Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Polynomial LongDivision What would be the result of $x^3-4x^2-5$ divided by $x-3$ ? I am getting $4$ as my solution can someone prove me wrong, this is very confusing.
| The remainder when dividing by $x-3$ is also the polynomial evaluated at $x=3$, which is $27-4\cdot 9-5=-14$.
Indeed,
$$\begin{align}x^3-4x^2-5&=x^2\cdot(x-3)-x^2-5\\&=(x^2-x)(x-3)-3x-5\\&=(x^2-x-3)(x-3)-14\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$, sum, area of convergence & center I want to find the
$$\sum _{n=0}^{\infty } \frac{x^n}{n+3}$$
now this is how I thought about doing it but I get stuck.
$$\sum _{n=0}^{\infty } x^n=\frac{1}{1-x}$$ given that absolute value of x is less then zero.
$$\int x^n \, dx=\frac{x^{n+1}... | I think this is a good way of handling these series:
Let $s(x)=\sum_{n=0}^{+\infty} \frac{x^n}{n+3}$ (defined for $|x|<1$). Then
$$
x^3s(x)=\sum_{n=0}^{+\infty}\frac{x^{n+3}}{n+3},
$$
and hence
$$
(x^3s(x))'=\sum_{n=0}^{+\infty}x^{n+2}=x^2\sum_{n=0}^{+\infty}x^n=\frac{x^2}{1-x}.
$$
Integrating from $0$ to $t$, we get
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sq... | Let's start from your last line:
$$\begin{align}
\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} &= \lim \frac{\sqrt x}{\sqrt x} \frac{\frac{1}{\sqrt x} + 1}{\sqrt{1 + \frac{1}{\sqrt x}} + \sqrt{1 - \frac{1}{ x}}} \\
&= \frac{1}{1 + 1} = \frac{1}{2}
\end{align}$$
where we note that everywhere we have ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? Trivially $n$ cannot be even , so this leaves us only with the possibilities $n \equiv1,3,5( \mod 6) $ , but I cannot r... | If $n\equiv1,5\pmod 6$ then $n^2\equiv 1\pmod 6$.
On the other hand, $2^n\equiv 2$ or $4\pmod 6$ depending on $n$ is odd or even. So if $n^2\equiv 1\pmod 6$, then
$$2^n+n^2\equiv 2+1\equiv 3\pmod 6$$
therefore $2^n+n^2$ is a multiple of three.
Notice the case $1^2+2^1$, that is the only exception to your statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The dimension of the SU(2) matrix group Let's take the matrix $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Consider its transpose $R^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$.
Then $RR^\dagger =1$ implies that $|a|^2 + |c|^2 = 1, |b|^2 + |d|^2 = 1, ab^* + cd^* = 0,$ and $a^*b + c^*d = 0$.
T... | The explanation is that the determinant is already constrained to be on the unit circle. We have $|\det(R)|^2 = \det(R)\overline{\det(R)} = \det(R)\det(R^\dagger) = \det(RR^\dagger) =\det(I) = 1$.
This means that the equation $a_0d_0 - a_1d_1 - b_0c_0 + b_1c_1 = 1$ implies $a_0d_1 + a_1d_0 - b_0c_1 - b_1c_0 = 0$, and ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
| It belongs to a family that starts with the familiar,
$$\begin{align}
&3^2+4^2 = 5^2\\
&10^2+11^2+12^2=13^2+14^2\\
&21^2+22^2+23^2+24^2 = 25^2+26^2+27^2\\
\vdots\\
&a^2 + (a+1)^2 + \dots + a_n^2 = b^2 + (b+1)^2 + \dots + b_{n-1}^2
\end{align}$$
where $b = a_n+1$ and $a = 2n^2-3n+1 = 0, 3, 10, 21, 36, 55, 78, 105, \col... | {
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Show that the substitution $t=\tan\theta$ transforms the integral ${\int}\frac{d\theta}{9\cos^2\theta+\sin^2\theta}$, into ${\int}\frac{dt}{9+t^2}$ To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:
$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\... | An alternative method. Let
$$I={\int}\frac{1}{9\cos^2x+\sin^2x}\,dx$$
Dividing the denominator and numerator by $\cos^2x$ we have
$$I=\int\frac{\sec^2x}{9+\tan^2x}dx$$
Make the substitution $t=\tan x$, so that $\sec^2x~dx=dt,$ and we are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Odd divisibility induction proof Prove that for odd $n>3$
$$64\ | \ n^4-18n^2+17$$
I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?
| If $n=2k+1$, we have
$$
\begin{align}
n^4-18n^2+17
&=(2k+1)^4-18(2k+1)^2+17\\[6pt]
&=16k^4+32k^3-48k^2-64k\\
&=384\binom{k}{4}+768\binom{k}{3}+320\binom{k}{2}-64\binom{k}{1}\\
&=64\left[6\binom{k}{4}+12\binom{k}{3}+5\binom{k}{2}-\binom{k}{1}\right]
\end{align}
$$
This indicates that $64$ divides $n^4-18n^2+17$ for all ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Comparing series Can anyone explain why if I compare the coefficient of $x^{n}$ of the equation
$$\sum_{k=0}^{\infty}a(n)x^n= \frac{1}{1-x}-\frac{x}{1-x^3}+\frac{x^2}{1-x^5}-\frac{x^3}{1-x^7}+...$$
I can get
$$a(n)=k_{1}(4n+1)-k_{3}(4n+1)$$ where $k_{i}(m)$ is the number of divisors of $m$ that are congruent to $j$ mod... | $$
\begin{align}
\sum_{n=0}^\infty(-1)^n\frac{x^n}{1-x^{2n+1}}
&=\sum_{n=0}^\infty\sum_{k=0}^\infty(-1)^nx^{n+(2n+1)k}\\
\end{align}
$$
The coefficient of $x^m$ is the the number of factors of $2m+1=(2n+1)(2k+1)$ that are $1$ mod $4$ ($n$ even) minus the number that are $3$ mod $4$ ($n$ odd). This confirms the formula... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is e... | Since there are an even number of terms:
$$\begin{align}1\cdot 2\cdots r &= (-1)(-2)\cdots (-r)\\
&\equiv (p-1)(p-2)\cdots(p-r)\pmod p\\
&=(r+1)(r+2)\cdots(p-1)
\end{align}$$
| {
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Solve the limit $\lim\limits _{x\to 0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}$ $$\lim _{x\to \:0}\frac{\sqrt{1-\cos\left(x^2\right)}}{1-\cos\left(x\right)}=\left|\frac{0}{0}\right|$$
I think you have to multiply by the conjugate. And then make the change equivalent small. Right?
| Remove the square root by observing that $1-\cos x\ge0$, so $1-\cos x=\sqrt{(1-\cos x)^2}$; hence you reduce to computing
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}
$$
and then take the square root of the result. Now
$$
\lim_{x\to0}\frac{1-\cos(x^2)}{(1-\cos x)^2}=
\lim_{x\to0}\frac{1-(1-x^4/2+o(x^4))}{(1-(1-x^2/2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of equations with radicals: $2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}$ and $2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}x^2}$
Solve the system of equations (in $\mathbb R$):
$$\begin{matrix}
2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}y^2}
\\
2\sqrt[4]{\frac{y^4}{3}+4}
=
1+\sqrt{\frac{3}{2}x^2... | As you say, you will have solution corresponding to $x=\pm y$. Now, if you expand the last equation, you have (if $x \geq 0$),$$-\frac{37 x^4}{12}+3 \sqrt{6} x^3+9 x^2+2 \sqrt{6} x-63=0$$ To get rid of the $\sqrt{6}$'s, define $x=\sqrt{6} z$ and the equation becomes $$-111 z^4+108 z^3+54 z^2+12 z-63=0$$ By inspection $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coefficient of $x^{30}$. Find the coefficient of $x^{30}$ in the given polynomial
$$
\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}\right)^5
$$
I don't know how to solve problems with such high degree.
| The coefficient attached to $x^{30}$ will be the number of ways you can add up to $30$ by using the numbers $0$-$12$ up to five times. (Here order matters)
For instance $1+1+2+10+6=30$ is one way.
$10+10+10+0+0=30$ is another and so is $0+10+10+10+0 = 30$.
The reason for this is more apparent for smaller polynomials.
F... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}... | The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have
$$
\frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots o... | Let $$\zeta_{11} = e^{2\pi i/11} = \cos \frac{2\pi}{11} + i \sin \frac{2\pi}{11}$$ be a primitive $11^{\rm th}$ root of unity; hence $$\zeta_{11}^0, \zeta_{11}^1, \ldots, \zeta_{11}^{10}$$ are the roots of $z^{11} - 1 = 0$, and the sum of these roots is therefore zero. Then
$$\begin{align*} \sum_{k=1}^{10} \left( \sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
| When you multiplied by $x^2 + y^2$, you transformed an exact differential equation to an inexact one.
If we rearrange the original equation, we get
$$\tag{*}\left(x + \frac{y}{x^2 + y^2}\right)\, dx + \left(y - \frac{x}{x^2 + y^2}\right)\, dy = 0$$
Now
$$\frac{\partial}{\partial y}\left(x + \frac{y}{x^2 + y^2}\right) ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it... | I think you're making a bit of confusion. On one hand you are never proving that the sequence is increasing, on the other hand your argument is a bit too complicated. Here is how I would go about proving it:
*
*$\mathbf{n = 1}$: First note that $a_1 < a_2$ if and only if
$$
a_1^2 = 2 < 2 + \sqrt{2} = a_2^2
$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Floor inequality with prime If $a$ and $b$ are positive integers and $a\ge b$ and $b$ is an odd prime, show that:
$$\left\lfloor \frac{6a-1}{b}\right\rfloor+\left\lfloor\frac{a}{b}\right\rfloor\ge \left\lfloor \frac{2a}{b}\right\rfloor+\left\lfloor \frac{3a-1}{b}\right\rfloor+\left\lfloor \frac{2a+1}{b}\right\rfloor$$
... | Let $a=nb+k$, with $0\le k < b$. Your inequality simplifies to
$$\left\lfloor\frac{6k-1}{b}\right\rfloor\ge\left\lfloor\frac{2k}{b}\right\rfloor+\left\lfloor\frac{3k-1}{b}\right\rfloor+\left\lfloor\frac{2k+1}{b}\right\rfloor$$
Now let's examine it case by case.
Case 1. $k = 0:$ Both sides of the inequality are $-1$, so... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
| From the given conditions we can write, $a=\sin \theta, b=\cos \theta,\ \theta\in (0,\pi/2)$. Then, the objective function becomes $$f(\theta)=25/2\sin 2\theta+25\sin \theta+10\cos\theta\\ f'(\theta)=25\cos 2\theta+25\cos\theta-10\sin \theta$$Equate this to $0$ to get a solution of $\theta$, I believe it pertains to so... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$ by induction How can I prove by induction that $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$? My guess is that there must be another form to express the sum of nested square roots, but I don't know how to find it.
| $$(1) \quad x=k+\sqrt x$$
$$x=\sqrt {k+\sqrt x}$$
continuing the recursion...
$$x=\sqrt {k+\sqrt {k+\sqrt {k+...}}}$$
Thus (1) is the equivalent expression.
Solve for x with $k=1$
$$x=\phi$$
Thus, x equals the golden ratio. Multiply the expression by c...
$$c \cdot x=c \cdot \sqrt {k+\sqrt {k+\sqrt {k+...}}}$$
$$c \cd... | {
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"timestamp": "2023-03-29T00:00:00",
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Can the given transformation possible for given determinant?
In forth step $(x-1)(x-2)$ is obtained by applying transformation R$1 \frac{1}{(x+1)}$ and R$2 \frac{1}{(x+2)}$.
But we get value of $x = -1$ or $ x = -2$ so $\frac{1}{(x+1)}$ and $\frac{1}{(x+2)}$ will be undefined because value of x can be -1 and -2.
S... | In Step four, $(x+1) , (x+2)$ is obtained not by applying transformation $R_{1\frac{1}{x+1}}$ and $R_{2\frac{1}{x+2}}$. Since determinant function is linear in each row, we have
\begin{align}det
\begin{bmatrix}
c.a_{1,1} & c.a_{1,2} & c.a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}
\end{bmatrix} ... | {
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Solve the recurrence of the alternating sum $R_n=R_{n-1}+(-1)^{n}(n+1)^{2}$ I have been trying to solve this recurrence for a few hours, but I haven't been able to find the solution yet:
$R_0=1$
$R_n=R_{n-1}+(-1)^{n}*(n+1)^{2}$.
I have been trying to substitute $T_n=(-1)^{n}*R_n$ and then solving for $T_n$ and got the ... | @hypergeometric's solution is my favorite, but another (very general) way of solving this recurrence is by a so-called generating function. Let us define $f(x) = \sum_{n=0}^\infty R_n x^n$. Then,
$$ R_n = R_{n-1} + (-1)^n(n+1)^2 \implies R_nx^n = x R_{n-1} x^{n-1} + (-1)^n (n+1)^2 x^n$$
Summing both sides from $n=1$ to... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
| Meat of induction step:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i^2} &= \frac{1}{(k+1)^2}+\sum_{i=1}^k\frac{1}{i^2}\\[1em]
&< \frac{1}{(k+1)^2}+2-\frac{1}{k}\tag{by ind. hyp}\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k+1}{k}-\frac{1}{k+1}\right)\\[1em]
&= 2-\frac{1}{k+1}\left(\frac{k^2-k}{k(k+1)}\right)\\[1em]
&< 2-\frac{1}{... | {
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"url": "https://math.stackexchange.com/questions/1251544",
"timestamp": "2023-03-29T00:00:00",
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Find $\sup_{x\in[0,1]} \frac{x}{x^2+n^2+1}$ We have $f_n:[0,1]\to \mathbb{R},\:f_n(x)=\frac{x}{x^2+n^2+1}$ and we need to prove that is uniform convergence using formula:
$\lim _{n\to \infty } \sup_{x\in[0,1]} |f_n(x)-f(x)| =0$
First step I prove that the sequence $f_n$ converges pointwise to $f(x)=0$. After it: $|... | $f_n(x)$ is a positive function over $(0,1)$ and by the AM-GM inequality
$$ f_n(x)=\frac{x}{x^2+(n^2+1)}=\frac{1}{x+\frac{n^2+1}{x}}\leq\frac{1}{2\sqrt{n^2+1}}\tag{1} $$
with equality attained in $x=\sqrt{n^2+1}$. It follows that $f_n$ is an increasing function over $[0,1]$ and its supremum is simply given by $f_n(1)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find Primitive Root for Polynomial Field Can someone help me get started on the problem below:
Recall that $\mathbf{F}_{p^k}$ can be realized as $\mathbf{F}_p[x]/P(x)
\cdot \mathbf{F}_p[x]$ where $P(x)$ is a polynomial of degree $k$ with
coefficients in $\mathbf{F}_p$ which is irreducible.
Problem Find primitive ro... | In the quotient $\mathbf{F}_8 \cong \mathbf{F}_2/(x^3+x+1)$ you have the relation
$$x^3+x+1=0 \iff x^3=-x-1=x+1,$$
since over $\mathbf{F}_2, -1=1$. You can use this relation to perform multiplications. In this case $\mathbf{F}_8^{\times} \cong \mathbf{Z}/7\mathbf{Z}$. By Lagrange the order of an element divides the ord... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding range of $||x| - |y||$ for the given conditions.
If $ z = x + iy$ and $ x^2 + y^2 = 16 $ , then the range of $||x|-|y||$ is...?
This is what I've tried yet:
Suppose $x = a\cos \theta$ and $y = b\sin \theta$, then we've :
$$\begin{align}
x^2 + y^2 = 16 \implies & a^2\cos^2 \theta + b^2\sin^2 \theta= 16 \\
\... | Hint:-
Put $x=4\sin\theta$ and $y=4\cos \theta$.
Solution
Then, $$||x|-|y||=4||\sin\theta|-|\cos \theta||$$Now, $$0\le|\sin\theta|\le 1$$and, $$-1\le-|\cos\theta|\le0$$which gives, $$||\sin\theta|-|\cos \theta||\le1\implies 4||\sin\theta|-|\cos \theta||\le4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimise $ab+bc+ac$ Let $a,b,c \in \mathbb R$, and $a^2+b^2+c^2=1$
How can I calculate the minimum value of $ab+bc+ac$? (i.e. most negative)
I've tried using the fact that $(a-b)^2+(b-c)^2+(a-c)^2 \ge 0$ but this gives an inequality in the wrong direction.
| Note that $f(a,b,c) = ab+bc+ac = {1 \over 2} ((a+b+c)^2-(a^2+b^2+c^2))$.
Hence $\min \{ f(a,b,c) | a^2+b^2+c^2=1 \} = {1 \over 2} \min \{ (a+b+c)^2-1 | a^2+b^2+c^2=1 \}$.
Note: See Barry's comment below for a simpler alternative to the following.
One way that doesn't use Lagrange multipliers is to change basis.
Let $u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}... | It should be a printing mistake. Your answer is correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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The complex roots of a biquadratic polynom In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.
Here is my answer:
First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:
$x^4-4x^2+16=(x^2-... | Hint:
with your substitution, the solutions that you have are:
$$
x^2=2(1\pm i\sqrt{3})=4 e^{\pm i \frac{\pi}{3}}
$$
and this gives immediately the autor's answer.
added:
The complex number $a+ib=2+2i\sqrt{3} \rightarrow $ can be write in polar form $\rho e^{i\theta}$ with:
$$
\rho=\sqrt{a^2+b^2}=\sqrt{4+12}=4
$$
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Differentiation method for evaluating $ \sum_{n=1}^\infty \frac{n^2}{3^n} $ I evaluated the following infinite sum (the original and broader question regarding this sum can be found at Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $).
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} $$
However, I'm getting the feeling that I made ... | You could introduce a step in which you write
$$
\begin{align}
f(x) & = \cdots \\
& = x \frac{d}{dx} \left( x \sum_{n=0}^\infty nx^{n-1} \right) \\
& = \color{red}{
x \frac{d}{dx} \left( x \sum_{n=1}^\infty nx^{n-1} \right)} \\
& = x \frac{d}{dx} \left( x \frac{d}{dx}
\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute limit using Taylor's expansion Using Taylor’s expansion, prove that the following limit exists and compute it.
$$\lim_{x \to 0}\left(\frac {x^2}{\frac {1}{1-x} - e^x}\right)$$
In this if I am using the taylor series expansion of $e^x$ then denominator has some value as $$ 1 + (1 + x + x^2 + x^3 + \ldots) + x ... | Hint: First we have
$$\frac{1}{1 - x} - e^x = \sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!} =\sum_{n=0}^\infty x^n \Big(1 - \frac{1}{n!}\Big) = \frac{x^2}{2} + \frac{5x^3}{6} + O (x^4) $$
Then
$$\lim_{x \to 0}\frac{x^2}{\frac{x^2}{2} + \frac{5x^3}{6} + O (x^4)}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solving a recurrence relation using generating functions My recurrence relation is
D(n) = D(n - 1) + D(n - 2) + 5(n - 1);
with the initial conditions D(2),D(3) being 6, 17 respectively.
The generating function G(z) for the sequence D(n) is given
I don't know how i got this. Please can anyone give explanation for this... | Define the generating function:
$$
G(z) = \sum_{n \ge 0} D(n) z^n
$$
Take your recurrence written as:
$$
D(n + 2) = D(n + 1) + D(n) + 5 (n + 1)
$$
Multiply by $z^n$, sum over $n \ge 0$ and recognize some sums:
$$
\frac{G(z) - D(0) - D(1) z}{z^2}
= \frac{D(z) - D(0)}{z} + G(z) + \frac{5}{(1 - z)^2}
$$
Solve for $G(z)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.
| $$3-2\cdot 3^{x+1} = (3^{x+1})^2$$
Set $z = 3^{x+1}$:
$$z^2 + 2\cdot z-3 = 0$$
Solve quadratic equation:
$$z = -1 \pm \sqrt{1 + 3} = -1 \pm 2$$
so $z \in \{-3,1\}$.
Compute $x$ from
$$x = \log_3(z) - 1.$$
For a real solution, you have to pick $z=1$, so
$$x = \log_3(1) - 1 = 0 - 1 = -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove non-existance of limit: $f(x,y) = \frac{xy\sin(\frac{x}{y})}{x^2 + |y|^3}$ I need to prove that $f(x,y) = \frac{xy^2\sin(\frac{x}{y})}{x^2 + |y|^3}$ does not tend to $0$ when $(x,y)$ approaches $(0,0)$.
In order to do so, I would need to find some direction $\alpha$ such that $f(\alpha(t))$ approaches to some val... | Used polar coordinates $x=r\cos\phi, y=r\sin\phi$ and compute the limit $r \rightarrow 0$. The result is
\begin{align}
\lim_{r\rightarrow 0} \frac{r^2\cos\phi\sin\phi \cdot \sin(\cot\phi)}{r^2 \cos^2\phi + |r\sin\phi|^3}
&=
\lim_{r\rightarrow 0} \frac{\cos\phi\sin\phi \cdot \sin(\cot\phi)}{\cos^2\phi + r|\sin^3\phi|}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$ It's straightforward to show that
$$\sum_{i=1}^ni=\frac{n(n+1)}{2}={n+1\choose2}$$
but intuitively, this is hard to grasp. Should I understand this to be coincidence? Why does the sum of the first $n$ natural numbers count the number of ways I can choose a pair ou... | The intuition is that for the pairs can be listed in the following way.
$$\begin{array}{ccccccc}
1,2 & & & & & & \\
1,3 & 2,3 & & & & & \\
1,4 & 2,4 & 3,4 & & & & \\
1,5 & 2,5 & 3,5 & 4,5 & & & \\
1,6 & 2,6 & 3,6 & 4,6 & 5,6 & & \\
1,\vdots & 2,\vdots & 3,\vdots & 4,\vdots & 5,\vdots &\ddots & \\
1,n+1 & 2,n+1 & 3,n+1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
Generating function of $a_{n} = n$ I want to find the closed form of this $\sum\limits_{n=0}^{\infty} {nx^{n}}$Is the following correct?
$(\sum\limits_{n=0}^{\infty} x^n)^{'}= (1)^{'}+(x)^{'}+(x^2)^{'}+(x^3)^{'}+...=$
$0 + 1 +2x + 3x^2 + ...=$
$=(\sum\limits_{n=1}^{\infty} {nx^{n-1}})=(\sum\limits_{k=0}^{\infty} {(k-1... | What you did is correct except that
$$
\sum\limits_{n=1}^{\infty} {nx^{n-1}}=\sum\limits_{k=0}^{\infty} {(k+1)x^k}
$$ and we are led to the correct result
$$
\sum\limits_{n=1}^{\infty} {nx^{n-1}}=\frac{1}{(1-x)^2}.
$$ or equivalently, multiplying by $x$:
$$
\sum\limits_{n=1}^{\infty} {nx^{n}}=\frac{x}{(1-x)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$
*
*It[Root] is clearly negative.
*It should have odd number of roots.
*Either One or Three or Five.
*It is e... | Consider the derivative
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{4!}(x^4 + 4x^3 + 4x^2) + \frac{1}{12}x^2 + \left(\frac{x^2}{4} + x + 1\right)\\
&= \frac{x^2}{24}(x+2)^2 + \frac{x^2}{12} + \left(\frac{x}{2}+1\right)^2\\
&> 0.
\end{aligned}$$
So the fifth-order Taylor polyno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Counterclockwise rotation matrix If I take the basis $(\vec{e_x},\vec{e_y})$ and make a rotation counterclockwise of angle $\theta$, I end up with two new vectors $(\vec{u},\vec{v})$ such that :
$\vec{u} = \cos\theta \vec{e_x} + \sin\theta \vec{e_y}$
$\vec{v} = \cos\theta \vec{e_x} - \sin\theta \vec{e_y}$
so
\begin{eq... | You can also do it in a more algebraic way. Since after rotation ($(x, y)$ is rotated to $(x', y')$), the length of the vector doesn't change, which means $n = \sqrt{x'^2 + y'^2} = \sqrt{x^2 + y^2}$ (see in figure attached).
Therefore we can get the following equation:
\begin{aligned}
y' & = n \cdot \sin(\theta + \al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What exactly IS a square root? It's come to my attention that I don't actually understand what a square root really is (the operation). The only way I know of to take square roots (or nth root, for that matter) it to know the answer! Obviously square root can be rewritten as $x^{1/2}$ , but how does one actually multip... | I don't believe any computer actually does it this way, but you can compute a continued fraction to get an exact representation of a square root. For example, since $10^2$ $<$ $111$ $<$ $11^2$:
\begin{eqnarray}
\sqrt{111}
&=& 10 + (\sqrt{111} - 10) \\
&=& 10 + \frac{11}{\sqrt{111} + 10} = 10 + \frac{1}{\frac{\sqrt{111... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
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"answer_id": 6
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Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
| $$\sqrt{x-4}+\sqrt{x-7}=1<=>$$
$$-11+2\sqrt{(x-7)(x-4)}=12-2x<=>$$
$$4(x-7)(x-4)=(12-2x)^2<=>$$
$$4x^2-44x+112=4x^2-48x+144<=>$$
$$4x-32=0<=>$$
$$4(x-8)=0<=>$$
$$x-8=0<=>$$
$$x=8$$
BUT THIS SOLUTION IS INCORRECT SO THERE ARE NO SOLUTIONS!!!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S =... | Hint: Notice that
$$
S = \sum_{i=0}^\infty \frac{2^{i+1}-1}{2^{2i}}
$$
and distribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Limit and Integral problem work verification-2 I have to calculate the following:
$$\large\lim_{x \to \infty}\left(\frac {\displaystyle\int\limits_{x^{2}}^{2x}t^{4}e^{t^{2}}dt}{e^{x}-1-x - \frac{x^2}{2}- \frac{x^3}{6}-\frac{x^4}{24}}\right)$$
My attempt:
Let $F(x)=\displaystyle\int\limits_0^xt^4e^{t^2}dt$. Then,
$$\l... | Hint: I strongly suggest the use of power series. We can write down the power series for $e^w$, substitute $t^2$, and integrate term by term to get a series for the top. The series for the bottom is easy to write down.
Remark: As mentioned in a comment, there is an error in the L'Hospital's Rule calculation. It is fixa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Need a more compact formula This is a part of solution of a programming contest problem
$$\sum_{i=0}^{k} {x-i \choose 2} $$ given $x-i \ge 2$ is always true.
for
$k=1$,$(x-1)^2$
$k=2$, $(x-1)^2+((x-2)*(x-3)/2)$
$k=3$, $(x-1)^2 + (x-3)^2$
and so on.
Is there a reduced form of this?
| $$\frac{1}{2}\sum_{i=0}^k (x-i)(x-i-1) = \frac{1}{2} \left( \sum_{i=0}^k (x^2-2xi+i^2 + x-i) \right) =$$
$$=\frac{1}{2} \left( \sum_{i=0}^k (x^2+ x) + \sum_{i=0}^k (-2xi-i) + \sum_{i=0}^k i^2\right)= \frac{1}{2} \left( (k+1)(x^2+x) -(2x+1)\left(\frac{k(k+1)}{2} \right) + \frac{k(k+1)(2k+1)}{6}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the soluti... | One solution is apparent for (a): $x=1$. There can be no other solutions, since the two sides' derivatives are $\frac{1}{x}\left(\frac{1}{\ln(2)}+\frac{1}{\ln(3)}\right)>\frac{1}{x}$ and $\frac{1}{x}\frac{1}{\ln(4)}<\frac{1}{x}$. This shows us that the left side is always growing at a faster rate than the right side, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find $S=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{1+n^2}$ How do I find the sum:
$$S=\sum_{n=-\infty}^{\infty} \dfrac{(-1)^n}{1+n^2}$$
I can't solve this can someone help me?
| $$S=\sum_{n \ \text{even}} \dfrac{1}{1+n^2}-\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}$$
Now, $$\displaystyle\sum_{n \ \text{even}} \dfrac{1}{1+n^2}=\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1+(2n)^2}=\dfrac{1}{4}\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}$$
Also,
$$\begin{align}\sum_{n \ \text{odd}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints wou... | we can use product to sum formula. we have
$$\begin{align}(1- 2\cos 3x)(-\cos 2x - \cos x) &= 2\cos 3x \cos 2x + 2 \cos 3x \cos x - \cos 2x - \cos x\\
&= (\cos 5x + \cos x) + (\cos 4x + \cos 2x) - \cos 2x - \cos x\\
&=\cos 5x+\cos 4x\end{align}$$
dividing it out by $1- 2\cos 3x$ gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction.
I ran into the above problem. The base case $n=1$ gives $21$ which is divisible by $7$.
Now assume it is true for $n$. Then for $n+1$, we have the expression
$$ 1 + 2^{(2^{n+1})} + 2^... | Consider the polynomial
$$f(x) = 1+x^{2^n} + \left(x^{2^{n}} \right)^2$$
We claim that $f(x) = (1+x+x^2)g(x)$, where $g(x)$ is also a polynomial for all $k \in \mathbb{Z}^+$. This is easy to prove, since $\omega = e^{2\pi i/3}$ and $\omega^2 = e^{4\pi i/3}$ are roots of $1+x+x^2$. We have $1+\omega^{2^n} + \left(\omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not abl... | Let the desired limit be denoted by $L$.
We have via LHR $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\tag{1}$$ and $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}} = \frac{1}{3}\tag{2}$$ and we also have $$\lim_{x \to 0}\frac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive... | One more simplest way:
Put $y=mx+2$ in the equation $x^2-2y^2=1$. Then it comes to a quadratic equation of $x$. From which we get two values of $x$. Since the line is tangent to the given hyperbola so, it can not intersect at two different points. So, the quadratic equation must give two identical values of $x$.
For th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Probability of $ax^2 + bx + c = 0$ having real solutions
$a$, $b$, $c$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $ax^2 + bx + c = 0$ has real solutions?
This is from a final high school math exam, and I don't know h... | To get a real solution, the discriminant $b^2-4ac\geq0$, so $b\geq 2 \sqrt{ac}$, and $\frac{b^2}{4}\geq ac$. All of our variables are real so there is no need to consider negatives.
Values of $a$ and $b$ can range from $1$ to $n$. The probability of each separate combination is $\frac{1}{n}*\frac{1}{n}$. Considering ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $ \int_0^\theta \cosh(a\sin x) dx$ The integral below seems quite simple, but I couldn't find anywhere the result.
$$ I = \int_0^\theta \cosh(a\sin x) dx$$
I tried to expand it into Taylor expansion series and successfully evaluate the integral, but it just got mess,
$$ I =\sum_{k=0}^{\infty} \frac{a^{2k}}{(... | $\int_0^\theta\cosh(a\sin x)~dx=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}~dx=\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$
For $n$ is any natural number,
$\int\sin^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $f(x,y)$ is continuous in $(0,0)$ Prove that $f(x,y)$ is continuous in $(0,0)$, where
\begin{equation}
f(x,y) = \begin{cases}
\frac{x^2y}{x^4+y^2}, & (x,y)\neq 0\\
0, & (x,y) = (0,0)
\end{cases}
\end{equation}
The solution I have is that f is not continuous in $(0,0)$. (The solution doesn't say more than tha... | What you're doing wrong is that you're separating both $r$ and $\theta$ variables. This is not following the continuity definition for functions with several variables.
Please recall the definition of continuity for a function of several variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$
How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
*
*I squared the whole denominator, but that didn't help.
*Also I searched for a propriety or identity like $A^2-B^2$, but I did... | The trick here is to multiply and divide by the conjugate.
For instance $$\frac{1}{1+\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{1-\sqrt{2}}{-1}$$
We can do the same here, but we will do it twice.
$$\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{((\sqrt{7}-2\sqrt{5})+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Decomposition into partial fractions to compute an integral I'm having problems with:
$$\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}dx$$
I was thinking: $\frac{x^4+1}{x^6+1}$ is an even function and the interval $(-\infty,\infty)$ is symmetric about 0, we could write the integral like:
$$2\int_{0}^{\infty}\frac{(x^4+1)}... | Another way,
Solution :
\begin{align}
\int_{-\infty}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx&=2\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,\mathrm dx\\[7pt]
&=2\int_{0}^{\infty}\frac{x^4\,\mathrm dx}{x^6+1}+2\int_{0}^{\infty}\frac{\mathrm dx}{x^6+1}\tag{$\color{red}{❤}$}\\[7pt]
&=\frac{\pi}{3\sin\left(\frac{5\pi}{6}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to find $\frac{0}{0}$ limit without L'Hôpital's rule I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.
$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$
I would appreciate any hint... | Hint: $$\frac{1}{1+x^3} = 1 - x^{3} + x^{6} - x^{9} + O(x^{12}) \,\,\, \text{and} \,\,\, \log (1 + 3x^3) = 3x^3 - \frac{9x^6}{2} - 9x^9 + O(x^{12})$$
$$\sin (3x^2) = 3x^2 - \frac{9x^6}{2} + O(x^{10}) \,\,\, \text{and} \,\,\, \arctan (2x^2) = 2x^2 - \frac{8x^6}{3} + O(x^{10}) $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find value of $xy\sqrt{y^2 - x^2}$ for the given differential equation.
If $(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0 $ , then prove that the value of $xy\sqrt{y^2 - x^2}$ is a constant.
This is what I've tried :
$$ y(y^2 - 2x^2) dx + x(2y^2 - x^2) dy = 0 \\
\cfrac{dy}{dx} = \cfrac{(2x^2 - y^2)y}{(2y^2 - x^2)x} \\
$$
I... | When you substitute the change of variables you obtain:
$$ xu' = u \left( \frac{2-u^2}{2u^2-1} -1 \right), $$
which upon integration and taking exponential in both sides gives (I used Mathematica):
$$ x e^C = \frac{1}{u^{1/3} ({1-u^2})^{1/6}}, $$ where $e^C :=K$ is a constant of integration. Substitute back $u = y/x$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{... | In the other way round, let
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) =y$$
$\implies -\dfrac\pi2\le y\le\dfrac\pi2\ \ \ \ (1)$ and $\tan y=\dfrac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}}$
Applying Componendo and dividendo, $$\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to show algebraically that $x^3 +3x +1$ is injective?
How to show algebraically that $$x^3 +3x +1$$ is injective?
Working with the usual method of assuming that $f(c)=f(d)$ and then seeing if $c=d$. I've tried several approaches, including factoring by the difference of cubes, followed by use of the quadratic for... | Let $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^3+3x+1$, let's suppose $x$ and $y$ are real numbers, then
\begin{align*}
f(x)=f(y)\quad \iff\quad x^3+3x+1&=y^3+3y+1\\
x^3-y^3+3(x-y)&=0
\end{align*}
We can factor $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$, so
\begin{align*}
f(x)=f(y)\quad \iff(x-y)(x^2+xy+y^2+3)&=0
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$
where ... | Despite the attempts in the other answers and comments, and the two ``proofs'' given there, the inequality does not hold in general. We give two proofs for its failure:
First Proof (involving huge numbers): Set $x=3/7$, $y=4/7$, $z=37/49$. Then $xy+yz+zx=1$. For $m=7$, $n=8$ set $A=\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Solving $\cos x+\sin x-1=0$ How does one solve this equation?
$$\cos {x}+\sin {x}-1=0$$
I have no idea how to start it.
Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?
Thanks in advance!
| Given $$\color{blue}{\cos x+\sin x-1=0} $$$$\cos x+\sin x=1 $$ Divide both sides by $\color{blue}{\sqrt{2}}$ we get $$\frac{1}{\sqrt{2}}\cos x+ \frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$ $$\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}=\cos\frac{\pi}{4}$$ Using formula $\color{purple}{\cos A\cos B+\sin A\sin B=\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 1... | Think of it like this.
$a^{2}$ means $a \times a$.
More generally, $a^{n}$ means $\underbrace{a \times a \times a \times \dots \times a}_{n \text{ times}}$.
So $2^{3}$ means $2 \times 2 \times 2$. That's $2 \times 2$, $3$ times.
Therefore $(2x^{2} + x)^{2}$ means $(2x^{2} + x) \times (2x^{2} + x)$.
Now, using your numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+... | I have differentiated your answer and the book's answer. They are both equal to the integrand. So your work is indeed correct.
In fact, the book's answer can be considered wrong since it lacks an absolute value around the argument to $ln$, whereas your answer does not need one there since the argument to $ln$ in your e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where t... | $$
\left(\frac{\sqrt 2}2\right)^6 = \left(\frac{\sqrt 2}{\sqrt 2\sqrt 2}\right)^6 = \left( \frac 1 {\sqrt 2} \right)^6 = \frac 1 {\sqrt{2}^6} =\frac 1 {(\sqrt 2^2)^3} = \frac 1 {2^3} = \frac 1 8.
$$
(But you have $\dfrac 1{4\pi}$ where you need $\dfrac 1 4 \pi$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Generating Functions Dice In how many ways can we roll a red die, a yellow die, and a black die, and get a sum of $9$?
(The dice are colored so that a red $2$, a yellow $3$, and a black $4$ is distinguished from a red $3$, a yellow $4$, and a black $2$, for example.)
Should I use generating functions somehow?
| Yes. Multiply out:
$$(x+x^2+x^3+x^4+x^5+x^6)^3$$
and find the coefficient of $x^9$. Do you see why this works?
EDIT: I'll elaborate more for the sake of completeness. It is because when you multiply:
$$\color{red}{(x+x^2+x^3+x^4+x^5+x^6)}\color{yellow}{(x+x^2+x^3+x^4+x^5+x^6)}(x+x^2+x^3+x^4+x^5+x^6)$$
Sorry for the yel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show $\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$
Show by means of the substitution $u = \cosh x$, that
$$\int \dfrac{\sinh x}{\cosh 2x}=\dfrac 1 {2\sqrt 2} \ln\left|\dfrac {\sqrt 2 \cosh x-1}{\sqrt 2 \cosh x +1}\right| + C$$
$$\int \dfrac{\... | I believe your mistake is missing the negative sign here:
$$\int \dfrac{\sinh x}{\cosh 2x}= -\dfrac {\sqrt 2}{2} \tanh ^{-1} (\sqrt 2) u + C$$
In lots more steps:
$$\begin{split}
\int \dfrac{\sinh x}{\cosh 2x}dx &= \int \dfrac{\sinh x}{2\cosh^2(x) - 1}dx \\
&= \int \dfrac{1}{2u^2 - 1}du \\
&= \dfrac{1}{\sqrt2}\int \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end th... | Another approach: consider the product
$$(1-x)(1+2x)\cdots(1+(-1)^nnx)=1+a_nx+b_nx^2+\cdots\ .$$
It is easy to see that the coefficient of $x$ in this expression is
$$\eqalign{a_n
&=-1+2-3+4-\cdots+(-1)^nn\cr
&=\cases{(-1+2)+(-3+4)+\cdots+(-(n-1)+n)&if $n$ is even\cr
-1+(2-3)+(4-5)+\cdots+((n-1)-n) &if $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that there are no positive integers $a$ and $b$, such that $b^4 + b + 1 = a^4$ So I've been trying to solve this problem for a couple of days now. What I've come up with is this:
By way of contradiction, assume that there are positive integers a and b such that $b^4 + b + 1 = a^4$.
Consider the case when $b \geq ... | In any case, $b^4-a^4+b+1=0$. Also,
$$
(b^4-a^4)=(b^2-a^2)(b^2+a^2)=(b-a)(b+a)(b^2+a^2)
$$
So,
$$
(b-a)(b+a)(b^2+a^2)= -b-1
$$
So $(b^2+a^2)$ now divides $b+1$. So $b^2+a^2\leq b+1$. The only way this will work is if $a=b=1$. Check that this doesn't satisfy the original equation. So we're done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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If for a prime p $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}$ then show that p divides a. Moreover if $p>3$ then $p^2$ divides a. Here is a problem from Hersteins Topics in Algebra:
If $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}=\frac{a}{b}$ for a prime $p$, then show that $p$ divides $a$. More... | The numerator is $(p-1)!\left ( \frac 11 + \frac 12 + ... + \frac 1{p-1}\right ) \equiv (p-1)!\left ( 1^{-1} + 2^{-1} + ... + (p-1)^{-1}\right ) \mod {p^2}$.
The Euler function for $p^2$ is $p(p-1)$. Thus, for any $k \lt p $ , $k^{-1} \equiv k^{p(p-1)-1}\mod p^2$.
Thus, $k^{-1} + (p-k)^{-1} \equiv k^{p(p-1)-1} + (p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Geometry Problem Isosceles Triangle
Given this isosceles triangle, find angle AMC.
| By using Ceva's theorem
$$\frac{AE}{EB} \cdot \frac{BQ}{QC} \cdot \frac{CP}{PA} = 1$$
$$\frac{\sin\angle ACE}{\sin\angle ECB}\cdot \frac{\sin 10^\circ}{\sin 40^\circ}
\cdot \frac{\sin 20^\circ}{\sin 30^\circ}=1$$
$$\frac{\sin(80^\circ-\angle ACE)}{\sin\angle ACE}=\frac{\sin 10^\circ}{\sin 40^\circ}\cdot \frac{\sin 20^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Difficult Integration $\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}$ I am trying to solve this integral:$$\int{\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}}dx$$
My idea was to replace the second term in the denominator as $u$ and therefore have $\int{\frac{1}{u^2-u}}dx$ but it looks still complicated. O... | Since I realized that I think I'm nearly done with a solution, here's the continuation:
For the record, with that $u$-substitution, the integral becomes:
$$u=(2x-1)^\frac{1}{4} \rightarrow x=\frac{u^4+1}{2}$$
$$du=\frac{1}{4}(2x-1)^\frac{-3}{4}2dx$$
$$\int \frac{1}{u^2-u}dx$$
$$\int \frac{1}{u^2-u}\left( 2(2x-1)^\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of a ... 3D object??? (Stupid question...)
Well we can represent a point as something like $P(a,b,c)$
We can represent a line as $\dfrac{x-a}{p}=\dfrac{x-b}{q}=\dfrac{x-c}{r}$
We can also represent a plane as $ax+by+cz=d$
So can we represent a $3D$ object also???
Or do we need to have have a 4D cartesian syst... | We can represent 3D objects as a function of three variables. Here are some simple, finite geometric shapes, defined implicitly in Cartesian coordinates. The coefficients are used to scale the shape, and can be set to any value. You will end up with tall or flat prisms and pyramids by adjusting them away from their cur... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral using contour integration Here is the integral I want to evaluate:
$$\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0$$
Apparently there are limitations as to what values $a, b$ are supposed to take but let us not concern about this. Since using the sub $u =\tan \frac{x}{2}$ (the Weiersstrass sub) results ... | Your transformation to a contour integral is correct; we note that
$$
\int_0^{2\pi}\frac{dx}{a+b\cos x}=\int_0^{2\pi}\frac{dx}{a+\frac{b}{2}(e^{ix}+e^{-ix})}
$$
and set $z=e^{ix}$ (a parametrization of the unit circle) so that $dz=ie^{ix}dx=izdx\to dx=\frac{dz}{iz}$. This yields
$$
\int_0^{2\pi}\frac{dx}{a+b\cos x}=\oi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
inequalities with fraction problem x $\frac{1}{x} > \frac{2x} {x^2 +2}$ solving this inequalities:
My long solution (wrong) :
multiplying $(x^2 + 2)^2 (x)^2 \dots$ (multiplying square of each denominator, getting rid of the > or < 0)
$x (x^2 +2)^2 > 2x(x^2) ( x^2 +2)$
$x(x^4+4x^2 +4)>2x^2(x^2+2)$
$x^5+4x^3+4x>2x^4+4x^2... | Since $x^2+2>0$,
$$\frac{1}{x}>\frac{2x}{x^2+2}\iff \frac{x^2+2}{x}=x+\frac{2}{x}>2x\iff x<\frac{2}{x}$$
$$\iff \begin{cases}\begin{cases}x>0\\ x^2<2\end{cases}\\\ \ \ \text{or}\\ \begin{cases}x<0\\ x^2>2\end{cases}\end{cases}\iff \begin{cases}\begin{cases}x>0\\ -\sqrt{2}<x<\sqrt{2}\end{cases}\\\ \ \ \text{or}\\ \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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The sum $\sum_{k=0}^\infty (k-1)/2^k$ is zero I am trying to prove an equation given in the CLRS exercise book (AP GP clrs appendix A.1-4). The equation is:
$$\sum_{k=0}^\infty (k-1)/2^k=0$$
I solved the LHS but my answer is 1 whereas the RHS should be 0 Following is my solution:
Let's say S = k/2^k = 1/2 + 2/2^2 + 3/... | You forgot to add the term $-1$ in $S_1$. It is achieved at $k=0$. So, $ S-S_1 = 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there any general method for solving $(a_1+a_2+..a_n)^2=a_1^3+a_2^3+...+a_n^3$ in positive integers $a_1,a_2,...a_n$? We know the identity $(1+2+...+n)^2=1^3+2^3+...+n^3$ . So I was thinking , for given $n\in \mathbb N$ , is there any general method for solving $(a_1+a_2+..a_n)^2=a_1^3+a_2^3+...+a_n^3$ in positive ... |
If $(a_1+a_2+\cdots+a_n)^2=a_1^3+a_2^3+\cdots+a_n^3$ and $m=\max a_k$, then $m\le n^2$.
Indeed, $m^3 \le \sum_{k=1}^n a_k^3 = \left( \sum_{k=1}^n a_k \right)^2 \le (nm)^2 = n^2m^2$.
This makes your last question easy to answer:
Indeed, if $(a+b+c)^2=a^3+b^3+c^3$ then $a,b,c \le 3^2=9$.
Testing all possibilities gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Finding minimal polynomial of big blocks diagonal matrix Consider the following matrix:
\begin{bmatrix}
-3 & 1 & -1 & & & & & \\
-7 & 5 & -1 & & 0 & & & 0\\
-6 & 6 & -2 & & & & & \\
& & & 4 & 0 & 1 & & \\
& 0 & & 0 & 1 & 0 & & 0\\
& & & 0 & 0 & 4 & & \\
& & & & & & -1 & -1\\
& 0 & ... | $p(A)=0$ is equivalent to $p(A_1)=\dots=p(A_n)=0$
So it is the minimum common multiple of the blocks minimum polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are the fields $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ and $\mathbb{Q}(\sqrt[7]{16})$ equal? I have trouble with these field extensions. Is field $\mathbb{Q}(\sqrt[7]{16}+3 \sqrt[7]{8})$ equal to field $\mathbb{Q}(\sqrt[7]{16})$?
We can $\sqrt[7]{16}+3 \sqrt[7]{8}$ express as $(\sqrt[7]{2}+3)(\sqrt[7]2)^3$.
| $$(\sqrt[7]{16})^6=8\sqrt[7]8$$
This shows that $\Bbb Q[\sqrt[7]{16}]\supseteq\Bbb Q[\sqrt[7]{16}+3\sqrt[7]8]$.
Now, consider the polynomials $x^4+3x^3$ and $9x^6+2x+6$. They are coprime, so we can apply the Bezout's identity (note that $\Bbb Q[x]$ is an Euclidean domain) to guarantee that there exist some polynomials ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find series power of $F(x) =e^{-x}x^{2}$ i need help for this problem; find a power series for $F(x) \text{=}e^{-x}x^{2} $ and derivate and prove this expression $$ \sum \limits^{\infty }_{n=1}\frac{(-2)^{n+1}(n+2)}{n!} =\text{4}$$
| For $f(x) = x^{2} \, e^{-x}$
\begin{align}
f(x) = x^{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n+2}}{n!}.
\end{align}
Differentiation leads to
\begin{align}
f'(x) = (2 - x) \, x \, e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, (n+2) \, x^{n+1}}{n!}
\end{align}
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$?
What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$
One the one hand, $\sqrt{x}\to 0$, but $\displayst... | I know we already have an answer, but I would argue this way:
for all $x$ and $n\ge 2$ we have
$$
\frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}\ln{n}}{n^2(1+x)}
$$
Then
$$
\sum_{n \ge 2} \frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{\ln{n}}{n^2} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Why does $\int_0^{2\pi} (1+2\cos(x))/(5+4\cos(x))\,dx$ vanish? The standard substitution $y=\tan(x/2)$ shows that
$$ \int_0^{2\pi} \frac{1+2\cos(x)}{5+4\cos(x)}\,dx = 0. $$
What is the "real explanation" for this fact? My guess is that the "book proof" involves contour integration; is this correct? Is there an elega... | Here is the desired elementary calculus argument. Note first that
$$
\int_0^{2\pi} \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; 2\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx.
$$
Let $u = \arccos\left(-\dfrac{4+5\cos x}{5+4\cos x}\right)$. Note that $u$ decreases continuously from $\pi$ to $0$ as $x$ goes from $0$ to $\pi$. I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2... | To provide a correction to your own work I would remove the $\lim$ at first because I want to simplifies to the maximum the expression and at the last the computation, as follows:
$${1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Diagonals of parallelepiped have length $2,3,5,11$ Is it possible that the four body diagonals (not face diagonals) of a parallelepiped have lengths $2,3,5$, and $11$?
I guess the answer is no, because it is hard for one diagonal to be longer than even the sum of the remaining diagonals ($11>2+3+5$). Maybe there is a c... | Yes, there is such an inequality. Place the parallelogram with one corner at the origin and the neighboring three corners at $a$, $b$, and $c$. The remaining four corners are at $a+b$, $a+c$, $b+c$ and $a+b+c$.
The length of the body diagonals are
$$ \begin{array}{c} D_0 = |a+b+c| \\ D_1 = |a+b-c| \\ D_2 = |a+c-b| \\ D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate the limit: $\lim _{ x\to1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $ How to calculate the limit?
$$\lim_{ x\to 1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $$
Is binomial theorem needed ?
| Suppose $n$ is a positive integer first. In that case we can put $x = 1 + h$ and let $h \to 0$. The limit evaluation is as follows
\begin{align}
L &= \lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{(e^{1 + h} - e)\sin\pi(1 + h)}\notag\\
&= -\frac{1}{e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
what is the value of $\int \sin(x)\cos(x)dx$? $\frac{\sin^2(x)}{2}$ or $\frac{-\cos^2(x)}{2}$ or $\frac{-\cos(2x)}{4}$ $\int \sin(x)\cos(x)dx = \frac{\sin^2(x)}{2}$ because
$$\frac{d}{dx}\frac{\sin^2(x)}{2}=\sin(x)\frac{\sin(x)}{dx}=\sin(x)\cos(x)$$
but also
$$\frac{d}{dx}\frac{-\cos^2(x)}{2}=-\cos(x)\frac{\cos(x)}{dx}... | $\displaystyle \int \sin x\cos xdx = \displaystyle \int \dfrac{\sin 2x}{2}dx = -\dfrac{\cos 2x}{4}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can I find the remainder? How can I find the remainder when
$$(12371^{56}+34)^{28}$$
is divided by $111$.
I have tried congruences modulo $111$ but without any success.
| $111=3\cdot 37$. You can use Euler's criterion (EC) with Quadratic reciprocity.
$$\left(\frac{13}{37}\right)=\left(\frac{37}{13}\right)=\left(\frac{-2}{13}\right)=-1$$
$$12371^{56}\equiv 13^{56}\equiv 13^2\left(13^{(37-1)/2}\right)^3\stackrel{\text{EC}}\equiv 21\left(-1\right)^3\equiv 16\pmod{\! 37}$$
$$\left(12371^{56... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer. Doesn't this problem seem a little out of the box? It seems beautiful, but I don't have an idea on how to start. Calculating the value does give me an integer-$32733$. Thanks.
| Let $\theta = 20^\circ = \frac{\pi}{9}$ and let $\omega_k = \tan(k\theta)$ for $k = 0,1,\ldots 8$. The sum we want can be rewritten as
$$\tan^6(20^\circ) + \tan^6(40^\circ) + \tan^6(80^\circ) = \omega_1^6 + \omega_2^6 + \omega_4^6$$
Since
$(\cos(k\theta) + i\sin(k \theta))^9 = e^{i9k\theta} = (-1)^k$,
the $\omega_k$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
| Let $P$ be the product, then: $P = 1^2-(\omega - \omega^2)^2=1-(\omega^2-2\omega^3+\omega^4)=1-\omega^2+2-\omega^4=3-\omega^2-\omega$. Observe that $1-\omega^3= (1-\omega)(1+\omega+\omega^2) = 0$ since $1 \neq \omega$. So: $-\omega - \omega^2 = 1$, and $P = 3-(-1) = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculus 2 - $\int(\sqrt{72+36x^2}dx$ I have done this problem several times and this is the only answer i ever come to. My schools webwork gives me incorrect for my answer (answer is not simplified but it should be accepted in this format). Did i do this correctly?
Here is my work:
\begin{align}
\int \sqrt{72+36x^2}\,... |
By the above diagram, we can show that
$$
\begin{aligned}
I &=6(\tan \theta \sec \theta+\ln |\sec \theta+\tan \theta|)+c \\
&=6\left(\frac{x}{\sqrt{2}} \cdot \frac{\sqrt{x^{2}+2}}{\sqrt{2}}+\ln \left|\frac{x}{\sqrt{2}}+\frac{\sqrt{x^{2}+2}}{\sqrt{2}}\right|\right)+c \\
&=3x \sqrt{x^{2}+2}+6 \ln \left|x+\sqrt{x^{2}+2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then ad... | Maybe more direct
$$
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}
=
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^{-1}} + \frac{\omega^{3}}{1 - \omega} + \frac{\omega^{-1}}{1 - \omega^{-2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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System of equations in a,b,c,d $a,b,c,d$ are complex numbers satisfying
\begin{cases}
a+b+c+d=3 \\
a^2+ b^2+ c^2+ d^2=5 \\
a^3+ b^3+ c^3+ d^3=3 \\
a^4+ b^4+ c^4+ d^4=9
\end{cases}
Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.
| $a^2+b^2=(a+b)^2-2ab=(3-(c+d))^2-2ab \implies ab=2-3(c+d)+(c+d)^2-cd$
$a^3+b^3=(a+b)^3-3ab(a+b)=(3-(c+d))^3-3(3-(c+d))(2-3(c+d)+(c+d)^2-cd)$
$c^3+d^3=(c+d)^3-3cd(c+d)$
$u=c+d,v=cd \implies (3-u)^3-3(3-u)(2-3u+u^2-v)+u^3-3uv=3 \implies v=\dfrac{u^3-3u^2+2u+2}{2u-3} $
$a^4+b^4+c^4+d^4=(a^2+b^2)^2-2(ab)^2+((c+d)^2-2cd)^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$5^m$, where m is any natural, can be expressed as the sum of two perfect squares? Prove that for all natural $m$, $5^m$ can be expressed as the sum of two perfect squares.
Also, prove that $5^m + 2$ can be expressed as the sum of three perfect squares.
| You can prove the first statement by induction on $m$. $m = 1$: $5^1 = 5 = 1+4 = 1^2+2^2$. Assume $5^m = a^2+b^2$, we have: $5^{m+1} = (1^2+2^2)(a^2+b^2) = (a-2b)^2+(b+2a)^2$ is a quite well-known identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Why is $\max(x, x')$ equivalent to $\frac{1}{2}( x + x' + |x - x' |)$? Why is it that
$$\max(x, x') = \frac{1}{2}( x + x' + |x - x'|)$$
is true? Is it supposed to be obvious? Because it seems to come out of thin air for me. Anyway, I've verified this by plotting it in matlab, so it works, but its hardly a satisfying p... | For determining the maximum of two arguments $x$, $x'$ we need the case distinction which of the two arguments is larger.
For determining the absolute value of one argument $x$ one again needs a case distinction, here if the argument is negative or not.
So it is not that surprising that one can express the maximum in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Show a function defined by summation is increasing, another is decreasing Problem: For real numbers $x\ge1$ and $k>0$, let $f:R\rightarrow R$ and $g:R\rightarrow R$ be defined as follows.
$f(x) = -\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2}$ ,
$g(x)=\frac{1}{2x^2}-\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2... | We can investigate the monotonicity of the functions $f$ and $g$ as you did because of the following
Lemma. [Fich, 435, Th.7 at p. 438] Let $\{u_n(x)\}$ be a sequence of functions on a segment $[a,b]$ such that for each $n$ a function $u_n’(x)$ is continuous. If a series $\sum_{n=1}^\infty u_n(x)$ converges to $u(x)$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^... | We look at the equation
$$ax^2-bx(x-1)+c(x-1)^2=0.\tag{1}$$
If $a\ne 0$, then $1$ is not a solution, so Equation (1) is equivalent to
$$a\left(\frac{x}{x-1}\right)^2-b\frac{x}{x-1}+c=0.\tag{2}$$
If we put $y=-\frac{x}{x-1}$ then the solutions of (2) are the solutions of $ay^2+by+c=0$. It follows that
$$\frac{-x}{x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrating $\sqrt{1-x^2}$ without using trigonometry I am a beginning calculus student. Tonight I had a thought. Maybe I could calculate $\pi$ using integration, but no trig.
The problem is that I don't really know where to start. I thought perhaps I could use the Chain Rule for derivatives.
$~~~~~~~~~$
| $\pi$ is a transcendental number. You can't get it in a closed form without using transcendental functions. It's not a value of any algebraic expression (polynomial, rational function, root of polynomial,...) evaluated at any algebraic number. So, you either have to bring $\pi$ into it from the boundaries of the integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to solve this combinations with repetitions problem using generating functions? Find the number of solutions to :
$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$
where none of the variables can be the number $3$.
I can solve this with Inclusion-Exclusion Principle, but I really love solving this kind of problem with generati... | You can express your close-to-geometric sum as the difference between sums, i.e.
$$f(x)=\sum_{k\ge 0} x^k -x^3=\frac{1}{1-x}-x^3=\frac{1-x^3+x^4}{1-x}$$
Then we have
$$F(x)=\left(\frac{1-x^3+x^4}{1-x}\right)^5=\sum_{j\ge0}\binom{j+4}{4}x^j\cdot\sum_{a+b+c=5}\binom{5}{a,b,c}(-1)^bx^{3b+4c}$$
and equating coefficients we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I prove that $\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$ $$\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$$
I believe this is correct since I couldn't find a directional limit that won't validate this.
From what I know, I have to prove that
$$\forall\epsilon\gt 0, \exists\delta\g... | hint: $$\cos(x^2+y^2) = 1 -\dfrac{(x^2+y^2)^2}{2!}+\dfrac{(x^2+y^2)^4}{4!}-\cdots$$ we are led to the answer as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?
$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$
Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2... | Continuing from where you've reached, you can conclude that $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4}$$
where as $x \to \infty$, we have $\frac{8}{x} \to 0$. The same goes for $\frac{1}{x^2} \to 0$, and $\frac{7}{x^4} \to 0$. We also have $\frac{11x}{x^4} \to 0$, so you can rewrite the above ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the minimum value of a function. Find the minimum value of the function:
$$f(x) = \frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2 - 2}{\left(x + \frac{1}{x}\right)^3 +\left(x^3 + \frac{1}{x^3}\right)}$$
for $x>0$.
I know that this function simplifies into something a little 'nicer' than what... | The numerator of $f(x)$ is $$
\begin{align}
\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2=&\Bigg(\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\Bigg)\times
\\&\Bigg(\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\Bigg)\,.
\end{align}$$
Hence, $f(x)=\left(x+\frac{1}{x}\right)^3-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that that the following three consecutive numbers $3^{2^{10}} − 1$, $3^{2^{10}}$,$3^{2^{10}}+ 1$ are the sum of two squares? Show that the following three consecutive numbers:
$$
3^{2^{10}} − 1, 3^{2^{10}} , 3^{2^{10}} + 1
$$
can be represented as sums of two integer squares.
| It is known that the set of sums of two squares is closed under the multiplication. On the other hand, it is clear that $3^{2^{10}}$ and $3^{2^{10}} + 1$ satisfy the statement.
Let us prove that in general
$3^{2^{n}} – 1$ is a sum of two squares.
By induction, we have $3^{2^{0}} – 1=2=1+1$ so assume that $$3^{2^{n}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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