For your first series $\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}$, apply the Divergence Test as you did: $$\lim_{x\to\infty}\frac{x^2}{x^2-2x+5}=1\implies\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}\space\text{diverges}$$
For your second series, $$\sum_{n=1}^{\infty}\frac{6\cdot2^{2n-1}}{3^n}=6\sum_{n=1}^{\infty}\frac{2^{2n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The limit of functions of two variables The limit of $f(x,y)=\frac{xy^2}{x^2+y^4}$ as $(x,y) \longrightarrow(0,0)$ is doesn't exist, because if we take two paths:
1) Along the path, $x=0$, $y\longrightarrow0$:
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$
2) Along the path, $x=y^2$, $y\longrightarrow0$:
$... | You need to be careful and consider all paths as well, not only paths with constant $\theta$.
$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$$
$$=\lim_{r\longrightarrow0}\left(r\color{blue}{\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}}\right)$$
Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac... | The main trick is here:
\begin{align}
\color{red} {x = {x+y\over2} + {x-y\over2}}\\[1em]
\color{blue}{y = {x+y\over2} - {x-y\over2}}
\end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $\color{red}x$ and $\color{blue}y,\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Inequality $|\cos(k)| \geq \frac{1}{2^k}$ for $k\geq 0$ My question : Is it true that $|\cos(k)| \geq \frac{1}{2^k}$ for all integers $k\geq 0$ ?
What I tried : I have checked with a computer that the inequality holds for
$0 \leq k \leq 4\times 10^5$. I can also show that the set of $k$ for which the inequality holds i... | I eventually found a proof, based on i707107's suggestion. By using a computer
program with sufficient precision, one can check that the inequality holds
for $k<316$. So it will suffice to show the inequality for $k\geq 316$.
Denote by $a_k\frac{\pi}{2}$ the integral multiple of $\frac{\pi}{2}$ that's closest to $k$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Analytical solution to $\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2} \cdot \sqrt{b^2-x^2}$ Can't find beauty analytical solution to such equation :
$$\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2}\cdot\sqrt{b^2-x^2}$$
Assuming $a,\ b \in \mathbb{N},\ a \le b,\ x \in \mathbb{R}$
Is it possible to find a solution f... | Hint:
Let $\sqrt{a^2-x^2}=y$. We have
$$y+\sqrt{b^2-a^2+y^2}=y\sqrt{b^2-a^2+y^2},$$
$$y=(y-1)\sqrt{b^2-a^2+y^2}.$$
Squaring, you will obtain a quartic equation, which doesn't seem to simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate $\lim_{x\rightarrow\infty}\frac{x^3}{3^x}$. What is wrong with my calculations? $\frac{x^3}{3^x}=\frac{\exp(3\cdot\ln(x))}{\exp(x\cdot\ln(3))}=\exp(3\ln(x)-x\ln(3))$
$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$
I will use L'Hospital
$\frac{\frac{18(\ln(x))}{x}+2(\ln(3))^2x}{\frac{3}{x}+... | This is incorrect:
$$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$
There should be a minus sign in the numerator:
$$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2-x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\fr... | An easy method is to consider the function $f(x)=\arctan{x}+\arccos{\frac{x}{\sqrt{x^2+1}}}$ then show that $f’(x)=0$ and deduce that $f(x)$ is a constant function and therefore $f(x)=f(0)=\pi/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Is The Series $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$ Divergent This is my solution for $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$. First I let the sequence in the series be labeled $A$, then I constructed a new sequence ($B$) that would have the similar behavior as $A$. This gives be the following:
$$A=\frac{... | For sufficiently large $N_0$, we have $n \ge N_0$, then $$n^4-3n+2\ge \frac12 n^4$$
and $$4n^5+7 \le 5n^5.$$
That is $n \ge N_0$, we have
$$\frac{n^4-3n+2}{4n^5+7}\ge \frac{\frac12n^4}{5n^5}=\frac1{10n}$$
$\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}= \sum_{n=1}^{N_0-1} \frac{n^4-3n+2}{4n^5+7} + \sum_{n=N_0}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$ Let $x,y,u,v \in \mathbb{R}.$ Prove that
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Proof 1:
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Square both side, we have
$$... | By Minkowski (the triangle inequality) we obtain:
$$\sqrt{x^2+xy+y^2}+\sqrt{u^2+uv+v^2}=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2}+\sqrt{\left(u+\frac{v}{2}\right)^2+\frac{3}{4}v^2}\geq$$
$$\geq\sqrt{\left(x+u+\frac{y+v}{2}\right)^2+\frac{3}{4}(y+v)^2}=\sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)... | As $\sin^2t=1-\cos^2t=(1-?)(1+?)$
$$\dfrac{\sin t}{1\pm\cos t}=\dfrac{1\mp \cos t}{\sin t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Determine $N$ of series $\sum_{n=1}^{N}\frac{n^n}{(2n+1)!}$ so that it differs from the actual sum by less than $\frac{1}{200}$ I can establish that:
$$\frac{n^n}{(2n+1)!}=\frac{n^n}{(2n)!(2n+1)}\le\frac{n^n}{n!}$$
But
$$\sum_{n=1}^{+\infty}\frac{n^n}{n!}$$
diverges (by the ratio test). And even if it converged I would... | Note that\begin{align}\frac{\dfrac{(n+1)^{n+1}}{(2n+3)!}}{\dfrac{n^n}{(2n+1)!}}&=\frac{(n+1)^{n+1}}{n^n}\times\frac{(2n+1)!}{(2n+3)!}\\&=(n+1)\left(1+\frac1n\right)^n\times\frac1{(2n+2)(2n+3)}\\&=\frac12\left(1+\frac1n\right)^n\frac1{2n+3}\\&<\frac1{2n+3}\text{ since $\left(1+\frac1n\right)^n<e<4$}\\&<\frac12.\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Conditions for $ax^2+ax+1$ to be a perfect square. I look for the conditions for $ax^2+ax+1$ to be a perfect square, like some sort of recursive algorithm. I have a fixed $a$ and I would like to express $x$ in terms of $a$ or any other form. (My peculiar case is $52x^2+52x+1$). Any help would be appreciated!
| if $a=4$, $x$ is any integer.
Assume $a \ne 0$,
$ax^2+ax+1 = 1$, solution $x = 0, -1$
$ax^2+ax+1 = 4$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+12}}}-1\right)$
$ax^2+ax+1 = 9$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+32}}}-1\right)$
$ax^2+ax+1 = 16$, solution $x = \frac{1}{2}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Value of $\lim\limits_{n\rightarrow \infty}(a_{1}+a_{2}+\cdots +a_{n})$
If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$
Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is
Options:
$(a)$ Does not exists
$(b)$ Greater than $\displaystyle ... | Using the Wallis formula
$$\dfrac\pi2 = \dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\dfrac{6\cdot6}{5\cdot7}\dots$$
in the form of
$$a_n=\left(\dfrac{1\cdot2\cdot3\dots n}{3\cdot 5\cdot7\dots(2n+1)}\right)^2 < \frac\pi{(2n+1)2^{2n+1}},$$
one can get
$$a_1+a_2+a_3+a_4+\dots < \frac19+\dfrac\pi{2^5}\left(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Checking differentiability of $e^\frac{-1}{x}$
Let $f:\mathbb R\to\mathbb R$ is defined by
$$f(x)=
\begin{cases}
e^\frac{-1}{x}&&x>0\\
0&&x\le0
\end{cases}$$
How do I check differentiability of $f(x)$ at $x=0$?
I have tried to use first principle but cannot proceed.
| Use L'Hôpital's rule repeatedly as you approach $x = 0$ from the right:
\begin{align*}
\lim\limits_{x \to 0^+}f'(x) &= \lim\limits_{x \to 0^+}\frac{e^{-\frac{1}{x}}}{x^2} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{1}{x^2}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx$ - correctness of the solution The task is to find the improper integral: $$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx.$$
My solution is (the main part):
$$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx = \int \limits_{-1}^1 \frac{1}{(x-1)^{\frac{2}{... | Note that your answer $$(-2)^{\frac{1}{3}} =-1.259921...$$ is a real number not a complex one.
In general, one of the cube roots of a real number is always real and in this case we just consider the real answer for the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$ $$\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$$
$$\lim_{n\to \infty} \sum_{r=0}^{n-1} \left[\frac{1}{\sqrt{n^... | This is the same as
$$\sum_{r=0}^n\left(\frac1{n\sqrt{1+\frac{r}n}}\right)-\frac1{n\sqrt{1+\frac{r}{n}}}=\frac1n\sum_{r=0}^n\left(\frac1{\sqrt{1+\frac{r}n}}\right)-\frac1{n\sqrt{1+\frac{r}{n}}}$$
Taking the limit gives
$$\lim_{n\to\infty}\left(\frac1n\sum_{r=0}^n\left(\frac1{\sqrt{1+\frac{r}n}}\right)-\overbrace{\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now,... | Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=\frac{2}{3}$ gives the last line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Minimum possible values of $\frac{x^2+y^2+z^2+1}{xy+yz+z}$ and $\frac{x^2+y^2+z^2+1}{xy+y+z}$ Let $k$ and $m$ be the minimum possible values of $$\frac{x^2+y^2+z^2+1}{xy+yz+z} \quad \text{and} \quad \frac{x^2+y^2+z^2+1}{xy+y+z}$$ respectively where $x,y,z$ are non-negative real numbers. What is the value of $km+k+m$?
I... | Here is a way to use AM-GM. In the first case, you can find $k$ if you can find suitable $\alpha, \beta$ s.t. the following AM-GMs can achieve equality simultaneously:
$$x^2+\alpha^2 y^2 \geqslant 2\alpha x = kxy\\ (1-\alpha^2)y^2+\beta^2z^2\geqslant 2\sqrt{1-\alpha^2}\beta yz = kyz \\ (1-\beta^2)z^2+1 \geqslant 2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using a Lyapunov function to classify stability and sketching a phase portrait
Consider the system
$$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
Where $k$ is a given positive integer.
a.) Find and classify according to stability the equilibrium solutions.
$\it{Hint:}$ Let $V(x,y) = x^2 + y^2$
b.) Sketch a phase port... | Phase portraits - a partial offering
Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $\dot{y}=0$ and $\dot{y}=0$.
$k = 1$
The linear system is
$$\begin{align}
\begin{split}
\dot{x} &= -x^{3} - xy^{2} = -x \left( x^{2} + y^{2} \right) \\
\dot{y} &= -y^{3} - x^{2}y = -y \left( x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show uniform convergence and pointwise convergence for $\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$
Consider the series:
$$\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$$
show this converges to:
(a) $\frac{1}{(1-z)^2}$ for $|z|<1$
(b) $\frac{1}{z(1-z)^2}$ for $|z|>1$
Finally, show that this ... | Use partial fraction decomposition and exploit a telescoping sum to obtain the pointwise convergence results in (a) and (b). For example, we have
$$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{z^{n-1}}{z^n - z^{n+1}}\left(\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} \right) \\ = \frac{1}{z(1-z)}\left(\frac{1}{1-z^n} - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^... | Let the induction statement be such that
$$P(n):\prod_{i=0}^{n} 1 + q^{2^i}=\dfrac{1-q^{2^{n+1}}}{1-q}$$
We show the base $(n=0)$ case, which states
$$P(0):\prod_{i=0}^{0} 1 + q^{2^0}=\dfrac{1-q^{2^{0+1}}}{1-q}$$
Working from the right hand side, we have
$$\dfrac{1-q^{2^{0+1}}}{1-q}=\dfrac{1-q^{2}}{1-q}=\dfrac{(1-q)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \}... | If $\root 3 \of 3$ and $\root 3 \of 9$ are algebraic, then so is $\root 3 \of 3 + \root 3 \of 9$. The polynomial for $\root 3 \of 3$ is obviously $x^3 - 3$, and $x^3 - 9$ for $\root 3 \of 9$.
I wonder if maybe we can just add up the polynomials? We thus get $2x^3 - 12$. By the fundamental theorem of algebra, this has t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $z=e^{\frac{2\pi i}m}$, then $\sum_{k=1}^mz^k=0$ ($m\neq1$) Suppose $z=e^{\frac{2\pi i}m}$ for $m\in\mathbb N$ and $m\neq1$.
Is the following equality hold?
$$\sum_{k=1}^mz^k=0\tag{1}$$
$(1)$ seems trivial geometrically ;
it says that the sum of all vectors with equal magnitudes and uniform angle differences should ... | $\sum\limits_{k=0}^{m-1}z^{k}=\frac {z^{m}-1}{z-1}=0$ because $z^{m}=1$. Now just multiply both sides by $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given that $d$ divides $n$, what is $d + \frac{n}{d} \pmod 4$ As the question titles states, how can I efficiently find whether $d + \frac{n}{d} \equiv 0 \pmod 4$, given that $d$ divides $n$? An example would be with $n = 35$ and $d = 5$.
$5 + \frac{35}{5} = 5+7 = 12 \equiv 0 \pmod 4$, so it "passes". I tried $d + \fra... | If $n$ is odd, then $d$ is odd. In that case $d+(n/d)\equiv 0$ iff $x^2\equiv -n$ because $d$ is a unit $\bmod 4$. But for odd $d$ we must have $d^2\equiv 1$, so the ordered pair $(n,d)$ passes for odd $n$ iff $n\equiv 3\bmod 4$.
If $n$ is even, then $d$ and $n/d$ must both be even (meaning $n$ is a multiple of $4$).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry substitution issue with sign When solving an integral such as $\displaystyle\int\frac{dx}{\sqrt{x^2+4}}$, you eventually end up with
$$ \ln\lvert\sec\theta+\tan\theta\rvert+C.$$
The next step is to rewrite this in terms of $x$. My book does the following: $x=2\tan\theta$, so by drawing a triangle, it can ... | When you make the substitution $x=2\tan \theta$, you have to be careful to specify the domain of $\theta$: the substitution is only valid if $\theta$ has a small enough domain for $\tan \theta$ to be continuous. The simplest possible choice of domain is probably $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. Note that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
| Hint. Note that
$$(2\pm\sqrt{2})=\sqrt{(2\pm\sqrt{2})^2}=\sqrt{6\pm 4\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove $(c^2-1)/4 \pm 1$ is not divisible by $2, 3, $ or $5$ if $c>3$ is an odd prime In this question asked the other day, RTn conjectured that, if $c>3$ is an odd prime,
then $k=\dfrac{c^2\pm4-1}4=\dfrac{c^2-1}4\pm1$ is prime. As the answers and comments there show,
this conjecture is not true; some numbers of that... | If $c>3$ is an odd prime, then $c$ is not divisible by $2, 3, $ or $5,$
unless $c=5, $ in which case the claim can be easily verified.
If $c$ is not divisible by $2,$ then $c^2-1=(c+1)(c-1)$ is the product of two consecutive
even numbers so divisible by $8,$ so $\dfrac {c^2-1}4$ is even, so $\dfrac{c^2-1}4\pm1$ is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{10... | We have
$$100! \times (1 + \sum_{n=1}^{100} \frac{(-1)^n(n^2 + n + 1)}{n!})$$
$$= 100! \times (1 + \sum_{n=1}^{100} (-1)^n (\frac{n}{(n-1)!} + \frac{n+1}{n!}))$$
because $\frac{n^2 + n + 1}{n!} = \frac{n}{(n-1)!} + \frac{n+1}{n!}$. Expanding, we have
$$= 100! \times (1 + [-\frac{1}{(1-1)!} - \frac{1+1}{1!} + \frac{2}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\i... | $$\lim_{n\to\infty}\left(\frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(\frac{4n^2+3n-10+(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
From $\lim_{n\to\infty} \frac{2n+4}{4n^2+3n-10}=0$,
$$\Rightarrow \lim_{n\to\infty}\left(1+\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the basis of GF(2) I do past papers and I stumbled upon this question in one of the papers. I know what is GF(2), but I have no idea how to find the basis from the given data.
\begin{bmatrix}1\\1\\0\\0\end{bmatrix}
\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
The question asks me t... | Unsophisticated method:
There are $16$ elements of GF$(2)^4:$
$\begin{bmatrix}0\\0\\0\\0\end{bmatrix}
\begin{bmatrix}0\\0\\0\\1\end{bmatrix}
\begin{bmatrix}0\\0\\1\\0\end{bmatrix}
\begin{bmatrix}0\\0\\1\\1\end{bmatrix}
\begin{bmatrix}0\\1\\0\\0\end{bmatrix}
\begin{bmatrix}0\\1\\0\\1\end{bmatrix}
\begin{bmatrix}0\\1\\1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding center and rotation angle of ellipse that contains three points Given three points $p_1, p_2, p_3 \in \mathbb{R}^2$, and an ellipse with shape parameters $(a,b)$ (the semi-major and semi-minor), is it possible to determine, if they exist, a center $c \in \mathbb{R}^2$ and a rotation angle $\theta \in [0, \pi]$,... | Let the three points be $A,B,C$. The idea of this solution is to transform $\triangle ABC$ into an equilateral triangle with vertices $(0,0), (1,0), (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) $
There are three possible arrangements for the vertices in terms of the pre-images of the equilateral triangle vertices:
$A , B , C$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculate a limit: $\lim_{t \to 0+} {1\over 2}\cdot({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}) $ Please calculate the limit $\lim\limits_{t \to 0+} {1\over 2}\cdot \left({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}\right)$ and provide the corresponding procedure.
T... | Use Taylor series of an exponential function so that a direct computation shows
\begin{align*}&\frac{1}{2} \bigg(\frac{\pi}{t}\frac{1+e^{-2\pi t} }{1-e^{-2pi t} } -
\frac{1}{t^2} \bigg)
\\&= \frac{1}{2t} \bigg\{ \pi \frac{2+(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }{
(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }(-1) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Name for this method of factoring quadratic and are there any textbooks that describe it? I remember learning this method of factoring quadratics in middle school or high school, but looking for a name or more information on it leads me to dead ends.
Given:
$ax^2+bx+c=0$
$d*e=a*c$
$d+e=b$
Then the factorization of the... | This is method is known to me as middle term factor.
Lets take an example $f(x)=x^2+6x+8$. We have to find two numbers such that their sum is their product is $8$ and the sum is $6$. So, factors are $(x+4)(x+2)$.
In general, $ax^2+bx-c$ here constant term $ac$ is negative so we have to find two numbers such that their ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve intersection of the straight line with equation $y=x-a$ and the circle equation $x^2 + y^2 = 4$
Find the coordinates of the points of intersection of the straight line with equation $y=x-a$ and the circle with equation $x^2 + y^2 = 4 $. Give the conditions on a for:
I did see the answer from the practice paper:... | Substituting $y = x - a$ into $x^2 + y^2 = 4$ gives $x^2 + (x - a)^2 = 4$, which expands to $2x^2 - 2ax + a^2 - 4 = 0$. This is a quadratic equation in $x$ that you can then use the quadratic formula to solve to get $x = \frac{2a \pm \sqrt{4a^2 - 4(2)(a^2 - 4)}}{4} = \frac{2a \pm \sqrt{-4a^2 + 32}}{4} = \frac{a \pm \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof-Verification:$\int x[3+\ln(1+x^2)]\arctan x{\rm d}x$. $$\begin{aligned}
&\int x[3+\ln(1+x^2)]\arctan x{\rm d}x\\
=&\int 3x\arctan x{\rm d}x+\int x\ln(1+x^2)\arctan x{\rm d}x\\
=&\int \arctan x{\rm d}\left(\frac{3x^2}{2}\right)+\int \ln(1+x^2)\arctan x{\rm d}\left(\frac{x^2}{2}\right)\\
=&\int \arctan x{\rm d}\lef... | We can verify your solution by simply finding its derivative. If $F(x)$ is the antiderivative of $f(x)$, then we have $F'(x)=f(x)$.
$$F'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan(x)\left(x^2+\frac{1}{2}\left(x^2+1\right)\ln\left(x^2+1\right)\right)-\frac{1}{2}x\ln\left(x^2+1\right)+C\right]$$
Let us find the deriv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x =... | Let $A = 18°$
$$5A = 90°$$
$$⇒ 2A + 3A = 90˚$$
Taking sine on both sides, we get
$$\sin 2A = \sin (90˚ - 3A) = \cos 3A $$
$$⇒ 2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$$
$$⇒ 2 \sin A \cos A - 4 \cos^3A + 3 \cos A = 0 $$
$$⇒ \cos A (2 \sin A - 4 \cos^2 A + 3) = 0 $$
Dividing both sides by cos A
$$⇒ 2 \sin A - 4 (1 - sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that
$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.
Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
(A) $... | $a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$ $a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$ $1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$ $a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$ $\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$ $\Rightarrow \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
A problem on number of solutions of a functional equation: $ f ( x ) = 8 f ( 2 x + 1 ) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that $ f ( 0 ) = 1 $ and for all $ x \ne - 1 $,
$$ f ( x ) = 8 f ( 2 x + 1 ) \text . $$
I have found only one solution: $ \frac 1 { ( x + 1 ) ^ 3 } $. The method was by iter... | Let $g(x) = f(x-1)$. Then we obtain $g(x) = f(x-1) = 8f(2x-1) = 8g(2x)$, and $g(-1)=1$. Note that values of $g$ for positive $x$ cannot be related to negative $x$. Clearly, $g(0)$ is arbitrary.
To find solutions for positive $x$, we let $h(x) = g(2^x)$, giving $h(x) = 8h(x+1)$. This has solutions $8^{-x} h_1(x)$ for an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
complete set of values of $a$ having modulus and linear terms
If $(9-x^2)>|x-a|$ has at least one negative real solution for $a\in\mathbb{R}.$ Then complete set of values of $a$ is
Plan
If $x>a$ Then $9-x^2>x-a\Rightarrow x^2+x-(a+9)<0$
If $x\leq a$ Then $9-x^2>a-x\Rightarrow x^2-x+a-9<0$
How do i solve these inequ... | For the case $x>a$:
$$x^{2} + x - (a+9) < 0 $$
The roots are:
$$\frac{-1 \pm \sqrt{4a+37}}{2} $$
Notice $a \ge -37/4$. Clearly the solution is:
$$\frac{-1 - \sqrt{4a+37}}{2} < x < \frac{-1 + \sqrt{4a+37}}{2} $$
The least possible value for $x$ is therefore $-1/2$, and clearly $-1/2 > -37/4$, so $a \ge -37/4$ does not v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
| Let $a=\sqrt{3}$.
\begin{align*}
\text{Then}\;\;&x^2+x-3-\sqrt{3}=0\\[4pt]
\iff\;&x^2+x=3+\sqrt{3}\\[4pt]
\iff\;&x^2+x=a^2+a\\[4pt]
\end{align*}
So by inspection, we get the solution $x=a$.
By Vieta's formula, the sum of the roots is $-1$, so the other root is $-1-a$, which is negative.
Thus, $x=\sqrt{3}$ is the only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Number of onto functions from $Y$ to $X$ (JEE Advanced 2018)
Let $X$ be a set with $5$ elements and $Y$ be a set with $7$ elements. If $\beta$ is the number of onto functions from $Y$ to $X$ then the value of $\dfrac{\beta}{5!}$ is?
My approach is:
First I give each element of $Y$ one element of $X$ which then leave... | Your method counts each surjective function that maps three elements of $Y$ to one element of $X$ three times, once for each way you could designate one of those three elements as the element of $Y$ that maps to that element of $X$.
For example, consider the surjective function $f: Y \to X$ defined by $f(1) = 1$, $f(2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$ this series was evaluated by Cornel Valean here using series manipulation.
I took a different path as follows:
using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
multiply both sides by $\ln^3x/x$ ... | This solution is by Cornel Valean.
Using the follwing identity: ( see Lemma $2(b)$ in this paper)
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
and since $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{n+k-1}\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}$$
Then $$\sum_{k=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$
Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$, with $a,b \in \Bbb Q$
How I can prove it? I tried elevating to cube but then?
$4 = a^3+ 3a^2b \sqrt[3]{2} + 3ab^2\sqrt[3]{4}+ 2b^3$
| Assume for contradiction that $\sqrt[3] 4 = a + b \sqrt[3] 2$. We then cube both sides, to obtain $4 = a^3 + 3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 + 2b^3$. We have that $4, a^3, 2b^3$ are all rational, so we also have $3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 = q$ for some rational $q$. But $\sqrt[3] 4 = (\sqrt[3] 2)^2$, so w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$ Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$
The original problem is :
$$\text{Evaluate} \iint _R ye^{-x^2-y^2}dxdy$$
Where $R=\left\{ (x,y) \vert x\geq0,y\geq0\right\}$
I sub-ed $x=r\cos\theta, y=r\sin\theta$ and this changed to
$\left(\int_0^{\frac{\pi}{2}}\sin\theta d\theta \r... | Let $z=\sqrt{2}r$ so that $dz=\sqrt{2}dr$ and $r^2=\frac{1}{2}z^2$. Then:
$$
\int_0^\infty r^2e^{-r^2}dr=\int_0^\infty\frac{1}{2}z^2e^{-\frac{1}{2}z^2}\frac{dz}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\int_0^{\infty}z^2e^{-z^2/2}dz=\frac{1}{4\sqrt{2}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz.
$$
The last equality above uses symmetry.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Convergence of moving point inside of unit disk. Suppose Set D :
$$D=\left\{(x, y)\vert x^2 + y^2 \leq 1 \right\}$$
And Point $P(0, 0)$ on coordinate plane.
Define 'Movement' :
For a point P, select any direction and move $\frac{1}{2^n}$ to straight (Sorry, my english is poor.) When 'Movement' Execute $n$-th times
Fo... | Short answer: Not only you can do this, it is even possible to move on one line connecting the origin and your point of interest. Maybe not surprisingly, this is related to the binary representation of a number.
Consider a point $p=re^{i\theta}=x+iy$, and let us limit ourselves to movement at angle $\theta$ only (i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show integral is 2 Pi Consider $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{0}^{\infty}\biggr(\frac{1}{\sqrt{(1-x)^2+y^2+z^2}}-\frac{1}{\sqrt{x^2+y^2+z^2}}\biggr)^2dxdydz$$
A numerical study suggest the integral converges to $2\pi$. See: Link to numerical studyHowever I am unable to show this. When y,z are ... | It's not a problem that the integral does not converge on one line $y=z=0$, because a contribution from a single line is negligible for an integral over a volume.
Let us make a change of coordinates:
$$ x=r\cos\theta, \qquad y = r\sin\theta\cos\phi, \qquad z=r\sin\theta\sin\phi$$
$$ 0\le r\le\infty, \qquad 0\le\theta\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to expand $\sum_{k=0}^\infty \frac{k^k}{k!}e^{-k\lambda}$ I know that $\sum_{k=0}^\infty \dfrac{k^k}{k!}\dfrac{e^{-k-k\tfrac{s}{n}}}{\sqrt{n}}$ can be expanded in powers of $\frac{1}{\sqrt{n}}$ to yield:
$\frac1{\sqrt{2s}}+\frac{1}{3\sqrt{n}}+\frac{\sqrt{2s}}{12n}+\dotso$
My question is twofold: (i) is there an ac... | We will assume $s > 0$ or the series diverges.
Recall for $|z| \le \frac1e$, the $0$-branch of the Lambert W function has following expansion:
$$W_0(z) = \sum_{k=1}^\infty \frac{(-k)^{k-1}}{k!} z^k$$
Pick any $\lambda_0 > 1$ and substitute $z$ by $-e^{-\lambda}$ for $\lambda > \lambda_0$. Above expansion converges unif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing aro... | We have
$$
\eqalign{
& \left( {a,b,c,d} \right)\quad \Rightarrow \quad 2^{\,a} + 2^{\,b} + 2^{\,c} = 4^{\,d} \quad \Rightarrow \cr
& \Rightarrow \quad 2^{\,2n} \left( {2^{\,a} + 2^{\,b} + 2^{\,c} } \right) = 2^{\,2n} 4^{\,d} \quad \Rightarrow \cr
& \Rightarrow \quad 2^{\,a + 2n} + 2^{\,b + 2n} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expr... | This problem is equivalent to
$$
\min(\max) a x^2+b x y + c y^2 \ \ \mbox{s. t.}\ \ x^2+y^2=1
$$
this is an homogeneous problem so calling $y = \lambda x$ and substituting we have equivalently
$$
\min(\max) f(\lambda) = \frac{a+\lambda b+\lambda^2c}{1+\lambda^2}
$$
and the extremals condition is
$$
f'(\lambda) = 0\Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1... | We can think at the problem as a stars and bar problem, that uses the inclusion-exclusion principle, as in the answer cited by IV_ and this one.
We interprete
\begin{equation}(1+x+x^2 + \cdots + x^{n})^{n-1} = \underbrace{(1+x+x^2 + \cdots + x^{n})(1+x+x^2 + \cdots + x^{n})\cdots (1+x+x^2 + \cdots + x^{n})}_{n-1\text{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before ... | We have that $$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)}$$
$$ = \left[\frac{xy}{z^2(x + y)} + \frac{1}{z}\right] + \left[\frac{yz}{x^2(y + z)} + \frac{1}{x}\right] + \left[\frac{zx}{y^2(z + x)} + \frac{1}{y}\right] - \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$
$$ = \frac{xy + yz +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2... | Since by PM $$\left(\frac{x^n+1}{2}\right)^k\geq\left(\frac{x^k+1}{2}\right)^n$$ is true for all $x\geq0$ and $n\geq k>0,$ it's enough to prove that
$$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$ or $f(x)\geq0,$ where
$$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{2m+1}\ln\left(x^{2m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Solve First Order Linear PDE with Method of Characteristic Given a problem
$x^2\,u_x+xy\,u_y=y\,u$
Then, the characteristic equation is
$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$
From the first two of those equation, i got
$C_1=\dfrac{y}{x}$
How to obtain the other $C_2$? So that i have implicit solution $F(C_1)=C... | $$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$$
From the first two ratio
\begin{align}
\dfrac{dx}{x^2} &= \dfrac{dy}{xy} \\
\implies \dfrac{dx}{x} &= \dfrac{dy}{y} \\
\implies \log x - \log y &= \log c_1 \\
\implies\frac{x}{y} &= c_1
\end{align}
From the last two
\begin{align}
\dfrac{dy}{xy} &= \dfrac{du}{yu} \\
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What's the mistake in $\sin(\frac{a}{2})=\pm\sqrt{\frac{1\pm \cos(a)}{2}}$? When I was encountered with the formula of $\sin(\frac{a}{2})$, I tried to derive it. First, I tried from the formula of $\cos(2a)$ and I successfully did that but after a while I was curious about if that can be derived from sine's double angl... | I think, it's better
$$\left|\sin\frac{\alpha}{2}\right|=\sqrt{\frac{1-\cos\alpha}{2}}$$ and
$$\left|\cos\frac{\alpha}{2}\right|=\sqrt{\frac{1+\cos\alpha}{2}}$$
Your mistake is that $\sqrt{x^2}=\pm x$ is wrong.
The right identity it's:
$$\sqrt{x^2}=|x|.$$
For example, after
$$\sin^4\frac{\alpha}{2}-\sin^2\frac{\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What property was used in this sine transformation? I have this expression:
$$
ψ(χ) = A\sin^3(\frac{πχ}{α})
$$
And somehow the book i read equalizes the previous equation to this one:
$$
ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})]
$$
What trigonometric identity was used to make this possible?
| The identity is
$$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$
Proof:
\begin{align*}
4\sin^3x-3\sin x &= \sin x(4\sin^2x-3)
\\
&=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity}
\\
&=\sin x(2\cos^2x-\cos(2x)-4\cos^2x) \quad \text{Double angle cosine}
\\
&=-\sin x(2\cos^2x+\cos(2x))
\\
&=-[(2\sin x\cos x)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$.
$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$
This proble... | Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x+y}{z}$ and by AM-GM twice we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sum_{cyc}\frac{xy}{\sqrt{y^2(y+z)^2+x^2(x+z)^2}}\leq$$
$$\leq\sum_{cyc}\frac{xy}{\sqrt{2xy(x+z)(y+z)}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{xy}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Matrix complicated equation Let $$A = \begin{bmatrix}
1 & 3 & 4\\
3 & 6 & 9\\
1 & 6 & 4
\end{bmatrix},$$
$B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$
What would be the value of
$ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?
| $B$ is a red herring here and might be replaced by any invertible $3\times3$-matrix.
Since $\operatorname{adj}(C)=\det(C)C^{-1}$ for invertible $C$ we have
$$\det(\operatorname{adj}(C)=\det(C^{-1}\cdot\det(C))=(\det(C))^3\det(C^{-1})=(\det(C))^2$$
if $C$ is of type $3\times3$.
Now happily compute
\begin{align}\det( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Minimal Rook Difference Grids In the below grid all 18 orthogonal differences are distinct, with a difference of 18 missing.
Could the highest number be 18? The resulting graph would have valence 4, making it an Eulerian Graceful graph with edges(mod 4)=2. Rosa (1967) proved Eulerian Graceful graphs must have edges... | I used a depth first search written in C to find the following:
$rdg(3,4)=30$, so the $3\times4$ rook graph is graceful.
\begin{array}{|c|c|c|c|}
\hline
0 & 1 & 9 & 30 \\
\hline
16 & 29 & 2 & 19 \\
\hline
22 & 3 & 27 & 7 \\
\hline
\end{array}
$rdg(4,4)=48$, so the $4\times4$ rook graph is also gra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This s... | $$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\
\\
let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\
\\
\therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\
\\
let\ \ G=\int_{0}^{\infty }\frac{ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to get the value of the root? I have this statement:
If $\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3}$, Which of
the following values are the closest to $\sqrt{21}$ ?
A) 68/15 B) 14/3 C) 19/4 D) 55/12 E) 9/2
My development was:
$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{35}-\sqrt{21}}{2} \appr... | From
$$\left(\frac23\right)^2\approx\left(\frac{\sqrt7}{\sqrt5+\sqrt3}\right)^2$$
we get
$$\frac49\approx\frac{7}{5+3+2\sqrt{15}},$$
hence $\sqrt{15}\approx31/8$.
Now we have
$$\frac{2\sqrt{3}}3\approx
\frac{\sqrt7\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt{21}}{\sqrt5+\sqrt3}.$$
Using the approximate value for $\sqrt{15}$ we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down be... | 1) When $S=x^2-2xy-y$ and $T = xy-2y^2-x$, then $$ 5|S-T =
(x-y)(x-2y+1) \ (a)$$
and $$ 5| S+T = (x-2y-1)(x+y)\ (b)
$$
Hence I know that $(x-2y+1)-(x-2y-1)=2$, by mathlove's comment.
Hence at least one of $(x-2y+1),\ (x-2y-1)$ can not be divided by $5$. So by $(a),\ (b)$ we have $5|x^2-y^2$.
2) When $U=2x^2+y^2+2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
| The triples $(2a-b-c,2b-a-c,2c-a-b)$ and $\left(\frac{1}{(b+c)^2},\frac{1}{(a+c)^2},\frac{1}{(a+b)^2}\right)$ are the same ordered.
Thus, by Chebyshov we obtain:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}\geq\frac{1}{3}\sum_{cyc}(2a-b-c)\sum_{cyc}\frac{1}{(b+c)^2}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^... | Just for the fun of it, you can re-write your inequality as:
$$\frac{1}{n}-\frac{1}{n+1}>\frac{1}{(n+1)^2}.$$
This can be re-writen as:
$$\int_{n}^{n+1}\frac{1}{x^2}dx>\frac{1}{(n+1)^2},$$
or, if $f(x)=\dfrac{1}{x^2}$:
$$\int_{n}^{n+1}f(x)dx>f(n+1),$$
which is obvious if you consider that $f$ is strictly decreasing and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
sum of series $\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $
The sum of series
$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $$
My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-... | The approximate value of sum of terms in $HP$ is $S_n\approx \frac{1}{d}\ln(\frac{2a+2(n-1)d}{2a-d})$ here $a$=reciprocal of first term $(a_1=4,a_2=3)$ respectively. $d$=difference between the reciprocal of two terms $(d_1=3,d_2=2)$ respectively
Note that the two HPs are $3+3(\frac{1}{4}+\frac{1}{7}+....)-2(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Modular arithmetic with Legendre symbol Let $n\in\mathbb{Z}_{>0}$ and let $p\neq3$ be a prime divisor of $n^2+n+1$. Show that $p\equiv1\mod3$.
I thought of trying to prove that $\left(\frac{p}{3}\right)=1$, since 1 is the only element of $\mathbb{F}_3$ that is a square modulo 3. I am supposed to use quadratic reciproci... | Since $4(n^2+n+1)=(2n+1)^2+3$ is divisible by $p$, we have $-3$ is a quadratic residue mod $p$. So
\begin{align*}
1=\left(\frac{-3}{p}\right)&=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)\\
&=\left(\frac{3}{p}\right)(-1)^{(p-1)/2}\\
&=\left(\frac{p}{3}\vphantom{\frac3p}\right)(-1)^{\frac{p-1}{2}\cdot\frac{3-1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
mod cancellation: compute $\, n/2\bmod 6\, $ from $\,n \bmod m\,$ for even $n$ I am using the C++ language.
I want to calculate these 2 expressions:-
In our case, $x= 100000000000000000$
Expression(1)
$$((3^x-1)/2)\mod7$$
The numerator $3^x-1$ is always divisible by $2$(basic number theory)
I calculated the above exp... | Dividing by $\boldsymbol{q}$ when the modulus is divisible by $\boldsymbol{q}$
Note that
$$
aq\equiv bq\pmod{pq}
$$
precisely when
$$
a\equiv b\pmod{p}
$$
Thus, for $n\gt0$,
$$
3^n-1\equiv\left\{\begin{array}{}2&\text{if $n$ is odd}\\8&\text{if $n$ is even}\end{array}\right.\pmod{12}
$$
implies
$$
\frac{3^n-1}2\equiv\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ . Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove:
$$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$
My solution in M&Y : (and I'm looking forward to s... | There is another solution by Rearrangement and AM-GM.
Indeed, let again $\{a,b,c\}=\{x,y,z\},$ where $x\geq y\geq z$.
Thus, $x+y+z=3,$ $xy+xz+yz=2$ and $$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x^2+2xz+z^2)=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\fr... | You could have quite good approximations using generalized harmonic numbers
$$S_n=\sum_{i=0}^n \frac 1 {\sqrt{n+i}}=H_{2 n}^{\left(\frac{1}{2}\right)}-H_{n-1}^{\left(\frac{1}{2}\right)}$$
Using the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{p}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove the following identity of Gamma : $\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$
Prove
$$\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$$
We know that :
$\Gamma(x+\frac{1}{2})=\frac{(2... | By the reflection formula, if $z\notin\mathbb{Z}$,
$$\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$$
so you can evaluate $\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})$ and $\Gamma(\frac{3}{8}) \Gamma(\frac{5}{8})$ and verify that
$$\frac{\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that
$$a^2 + b^2 + c^2=d$$
then $a+b+c+d$ equals?
$$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$
Here is what I tried: assu... | Let $x$ be the common difference.
From $a^2+b^2+c^2=d$, we get $d > 0$,
Since $a,b,c,d$ are distinct, we get $x\ne 0$.
If $x < 0$, then $c > d$, hence
$$c^2 > d^2\ge d=a^2+b^2+c^2 > c^2$$
contradiction.
Hence $x > 0$.
\begin{align*}
\text{Then}\;\;&a^2+b^2+c^2=d\\[4pt]
\implies\;&(b-x)^2+b^2+(b+x)^2=b+2x\\[4pt]
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\c... | If you expand the squares, then you get
$$\begin{split} \sum_{k=1}^{n-1} \Big(a+\frac{k}{n}\Big)^2&=\sum_{k=1}^{n-1} \Big(a^2+2a\frac{k}{n}+\frac{k^2}{n^2}\Big)
\\
&=(n-1)a^2+2a\cdot \frac{1}{n}\cdot\frac{n(n-1)}{2}+\frac{1}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}.
\\
&=(n-1)a^2+a(n-1)+\frac{(n-1)(2n-1)}{6n},
\end{split}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$
I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$
I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving... | For $x>0$ we have:$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu\overset{vu=t}=\frac14\int_0^1\frac{1}{u}\int_0^u \frac{1}{\sqrt{t}\sqrt{1-xt}}dtdu$$
$$=\frac12\int_0^1 \frac{1}{u}\frac{\arcsin \sqrt{xt}}{\sqrt{x}}\bigg|_0^udu=\frac1{2\sqrt x} \int_0^1 \frac{\arcsin\sqrt{xu}}{u}du\overset{xu=t}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{... | In this answer I will try to consider generalizations, and organize things better. My first answer was a proof discovered by ad-hoc methods.
Lemma
If $x \equiv y \pmod p$, then $x^{p^n} \equiv y^{p^n} \pmod {p^{n+1}}$ for all $n \geq 0$.
Proof
The case $n = 0$ is trivial. We now prove that if $x \equiv y \pmod {p^k}$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Other methods for a limit I know that I can compute the limit
$$
\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}=\frac{N(N+1)}{2}
$$
using L'Hospital's rule (not one but two times) but I am looking for other ways. Are there any of them?
p.s.: the limit follows from a shortcut used in order to find the value of $$
\sum_... | You can simplify your fraction as
\begin{align}
\frac{Nx^{N+1}-Nx^N+1-x^N}{(1-x)^2} &=\frac{-Nx^N(1-x)+(1-x)(1+x+\dots+x^{N-1})}{(1-x)^2} \\
& = \frac{-Nx^N+1+x+\dots+x^{N-1}}{1-x} \\
& = \frac{1+x+\dots+x^{N-1}+x^N-(N+1)x^N}{1-x}
\end{align}
For $x \to 1$ the numerator tends to $0$, so this is still a form $0/0$ and y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$a_n=\min\limits_{x+y=1}(x^n+y^n)=\frac{1}{2^{n-1}}$ We must prove :$\min\limits_{x+y=1}(x^n+y^n)=\frac{1}{2^{n-1}} $ for all $n \in \mathbb{N}_{>0}$
And to prove this we can use the inequality: $\frac{x+y}{2}\leq (\frac{x^n+y^n}{2})^{\frac{1}{n}}$
where equality is satisfied if $x=y=1/2$
My question is how we can prov... | This is just convexity of the function $f(x)=x^{n}$. The inequality simply says $f(\frac {x+y} 2) \leq \frac {f(x)+f(y)} 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sum_\text{cyc}\frac{1}{3-ab}\le\frac{3}{2}$
Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le\frac{3}{2}$$
By Am/Gm we have: $$ {1\over 3-ab}\leq {2\over 6-a^2-b^2} = {2\over 5+c^2}<{2
\over 5}$$
so
$$\sum_{cyc} {2\over 5+a^2} < {6\over 5} <{3\over 2}$$
So ... | Your calculation is correct: $m = \frac 65$ is an upper bound. It is not the best upper bound though, because the estimates $\frac{2}{5+c^2} < \frac 25$ are not sharp for all three variables simultaneously.
The best upper bound is $\color{red}{m=\frac 98}$. Proof: The function $f(t) = \frac{1}{3-\sqrt t}$ is increasing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Showing $f(x) = \frac{x^2}{\sin(x)}$ is analytic near $0$ Problem
Show the function
$$
f(x) = \frac{x^2}{\sin(x)}
$$
is an analytic about $x=0$.
Try
We have
$$
f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }
$$
Letting $f(x) = \sum_{n=0}^\infty a_n x^n $, we have
$$
a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 1/6, a_4 = 0, ... | We have $f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }=\frac{x}{1 - x^2/3! + x^4/5! - \cdots }.$
The nominator and the denominator are power series with radius of convergenc $= \infty$. The power series in the denominator is $ \ne 0$ at $x=0.$ Hence there is $ r>0$ such that
$f(x)= \sum_{n=0}^\infty d_nx^n$ for $|x|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}... | $\log_3 x=4\implies x=3^4=81,$ not $x=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
An equality from the line through the centroid of a triangle $G$ is the centroid of $\triangle\mathit{ABC}$. $X$, $Y$, and $Z$ are points on the sides (or lines through the sides) $\overline{\mathit{BC}}$, $\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$, respectively, such that $G$, $X$, $Y$, and $Z$ are colline... | Let $Z$ be placed on the line $AB$ such that $A$ is placed between $Z$ and $B$, $X$ be placed on the side $BC$,
$Y$ be placed on the side $AC$, $BD$ be a median of $\Delta ABC$, $GX=x$, $GY=y$ and $GZ=z$.
Thus, $$[Z,Y,G,X]=[A,Y,D,C],$$ which says
$$\frac{ZG}{ZX}:\frac{YG}{YX}=\frac{AD}{AC}:\frac{YD}{YC}$$ or
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum whe... | Hint: Write $$d=\sqrt{(x-a)^2+y^2}$$ where $$y^2=4-\frac{4}{9}x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Why do the digits of a number squared follow a similar quotient? I realized that, any number $k$ with $n$ digits, and when $k$ is squared, i.e $k^2$ will have $2n-1$ or $2n$ digits.
Per example, let $k = 583$, thus $n = 3$ and the digits of $583^2$ are $2n$.
But I started thinking, how can I narrow the result to be mor... | A (natural) number $k$ has $m$ digits if an only if $10^{m-1} \le k < 10^m$
So $10^{2m-2} \le k^2 < 10^{2m}$.
If $10^{2m-2} \le k^2 < 10^{2m-1}$ then $k^2$ will have $2m-1$ digits.
Taking the square roots we see This happens when $10^{m-1} \le k < \sqrt{10^{2m-1}}$
$\sqrt{10^{2m-1}} = \sqrt{10*10^{2m-2}} = 10^{m-1}*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Find all pairs $(k, n)$ of positive integers such that $k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$ Find all pairs $(k, n)$ of positive integers such that
$$k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$$
I tried to solve this problem but only found one solution $(1,1)$. Please help me to solve thi... | Via Legendre's formula, we have for any prime number $p$
$$ \nu_p(k!)=\sum_{i=1}^{\infty}\left \lfloor \frac{k}{p^i} \right \rfloor
\le \sum_{i=1}^{\infty} \frac{k}{p^i} =\frac{k}{p-1} $$
Note that $\nu_2(2^n-2^i)=i$, and hence
$$\nu_2(k!)=\nu_2\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)=\sum_{i=0}^{n-1}i=\frac{n(n-1)}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
Prove that
$$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$
I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} ... | Here is a different solution using complex method. We can notice that
\begin{align*}
I = &\int_0^{\frac \pi 4} \ln\left(1+\tan x + \sqrt{2\tan x} \right) dx + \int_0^{\frac\pi 4} \ln(\cos x) dx \\
= &I_1 + I_2.
\end{align*}
Evaluation of $I_1$ : Let $\displaystyle\omega = e^{\frac {\pi i} 4} =\frac{1+i}{\sqrt 2}$ and m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | No need to find the intersection, that is no need for quadratics. The normal to $y=-\frac34x+3$ through the origin is given by the equation $y=\frac43x$; determine the intersection point of both to find that the distance from the line to the origin is $\frac{3}{\sqrt{1+\left(-\frac34\right)^2}}=\frac{12}{5}$. Hence t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
A lucky proof for the Basel problem. I'll modify this part since I want the proof to be here.
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac43\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=-\frac43\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac43\int_0^1 \frac{\ln x}{x^2-1}dx$$
$$\int_0^1 \frac{\ln x}{x^2-1}dx\overset{x\rightarrow \fra... | Fixing $a >0$, we see that the integral
\begin{align*}
f(z) := \int_0^\infty \frac{\log x}{(x+a)(x+z)} dx.
\end{align*} converges absolutely for $z$ away from $(-\infty,0]$. So $f$ defines an analytic function on $\mathbb C \setminus (-\infty,0]$ (which can be checked by Morera's theorem and Fubini's theorem, for examp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 2
} |
Is the result for $3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}$ known in the literature? I was able to get the following result
$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}=11\zeta(3)\zeta(6)+\frac52\zeta(4)\... | In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al.
The results are:
$$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Why $xy = 100$ does not represent a direct variation? $xy = 100 $
$y = 100/x$
$x$ is not equal to $0$ and can be represented as $1\cdot x$
However I still do not understand on why $100/x$ does not represent a direct variation. Is it because of the use of division within the right side of the equation is not valid?
| The relation is between the two variables $x$ and $y$. A direct variation means that when $x$ increases $y$ will increase too by an amount specified by the equation, for example:
$y=x$. If $x = 1$ then $y = 1$ and if $x$ increases to be $x=2$ then $y=2$ this is a direct variation and it may be represented like the $f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $2x^6+12x^5+30x^4+60x^3+80x^2+30x+45=0$ has no real roots I tried solving the above question but was unable to prove it.
I used Descartes rule of sign, factorisation techniques, and many other things but could not figure out the solution.
| Factor $x^4$ and separate a non-negative part of the expression covering completely the terms $x^6$ and $x^5,$ then factor $x^2,$ ...
$$\begin{aligned}P(x)=&2x^6+12x^5+30x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+12x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+30x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$, compute: $\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$ Be $f(x)$ a polynomial in $\Bbb R$ such that:
If$$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$$
Compute: $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$$
I noticed that, if the first limit exists, the... | Given that $$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3\tag1$$
Take $~f(x)=3x-5~$, then the above limit is satisfied.
Now $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}=\lim \limits_{x \to -1} \frac{(-3x-5)^2-4}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{(3x+3)(3x+7)}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$,... | Try moving all the terms raised to the power of 3/2 onto one side of the equation only, and move all other terms to the other side of the equation. Then square both sides so that the power of 3/2 becomes 3, and use algebra from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"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.