Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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solving Differential Equation $y''+x^2 y'+(2x+1)y=0$ I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
| Plugging in $y = \sum_{n=0}^\infty a_nx^n$ gives
$$ \sum_{n=2}^\infty n(n-1)a_nx^{n-2} + \sum_{n=0}^\infty (n+2)a_nx^{n+1} + \sum_{n=0}^\infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$ I'm looking for a simple proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2}-\frac{1}{2!}\frac{x^4}{n^... | It is $e^t \geq t + 1$ for all $t \in \mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} \geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n \in \mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 \geq 1 - x^2/n^2 \Leftrightarrow x^2/n^2 + 1/n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?
My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\im... | Finishing what you did
$$\sin(4x)>0\iff$$
$$2k\pi <4x<(2k+1)\pi \iff$$
$$\frac{k\pi}{2}<x<\frac{k\pi}{2}+\frac{\pi}{4}$$
$k=0$ gives $0<x<\frac{\pi}{4}$
and
$k=1$ gives $\frac{\pi}{2}<x<\frac{3\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not p... | You may set $x = -\frac{1}{t}$ and consider the limit for $\stackrel{t\rightarrow 0+}{\longrightarrow}$:
$$\begin{eqnarray*} \sqrt{x^2+5x+3}+x
& \stackrel{x = -\frac{1}{t}}{=} & \frac{\sqrt{1-5t +3t^2} - 1}{t} \\
& \stackrel{t\rightarrow 0+}{\longrightarrow} & f'(0) = -\frac{5}{2}\mbox{ for } f(t) = \sqrt{1-5t +3t^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implie... | $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt x) = 1$
implies
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = 1$
or
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = -1$
in either case
$\sqrt{x-4}, \sqrt x$ are both integers. In fact, they are both even, or they are both odd. But that isn't entirely relevant.
There is only one pair of inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Positive divisors of n = $2^{14} \cdot 3^9 \cdot 5^8 \cdot 7^{10} \cdot 11^3 \cdot 13^5 \cdot 37^{10}$ How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2
The answer is (1)(1)(2)(2)(1)(1)(3) = 12
I don't understand though because I would have don... | Since we are required to be a multiple of $2^{10}\cdot3^9\cdot5^2\cdot7^5\cdot11^2\cdot 13^2\cdot37^2$ and also a perfect cube, we know that whatever our divisor is, it must be divisible by $2^{12}\cdot3^9\cdot5^3\cdot7^6\cdot11^3\cdot 13^3\cdot37^3$. Now when you run your counting argument, the permissible ranges of e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving a difference equation for coin toss sequence probabilities I want to solve the following difference equation:
$$b_n-b_{n-1} = \frac{1}{8}(1-b_{n-3})$$
I tried to solve it similar to the solution of Fibonacci sequence here, but when I try to assume the solution will be of the form $l^n$, I end up with the follow... | I have another approach to this I just worked out. Here, we try to use express the recurrence as a system of linear equations. Then, we leverage the eigen values of the matrix associated with the matrix of this linear system.
So far, we have only one equation. This is:
$$b_n -b_{n-1} + \frac{1}{8}b_{n-3} = \frac{1}{8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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The series $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$ Consider the expression $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$
Denote the numerator and the denominator of the $j^\text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1\qqua... | $D_j=2D_{j-1}+1\\\ \ \ \ =2(2D_{j-2}+1)+1\\\ \ \ \ =4D_{j-2}+1+2\\\ \ \ \ =4(2D_{j-3}+1)+1+2\\\ \ \ \ \ \ \ \ \vdots\\\ \ \ \ =2^kD_{j-k}+2^k-1\\\ \ \ \ =2^{j-1}D_1+2^{j-1}-1\\\ \ \ \ =3\cdot2^{j-1}-1$
$s_n=N_n/D_n=\displaystyle\frac n{3\cdot2^{n-1}-1}, n\ge1$
$s_n<\displaystyle\frac n{3\cdot2^{n-1}-2^{n-1}}=\frac n{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The li... | $(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n}) \ge \sqrt{1}\cdot\sqrt{2}\cdots \sqrt{n}= \sqrt{n!}$, hence
$0 \le \frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} \le \frac{1}{\sqrt{n}}.$
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find $\lim_{n\to\infty} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$
Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2
\tex... | $$\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})=2\cos(\frac{x}{2})2\sin(\frac{x}{4})\cos(\frac{x}{4})$$
By repeated application of the half angle formula we find
$$\sin(x)=\lim_{n\to\infty}2^n\sin(\frac{x}{2^n})\prod_{k=1}^{n}\cos(\frac{x}{2^{k}})$$
By expanding $\sin(\frac{x}{2^n})$ into its taylor series we can easily ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Find all matrices which satisfy $M^2-3M+3I = 0$ I am trying to find all matrices which solve the matrix equation
$$M^2 -3M +3I=0$$
Since this doesn't factor I tried expanding this in terms of the coordinates of the matrix. It also occurs to me to put it into "vertex" form:
$$M^2 - 3M + \frac{9}{4}I+\frac{3}{4}I=0$$
$... | $m^2 - 3m + 3 = 0\\
\lambda = \frac {3}{2} \pm i\frac {\sqrt {3}}{2}$
You could say that it is all matrices with eigenvalues equal to $\frac {3}{2} + i\frac {\sqrt {3}}{2},\frac {3}{2} - i\frac {\sqrt {3}}{2}$
If we restrict our universe to real $2\times 2$ matrices.
Then it would be all matrices with characteristic eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + ... | We don’t need l’Hopital, indeed by standard limits
$$\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\frac{x \cos\left(\frac{1}{x}\right)}{\frac{\sin(x)}x}\to \frac01=0$$
indeed by squeeze theorem
$$\left|x \cos\left(\frac{1}{x}\right)\right|\le |x|\to0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1... | At line $3$ you should have
$$\frac{x^2+1-(x-1)^2}{\sqrt{x^2+1}+x-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How many sequences can be made with 5 digits so that the difference between any two consecutive digits is $1$? Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten-digit sequences can be written so that the difference between any two consecutive digits is $1$?
I was wondering if my solution is right.
Let $a(n)$ ... | Your approach looks good to me but you have to check the final result somehow. You can do that with 15 lines of code:
#include <iostream>
using namespace std;
int count(int startDigit, int length) {
if(startDigit < 0 || startDigit > 4) {
return 0;
}
if(length == 1)
return 1;
return coun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Distance from meeting the bisectors to the side of the quadrilateral In a quadrilateral $ABCD,\;AC$ and $BD$ are bisectors of $\angle BAD$ and $\angle ADC$. If $AC$ intersects $BD$ at $P,\;AB=6,\;CD=3$ and $\angle APD= 135º$, calculate the distance from $P$ to $AD$.
I even designed it but I did not find the solution;
|
Let $BE = b, EA = a, BA=c$ and the distance from $P$ to the three edges of $\triangle ABE$ be $r$. Then
$$r = \frac{b+a-c}{2}\tag{1}.$$
Since $BC$ is the angle bisector of $\angle EBA$,
$$\frac{a-6}{b} = \frac{6}c,\text{ or } (a-6)c = 6b\tag{2}.$$
Similarly
$$\frac{b-3}{b} = \frac3c,\text{ or }(b-3)c = 3a\tag{3}.$$
Ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve the system of equations in the set of real numbers. Solve the system of equations in the set of real numbers:
$$\begin{cases}
\frac1x + \frac1{y+z} = \frac13 \\
\frac1y + \frac1{x+z} = \frac15 \\
\frac1z + \frac1{x+y} = \frac17
\end{cases}$$
I got:
$$\begin{cases}
3(x+y+z)=x(y+z) \\
5(x+y+z)=y(x+z) \\
7(x+y+z)=... | Let $x+y+z=m$
Adding all the equations, we get,
$xy+yz+zx=\frac{3m+5m+7m}{2}=\frac{15m}{2}$
Subtracting each equation one by one from this, we get,
$xy=\frac{m}{2}$
$yz=\frac{9m}{2}$
$zx=\frac{5m}{2}$
Dividing by $ xyz$, we get, $$\frac{1}{x}:\frac{1}{y}:\frac{1}{z}=9:5:1$$
$$\Longrightarrow x:y:z=\frac{1}{9}:\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix}
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(... | Can you try using the exponent formula for matrices? It works the same way: $$e^{A} =1+A + \frac{1}{2!} A^2+\dots \in \text{GL}(2, \mathbb{R})$$
The exponent of any matrix is an invertible $3\times 3$ matrix.
if we need a tangent function defined on matrices all of $\tan, \sin, \cos$ can be done with exponent function:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to show that $\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$ Given:$$\sum_{n=1}^{\infty}\frac{\phi^{2n}}{n^2{2n \choose n}}=\frac{9}{50}\pi^2$$
Where $\phi=\frac{\sqrt{5}+1}{2}$
How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how t... | Consider the well-known series expansion of the squared $\arcsin$ function. Namely
$$2\arcsin^2\left(\frac x2\right)=\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2\binom{2n}n}$$
As you can see the RHS equals your given sum for the case that $x=\phi$. Plugging this in yields to
$$\begin{align}
2\arcsin^2\left(\frac \phi2\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluate:$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$ If$$S_{n}=\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{n-3}-\binom{n-3}{n-6}+.......$$
Does $S_{n}$ have a closed form.
My Attempt
$$S_{n}=\binom{n}{0}-\binom{n-1}{n-2}+\binom{n-2}{2}-\binom{n-3}{3}+.......$$
$$S_{n}=\binom{n}{n}-\binom{n... | If $S_{n} = \sum_{j\geq 0} (-1)^{j}\binom{n-j}{j}$, then
\begin{align}
S_{n+1} &= \sum_{j\geq 0} (-1)^{j} \binom{n+1-j}{j} \\
&= \sum_{j\geq 0} (-1)^{j} \left[\binom{n-j}{j} + \binom{n-j}{j-1}\right] \\
&= \sum_{j\geq 0} (-1)^{j} \binom{n-j}{j} - \sum_{j\geq 1} (-1)^{j-1}\binom{n-1-(j-1)}{j-1} \\
&= S_{n} - S_{n-1}
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 = \frac{1}{4}$
From here I want to c... | $$x-x^2 = \frac{1}{4} \implies x^2-x+\frac{1}{4} = 0$$
Here, if you notice, you can see we have two perfect squares: $x^2$ and $\frac{1}{4}$. Also, twice the product of their square roots, which is $2\cdot x\cdot \frac{1}{2}$, gives the coefficient of the middle term. You probably know
$$a^2\pm 2ab+b^2 = (a\pm b)^2$$
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 0
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Using the binomial expansion to show $\sum_{k=0}^n {n \choose k} \frac{1}{2^k} = (3/2)^n$ I have to show $$\sum_{k=0}^n {n \choose k} \frac{1}{2^k} = (3/2)^n$$ using the binomial theorem. I haven't had any practice on these types of questions so i'm unsure as to how to proceed. Would anybody be able to what the method... | There is a theorem (Binomial theorem, demostrable by induction) that shows you:
\begin{equation}
(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}
\end{equation}
Then if you use the $\displaystyle a=\frac{1}{2}$ and $b=1$, you obtain:
$$\sum_{k=0}^n \binom{n}{k} \frac{1}{2^k} = (3/2)^n$$
Proof:
The demostration of the B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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System of equations has no solution
If the system of linear equations,
\begin{cases}
x &+ ay &+ z &= 3\\
x &+ 2y &+ 2z &= 6\\
x &+ 5y &+ 3z &= b\\
\end{cases}
has no solution, then:
*
*$a=-1,b=9$
*$a=-1,b \ne 9$
*$a\ne-1,b = 9$
*$a=1,b \ne 9$
I really fail to understand why the answer cannot be... | I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
$$
\begin{pmatrix}
1 & a & 1 & 3\\
1 & 2 & 2 & 6 \\
1 & 5 & 3 & b
\end{pmatrix}
\to
\begin{pmatrix}
1 & a & 1 & 3\\
0 & 2-a & 1 & 3 \\
0 & 5-a & 2 & b-3
\end{pmatrix}
\to
\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
My Attempt
$$
b=\frac{\sqrt{3}-1}{2}a\quad\& \quad \frac{A-B}{2}=B\quad\&\quad\frac{A+B}{2}=2B\\\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2}}{\tan\frac{A+B}{2}}\implies \frac{3-\sqrt{3}}{\sqrt{3}+1}=\f... | In the standard notation by law of sines we obtain:
$$\frac{\sin\beta}{\sin3\beta}=\frac{\sqrt3-1}{2}$$ or
$$\frac{1}{3-4\sin^2\beta}=\frac{\sqrt3-1}{2}$$ or
$$3-4\sin^2\beta=\sqrt3+1$$ or
$$8\sin^2\beta=(\sqrt3-1)^2$$ or
$$\sin\beta=\frac{\sqrt3-1}{2\sqrt2},$$ which gives $$\beta=15^{\circ}.$$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3062125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\le... | $\newcommand{\Li}{\operatorname{Li}}$Starting with a generating function for the Harmonic Numbers
$$-\frac{\log(1-x)}{1-x}~=~\sum_{n=1}^\infty H_nx^n\tag1$$
we divide both sides by $x$ and integrate afterwards to get
$$\small\begin{align*}
\int-\frac{\log(1-x)}{x(1-x)}\mathrm dx&=\int\sum_{n=1}^\infty H_nx^{n-1}\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
solving $\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}$ To investigate the convergence of a series I have to solve the folliwing limit:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
\end{equation}
It should be $\fra... | $$\lim_{x\to\infty}\frac{(x^2-1)\sqrt{x+2}-x^2\sqrt{x+1}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac{x^2\big[\sqrt{x+2}-\sqrt{x+1}\big]}{x\sqrt{x+1}}-\frac{\sqrt{x+2}}{x\sqrt{x+1}}$$
The latter goes to $0$ since $$\lim_{x\to\infty}\frac{\sqrt{x+2}}{x\sqrt{x+1}}=\lim_{x\to\infty}\frac1x\cdot\sqrt{\frac{x+2}{x+1}}=\lim_{x\to\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to m... | \begin{align}
\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} &= \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{1}{a^7n^{2} mc^2}\\
& = \frac{\sqrt{mn^3}}{a^2} \cdot \frac{\sqrt{c^{3}}}{a^7n^{2}mc^2} \\
& = \frac{\sqrt{m}n\sqrt{n}c\sqrt{c}}{a^9 n^{2}m c^2} \\
& = \frac{\sqrt{nmc}}{a^9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Associated elements in $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ I got four elements and want to check whether these elements are associated or not. For
$\alpha = \frac{1+\sqrt{-3}}{2}$
my elements are:
$a_1 = 2 - \alpha = \frac{3 - \sqrt{-3}}{2}$
$a_2 = 1 - 2\alpha = -\sqrt{-3}$
$ a_3 = 3 + 2\alpha = 4 + \sqrt{-3}$
$ a_4 = ... | I would just compute the norms first. We have
$$N(a_1)=\frac{3^2+3}{4}=3,$$
$$N(a_2)=3,$$
$$N(a_3)=4^2+3=19,$$
and
$$N(a_4)=2^2+3=7.$$
So the only candidates to be conjugates are $a_1$ and $a_2$, which you have verified to be the case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to compute the following floor sum? I want to compute
$$S = \sum_{k=0}^{m} \left\lfloor \frac{n-5k}{2}\right\rfloor.$$
This sum was motivated by the need to compute the number of solutions to
$$x_1+2x_2 +5x_3 = n (*)$$
In particular we observe that the number of solutions to $x_1+2x_2=n$ is $\left\lfloor\frac{n}{2... | First, suppose $n$ and $m$ are even. Then the sum of the even terms of $S$ is
$$S_2 = \sum\limits_{k=0}^{m/2}\frac{n-10k}{2} = \frac{nm}{4} - \frac{5m(m+2)}{8}.$$
and similarly, if $n$ is even and $m$ is odd, the sum of the odd terms of $S$ is
$$S_1 = \sum\limits_{k=0}^{(m-1)/2}\frac{n - 6 - 10k}{2} = \frac{(n-6)(m-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I calculate $\int\frac{x-2}{-x^2+2x-5}dx$? I'm completely stuck on solving this indefinite integral:
$$\int\frac{x-2}{-x^2+2x-5}dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-\int\frac{x}{x^2-2x+5}dx -\int\frac{2}{(x-1)^2 + 4}dx$$
The second one is tri... | The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
\begin{align}\int\frac{x-2}{-x^2+2x-5}dx &= - \int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Envelope Question: Five letters addressed to individuals 1-5 are randomly placed in five addressed envelopes, one letter in each envelope. I'm trying to find the probability of:
*
*Exactly three letters go in the correct envelopes.
*Exactly two letters go in the correct envelopes
*No letters go in the correct enve... | Note the use of the word exactly. You did not take that into account.
Probability that exactly three letters are placed in the correct envelopes
There are indeed $\binom{5}{3}$ ways to select which three letters are placed in the correct envelopes. That leaves two envelopes in which to place the remaining two letters... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}0$. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} < K$ for all $x>0$.
My try:
U... | Using $$ \sqrt{a}-\sqrt{b} = {a-b\over \sqrt{a}+\sqrt{b}}$$
so $${(x^2+ax)-(x^2+bx)\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$
thus $${10x\over \sqrt{(x^2 + ax)} + \sqrt {(x^2 +bx)}} < K$$
taking $x\to \infty$ we get $$\lim _{x\to \infty}{10\over \sqrt{1 + {a\over x}} + \sqrt {1 +{b\over x}}} = 5 \leq K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Primitive of a function with $\sin \frac{1}{x}$ I have the next integral:
$$\int\biggl({\frac{\sin \frac{1}{x}}{x^2\sqrt[]{(4+3 \sin\frac{2}{x})}}}\biggr)\,dx ,\;x\in \Bigl(0,\infty\Bigr)$$
I used the substitution $u=\frac{1}{x}$ and I got $$-\int\biggl({\frac{\sin u}{\sqrt[]{(4+3 \sin2u)}}}\biggr)\,du$$
Can somebody ... | The substitution $u=1/x$ yields $dx=-\frac{1}{u^2}\,du$, so the integral becomes
$$
\int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du=
\int\frac{-\sin u}{\sqrt{4+3\sin2u}}\,du
$$
This can be improved by setting $u=\pi/4-v$, so we get
$$
\frac{1}{\sqrt{2}}\int\frac{\cos v-\sin v}{\sqrt{4+3\cos2v}}\,dv=
\frac{1}{\sqrt{2}}\biggl(
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$? This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to che... | As a check:
Partial answer, completing the square.
$x,y\ge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4\ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=\sqrt{1/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Can we have two different polynomials of the same degree $d$ here in the factorisation of $x^{p^n} -x$? In the proposition "The polynomial $x^{p^n} -x$ is precisely the product of all the distinct irreducible polynomials in $\Bbb F_p[x]$ of degree $d$ where $d$ runs through all divisors of $n$."
Can we have two differe... | Well, another example is
$$x^{16}-x = x(x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1),$$
where the polynomials $x^4+x+1$ and $x^4+x^3+1$ are primitive and conjugate and the polynomial $x^4+x^3+x^2+x+1$ is not primitive (the roots are 5th roots of unity).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$
Find any local max or min of
\begin{align}
f(x,y,z)=x^2+y^2+z^2 && (1)
\end{align}
such that
\begin{align}
x+y+z=1 && (2)\\
3x+y+z=5 && (3)
\end{align}
My attempt. Let
$L(x,y,z,\lambda_1, \lambda_2)= f(x,y,z)+\lambda_2 (x+y+z-1) + \lambda_... | Your result is correct. This is an alternative approach.
Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $t\in \mathbb{R}$.
Hence we reduce the problem to a $1$-variable case,
$$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$
The above quadratic polynomial has just one local/global m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Not possible to find non-zero terms of series expansion? I've been asked to compute the first 3 nonzero terms of a power series expansion about x=0 for two linearly independent solutions to the ODE:
$$(1+x^3)y''- 6xy =0 $$
I have tried to solve this many different ways and continue to get the same solution, which does ... | To verify the correctness of your solution I made the following MATHEMATICA script with the results shown below.
Operator = (1 + x^3) D[#, {x, 2}] - 6 x # &;
n = 20;
Sumb = Sum[Subscript[alpha, k] x^k, {k, 0, n}];
res = Operator[Sumb] /. {Subscript[alpha, 0] -> Subscript[a, 0],
Subscript[alpha, 1] -> Subscript[a, 1]};... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Definite integration when denominator consists of $x\sin x +1$ $$\int_0^\pi\frac{x^2\cos^2x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$$
The answer is $0$. I tried and made $(x\sin x +1)^2 $ in numerator and proceed,
but not able to do any further.
| First, let us try and simplify the integrand a bit. As
\begin{align}
\frac{x^2 \cos^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2} &= \frac{x^2 - x^2 \sin^2 x - x \sin x - \cos x - 1}{(1 + x \sin x)^2}\\
&= \frac{x^2 + x \sin x - \cos x - (1 + x \sin x)^2}{(1 + x \sin x)^2}\\
&= -1 + \frac{x^2 + x \sin x - \cos x}{(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove combinatorial $\sum_{k=0}^{n/2} {n\choose2k} = \sum_{k=0}^{n/2 - 1} {n\choose2k+1} = 2^{n-1}$ I have problems solving the following formula for even positive integers $n$:
$$\sum_{k=0}^{n/2} {n\choose 2k} = \sum_{k=0}^{n/2 - 1} {n\choose 2k+1} = 2^{n-1}$$
I tried to prove it by induction but it didn't work... | For brevity we prove $$\sum_{k=0}^{n} \binom{2n}{2k} = \sum_{k=0}^{n-1} \binom{2n}{2k+1} = 2^{2n-1},\hspace{2cm}(*)$$ for $n\in\mathbb{N}$. It is sufficient to show that $\sum_{k=0}^{n} \binom{2n}{2k}=2^{2n-1}$ since we have
$$\sum_{k=0}^{n} \binom{2n}{2k} +\sum_{k=0}^{n-1} \binom{2n}{2k+1}=\sum_{k=0}^{2n} \binom{2n}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that this matrix is unitary I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=\begin{pmatrix} -1+\... | $ U^{*} = U = \begin{pmatrix} \frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \\ \vdots & \vdots & \ddots & \vdots \\
\frac{2}{N} & \frac{2}{N} & ... & \frac{2}{N} \end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$
Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$ where
$$C:x^2+y^2+z^2=a^2, x+ y=0, a \gt 0$$
At first I thought to parametrize this as: $x=a \cos t , y=a \sin t, z =0$, but then the integral will result in $0$ and this might not be true.
| Ok, so lets start with $x + y = 0$ since it has been given to you. This implies that $x^2 = y^2$. From this result we can simplify the following:
\begin{align*}
C: 2x^2 + z^2 = a^2
\end{align*}
As you can see this implies more of an elliptical curve. Your initial parametrization assumed a circular curve. Also, it is ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integral $\int\frac{1}{1+x^3}dx$
Calculate$$\int\frac{1}{1+x^3}dx$$
After calculating the partial fractions I got:
$$\frac{1}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx=\frac{1}{3}\ln(x+1)+\frac{1}{3}\int\frac{2-x}{x^2-x+1}dx$$
I have no idea on how to proceed. Am I missing a substitution or something... | Hint: You can break up your second calculated indefinite integral into two components as so:
$$\frac{2-x}{x^2-x+1} = -\frac{1}{2}(\frac{2x -1}{x^2-x+1}) + \frac{3/2}{(x-\frac{1}{2})^2 + 3/4}$$
Note that $x^2-x+1 = (x - \frac{1}{2})^2 + \frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\begin{bmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{bmatrix} = \begin{bmatrix}1&1\\1&0\end{bmatrix}^n$ for all $n ∈ N$. The Fibonacci numbers $F_n$ are recursively defined by
$F_0 = 0, F_1 = 1$
$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$
i) Show that $\begin{bmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{bmatrix} = \begin{bmatrix}... | i) Diagonalize the RHS matrix:
$$\begin{pmatrix}1&1\\ 1&0\end{pmatrix}^n=
\begin{pmatrix}\psi&\phi\\ 1&1\end{pmatrix}
\begin{pmatrix}\psi^n &0\\ 0&\phi^n\end{pmatrix}
\begin{pmatrix}-\frac{1}{\sqrt{5}}&\frac{\phi}{\sqrt{5}}\\ \frac1{\sqrt{5}}&-\frac{\psi}{\sqrt{5}}\end{pmatrix}=\\
\begin{pmatrix}\psi^{n+1}&\phi^{n+1}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.
| $$6^5-1=5(6^4+6^3+6^2+6+1)$$ and since
$$6^4+6^3+6^2+6+1\equiv(1+1+1+1+1)(\mod5)=5,$$ we obtain $6^6-1$ is divisible by $25$.
Also, $6^n$ is divisible by $4$ for all $n\geq2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\se... | We substitute $$t=\frac{\tan(u)}{2}$$ then$$dt=\frac{\sec^2(u)}{2}du$$ and
$$\sqrt{4t^2+1}=\sqrt{\tan^2(u)+1}=\sec(u)$$ and our integral will be
$$\frac{1}{2}\int\frac{1}{4}\tan^2(u)\sec^3(u)du$$ and this is
$$\frac{1}{8}\int\sec^3(u)(\sec^2(u)-1)du$$ and then we need the formula
$$\int\sec^m(u)du=\frac{\sin(u)\sec^{m-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$
Prove that if $0 < a < b$ then
$$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides ... | You just need to note that every step you did is reversible. For the conclusion, from $a\le\sqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=\sqrt{a}$ and $y=\sqrt{b}$. You need to show that
$$
y-x<\sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.
Using double-angle formula, we get
$$\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)}=\dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}.$$
Dividin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the three cube roots of $z = -2+2i$ $n$th roots of a complex number:
For a positive integer $n$, the complex number $z=r(\cos(\theta) + i\sin(\theta))$ has exactly $n$ distinct $n$th roots given by
$$\sqrt[n]r\left(\cos\left(\frac{\theta + 2\pi k}n\right) + i\sin\left(\frac{\theta + 2\pi k}n\right)\right)$$
where ... | You have $r=\sqrt{8}$ and then you have to compute $\sqrt[3]{r}=\sqrt[3]{\sqrt{8}}=\sqrt[6]{8}$. This is the basic property that you have to use: if $a\geq 0$ and $n,m\in \mathbb{N}$, with $n,m\geq 1$, then $\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find rational numbers $\alpha $ and $\beta$ in $\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$ How should we find two rational numbers $\alpha$ and $\beta$ such that
$$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$$
The answer I got $\alpha = 1 $ and $\beta = 1$. If I'm wrong, please correct me. Thank you
| Apply the detesting formula
$$\sqrt[3]{a+b \sqrt R}=\frac12\sqrt[3]{3bs-a}\left(1+\frac1s \sqrt R\right)\tag1$$
where $s$ is the solution of
$$s^3-\frac{3a}b s^2+3R s-\frac{a}b R =0$$
So, for $\sqrt[3]{7+5\sqrt{2}}$, solve for $s$ in
$$s^3-\frac{21}5s^2+6s-\frac{14}5=\frac15(s-1)(5s^2-16s+14)=0$$
which yields $s=1$. Pl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3099920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Determine fair price of a digital option
A digital option pays one dollar at time $t = T$ if the asset price is
above a fixed level (strike) $K$ and is worthless otherwise.
Consider the following
model, with $r = 0$:
\begin{array}{|c|c|c|}
\hline
\omega& S(0) & S(1) & S(2) \\ \hline
\omega_1&6 & 10&12 \\ \hline
... | The elementary events $\{\omega_j\}$ correspond to paths on a binomial lattice with transition probabilities $p_1, \, p_2,$ and $p_3$ as shown:
$$\omega_1: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{p_2} S_{uu}=12\\\omega_2: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{1-p_2} S_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How does $d_1$ equal $\frac{1}{2}\sqrt{\left(x_2-x_1\right)^2 + \left(y_2-y_1\right)^2}$? I'm going over the proof of the midpoint formula and the solution in my textbook solves its first distance as follows
$$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}$$
$$=\frac{1}{2}\sqrt... | $$d_1 = \sqrt{\left(\frac{x_1+x_2}{2}-x_1\right)^2 + \left(\frac{y_1+y_2}{2}-y_1\right)^2}
=\sqrt{\frac1{\color{red}4}\left(x_1-x_2\right)^2+ \frac1{\color{red}4}\left(y_1-y_2\right)^2}\\
=\frac12\sqrt{\left(x_1-x_2\right)^2+ \left(y_1-y_2\right)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$ $a_{0},a_{1},...,a_{k}$ are real numbers and $a_{0}+a_{1}+...+a_{k}=0$
$$L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$$
$L=?$
My attempt:
$a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $L=\lim_{n\ri... | We may expand each term in the sum using the binomial theorem (or, equivalently, as a Taylor series):
$$\sqrt[3]{n+j}=\sqrt[3]{n}\,\left(1+\frac{j}{n}\right)^{1/3}\!\!=\sqrt[3]{n}\left(1+\frac{j}{3n}-\frac{j^2}{9n^2}+\cdots\right).$$
Therefore
$$S_n \doteq a_0 \sqrt[3]{n} + a_1 \sqrt[3]{n+1} + \cdots + a_k \sqrt[3]{n+k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ wi... | You can't plug it in since the function $f(x) = {x^2+x\over x^2-x-2}$ is not continuous at $x=-1$.
You have to cancel it first by $x+1$, then you can plug it in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Integral $\int_0^1 \frac{\arctan x}{x^2-x-1}dx$ After seeing this integral I've decided to give a try to calculate:
$$I=\int_0^1 \frac{\arctan x}{x^2-x-1}dx$$
That is because it's common for many integrals to have a combination of a polynomial in the denominator and a logarithm or an inverse trig function in the numera... | Here is my attempt. However, it does not result in a typical "nice, closed form."
Let $I$ be the original integral. Let $x^2-x-1 = 0$. Solving for $x$ using the quadratic formula gives us $x \in \left\{\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right\}$, which means we can rewrite $x^2-x-1$ as $\left(x-\frac{1+\sqrt{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proving convergence of $ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $ $$ \sum_{n=1}^{\infty} \left({n \over n+1}\right)^{n^2} $$
We have :
$$ \left({n \over n+1}\right)^n = \left({1 \over 1+{1\over n}}\right)^n =
{1 \over \left( 1+ {1 \over n}\right)^n}$$
So,
$$ \lim_{n \rightarrow \infty} \left({n \over n... | By comparing this series with respect to this geometric series we found that, your series converge to the a number less than $1$:
$$\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}=\frac{1}{2}+(\frac{2}{3})^4+(\frac{3}{4})^9+(\frac{4}{5})^{16}...< \frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+(\frac{1}{2})^4+...=1$$
If we could ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Constrained (and non) extrema of $f(x,y)=2x^2-2xy^3+3y^2$ I need to find the critical points on the boundary,inside $D$,outside $D$, and find the image of this function (constrained on $D$).
$f(x,y)=2x^2-2xy^3+3y^2$
$D=\{2x^2+3y^2\le 9\}$
Critical points non-constrained:
$f_x=4x-2y^3=0$
$f_y=-6xy^2+6y=0$ --> $6y(-xy+1... | By AM-GM
$$2x^2+3y^2-2xy^3\leq9+|2xy^3|=9+\frac{1}{\sqrt2}\sqrt{2x^2(y^2)^3}\leq$$
$$\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{2x^2+3y^2}{4}\right)^4}\leq9+\frac{1}{\sqrt2}\sqrt{\left(\frac{9}{4}\right)^4}=9+\frac{81}{16\sqrt2}.$$
The equality occurs for $2x^2=y^2$, $-xy^3=|xy^3|$ and $2x^2+3y^2=9,$ which gives
$$(x,y)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$ efficiently with combinatorics? To find the coefficient of $x^6$ in $(1+x+\frac{x^2}{2})^{10}$,
I used factorization on $(1+x+\frac{x^2}{2})$ to obtain $\frac{((x+(1+i))(x+(1-i)))}{2}$, then simplified the question to finding the coefficient of $x^6... | The number of ways to partition 6 with just 2, 1, and 0 are
\begin{align*}
6 &= (3, 0, 7)\cdot(2, 1, 0)\\
&= (2, 2, 6)\cdot(2, 1, 0)\\
&= (1, 4, 5)\cdot(2, 1, 0)\\
&= (0, 6, 4)\cdot(2, 1, 0)
\end{align*}
where $\cdot$ indicates the dot product. These partitions represent the choices of $x^2, x^1, x^0$ in the expansion ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the area of the shaded region of two circle with the radius of $r_1$ and $r_2$ In the given figure , $O$ is the center of the circle and $r_1 =7cm$,$r_2=14cm,$ $\angle AOC =40^{\circ}$. Find the area of the shaded region
My attempt: Area of shaded region $=\pi r^2_2 - \pi r^2_1= \pi( 196-49)= 147\pi$
Is it t... | The area of a sector of a circle with angle $\theta$ is $$ \frac{\theta}{360}\pi r^2$$
For your smaller circle, the shaded area is \begin{align}\frac{360-\theta}{360}\pi r_1^2&= \frac{360-40}{360}\pi 7^2\\
&= \frac{320}{360}\times 49\pi\\
&= \frac 89 \times 49\pi\end{align}
For the larger circle we want to calculate t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Determinant of a particular matrix. What is the best way to find determinant of the following matrix?
$$A=\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right)$$
I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please ... | \begin{align}
&|A|\\
&=\det\left(\begin{matrix}
1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2
\end{matrix}\right) \\
&=\begin{vmatrix}
1&ax&a^2\\1&ay&a^2\\ 1&az&a^2
\end{vmatrix}+\begin{vmatrix}
1&ax&x^2\\1&ay&y^2\\ 1&az&z^2
\end{vmatrix} \tag{multilinearity on 3rd column} \\
&=0+a\begin{vmatrix}
1&x&x^2\\1&y&y^2\\ 1&z&z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\cos^3x$ by parts I've converted $\cos^3(x)$ into $\cos^2(x)\cos(x)$ but still have not gotten the answer.
The answer is $\dfrac{\sin(x)(3\cos^2x + 2\sin^2x)}{3}$
My answer was the same except I did not have a $3$ infront of $x$ and my $2\sin^2x$ was not squared.
Help!
| $$
\begin{align}
\int\cos^3{x}\,dx
&=\int\cos^2{x}\cdot\cos{x}\,dx\\
&=\int\cos^2{x}(\sin{x})'\,dx\\
&=\cos^2{x}\sin{x}+2\int\sin^2{x}\cos{x}\,dx\\
&=\cos^2{x}\sin{x}+2\int(1-\cos^2{x})\cos{x}\,dx\\
&=\cos^2{x}\sin{x}+2\int\cos{x}\,dx-2\int\cos^3{x}\,dx\\
&=\cos^2{x}\sin{x}+2\sin{x}-2\int\cos^3{x}\,dx
\end{align}
$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Understanding a proof from the APMO 1998 on inequalities. I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung.
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{xyz} }$
I know that the AM-GM inequality must be applied but I... | You can get the Pham Kim Hung's proof by the following way.
We'll try to find non-negatives $a$, $b$ and $c$ such that $a+b+c=1$ and
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\frac{x}{\sqrt[3]{xyz}}$$
Now, by AM-GM
$$a\cdot\frac{x}{y}+b\cdot\frac{y}{z}+c\cdot\frac{z}{x}\geq\left(\frac{x}{y}\right)^a\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3116317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$ The original exercise is to
Prove that
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$
Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.
I've tried rewriting the left side in te... | The obvious solution
$\cos3(20^\circ)=4\cos^320^\circ-3\cos20^\circ$
and $\sin3(10^\circ)=3\sin10^\circ-4\sin^310^\circ$
and $\cos3(20^\circ)=\sin3(10^\circ)$
Alternatively, dividing both sides by $\cos20^\circ+\sin10^\circ$ and replacing $\cos20^\circ=1-2\sin^210^\circ$
with $\sin10^\circ=s,3s-4s^3=\sin3(10^\circ)=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Disprove or prove using delta-epsilon definition of limit that $\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$ I want to prove if the following limit exists, using epsilon-delta definition, or prove it doesn't exist:$$\lim_{(x,y) \to (0,0)}{\frac{x^3-y^3}{x^2-y^2}} = 0$$
My attempt:
First I proved some direction... | OK, here is a completely different answer using $(x,y)$ coordinates only and no parameters.
Basically, try a power curve $y=x^n$.
BUT this gives you a path approaching the origin near the $x$ or $y$ axis, which is not a problem. So we need to modify it to place it near the "problem" line $y=-x$.
Consider the curve
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding out the remainder of $\frac{11^\text{10}-1}{100}$ using modulus
If $11^\text{10}-1$ is divided by $100$, then solve for '$x$' of the below term $$11^\text{10}-1 = x \pmod{100}$$
Whatever I tried:
$11^\text{2} \equiv 21 \pmod{100}$.....(1)
$(11^\text{2})^\text{2} \equiv (21)^\text{2} \pmod{100}$
$11^\text{4} \... | A really quick way to "see" the answer is with binomial theorem.
$11^{10} = (10+1)^{10} = 10^{10} + k_1\cdot 10^9 + k_2 \cdot 10^8 + ... + 10\cdot 10^1 + 1$
where the $k$'s represent various combinatorial constants. The values are unimportant. What's important is that when we take the whole thing modulo $100$, the expr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limiting value of a sequence when n tends to infinity Q) Let, $a_{n} \;=\; \left ( 1-\frac{1}{\sqrt{2}} \right ) ... \left ( 1- \frac{1}{\sqrt{n+1}} \right )$ , $n \geq 1$. Then $\lim_{n\rightarrow \infty } a_{n}$
(A) equals $1$
(B) does not exist
(C) equals $\frac{1}{\sqrt{\pi }}$
(D) equals $0$
My Approach :- I am n... | The hint:
$$0<a_n=\prod_{k=1}^n\left(1-\frac{1}{\sqrt{k+1}}\right)<\prod_{k=1}^n\left(1-\frac{1}{k+1}\right)=\frac{1}{n+1}\rightarrow0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Formulas for sequences and then how to find the limsup and liminf For each of the following sequences, calculate the limit superior and the limit inferior. If the sequence converges, calculate its limit. Justify your answers.
A) $\ \ \left\{\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{3},\frac{2}{3},-\frac{2}{3},\frac... | A) Limes superior is $1$, limes inferior is $-1$.
The important subsequence for proving lim sup is
$$a_n =\left\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\ldots\right\}$$
as its limit is $1$, and for every its element are two previous elements in the original sequence smaller.
For lim inf the reasoning is sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove/show this actually defines a homomoprhism
We define the homomorphism $f: \text{SL}_2(\mathbb Z / 2 \mathbb Z) \to \text{SL}_2(\mathbb Z / 2 \mathbb Z)$ that maps the generators to:
$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$ \begin{pmatrix} 0 & 1 \... | A very concrete way to verify that this is a homomorphism, despite the fact that the generators don't commute, is to find another relation between the generators. This approach turns out to be effective quite often. In this case, setting
$$T:=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}
\qquad\text{ and }\qquad
S:=\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
About an integral problem (Leibniz integral rule)? I want to evaluate this integral $\int_{0}^\frac{\pi}{2} \ln(1-a^2\sin^2x)dx$, where $|a|<1.$
First, I differentiate with respect to $a.$ Then, it would become a terrible integral. That is $$\int_{0}^\frac{\pi}{2} \frac{-2a\sin^2x}{1-a^2\sin^2x}dx.$$ After that, I thin... | $$F(a)=\int_0^{\pi/2}\log\left[1-a^2\sin(x)^2\right]\mathrm dx\Rightarrow F(0)=0$$
$$F'(a)=\frac2a\int_0^{\pi/2}\frac{-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx$$
$$F'(a)=\frac2a\int_0^{\pi/2}\frac{1-a^2\sin(x)^2}{1-a^2\sin(x)^2}\mathrm dx-\frac2a\int_0^{\pi/2}\frac{\mathrm dx}{1-a^2\sin(x)^2}$$
$$F'(a)=\frac\pi a-\frac2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\max\{y-x\}$
If $x+y+z=3, $ and $x^2+y^2+z^2=9$ , find $\max\{y-x\}$.
I tried to do this geometrically, $x+y+z=3$ is a plane in $\Bbb{R}^3$ and $x^2+y^2+z^2=9$ is a ball with radius 3 and center of origin . So the candidate points for $y-x$ are on the intersection of the plane and the ball. But now I am confu... | By taking the rotation $X=(y-x)/\sqrt{2}$ and $Y=(y+x)/\sqrt{2}$ we have that the equations become
$$\begin{cases}
X^2+Y^2+z^2=9\\
\sqrt{2}Y+z=3\end{cases}$$
Hence $z=3-\sqrt{2}Y$ and
$$X^2=9-(3-\sqrt{2}Y)^2-Y^2=3Y(2\sqrt{2}-Y)\leq 6$$
with equality for $Y=\sqrt{2}$ (and $z=1$).
It follows that
$$X=\frac{y-x}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding integer solutions out of a a | b Determine all positive integer values of (n) such that
$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$
What is the sum of all these values?
CURRENT PROGRESS:
I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6... | So $(n+1)(n^2 -n + 6) = 3\cdot 2^{2009}$ so
So $n+1 = 3^t2^s; n^2 -n+6 = 3^r2^w; t+r \le 1; s + w \le 2009$.
Case 1:$t= 0$.
$n = 2^s -1$ and $n^2 - n + 6 = 2^{2s} - 2^{s+1} -2^s + 6$.
If $s = 0$ we have $n=0$ and that's a solution (if we assume ${0\choose k}=0$ for $k \ne 0$). If $s \ge 1$ then
$n^2 -n +6 = 2(2^{2s-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate :$\int_{-1}^{1} 2\sqrt{1-x^2} dx $
Evaluate: $$\int_{-1}^{1} 2\sqrt{1-x^2} dx $$
The answer is $\pi$
My attempt
$x = \sin(u), dx = \cos(u)du$
$$\int_{-1}^{1} 2 \sqrt{1-\sin^2(u)}\cos(u)du = \int_{-1}^{1} 2 \cos^2(u)du =\int_{-1}^{1} \frac{1}{2}(1+\cos(2u))du = \bigg(\frac{u}{2} + \frac{1}{2}\sin(2u) \bigg)\B... | Before evaluating the definite integral, you need to make the following back-substitution:
$$
u=\arcsin{x}
$$
And use this trigonometric identity ($-\frac{\pi}{2}\le u\le \frac{\pi}{2} \implies \cos{u}\ge0$):
$$
\sin{2u}=2\sin{u}\cos{u}=2\sin{(\arcsin{x})}\sqrt{1-\sin^2{(\arcsin{x})}}=2x\sqrt{1-x^2}
$$
You also forgot ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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What is value of $\prod_{x\geq 2} \frac{x^2+1}{x^2-1}\;?$ For $x \in \mathbb{Z^+}$ what is:
$$
\prod_{x\geq 2} \frac{x^2+1}{x^2-1}\;?
$$
Numerically it seems to be approximately $3.676$.
| Very partial answer:
$$\prod_{x\geq2} \frac{x^2+1}{x^2-1} =
\prod_{x\geq2} \frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}.$$
Then Maple says that both the numerator and denominator converge. It gives
$$\prod_{x\geq2} 1-\frac{1}{x^2} = \frac{1}{2}$$
and
$$\prod_{x\geq2} 1+\frac{1}{x^2} = \frac{\sinh \pi}{2\pi}.$$
So your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sqrt{a}-\sqrt{b}$ is a root of a polynomial with integer coefficients, then so is $\sqrt{a}+\sqrt{b}$.
If $\sqrt{a} - \sqrt{b}$, where $a$ and $b$ are positive integers and non-perfect squares, is a root of a polynomial with integer coefficients, then $\sqrt{a} + \sqrt{b}$ also is.
It seems to hold some relation... | (Let's agree first that $a, b \ge 0$, so that $\sqrt{a}$ and $\sqrt{b}$ are real and sensible)
This statement:
For any polynomial $p(x) \in \Bbb{Z}[x]$, if $p(\sqrt{a} - \sqrt{b}) = 0$ then $p(\sqrt{a} + \sqrt{b}) = 0$
is in general false, with the easiest counter example probably being $a = b = 1,\ p(x) = x$. But th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place Prove, that for every real numbers $ x \ge y \ge z > 0 $, and $x+y+z=\frac{9}{2}, xyz=1$, the following inequality takes place:
$$ \frac{x}{y^3(1+y^2x)} + \frac{y}{z^3(1+z^2y) } + \frac{z}{x^... | With the two equations $$x+y+z=\frac{9}{2}$$ and $$xyz=1$$ we can express the variables $$x,y$$ by $z$ for instance and your inequality problem reduces to a one variable problem
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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waiting for TT and H? We toss a fair coin until we see one consecutive TT pattern and one H, their order is irrelevant.
For instance, HHTHTHTT, or TTH.
Let X be the number of tossed required. What is E(X)?
Here is my solution: Let
$X_H$= number of tosses required to get H.
$X_{TT}$= number of tosses required to get T... | Look at this flow chart (each arrow has probability $\frac{1}{2}$ except the last absorbing state has probility $1$ in its self-loop):
Your answer $6.5$ is confirmed by calculating the expected number of steps for this absorbing Markov chain starting from the empty state and reaching the target state $(\text{has seen ... | {
"language": "en",
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"source": "stackexchange",
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$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
$ f(\frac{x^2+1}{x^2})=x^2+\frac{1}{x^2}-2 \Rightarrow f(x)=?$
I'm getting a wrong answer even though my solution looks valid to me:
Inverse of $(1+\frac{1}{x^2})$ is $(\frac{1}{{\sqrt{x-1}}})$, so I plug this expression in wherever I see $x$'s
$$f(x)=(\... | In my opinion your answer is correct. Mimicking your computations I got the same result.
Defining $t=\dfrac{x^2+1}{x^2}$ we got that $$x^2=\dfrac{1}{t-1}.$$ So
$$f(t)=f\left(\dfrac{x^2+1}{x^2}\right)=\dfrac{1}{t-1}+t-1-2=\frac{(t-2)^2}{t-1}.$$
This holds for any $t\in (1,\infty)$ assuming $x\in \mathbb R\setminus \{0\}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differential Equations - Exact ODE's
Hi all, I've attempted the question part b) after rearranging the 1st expression to :
$$v(x,y)\, dx - u(x,y) \, dy = 0.$$
After this I tried using the method of differentiating each expression and trying to calculate if they were exact. It was very lengthy so I am unable to post on... | $$(-v)dx+(u)dy=\frac{2xy}{(x^2+y^2)^2}dx+\left(1+\frac{y^2-x^2}{(x^2+y^2)^2} \right)dy=0$$
$$\frac{\partial (-v)}{\partial y}= \frac{\partial }{\partial y}\left(\frac{2xy}{(x^2+y^2)^2} \right)= \frac{2x}{(x^2+y^2)^2}-\frac{8xy^2}{(x^2+y^2)^3} =\frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$$
$$\frac{\partial u}{\partial x}= \frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why don't we have to use polynomial long division on quadratic irreducible partial fractions in this case? For $(x^2+3x+1)
/(x^2+1)^2 $
Why don't we have to use polynomial long division in this case since the degree of the denominator has the same magnitude of degree as the numerator before doing partial fraction decom... | You can use long division in this particular case, where the denominator is the power of a quadratic. But it is not much help in other cases.
\begin{eqnarray}
\frac{x^2+3x+1}{(x^2+1)^2}&=&\frac{x^2+3x+1}{x^2+1}\cdot\frac{1}{x^2+1}\\
&=&\left(1+\frac{3x}{x^2+1}\right)\cdot\frac{1}{x^2+1}\\
&=&\frac{1}{x^2+1}+\frac{3x}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the fact that $\frac{a^2}{b} + \frac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$ There's a math competition I participated yesterday (19/3/2019).
In these kinds of competitions, there will always be at least one problem about inequalities.
Now this year's problem about inequality is very easy. I am more interest... | By the AM-GM inequality, we have that
$$\begin{align*}
\dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) &= \dfrac{a + b}{ab} \cdot (a^2 + b^2 - ab + 7ab)\\
&= \dfrac{a + b}{ab} \cdot [(a + b)^2 + 4ab]\\
&\ge \dfrac{a + b}{ab} \cdot 4(a + b)\sqrt{ab}\\
&= \dfrac{(a + b)^2}{2\sqrt{2ab \cdot (a^2 + b^2)}} \cdot 8\sqrt{2(a^2 + b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)$ definition question I know the definition of $\mathbb{Q}\!\left(\sqrt{2}\right)=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Why can't you similarly say $\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)=\{a+b\!\left(\sqrt{2}+\sqrt{3}\right)\mid a,b\in\mathbb{Q}\}$? Any intuitive answers... | Recall polynomials with rational coefficients is the set
$$\mathbb{Q}[x] = \{q_0 + q_1x + q_2x^2 + \dots q_nx^2 : q_i \in \mathbb{Q} \}$$
So if we let $x = \sqrt{2}$, we would get
\begin{align}
\mathbb{Q}[\sqrt{2}] &= \{q_0 + q_1\sqrt{2} + q_2\sqrt{2}^2 + \dots + q_n\sqrt{2}^n \}\\
&\approx\{q_0 + q_1\sqrt{2} \}
\end{a... | {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\frac{1}{\sqrt{x^2+x+1}}$
How to integrate $$\frac{1}{\sqrt{x^2+x+1}}$$
I tried to solve this integral as follows
$\displaystyle
\int \frac{1}{\sqrt{x^2+x+1}} \ dx=
\int \frac{1}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac{1}{\sqrt{(\frac{2x+1}{2})^2+\frac{3}{4}}} \ dx=
\int \frac... | Here $$\int \frac{1}{\sqrt{g^2-1}} \ dg=...$$
Correct is:
$$\int \frac{1}{\sqrt{1-g^2}} \ dg=
\arcsin g $$
| {
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"source": "stackexchange",
"question_score": "3",
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For which $a>0$ series is convergent?
For which $a>0$ series $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?
My try:From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{... | The answer is indeed as you concluded $a > \frac{1}{4}$.
Note that (from the first line in your attempt)
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$
for $n$ sufficiently large, and that is all you need to conclude
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integers $n$ satsifying $\frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}}$
If $\displaystyle \frac{1}{\sin \frac{3\pi}{n}}=\frac{1}{\sin \frac{5\pi}{n}},n\in \mathbb{Z}$, then number of $n$ satisfies given equation ,is
What I tried:
Let $\displaystyle \frac{\pi}{n}=x$ and equation is $\sin 5x=\sin 3x$
$\di... | I'm afraid you have used the wrong property: $\sin A-\sin B=2\cos\left(\dfrac{A+B}2\right)\sin\left(\dfrac{A-B}2\right)$. So you get$$2\cos(4x)\sin x=0$$Thus, $x=k\pi\vee4x=(2k+1)\pi/2$ for $k\in\Bbb Z$. For the former, $x=\pi/n=k\pi\implies nk=1$ which is true iff $n=k=1\vee n=k=-1$. In the latter case, $(2k+1)\pi/8=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to find an exact solution to a certain linear second-order non-homogeneous differential equation The equation below came up in some work I'm doing and I'm at a loss as to how to get a non-numerical solution:
$\ddot x + k_1 \tan( k_1 t) \sec(k_1 t)\dot x+k_2\cos(k_1t) x=k_3\cos(k_1t)$
| Assume $k_1,k_2\neq0$ for the key case:
$\ddot x+k_1\tan(k_1t)\sec(k_1t)\dot x+k_2\cos(k_1t)x=k_3\cos(k_1t)$
$\dfrac{d^2x}{dt^2}+\dfrac{k_1\sin(k_1t)}{\cos^2(k_1t)}\dfrac{dx}{dt}+\cos(k_1t)(k_2x-k_3)=0$
$\cos^2(k_1t)\dfrac{d^2x}{dt^2}+k_1\sin(k_1t)\dfrac{dx}{dt}+\cos^3(k_1t)(k_2x-k_3)=0$
Let $u=x-\dfrac{k_3}{k_2}$ ,
Th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.
Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3... | Rearranging and squaring both sides of the original equation gives
$$4(3x+2)^2(x+3)=(13x+42)^2$$
$$36x^3+156x^2+160x+48=169x^2+1092x+1764$$
$$36x^3-13x^2-932x-1716=0$$
$$(x-6)(4x+11)(9x+26)=0$$
$$x=6, -\frac{11}{4}, -\frac{26}{9}$$
But $x=6$ is an extra solution created due to squaring both sides and will not work in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find number of ways to choose $3n$-subset with repetitions from set $\left\{A,B,C\right\}$ Find number of ways to choose $3n$-subset with repetitions from set $\left\{A,B,C\right\}$ such that:
1. Letter $A$ occur at most $2n$
2. Letter $B$ occur at most $2n$
3. Letter $C$ occur odd times
Approach
I want to use there e... |
We obtain for $n\geq 1$:
\begin{align*}
[x^{3n}]&\left(\frac{1-x^{2n+1}}{1-x}\right)^2\frac{x}{1-x^2}\tag{1}\\
&=[x^{3n-1}]\frac{1-2x^{2n+1}}{(1-x)^2\left(1-x^2\right)}\tag{2}\\
&=\left([x^{3n-1}]-2[x^{n-2}]\right)\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\sum_{j=0}^\infty x^{2j}\tag{3}\\
\end{align*}
Comment:
*
*In (2... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$
Prove that if $p$ is prime and $a^7-b^3=p^2$ then $\text{gcd}(a,b)=1$.
Any help is appreciated.
| Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....
What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.
For each of these choices what do you get when you look at $\frac {a^7}{hcf(a,b)} - \frac {b^3}{hcf(a,b)} = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding intersection of a sphere and a plane Suppose I have the sphere $\ x^2 + y^2 + z^2 = 4 $ and the plane $\ x-y\sqrt{3} =0. $
How do I find the intersection curve and write it in polar terms?
In polar coordinates the sphere is just $\ r = 2 $
substituting $\ x = \sqrt3 y $ in ball equation:
$$ \ (\sqrt3 y)^2... | Find orthogonal vectors each of length (radius) $2$ in the plane given:
$(1, \sqrt{3},0)$ and $(0,0,2)$
Now create a trigonometric (polar) sum of these vectors:
$$\cos \theta (1, \sqrt{3}, 0) + \sin \theta (0,0,2) = (\cos \theta, \sqrt{3} \cos \theta, 2 \sin \theta)$$
In traditional polar coordinates:
$$(r, \theta, \p... | {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify the the binomial coefficient in the binomial series? Use binomial series to expand the function $\frac{5}{(6+x)^3}$ as a power series.
I understand the process to get the following summation:
$\frac{5}{6^3}\sum_{n=0}^{\infty} {-3 \choose n} (\frac{x}{6})^n $
However, I am stuck on seeing what's going on... | We have the following:
\begin{align}
\frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n+2)}{n!}\left(\frac{x}{6}\right)^n &= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n (3)(4)(5)\ldots(n)(n+1)(n+2)}{(1)(2)(3)\ldots(n)}\left(\frac{x}{6}\right)^n\\
&= \frac{5}{6^3}\sum_{n=0}^\infty \frac{(-1)^n(n+1)(n+2)}{(1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}$
Evaluate:$$
\lim_{x\to 1}\frac{(x^n -1)(x^{n-1}-1)\cdots(x^{n-k+1}-1)}{(x-1)(x^2-1)\cdots(x^k-1)}
$$
I'm trying to spot an error in my calculations. It is known that $x^n - 1$ may be factored out as $(x-1)(1+x+x^2+\cdots+... | You are indeed missing a $+1$ on the top. $x^2-1=(x-1)(x+1)$. You divide by $(x-1)$ and set $x=1$ and you get $2$, not $n-1=2-1=1$. In the $1+x+...+x^{n-1}$ there are $n$ terms, since you have $x^0$ as well. Same at the bottom, you go up to $k$, not $k-1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ when $e^{2x}-3e^x=4$
Solve for $x$ when $e^{2x}-3e^x=4$
This is what I've gotten so far:
\begin{array}
he^{2x}-3e^x&=&4\\
\ln(e^{2x}-3e^x)&=&\ln(4) \\
\dfrac{\ln(e^{2x})}{\ln(3e^x)}&=&\ln(4)\\
\dfrac{2x}{\ln(3) + \ln(e^x)}&=&\ln(4) \\
\dfrac{2x}{\ln(3) + x}&=&\ln(4) \\
x= \dfrac{\ln(3) \cdot \ln(4)}{2-\... | Let, $e^x=y$ then the equation $e^{2x}-3e^x=4$ becomes $y^2-3y=4$
Solving $y^2-3y-4=0$ we get, $$\Delta=b^2-4ac=9+16=25$$ $$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\implies y=\frac{3\pm\sqrt{25}}{2}$$ $$\therefore y=-1,4$$
Substituting $y=e^x$
$e^x=-1\implies$ No Solution Exist $\forall x\in\mathbb R$ because $e^x>0$
So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.
If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.
I recently came across a question in which we had to prove the above inequality using the given condition as mentioned above. Here $a,b,c$ are distinct positive integers and $k$ is als... | From $\displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\bigg[(a-b)^2+(b-c)^2+(c-a)^2\bigg]\geq 3.$
Because $a,b,c$ are distinct integers.
So here we have taken $a,b,c$ as $3$ consecutive integers.
And $\displaystyle (a+b+c)^2-3(ab+bc+ca)\geq 0\Rightarrow (a+b+c)^2\geq 3(ab+bc+ca)=3(3k^2-1)$
So $\displaystyle (a+b+c)\geq \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does this infinite series $1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots$ simplify to an integral $\int_0^1\frac{dx}{1+x^3}$? How does the infinite series below simplify to that integral?
$$1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots=\int_0^1\frac{dx}{1+x^3}$$
I thought of simplifying the series to the sum to inf... | $$\int_{0}^{1}{\frac{dx}{1-(-x)^3}=\int_{0}^{1}{\sum_{n=0}^{\infty}{(-x)}^{3n}}}dx=\sum_{n=0}^{\infty}{(-1)^{3n}\int_{0}^{1}{x^{3n}}dx}$$
$$=\sum_{n=0}^{\infty}{\frac{(-1)^{3n}}{3n+1}}= 1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Delta, given Epsilon lim x->0 root (x+1) = 1. epsilon = 0.1 Can I solve this using rationalization and then assuming delta <= 1? If so, why is my answer wrong?
we know |√(x+1)-1| < 0.1
rationalization of LHS,
x<0.1*(|√(x+1)+1|)
hence, delta = 0.1*(|√(x+1)+1|)
to find x, consider delta <= 1, x <= 1 hence √(x+1)+1 <... | There are two mistakes you are making:
*
*The $\delta$ you are choosing in the first place is not independent of $x$.
*In trying to rectify this, you find an upper bound $c$ for your expression $f(x)$ for $\delta$ that is dependent on $x$ and then choose $\delta$ equal to that upper bound $c$. However, you previous... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving polynominals equations (relationship of roots)
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
Find (evaluate):
$$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$
So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\... | Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 \Rightarrow (x+1)(x-2)(x-3)=0 \Rightarrow \\
\alpha =-1, \beta =2,\omega=3.$$
Hence:
$$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}=\\
\frac{-1+ 2}{3} + \frac{2 + 3}{-1} + \frac{-1 + 3}{2}=\\
\frac13-5+1=\\
-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Given $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$ Find $\frac{a_7}{a_{13}}$ Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$
My try:
I assumed $A=2x^2$,$B=3x$ and $C=4$
Then we have the following cases to collect coefficient of $x^7$:
Case $1.$ $A^3 \times B^1 \times C^6$
Case $2.$ $A^2 ... | Given $$(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}x^k$$
Now using $\displaystyle x\rightarrow \frac{2}{x}$
$$\Rightarrow 2^{10}(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k}x^{20-k}$$
$$\Rightarrow (2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k-10}x^{20-k}$$
Now Comparing Coefficient of $x^{7}$ on both side, we get
$$a_{7}=a_{13}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$ While solving a problem I came across this task, minimizing
\begin{align}
\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2.
\end{align}
One can eas... | Here is a solution using AM-GM and the double-angle formulae
*
*$(\star):\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$
\begin{eqnarray*}\left ( \sin^2 x + \frac{1}{\sin^2 x} \right )^2 + \left ( \cos^2x + \frac{1}{\cos^2x} \right )^2
& \stackrel{AM-GM}{\geq} & 2\left ( \sin^2x + \frac{1}{\sin^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
$p^2-p+1=P^3$ prove how many $p$, $P$ pairs are there Determine with a proof all prime numbers $p$ such that $p^2-p+1$ is a cube of a prime number.
$19^2-19+1=7^3$
But is it the only $p$?
How should I prove it?
| We will use here: If $a$ and $b$ are positive integers then $$\boxed{a\mid b\implies a\leq b}$$
Rename $q=P$. We can assume that $p\geq 23$ and so $q\geq 8$. We have $$ p(p-1) = (q-1)(q^2+q+1)$$
*
*If $p\mid q-1$ then $q^2+q+1\mid p-1$. So $p\leq q-1$ and $q^2+q+1\leq p-1$ so we have $$q^2+q+1\leq q-2\implies q^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Check for equality of the integrals I need to show the following identity
$$
\int_a^b\frac{dx}{\sqrt{x-x^2}}=\int_{\tfrac{1-\sqrt{1-a}}{2}}^{\tfrac{1-\sqrt{1-b}}{2}}\frac{dx}{\sqrt{x-x^2}}+\int_{\tfrac{1+\sqrt{1-b}}{2}}^{\tfrac{1+\sqrt{1-a}}{2}}\frac{dx}{\sqrt{x-x^2}}
$$
the sum of the integrals on the right side
$\arc... | The integrals in the right-hand side sum to
$$
2\arcsin\sqrt{1-a}-2\arcsin\sqrt{1-b}
$$
If you look at https://math.stackexchange.com/a/3211273/62967, you'll see that
$$
\arcsin(2x-1)=2\arcsin\sqrt{x}-\frac{\pi}{2}
$$
For $x=1-a$, we have $2x-1=2-2a-1=1-2a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3... | When $N = 1 + 2k$;
\begin{align}2^{1+2k} +1 \pmod 3 & \equiv 2(2^{2k})+1 \pmod 3\\
&\equiv 2(4^{k})+1 \pmod 3\\
&\equiv 2+1 \pmod 3\\
&\equiv \pmod 3
\end{align}
When $N = 2k$
$$2^{2k} +1 \pmod 3\equiv 2 \pmod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | None of the other answers so far has really addressed the error that you made in your attempt. After rearranging the original equation into $$\frac{1}{x^2}-\frac{1}{x}-4=0,$$ you then decided that $a=1/x^2$, $b=-1/x$ and $c=-4$. Substituting these names for the corresponding values in this equation gives you $$a+b+c=0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
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.