Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $\lim \limits_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n} ~~~ (a, b>0)$
$$\lim_{n \to \infty}\left(\frac {\sqrt[n]a + \sqrt[n]b}2\right)^{n} ~~~ (a, b>0)$$
I extended its domain and applied L'Hopital's rule to get the answer $\sqrt{ab}$.
However, is it possible to avoid using L'Hopital's ru... | If $a=b$, then it is trivial.So assume without loss of generality. $x = \frac ab>1.$
Then it is the same as:
$$\left(\dfrac{\sqrt[n]{x}+1}{2}\right)^n\to\sqrt{x}$$.For simplicity, denote $\sqrt[2n]{x} = y >1.$ Then, we have $$\left(\dfrac{y^2+1}{2}\right)^n - y^n = y^n\left[\left(1+\dfrac{(y-1)^2}{2y}\right)^n-1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Complex number problem $i \tan (\theta)$ proof.
Given $z = \cos (\theta) + i \sin (\theta)$,
prove $\dfrac{z^{2}-1}{z^{2}+1} = i \tan(\theta)$
I know $|z|=1$ so its locus is a circle of radius $1$; and so $z^{2}$ is also on the circle but with argument $2\theta$; and $z^{2}+1$ has argument $\theta$ (isosceles tria... | For $z = \cos \theta + i \sin \theta$ you have
$$
\frac{z^2-1}{z^2+1} = \frac{ \cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta -1}{\cos^2 \theta - \sin^2 \theta + 2i \cos\theta \sin \theta +1 } \, .
$$
Now substitute $1 = \cos^2 \theta + \sin^2 \theta $ in both numerator and denominator, and collect terms:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
| We have
$$2x+\frac{6}{(x+2)^2}-\frac{2}{x+2}=\frac{a}{(x+2)^2}$$
$$2x(x+2)^2+6-2(x+2)=a$$
$$2x^3+4x^2+4x+6-2x-2=a$$
$$2x^3+4x^2+2x+4=a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Mean of two numbers by infinite sequences Consider two numbers $a$ and $b$, and the following sequence alternating between even and odd positions:
$$
a+2b+3a+4b+5a+6b\ldots,
$$
If we ''normalize''
$$
\frac{a+2b+3a+4b+\ldots}{1+2+3+4+\ldots},
$$
it turns out this ratio approaches the mean value of $a$ and $b$: $(a+b)/2$... | Some thoughts; not sure if this fully answers "why" but an attempt:
The difference in coefficients of $a$ and $b$ goes roughly like the last coefficient term (whichever one's ahead is ahead by around that much). In the polynomial case this difference becomes small relative to the sums; in the geometric case, this stays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find $\sin81^\circ$ given $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$
If $\sin18^\circ=\dfrac{\sqrt{5}-1}{4}$, then $\sin81^\circ$ is equal to
\begin{align}
&a)\quad\frac{\sqrt{5}+1}{4}\\
&b)\quad\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}\\
&c)\quad\frac{\sqrt{10+2\sqrt{5}}}{4}
\end{align}
$$
\cos18^\circ=\sqrt{1-\frac... | If you suspect (b) is the answer, just square it. You'll get
$$
\frac1{16}\left(3 +\sqrt 5+2\sqrt{(3+\sqrt 5)(5-\sqrt 5)} + 5 - \sqrt 5\right)
$$
which simplifies to
$$
\frac1{16}\left(8 + 2\sqrt{10+2\sqrt 5}\right),
$$
the square of your answer. Your answer must then agree with (b), since both answers are positive (as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving $\lim_{n\rightarrow-\infty}\frac{3x^2+x}{2x^2+1}=\frac{3}{2}$ Using Definition Definition of Limit of Function As n $\rightarrow\infty$: Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a function and let $B\in\mathbb{R}$. If for all $\epsilon>0$, there exists $N>0$ such that $x<-N\Rightarrow |g(x)-B|<\epsilon$, we w... | You want the $x^2$ in the bottom to become dominant. And the negatives are confusing things. Bout $x < -\frac 32$ so $-x \ge \frac 32 > 0$ so
$\frac {-x + \frac 32}{2x^2} = \frac {1 + \frac 3{-2x}}{-2x}$
Replace $y = -x$ and $y > \max(\frac 32, \frac 1{2\epsilon})$ and $\frac 1y < \min (\frac 23, 2\epsilon)$.
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | By induction we can prove the stronger
$$\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}\right)$$
indeed
1. base cases: by inspection the inequality is satisfied for for
$n=1,2, 3$
2. induction step:
*
*assume true that (Ind. Hyp.): $\prod_{k=1}^n\left(1+\frac{1}{k^3}\right)<3\left(1-\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Integrating $\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$ Here is the problem
$$\int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx$$
I have tried diving by $\cos^2 x$ and using partial fractions. Also substituting $\tan$ formulas or separating and partial fraction mess up the limits for me.
So can... | Use your trig identities to create a $u$-substitution. We have
$$
[\sin(x) + \cos(x)]dx = -d[\cos(x) - \sin(x)]
$$
and
$$
\sin(x)^4 + \cos(x)^2 = 1 - \sin(x)^2\cos(x)^2 = 1 - \frac{1}{4}\left(1 - [\cos(x) - \sin(x)]^2\right)^2,
$$
so
$$
\int_0^{\pi/4}\frac{\sin(x)+\cos(x)}{\sin(x)^4 + \cos(x)^2}dx = 4\int_0^1\frac{du}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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I would really appreciate some help solving this induction problem!!
$3.$ for all $n\ge1$, $\displaystyle\sum_{i=1}^n(2i)^2=\frac{2n(2n+1)(2n+2)}{6}$
I have
$$\frac{2n(2n+1)(2n+2)(12(n+1)^2)}6= \frac{2n(2n+1)2(n+1)12(n+1)(n+1)}6.$$
I think I need to do something with the $(n+1)$.
However, I'm not sure where to go fro... | I'm not sure where you got
$$
\frac{2n(2n+1)(2n+2)(12(n+1)^2)}{6}
$$
from. It should be
$$
\frac{2n(2n+1)(2n+2)\color{red}{+}12(n+1)^2}{6}
$$
and with this fix you should be able to finish up.
However, it's better if you do some simplifications, to get rid of useless factors.
The induction step consists in writing
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Trigonometric Identities: Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$
Given that $2\cos(3a)=\cos(a)$ find $\cos(2a)$.
$2\cos(3a)=\cos(a)$
I converted $\cos(2a)$ into $\cos^2(a)-\sin^2(a)$
Then I tried plugging in. I know this is not right, but I have no clue how to solve this. Hints please?
edit:
Because I got th... | I like to split $\color{blue}{\cos(3 a) = \cos(a+2a) = \cos(a) \cos(2 a) -\sin(a) \sin(2 a)}$. This makes the expression
$$ 2 \left( \cos(a) \cos(2 a) -\sin(a) \sin(2 a) \right) = \cos(a) $$
or (re-arrange and divide by $2\cos(a)$)
$$ \frac{2 \cos(a) \cos(2 a)}{2 \cos(a)} = \frac{\cos(a)}{2 \cos(a)} + \frac{2\sin(a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Examining whether $\sum \limits_{n=0}^\infty\frac{(-1)^{n+1}}{5n+1}$ is convergent, absolute convergent or divergent Everything in red is edited
To show, that the series is convergent we show at first, that $\color{red}{\lim \limits_{n \to \infty} \left(\dfrac{1}{5n+1}\right)}=0$.
$\color{red}{\lim \limits_{n \to \inft... | You first part is fine for the second that step is wrong
$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}
\color{red}{=\sum \limits_{n=1}^\infty \frac{1}{5n}}$$
we can simply refer directly to limit comparison test with $\sum \frac 1n$ and conclude for divergence indeed
$$\frac{\frac{1}{5n+1}}{\frac1n}=\frac{n}{5n+1}\to \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $\lim_{x\to 0}\frac{e^{(x+1)^{1/x}}-(x+1)^{e/x}}{x^2}$ By L'Hôpital's rule, it gets $$\lim_{x\to 0}-\frac{\left(e^{(x+1)^{1/x}} (x+1)^{1/x}-e (x+1)^{e/x}\right) ((x+1) \log (x+1)-x)}{2x^3(1+x)},$$which becomes more complicated.
What could I do then?
| We have that
$$(x+1)^{\frac{1}{x}}=e^{\frac1x \log (1+x)}=e^{1-\frac12 x+\frac13x^2+o(x^2)}=e \cdot e^{-\frac12 x+\frac13x^2+o(x^2)}=e\left(1-\frac 1 2 x+\frac 1 3 x^2 +\frac12\left(-\frac12 x+\frac13x^2\right)^2+ o(x^2)\right)
=e\left(1-\frac 1 2 x+\frac{11}{24}x^2+o(x^2)\right)$$
$$e^{{(x+1)}^{\frac{1}{x}}}=e^{e}e^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Calculating coefficient I have a generating function,
$$ \frac{(1-x^7)^6}{(1-x)^6} $$
and I want to calculate the coefficient of $x^{26}$
Solution for this is,
$$ {26+5 \choose 5} - 6{19+5 \choose 5} + 15{12+5 \choose 5} - 20{5+5 \choose 5} $$
Is there formula for this? If there is, what is called?
If there is no formu... | Use the Negative binomial series:
$$\begin{align}[x^{26}]\frac{(1-x^7)^6}{(1-x)^6}&=[x^{26}](1-x^7)^6\cdot (1-x)^{-6}=\\
&=[x^{26}]\sum_{i=0}^6 {6\choose i}(-x^7)^i\cdot \sum_{j=0}^{\infty} {-6\choose j}(-x)^j=\\
&=[x^{26}]\sum_{i=0}^6 {6\choose i}(-x^7)^i\cdot \sum_{j=0}^{\infty} {6+j-1\choose j}x^j=\\
&=[x^{26}]\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove $\left(1-\frac{1}{n^2}\right)^n\times\left(1+\frac 1n\right)<1$ for every natural $n > 0$ by induction Consider this inequality
$$\left(1-\dfrac{1}{n^2}\right)^n\times\left(1+\dfrac 1n\right)<1$$
which is meant to be valid for any nonzero natural number $n$.
It is asked to prove it by induction. I haven't made ... | The inequality
$\left(1-\dfrac{1}{n^2}\right)^n\cdot\left(1+\dfrac 1n\right)
=\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^{n+1}
=\dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\left(1+\dfrac{1}{n-1}\right)^{n}}<1$
is equivalent to the sequence
$\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$
is strictly dec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Integral using polar coordinates Let $X=\{(x,y)\in\mathbf{R}^2\mid x^2+y^2\leqslant 1,x,y\geqslant 0 \}$. Calculate $\int_X xye^{x^2+y^2}\,dx\,dy$.
By using polar coordinates, I get $=\int_0^{\pi/2}\int_0^1e \cos\theta\sin\theta\,dr\,d\theta=\int_0^{\pi/2}\frac{1}{2}e\sin 2\theta\,d\theta=\left.-\frac{1}{4}e\cos 2\the... | Your polar integral is wrong; it should be
$$\int_0^{\pi/2}\int_0^1r^2\cos\theta\sin\theta\cdot e^{r^2}r\,dr\,d\theta$$
where the $r^2\cos\theta\sin\theta$ comes from rewriting $xy$ and the extra $r$ factor is the Jacobian of the transformation to polar coordinates. Continuing the evaluation, we get
$$=\int_0^{\pi/2}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this nonlinear equation? This is the system of equations:
$$xy + z = -30\\
yz + x = 30\\
zx + y = -18$$
I have also done some work: if we add the first two equations we'll get:
$$xy + z + yz + x = 0\\
\implies y (x+z) + z + x = 0\\
\implies (x+z)(y+1) = 0$$
Now, a product equals zero if at least one of th... | As you observe, from the first two equations, $y=-1$ or $z=-x$. If $y=-1$, then we have two equations to solve for $x$ and $z$, namely, $x-z=30$ and $xz=-17$. Therefore, $x$ and $-z$ are the roots of the quadratic polynomial $t^2-30t+17$, whereby it follows that $$(x,y,z)=(15-4\sqrt{13},-1,-15-4\sqrt{13})$$
and
$$(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{2n+1}\left(\frac{n+1}{2n+1}\right)$ Does the following sum equal 1 (or some amount less than 1)?
$$S\equiv\sum_{n=0}^{\infty}C_n\left(\frac{1}{2}\right)^{\!2n+1}\!\!\left(\frac{n+1}{2n+1}\right)=\sum_{n=0}^{\infty}\left(\frac{(2n)!}{(n+1)!\cdot n!}\right... | Simplifying a bit the notation, since $C_n=\frac{1}{n+1}\binom{2n}{n}$, we want to compute
$$ S = \frac{1}{2}\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\int_{0}^{1}x^{2n}\,dx \tag{1}$$
and we may easily recognize the Maclaurin series of $\frac{1}{\sqrt{1-x^2}}$, leading to:
$$ S = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which values of $a$ is $f$ primitivable?
Let $a \in \mathbb{R}$ and $p,q$ be natural numbers with $p \geq q+2.$ For which values of $a$ is the function
$$f(x) =
\begin{cases}
\frac{1}{x}\sin \frac{1}{x^p}\sin \frac{1}{x^q}, &x \neq 0 \\
a, &x=0
\end{cases}
$$
primitivable?
I noticed that $\frac{1}{x}\sin \f... | Take $a=0$. Consider the function
$$
G(x):=x^{p}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}.
$$
Then for $x\neq0$,
\begin{align*}
G^{\prime}(x) & =px^{p-1}\cos\frac{1}{x^{p}}\sin\frac{1}{x^{q}}-px^{-1}%
\sin\frac{1}{x^{p}}\sin\frac{1}{x^{q}}\\
& -qx^{p-q-1}\cos\frac{1}{x^{p}}\cos\frac{1}{x^{q}}\\
& =:h(x)-pf(x),
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse... | I found an algorithm that I hope it help you. Let the inverse be$$A^{-1}=\begin{bmatrix}c_1&c_2&\cdots&c_n\end{bmatrix}$$In other words, $c_i$s are the columns of $A^{-1}$. We build the $c_i$s recursively as following:
$$c_{2}=e_1\\c_{2k+2}=-c_{2k}+e_{2k+1}\qquad,\qquad 1\le k\le {n\over 2}-1$$also $$c_{n-1}=e_n\\c_{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How does one solve $\sin x-\sqrt{3}\ \cos x=1$? I thought this one up, but I am not sure how to solve it. Here is my attempt:
$$\sin x-\sqrt{3}\ \cos x=1$$
$$(\sin x-\sqrt{3}\ \cos x)^2=1$$
$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$$
$$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$$
$$2\cos^2x-2\sqrt{3}\sin x\cos x=0$$
$$2\c... | Hint :
$$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$
Note : You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbe... | If you can drop the method by induction you may also proceed as follows:
For $|x| < 1$ you have
$$\frac{1}{1-x}= \sum_{n=0}^{\infty} x^n \Rightarrow \frac{1}{(1-x)^2}= \sum_{n=1}^{\infty} n \cdot x^{n-1} \Rightarrow \color{blue}{\frac{x}{(1-x)^2}= \sum_{n=1}^{\infty} n \cdot x^{n}}$$
For $x=\frac{1}{2}$ you have
$$\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992278",
"timestamp": "2023-03-29T00:00:00",
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Prove that number $1^{2005}+2^{2005}+\cdots+n^{2005}$ is not divisible by number $n+2$ Prove that number $1^{2005}+2^{2005}+\cdots+n^{2005}$ is not divisible by number $n+2$ for every $n\in \mathbb N$
I have solution $2(1^{2005}+2^{2005}+ \cdots +n^{2005})=2+(2^{2005}+n^{2005})+(3^{2005}+(n-1)^{2005})+\cdots+(n^{2005}+... | The solution in your statement is using the following fact:
Let $k$ be an odd natural number. Then $x+y$ divides $x^k+y^k$.
The simplest way to get this is perhaps to view this as a polynomial in $x$ and then if we substitute $x=-y$, we get $x^k+y^k=x^k+(-x)^k=0$.
In your question each of the term: $(m^{2005}+(n-m+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993618",
"timestamp": "2023-03-29T00:00:00",
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Can you prove this equality? In their book An Introduction to Optimization, on the chapter on gradient algorithms, to prepare for discussing convergence properties of the descent methods, authors Chong and Zak have following:
$f(x) = \frac{1}{2}x^TQx-b^Tx$, where $Q$ is symmetric and $Q>0$.
...
$V(x)=f(x)+\frac{1}{2}(... | \begin{align}
V(x)=&f(x)+\frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-b^Tx+ \frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-(Qx^*)^Tx+ \frac{1}{2}(x^*)^TQx^* \\
=&\frac12 x^TQx-(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^* \quad(\because Q^T=Q)\\
=&\frac12 x^TQx-\frac12(x^*)^TQx-\frac12(x^*)^TQx+ \frac{1}{2}(x^*)^TQx^*\\
=&\frac12 x^TQx-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$ Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$
I have a solution
Let $f(x)=\frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for... | If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.
Assume $1<t$, then you can have infinite negative solutions as per the path
shown. You cannot have a positive value for any one of the unknowns.
Only if $0<t<<1$ you may find infinite positive solutions starting with a high v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find largest eigenvalue of specific nonnegative matrix I look for the largest eigenvalue of the following matrix (or at least a small upper bound).
The only thing I know is that the eigenvalue is smaller than 1 and converges to $\cos(\frac{\pi}{2n+1})$ with growing n.
In general, it is very hard to compute the characte... | An asymptotic expansion for large $n$ can be obtained as follows. Expanding $A - \lambda I$ by the first row and expanding one of the resulting matrices by the first column gives
$$\det(A - \lambda I) =
\left( \frac {n + 1} {2 n + 1} - \lambda \right) \det \tilde A_{n - 1} -
\frac 1 4 \det \tilde A_{n - 2},$$
where $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$x$ intercept problem How would I find the $x$ intercept of $x^5-x^3+2=0$? I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.
| My answer is not totally correct but I hope it helps a little :
Consider $f(x)=x^5-x^3+2$ . we have $f'(x)=5x^4-3x^2$ and $f'(x)=0$ if $x=0$ or $x=\pm \sqrt{\frac{3}{5}}$ and it is also possible to determine the sign of $f'(x)$ for all other $x$.
Note that $5x^4-3x^2>0$ if $5x^4>3x^2$ if $x^2>3/5$ . that is, $x >+ \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A differentiation/derivative/calculus problem The question is as follows:
$$y=x^2/(x+1)$$
The normal to this curve at $x=1$ meets the $x$-axis at point $M$.
The tangent to the curve at $x=-2$ meets the $y$-axis at point $N$.
Find the area of triangle $MNO$, where $O$ is the origin.
PS- this is not a school h.w so d... | $y(x) = \dfrac{x^2}{x + 1} = (x + 1)^{-1}x^2; \tag 1$
$y'(x) = -(x + 1)^{-2}x^2 + 2x(x + 1)^{-1} = -(x + 1)^{-2}x^2 + 2x(x + 1)(x + 1)^{-2}$
$= -(x + 1)^{-2}x^2 + (2x^2 + 2x)(x + 1)^{-2} = (x^2 + 2x)(x + 1)^{-2} = \dfrac{x^2 + 2x}{(x + 1)^2}; \tag 2$
$y(1) = \dfrac{1}{2}; \; y'(1) = \dfrac{3}{4}; \tag 3$
$y(-2) = -4; \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the polynomial is $g(x,y)(x^2 + y^2 -1)^2 + c$ This is from a Brazilian math contest for college students (OBMU):
Let $f(x,y)$ be a polynomial in two real variables such that the polynomials
$$\frac{\partial f}{\partial x}(x,y)$$
$$\frac{\partial f}{\partial y}(x,y)$$
are divisible by $x^2+y^2-1$. Prove tha... | Treating $f$ as a polynomial in $(\mathbb{R}[y])[x]$, there exist polynomials $p(x,y) \in (\mathbb{R}[y])[x], q(y), r(y) \in \mathbb{R}[y]$, such that $f(x,y)=(x^2+y^2-1)p(x,y)+xq(y)+r(y)$.
We have that $\frac{\partial f}{\partial x}(x,y)=(x^2+y^2-1)\frac{\partial p}{\partial x}(x,y)+2xp(x,y)+q(y)$ and $\frac{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\s... | Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$\lim_{x \to 0} \tan x= \frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . .$$
$$\lim_{x \to 0} \sin x= \frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$\lim_{x \to 0} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to calculate $\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} $ How to calculate $\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} $
I know that result $\frac{331}{910}$ because I checked it in Mathematica but I have troubles with calculate that.
$$\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n} = ... = $$
$$ \sum_{n=... | For $|p|<1$, $\sum_{n=1}^{\infty}p^n=\frac{p}{1-p}$, so
$$\sum_{n=1}^{\infty}\frac{(2^n+(-1)^n)^2}{11^n}=\sum_{n=1}^{\infty}\frac{4^n}{11^n}+2\sum_{n=1}^{\infty}\frac{(-2)^n}{11^n}+\sum_{n=1}^{\infty}\frac{1}{11^n}\\
= \frac{\frac{4}{11}}{1-\frac{4}{11}}-2\frac{\frac{2}{11}}{1+\frac{2}{11}}+\frac{\frac{1}{11}}{1-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the integral $\int _{1}^{e} (x \ln x)^2 dx$.
Find : $$\int _{1}^{e} (x \ln x)^2 \;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (\ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?... | Let $u=(\ln x)^2$ and $dv=x^2 \,dx$. Then $du=2(\ln x)\frac{1}{x} \, dx$ and $v=\frac{x^3}{3}$. So
$$I=\int_1^e(x \ln x)^2 \, dx=(\ln x)^2 \frac{x^3}{3}\Big|_{1}^{e}-\frac{2}{3}\int_1^e x^2 \ln x \, dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=\ln x$ and $dv=x^2 \, dx$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$
My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?
| Let $t=\tan(\frac x2)$.
use the identities
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2},$$
and
$$\tan(x)=\frac{2t}{1-t^2}.$$
thus
$$3\sin(x)+\cos(x)=2\implies$$
$$6t+1-t^2=2(1+t^2) \implies$$
or
$$3t^2-6t+1=0$$
hence
$$\frac{3\tan(x)}{4\tan(x)+3}=$$
$$\frac{6t}{8t+3(1-t^2)}=$$
$$\frac{6t}{2t+4}=3-\frac{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
Prove $a_n = \sqrt{n+1}-\sqrt{n}$ is monotone $n \ge 0$.
To be monotone it must be either increasing or decreasing, so:
$a_n \ge a_{n-1}$ or $a_n \le a_{n-1}$
$\sqrt{n+1}-\sqrt{n} \ge? \sqrt{n}-\sqrt{n-1} $
$\sqrt{n+1}+\sqrt{n-1} \ge? \sqrt{n} + \sqrt{n}$
I k... | $$
\\a_{n+1}\le a_{n}<=>
\\\sqrt{n+1}+\sqrt{n-1}\le2\sqrt{n}<=>
\\(\sqrt{n+1}+\sqrt{n-1})^2\le(2\sqrt{n})^2<=>
\\\sqrt{(n-1)(n+1)}=\sqrt{n^2-1}\le \sqrt{n^2}=n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$ $$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$$
My try:
The limit can be written as follows:
$$\lim_{n\to\infty}\... | We have that by Stolz-Cesaro
$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=\frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=\frac{n+2}{2n+1}\left(1+\frac{1}{n+1}\right)^{n} \to \frac e 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Evaluate integral $\int_{-2}^0 x^2+x\ dx$ using Riemann Sum Consider the integral $$\int_{-2}^0 x^2+x\ dx.$$
The question says to use Riemann Sum theorem which is $$\sum_{i=1}^nf(x_i)\delta x$$
I know that $\delta x= \frac{-2}{n}$ and that $x_i=-2+(\frac{2}{n}i)$
After i plug everything in I get $$\sum_{i=1}^n\frac{2}{... |
I know that $\delta x= \frac{-2}{n}$
The $\delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-\frac{2i}{n}$. So you have,
\begin{align*}
&\int_{-2}^0 x^2+x\ dx\\
=&\lim_{n\rightarrow+\infty}
\sum_{i=1}^n \left(\left(-\frac{2i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$
Note that $(x^4+x^3+x^2+x+1)(x-1) = x^5 - 1$.
I tried reducing the equation $\mo... | *
*Notice that $x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1}$. Now, $\frac{x^5-1}{x-1} \equiv 0 \pmod p \implies x^5\equiv 1 \pmod p$. It follows that the order of $x$, call it $r$, $\pmod p$ divides $5$. But $5$ is a prime number, hence $r=1,5$.
*We know that The order of any number $\pmod p$ divides $\phi(p)$. If the order ... | {
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"url": "https://math.stackexchange.com/questions/3042635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mysterious polynomial sequence Can someone identify this polynomial sequence? Is it known in mathematics? I'm interested in various properties of this sequence.
I'd like to find $P(n)$, $n\in \mathbb{Z}^+$
\begin{align}
P(0)&= 1\\
P(1)&= a\\
P(2)&= a^2+b\\
P(3)&= a^3+2ab\\
P(4)&= a^4+3a^2b+b^2\\
P(5)&= a^5+4a^3b+3ab^2\... | Hint. Note that the following recurrence holds: for $n\geq 2$,
$$P(n)=aP(n-1)+bP(n-2).$$
They are related to the Fibonacci polynomials. The wiki page gives a list of properties. For example we have that
$$P(n)=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}a^{n-2k}b^k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be... | We need to find two numbers with product $2(-3) = -6$ and sum $5$. You are correct that the numbers are $-1$ and $6$. Splitting the linear term yields
\begin{align*}
2x^2 + 5x - 3 & = 2x^2 - x + 6x - 3 && \text{split the linear term}\\
& = x(2x - 1) + 3(2x - 1) && \text{factor by grouping}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, ... | Hints.
$1) x-\frac1x=3\implies x^2+\frac1{x^2}-2=9\\2)\frac y{x+y}+\frac x{x+y}=1\\3)x^4+y^4=6x^2y^2\\\implies x^4+y^4+2x^2y^2=(x^2+y^2)^2=8x^2y^2\\\implies x^4+y^4-2x^2y^2=(x^2-y^2)^2=4x^2y^2$
Divide the two and take the square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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I calculated $\sin 75^\circ$ as $\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$, but the answer is $\frac{\sqrt{2}+\sqrt{6}}{4}$. What went wrong? I calculated the exact value of $\sin 75^\circ$ as follows:
$$\begin{align}
\sin 75^\circ &= \sin(30^\circ + 45^\circ) \\
&=\sin 30^\circ \cos 45^\circ + \cos 30^\circ... | They’re the same value. Multiply the numerator and denominator of your answer by $\sqrt 2$ to see why.
$$\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2}{\sqrt 2}\cdot\frac{1+\sqrt 3}{2\sqrt 2} = \frac{\sqrt 2+\sqrt 6}{4}$$
You can also use $\sin 45 = \cos 45 = \frac{\sqrt 2}{2}$ (rationalizing $\frac{1}{\sqrt 2}$) to get t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Isn't $\frac{ax+bi}{ax+bi}$ equal to $1$? Isn’t $\dfrac{ax+bi}{ax+bi}$ equal to $1$?
Here, $i=\sqrt{-1}$, & $a$,$b$ & $x$ $\in$ $R$
| Division in the field of complex numbers is defined in the following manner:
$\frac{ax+bi}{ax+bi}=\frac{(ax+bi)(ax-bi)}{(ax+bi)(ax-bi)}$, so
$\frac{ax+bi}{ax+bi}=\frac{a^2x^2+b^2}{a^2x^2+b^2}$, which is a real number and equal to 1. Hence your intuition is correct. But note that this is not true for $b=0$ and either $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?
$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$
I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.
| Not a full answer (yet)
I'll try to provide a solution which doesn't use any special functions, only the well known "classic" series.
$$I=\int_0^1 \frac{\ln{(x^2+1)}}{x+1}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{k+1} \int_0^1 x^{2(k+1)+n}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{(k+1)(2k+n+3)}$$
Let's use partial fracti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving a Diophantine Equation dealing with powers of 2 Let x and y be positive integers with no prime factors larger than 5. Find all such $x$ and $y$ which satisfy $$x^2-y^2=2^k$$
where the range of k is $1\leq k\leq2019$.
I managed to solve the equation for general solutions where $(x, y)=(3 \times 2^z, 2^z)$ and $(... | First, we have the equation:
$$x^2 - y^2 = 2^k$$
We define $\nu_2(n)$ to be the power of $2$ that divides $n$.
Now, assume that $\nu_2(x) \neq \nu_2(y)$. We divide our equation by $2^{2\min(\nu_2(x),\nu_2(y))}$. Then, we will have the LHS as an odd value, as one term would be even, and the other would be odd. Thus, we... | {
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"source": "stackexchange",
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Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\beg... | Let's do it by induction:
$0 \le a_n < 2$
So $6= 6- 0 \ge 6- a_n > 6-2 = 4$ and $\frac 16 \le \frac 1{6-a_n} < \frac 14$
$6 + a_0 \ge 6 > 0$ so
$\frac {6 + a_n}6 \le \frac {6+a_n}{6-a_n} < \frac{6+a_n}4$.
And $6+a_n \ge 6+0 = 6$ and $6+a_n < 6+2 = 8$ so
$0 \le 1 = \frac 66 \le \frac {6+a_n}6 \le \frac {6+a_n}{6-a_n} ... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to c... | $$x-x^2=x(1-x)\leq\left(\frac{x+1-x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=1-x$ only, which gives $x=\frac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
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Does the constant $C$ in this solution to a differential equation equal infinity? The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$. My solution is
$$\begin{align}
y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\
&\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' ... | For non-linear equations like this, it is valid to take $C\to\infty$. You can also write $c = \frac{1}{C}$ to get an alternate form
$$ y(t) = \frac{1}{t} + \frac{2ct}{1-ct^2} $$
| {
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Is my approach correct to this equation? The problem is the following:
Does $a \in \mathbb{R}$ exist such that $[a + \sqrt{2n + 1}] = [a + \sqrt{2n + 2}]$ for all $n \in \mathbb{N}$? ($[x]$ denotes the whole part of $x$).
Note: I will also use $\{x\}$ to denote the fractional part of $x$. Also that $[x] + \{x\} = x$.... | Your idea of looking at $\alpha=\{a\}$ is a good one.
If $\alpha \lt 2-\sqrt 3\approx 0.268$ we choose $n=1$ and note that $\sqrt {2n+1}+\alpha=\sqrt 3+\alpha$ has floor $1$ while $\sqrt {2n+2}=\sqrt 4=2$
If $\alpha \ge 2-\sqrt 3$ we want to choose $n$ so that $$\lfloor\alpha+\sqrt {2n+2}\rfloor = k \gt \lfloor \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$ Reduction Formula I'm having trouble proving the following reduction formula:
If
$$I_{m,n}=\int\frac{x^m}{(ax^2+bx+c)^n}dx$$
then
$$\int\frac{x^m}{(ax^2+bx+c)^n}dx=-\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}}-\frac{b(n-m)}{a(2n-m-1)}I_{m-1,n}+\frac{c(m-1)}{a(2n-m-1)}I_{m-2,n}$$
... | So this means you are using the integration by parts in the wrong direction or with a wrong choice. In the reduction formula you are trying to achieve, you can notice that $m$ is decreasing while $n$ stays constant. Thus, you need a choice of $u,v$ such that:
$$udv = \dfrac{x^mdx}{f^n(x)},$$
where $f(x) = ax^2+bx+c$ an... | {
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"source": "stackexchange",
"question_score": "2",
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Laplace's equation: separation of variables Question:
Let $(r,\theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k \in \Bbb R$ for which
$$\nabla^2 u=0 \qquad 1≤r≤2 \\ ku + \frac{\partial u}{\partial r}=0 \qquad r=1,2$$
has a non-trivial solution of the form $u(r,\theta) = f(r)g(\th... | From the periodicity condition, we get a general solution
$$ u_n(r,\theta) = f(r)g(\theta) = (Ar^n + Br^{-n})(C\cos n\theta + D\sin n \theta) $$
where $n=1,2,3,\dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
\begin{align}
k(A+B) + n(A-B) = 0 ... | {
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Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly... | 1) The sum of the digits of a multiple of $9$ will be a multiple of $9$.
Why? If $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.
If $N = \sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a mu... | {
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"source": "stackexchange",
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Prove $\sum_{k=1}^m \cot^2 k\pi/(2m+1)=m(2m-1)/3$
Prove that $$ \sum_{k=1}^m \cot^2 \frac{k\pi}{2m+1}=\frac{m(2m-1)}{3} $$
I have tried to use $$\sin\left((2m+1)x\right)=
\left(\sin^{2m+1}x\right) \cdot \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}\left(\cot^2x\right)^{m-j}\right)$$
and induction without any success. ... | Further to my last comment and given you used
$$\sin\left((2m+1)x\right)=
\left(\sin^{2m+1}x\right) \left(\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}(\cot^2x)^{m-j}\right)=
\left(\sin^{2m+1}x\right) \cdot P\left(\cot^2{x}\right)\tag{1}$$
where
$$P(x)=\sum_{j=0}^m (-1)^j \binom{2m+1}{2j+1}x^{m-j}=\binom{2m+1}{1}x^m-\binom{2... | {
"language": "en",
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"source": "stackexchange",
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Simplify $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$ I am to simplify $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$
into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$
I am able to get to $\frac{x+4\sqrt{y}\sqrt{2}}{2}$ but cannot arrive at the provided solution.
Here is my working:
$\frac{x\sqrt{... | $$\frac{x \sqrt{64y} + 4 \sqrt{y}}{\sqrt{128 y}}$$
Factor our common factor $\sqrt{y}$ from numerator and denominator.
$$\frac{\sqrt{y}(x \sqrt{64} + 4)}{\sqrt{y}(\sqrt{128})}$$
Notice $\sqrt{64} = \sqrt{8^2} = 8$ and $\sqrt{128} = \sqrt{64 * 2} = \sqrt{8^2 * 2} = 8\sqrt{2}$.
$$\frac{8x + 4}{8\sqrt{2}}$$
Factor out $4$... | {
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Prove that $N - \lfloor{N/p}\rfloor = \lfloor{\frac{p-1}{p}\left({N + 1}\right)}\rfloor$ for positive $N$ and prime $p$ I am counting the number of positive integers less than or equal to some positive integer $N$ and not divisible by some prime $p$. This gets generalized for $k$ primes where I use the principle of in... | let
$$
f(N) = N - \lfloor \frac{N}{p} \rfloor
$$
then
$$
f(N+1) - f(N) = 1 - \bigg(\lfloor \frac{N+1}{p} \rfloor - \lfloor \frac{N}{p} \rfloor \bigg)
$$
thus when $N$ is incremented by 1, $f(N)$ is also incremented by 1 except when $N \equiv_p -1$. in this case $f(N+1) = f(N)$
if now we let
$$
g(N) = \lfloor{\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show this inequality with $\sum_{i=1}^{n}a_{i}=n$
Let $n\ge 3$ be postive number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that
$$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{3}\cdots a_{n}+a_{n}a_{1}\cdots a_{n-2})$$
it seem can use indution to... | The hint.
Prove this inequality for $n=3$ and for $n=4$.
But for all $n\geq5$ by AM-GM we obtain:
$$\sum_{k=1}^na_k^3a_{k+1}\geq n\sqrt[n]{\prod_{k=1}^na_k^4}.$$
Thus, it's enough to prove that:
$$n\sqrt[n]{\prod_{k=1}^na_k^4}+n\geq2\prod_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}.$$
Now, let $\prod\limits_{k=1}^na_k=const$ a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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integral with parameter and 2 denominators Hello I want to solve following integral
$\displaystyle\int \frac{x^22t}{(1+x^2)(1+t^2x^2)}\,dx=-\dfrac{2\arctan(tx)-t \arctan(x)}{t^2-1}$
My Problem is that I just do not know how to start solving this problem .
| Partial fraction decomposition gives$$\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}=\frac {x^2}{(1+x^2)(1+t^2x^2)}$$Hence, the integral now becomes$$\begin{align*}\mathfrak{I} & =2t\int\mathrm dx\,\left[\frac 1{(1-t^2)(1+t^2x^2)}-\frac 1{(1-t^2)(1+x^2)}\right]\\ & =\frac {2t}{1-t^2}\int\frac {\mathrm dx}{1+t^2x^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find shortest distance from the parabola $y=x^2-9$ to the origin.
Find shortest distance from the parabola $y=x^2-9$ to the origin.
First, I find minima of $\sqrt{x^2+(x^2-9)^2}$, so use derivative and ...
Is have an easier way?
| $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+\frac{1}{4}+\frac{35}{4}=\left(x^2-9+\frac{1}{2}\right)^2+\frac{35}{4}\geq\frac{35}{4}.$$
The equality occurs for $x^2=\frac{17}{2},$ which says that $\frac{\sqrt{35}}{2}$ is a minimal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve differential equation $xyy'=x^4+y^4$ How to find general solution to this differential equation (if it exists):
$$ xyy'=x^4+y^4 ?$$
I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=\dfrac{x^4+y^4}{xy}$$... | Observing that $y y' = \left(\frac{1}{2}y^2\right)'$ we define the new dependent variable $z := \frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$.
This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equati... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Counting, Probability and Binomial Coefficients If $$P_{2n+2}=\sum_{k=n+2}^{2n+2}{2n+2 \choose k}p^kq^{2n+2-k}$$
and,
$$P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}$$
where $0<p<q<1$ and $q=1-p$
Prove that
$$P_{2n+2}=P_{2n}+{2n \choose n}p^{n+2}q^n-{2n \choose {n+1}}p^{n+1}q^{n+1}$$
$\mathbf {Inspiration:}$ A and... | We consider $(p+q)^2P_{2n}$ instead of $P_{2n}$ (because then our identity will be homogeneous). Then, we will simply compare coefficents of monomials $p^kq^{2n+2-k}$ in both sides of identity.
Note that (from equality $p+q=1$)
$$
P_{2n}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n-k}=\sum_{k=n+1}^{2n}{2n \choose k}p^kq^{2n... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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How to apply CRT to a congruence system with moduli not coprime? $x=1 \pmod 8$
$x=5 \pmod{12}$
8 and 12 are not coprime, I could break it to:
$x=1 \pmod 2$
$x=1 \pmod 4$
and
$x=5 \pmod 3$
$x=5 \pmod 4$
But what are the next steps to solve it? By the way, $x$ should be $17$ not sure how to get that number ...
Thanks in... | Alternatively:
$$\begin{cases}x\equiv 1\pmod{8}\\ x\equiv 5\pmod{12}\end{cases} \Rightarrow \begin{cases} x=8n+1\\x=12m+5\end{cases} \Rightarrow 8n+1=12m+5 \Rightarrow \\
2n-3m=1 \Rightarrow \begin{cases}n=2+3k\\m=1+2k\end{cases} \Rightarrow \begin{cases} x=8(2+3k)\\ x=12(1+2k)+5\end{cases} \Rightarrow \\
x=24k+17 \Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075823",
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"source": "stackexchange",
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How to prove this closed formula for Cantor set?
Let $C_0=[0,1]$ and $C_{n+1} = \dfrac{C_n}{3} \bigcup\left(\dfrac{2}{3}+\dfrac{C_n}{3}\right)$.
Theorem: $$C_n=\bigcap_{m=0}^{n}\bigcup_{k=0}^{\left\lfloor \frac{3^{m}}{2}\right\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]$$
I have tried to prove this assertion ... | Notice that
$$\bigcap_{m=0}^{n+1}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]=\left(\bigcap_{m=0}^{n}\bigcup_{k=0}^{\lfloor 3^m/2\rfloor}\left[\frac{2k}{3^m},\frac{2k+1}{3^m}\right]\right)\cap \bigcup_{k=0}^{\lfloor 3^{n+1}/2\rfloor}\left[\frac{2k}{3^{n+1}},\frac{2k+1}{3^{n+1}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration by substitution to take out square root
Find $$\int (x+1)\sqrt{x^2+1}\,dx .$$
In order to not bother with the square root I thought of doing this:
$let$ $ x^2+1=(x+t)^2$ $\Rightarrow$ $x=\frac{1-t^2}{2t}$ $\Rightarrow x-t=\frac{1+t^2}{2t}$ $(1)$
$\Rightarrow$ $t=\sqrt{x^2+1}-x$ $(2)$
And
$dx=-\frac{t^2+1}... | After the rationalizing substitution, I think you made a few errors. It should be
$$\int (x+1)\sqrt{x^2+1}\,dx=\int\frac{t^2-2t-1}{2t}\cdot \frac{t^2+1}{2t}\cdot \frac{t^2+1}{2t^2}dt.$$
Alternative approach by using hyperbolic functions.
Let $x=\sinh(t)$ then $\cosh^2(t)-\sinh^2(t)=1$ and
$$\begin{align*}\int (x+1)\sq... | {
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Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x... | Writing your inequality in the form $$\frac{\ln\left(6+\frac{2}{x}\right)}{\ln(5)}+\frac{\ln\left(1+\frac{x}{10}\right)}{-\ln(5)}\le 1$$ we get the inequalities
$$6+\frac{2}{x}>0$$ and $$1+\frac{x}{10}>0$$ and
$$\frac{6+\frac{2}{x}}{1+\frac{x}{10}}\le 5$$ we get
$$1-\sqrt{5}\le x<-\frac{1}{3}$$ or $$x\geq 1+\sqrt{5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solution of an algebraic equation with integers Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
| Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divis... | {
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"source": "stackexchange",
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Evaluate $\sum_{n=1}^{\infty}\frac{\psi^{''}(n)}{2n-1}$, where $\psi^{''}(n)$ is 2nd derivative of digamma function. Does the following sum have a closed form?
$$\sum_{n=1}^{\infty}\frac{\psi^{"}(n)}{2n-1},$$ where $\psi^{"}(n)$ is 2nd derivative of digamma function.
| The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
$$\psi^{(m)}(x) = (-1)^{m+1} \int_0^\infty \frac{t^m}{1-e^{-t}} e^{-xt} \, \mathrm{d} t.$$
Using the monotone convergence theorem, we get
$$\label{1} \tag{1}\sum_{n=1}^\infty \frac... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$ $$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$
I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.
| An elementary approach: "recognize the derivative."
You can rewrite, for $x\notin\{-3,3\}$,
$$
\frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}
= 1+ \frac{\sqrt{x+6}-3}{x^{2}-9}
= 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)}
= 1+ \frac{\sqrt{x+6}-3}{(x-3)(x+3)} \tag{1}
$$
so the question boils down to computing $\lim_{x\to 3} \frac{\sq... | {
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Prove that $\sum_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor (n-k)/2 \rfloor}=\binom{2n+1}{n}$ Where the thing that looks like a floor function is the floor function. This is an interesting result which I hoped to prove by induction, but ran into trouble applying the inductive hypothesis. The base case is trivial. Her... | There are $ 2n+1 \choose n $ strings consisting of $n$ ones and $n+1$ zeroes.
Given any such string $S$, let $S_1$ consist of the first $n$ digits of the string $S$ and let $S_2$ consist of the last $n+1$ digits of $S$. Suppose that there $k$ values of $i$ such that the $i^{th}$ element of $S_1$ is different from the $... | {
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Find solutions of the given equation in the form of power series Find solutions of the given equation in the form of power series
$$y'' +2xy' = 0$$
My approach:
$$2xy' = 2a_1x + 2\times2a_2x^2 + 2\times3a_3x^3 + ... = \sum 2ia_{i}x^{i}$$
$$y'' = 2a_2 + 3\times2a_3x + 4\times3a_4x^2 + ... = \sum (i+1)(i+2)a_{i+2}x^{i}$$... | Let the solution of the given ODE be $y=\sum c_{s}x^{s}$.
Then this satisfies the equation $y^{''}+2xy^{'}=0$.
Now, $y^{'}=\sum sc_{s}x^{s-1}$ and $y^{''}=\sum s(s-1)c_{s}x^{s-2}$. Substituting these values in the differential equation.
$\sum s(s-1)c_{s}x^{s-2} + 2x \sum sc_{s}x^{s-1}=0$
$=>\sum s(s-1)c_{s}x^{s-2} + 2... | {
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"url": "https://math.stackexchange.com/questions/3089497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?
This is what I did:
I try to multiply by the conjugate. Its value I believe is technically ... | You basically used the formula $$a+b=\frac{a^2-b^2}{a-b}$$
in an indirect way.
The problem gave you $a-b$ and, since $a,b$ are square roots of simple expressions, it is easy to calculate $a^2-b^2$.
And for the same reason, the same trick can be used to calculate
$$\sqrt{\mbox{nice}}\pm \sqrt{\mbox{nice}}$$
whenever w... | {
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} |
A series question whose indices involves greatest integer function Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $a$ and $n$ be nonnegative integers such that $n\geq a+1$. Prove that
$$ \sum_{k=a}^{n} f(k) - \sum_{k=a}^{n} (-1)^{k} f(k) = 2 \sum_{k=\lfloor{\... | The index of summation is ranging as
$$
a \le k \le n
$$
Replace it with its even and odd components
$$
k = 2j - i\quad \left| {\,i = 0,1} \right.
$$
Then for the even component it shall be
$$
\eqalign{
& i = 0\quad \Rightarrow \quad a \le 2j \le n\quad \Rightarrow \quad \left\lceil {{a \over 2}} \right\rceil \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see... | $\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$
Or $$(\sqrt{2x^2 + 3x +5}-7) + (\sqrt{2x^2-3x+5}-5)=3x-12$$
Or $$\frac{\left(x-4\right)\left(2x+11\right)}{\sqrt{2x^2+3x+5}+7}+\frac{\left(x-4\right)\left(2x+5\right)}{\sqrt{2x^2-3x+5}+5}=3(x-4)$$
It is not difficult to prove $$\frac{2x+11}{\sqrt{2x^2+3x+5}+7}+\frac{2x+5}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Combinatorics - Limited Replacement A man is selling $15$ different items, and has exactly $3$ of each item. They all cost the same and we have enough to buy $7$ items in total. How many combinations of items can we buy? I know that if he had unlimited of each item, the formula would be ${n+k-1\choose k}$, so ${15+7-1\... | If we let $x_j$ denote the number of items of the $j$th type, $1 \leq j \leq 15$, that are selected, then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} + x_{15} = 7 \tag{1}$$
A particular solution of the equation in the nonnegative integers corresponds to the placem... | {
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"url": "https://math.stackexchange.com/questions/3094308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $2\sin\theta\cos\theta + \sin\theta = 0$
The question is to solve the following question in the range $-\pi \le \theta \le \pi$
$$2\sin\theta\cos\theta + \sin\theta = 0$$
I missed the obvious sin factorisation so proceeded as below. I see the correct solutions should be $\pm2/3\pi$ and the values when $\sin\... | For $-\pi\leq\theta\leq\pi$:
\begin{align*}
2\sin{\theta}\cos{\theta}+\sin{\theta}&=0 \\
\sin{\theta}\big(2\cos{\theta}+1\big)&=0 \\
\end{align*}
Thus, $\sin{\theta}=0$ or $2\cos{\theta}+1=0$ .
*
*If $\sin{\theta}=0$, then $\theta\in\{-\pi,0,\pi\}$ .
*If $2\cos{\theta}+1=0$, then $\cos{\theta}=-1/2<0$, then $\the... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$, then what is $\lfloor S \rfloor$?
If
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\cdots$$
then $$\lfloor S \rfloor = \text{?}$$
What I tried:
I know that
$$S=1+\frac{1}{2^4}+\frac{1}{3^4}+\cdots=\zeta(4)=\frac{\pi^4}{90}\approx 1.1$$
then $\lfloor ... | Note that $$\frac{1}{k^4}=\int_{k-1}^{k}\frac{1}{k^4}\,dx\le \int_{k-1}^k\frac{1}{x^4}\,dx$$
Therefore, $$\sum_{k=2}^n \frac{1}{k^4}\le \sum_{k=2}^n\int_{k-1}^k\frac{1}{x^4}\,dx=\int_1^n \frac{1}{x^4}\,dx=\frac{1}{3}\left[1-\frac{1}{n^3}\right]$$
Taking limit as $n\to\infty$, we get $$\sum\limits_{k=2}^\infty \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$? For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$
I checked in very many cases. Example :$c=1,
a=2,b=\frac{1}{2}...$ then it’s true, but cannot prove that
My attempts:
I conside... | I consider function $ f(x)=\sqrt{24x^2+25}$
And $f’(x)=\frac{24x}{\sqrt{24x^2+25}}$,
$f’’(x)=\frac{600}{(24x^2+25)\sqrt{24x^2+25}}>0$
So $f(x)+f(y)+f(z)\geq 3f(\frac{x+y+z}{3})$
(inequalyti Jensen’s)
But can not prove that $\sqrt{ab}+\sqrt{bc}
+\sqrt{ca}\geq 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that for all $n \in \Bbb{N}^*$, $\frac{1}{n+1} \le u_n$ $$u_n:= \int_{1}^{e}x^{1/3}(1-\ln(x))^n \, \mathrm{d}x$$
*
*I showed that $u_n$ is decreasing.
*I showed that for all $n \in \Bbb{N}^*$, $u_{n+1} = -\dfrac{3}{4} + \dfrac{3}{4}(n+1)u_n$.
*I have to show that for all $n \in \Bbb{N}^*$, $\dfrac{1}{n+1} \... | Taking the $Z-$transform for the equation that you proved, i.e.
\begin{equation}
u_{n+1}=-\frac{3}{4}+\frac{3}{4}(n+1)u_n
\end{equation}
\begin{equation}
zU(z)-zu(0)=-\frac{3}{4}\frac{z}{z-1}-\frac{3z}{4}\frac{\text{dU(z)}}{\text{dt}}+\frac{3}{4}U(z)
\end{equation}
Assuming $u(0)=1$, and after some manipulation of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do the chances of winning Game A compare to the chances of winning Game B?
A coin is biased in such a way that on each toss the probability of heads is $\frac{2}{3}$ and the probability of tails is $\frac{1}{3}$. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Ga... | For Game A, winning corresponds to getting $3$ heads in a row or $3$ tails in a row. These have probabilities $(2/3)^3$ and $(1/3)^3$, respectively. This means that the probability of winning Game A is, as you correctly computed,
$$\left(\frac{2}{3}\right)^3 + \left(\frac{1}{3}\right)^3 = \frac{1}{3}$$
For Game B, win... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In a right angled $\triangle ABC$, $DE$ and $DF$ are perpendicular to $AB$ and $BC$ respectively. What is the probability of $DE\cdot DF>3$?
In a right angled $\triangle ABC$, $\angle B = 90^\circ$, $\angle C = 15^\circ$ and $|AC| = 7.\;$ Let a point $D$ (Random Point) be taken on $AC$ and then perpendicular lines $DE... |
Let $DC = x , DF = o , DE = a $ and $ DA=y $ .
We have :- $$ o = x \sin 15 \tag{1}$$ $$a= y \ cos 15 \tag{2} $$ Multiplying $(1),(2)$ , we get :- $$ o\cdot a = xy \sin 15 \cos 15 = \frac{xy}{2} sin 30 = \frac{xy}{4} > 3 $$ ( $\because 2 \sin \theta \cos \theta = \sin 2\theta $)
Hence , we need $xy = x(7-x) > 12$ or $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2... | Here you are solving:
\begin{equation}
I(a,b)=\int_0^\infty\frac {\ln\left(a^2+x^2\right)}{\left(b^2+x^2\right)}dx
\end{equation}
The method you have taken is perfectly fine. Here I will employ Feynman's Trick and introduce a new parameter:
\begin{equation}
J(t;a,b)=\int_0^\infty\frac {\ln\left(a^2+tx^2\right)}{\left(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Use the definition of limit to show that $\lim_{x \to -1} \frac{x+5}{2x+3}=4$
Use the definition of limit to show that
$\lim_{x \to -1} \frac{x+5}{2x+3}=4$
We have $|\frac{x+ 5}{2x+3} -4|= |\frac{x+5-8x-12}{2x+3}|=|\frac{-7x-7}{2x+3}|=\frac{7}{|2x+3|}|x+1|$
To get a bound on the coefficient of $|x+1|$, we restrict $x... | Yes that's right. From this step on, as $x\to -1$ you can write$$-\epsilon<x+1<\epsilon$$ for some small choice of $\epsilon>0$ therefore $$1-2\epsilon<2x+3<1+2\epsilon$$since for small enough $\epsilon>0$ we have $1-2\epsilon >{1\over 2}$ we can conclude that $$0\le {|1+x|\over |2x+3|}<{\epsilon \over {1\over 2}}=2\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$ How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$
I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it ... | With $\zeta = \exp(2\pi i/n)$ and
$$f(z) = \frac{1}{1-z} \frac{n/z}{z^n-1}$$
we have for $1\le k\le n-1$
$$\mathrm{Res}_{z=\zeta^k} f(z)
= \frac{1}{1-\zeta^k}$$
so that
$$S = \sum_{k=1}^{n-1} \frac{1}{1-\zeta^k}
= \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$
Residues sum to zero and the residue at infinity is zero... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluate the values of $x$ in $\sqrt{2x-5} = \sqrt[3]{6x-15}$
$$\sqrt{2x-5} = \sqrt[3]{6x-15}$$
*
*Evaluate the values of $x$
I believe that there would be an easier approach to this problem because I will have to expand 3th degree binomial as shown below
$$(2x-5)^3 = (6x-15)^2$$
What am I missing?
Regards
| $$(2x-5)^3 = (6x-15)^2$$
$$(2x-5)^3 = (3(2x-5))^2$$
$$(2x-5)^3 = 9(2x-5)^2$$
This may help :)
Then: $(2x-5)=0\lor (2x-5)=9$
Thus: $x=2\frac{1}{2}\lor x=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x... | $$x^3+y^3 = x^3+\color{red}{x^2y-x^2y}+y^3 $$
$$ = x^2(x+y)-y(x^2-y^2)$$
$$ = x^2(x+y)-y(x-y)(x+y)$$
$$ = (x+y)(x^2-y(x-y))$$
$$ = (x+y)(x^2-xy+y^2)$$
We can proceede similary for arbitrary odd $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove that $\frac{m^{n+1}}{n+1}<1^n+2^n+\dots+m^n<(1+\frac{1}{m})^{n+1}\frac{m^{n+1}}{n+1} $? The first part of the problem is:
Prove that for all integers $n \ge 1$ and real numbers $t>1$,
$$ (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1)$$
I have done the first part by induction on $n$ for any real $t>1$.
However, I don't... | Let $S=1^n+2^n+3^n+\cdots+m^n$.
From a quick graph sketch it is clear that:
$$\begin{align}
\int_0^m x^n dx \;\;&<\qquad \qquad \quad S &&< \int_0^m (x+1)^n dx\\
\left[\frac {x^{n+1}}{n+1}\right]_0^m\;\;&<\qquad \qquad \quad S &&<\;\;\left[\frac {(x+1)^{n+1}}{n+1}\right]_0^m\\
\frac {m^{n+1}}{n+1}\;\;&<\qquad \qquad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
solve an equation in complex plane $a(5-i)+b=ai-3$, $a$ and $b$ are conjugate complex number.
Find $a$ and $b$.
I have tried several methods to solve it but it stuck.
How find the relationship of a and b in the equation with the complex number?
| $a(5-i)+b=ai-3$
so
$b=a(2i-5)-3$
but
$b=x+iy$
and
$a=x-iy$
$x+iy=(x-iy)(2i-5)-3=2ix-5x+2y+5iy-3=$
$=(2y-5x-3)+i(2x+5y)$
So you must solve the system
$x=2y-5x-3$
and
$y=2x+5y$
$x=\frac{y}{3}-\frac{1}{2}$
$y=\frac{2y}{3}-1+5y=\frac{17}{3}y-1$
To sum up
$y=\frac{3}{14}$
$x=-\frac{3}{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let a,b and c be the side lengths of triangle ABC respectively...find the greatest value of b*c. Let a,b and c be the side lengths of triangle ABC respectively. If the perimeter of $\Delta$ABC is 7, and that $\cos A=-\frac{1}{8}$, find the greatest value of $b*c$.
This is how I start the solution:
$$a+b+c=7 \impli... | $$ a^2=b^2+c^2+\frac{bc}{4}$$
so $$ (7-b-c)^2 = b^2+c^2+bc/4$$
so $$ 49-14b-14c+2bc = bc/4$$
so $$ 7bc= 56(b+c)-196 \geq 112\sqrt{bc}-196$$
Put $ x = \sqrt{bc} \;\;\;\;(\leq {b+c\over 2} < {7\over 2})$ then $$7x^2-112x+196\geq 0$$
so $$x^2-16x+28\geq 0$$
so $$x\in (0,2]\cup[14,\infty)$$
and thus $x_{\max} \leq 2$ (an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \fra... | Keep in mind that it is a limit at $-\infty$. Therefore, the way you should deal with the numerator is$$\sqrt{4x^2-2x}=-x\frac{\sqrt{4x^2-2x}}{-x}=-x\sqrt{4-\frac2x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
If $a, b, c, d$ exists s.t. $p=a^2+kb^2$, $pn=c^2+kd^2$, proof that integer $x, y$ such that $n=x^2+ky^2$ exists. Question. $p$ is a prime, $k$ is a given natural number. If $a, b, c, d$ exists s.t. $p=a^2+kb^2$, $pn=c^2+kd^2$, proof that integer $x, y$ such that $n=x^2+ky^2$ exists.
My approach. Let $n=x^2+ky^2$.
$pn=... | feb. 21 2019
LEMMA: with integers, if $v^2 | w^2,$ then $v | w.$
With $b \neq 0,$ since $p$ is a prime rather than a square,
$$ p = a^2 + k b^2 $$
$$ np = c^2 + k d^2 $$
$$ kb^2 = p - a^2 $$
$$ b^2 np = b^2 c^2 + (k b^2) d^2 = b^2 c^2 + (p-a^2)d^2 = (b^2 c^2 - a^2 d^2) + p d^2 $$
Thus
$$ p | (bc-ad)(bc+ad) $$
and $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the Exact X-Coordinates of the Points on the Graph f(x) for Which the Tangent Line is Parallel to the Line g(x) I cannot seem to solve the following situation:
When $f'(x)$ is equal to the slope of $g(x)$ over the interval of $\frac{\pi}{2}\le x \le \pi $ when
.
$$f'(x)=\frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} ... | Since $g^\prime(x)=\dfrac{2}{3\sqrt[3]{6}}$ we wish to find a solution for
$$ \frac{2\cos(2x)}{3\sin(2x)^{\frac{2}{3}} }= \dfrac{2}{3\sqrt[3]{6}}$$
in the interval $\frac{\pi}{2}\le x \le \pi$. Simplifying and cubing both sides gives
\begin{eqnarray}
\frac{\cos^3(2x)}{\sin^2(2x)}&=&\frac{1}{6}\\
6\cos^3(2x)-\sin^2(2x)&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that sequence is a Cauchy sequence Prove that given sequence $$\langle f_n\rangle =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{(-1)^{n-1}}{n}$$
is a Cauchy sequence
My attempt :
$|f_{n}-f_{m}|=\Biggl|\dfrac{(-1)^{m}}{m+1}+\dfrac{(-1)^{m+1}}{m+2}\cdots\dots+\dfrac{(-1)^{n-1}}{n}\Biggr|$
using $ m+1>m \impl... | If you ignore the signs of the terms,
the result diverges.
So you can't do that.
$f_n
=\sum_{k=1}^n \dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=\sum_{k=m+1}^n \dfrac{(-1)^k}{k}
=\sum_{k=1}^{n-m} \dfrac{(-1)^{k+m}}{k+m}
=(-1)^m\sum_{k=1}^{n-m} \dfrac{(-1)^{k}}{k+m}
$.
If
$n-m$ is even,
so $n-m = 2j$,
then
$\begin{arra... | {
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"url": "https://math.stackexchange.com/questions/3123120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The intersection of a sphere and the xy-plane is the circle $(x-1)^2+(y-2)^2=5$ and point $(1,2,1)$ is on the sphere The intersection of a sphere and the $xy$-plane is the circle $$(x-1)^2+(y-2)^2=5$$ and point $(1,2,1)$ is on the sphere. Find the center and radius of the sphere.
When the sphere and $xy$-plane intersec... | Let $O(a,b,c)$ be center of sphere, $A(1,2,0)$ be the center of intersection circle, $B(1,2+\sqrt{5},0)$ be the point of the sphere and intersection circle, $C(1,2,1)$ be the point on the sphere.
Then the radius of the sphere is $R=OC=OB=AO+AC=AO+1$.
The triangle $AOB$ is right, so:
$$AO^2+AB^2=OB^2 \Rightarrow AO^2+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Prove that $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$ is decreasing Let $\ a>0$ and the sequence $(x_{n})_{n>=0}$ defined by $\ x_{n}=\int_{n}^{2n} \frac{x+a}{x^{3}+2a}dx$. Prove the sequence is monotonically decreasing and $0<x_n<\frac{4+3a}{8}$, for any $n>0$. I've made little progress towards proving that $x_{n}... | First, let's note that
$$ \int_n^{2n} \frac{x+a}{x^3+2a}dx \leq \int_n^{2n} \frac{x+a}{x^3}dx.$$
Evaluating the integral on the right, we have
$$ \int_n^{2n} \frac{x+a}{x^3}dx = \int_n^{2n} \frac{1}{x^2}dx + a\int_n^{2n} \frac{1}{x^3}dx.$$
These are easy integrals to evaluate. This gives us that
$$ \int_n^{2n} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why does the domain of convergence change after this integration? So, while trying to derive the Taylor expansion for $\ln(x+1)$ around $x=0$, I started by using the well-known power series:
$$\frac{1}{1-x}=1+x+x^2 +x^3 +... \qquad \text{when } \lvert x \rvert < 1$$
Letting $x \mapsto-x$, we get:
$$\frac{1}{1+x}=1-x+x^... | Writing
$$\tag{*}\ln(1+x) = \int_0^x \frac{dt}{1+t} =\int_0^x (1-t+t^2-t^3+...)\,dt \\ = x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^4}{4} + \ldots $$
is true for $x < 1$ because the power series converges uniformly on the compact interval $[0,x]$ (within the interval $(-1,1)$ bounded by the radius of convergence) and term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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This polynomial is divisible by $(x-2)$ so the remainder is $0$. How do we evaluate the product $ab$? $$P(x) = x^{2a+b-1} +x^{a-2b+5}-2x^{a+b-1}$$
This polynomial is divisible by $(x-2)$ so the remainder is $0$. How do we evaluate the product $ab$?
My attempt:
This polynomial is divisible by $(x-2)$ so the remainder ... | We continue from the last line, since it has been pointed out in comments that $u\ne0$:
$$0=um^3+64-2m^3=(u-2)m^3+64$$
$$64=(2-u)m^3$$
Given that $u,m$ are powers of 2, $2-u$ is positive, but $u>0$ so $u=1$ and $m=4$, i.e. $a=0,b=2$ and $ab=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Coefficients of $1,x,x^2$ in $((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Finding coefficients of $x^0,x^1,x^2$ in
$((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
where there are $k$ parenthesis in the left side
Try:
Let $P(x)=((\cdots (x-2)^2-2)^2-2)^2\cdots )-2)^2$
Let we assume
$P(x)=a_{k}+b_{k}x+c_{k}x^2+d_{k}x^3+\cdots\c... | Let the $x^2,x,1$ coefficients of the polynomial formed from the expression with $k$ brackets be $a_k,b_k,c_k$ respectively. We have
$$a_1,b_1,c_1=1,-4,4$$
$$a_{k+1},b_{k+1},c_{k+1}=2a_k(c_k-2)+b_k^2,2b_k(c_k-2),(c_k-2)^2$$
We first notice that $c_k=4$ for all $k\ge1$, so we reduce the recurrences left to solve to
$$a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$ = $21\sqrt{6}$ but I get $207\sqrt{6}$ I'm asked to simplify $6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}$
The provided solution is $21\sqrt{6}$ but I arrive at a different amount.
Here is my working, trying to understand where I went wrong:
First expression:
$6\sqrt{24}$ = $6\sqrt{4}$ *... | $$6\sqrt{24} + 7\sqrt{54} - 12\sqrt{6}=6\cdot2\sqrt6+7\cdot3\sqrt6-12\sqrt6=21\sqrt6.$$
I used the following law.
$$a(bc)=(ab)c.$$
For example,
$$6\cdot2\sqrt6=(6\cdot2)\sqrt6=12\sqrt6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Inductive proof of $D_N=-H_n\cdot R_N \cdot H_n $ (Grover Iterator) I am currently working on an inductive proof $D_N=-H_n\cdot R_N \cdot H_n $ (H is the Hadamard matrix for n Bits), the induction assumption (base case) and the induction condition have been done by me, but I can not go further in the Inductive step.
F... | You have to notice that $N=2^{n-1}$ and you are doing induction on $n$. The induction step is that
$$-H_nR_{2^{n-1}}H_n=D_{2^{n-1}}.$$
Also notice that $R_2^{n}$ can be writen as a block matrix in the following way:
$$ R_{2^n}=\left(\begin{array}{c|c}
R_{2^{n-1}} & 0\\
\hline
0 & I_{2^{n-1}}
\end{array}\right). $$
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | If there are an even number of heads then there must be an odd number of tails. But heads and tails are symmetrical, so the probability must be $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 0
} |
Integral involving incomplete beta function I have the following integral,
$$\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)dx$$
where $B_x(c,d) = \int_{0}^xt^{c-1}(1-t)^{d-1}dt$ is the incomplete beta function, and $a,b,c,d>0$.
Question: Does this have a closed form?
My attempt:
*
*First, playing around in Wolfram Alpha mak... | Using the hypergeometric representation of the incomplete Beta function
\begin{equation}
B_x\left( c,d \right)=\frac{x^c}{c}{}_2F_1\left( c,1-d;1+c;x \right)
\end{equation}
the integral can be written as
\begin{align}
I\left( a,b,c,d \right)&=\int_{0}^1x^{a-1}(1-x)^{b-1}B_x(c,d)\,dx\\
&=\frac{1}{c}\int_{0}^1x^{a+c-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $ Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $.
I tried multiplying by the conjugate and I get to an ugly expression. What should I do?
| Hello Taylor my old friend...
$\begin{array}\\
\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3}
&=\left(x+x^{-1/2}\right) ^{3/2} - x^{3/2}\\
&=x^{3/2}\left(\left(1+x^{-3/2}\right) ^{3/2} - 1\right)\\
&=x^{3/2}\left(\left(1+\frac32x^{-3/2}+O(x^{-3})\right) - 1\right)\\
&=x^{3/2}\left(\frac32x^{-3/2}+O(x^{-3})\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.
I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,57... | FYI,
I noticed that
$\dfrac{a_{n+1}+a_{n-1}}{a_n}=6 \hspace{1cm}\forall n \gt1\tag1$
(unfortunately I could not prove it so far).
Using this recursion formula easy to determine the general formula of $a_{n+1}$.
Searching the general formula in the form
$a_n=c_1 r_1^n+c_2 r_2^n\tag2$
where $r_1, r_2$ are the roots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.