Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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To find trace and determinant of matrix
Possible Duplicate:
Computing the trace and determinant of $A+B$, given eigenvalues of $A$ and an expression for $B$
Let $A$ be a $4\times 4$ matrix with real entries such that $-1,1,2,-2$ are its eigenvalues. If $B=A^{4}-5A^{2}+5I$, where $I$ denotes the $4\times 4$ identity ... | The characteristic polynomial of $A$ is
$$(t+1)(t-1)(t+2)(t-2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$$
Therefore, by the Cayley-Hamilton Theorem,
$$A^4 - 5A^2 + 4I = 0.$$
In particular, $B= A^4 - 5A^2 + 5I = (A^4-5A^2+4I)+I = I$.
So $B=I$, $A+B=A+I$, and $A-B=A-I$.
The eigenvalues of $A+\mu I$ are of the form $\lambda+\mu... | {
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"url": "https://math.stackexchange.com/questions/148074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int\frac{x^{1/2}}{1+x^2}\,dx.$
Compute $$\int\frac{x^{1/2}}{1+x^2}\,dx.$$
All I can think of is some integration by substitution. But ran into something scary. Anyone have any tricks?
| $$I=\int\frac{x^{\frac{1}{2}}}{1+x^2}$$
$x=t^2$, we get,
$$=\int\frac{2t^2dt}{1+t^4}$$
$$=\int\frac{t^2+1}{1+t^4}dt + \int\frac{t^2-1}{1+t^4}dt$$
upon dividing by $t^2$, we get
$$=\int\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt +\int\frac{1-\frac{1}{t^2}}{(t+\frac{1}{t})^2-2}dt$$
All set now, lets integrate
$$\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate square root of $i \Leftrightarrow z^2=i$ Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:
$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ ... | $i=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$ and we want $z^2=i$, so $r^2(\cos(2\theta)+i\sin(2\theta))=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$. Then use that $r_1(\cos\theta_1+i\sin\theta_1)=r_2(\cos\theta_2+i\sin\theta_2)\implies r_1=r_2\wedge\theta_1=\theta_2+2\pi k,k\in\mathbb{Z}$.
This gives $r^2=1\wedge2... | {
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Prove $\gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $\gcd(a,b) = 1$
Assuming that $\gcd(a,b) = 1$, prove that $\gcd(a+b,a^2+b^2) = 1$ or $2$.
I tried this problem and ended up with
$$d\mid 2a^2,\quad d\mid 2b^2$$
where $d = \gcd(a+b,a^2+b^2)$, but then I am stuck; by these two conclusions how can I conclude $d=1$ or $2$?
And... | If $d$ divides $a+b$ then it divides $a(a+b)=a^2+ab$ so if it also divides $a^2+b^2$ then it divides $a^2+ab-(a^2+b^2)=ab-b^2=b(a-b)$. A similar calculation shows it divides $a(a-b)$. Then from $\gcd(a,b)=1$ we get $d$ divides $a-b$. But then it divides $(a+b)+(a-b)=2a$ and $(a+b)-(a-b)=2b$, so it's 1 or 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
| Hints
$$3x^{3}=24\Leftrightarrow x^{3}-8=0$$
$$\frac{x^{3}-8}{x-2}=x^{2}+2x+4$$
--
Added: The second equation means that
$$x^{3}-8=\left( x-2\right) \left( x^{2}+2x+4\right) .$$
Thus we have
$$x^{3}-8=0\Leftrightarrow \left( x-2\right) \left( x^{2}+2x+4\right) =0,$$
whose solutions are the values of $x$ that make the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$... | A 'backwards' answer:
The general polar equation of a circle of radius $\rho$ centered at $(r_0,\theta_0)$ is
$$r^2-2 r r_0 \cos(\theta-\theta_0) + r_0^2 = \rho^2.$$
When $r_0 = \rho$, this reduces to (ignoring the $r=0$ solution)
$$r = 2(r_0 \cos \theta_0) \cos \theta + 2(r_0 \sin \theta_0) \sin \theta,$$
choosing $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
} |
How does he get a perfect swap numerator and denominator. I'm going through a exercise, in which all the answers are given, but the tutor makes a step and I can't follow at all. A massive jump with no explanation.
Here is the question:
$\lim_{x \to 2} \frac{\frac{1}{2}-\frac{1}{x}}{x-2}$
He then simplifies:
$ \frac{x-2... | Note that
\begin{align*}
\frac{\frac{1}{2}-\frac{1}{x}}{x-2} &= \frac{\frac{x-2}{2x}}{x-2} \\\ &= \frac{2x \cdot \frac{x-2}{2x}}{2x \cdot (x-2)} \qquad \Bigl[\small\text{Multiplying numerator and denominator by} \ 2x \Bigr] \\\ &= \frac{(x-2)}{2x \cdot (x-2)} \\\ &= \frac{1}{2x}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | We can write : $M= \frac 12 (I + J)$ were $I$ the unit matrix and $J$ the matrix who all entries equal $1$.
$J$ is diagonalizable as symetric real matrix with propre value $0$ , $n-1$ times because the rank of $J$ is $1$ and $n$ one time because $JX=nX$ where $X$ is the vector who has all c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)
$$\begi... | Maybe, you could try something with this
\begin{equation}
\sum_{k=-\infty}^{\infty}\frac{1}{|x-kx_0|} = \sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=0}^{\infty}\frac{1}{|x-kx_0|} =
\end{equation}
\begin{equation}
= \frac{1}{|x|}+\sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=1}^{\infty}\frac{1}{|x-kx_0|} = \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Orthogonal Trajectory of $x^2 + 2y^2 = k^2$ $x^2 + 2y^2 = k^2$
I first take the derivative like the instructions say.
$2x + 4y \frac{dy}{dx} = 0$
I am not entirely sure why a dy and dx appears but it does in the instructions so I go with it.
Now I need to solve for $y'$
$ + 4y \frac{dy}{dx} = -2x$
$ \frac{dy}{dx} = \f... | From the line David noted you; You would have $\frac{dy}{2y}=\frac{dx}{x}$. So by integrating from both sides, you get $\frac{1}{2}$Ln$|y|$=Ln$|x|+c$:
$$\int\frac{dy}{2y}=\int\frac{dx}{x}$$ $$\frac{1}{2}\int\frac{dy}{y}=\int\frac{dx}{x}$$ $$\frac{1}{2}Ln|y|=Ln|x|+c$$ wherein $c$ is a constant. You can simplify the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
| The integration is obtained as follows:
$$\int 2\frac{x+2}{5}dx=\frac{2}{5}\int (x+2)d(x+2)=\frac{2}{5}\int udu=\frac{2}{5}\frac{u^2}{2}=\frac 1 5 (x+2)^2$$
Since $\frac 1 5 (x+2)^2$ is a primitive of $2\frac{x+2}{5}$ we can use FTCII, and get
$$\int 2\frac{x+2}{5}dx=\frac{(\color{red}{1}+2)^2}{5}-\frac{(\color{red}{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:
Letting
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}... | This is almost correct, but I believe the original sum needs to range from $-\infty$ to $\infty$ instead of $0$ to $\infty$. The solution that follows considers the sum $\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}$, and throughout I will write $\sum_{n=-\infty}^\infty f(n)$ to mean $\lim_{N\rightarrow \infty}\sum_{n=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 1
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Maximum point of a polar function I have a curve C with polar equation
$$r^2 = a^2\cos{2\theta} $$
And I am looking to find the length $x$ when $r=max$
Judging from the equation:
$$r = \sqrt{a^2\cos{2\theta}} $$
R will be maximum at $\cos{2\theta}=1$
So the maximum value of $r$ is:
$$r = \sqrt{a^2} =a$$
However the d... | Differentiating $r$ wrt to $\theta$ gives $\theta \mapsto |a|\frac{\sin 2 \theta}{\cos 2 \theta}$, which is zero when $\cos 2 \theta = \pm 1$. No disagreement there!
From this compute $x = r \cos \theta$. Since $\theta = 0$ maximizes $r$, the corresponding $x = |a|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/161386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference ... | Representing natural logarithm as an integral and changing the order of integration we obtain:
$$\ldots = \int_0^1 \frac{x^4}{x^2 - 1} \, dx \int_1^x \frac{dt}{t} = \int_0^1 \frac{dt}{t} \int_0^t \frac{x^4}{x^2 - 1} \, dx \\= \int_0^1 \frac{t + \frac{1}{3} t^3 - \tanh^{-1} t}{t}\, dt = \frac{10}{9} - \int_0^1 \frac{\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/161749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot ... | Looks about right. In your expression, multiply top and bottom of the thing inside the log by $\sqrt{x^2+x+1}-(x-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Error Term in Passing from Summation to Integral I encountered the following in a paper and do not understand how the error term is being bounded. In what follows, $n$ and $k$ are large integer constants.
$$ \sum_{i=0}^{n-1} \ln\left(1 - \frac{i}{k}\right) = \int_0^n
\ln\left(1 - \frac{x}{k}\right) dx \pm e(n,k) $$ w... | $$0 \le e(n,k) = \int_0^n \left(\ln \left(1 - \dfrac{\lfloor x \rfloor}{k} \right) - \ln \left(1 - \dfrac{x}{k}\right)\right) dx \le \int_0^n \left(\ln \left(1 - \dfrac{ x-1}{k} \right) - \ln \left(1 - \dfrac{x}{k}\right)\right) dx $$
So the inequality is true if $$\ln \left(1 - \dfrac{ x-1}{k} \right) - \ln \left(1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Want to test the series for convergence/divergence $\sum_n \frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.
My attempt is
$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish ... | Hint: $\sqrt{n+1}-1\sim\sqrt n$ and $(n+2)^3-1\sim n^3$ (when $n\to +\infty$). Since the series have non-negative terms, we just have to deal with the convergence of $\sum\limits_{n\geq 1}\frac{\sqrt n}{n^3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Closed form solutions of $\ddot x(t)-x(t)^n=0$ Given the ODE:
$$\ddot x(t)-x(t)=0$$
the solution is:
$$x(t)=C_1\exp(-t)+C_2\exp(t)$$
If we square the $x(t)$ we have:
$$\ddot x(t)-x(t)^2=0$$
and the solution is given by:
$$x(t)=6\wp(t+C_1;0,C_2)$$
and so for $x(t)^3$ which gives:
$$x(t)=\operatorname{sn}\left(\left(\fra... | $$\begin{align*}
y''-y^n&=0\\
y'y''&=y'y^n\\
\int y'y'' dx &=\int y'y^n dx\\
\frac{(y')^2}{2} &=\frac{y^{n+1}}{n+1} +c_1\\
y' &=\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }\\
\int \frac{dy}{\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }} &=\int dx =x+a\\
\frac{1}{(2c_1)^{1/2}}\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=x+a\\
\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/166981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Squares of the form $x^2+y^2+xy$ How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
| Let $c^2+a^2+ca= (c+na)^2$ where $n$ is an integer $\implies a=\frac{(2n-1)c}{1-n^2}$.
Let $a^2+b^2+ab =(a+mb)^2$ where $m$ is an integer $\implies b=\frac{(2m-1)a}{1-m^2}=\frac{(2n-1)(2m-1)}{(1-n^2)(1-m^2)}c$.
If $c|(1-n^2)(1-m^2)$,
$c=r(1-n^2)(1-m^2)$ (say, where $r$ an integer),
then, $b = r(2n-1)(2m-1)$
and $a = r... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there does not exist any positive integers pair $(m, n)$ satisfying: $n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$ How to prove that, there does not exist any positive integers pair $(m,n)$ satisfying:
$n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$.
| This is an edited version of a partial answer that I posted sometime ago and subsequently deleted (not sure if resurrecting an answer is the correct thing to do after it has been up-voted and then deleted, perhaps someone will advise). If anyone can suggest where any of this can be improved, or point out any mistakes, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove $\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$ and $\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$ Can anyone help me solve the following trig equations.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$$
My work thus far
$$\frac{\frac{1}{\cos{A}}+\frac{1}{\sin{A}}}{\frac{\sin{A... | Try waiting until the last minute before converting to sines and cosines.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}$$
Recall that $\tan A\cot A = 1$ and $\tan A = \frac{\sec A}{\csc A}$
$$=\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}\cdot\frac{\tan A}{\tan A} $$
$$=\frac{\sec{A}\tan A+\csc{A}\tan A}{\tan^2{A} + 1} $$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/171675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible
| let y=$\frac{x^2+x+1}{x^2-x+1}$
=>$x^2(y-1)-x(y+1)+(y-1)=0$
As x is real, the discriminant= $(y+1)^2-4(y-1)^2≥0$
=>$(y-3)(y-\frac{1}{3})≤0$
=>$\frac{1}{3}≤y≤3$
| {
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"url": "https://math.stackexchange.com/questions/174905",
"timestamp": "2023-03-29T00:00:00",
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Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
| $$ax^2+bx+c=0 - \text{divide by $a$ because $a\neq 0$ }$$ we get
$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$ I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$
Proof: ... | This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as
$$
(1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x,
$$
so there is no need of the fifth line. Or even
$$
(1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 0
} |
How to solve $1/2 \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0$? How to solve trigonomtry function involving $\sin x \cos x$ and $\sin 2x$:
$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$
| Hint:
Using the identity $\sin(2x) = 2 \sin x \cos x$ we have
$$ \sin x \cos x + \sin x + 2\cos x + 2 = 0$$
Factor
$$ (1 + \cos x) \sin x + 2(1 + \cos x) = 0 \\
(1 + \cos x)(2 + \sin x) = 0
$$
So either $1 + \cos x = 0$ or $2 + \sin x = 0.$
Solve for $x$ in each case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I want to find out the angle for the expression $a^3 + b^3 = c^3$. like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.
| There is no single angle corresponding to the relationship $a^3+b^3=c^3$.
Suppose that a triangle has sides of lengths $a,b$, and $c$ such that $a^3+b^3=c^3$. We know from the law of cosines that if $\theta$ is the angle opposite the side of length $c$, then $c^2=a^2+b^2-2ab\cos\theta$, so
$$\cos\theta=\frac{a^2+b^2-c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series? The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that ... | If you want to approximate the value of $\arctan(2)$, then you can expand $\arctan(x)$ at a point close to $2$. For example you can have an expansion at the point $1$, but if you approximate at a point that is closer to 2 is better. Here is the Taylor series at the point $x=1$
$$ \arctan(x) = \frac{\pi}{4} +{\frac {1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the... | *
*$n=2k+1$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16$$
Now, below is square of an odd integer hence it can be reperesented as (8a+1).
$$n^2$$
Therefore $$8a(8p+6) + 16\\= 64ap + 48a+16\\=16(4ap + 3a + 1)$$
This proves that when n is an odd integer it will be divisible by 16
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, ... | You could use induction.
Assuming that $$S_{2n}+4S_n=n(2n+1)^2$$ then add terms to both sides so that the left side increments its index:
$$
\begin{align}
&S_{2n}+4S_n+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
S_{2(n+1)}+4S_{n+1}&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=(n+1)(2n+1)^2+4(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Seifert matrices and Arf invariant -- Cinquefoil knot I have computed the following Seifert matrix for the Cinquefoil knot:
$$ S = \begin{pmatrix} 1 & -1 & -1 & -1 \\ 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\ 0 & 1& 1 & 1\end{pmatrix}$$
I also found a Seifert matrix for this knot on the internet but I still don't kn... | What makes you think that $x_i$ form a symplectic basis for $S$? In fact they don't. Also, you don't want a symplectic basis for $S$ but rather for $I = S^T - S$.
Your Seifert matrix must be wrong since if I compute the Arf invariant using the $S$ with a symplectic basis
$$ e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\1 \end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Closed form for a sum of values of a quadratic? Today in class we were analyzing the number of half-spaces created by $n$ number of planes. For two planes there are 4 spaces, 3 there are 8, 4 there are 15, etc. our teacher challenged us to find the formula for $n$ planes. Me and my friend came up with
$$
1+\sum^n_{x=1}... | The reasoning is basically right, with slight glitches in the details. The desired number is
$$2+\sum_{x=1}^{n-1} \left(\frac{x(x+1)}{2}+1\right).\tag{$1$}$$
Or else, if you want to sum from $1$ to $n$, the desired number is
$$1+\sum_{x=1}^{n} \left(\frac{x(x-1)}{2}+1\right).\tag{$2$}$$
Note that $(x+1)^3-x^3=3x^2+3x+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Counting Hexagons in Triangle Grid Recurrence? (This is from a long finished programming competition)
Consider a triangle grid with side N. How many hexagons can fit into it?
This diagram shows N = 4:
I need a recurrence for it:
I tried the following:
$T_1 = 0$
$T_2 = 0$
$T_3 = 1$
$T_4 = 7$
$T_n = ???$
Using inclusio... | Starting from your idea: By inclusion-exclusion, for $n>3$
$$\tag{1}T_n = X_n +3T_{n-1}-3T_{n-2}+T_{n-3},$$
where $X_n$ is the number of "full-size" hexagons, i.e. those that touch all three sides. A full-size hexagon is determined by three positive integers $a,b,c$ (the sizes of the small triangles chopped off) with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Inequality. $a^7b^2+b^7c^2+c^7a^2 \leq 3 $
Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that
$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$
| Using AM-GM inequality, we get
$$
3 = \frac {(a^6 + b^6 + c^6)^2} 3 = \sum_{cyc} a^6 \frac {a^6 + 2b^6} 3 \geq \sum_{cyc} a^6\sqrt[3]{a^6 b^{12}} = a^8b^4 + b^8c^4 + c^8a^4
$$
Now, by means of Cauchy–Schwarz inequality we complete the proof
$$
a^3 \cdot a^4 b^2 + b^3 \cdot b^4 c^2 + c^3 \cdot c^4 a^2 \leq \sqrt{a^6 + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Polynomial factors Why must $x^2 + x + 1$ be a factor of $x^5+x^4+x^3+x^2+x+1$?
I know that when we divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ we get $x^2 + x + 1$, but is there an argument/theorem or anything that could tell that $x^2+x+1$ must divide $x^2 + x + 1$?
| It depends on what you count as telling. It follows from an easy, basic manipulation of familiar identities, but I don’t see any way to recognize this without appealing to something at least somewhat computational. I look at the sum $x^5+x^4+x^3+x^2+x+1$ and immediately think of $x^6-1$, and from there it all falls out... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric identity and roots of a polynomial. Prove that
$$(\operatorname{cosec} A–\sin A) (\sec A–\cos A) = \frac {1}{\tan A + \cot A} $$
Also help me with this question please
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2–2x–15$ then find a quadratic polynomial whose series [roots?] are $2\alpha$ a... | $\csc A -\sin A=\frac{1-\sin^2A}{\sin A}=\frac{\cos^2A}{\sin A}$
$\sec A -\cos A=\frac{1-\cos^2A}{\cos A}=\frac{\sin^2A}{\cos A}$
Multiplying we get, $\sin A \cos A$
Now $\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin^2A+\cos^2A}{\cos A\sin A}=\frac{1}{\cos A\sin A}$
So, $\alpha+\beta=\frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality under condition $a+b+c=0$ I don't know how to prove that the following inequality holds (under condition $a+b+c=0$):
$$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$
| The following proof, as it is, only works for $a,b,c \not\in (-2,0)$.
Let $|a| \geq |b| \geq |c|$
$$
\frac{(2x+1)^2}{2x^2+1} = 1 + \frac{2x^2+4x}{2x^2+1}
$$
Then it follows that
$$
\text{left-hand side}
= 3 + \frac{2a^2+4a}{2a^2+1} + \frac{2b^2+4b}{2b^2+1} + \frac{2c^2+4c}{2c^2+1}
\geq 3 + \frac{2(a^2+b^2+c^2)+4(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/200327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Relations between the roots of a cubic polynomial How do I solve the last two of these problems?
The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\... | For the first, as Avatar described, you use that $\frac{1}{\alpha+1}$ satisfies the polynomial relation $6y^3-7y^2+3y-1=0$ to express its third power in lower powers of itself.
For the second, you rewrite it simply as $\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\alpha+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\beta+1)^3}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/201191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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The sum of the different points on $x$-axis What is the sum of the different points on $x$-axis and that intersect with the two curves :
$$x^2=x+y+4 , y^2=y-15x+36$$
| As the comments have said, we set $y=0$. For the curve $y^2=-15x+36$, we get $x=36/15$. For the curve $x^2=x+y+4$ we get $x^2-x-4=0$.
By the Rational Roots Theorem, the only conceivable rational roots of $x^2-x-4=0$ are integers that divide $4$. None of these is in fact a root, but that is irrelevant, the point is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/201302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$... | You can collapse the left-hand side into a single sine function:
$$\sin(x)+\sqrt3\cos(x) = 2\sin(x+\pi/3)$$
Then, dividing by two, all that remains is to solve the following:
$$\sin(x+\pi/3) = \frac{1}{2}$$
Wikipedia has an article on useful trigonometric identities, including linear combinations of sin and cos.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/201399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$
So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to us... | Here is a rather direct calculation of the limit using squeezing. It needs
*
*$\ln k < \int_k^{k+1}\ln x \; dx < \ln (k+1)$
*$ \int \ln x \; dx = x(\ln x - 1) \color{grey}{+C} $
*$\lim_{n \rightarrow \infty}\frac{\ln (n+1)}{n} = 0$
Set
$$
x_n = \ln \frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^n \ln k - \ln n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/201906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
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Proof that $\sqrt[3]{3}$ is irrational Proof that $\sqrt[3]{3}$ (Just need a check)
Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function i... | No, $3b^3=a^3$ does not imply that $a=3b$. In fact if $a=3b$, then $a^3=(3b)^3=3^3b^3=27b^3$. What $3b^3=a^3$ does imply is that $3\mid a^3$, which then further implies that $3\mid a$. Then you have $a=3m$ for some integer $m$, and $3b^3=a^3=(3m)^3=27m^3$. Divide through by $3$ to get $b^3=9m^3$. Now $3\mid b^3$; what ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/203274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Check my workings: Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition. Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.
Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that
$$0<|x+... | Here's how I would do it:
For any $\epsilon>0$, choose $\delta=\min\left(1,\dfrac{\epsilon}{11}\right)$. Then:
$$\begin{align*}
0<|x+2|<\delta\ \ \implies\ \ |(3x^2+4x-2)-2| &= |3x^2+4x-4| \\
&= |3x-2||x+2| \\
&= 3\left|x-\frac{2}{3}\right||x+2| \\
&= 3\left|x+2-\frac{8}{3}\right||x+2| \\
&\le 3\left(|x+2|+\left|\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| The main question is how to write $16-x^2$ as a square. Note that $\cosh^2 \theta-\sinh^2 \theta = 1$ hence $1-\tanh^2\theta = \text{sech}^2 \theta$ thus $16-16\tanh^2\theta = 16\text{sech}^2 \theta$. This suggests a $x=4\tanh \theta$ substitution. Observe $dx = 4\text{sech}^2(\theta) d\theta$. Putting this all togethe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the derivative of $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ Find the derivative
How do I tackle this? My answer is totally different from the correction model, but I have tried for half an hour to show my answer in lateX but I don't know how to, it's too complicated, so, can someon... | Alternatively, whenever you have a quotient, you can turn it into a product. This doesn't always make things easier, but sometimes it does. You be the judge in this case. For example
$$h(x) = \frac{2\sqrt{x}}{x^2+2} = 2\sqrt{x} (x^2 + 2)^{-1}$$
So, now we can do the product rule to find the derivative
$$\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find all real numbers $t$ such that the quadratic form $f$ is positive definite. Where $$f(x_1,x_2,x_3)=2x_1^2+x_2^2+3x_3^2+2tx_1x_2+2x_1x_3$$.
This is a problem in my Matrix Analysis homework. Below is my effort.
Let $x=(x_1,x_2,x_3)^T$, then we have $$f=x^*Sx$$, in which $$S=\left(\begin{matrix}2&t&1\\t&1&0\\1&0&3\e... | We can solve it writing $x^tSx$ as a sum of squares. We have
\begin{align}
x^tSx&=(x_2+tx_1)^2-t^2x_1^2+2x_1^2+3x_3^2+2x_1x_3\\
&=(x_2+tx_1)^2+3\left(x_3^2+\frac 23x_1x_3\right)+(2-t^2)x_1^2\\
&=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2-3\frac{x_1^2}9+(2-t^2)x_1^2\\
&=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2+\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum of a floor function Given that $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor = Nn-\frac{N(N-1)(2N+5)}{6}$ where $N=\lfloor \sqrt{n} \rfloor$, for which values of N is the sum $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor$ divisible by N?
| The question amounts to asking for what values of $N$ the expression $$Nn-\frac{N(N-1)(2N+5)}{6}$$ is divisible by $N$. Since $Nn$ is certainly divisible by $N$, this is the same as asking when $$\frac{N(N-1)(2N+5)}{6}=N\cdot\frac{(N-1)(2N+5)}6$$ is divisible by $N$. Clearly this is the case precisely when $(N-1)(2N+5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Inverse function of a polynomial What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!
| This is an experimental way of working out the inverse.
We can treat the polynomial like an expansion
\begin{equation}
f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots
\end{equation}
then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation}
f^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 2
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about a series $\sum_{n \geq1}{\frac{2n^2}{3^n}} $ How is this series ?
$$\sum_{n \geq1}{\frac{2n^2}{3^n}} ?$$
How I made:
$$a_{n}=\frac{2n^2}{3^n}$$ and then $$\frac{a_{n+1}}{a_n} \leq 1$$ so the series is convergence .
Is ok ?
thanks :)
| Since $2n^2 < 4\cdot 2^n$, the series is dominated by a geometric series, so it is convergent. Moreover, we have:
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{x^n}{3^n} = \frac{x}{3-x}, $$
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{x}{3-x}\right) = \frac{3x}{(x-3)^2}$$
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have
\begin{align}
L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\
& = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
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primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$
| The number $32$ is small, so it is easy to show by computing that $5$ has order $8$. Note that we only need to compute $5^{2^k}$ modulo $32$. For $\varphi(32)=16$, so the order of $5$ must be a power of $2$.
The following is severe overkill. We show that for any $n\ge 3$, $\mathbb{Z}^\ast_{2^n}$ is generated by $5$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/223925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \fr... | Each unit of $C$ contains $\frac15$ unit of $A$ and $\frac45$ unit of $B$. Each unit of $D$ contains $\frac47$ unit of $A$ and $\frac37$ unit of $B$. Thus, $c$ units of $C$ and $d$ units of $D$ contain $\frac{c}5+\frac{4d}7=\frac{7c+20d}{35}$ units of $A$ and $\frac{4c}5+\frac{3d}7=\frac{28c+15d}{35}$ units of $B$. (Sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\int x^n \sin x \, dx
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{(n-1)}{k}{n-2k-1}
$$
I started with differentiatio... | Well.. I think I have a different way
$$ \int_a^b e^{izt} dz = \frac{e^{izt} }{it} |_{z=a}^{z=b}$$
Differentiate both sides with $t$ n times (*) and apply leibniz product rule(**) and some rearranging::
$$ \int_a^b z^n e^{izt} dz = \frac{1}{i^{n+1}} \sum_{k=0}^n \frac{1}{t^{k+1}} \binom{n}{k} (-1)^k k! (iz)^{n-k} e^{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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The roots of $x^3+4x-1=0$ are $a$, $b$, $c$. Find $(a+1)^{-3}+(b+1)^{-3}+(c+1)^{-3}$ This is a question in A level Further Pure mathematics pastpaper Nov 2010.
The roots of the equation $x^3+4x-1=0$ are $a$, $b$ and $c$.
i) Use the substitution $y=1/(1+x)$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $1/(a+1... | Let the equation
$$6y^3 - 7y^2 + 3y - 1 = 0\tag{*}$$
have roots $\alpha$, $\beta$ and $\gamma$.
Using Vieta's formulas:
$$\sum \alpha = \alpha + \beta + \gamma = - \frac{-7}{6} = \frac{7}{6}\tag{1}$$
$$\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{3}{6} = \frac{1}{2}$$
Using the identity
$\sum \al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq... | We need to prove that:
$$\sum_{cyc}\frac{a}{a+bc}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{a}{a+bc}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a-bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{a(a+b+c)-3bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a+3c)-(c-a)(a+3b)}{a+bc}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a+3c}{a+bc}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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probability problem and There are three caskets of treasure. The first casket contains 3 gold coins, the second casket contains 2 gold coins and 2 bronze coins, and the third casket contains 2 gold coins and 1 silver coin. You choose one casket at random and draw a coin from it. The probability that the coin you drew i... | If we draw from the first, our probability of gold is $\dfrac{3}{3}$. If we draw from the second, our probability of gold is $\dfrac{2}{4}$. And if we draw from the third, our probability of gold is $\dfrac{2}{3}$. So the required probability is
$$\frac{1}{3}\cdot\frac{3}{3}+\frac{1}{3}\cdot \frac{2}{4}+\frac{1}{3}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/235735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the integer part of $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +\cdots + \frac{1}{\sqrt{(2n+1)^2}}$ I tried to solve the following problem.
What is the integer part of
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{(2n+1)^2}}=\sum_{k=0}^{2(n^2+n)} \frac 1{\sqrt{2k+1}} ?$$
I tried using some ... | A more general sum can be bounded as follows:
$$
\int_0^{n+1} \frac{1}{\sqrt{2 x + 1}} dx \leq \sum_{k=0}^n \frac{1}{\sqrt{2 k + 1}} \leq 1 + \int_0^n \frac{1}{\sqrt{2 x + 1}}dx
$$
or, computing the integrals
$$
\sqrt{2 n + 3} -1 \leq \sum_{k=0}^n\frac{1}{\sqrt{2 k + 1}} \leq \sqrt{2 n + 1}.
$$
The second inequality i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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solving the equation $x^4-5x^3+11x^2-13x+6=0$ by given condition 1.(a) solve the equation $x^4-5x^3+11x^2-13x+6=0$ , given that two of its roots $p$ & $q$ are connected by the relation $3p+2q=7$
(b) solve the equation $x^4-5x^3+11x^2-13x+6=0$ which has two roots whose difference is $1$
did I need to solve these problem... | $$x^4-5x^3+11x^2-13x+6=0$$
$$x^4-x^3-4x^3+4x^2+7x^2-7x-6x+6=0$$
$$(x-1)x^3-(x-1)4x^2+(x-1)7x-(x-1)6=0$$
$$(x-1)(x^3-4x^2+7x-6)=0$$
$$(x-1)(x^3-2x^2-2x^2+4x+3x-6)=0$$
$$(x-1)\left( (x-2)x^2-(x-2)2x+(x-2)3 \right)=0$$
$$(x-1)(x-2)(x^2+2x+3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/241118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Inverse Laplace Transform Assistance How do I compute the following transform?
$$\frac {s-1}{2s^2+s+6}$$
I've gotten this far:
$$\frac {1}{2}\cdot \frac {s-1}{\left(s+\frac{1}{4}\right)^2 + \frac{47}{16}}$$
| $$ F(s)=\frac {s-1}{2s^2+s+6}=\frac {s-1}{2(s^2+\frac{s}{2} +3)} $$
$$ ax^2+bx=a\left[ (x+\frac{b}{2a})^2-(\frac{b}{2a})^2 \right] $$
So:
$$ 2(s^2+\frac{s}{2} +3)=2 \left[ (s+\frac{1}{4})^2-(\frac{1}{4})^2 +3 \right]=2 \left[ (s+\frac{1}{4})^2+(\frac{47}{15}) \right] $$
therefor:
$$ F(s)=(\frac{1}{2}) \frac{s-1 {\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the sum? problem.. I need to evaluate the sum: can someone help?
The series is as follows:
$$
\frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x}
$$
| We know, $$\sum_{0\le s\le n-1}y^s=\frac{y^n-1}{y-1}$$
Integrating either sides wrt $y$, we get $$\sum_{1\le r\le n}\frac{y^r}r=\int \left(\frac{y^n-1}{y-1}\right) dy$$
If $\sum_{1\le r\le n}\frac{y^r}r=S(y,n),$
$\frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x}=-S(-\frac14, x+1)$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why are these two expressions different in this induction problem? Prove with $n \ge 1$:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$
First, I prove it for $n=1$:
$$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \f... | There is a flaw in the logic in this posting:
begin quote:
First, I prove it for $n=1$:
$$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$
Which is true.
end of quote
This is not valid reasoning. You're saying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral by partial fractions
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)
factor out the 5, and use partial fraction
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$.
$A\left... | Probably they missed include a constant in book's answer.
If we include a constant $k$, the book's answer will change to:
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k$$
But with some algebra we get
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k=$$ $$=5(\frac{(x-5)}{x-5}\ln|x-5|-\frac{x}{x-5})+k= 5(\ln|x-5|-\frac{x}{x-5}+1)-5+k=$$ $$=5(\ln|x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/247178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Converting $x = \sin \frac{t}{2}, y = \cos \frac{t}{2}$ to Cartesian form How can we transform these parametric equations to Cartesian form?
$$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$
| $$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$
$$x^2+y^2=(\sin \frac{t}{2})^2+(\cos \frac{t}{2})^2=1$$ so
$$x^2+y^2=1$$ is equation of some circle
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/248920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Computation of a certain integral I would like to compute the following integral. This is for a complex analysis course but I managed to around some other integrals using real analysis methodologies. Hopefully one might be able to do for this one too.
$$\int_{0}^{2\pi} \frac{1}{a-\cos(x)}dx, \text{ with } a > 1.$$
Any ... | Let $\tan(x/2) = t$. We then get that $$\sec^2(x/2) dx = 2dt \implies dx = \dfrac{2dt}{1+t^2}$$
Also, $\cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$.
Hence,
\begin{align}
\dfrac{dx}{a- \cos(x)} & = \dfrac{2dt}{1+t^2} \dfrac1{a - \dfrac{1-t^2}{1+t^2}}\\
& = \dfrac{2dt}{a(1+t^2) - (1-t^2)}\\
& = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I wa... | We have$:$ $$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{
2} \right) -abc \left( a+b \right) \left( b+c \right) \left( c+a
\right) $$
$$=ac \left( {b}^{2}+{c}^{2} \right) \left( a-b \right) ^{2}+ab \left( {a}^{2}+{c}^
{2} \right) \left( b-c \right) ^{2} +bc \left( {
a}^{2}+{b}^{2} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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Does multiplying polynomials ever decrease the number of terms? Let $p$ and $q$ be polynomials (maybe in several variables, over a field), and suppose they have $m$ and $n$ non-zero terms respectively. We can assume $m\leq n$. Can it ever happen that the product $p\cdot q$ has fewer than $m$ non-zero terms?
I ask this ... | Here's an elementary example. Start with the well-known identity $x^n - 1 = (x-1) (x^{n-1} + x^{n-2} + \ldots + x + 1)$. If $n$ is odd, we can factor $x^n+1$ in a similar way by flipping the signs: $x^n + 1 = (x+1) (x^{n-1} - x^{n-2} + \ldots - x + 1)$. Now mix and match the two:
$$\begin{align*}
x^{2n} - 1 &= (x^n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 3,
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Subsitution in Integrals Okay, so I'm working on definite integrals and I calculated the indefinte integral of $\frac{x}{\sqrt{x+1}}$ to be $\frac{2}{3}u^{3/2}-2\sqrt{u}$ where $u=x+1$. The definite integral is on the interval $[0,5]$ so I used my $u$-subsitution equation on the interval $[1,6]$. I keep getting $7.9981... | Same answer, but with the useful trick of "adding $0$":
\begin{align}
\int \frac{x}{\sqrt{x+1}}dx = \int \frac{x+1 -1}{\sqrt{x+1}}&= \int \left(\frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}} \right)dx
\\
&= \int\left( (x+1)^{\frac{1}{2}} - (x+1)^{- \frac{1}{2}}\right)dx
\\
&= \frac{2}{3} (x+1)^{\frac{3}{2}} - 2 (x+1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Number of ordered sets of integers How many ordered sets of integers $(x,y,z)$ satisfying $$x,y,z \in [-10,10]$$ are solutions to the following system of equations:
$$x^2y^2+y^2z^2=5xyz$$
$$y^2z^2+z^2x^2=17xyz$$
$$z^2x^2+x^2y^2=20xyz$$
By... | Let $(x,y,z)$ be a solution. Then the three numbers
$$
u_1=x^2y^2+y^2z^2-5xyz, \ u_2=y^2z^2+z^2x^2-17xyz, \ u_3=z^2x^2+x^2y^2=20xyz
$$
are all zero.
Now
$$
\begin{array}{lcl}
(1) \ -u_1+u_2+u_3 &=& 2x^2z^2-32xyz=32xz\bigg( \frac{xz}{16}-y\bigg) \\
(2) \ u_1-u_2+u_3 &=& 2x^2y^2-8xyz=8xy\bigg( \frac{xy}{4}-z\bigg) \\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| Suppose that the positive integer $d$ divides both $a+b$ and $a^2-ab+b^2$. You showed that $d$ divides $3ab$.
Note that $d$ is relatively prime to $ab$. Otherwise, there is a prime $p$ that divides both $d$ and $ab$. Since $p$ divides $ab$, it follows that $p$ divides one of $a$ or $b$, say $a$. But since $p$ divides $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
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How to compute $7^{7^{7^{100}}} \bmod 100$? How to compute $7^{7^{7^{100}}} \bmod 100$? Is $$7^{7^{7^{100}}} \equiv7^{7^{\left(7^{100} \bmod 100\right)}} \bmod 100?$$
Thank you very much.
| $7^4\equiv1\pmod{100}$
Let $$y=7^{7^{7^{100}}}$$
So, $7y=7^{(7^{7^{100}}+1)}\equiv1\pmod{100}$ as $(7^{7^{100}}+1)$ is divisible by $(7+1)=8$
Let us use the property of the successive convergents of a continued fraction.
Now, $\frac{100}7=14+\frac27=14+\frac1{\frac72}=14+\frac1{3+\frac12}$
So, the previous convergent ... | {
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"url": "https://math.stackexchange.com/questions/259870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
| It’s easier to verify that it multiplies back together correctly:
$$x\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=x^n+\color{red}{x^{n-1}y+\ldots+x^2y^{n-2}+xy^{n-1}}\;,\tag{1}$$
and
$$y\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=\color{red}{x^{n-1}y+x^{n-2}y^2+\ldots+xy^{n-1}}+y^n\;.\tag{2}$$
The tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
Find the rotation axis and angle of a matrix $$A=\frac{1}{9}
\begin{pmatrix}
-7 & 4 & 4\\
4 & -1 & 8\\
4 & 8 & -1
\end{pmatrix}$$
How do I prove that A is a rotation ? How do I find the rotation axis and the rotation angle ?
| You have $A^T A = I$. Hence $A$ is a rotation. Since $\det A = 1$, it is proper.
By inspection, $A \begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\2\end{bmatrix}$, which gives the axis of rotation.
Inspection also shows that $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 4 \\... | {
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"url": "https://math.stackexchange.com/questions/261617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 3
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$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$ $$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$$
Is that correct?
I would've done
$$
2^{10 - x} \cdot 2^{10 - x}\;\; = \;\;
(2)^{10 - x + 10 - x} \; = \; (2)^{2 \cdot (10 - x)} \;=\; 4^{10 - x}\tag{1}
$$
Is that allowed?
If so, can I say that
$$
\frac{4^x}{2^y} = 2^{x - y} \tag{2}
$... | $$\text{Yes:}\;\;\;\large 2^{10 - x} \cdot 2^{10 - x} =\; 4^{10-x}\tag{1.a}$$
$$\underbrace{\large 2^{10 - x} \cdot 2^{10 - x}
= (2)^{10 - x + 10 - x} = (2)^{2 \cdot (10 - x)} = 4^{10 - x}}\tag{1.b}
$$
$$\text{Yes, that is that is allowed.}$$
$$\underbrace{
\large \frac{4^x}{2^y} = 2^{x - y}}\;\;\; ?\tag{2 ?}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Convergence of parameterized series $\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$ $$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$$
I guess that more useful form is:
$$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} ... | HINT: Rewrite the series as
$$ \begin{eqnarray}
\mathcal{S} &=& \sum_{n=2}^\infty \left( \sqrt{n+1}-\sqrt{n} \right)^p \cdot \ln \left(\frac{n-1}{n+1} \right) \\ &=& \sum_{n=2}^\infty \left( \frac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}} \right)^p \cdot \ln \left(1-\frac{2}{n+1} \right)\\ &=& \sum_{n=2}^\infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/262791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
| Let $\sin(y) = t$. Recall that $\cos^2(y) = 1-t^2$. Then $\cos(y) dy = dt$. Hence,
$$I = \dfrac14 \int_0^1 \dfrac{\ln(\sin(y)) \ln(\cos^2(y)) \cos(y) dy}{\sin(y) \cos^2(y)}$$
$$4I = \int_0^1 \dfrac{\ln(t) \ln(1-t^2) dt}{t (1-t^2)}$$
$$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\dfrac{\ln(t)}{t(1-t^2)} \sum_{k=1}^{\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
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What is the value of $N$? $N$ is the smallest positive integer such that for any integer $n > N$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $N$?
Note: $N$ is a single digit number.
|
$\text{We have}\quad f(n)=n^3-7n^2+11n-5\tag{1}.$
$\textrm{We want to solve for}\;\; n\;\textrm{ when}\;\;f(n) = n^3 - 7n^2 + 11n - 5 = 0\tag{2}$
Note that replacing $n$ with $1$ in equation $(2)$, and then summing, gives us:
$$1(1^3) - 7(1^2 + 11(1) - 5 = 1 - 7 + 11 - 5 = 0,$$ and so $n = 1$ is a solution to $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We the... | Regarding the interpretation of the matrices:
Let
$$X =
\left(\begin{array}{c}
1 \\
x \\
\vdots \\
x^n
\end{array}\right)
\qquad \textrm{and}\qquad
A =
\left(\begin{array}{c}
a_0 \\
a_1 \\
\vdots \\
a_n
\end{array}\right)$$
so
$$X^T A = \sum_{k=0}^n a_k x^k.$$
(I've written $X^T A$ instead of the preferred... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 3
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How to recognize, where function has no derivative? For example this function: $f(x) = (x + 1)|x + 3| + 2$ it has no derivative in $x = -3$, but how can I discover it?
| Recall that the $\vert y \vert$ is not differentiable at $y=0$. Hence, in your case, $f(x) = (x+1) \vert x + 3\vert + 2$ i.e.
$$f(x) = \begin{cases} (x+1)(x+3) + 2 & x \geq -3\\ -(x+1)(x+3) + 2 & x \leq -3 \end{cases}$$
Hence, the only point where derivative need not exist is at $x+3 = 0$ i.e. $x=-3$, since away from $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Smallest possible value on Fibonacci Function Suppose $f$ is a polynomial with integer coefficients, such that for all non-negative integers $n$ the $n$-th Fibonacci number $u_n$ divides $f(u_{n+1})$. Find the smallest possible positive value of $f(4)$.
| This is a really nice problem! The answer is $255$. First note that (by taking integer linear combinations of polynomials with the given property) the problem is equivalent to "find the GCD of all possible nonzero values of $f(4)$". Now observe that $g(x)=x^4-1$ is one such polynomial (left as an exercise! see hint ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/266337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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for a $3 \times 3$ matrix A ,value of $ A^{50} $ is I f
$$A= \begin{pmatrix}1& 0 & 0 \\
1 & 0 & 1\\
0 & 1 & 0 \end{pmatrix}$$
then $ A^{50} $ is
*
*$$ \begin{pmatrix}1& 0 & 0 \\
50 & 1 & 0\\
50 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
48 & 1 & 0\\
48 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0... | Simplify your problem as follows. The Jordan Normal form of your matrix $A$ is
$$J = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1& 1 \\ 0& 0& 1 \end{pmatrix},$$
where the matrix $P$ is such that $P^{-1}AP = J$ is given by
$$P = \begin{pmatrix} 0 & 0& 2 \\ -1 & 1 & 1 \\ 1 & 1 &0 \end{pmatrix}.$$
Then $A^{50} = PJ^{50}P^{-1}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/267492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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correct combination of differentiation rules I am trying to calculate the derivative of a rather complex function for my homework. I think I have found the solution, it just seems too bulky for my taste. See the bottom for specific questions I have regarding my solution.
$ f(x)=\frac{\overbrace{\sin x}^\text{u(x)}\cdot... | Your solution is pretty close, but remember that you're not multiplying $\sin x$ by $(e^x + x^3)$ in your numerator, as you do when you try to multiply $u'(x)$ by $v'(x).$ Instead, try differentiating your numerator in the following form: $e^x\sin x + x^3.$ I believe the rest will follow from the quotient rule.
$n(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$(\tan^2(18^\circ))(\tan^2(54^\circ))$ is a rational number Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks!
| $$\tan18^\circ\tan54^\circ=\frac{2\sin18^\circ\sin54^\circ}{2\cos18^\circ\cos54^\circ}$$
$$=\frac{\cos36^\circ-\cos72^\circ}{\cos36^\circ+\cos72^\circ}$$ (applying $\cos(A\pm B)$ formulae)
$$=\frac{\frac{\sqrt5+1}4-\frac{\sqrt5+1}4}{\frac{\sqrt5+1}4+\frac{\sqrt5+1}4}$$ as $$\cos 72^\circ=2\cos^236^\circ-1=\frac{\sqrt5-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Proving $f(x+y)=f(x)f(y)$ for exponential function $f$ I'm sorry if the title was a bit misleading but I don't know how else to phrase it.
I'm having trouble understanding some lecture notes and I couldn't find extra information by Googling.
Anyways, the exponential function is represented as
$$\lim_{n}\left(1+\frac{x}... | We have
$$\begin{align*}
&\frac{xy}{n^2}\left[\sum\limits_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\right]\\
&\qquad=\frac{xy}n\cdot\frac1n\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\;.
\end{align*}$$
Your first ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} =
-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
| \begin{align*}
\sum_{n=2}^\infty \frac{2}{3^n( n^3 - n) } &= \sum_{n=2}^\infty \frac{1}{3^n}\left( \frac{-2}{n} + \frac{1}{n+1} + \frac{1}{n-1} \right) \\
&= \sum_{n=1}^\infty \frac{1}{3^{n+1}n} - \sum_{n=1}^\infty \frac{2}{3^{n+1} (n+1)}+\sum_{n=1}^\infty \frac{1}{(n+2)3^{n+1}} \\
&= \frac{1}{3}\sum_{n=1}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/276614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Cardano's Formulas help I am working on solving this cubic: $x^3 +x^2 - 2 = 0$ using Cardano's explicit formulas:
$$
A = \sqrt[3]{{-27\over 2}q + {3 \over 2} \sqrt{-3D}} \qquad B = \sqrt[3]{{-27\over 2}q - {3 \over 2} \sqrt{-3D}}
$$
where for a negative discriminant, I will get a real root $\alpha = {A+ B \over 3}$ f... | It should be $y=x+\frac{1}{3}$. Then after we have found a solution $y_0$ of the cubic in $y$, the corresponding $x_0$ is $y_0-\frac{1}{3}$. That's where the $-1$ comes from.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the following identity I am having some trouble proving following identity without use of induction, with which it is trivial.
$$\sum_{n=1}^{m}\frac{1}{n(n+1)(n+2)}=\frac{1}{4}-\frac{1}{2(m+1)(m+2)}$$
I did expand the expression:
$$\sum_{n=1}^{m}\left( \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \right)$$
I have ... | Hint: by partial fraction
$$\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}\Longrightarrow$$
$$1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$
Choosing smartly values for $\,n\,$ above , we get
$$1=2A\Longrightarrow A=\frac{1}{2}\;\;,\;\;1=-B\Longrightarrow B=-1\;\;,\;\;1=2C\Longrightarrow C=\frac{1}{2}$$
so
$$\frac{1}{n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that: Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
*
*$|a_{n}|+|a_{n-1} ... | Part (a)
By replacing $f(x)$ with $\pm f(\pm x)$ if necessary, we may assume that $a_n, a_{n-1}$ are both non-negative. Let $T_n(x)$ be the Chebyshev polynomial of the first kind, of degree $n$, i.e. it satisfies
$$T_n(\cos \theta) = \cos (n\theta)$$
Notice that $T_n(x)$ is a polynomial of degree $n$ with leading coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
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Probability of number of events where $p_{n+1} = \frac{1}{5}p_n$ For $ n \ge 0, p_{n+1} = \frac{1}{5}p_n $. Where $p_n$ is the probability of an envent occuring n times in a period.
What is the probability that more than one event occurs?
| $1= \sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0 $
$(\frac{1}{5})(1) = (\frac{1}{5})\sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0 $
$1 - (\frac{1}{5})(1) = p_0 (\frac{1}{5})^{0}$
$ \frac{4}{5} = p_0$
using the formula in the question,
$p_1 = \frac{1}{5}(\frac{4}{5})$
$p_1 = \frac{4}{25} $
$P(n\ge 2) = 1 - P(n=0) - P... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$
I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{... | The $r$th term is $\frac{r^2-1}{r^2}=\frac{(r-1)}r\frac{(r+1)}r$
So, the product of $n$ terms is $$\frac{3.1}{2^2}\frac{4.2}{3^2}\frac{5.3}{4^2}\cdots \frac{(n-1)(n-3)}{(n-2)^2}\frac{(n-2)n}{(n-1)^2}\frac{(n-1)(n+1)}{n^2}$$
$$=\frac12\frac32\frac23\frac43\cdots\frac{n-2}{n-1}\frac n{n-1}\frac{n-1}n\frac{n+1}n=\frac12\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/286798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Equations over permutations Let $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} \in S_4$ and $\theta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{pmatrix} \in S_4$.
Solve the following equations $(x \in S_4)$:
a) $x \sigma = \sigma x$;
b) $x^2 = \sigma$;
c) $x^2 = \theta$.
I'm writing here... | c) Let $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ a & b & c & d \end{pmatrix} \in S_4$.
Case 1. If $a=1$, then $x(1)=1$, which implies that $x(x(1))=x(1)=1$. On the other hand, $\theta(1)=2$.
We obtained that $x^2(1) \ne \theta(1)$. Therefore, in this case, the equation $x^2=\theta$ has no solutions.
Case 2. If $a=2$, then $... | {
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"url": "https://math.stackexchange.com/questions/287339",
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"source": "stackexchange",
"question_score": "1",
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How to show $AB^{-1}A=A$
Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$
where $a \neq b$ and $a + (n - 1)b = 0$.
Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n ... | Note that
$$A = (a-b)I + b e e^T$$
where $I$ is the identity matrix, $e = \begin{bmatrix} 1&1&1& \cdots &1 \end{bmatrix}^T$. We have
$$B = A+ \dfrac{ee^T}n = (a-b)I + \left(b + \dfrac1n\right)ee^T$$
Now make use of Sherman-Morrison Woodbury formula to compute $AB^{-1}A$ and conclude what you want.
I have added the deta... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show $\sum_{i=1}^N {\frac{i}{p^i}}\le \frac{p}{(p-1)^2}$ Let $p$ be a prime number. How can I show that, for any positive integer $N$,
$$\sum_{i=1}^N {\frac{i}{p^i}}\le \frac{p}{(p-1)^2}?$$
I can see that
$$\sum_{i=1}^N {\frac{1}{p^i}}\lt \sum_{i=1}^\infty {\frac{1}{p^i}} = \frac{1}{p-1}$$
by the infinite sum of a geom... | Note that if $|x| <1$ $$\begin{align}\sum_{i=0}^{\infty}i\cdot x^i &=x+2x^2+3x^3+4x^4+\cdots \\ &=(x+x^2+x^3+x^4+\cdots)+(x^2+x^3+x^4+\cdots)+\cdots \\ &=\frac{x}{1-x}+\frac{x^2}{1-x}+\cdots \\&=\frac{1}{1-x}(x+x^2+x^3+\cdots)\\&=\frac{1}{1-x}\frac{x}{1-x}\\&=\frac{x^2}{1-x}\end{align}$$
Here $x=\frac{1}{p-1}<1$ So $\s... | {
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"source": "stackexchange",
"question_score": "3",
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Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which
$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$
Is your answer the only solution? If so, show why.
I was surprised that a teacher would assign this kind of problem to a 5th gr... | I am posting another solution since one of my students found this method.
(He is 16 years old but as you see he is gifted)
Start with $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$. No write $\frac{1}{6}=\frac{3}{18}=\frac{1}{18}+\frac{2}{18}=\frac{1}{9}+\frac{1}{18}$ and you get $\frac{1}{2}+\frac{1}{3}+\frac{1}{9}+\frac{1}{18... | {
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"source": "stackexchange",
"question_score": "373",
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Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\fr... | We need to prove that
$$\sum_{cyc}\frac{x^3}{3-x^2}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{x^3}{3-x^2}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x^2+3x+3)}{3-x^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x^2+3x+3)}{3-x^2}-2(x^2-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(2x^2+6x+3)}{3-x^2}\geq0.$$
Do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/290844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$ How can i calculate the Given limit
$\displaystyle \lim_{x\rightarrow 0}\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\dots\sin (nx)}{x^{n+2}}\;\;,$ where $n\in\mathbb{N}$
| We know for small $y,$ $$\sin y=y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots=y\left(1-\frac{y^2}{3!}+\frac{y^4}{5!}-\cdots\right)$$
So, $$\prod_{1\le r\le n}\sin rx=\prod_{1\le r\le n}rx\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right)=n!x^n\prod_{1\le r\le n}\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Integrate: $\int x(\arctan x)^{2}dx$ I'm not sure how to start I think we have to use integration by parts
$$\int x(\arctan x)^{2}dx$$
| Using integration by parts, we get
$$\begin{align} \int dx \: x (\arctan{x} )^2 &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \frac{x^2}{1+x^2} \arctan{x} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \arctan{x} + \int dx \: \frac{\arctan{x}}{1+x^2} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - x \arctan{x} + \int d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How can I evaluate $\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$ without invoking l'Hôpital's rule? In the math clinic I work at, somebody in a Calculus 1 class asked for help with this limit problem. They have not covered basic differentiation techniques yet, let alone l'Hôpital's rule.
$$\lim_{x\to1}\frac{\sqrt{5-x... | $$\begin{align}
& \hphantom{=} \lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1} \\
& = \lim_{x\to 1} \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}\frac{\sqrt{2-x}+1}{\sqrt{2-x}+1} \\
& = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x} \\
& = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x}\frac{\sqrt{5-x}+2}{\sqrt{5-x}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Missing and parasite roots in the trigonometric equation. I have this equation:
$$\boxed{\cos(2x) - \cos(8x) + \cos(6x) = 1}$$
RIGHT
And its right solution from the textbook is:
$$
\begin{align}
\cos(2x)+\cos(6x)&=1+\cos(8x)\\\\
2\cos(4x)*\cos(2x)&=2\cos^2(4x)\\\\
\cos(4x)*(\cos(2x)-\cos(4x))&=0\\\\
\cos(4x)*2\sin(3x)... | If $\cos 2x=1\implies 2x=2n_1\pi,x=n_1\pi$
If $\cos2x=-\frac12=\cos \frac{2\pi}3\implies 2x=2n_2\pi\pm \frac{2\pi}3= \frac{2\pi}3(3n_2\pm1)\implies x=\frac\pi3(3n\pm1)$
If we compare these $x=n_1\pi,\frac\pi3(3n_2\pm1)$ with $x=k_1\pi,k_2\frac\pi3,$
the apparent mismatch is when $3\mid k_2$
But, in that case the resu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/292559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $P[X^2> Y^2]$
Let $X$ and $Y$ have the joint probability density function
$$f(x,y)= \begin{cases}cxye^{-(x^2+2y^2)}, & x>0,y>0 \\ 0, &\text{otherwise} \end{cases}$$
Evaluate $c$ and $P[X^2> Y^2]$.
Trial: We know that
\begin{align}
c\int_{0}^{\infty}\int_{0}^{\infty}xye^{-(x^2+2y^2)}\,dx\,dy&=1\\ \imp... | The transformation you want is $x=r \cos{\theta}$, $y= \frac{r}{\sqrt{2}} \sin{\theta}$, and $dx \, dy = \frac{r}{\sqrt{2}} dr \, d \theta$. As far as the integration limits go, $x^2 > y^2$ implies $0 < \tan{\theta} < \sqrt{2}$, while $r \in [0,\infty)$.
The integral then becomes
$$\frac{8}{2} \underbrace{\int_0^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/295209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Closed formula for linear binomial identity I have the following identity:
\begin{equation}
m^4 = Z{m\choose 4}+Y{m\choose 3}+X{m\choose 2}+W{m\choose 1}
\end{equation}
I solved for the values and learned of the interpretation of W, X, Y, and Z in my last post:
Combinatorial reasoning for linear binomial identity
Now, ... | If $r>1$:
$\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)=\sum_{k=1}^{n}\left(\begin{array}{c}
k-1\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k-1\\
r-1
\end{array}\right)=\\
=\sum_{k=0}^{n-1}\left(\begin{array}{c}
k\\
r
\end{array}\right)+\sum_{k=0}^{n-1}\left(\begin{array}{c}
k\\
r-1
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/299758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defi... | Just a comment on such similar cases. From this short answer,
$$x+2^n=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\tag{1}$$
Let $N=\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$. Then
$$\begin{aligned}
N &< \sqrt{1+\sqrt{2+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}} \\
N &<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/300299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"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.