Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$ Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$
I used the R method to simplify it -
$ \sqrt{80} \cos (x-26.565) $
Minimum value of that =
$ \sqrt{80} \cos (x-26.565) = - \sqrt{80}$
$ \cos (x-26.565) = -1$
This c... | Since
$$
8^2+4^2=80
$$
you know that $8\cos x+4\sin x=\sqrt{80}\cos(x-\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\pi/2$:
\begin{align}
8&=\sqrt{80}\cos\alpha\\
4&=\sqrt{80}\sin\alpha
\end{align}
Thus the angle $\alpha$ is in the first quadrant and so
$$
\alpha=\arcsin\frac{4}{\sqrt{80}}=\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Condition for two cubic equations to have two common roots The equations $x^3-x^2+bx+c=0$ and $x^3+cx^2+bx-d=0$ have two common roots.The question is to show that $b^2=d$
I couldn't get how to approach this problem.Any help would be appreciated. thanks
| Let $\alpha,\beta,\gamma$ be the roots of $x^3-x^2+bx+c$.
Also, let $\alpha,\beta,\omega$ be the roots of $x^3+cx^2+bx-d$.
By Vieta's formulas, we get
$$\alpha+\beta+\gamma=1,\qquad \alpha+\beta+\omega=-c\tag1$$
Since we get
$$\alpha^3-\alpha^2+b\alpha+c=0$$
$$\alpha^3+c\alpha^2+b\alpha-d=0$$
subtracting the latter fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the Integral $ \iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz$ I am attempting to evaluate the following triple integral using spherical coordinated:
$$
\iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz \\
W = \{(x,y,z) \in \mathbb{R}^3 : 1 \le x^2+y^2+z^2 \le 9; 0 \le x \le y, z \ge 0\}
$$
This tells me that I have the ... | $x\geq0, y\geq0, z\geq0$ show $0\leq\theta\leq\dfrac{\pi}{2}$ and $0\leq\phi\leq\dfrac{\pi}{2}$. Also $x\leq y$ states $r\cos\theta\sin\phi\leq r\sin\theta\sin\phi$ or $\tan\theta\geq1$ so $\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{2}$ then
$$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _1^3\frac{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$ $$L=\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$$
I am not sure of my answer. Please help me.
$$L=\lim_{x\to0}\frac{1+x}x\frac{\ln(1+x)}x-\frac1x=\lim_{x\to0}\frac{1+x-1}x=1$$
| You are missing a factor of $\frac12$, so I recommend you be careful and write your steps out.
Using a MacLaurin Expansion: $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{\ln(1+x)}{x} - \frac1{x} = $$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{x - \frac{x^2}2 + O(x^3)}{x} - \frac{1}{x} =$$
$$\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to Simplify $ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)} $
Simplify $$ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)}.$$
The solution is : $-\cot(x)$
I tried to: $$\frac{\sin(2x)\cos(x)+\cos(2x)\sin(x)+\sin^3(x)}{\cos(2x)\cos(x)+\sin(2x)\sin(x)-\cos^3(x)}.$$
| With one step further you have
$$\sin(3x) = - 4 \sin^3 x + 3 \sin x~~~and~~~\cos 3x = -3\cos x+4\cos^3x$$that is
$$\sin(3x) +\sin^3 x = 3\sin x( 1-\sin^2 x ) = 3\sin x\cos^2x ~~~$$and$$~~~\cos 3x-\cos^3x = -3\cos x( 1-\cos^2x)= -3\cos x\sin^2x$$
Then $$\frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)} =\frac{3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Time and Work. How much work does C do per hour?
A, B and C need a certain unique time to do a certain work. C needs 1
hours less than A to complete the work. Working together, they require
30 minutes to complete 50% of the job. The work also gets completed if
A and B start working together and A leaves after 1 ... | Let $A$, $B$ and $C$ be the rates measured in an amount of work per hour $\left(\frac{work}{hour}\right)$ at which workers $A$, $B$ and $C$ accomplish work. If $C$ is the amount of work worker $C$ can do in one hour and $1$ ($100\%$) is the amount of work of the entire job, then $t$ is the time it takes him to bring it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Minimum values of the sequence $\{n\sqrt{2}\}$ I have been studying the sequence
$$\{n\sqrt{2}\}$$
where $\{x\}:= x-\lfloor x\rfloor$ is the "fractional part" function. I am particularly interested in the values of $n$ for which $\{n\sqrt{2}\}$ has an extremely small value - that is, when $n\sqrt 2$ is extremely close ... | This is about Pell type equations. Absolutely complete detail would be a bit long. I have posted many times about solving $x^2 - n y^2 = T,$ where $T$ is some target number, and $n$ is positive not a square.
Your sequence is this: given positive integer $x,$ let $v = \lfloor x \sqrt 2 \rfloor.$ Your numbers will be
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
} |
Arclength Parameterization of the Trefoil Knot I would like to find an arclength parameterization of the trefoil knot
The parameterizations I can find are:
$(sin(t) + 2sin(2t),$ $cos(t) - 2cos(2t),$ $sin(3t))$
and
$((2+cos(3t))cos(2t),$ $(2+cos(3t))sin(2t),$ $sin(3t))$
for $t \in [0,2\pi)$
writing $t = t(\theta)$, ... | Since both integrands here are square roots of quadratic polynomials in $\cos3t$, they are reducible to elliptic integrals through the method I detailed here. There will be an extra algebraic/trigonometric term here, but it simplifies quite a bit because of preceding substitutions. I did all of the work in Mathematica ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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How can i solve $5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$? How can i solve it?
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$
I don't have idea how to solve it..
| Taking uldek's comment in the form of an answer.
Let $A=5^{2x-{x^2}/3}$. Note that $A>0$ for any $x$.
Notice that A appears as is on the LHS.
Notice that A is hidden on the RHS, as $5^{2-2x}*(5^{1/3})^{x^2}=5^2*5^{-2x+{x^2}/3}=5^2*A^{-1}$.
Now the inequality is $A < 25*A^{-1} +24$;
Take all the terms on the same sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving the integral I'm trying to calculate the following integral :$\int \sqrt x (2-3x^2)^2dx$, but somehow I can't get it right, here's what I did:
First I expanded the expression:
\begin{align}&\int \sqrt x (4-12x^2 + 9x^4)
\\&= \int(4\sqrt x-12x^2\sqrt x + 9x^4\sqrt x) \end{align}
Then I evaluated each term indivi... | You just forget the constant (and the minus sign as commented), other than that, your answer is correct.
We can verify the solution by differentiating the solution.
\begin{align}\frac{d}{dx}\left(\frac{8x^\frac{3}{2}}{3} - \frac{24x^\frac{7}{2}}{7} + \frac{18x^\frac{11}{2}}{11}+c\right) &=\left( \frac32\right)\frac{8x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
system of equations with 3 variables in denominator I'm having problems with this system of equations:
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$$
I've tried substitution method but that took me nowhere, I'm not sure I can solve ... | I hope you mean the following.
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2},$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3},$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}.$$
If so, let $\frac{1}{x+y}=c$, $\frac{1}{x-z}=b$ and $\frac{1}{y-z}=a$.
Thus,
$$3c+2b=\frac{3}{2},$$ $$c-10a=\frac{7}{3}$$ and
$$3b+5a=-\frac{1}{4}.$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \... | note that $$\frac{1-2x}{2x+3}\geq 0$$ is hold if $$1-2x\geq 0$$ and $$2x+3>0$$ or
$$1-2x\le 0$$ and $$2x+3<0$$ you have to solve These two cases
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? Given the inequality:
$\frac{1}{x} < 5$
In order to find a solution, I would normally multiply both sides by $x$:
$1 < 5x$
Then I would divide by $5$
$\frac{1}{5} < x$
To obtain the solution: $x > \frac{1}{5}$.
Now, the thing is, the solut... | $$\frac { 1 }{ x } <5\quad \Rightarrow \frac { 1 }{ x } -5<0\quad \Rightarrow \quad \frac { 1-5x }{ x } <0\quad \Rightarrow \frac { x\left( 1-5x \right) }{ { x }^{ 2 } } <0\\ x\left( 5x-1 \right) >0\quad \Rightarrow x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 5 } ,+\infty \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
If $\vert{z_{1}+ \cdots + z_{n}}\vert$ = $\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert$ then $z_{j} = c_{j}z_{1}$ Could someone give me just one suggestion to solve this problem?
Let $z_{1}, \ldots z_{n} \in \mathbb{C}$, with $z_{1}\neq 0$. Prove that if
$\vert{z_{1}+ \cdots + z_{n}}\vert$ =
$\vert{z_{1}}\vert... | Here's the base case.
If
$|(a+bi)+(c+di)|
=|a+bi|+|c+di|
$
then,
$\sqrt{(a+c)^2+(b+d)^2}
=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}
$.
Squaring,
$(a+c)^2+(b+d)^2
=a^2+b^2+c^2+d^2+2\sqrt{(a^2+b^2)(c^2+d^2)}
$
or
$2ac+2bd
=2\sqrt{(a^2+b^2)(c^2+d^2)}
$.
Dividing by 2
and squaring again,
$a^2c^2+2abcd+b^2d^2
=a^2c^2+a^2d^2+b^2c^2+b^2d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the MSE of the MLE
Let $X_1,\ldots,X_n$ be i.i.d exponentially distributed r.v's with pdf:
$$ f_\theta(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$$
such that $x\ge 0$.
I have found that the MLE is given by $$ \hat{\theta}(X)= \frac{1}{n}\sum_{i=1}^n X_i.$$
Note that $\hat{\theta}$ is unbiased as
$$ E(\hat... | It's not true that
$$ E(\hat{\theta^2}) = E\left(\frac{1}{n^2}\sum_{i=1}^n X_i^2\right).$$
Rather, one has
\begin{align}
E(\hat{\theta^2}) &= E\left[\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\right] \\
&= E\left[\frac{1}{n^2}\sum_{i,j} X_iX_j \right] \\
&= \frac{1}{n^2}E\left[\sum_{i=1}^n X_i^2 + \sum_{i\ne j} X_iX_j \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}$ using AM-GM? In this question Prove that, if $g(x)$ is concave, for $S = {x : g(x) > 0}$, $f(x) = 1/g(x)$ is convex over $S$. , in the proof of Math536,
how to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}?$
This is the proof of Math536:
$$\begin{align}
1 &= (a... | Don't worry about functions.
For any two $x,y$ then $\frac {g(x)}{g(y)} = k$ will be a positive number, and $\frac {g(y)}{g(x)} = \frac 1k$.
So by AM-GM. $\frac {k + \frac 1k}2 \ge \sqrt {k*\frac 1k}=1$.
And that's it.
....
In fact this basically is the proof of the AM-GM.
$(k - 1)^2 \ge 0$ so
$k^2 + 1 \ge 2k$ so
$k +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have: $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
... | There are an infinite number of such integers.
One way to show this is to start from the fact that there are an infinite number of Pythagorean triples $a^2 + b^2 = c^2$. Given any such triple, let $N$ be given by:
$$N = (ab)^4 + (bc)^4 + (ac)^4 + c^4$$
By an identity of Fauquembergue (1) we have (given $a^2 + b^2 = c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$ $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$$
I'm surprisingly struggling with this limit, could you give me a hint how to handle it (no L'Hospital and no prior knowledge what the limit ... | $$\lim _{ n\rightarrow \infty } n\left( \cos \left( \frac { 1 }{ \sqrt { n } } \right) -1 \right) =\lim _{ n\rightarrow \infty } n\left( \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) - } } \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
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Translate and rotate the coordinate axes to put the quadric in standard position, $2xy+2xz+2yz-6x-6y-4z=-9$
Referring to the image attached. Is my working correct?
The answer gives the coefficient on the other side of the equation is -1. Please show where went wrong.
Thanks.
| Let $A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix} \, .
$
To write the quadric in standard form, we must write it with respect to an orthonormal set of eigenvectors, which amounts to orthogonally diagonalizing $A$.
As mentioned in the comments, you need to apply Gram-Schmidt orthogonalization to o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
| This is really the same as aid78's method.
$$(3+4i)e^{ix}=(3\cos x-4\sin x)+i(3\sin x+4\cos x)
=2+iA$$
for $A=3\sin x+4\cos x$. As $|(3+4i)e^{ix}|^2=3^2+4^2=25$
then $25=4+A^2$, so $A=\pm\sqrt{21}$ (both are possible).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
... | $$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=1}^{\infty} \frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=2}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}$$
$$Sum=\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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In which interval lies the minimum? Let $a_1,a_2,a_3,a_4$ be real numbers such that $a_1+a_2+a_3+a_4 =0 $ and $a_1^2+a_2^2+a_3^2+a_4^2=1$.
Then in what interval does the smallest possible value of the following expression lies?
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$$
Here lies the actual question with the ... | Let $a_1=0$, $a_2=\frac{1}{\sqrt2},$ $a_3=0$ and $a_4=-\frac{1}{\sqrt2}.$
Hence, we get a value $2$.
We'll prove that it's a minimal value.
Indeed, let $a_1=a$, $a_2=b$, $a_3=c$.
Thus, $a_4=-a-b-c$, $a^2+b^2+c^2+(a+b+c)^2=1$ and we need to prove that
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\geq2$$ or
$$(a-b)^2+(b-c)^2+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the derivative of $\frac{2}{x^2}$ directly from definition I have to find the derivative of $\frac{2}{x^2}$ using only the definition. I have tried to do this several times but I always get the wrong answer - could you tell me where I made an error?
$$\frac{d}{dx}[\frac{2}{x^2}] = \lim_{h \to 0} \frac{\frac{2}{(x+... | You missed an $x$$$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$ It should have been $$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4x}{x^4}= \frac{-4}{x^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
When are these eigenvalues non-negative? I'm trying to find a pair of real numbers $(a,b)$ which ensure that some matrix is strictly positive semi-definite.The eigenvalues of this matrix are $$\lambda=1 + a \pm \sqrt{(c+b)^2+2(x'-ax)^2}$$ and $$\lambda=1 - a \pm \sqrt{(c-b)^2+2(x'-ax)^2}.$$
I therefore need one of the... | I guess you know if we put $a=m+n$, $b=m-n$, $t=x’-ax$, $a=x$, $b=x’$ then we obtain this problem with additional conditions that $0\le a,b\le 1/\sqrt{2}$ and one of $\lambda$’s is $0$. Nevertheless, these conditions turn us to a very different way.
Let’s start. The required $a$ and $b$ exist iff all the following con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive m... | The ratio of the consecutive terms of the sequence $a_n=\frac{n}{\sqrt[n]{n!}}$ is greater than $1$.
Therefore, $\frac{n}{\sqrt[n]{n!}}$ is increasing.
$$
\begin{align}
\left(\frac{a_{n+1}}{a_n}\right)^{n(n+1)}
&=\left[\frac{\frac{\color{#C00}{n+1}}{\color{#090}{\sqrt[n+1]{(n+1)!}}}}{\frac{\color{#C00}{n}}{\color{#090}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Numerical radius of a pair of operators in Hilbert spaces Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=\displaystyle\sup_{\|x\|=1}\left(|\langle Cx,x \rangle|^2+|\langle Dx,x \rangle|^2\right)^{1/2}.$$
Moreover, the following in... | For your first question, consider
$$ C = D = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ \frac{\sqrt{2}}{4}\| 2\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}\| = \frac{\sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = \sqrt{2} \sup_{\|x\|=1} |\langle \begin{pmatrix}0 \\ x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$... | I Assume that all angles are acute. Now its wise to use concavity of $\cos$ to say (from Jensen's Inequality)
$$\cos(A)+\cos(B) + \cos(C) \ge 3\cos(\frac{A+B+C}{3}) = 3 \cos(\frac{\pi}{3})=\frac{3}{2}$$
equality occurs for $A=B=C = \pi / 3$.
I don't know how to handle obtuse triangles with this..,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$ I have the following series:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$$
But nothing se... | $$\sum_{k=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}=\frac{1}{3}\sum_{k=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\left(1-\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{7}+...\right)=$$
$$=-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)+\frac{2}{3}\sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Simple ODE Analytical Solution Question: $\frac{dx}{dt} = ax + bu$ my question is in regards to the following formula:
\begin{align}
\frac{dx}{dt}=ax+bu
\end{align}
where x(0)=0 & u=constant and a & b are scalar
The answer to this or at least what I was given as the answer is the following:
\begin{align}
x=-\frac{bu}... | $$\begin{align}
x = -\frac{b}{a} \frac{e^{at}e^{Ct}}{a}
\end{align}$$
This line isn't correct, it should be:
$$\begin{align}
x = -\frac{b}{a} +\frac{e^{(t+C)a}}{a}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculating limit $\lim\limits_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$ for an unknown function. Given that $f(x)$ is a continuous function and satisfies $f'(x)>0$ on $(-\infty,\infty)$ and $f''(x)=2 \forall x \in(0,\infty)$.We need to find the limit
$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f... | Since $f''(x)=2~~for ~~x>0$, Therefore $f$ has the form $f(x)=x^2+bx +c~~~for~~~x>0$
Since the limit is at $+\infty$ it suffices to consider $x>0$
$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)} = \lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-8x-4b}{x^2+bx +c} =3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$ What is the sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots?$$
I know how to check if a series is convergent or not.Is there any technique to find out t... | My solution is adapted from another similar solution.
\begin{align}
\sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{n!10^n}
&=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{n!\left(10/3\right)^n}\\
&=\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}\left(\frac{3}{10}\right)^n\\
&=\left[\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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An alternative way to find the sum of this series? $\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now I have tried to solve this in a usual way, first find the nth term $t_n$.
$t_n$= $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$) ... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$
But know, I don't know how to continue for ge... | $1+7+7^2+7^3=400$, so $$\begin{array}{rcl}7+7^2+7^3+\cdots+7^{4k}&=&(1+7+7^2+7^3)(7+7^5+7^9+\cdots+7^{4k-3})\\&=&100\cdot 4(7+7^5+7^9+\cdots+7^{4k-3})\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | $$2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)=0,$$ which says $xy+xz+yz=0$.
Thus, $$1=(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz=1-3xyz$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 4
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Determinant equal to zero : elements are angles Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle.
I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$
We have the following:
\begin{align*}&\det \be... | If a point $P$ has trilinear coordinates $[\cos B,\cos A,-1]$ it lies on the height from vertex $C$, since the circumcenter and the orthocenter are isogonal conjugates and the distances of the circumcenter from the $BC$ and $AC$ sides are exactly given by $R\cos A$ and $R\cos B$. The claim turns out to be equivalent to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the range of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$
Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$
MY App... | Using Cauchy root test we have:$$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{1+\frac{2}{n}}}|4x-12|\frac{1}{(n^2+1)^{\frac{1}{n}}}=\frac{1}{3}|4x-12|$$then for $|4x-12|<3$ the series converges and for $|4x-12|>3$ it diverges. Also for $|4x-12|=3$ we have:$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let $f\left(x\right)=\begin{cases} 0 & -1\leq x\leq0\\ x^4 & 0
Question Let $f\left(x\right)=\begin{cases}
0 &\text{if } {-}1\leq x\leq0\\
x^4 &\text{if } 0<x\leq1
\end{cases}$. If $f(x) = \sum_{k=0}^{n}$$\frac{f^{(k)}\left(0\right)}{k!}x^{k}+\frac{f^{(n+1)}\left(\xi\right)}{\left(n+1\right)!}x^{n+1}$
is the Taylor's... | First, let us find the value of $n$. Your function $f$ is only 3 times differentiable at $x = 0$, because
\begin{align}
f'(x) &=
\begin{cases}
0 &\text{if } {-}1 \leq x \leq 0 \\
4x^3 &\text{if } 0 < x < 1
\end{cases}
&
f''(x) &=
\begin{cases}
0 &\text{if } {-}1 \leq x \leq 0 \\
12x^2 &\text{if } 0 < x < 1
\end{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Equation of parabola which touches a line and coordinate axis
Equation of parabola which touches $y=x$ line at $(1,1)$ and touches $x$ axis at $(1,0)$
Try: let focus of parabola be $S(p,q)$ and equation of directrix be $y=mx+c$ and a point $P(x,y)$ on parabola.
The definition of parabola $$(x-p)^2+(y-q)^2=\displaysty... | When you render the parabola in the form
$(ax^2+bxy+cy^2)+(dx+ey)+f=0$
the quadratic terms make a completed square ($b^2-4ac=0$). Exploiting this we may use a simplified form, noting the coefficient of $x^2$ must be nonzero since the parabola does not have a horizontal symmetry axis:
$(x+ay)^2+(bx+cy)+d=0$; $x+ay$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find k for Positive Definite Quadratic Form I have two quadratic forms and I need to find $k$ (different $k$ for each possibly) that makes them positive definite. Here are the two:
*
*$Q(y)=5y_1^2+y_2^2+ky_3^2+4y_1y_2=2y_1y_3-2y_2y_3$
*$Q(y)=ky_1^2+ky_2^2+ky_3^2+2y_1y_2+2y_1y_3-2y_2y_3$
What I would like to do is... | For the first we need to find a value of $k$ for which
$$5a^2+b^2+kc^2+4ab-2ac-2bc\geq0$$ or
$$kc^2-2(a+b)c+5a^2+b^2+4ab\geq0$$ for which we need
$k>0$ and since it's a quadratic inequality of $c$, we need also $\Delta\leq0$, which is $$(a+b)^2-k(5a^2+b^2+4ab)\leq0$$ or
$$(5k-1)a^2+2(2k-1)ab+(k-1)b^2\geq0,$$ for which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sq... | The third line which is
$$\lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
should be
$$\lim_{n \to \infty}\left( \frac{{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integration by Parts - $\int \frac{x}{(x+1)^2} dx$ I have been trying to evaluate the following integral by using integration by parts, but I contine to yield the incorrect answer.
$$\int \frac{x}{(x+1)^2} dx$$
I choose $u=x$, $dv=\frac{1}{(x+1)^2}dx => du=dx$, $v=\frac{-1}{x+1}$.
The integration by parts formula, $\in... | $$g'= (\frac{-x}{x+1}+ ln(x+1)+C)'= \frac {-1(x+1)+x}{(x+1)^2}+\frac 1 {x+1}=\frac x {(x+1)^2}$$
Your answer is correct..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$ now find the $\text{min}(\dfrac{x}{z})=?$
Let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$. Find $\text{min}\bigg(\dfrac{x}{z}\bigg)= \ ?$
I found $(x-y+z-2\sqrt{xz})(x-y+z+\sqrt{xz... | Since $$2(xy+xz+yz)-x^2-y^2-z^2)=$$
$$=(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{x}-\sqrt{y}+\sqrt{z})(\sqrt{y}+\sqrt{z}-\sqrt{x})\leq0,$$
We obtain $$\sqrt{y}+\sqrt{z}-\sqrt{x}\leq0.$$
Thus, $$\frac{x}{z}\geq\frac{(\sqrt{y}+\sqrt{z})^2}{z}\geq4.$$
The equality occurs for $y=z$, which says that we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. I have started by assuming that $x + \sqrt{3} = \frac{a}{b}$ and substituting $x = \frac{a}{b} - \sqrt{3}$ into $x^2 + \sqrt{3}$. It led me to $\f... | Well, since $x^2 + \sqrt{3}$ is assumed to be rational, then your work implies that
$$b - 2a = 0.$$
Make sure that you understand why this is the case. Once you have this, $2a = b$ and
$$x = \frac 1 2 - \sqrt{3}$$
is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I integrate $\sqrt{ \frac{x}{ax^3 + b}}$ analytically? How can I calculate $$\int^1_0\sqrt{\frac{x}{ax^3 + b}} \mathrm dx$$ analytically?
I searched my integral table, but I haven't found the solution.
But using WolframAlpha, I could find the analytic result like this,
$$\int \sqrt{\frac{x}{ax^3 + b}} \mathr... | Hint...substitute $u^2=x^3$ and you have a standard $\operatorname{arsinh}$ integral
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by pa... | Let $a\neq0$, $n\in\mathbb{N}$ and define, $$I_n:=\int{\frac{dx}{(x^2+a^2)^n}}$$
So, If we use integral by parts, we have $$I_n = \frac{x}{(x^2+a^2)^n}+2n\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}\qquad(*)$$
On the other hand, is clear that $$I_n=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2\int{\frac{dx}{(x^2+a^2)^{n+1}}}=\int{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the ... | HINT
Note that $$a_n= \sum_1^{T(n)} k^3-\sum_1^{T(n-1)} k^3$$
where $T(n)=\frac{n(n+1)}{2}$ are triangular numbers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1... | I want to show a slight variation of Aretino's nice proof that does not presuppose knowledge of the identity we want to prove, but rather derives it from the given equation. We need the identity $\cos(\sin^{-1}(a)) = \sqrt{1-a^2}$ as well as
$$ \sin(a+b+c) = \sum\limits_{cyc} \Big(\sin(a)\cos(b)\cos(c)\Big)-\sin(a)\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
FInd the limit without the l’Hospital’s rule FInd the limit without the l’Hospital’s rule:
$$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x}-\sqrt{x+2}}$$
I have tried multiplying the equation with:
$\frac{\sqrt{2x}+\sqrt{x+2}}{\sqrt{2x}+\sqrt{x+2}}$ and got stuck with
$\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{2x-... | Your method is correct, but there a sign error in the numerator of your first fraction. Doing what you were, it should have been:
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}\color{red}+\sqrt{x+2})}{2x-(x+2)}$$
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}+\sqrt{x+2})}{2x-(x+2)}\cdot\color{blue}{\frac{(3+\sqrt{x^2+5})}{(3+\sqrt{x^2+5})}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0... | The function $\phi(x):=|x+a|-|x+2|$ is $\ \equiv a-2$ when $x\gg1$ and is $\ \equiv2-a$ when $x\ll-1$. On the other hand $|x|\gg1$ implies ${\rm sgn}\bigl(b|x+5|\bigr)={\rm sgn}(b)$ when $b\ne0$, which is forbidden for an odd function. It follows that necessarily $b=0$.
The graph of $\phi$ has corners at $x=-a$ and at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Unable to prove that $n^5-n$ is a multiple of 30. It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.
Basically, the issue is how to show that :
$(n-1)(n)... | You can rearrange:
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$
$$(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1).$$
Both terms are divisible by $2,3,5$, hence by $30.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the functions satisfying $ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } $
Find all the functions $ f : \mathbb R \to\mathbb R $ such that
$$ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } \text . $$
I thin... | First of all, note that you have $ f : \mathbb R \to \mathbb R $, and in particular $ 0 $ is in the domain of $ f $. But the functional equation
$$ \frac { f ( x y ) } { x y } = \frac { f ( x + y ) } { x + y } \cdot \frac { f ( x - y ) } { x - y } \tag 0 \label 0 $$
can only hold when $ x , y \ne 0 $ and $ x \ne \pm y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$T(n) = T(n-1) + T(n-2) + n \in \Theta(2^n)$ I've tried enclosing $T(n)$ between two functions that represent the lower bound and upper bound, respectively, and prove that both of them are in $\Theta(2^n)$
I've noticed a good candidate for the upper bound would be $S(n) = S(n-1) + S(n-1) + n = 2S(n-1) + n = 4S(n-2) + 2... | Note that $$T(n)+n+3=T(n-1)+n+2+T(n-2)+n+1$$ hence $S(n)=T(n)+n+3$ solves the recursion $$S(n)=S(n-1)+S(n-2)$$ The characteristic equation reads $$x^2=x+1$$ hence, except for specific initial conditions, $S(n)=\Theta(\lambda^n)$ with $$\lambda=\frac{1+\sqrt5}2\approx1.618$$ Since $\lambda>1$, one gets $$T(n)\in\Theta(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for a... | Let us find the partial fraction decomposition of the rational function
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)}, \quad n = 0,1,2,\ldots$$
using the Heaviside cover-up method.
As we have $n + 1$ distinct linear factors we can write the rational function as
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \frac{A_0}{x} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex integral $\int^{2\pi}_0 e^{i\tan^{-1}\frac{r\sin(t-t_0)}{1+r\cos(t-t_0)}-i\tan^{-1}\frac{1-r \cos( t-t_0)}{r\sin(t-t_0)}}dt$ Is there a way to evaluate the following integral by simplifying the exponent?
$$\int^{2\pi}_0 \exp \left(i\tan^{-1}\frac{r\sin(\phi-\phi_0)}{1+r\cos(\phi-\phi_0)}-i\tan^{-1}\frac{1-r \co... | $$
\arctan u + \arctan v = \arctan \frac{u+v}{1-uv} \tag 1
$$
\begin{align}
& \arctan \frac A {1+B} - \arctan\frac{1-B} A = \arctan \frac{\frac A{1+B} - \frac{1-B} A}{1 + \frac{A(1-B)}{A(1+B)}} \\[10pt]
= {} & \arctan\frac{A^2 - (1-B^2)}{A(1+B) + A(1-B)} = \arctan \frac{A^2+B^2 - 1}{2A}
\end{align}
In your case $A^2+B^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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$\tan(\theta) = \frac{a}{b}$ s.t. $\gcd(a,b) = 1$, $a+b$ is odd. Show $\theta$ can be trisected iff $a^2 + b^2$ is a perfect cube I've been stuck on this problem for quite some time.
Let tan$(\theta) = \dfrac{a}{b}$ s.t. $a,b \in \mathbb{Z}$, $b \neq 0$, $\gcd(a,b) = 1,$ and $a+b$ is odd. Show $\theta$ can be trisecte... | Since $R=\mathbb Z[i]$ is a UFD, you can factorize $a+bi$ as a product of primes raised to certain exponents. Classifying exponents according to their value modulo $3$ (which can be $0,1$ or $2$), you can write $a+bi=x^3yz^2$ with $x,y,z\in R$ and $y$ and $z$ are coprime in $R$.
Suppose that the norm $N(a+bi)=a^2+b^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $
For positive real $a, b$, $$ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right).$$
I know the inequality $ (a+b)^2 \le 2... | The inequality you know gives you
$$\frac{(a + b)^2}{\sqrt{(a + b)^2 + 1}} \le 2\left(\frac {a^2}{\sqrt{(a + b)^2 + 1}} + \frac{b^2}{\sqrt{(a + b)^2 + 1}}\right).$$
Now just remember that $(a + b)^2 > a^2$ as well, so
$$\frac{1}{\sqrt{(a + b)^2 + 1}} < \frac{1}{\sqrt{a^2 + 1}}$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Suppose $a,b,c \neq 0$. Show that $\det(1+a,1,1),(1,1+b,1),(1,1,1+c) = abc\left(1 + \frac1a + \frac1b + \frac1c\right)$ Suppose $a,b,c \neq 0$ .Show that the \begin{eqnarray*} \det \begin{bmatrix} 1+a &1 &1 \\1 &1+b & 1 \\ 1 & 1 & 1+c \end{bmatrix} = abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right). \end{eq... | $$abc(1 + {1\over a} + {1\over b} + {1\over c})= abc \cdot 1 + abc \cdot {1\over a}+ abc \cdot {1\over b}+ abc \cdot {1\over c}= abc + bc + ac + ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar{a}\cdot d) \cdot e$, then $(G, \cdot)$ is a group
Let $G$ be a non-empty set, $"\cdot"$ be a binary operation on $G$ and a single operation $x \mapsto \bar{x}$ such that $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar{a}\cdot d) \cdot e, \forall ... | For all $a,b,c,d,e \in G$, $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar a\cdot d) \cdot e. \tag 1$$
Lemma 1: For all $x \in G$, we have $\bar x \cdot x$ is a left identity.
Proof: If we take $a=c=d=x$ and $b=e=y$ in $(1)$ we obtain,
$$x\cdot(y \cdot x) = x \cdot (y \cdot x) \implies y = (\bar x\cdot x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(... | substitute $$x=2\sqrt{2}\tan(u)$$ then we get
$$dx=2\sqrt{2}\sec^2(u)du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equa... | By replacing $\tan^2 x=\sec^2x-1$ we have$$4\sec^2x-5\sec x=5$$which yields to $$\sec x=\dfrac{5\pm\sqrt{25+80}}{8}=\dfrac{5\pm\sqrt{105}}{8}$$where only $\sec x=\dfrac{5+\sqrt{105}}{8}$ is acceptable. Also the equation $\sec x=p>1$ has exactly two roots in $[0,2\pi)$ so for having 7 distinct roots we must be in the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Summation of primes
Let $P(n)=\displaystyle\sum^{n}_{i=1}p_i$, where $p_i=i^{th}$ prime $\geq 2$. Does exist $k\in \mathbb{N}$, such that, $P(k)$ and $P(k+1)$, both are squares?
Please provide any clue.
| No, there is not such $k$.
If such integer $k$ exists then there are positive integers $a$ and $b$ with $a>b$ such that
$$p_{k+1}=P(k+1)-P(k)=a^2-b^2=(a+b)(a-b)\implies a-b=1,\; a+b=p_{k+1}.$$
Hence $b=(p_{k+1}-1)/2$ and
$$\sum_{i=1}^{k}p_i=P(k)=b^2=\left(\frac{p_{k+1}-1}{2}\right)^2.$$
Now we show that the above equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove polynomial identity $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots+1).$ I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below:
$(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$
| It's like this: $(x-1)(x^{k-1}+x^{k-2}+\ldots+1)$
$=x(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)$
$=x^{k-1+1}+x^{k-2+1}+x^{k-3+1}+x^{k-4+1}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x-1$
$=x^{k}+x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x^2-x-1$
$=x^{k}+(x^{k-1}+x^{k-2}+x^{k-3}...+x^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the kernel of a 4x4 matrix $$
\begin{pmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
... | Yes it is correct, in case of doubt you can check it directly by simple multiplication for RREF matrix
$$\begin{bmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}\begin{bmatrix}
1\\
-2\\
1\\
0\\
\end{bmatrix}=0$$
$$\begin{bmatrix}
1 & 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2... | HINT
Note that
*
*$n^2-2n+7=n^2-2n+1+6=(n-1)^2+6$ is even $\iff$ $n-1$ is even
and
*
*$n+1=(n-1)+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Binomial Coefficient deck of cards probability question A regular deck of 52 playing cards has 13 ranks in 4 suits. The ranks of Jack, Queen, King and Ace of each suit are top cards. Suppose you are randomly dealt seven cards. What is the probability of getting three top cards in the same suit and any four cards in ano... | There are $\binom{4}{1}$ ways to choose the suit from which the three top cards are drawn, $\binom{4}{3}$ ways to choose three top cards of that suit, $\binom{3}{1}$ ways to choose a suit from which four cards are drawn, and $\binom{13}{4}$ ways to choose four cards of that suit. Hence, the number of favorable cases i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluation of $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration I don't succed to evaluate this sum $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration because using fraction method is too hard for me and i have got that is equal $\frac 3 8$ , Is there any way to use integral for Evaluation ? ... | It's telescopic. By Sophie Germain's identity $$n^4+4 = (n^2+2)^2-(2n)^2 = (n^2-2n+2)(n^2+2n+2),$$
so
$$ \frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2} = \frac{1}{f(n)}-\frac{1}{f(n+2)} = \frac{4n}{n^4+4} $$
and
$$ \sum_{n\geq 1}\frac{n}{n^4+4}=\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{f(n)}-\frac{1}{f(n+2)}\right) = \frac{1}{4}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please gi... | From the original equation, we get:
$$abc=(a+b+c)(ab+bc+ca)$$
which is equivalent to
$$a^2(b+c)+bc(b+c)+ab(b+c)+ca(b+c)=0$$
$$\implies(b+c)(a+c)(a+b)=0$$
Then, obviously any one of the following must hold:
$$a=-b\\b=-c\\c=-a$$
In any case we can prove the equation $$\frac{1}{a^n+b^n+c^n}=\frac{1}{a^n}+\frac{1}{b^n}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ $A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$
here after row reduction
$\begin{bmatrix}1&0&2&1\\0&1&-1&2\\0&0&1&2\\0&0&0&0\end{bmatrix}$
clearly determinant is zero
but how can I find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$... | No need for row reduction.
From $$A^2=\begin{pmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{pmatrix}$$ you can immediately see that $A^2 e_2 = 3e_1$ and $A^2e_4 = 2e_1$.
Therefore $$A^2e_2 = \frac32 A^2 e_4 \implies 0 = A^2\left(e_2 - \frac32 e_4\right) = A^2 \begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix}$$
so $\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many $k$ satisfy the equation $(p \cdot k)^2 \equiv 0 \pmod{p^n}$ where $k < p^n$ and $p$ is prime How many $k$ satisfy the equation $(p \cdot k)^2 \equiv 0 \pmod{p^n}$ where $k < p^n$ and $p$ is prime?
I attach very simple code:
#include <stdio.h>
#include <math.h>
unsigned long long int ipow( unsigned long long ... | I believe your program is wrong, for example for $p=3$, $n=3$ and $k<3^3$
$$3^3 \mid (3 \cdot 3 \cdot 1)^2$$
$$3^3 \mid (3 \cdot 3 \cdot 2)^2$$
$$...$$
$$3^3 \mid (3 \cdot 3 \cdot 8)^2$$
where $k=3\cdot 8 < 3^3$, so there are $8$ cases, not $3$.
Generally (using Euclid lemma)
$$p^n \mid p^2 k^2 \Rightarrow p^{n-2} \... | {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to diophantinte equation $x^4+2y^4=z^2$ It is well known that the equation $x^4+y^4=z^2$ has no non-trivial solutions. The same holds also for the equation $x^4+2y^4=z^4$. What about the equation
$x^4+2y^4=z^2$?
| The classic book Diophantine Analysis by R. D. Carmichael says on page 17 that the equation $x^4+2y^4=z^2$ has no non-trivial solutions. Here is the original text, slightly edited:
*The equation $x^4+2y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which
are different from zero.
Suggestion. This may ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Extrema of a funcion of $x$ depending on $n$. I have this function:
$$f(x)=\frac{x^{3}+3(\sqrt{n}-2)x^{2}-24\sqrt{n}x-2}{n^{2}}$$
I have to find the critical point of this funcion depending on $n\ge1\in \mathbb N$.
So I have computed the first derivative
$$f'(x)=\frac{3x^{2}+6(\sqrt{n}-2)x-24 \sqrt{n}}{n^{2}}$$
So the ... | You have obtained a quadratic equation $ax^2+bx+c=0$ then, as a standard method, we can apply the formula for the roots
$$x_{1,2}=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
to $$x^2+(2\sqrt n-4)x-8\sqrt n=0$$
and obtain
$$x_{1,2}=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{-(2\sqrt n-4)\pm \sqrt{4n+16-16 \sqrt n + 32\sqrt n}}{2... | {
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Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...
I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
| Hint: $$ [1+2\cdot( 1+ 2+\ldots +(n-1))]\cdot n + 2(1+2+\ldots +(n-1)) $$
You can then use Gauss formula for $1+ 2+\ldots +(n-1)$.
As in $n=4$:
$$13 + (13 + 2) + (13 +4) + (13 +6) = 13\cdot 4 + 2\cdot(1+2+3)$$
and
$$ 13 = 1 + 2\cdot( 1+ 2 + 3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Summation of binomial coefficients upon $(r+2)$
$$\sum^{50}_{r=0}(-1)^r \dfrac{\dbinom {50}r}{r+2}= ?$$
Attempt:
$(1-x)^{50}= \sum (-1)^r \dbinom{50}r x^{n-r}$
Integrating both sides and then placing limits 0 to 1:
$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{52-r}$
So the answer should b... |
OP's expression is a special case of a formula for the reciprocal binomial coefficient using the Beta function
\begin{align*}
\color{blue}{\binom{n}{k}^{-1}}&\color{blue}{=(n+1)\int_0^1 z^k(1-z)^{n-k}\,dz}\\
&=(n+1)\int_0^1z^k\sum_{r=0}^{n-k}\binom{n-k}{r}(-z)^r\,dz\\
&=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}(-1)^r\int_0^... | {
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"source": "stackexchange",
"question_score": "5",
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Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$ Question: Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$
My attempt:
If $x>y$ then
$|x-y|=x-y$
Thus now,
$2xy\le x-y\le x^2+y^2$
If $x-y\geq 2xy$ then let $x-y=k$ ($k$ is ... | $$2xy\le |x-y|\le x^2+y^2$$
First, consider the inequality on the right.
Depending on whether $x$ or $y$ is the larger of the two the second inequality resolves into the two inequalities
$$ \left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\ge\left(\frac{\sqrt{2}}{2}\right)^2 $$
or
$$ \left(x+\frac{1}{2}\right)^... | {
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A difficult improper integral
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$
I got $2$ as a result. Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance
| We are dealing with
$$ \int_{1}^{+\infty}\frac{1-\tanh(x/2)}{2x^2}\,dx=\frac{1}{2}-\frac{1}{4}\int_{1/2}^{+\infty}\frac{\tanh(z)}{z^2}\,dz=\frac{1}{2}-\int_{1/2}^{+\infty}\sum_{n\geq 0}\frac{2\,dz}{z(4z^2+(2n+1)^2\pi^2)}$$
or
$$ \frac{1}{2}-\sum_{n\geq 0}\frac{\log(1+(2n+1)^2 \pi^2)}{(2n+1)^2 \pi^2}=\frac{1}{2}+\frac{\... | {
"language": "en",
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Finding the limit: $L = \lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}$ I'm trying to solve the following question. Have tried taking $\log$ on both sides, but didn't get very far. Seems difficult to apply L'Hopital's rule.
$L = \lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}$
... | $$\prod_{r=3}^{n}\frac{r^3-8}{r^3+8}=\prod_{r=3}^{n}\frac{r-2}{r+2}\prod_{r=3}^{n}\frac{(r+1)^2+3}{(r-1)^2+3}=\frac{\prod_{k=1}^{n-2}k}{\prod_{k=5}^{n+2}k}\cdot\frac{\prod_{h=4}^{n+1}(h^2+3)}{\prod_{h=2}^{n-1}(h^2+3)}$$
for any $n\geq 7$ can be written as
$$ \frac{4!}{(n+2)(n+1)n(n-1)}\cdot\frac{((n+1)^2+3)(n^2+3)}{(2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679406",
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$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ without 'Hopital Rule Compute $$\lim_{x\rightarrow 0} \frac{(x+2)\cdot \ln(1+x)-2x}{x^3}$$ without L'Hopital Rule.
I proved before that that $$\lim_{x\rightarrow 0} \frac{ \ln(1+x)-x}{x^2}=-\frac{1}{2}$$
I tried to use it and I have to compute $$\lim_{x\righta... | Use that
$$
2\ln(1+x)=2x-x^2+2\frac{x^3}{3}+o\left(x^3\right)
$$
Hence
$$\frac{2\ln(1+x)+x^2-2x}{x^3}=\frac{2x^3+o\left(x^3\right)}{3x^3}\underset{x \rightarrow 0}{\rightarrow}\frac{2}{3}$$
| {
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prove that if $p$ is a prime number, then $\sqrt{p}$ is an irrational number.
*
*$\sqrt{p}$ is rational.
*$\sqrt{p}=\frac{a}{b}$, where $a,b$ are integers with $\gcd(a,b)=1$.
*$a^2=b^2p$.
Since $p$ divides $a^2$, $p$ divides $a$.
*
*$a=kp$.
*$a^2=k^2p^2=b^2p$
*$p=\frac{b^2}{k^2}\Rightarrow\sqrt{p}=\frac{... | By way of contradiction, assume $\sqrt{p}$ is rational. Then there exist $a, b \in \mathbb{Z}$ with $b\neq 0$ such that $\sqrt{p} = \frac{a}{b}$. Without loss of generality, we may assume $\text{gcd}(a,b) \neq 1$
We can make this assumption, because we still lose no generality.
Now using $\text{gcd} (a,b) = d \neq 1$.... | {
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Solving a cubic modular congruence I need to solve this congruence:
$$x(x^2-1) \equiv 34 \pmod {35}$$
I noticed that the left hand side is divisible by $3$, namely, $x(x-1)(x+1) \equiv0 \pmod 3$. Also, I thought that using the Chinese remainder theorem might work here. However, I can't find a good way to apply this th... | Well $x(x+1)(x-1)$ being the product of three consecutive numbers is pretty cute if we are going to take modulo low terms.
So by CRT we want to solve
$x(x+1)(x-1) \equiv 4\equiv -1 \pmod 5$ and if we have $x\equiv 0,1,4$ we have $x(x+1)(x-1)\equiv 0 \pmod 5$ so we must have $x\equiv 2,3$ so $x(x+1)(x-1) \equiv$ either ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac... | $$S-i=-i\cdot (\sum_{k=1}^{10}({e^{\frac {2\pi i}{11}}})^k+1)=-i\cdot (\frac{1-({e^{\frac{2\pi i}{11}}) ^{11}}}{1-e^{\frac{2\pi i}{11}}})=-i\cdot 0=0$$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the smallest distance between point and ellipsoid
Find the smallest distance between the points on the ellipsoid $x^2+2y^2+z^2=16$ and the point $(0,0,1)$.
Answer:
Let $(x,y,z)$ be the closest point of the ellipsoid. The distance is
$$\sqrt{x^2+y^2+(z-1)^2}$$
Let $f(x,y,z)=x^2+y^2+(z-1)^2$. Construct the Lagrang... | If $\lambda=-\frac12$, $z=2$ and from the original constraint $x^2+2y^2+z^2=16$ (where $x=0$ as shown prior) we get $y=\pm\sqrt 6$. In this case, the distance is $d=\sqrt{0^2+\sqrt6^2+(2-1)^2}=\sqrt7$.
If $y=0$, from the original constraint we have $z=4\lor z=-4$, which give $d=\sqrt{0^2+0^2+(4-1)^2}=3$ and $d=\sqrt{0^... | {
"language": "en",
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Find $y$ for $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$ where $y$ is a positive real number Here is the answer in the textbook with which I disagree / don't understand:
$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$
$$y^{y^{\frac{3}{2}}} = (y^{\frac{3}{2}})^y = y^{\frac{3}{2}y}$$
Comparing indices, we get:
$$y^{\frac{3}{2}} = \frac{3}... | For positive $y$ you have $(\sqrt[y]{y})^y = y$
so $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y\implies y^{\sqrt[y]{y}} = y \implies y^{\sqrt[y]{y}-1} = 1$
so either $y=1$ or $\sqrt[y]{y}-1=0$
in both cases implying $y=1$
Alternatively if the original question is actually $y^{y \sqrt{y}} = (y \sqrt{y})^y$
Then this indeed gi... | {
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Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can... | Starting from
$$
\left\{ \matrix{
{{x^{\,2} } \over 9} + {{y^{\,2} } \over 4} = 1 \hfill \cr
A = 4xy \hfill \cr} \right.
$$
which gives the intercepts between the ellipse and a hyperbola,
you could have simply substitute $y=A/(4x)$, to get
$$
{{x^{\,2} } \over 9} + {{A^{\,2} } \over {64x^{\,2} }} = 1
$$
i.e.
$$
6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdo... | They have to be divisible by $15$, $2$, and $3$, or divisible by $30$, but from your method, the divisors can be divisible by $10$. What I mean mathematically is:
$$30(2^2\cdot3^3\cdot5^2\cdot7\cdot11)\ne 2079000$$
Any even number must be divisible by $2$, and any number divisible by $15$ and $2$ must be divisible by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How I can solve $s(n)=n+s(n-1)$ by iteration method? $$
s(n)=
\begin{cases}
0, \text{if $n=0$}\\
n+s(n-1), \text{if $n>0$}\\
\end{cases}
$$
Using the relation
\begin{align}
s(n) &= n+s(n-1) \\
&= n+n-1+s(n-2) \\
&= n+n-1+n-2+s(n-3) \\
&= \dots \\
&= n+n-1+n-2+\dots+1 \\
&= \dfrac{n^2+n}2 \\
&= \theta(n^2)
\end{align}
I... | I like to turn
recurrences to
telescoping sums
whenever possible.
From
$s(n) = n+s(n-1)
$
I get
$s(n)-s(n-1) = n$.
Summing both sides from $1$ to $m$,
$\sum_{n=1}^m s(n)-s(n-1) = \sum_{n=1}^m n $.
The left side telescopes into
$s(m)-s(0)$
and the right side is just
$\dfrac{m(m+1)}{2}
$.
Note that if
the first value is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Haggling problem Adam is trying to sell Bob a bike for $a$ dollars.
Bob does not agree on the price $b$ ($0 < b < a$).
Adam does not agree on this price but does lower his initial price to $\dfrac{a+b}{2}$ dollars.
Bob responds by offering $\dfrac{b+\frac{a+b}{2}}{2}$ dollars.
They continue haggling this way, each tim... | *
*We first write the series of bids: Start with $a$ and then
$$
\tfrac{b}{1} \to \tfrac{a+b}{2} \to \tfrac{2b+(a+b)}{4}=\tfrac{a+3b}{4} \to \tfrac{2(a+b)+(a+3b)}{8}=\tfrac{3a+5b}{8} \to \tfrac{2(a+3b)+(3a+5b)}{16}=\tfrac{5a+11b}{16}\to\dotsb \tag{*}
$$
*The denominators in (*) are the series $\{ 2^0, 2^1, 2^2, 2^3, ... | {
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Number of arrangements of four blue, three green, and two red balls in which no two blue balls are adjacent Find the number of ways of arranging four blue, three green and two red balls such that no two blue balls are adjacent.
I have done this question and got the answer which is
$$\binom{6}{4} \cdot \frac{5!}{2!\cdo... | There are $4 + 3 + 2 = 9$ positions to fill. We fill four of them with blue balls, three of the remaining five positions with green balls, and the remaining two positions with red balls. Therefore, the number of possible arrangements of the balls is
$$\binom{9}{4}\binom{5}{3}\binom{2}{2} = \frac{9!}{4!5!} \cdot \fra... | {
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Find a basis for $W$ and identify $\dim(W)$. Let $\mathbf{R}^{3\times3}$ be the linear space of all $3 \times 3$ matrices.
Let $W$ be the set of all symmetric $3\times3$ matrices. Then
$W$ is a linear subspace of $\mathbf{R}^{3×3}$. Find a basis for $W$ and identify $\dim(W)$.
Would the $\dim(W)$ be $3$ since it is a ... | The $3\times3$ symmetric matrices are those of the form$$\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}.$$They form a $6$-dimensional space. A basis of this space is$$\left\{\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix},\... | {
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Number of 4x4 matrices such sum of all rows and columns is zero. The number of matrices $A = [a_{ji}], 1 \le i,j \le 4$ such that $a_{ij} = +- 1$ and $\sum_{i = 1}^{4}a_{ij} = \sum_{j=1}^{4}a_{ij} = 0$ is...
I solved it by first selecting two values of “i” which is $\binom{4}{2}$ and two values of “j” similarly. Now if... | You are both over-counting and under-counting!
First, you are over-counting:
Suppose you first pick $i=1,2$ and $j=1,2$. So, $a_{11}$, $a_{12}$, $a_{21}$, and $a_{22}$ are all set to $1$. But that forces $a_{33}$, $a_{34}$, $a_{43}$, and $a_{44}$ to be a $1$ as well.
OK, but now start with $i=3,4$ and $j=3,4$. Then... | {
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Showing a function grows like a certain power of $n$ Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?
|
Letting $n=2m+1, m\in\mathbb{N}$ we obtain
\begin{align*}
\color{blue}{f(2m+1)}&=\frac{1}{2^{2m+1}(2m+1)}\binom{2m+1}{m+1}\\
&=\frac{1}{2^{2m+1}(m+1)}\binom{2m}{m}\tag{1}\\
&\,\,\color{blue}{=\frac{1}{2^{2m+1}}C_m}\tag{2}
\end{align*}
with $C_m$ the $m$-th Catalan-number.
In (1) we use the binomial identity $\bin... | {
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Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose... | For $0< x< 1$ it holds: $1+x^2<1+x+x^2<3$. Therefore
$$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Power Series Representation of a Function, (differentiation) Suppose that $f(x) =\frac{x^2+x}{(1-x)^3}$. What would the power series representation for this function?
I have found the power series representation for $\frac{1}{(1-x)^3}$ which is $\sum_{n=2}^{\infty} \frac{n(n-1)x^{n-2}}{2}$. Would it be allowed to multi... | This is indeed correct and the result is then given by $\sum_{n=0}^{\infty}n^2x^n$, as stated in M. Iwaniuk's comment.
$$\left(x^2+x\right)\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=x^2\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}+x\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=\sum_{n=2}^{\infty}\frac{n(n-1)x^{n}}{2}+\sum_{n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4... | It is a matrix such that $P^4= P^2$ and some additional relations.
We see that
$$ 0 = P^4 - P^2 = P^2(P^2 - I) = P^2(P-I)(P+I)$$
so the minimal polynomial $\mu_P(x)$ divides $x^2(x-1)(x+1)$.
However, the minimal polynomial is not any of $$x, x^2, x-1, x+1, \underbrace{x(x-1)}_{P^2 \ne P}, \underbrace{x(x+1)}_{P^2 \ne ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 0
} |
Computing a series rised from definite integral of a floor function I interested to compute $$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx.$$
So, I did like this:
$$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx=\sum_{n=1}^\infty
\int_{1/(n+1)^2}^{1/n^2} ndx=\\
\sum_{n=1}^{\infty}\frac{2n+1}{n(n+1)^2}=\\
=\sum_{n=1}^\infty \... | $\sum_{n=1}^\infty \frac{1}{n(n+1)^2}=\sum_{n=1}^\infty \frac{n+1-n}{n(n+1)^2}=\sum_{n=1}^\infty \frac{1}{n(n+1)}-\sum_{n=1}^\infty \frac{1}{(n+1)^2}=1-\sum_{n=1}^\infty \frac{1}{(n+1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck on this square root conjugate problem I have to multiply the following:
$((x+h)\sqrt{x+h} - x\sqrt{x}) * ((x+h)\sqrt{x+h} + x\sqrt{x})$
I tried to use $(a - b)(a + b) = a^2 + b^2$ so:
$((x+h)\sqrt{x+h} - x\sqrt{x})((x+h)\sqrt{x+h} + x\sqrt{x}) = (x^2 + h^2)(x + h) - x^2(x)$
and then:
$x^3 + x^2h + h^2x + h^3 - x^... | Notice that $$\begin{align} (x+h)^3 &= x^3 + h^3 + 3xh(x+h) \\ &= x^3 + h^3 + 3x^2h + 3xh^2.\end{align}$$ Therefore, the book's answer is equivalent to $$(x+h)^3 - x^3 = (x+h)^2(x+h) - x^2(x),\tag1$$ but the answer you obtained was $(x^2 + h^2)(x+h) - x^2(x)$, and $x^2 + h^2\neq (x+h)^2$. From this observation, it is c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$5$ points on ellipse I want to take five points and locate them on ellipse so that the distance between the point $1$ and point $2$, the point $2$ and point $3$, the point $3$ and point $4$, the point $4$ and point $5$, the point $5$ and point $1$ were equal to each other. I think it will look like a polygon with equa... | Equation of the ellipse :
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$
with given values $a,b$.
They are many solutions, depending where we put the first point on the ellipse.
If you want only one solution, one of the simplest is obtained in putting the first point A on $(x=a,y=0)$.
See Figure below. Wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Algebraic Proof on equivalence $$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$
Demonstrate the identities above considering a, b & c real numbers and distinct from each other.
I'm stucked on this problem. Take a look on what I did:
$x=(a-b)\therefore -x=(b-a)$; $y=(a-c);-y=(c-a)$;... | Put the LHS over a common denominator, getting
$$\frac{a^3(b-c)}{(a-c)(b-c)(a-b)}-\frac{b^3(a-c)}{(a-c)(b-c)(a-b)}+\frac{c^3(a-b)}{(a-c)(b-c)(a-b)}=$$
$$=\frac{a^3b-a^3c-ab^3+ac^3+b^3c-bc^3}{(a-c)(b-c)(a-b)}=\frac{(a+b+c)(a-c)(b-c)(a-b)}{(a-c)(b-c)(a-b)}=a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
parabola locus problem If $Q_1$ and $Q_2$ be the angle made by tangents to the axis of $y^2=4x$ from point $P$ and if $Q_1+Q_2=45^{\circ}$ then locus of $P$ is
for options see here
| Let $P(X,Y)$ be the pole.
The equation of polar or equivalently the chord:
$$Yy-2(x+X)=0 \tag{1}$$
Substitute $x=\dfrac{y^2}{4}$ into $(1)$,
\begin{align}
y^2-2Yy+2X &= 0 \\
y_1+y_2 &= 2Y \\
y_1 y_2 &= 4X \\
m_1 &= \frac{2}{y_1} \tag{$2yy'=4$} \\
m_2 &= \frac{2}{y_2} \\
\frac{m_1+m_2}{1-m_1 m_2} &= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\gcd(m,15)=\gcd(n,15)=1,$ show that $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. If $\gcd(m,15)=\gcd(n,15)=1$ show $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$.
This is what I have so far but I'm stuck on essentially the last step.
Proof: Assume $\gcd(m,15)=\gcd(n,15)=1$. By the Euler's Phi Function: If $\gcd(a,m)=1$ th... | Since $\gcd(m,5) = \gcd(n,5)=1$ we have by Fermat little theorem:
$$ m^4\equiv 1\equiv n^4 \pmod 5$$ so $5\mid m^4-n^4$.
On the other hand since $\gcd(m,3) = \gcd(n,3)=1$ we have again by Fermat little theorem:
$$ m^2\equiv 1\equiv n^2 \pmod 3$$ so $3\mid m^2-n^2$ and so $3\mid m^4-n^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finite Complex Exponential Geometric Series with Negative Exponents So I want to know if I am doing my manipulations correctly. I have the following expression:
$$ X(\omega) = A\frac{1 - e^{-j \frac{1}{2} \omega N}}{1 - e^{j
\frac{1}{2} \omega}} $$
Using the geometric series formula:
$$ \displaystyle\sum\limits_{n=0}^... | Addendum
Let me try and clear your doubts about summing definition and handling.
There are fundamentally three ways to express a sum.
a) over a set
e.g.
$$
\sum\limits_{k\, \in \,\left\{ {1,2,3} \right\}} k = 1 + 2 + 3 = 2 + 1 + 3 = \ldots
$$
b) under a condition on the index
e.g.
$$
\sum\limits_{1\, \le \,k\; \le \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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.