Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Trigonometric equation, find $\sin \theta $ Find $\sin \theta $ if $a$ and $c$ are constants
$$ 1-\left(c-a\tan\theta\right)^2=\frac{\sin^2\theta\cos^4\theta }{a^2-\cos^4\theta } $$
| I think this relust is ugly.let
$$t=\sin{x}$$
$$1-\left(c-a\cdot\dfrac{t}{\pm \sqrt{1-t^2}}\right)^2=\dfrac{t^2(1-t^2)^2}{a^2-(1-t^2)^2}$$
so
$$(1-t^2)-(\pm c\sqrt{1-t^2}-at)^2=\dfrac{t^2(1-t^2)^3}{a^2-(1-t^2)^2}$$
then
$$[a^2-(1-t^2)^2][(1-t^2)-(c^2(1-t^2)-\pm 2act\sqrt{1-t^2}]=t^2(1-t^2)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Probability of choosing a defective appliance Amongst $15$ appliances there are $10$ functional and $5$ defective. We choose $3$ appliances at random. What is the probability of
A) all of them being functional
B) exactly one being defective
The total number of possible choices is $15\cdot 14\cdot 13$, number of choic... | The number of ways to choose $1$ out of $5$ defective appliances is $\binom{5}{1}=5$
The number of ways to choose $2$ out of $10$ functional appliances is $\binom{10}{2}=45$
The number of ways to choose $3$ out of $15$ any type of appliances is $\binom{15}{3}=455$
Hence the probability of choosing $1$ defective and $2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the radius of convergence of the Taylor series of $f(x) = \frac{x-3}{x+2}\ln(5+x)$ at $x=0$ I need find radius of convergence for Taylor series in $x = 0$ (over $\mathbb{R}$) and find $x$'s at which series converges to $f$
$$f(x) = \frac{x-3}{x+2}\ln(5+x)$$
My solution
$\ln(5+x) = -\ln(5)\ln(1+\frac{x}{5}) = -\ln(... | There are several mistakes in your computation. First of all
$$
\log(5+x)=\log5+\log\Bigl(1+\frac x5\Bigr).
$$
The Taylos series of $\log(1+x/5)$ converges for $|x/5|<1$, that is, $-5<x<5$. Similarly the Taylor series of $1/(x+2)$ converges for $|x|<2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$ What is the value of $$\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?$$
EDIT
I bumped into this series when inserting $\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)$ and $\overrightarrow{r}=\left(\begin{array... | $$
\begin{align}
\sum\binom{2n}{n}x^n
&=\sum\frac{1}{n!}\frac{(2n)!}{n!}x^n\\
&=\sum\frac{1}{n!}2^n(2n-1)(2n-3)\cdots(3)(1)x^n\\
&=\sum\frac{1}{n!}\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\cdots\left(\frac32\right)\left(\frac12\right)(4x)^n\\
&=\sum\frac{1}{n!}\left(-{\frac{2n-1}2}\right)\left(-{\frac{2n-3}2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Inequality $\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \frac{32}{7}$ If $a,b,c,d$ are positive real numbers such that $a^2+b^2+c^2+d^2 = 1$,
Prove that: $$\frac{1}{1-abc} + \frac{1}{1-bcd} + \frac{1}{1-cda} + \frac{1}{1-dab} \le \dfrac{32}{7}$$
I saw this problem is very similar to the pr... | We make use of the inequality: $\displaystyle \sum\limits_{cyc} abc \le \frac{1}{16}\left(\sum\limits_{cyc} a\right)^3$ several nice proofs are given here.
(The cyclic sum is taken over $a,b,c,d$)
We have: $\displaystyle \sum\limits_{cyc} (abc)^2 \le \frac{1}{16}\left(\sum\limits_{cyc} a^2\right)^3 = \frac{1}{16}$
and ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Is there any formula for $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$? I need to turn this $16x^2 - 8x + 1$ to $(4x-5)(4x+3)$ and I completely lost my logic on this one.
When trying to get $D$ with $D=b^2-4ac$ formula, I get $D < 0$.
Can someone explain me how to get these $-5$ and $+3$. Should I do this in my mind or is there a... | Hint: Completing the square you have
$$16\Big[x^2 - \frac{1}{2}x\Big] +1 = 16\Big[\left(x - \frac{1}{4}\right)^2 - \frac{1}{16}\Big] +1 = 16\left(x - \frac{1}{4}\right)^2 - 1 +1 = (4x - 1)^2$$
Now think of $(4x -1)^2 - 4^2$ and use $a^2-b^2 = (a - b)(a+b)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Minimizing Sum of Reciprocals Find the minimum value, in terms of $k$ of $\frac{1}{x_1}+…+\frac{1}{x_n}$ if $x_1^2+x_2^2+…+x_n^2=n$ and $x_1+x_2+…+x_n=k$, where $\sqrt{n} < k \leq n$.
I tried the am-hm, but how to relate with the sum of squares?
| Given that $\sum\limits_{j=1}^nx_j^2=n$ and $\sum\limits_{j=1}^nx_j=k$
we want to minimize $\sum\limits_{j=1}^n\dfrac1{x_j}$.
That is, we need to find $x_j$ so that for all $\delta x_j$ where $\sum\limits_{j=1}^nx_j\,\delta x_j=0$ and $\sum\limits_{j=1}^n\delta x_j=0$, we also have $\sum\limits_{j=1}^n\dfrac{\delta x_j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Integrate using residue theorem This was a question on my complex analysis take home final. Since the semester is over and grades have been posted I believe I can post it now.
Let $a > 0$ and $b > 0$. Verify that
$$\int_{-\infty}^{\infty} \frac{dx}{e^{a x}+e^{-b x}} = \frac{\pi}{(a+b) \sin{\left (\frac{a \pi}{a+b} \ri... | Rewrite the integrand as
$$\frac{e^{\left (\frac{b-a}{2} x \right )}}{2 \cosh{\left (\frac{a+b}{2} x \right )}} $$
Thus consider the contour integral
$$\oint_C dz \frac{e^{\left (\frac{b-a}{2} z \right )}}{2 \cosh{\left (\frac{a+b}{2} z \right )}} $$
where $C$ is the rectangle with vertices $\pm R \pm i 2 \pi/(a+b)$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Value of $\lim\limits_{z \to 0}\bigl(\frac{\sin z}{z}\bigr)^{1/z^2}$ Find the value of $$\lim\limits_{z \to 0}\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$
So I took a log: $$\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ If I could expand it something like $\log(1+x)$ .. any hints ?
| You were in the good track. Starting with $$A=\left(\dfrac{\sin z}{z}\right)^{1/z^2}$$ $$\log(A)=\frac{1}{z^2}\log\left(\frac{\sin z}z\right)$$ Now, as Ian did, use the Taylor expansion $$\sin(z)=z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+O\left(z^8\right)$$ $$\frac{\sin z}z=1-\frac{z^2}{6}+\frac{z^4}{120}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How many lines bisect both perimeter and area of a 3-4-5 triangle?
How many lines exist that divide both the perimeter and the area of a triangle with sides $3$, $4$ and $5$ into half?
| If the line passes through the sides of length $3$ and $4$,
and its intersection with side $3$ is $x$ units from the acute angle on that side,
then the line cuts off a right triangle of base $3-x$ and height $3+x$.
The area of this triangle is $\frac 12 (9-x^2) \leq \frac 92.$
Setting this equal to $3$, we would have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Arithmetic Error in Calculation of the Limit of a Given Function I consider a function $f(x)$ which is equal to $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6}$
While trying to evaluate the $\lim_{x \to 6} f(x)$
It is true that $\dfrac{\frac{3}{x}-\frac{1}{2}}{x-6} = \dfrac{\frac{3}{x}}{x-6}-\dfrac{\frac{1}{2}}{x-6}$
It is also ... | it is equivalent to $\frac{\frac{6-x}{2x}}{x-6}=-\frac{x-6}{2x(x-6}$
since $\frac{3}{x}-\frac{1}{2}=\frac{6}{2x}-\frac{x}{2x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Three digit number $ABC$ with $ABC = A + B^2 + C^3$ Is there a trick for solving this problem about number of digits?
$ABC$ is a three-digit natural number, such that $ABC = A + B^2 + C^3$.
According to above equation what is $ABC$ ?
| The following is not really a trick, but at least it's a computation which can be done by hand.
First note that $10B\ge B^2$ (since $0\le B\le9$). Hence
$$100A+10B+C=A+B^2+C^3\implies 99A+C<C^3$$
and since $A\ge1$, this gives us immediately that $C\ge5$.
Now, assume that $B=0$. Then we have $100A+C=A+C^3$, but since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given state... | You could use this
$$\eqalign{
& {(\sin a + cosa)^3} = \sin {a^3} + cos{a^3} + 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr
& \Rightarrow {\sin ^3}a + co{s^3}a = {(\sin a + cosa)^3} - 3\sin a \cdot cosa\left( {\sin a + cosa} \right) \cr
& and \cr
& {(\sin a + cosa)^2} = 1 + 2\sin a \cdot cosa ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't g... | Since $x\ge0$, $\sqrt{x^2+x}\ge x$, and since $x\ge1$, $\sqrt{x^2+x}\ge \sqrt2$. Thus,
$$
\frac{x+y}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1
$$
and
$$
\frac1{\sqrt{x^2+x}+\sqrt{y^2+y}}\le\frac1{2\sqrt2}
$$
we have
$$
\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1+\frac1{2\sqrt2}\lt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving for $r$ in ${12\choose{r}}=924$ I can solve the equation $_{12}C_r=924$ fairly easily by guess and test because there are so few possible $r$ values, but is there a clean way to solve an equation of this format algebraically? I can't seem to simplify the combination expression without using a large number of c... | Use the definition $$_nC_r = \frac{n!}{r!(n-r)!}$$ Plug in $n=12$ and set everything equal to $924$. With a little algebra you'll get $$\frac{12!}{924} = r!(12-r)!$$ Then write $924 = 2^2\cdot 3 \cdot 7 \cdot 11$ which means $$\frac{12!}{924} =\require{cancel} \frac{1\cdot 2\cdot \cancel{3} \cdot \cancel{4} \cdot 5 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me... | You have terms like $(\frac{1}{2^n} + \frac{1}{3}\cdot \frac{1}{2^{n-1}} + \cdots + \frac{1}{3^n})$ for each $n$. We can collect terms in $n$ instead of $k$ - we get a series $\frac{1}{3^k} \sum_{i = 0}^\infty 2^{-i}$ for each $k$.
E.g., pick a value of $k$. For $\frac{1}{3^k}$, there exists exactly one term also conta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Limit $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$ I try to calculate $\lim \limits_{x \to 2}{\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4}}$. So, $\frac{\sqrt{x^3 - 3x^2 + 4}-x +2}{x^2 - 4} = \frac{(x-2)x}{(x+2)(x+\sqrt{(x-2)^2(x+1)}-2)}$ but I don't know what to do next.
| HINT: Take $x=t+2$ and take limit as $t\to 0$ also notice that $\sqrt{t^2}=|t|$ and than take $\lim_{t\to0^+}$ and $\lim_{t\to 0^-}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Proving $\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$ As the title,
By considering $\bigtriangleup$ABC, Prove
$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C$$
Thanks
| $$\cos^2B+\cos^2C+2\cos A\cos B\cos C =1+\cos^2B-\sin^2C+2\cos A\cos B\cos C$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $$\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
Again, $\cos(B+C)=\cos(\pi-A)=-\cos A$
Using Werner Formula,
$$2\cos A\cos B\cos C=\cos A[2\cos B\cos C]$$
$$=\cos A[\cos(B-C)+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Easiest way to calculate determinant 5x5 witx x I would like to calculate this determinant:
\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix}
| The matrix is tridiagonal, and the determinant can be computed recursively as follows.
Let $f_n$ be the determinant of the $n\times n$ submatrix in the upper left of your matrix, and let $f_0=1$. Write the matrix as
\begin{vmatrix}a_1&b_1&0&0&0\\c_1&a_2&b_2&0&0\\0&c_2&a_3&b_3&0\\0&0&c_3&a_4&b_4\\0&0&0&c_4&a_5\end{vma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Invert a $2\times 2$ Matrix containing trig functions Invert the $2\times 2$ matrix:
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
My thought was to append the $2\times 2$ identity matrix to the right of the trig matrix and use row operations to get the answer.
I have to show... | \begin{align}
\left[\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right]^{-1} & = \frac{1}{{\left|\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Does the limit $\lim\limits_{x \to\infty}x^2 \sin\frac{1}{x}$ not exist? I used $u$ substitution for the limit $$\lim_{x \to\infty}x^2 \sin\frac{1}{x}$$ and got the limit does not exist by saying $u=\frac{1}{x}$. Is this correct and if so would that mean $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2}$ and $\lim\limits... | Since $\sin\frac{1}{x}\to0$ and $x^2\to\infty$, we use l'Hôpital's theorem:
$$\lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x^2}}=\lim_{x\to\infty}\frac{-\frac{1}{x^2}\cos\frac{1}{x}}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{x}{2}\cos\frac{1}{x}=\infty$$
since $\cos\frac{1}{x}\to1$ and $\frac{x}{2}\to\infty$, as $x \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Find the general solution of the following first order differential equation $\frac{dy}{dx}=\frac{8x^3+3x^4}{y^4}$ I have multiplied both sides by $y^4$
which gives me $\frac{dy}{dx}y^4=x^3(3x+8)$
Then do I integrate both sides with respect to x?
$\int\frac{dy}{dx}yx^4dx=\int{x^3(3x+8)dx}$
Am I still on the right track... | By separation of variables, you get $$y^4 \,dy = (3x^4 + 8x^3)\,dx$$
Now, integrate each side of the equation:
$$\int y^4 \,dy = \int (3x^4 + 8x^3)\,dx$$
$$\iff \frac 15 y^5 = \frac 35x^5 + 2x^4 + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modular exponentiation How do you solve:
$$5^{{9}{^{13}}^{17}} \equiv x\pmod {11}$$
I've been trying with this but no luck.
I get to ${{9}{^{13}}^{17}} \equiv x\pmod {11}$ from $5^3 * 5^3 * 5^3 = 64 \equiv 9\pmod {11}$. But volfram alpha disagrees with all that. So just a nudge in the right direction please.
thanks!
| As an alternate way $$5^5=3125$$ $3124\equiv 0\pmod {11}$ (use the criterion for divisibility with $11$), so $$5^5\equiv 1 \pmod {11}$$ On the other hand, $9\equiv 4\pmod 5$ and $13\equiv 1\pmod 2\Rightarrow 13^{17}\equiv 1\pmod 2$, so $9^{13^{17}}\equiv 4\pmod 5$. It follows that $$5^{9^{13^{17}}}\equiv 5^4\pmod {11}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Area between two functions My question is from Apostol's Vol. 1: One-variable calculus with introduction to linear algebra textbook.
Page 94. Exercise 16. Let $f(x)=x-x^2$, $g(x)=ax$. Determine $a$ so that the region above the graph of $g$ and below the graph of $f$ has area $\frac{9}{2}.$
My attempt at a solution. I... | By graphing the two functions, we can see that the region represented by the volume is:
$$f(x) - g(x) = x-x^2 - ax$$
In order to find the points of intersection, we must set the two graphs equal to one another:
$$x-x^2 = ax \Leftarrow\Rightarrow \ x^2 + x(a - 1) = 0 \Leftarrow\Rightarrow\ x(x + a - 1) =0$$
Therefore, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to show without calculator that $\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor$ By wolfram alpha, I get
$\left\lfloor\, \log_{10}{999^{999}}\right\rfloor =\left\lfloor\, \log_{10}{999^{999}}+\log_{10}2\right\rfloor=2996$.
How to prove that $\left\lfloo... | Prove that:
$$\log_{10}999^{999}+\log_{10}2<\log_{10}1000^{999}=2997$$
In other word:
$$\log_{10}2<\log_{10}1000^{999}-\log_{10}999^{999}=\log_{10}\left(\frac{1000}{999}\right)^{999}$$
so:
$$2<\left(\frac{1000}{999}\right)^{999}=\left(1+\frac{1}{999}\right)^{999}$$
It's true by Bernoulli's inequality.
Next we should pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+... | Hint. Recall that, as $n \to +\infty$, we have
$$1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\sim \ln n$$
then
$$n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)\sim n\ln n$$
and
$$
\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)} \sim \sum \frac{1}{n \ln n},
$$ y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1119733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to show $\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\pi/2$ How to show that
$$\int_0^\infty\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi}{2}$$
The problem is that I don't know what is
$$\lim\limits_{x\to\infty}{\mathrm{arcsec}\ x}$$
| Assuming that you intended to write
$$
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx=\frac{\pi}{2}
$$
First note that
$$
\int_1^\infty \frac{1}{x\sqrt{x^2-1}}dx= \int_1^\infty \frac{x}{x^2\sqrt{x^2-1}}dx
$$
Let $u=\sqrt{x^2-1}$, then $du=\frac{x}{\sqrt{x^2-1}}dx$. So now
$$
\int_0^\infty \frac{1}{u^2+1}du
$$
$$=\lim\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How do I simplify this expression about factorization? I am trying to simplify this
$$\frac{9x^2 - x^4} {x^2 - 6x +9}$$
The solution is
$$\frac{-x^2(x +3)}{x-3} = \frac{-x^3 - 3x^2}{x-3} $$
I have done $$\frac{x^2(9-x^2)}{(x-3)(x-3)} = \frac{x^2(3-x)(3+x)}{(x-3)(x-3)} $$
but I do not find a way to simplify
How can I... | Notice that $(3-x)=-(x-3)$. Simplify accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$ Prove that if $ a,b,c $ are non-negative real numbers such that $a+b+c = 3$, then $$ abc(a^2 + b^2 + c^2) \le 3 $$
My attempt :
I tried AM-GM inequality, tried to convert it to $a+b+c$, but I think I cannot ... | Let:
$$ M_p = \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p} $$
for $p>0$ and $M_0 = \sqrt[3]{abc}$. Since $M_p$ is a log-concave and increasing function,
$$ M_0^3 M_2^2 \leq M_{\frac{4}{5}}^5 \leq M_1^{5} = 1 $$
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What bound does the Hamming bound give you for the largest possible size of a $t$-error-correcting code of length $2t + 1$? Let $\mathbb{A}$ = $\{0, 1\}$ and suppose $t$ is a positive integer. What bound does the Hamming bound give you for the largest possible size of a $t$-error correcting code of length $2t+1$?
I hav... | In the equation you gave for the Hamming bound, $n$ is the length of the code and $q$ is the size of the alphabet. In your case, $n=2t+1$ and $q=|\mathbb{A}|=2$. Substituting $q=2$ in the bound:
$$
|C|
\le \frac{2^n}{\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{t}}
= \frac{2^{n}}{\binom{2t+1}{0} + \binom{2t+1}{1} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rolling dice probability by solving inequlity I was trying to solve a problem where I have to find the probability of the sum of $\mathcal 3$ rolls of a die being less than or equal to $\mathcal 9$.
In order to solve the problem I try first to find the number of non-negative integer solutions to the following inequalit... | An alternative solution (not a direct answer to your question):
Sums equal to $3$:
*
*We have $1$ permutation of $1,1,1$
Sums equal to $4$:
*
*We have $3$ permutations of $1,1,2$
Sums equal to $5$:
*
*We have $3$ permutations of $1,1,3$
*We have $3$ permutations of $1,2,2$
Sums equal to $6$:
*
*We h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how to find $a_{50}$ from a recursive term Given $a_{n+1}=a_n+2n+3,a_1=3$
How can I find $a_{50}$?
I can compute $a_2,a_3,...,a_{50}$
But it's a long way. Is there any smart technique to compute?
Thanks.
| Other then methods above here is a general method which can work in many cases characteristic equation,but first you have to transform $a_{n+1}$ as combination of $a_k$'s without the constant terms and $n$ terms
$$a_{n+1}=a_n+2n+3\\a_n=a_{n-1}+2n+1\text{ I plugged n=n-1 into first equation}\\a_{n+1}-a_n=a_n-a_{n-1}+2\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $
$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle ... | Note that $(1+x+x^2+x^3+x^4+x^5+x^6)$ has $z, z^2,z^3,z^4,z^5,z^6$ as roots, thus f(x) should have atmost one real root (MVT)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to vari... | $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca)$
therefore $$(ab+bc+ca)=((a+b+c)^2-1)/2$$ so the value is not fixed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Proving $ab(a+b)+ac(a+c)+bc(b+c)$ is even
Prove that $\forall a,b,c\in \mathbb N: ab(a+b)+ac(a+c)+bc(b+c)$ is even
I tried to simplify the expression to something that would always yield an even number: $ (a+b+c)(ab+ac+bc)-3abc$ but that's just a sum of numbers that are divisible by $3$...
Is there a way to do this w... | $a$ and $a^2$ are either both even or both odd. So you don't change parity when you replace $a^2$ with $a$:
$$\begin{align}
ab(a+b)+bc(b+c)+ca(c+a)
&=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\\
&\equiv ab+ab+bc+bc+ac+ac\\
&= 2(ab+bc+ac)\\
\end{align}$$
which is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find coefficient of x in a generating function The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
I factored out some $x$'s, to get $x^3(1+x^2+x^3)x^4(1+x+x^3)(1+x^5+x^{10}+x^{15}+...)$and then combined the factored terms to get $... | Multiplying the first two factors you find:
$x^7+ x^8+x^9+2x^{10} + $ other monomials of degree $n>10$ and, when you multiply such polynomial with the third factor, you see that the only monomial in $x^{10}$ has coefficient $2$ since all other terms have exponents $n=7,8,9$ or $n>10$.
Hint:
The OP has changed: so for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Line integral along the curve $\gamma(t)=(4 \cos t, 4 \sin^2 t) $ Let us consider the vector field in the plane:
$$\vec{F}=\left(x \frac{e^{x^2+y^2} - e}{x^2+y^2},y\frac{e^{x^2+y^2} - e}{x^2+y^2}\right)$$
calculate the line integral along the curve defined by:
$$\gamma:
\begin{cases}
x=4 \cos t \\
y=4 \sin^2 t\\
\end{c... | Here is a direct evaluation using this link
$\begin{align}
dx&=-4\sin t \\
dy&=8\cos t \sin t \\
F&=\Big(\frac{2[\exp{(2\cos 4t +14)}-e] \cos t }{7+\cos 4t},\frac{2[\exp{(2\cos 4t +14})-e] \sin^2 t }{7+\cos 4t}\Big)
\end{align}$
Your integral will hence be
$\begin{align}I&=-\int_0^{\frac{\pi}{2}}\frac{2[\exp{(2\cos 4t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem while calculating a limit with tangens I'm having trouble calculating this limit:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x})$$
I've tried using the gact that $\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1$, but that obviously doesn't work here.
Neither does the l'Hospital's rule.
Could you g... | apply l'Hospital's rule:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x}) = \lim _{x \rightarrow 0} (\frac{tanx - x}{2x^2 \tan x}) = \lim _{x \rightarrow 0} (\frac{\frac{1}{cos^2x} - 1}{4xtan x + \frac{2x^2}{cos^2x}})
= \lim _{x \rightarrow 0} (\frac{1 - cos^2x}{2x^2 + 4xsinxcosx}) = \lim _{x \rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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A chess player, X, plays a series of games against an opponent, Y A chess player $X$ plays a series of games against an opponent $Y$. For each game, the probability that $X$ wins is $p$, independently of the results of other games. If $X$ plays 4 games against $Y$, show that the probability that this series of games co... | Ok, this is simpler ...
You want all cases with a "WW" in them. e.g. WWLL, WWWL, etc. Let "A" mean either win or lose Prob(A)=1 of course. The cases are:
WWAA: $p^2$
LWWA: $(1-p) \times p^2$
ALWW: $(1-p) \times p^2$
$= 3 p^2 -2 p^3$
$= p^2 (3 -2 p)$
Note - to avoid counting cases twice, I listed the case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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For what values of $a$, $\sum_{n=1}^{\infty}\frac{(\cos n)(\sin na)}{n}$ converges? For what values of $a$, $$\sum_{n=1}^{\infty}\dfrac{(\cos n)(\sin na)}{n}$$ converges? Every hint is appreciated.
I know that $(\cos n)(\sin na)=\dfrac{1}{2}(\sin (n+1)a+\sin (n-1)a)$.
| Assuming $a$ is real, the series converges for all $a$. It follows from the fact that $\sum_{n = 1}^\infty \sin[n(a+1)]/n$ and $\sum_{n = 1}^\infty \sin[n(a-1)]/n$ converge for all $a$. Let's consider the latter series. If $a - 1$ is an integral multiple of $2\pi$, then each summand is $0$, so the series converges to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+... | HINT : If $u,v,w$ are functions of $x$
Then $$\frac{d(uvw)}{dx}=uv\frac{dw}{dx}+uw\frac{dv}{dx}+vw\frac{du}{dx}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me ... | Your mistake is the partial fraction expansion. It should be $$\frac{x^3+2}{(x-1)^2}=x+2+\frac{3}{x-1}+\frac{3}{(x-1)^2}$$ Then the integral becomes $$x^2/2+2x+3\ln (x-1)-3(x-1)^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$? Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come t... | When you went from $$\sqrt{4-x^2} - {{2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$
to $$ {{4 - x^2 - 2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$ you slipped a sign. The last term in the numerator should be $+4x$ instead of $-4x$.
Never mind. That error just reverses a previous erroneous sign. Your real error is in your first simplifi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-... | I have notice that if $A(x^2,x),B(1,2),C(5,4)$, so $f(x)=AC-AB$.
Also $AC-AB\leq BC$. so the max of $f(x)$ will be $BC=\sqrt{20}$.
I checked and it's the right value by wolfram.
The problem is that i can't justify it.
For example if i take: $g(x)=\sqrt{(x-4)^2+(x-5)^2}-\sqrt{(x-2)^2+(x-1)^2}$
And choose: $A(x,x),B(1,2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | For the last question, you have a triangle with side lengths $1,x,\sqrt{x^2+1}$ to have tangent $x$. That triangle has cosine $\frac{1}{x^2+1}$. Hence $$\cos \arctan x=\frac{1}{\sqrt{x^2+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Proof by induction and combinations I think I am stuck on this, I am not sure if I'm going down the correct path or not. I am trying to algebraically manipulate $p(k+1)$ so I can use $p(k)$ but I am unable to do so, so I am not sure if my math is bad or I am going about it the incorrect way.
15b.) Prove by induction t... | You dont need to use mathematical induction on proving your claim. It will follow directly from the definition of the binomial coefficient.
Note that: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.
While $\binom{n}{k-1}=\frac{n!}{(k-1)!(n-(k-1))!}$ which reduces to $\frac{n!}{(k-1)!(n-k+1)!}$.
By multiplying $\frac{n-k+1}{k}$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Congruence modulo a power of two let $a \geq 3$. Let $2^a \mid c^2-1$.
By simple divisibility rules I get
$$c \equiv 1 \mod 2^a \quad \text{or} \quad c \equiv -1 \mod 2^a $$
or
$$c \equiv 1 \mod 2^{a-1},\quad c \equiv -1 \mod 2 $$
or
$$c \equiv -1 \mod 2^{a-1},\quad c \equiv 1 \mod 2 $$
Can you explain me how to... | We want to solve the congruence $(c-1)(c+1)\equiv 0\pmod{2^a}$, where $a\ge 3$. Note that $c$ must be odd, and that $c-1$ and $c+1$ are two consecutive even numbers.
Since $c-1$ and $c+1$ are consecutive even numbers, one of them is congruent to $2$ modulo $4$, and therefore only has one $2$ to contribute to the produc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$ Ok so I'm kind of struggling with this:
The question is:
"Use mathematical induction to prove that
1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"
Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2
Assuming P(k) is ... | Here's the meat of the argument:
\begin{align}
1\cdot3+2\cdot 4+\cdots+(k+1)(k+3)&\geq \frac{k^3+5k}{3}+(k+1)(k+3)\tag{ind. hyp.}\\[1em]
&= \frac{k^3+5k+3(k^2+4k+3)}{3}\tag{simplify}\\[1em]
&= \frac{k^3+3k^2+17k+9}{3}\tag{simplify}\\[1em]
&\geq \frac{k^3+3k^2+8k+6}{3}\tag{since $k\geq 1$}\\[1em]
&= \frac{(k+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solving the ODE $y+xy'=x^4 (y')^2$ I am trying to get to the solution which is $$y=c^2 +\frac{c}{x}$$
How would I go about solving this?
| $$y+x\frac{dy}{dx}=x^4\left(\frac{dy}{dx}\right)^2$$
Change of variable : $x=\frac{1}{X}$
$\frac{dy}{dx}=\frac{dy}{dX}\frac{dX}{dx}=-X^2\frac{dy}{dX}$
$$y-xX^2\frac{dy}{dX}=x^4\left(X^2\frac{dy}{dX}\right)^2$$
$$y-X\frac{dy}{dX}=\left(\frac{dy}{dX}\right)^2$$
On can see immediately that $y=aX+b$ is convenient :
$(aX+b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Evaluate $\int\frac{x^3}{(x^2+1)^3}dx$ $\int\frac{x^3}{(x^2+1)^3}dx$ what is my $z$? I know that $x=z^n$ and $n$ is the LCD of the exponents, but my lcd here is just one? so my $z$ is gonna be $z=x^2+1$? but I saw the notes of my classmate its $z^2=x^2+1$ can someone explain this thoroughly please
| If you substitute $z^2=x^2+1$ , $2zdz=2xdx$ then $$\int\frac{x^3}{(x^2+1)^3}dx =\int\frac{z^2-1}{z^6}zdz = \int(\frac{1}{z^3}-\frac{1}{z^5})dz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strong induction with Fibonacci numbers I have two equations that I have been trying to prove. The first of which is:F(n + 3) = 2F(n + 1) + F(n) for n ≥ 1.For this equation the answer is in the back of my book and the proof is as follows:1) n = 1: F(4) = 2F(2) + F(1) or 3 = 2(1) + 1, true.2) n = 2: F(5) = 2F(3) + F(2)... | The reason for having two initial cases is that the recurrence defining the Fibonacci numbers defines each of them in terms of the two preceding Fibonacci numbers. The proof by induction uses the defining recurrence $F(n)=F(n-1)+F(n-2)$, and you can’t apply it unless you know something about two consecutive Fibonacci n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Using complex analysis to evaluate $\int_0^\infty\frac{(\ln x)^3}{1+x^2}d x$ Here is my attempt:
Let $R>1>r$ and $C$ be the closed curve in $\mathbb{C}$ consists of the following pieces:
$$C_1=\{Re^{it}: t\in(0,\pi)\},\quad C_2=[r,R],\quad C_3=\{re^{it}: t\in(0,\pi)\},\quad C_4=[-R,r]$$
all curves are oriented counter... | Using the contour I used here, we have that
$$
\int_0^{\infty}\frac{\log^3(x)}{1+x^2}dx = \int_{\Gamma}f(z)dz + \int_{-\infty}^{-\epsilon}f(z)dz + \int_{\gamma}f(z)dz + \int_{\epsilon}^{\infty}f(z)dz
\tag{1}
$$
where $f(z) = \frac{\log^3(z)}{z^2 + 1}$, $\Gamma$ is large semi circle, and $\gamma$ is small semi circle. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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induction exercise - having struggles with $1^3 + 2^3 + ... + (n+1)^3 = [(n(n+1))/2]^2$ I'm trying to solve by induction that $1^3 + 2^3 + ... + (n+1)^3 = [(n(n+1))/2]^2$
However, I have a lot of trouble, and it must be said that I don't do a great deal of mathematics.
I keep getting $(k+1)^3 = (k+1)^2$ every time I ... |
Please point out where I am going wrong.
$[ (k(k+1))/2]^2 + (k+1)^3 = [ (k(k+1))/2 + (k+1) ]^2$
$(k^2(k+1)^2)/4 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^2$
You did this on the RHS:
$$\left(\frac{k(k+1)}2+(k+1)\right)^2 = \frac{k^2(k+1)^2}4 + (k+1)^2.$$
This is not correct. You have $(a+b)^2=a^2+2ab+b^2$, not $(a+b)^2=a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the ri... | $\begin{aligned}& \prod_{2 \le k \le n} \bigg(1-\frac{1}{k^2} \bigg) = \exp\bigg[\ln{\prod_{2\le k\le n}\bigg(1-\frac{1}{k^{2}}\bigg)}\bigg] = \exp\bigg[\sum_{2 \le k \le n} \ln{\bigg(\frac{k^2-1}{k^2} \bigg)}\bigg] = \exp(S_{n}). \\& \begin{aligned} & \begin{aligned} S_{n} & = \sum_{2 \le k \le n}\ln(k-1)+ \sum_{2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^... | Hint
You have $$\frac{3 x^2+17 x}{x^3+3 x^2-6 x-8}=\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$$ So, write $$f(x)=3x^2+17x-\Big(A(x-2)(x+4)+B(x+1)(x+4)+C(x+1)(x-2)\Big)=0$$ Now $$f(2)=3\times 2^2+17 \times 2-B\times(2+1)\times(2+4)=46-18B=0$$ Repeat for $f(-1)$ and $f(-4)$; you just have one linear equation at the ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you factorise $x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3$? I'm trying to factorise
$$ x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3 $$
into four linear factors. By plugging it into WolframAlpha I've learned that it's
$$-(x-y)(x-z)(y-z)(x+y+z)$$
My question is: what are the steps involved in factorising the expression? ... | Here's the way we did this when I was beginning high school: as it's a symmetric function of 3 variables, at some point we have to break the symmetry. All we meed is remarkable identities:
The expression can be rewritten as
\begin{align*}
(x^3-y^3)z&+(z^3-x^3)y+(y^3-z^3)x =(x^3-y^3)z+(z^3-y^3+y^3-x^3)y+(y^3-z^3)x\\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expressing in terms of symmetric polynomials. How to express $$a^7+b^7+c^7$$ in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?
| Hint: Expand $(a+b+c)^7=\sigma_1^7$ with Newton to find the terms you need to subtract from it to get $a^7+b^7+c^7.$ I think it's not too high of a power to do that, since its expansion has $\frac{8\cdot9}{2}=36$ terms, arguably not too many, but after all one can also compute $(a+b+c)^6$ by squaring $(a+b+c)^3$ and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Describe the image of the set $\{z:|z|<1, Im(z)>0\}$ under the mapping $w =\frac{2z-i}{2+iz}$ Describe the image of the set $\{z:|z|<1, Im(z)>0\}$ under the
mapping $w =\frac{2z-i}{2+iz}$
First I need to find the inverse which is $z=\frac{2w+i}{2-iw}$.
Now let $w=u+iv$, we have
$$z=\frac{2w+i}{2-iw}=\frac{2u+2iv+i}{2... | You have errors in your calculations of $T(-1), \space T(0)$
First verify that you can get $$T(-1)=-\frac{3}{5}-\frac{4}{5}i \quad T(0)=-\frac{i}{2}\quad T(1)=\frac{3}{5}-\frac{4}{5}i\quad T(i)=i$$
Next since you're looking for the mapping of the area between the real axis and the upper half unit disk, you first need t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle. Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.
| i will take the equilateral triangle base $ABC$ to have sides of length $2\sqrt3$ and the edges $AD= a, Bd = b, CD = c.$ set up the coordinate system so that $$A=(2,0,0),\, B=(-1, \sqrt 3, 0),\, C = (-1,-\sqrt 3, 0), D=(x,y,z) $$ we will derive a formula for $x,y,z$ in terms of $a,b,c.$
on equating the squares of leng... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\int _0^{\frac{1}{3}} \frac{e^{-x^2}}{\sqrt{1-x^2}}dx$ can someone give me a hint on how to calculate this integral?
$\int _0^{\frac{1}{3}} \frac{e^{-x^2}}{\sqrt{1-x^2}}dx$
Thanks so much!
| $\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^{\sin^{-1}\frac{1}{3}}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)$
$=\int_0^{\sin^{-1}\frac{1}{3}}e^\frac{\cos2x-1}{2}~dx$
$=e^{-\frac{1}{2}}\int_0^{2\sin^{-1}\frac{1}{3}}e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)$
$=\dfrac{e^{-\frac{1}{2}}}{2}\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method t... | Hint $\ $ By CRT the square roots mod $5$ and $13$ lift to $4$ square roots mod $65$. There is an obvious root $\,{\rm mod}\ 65\!:\ x^2\equiv -1\equiv 64\,$ if $\,x\equiv \pm8\equiv \pm (2,5)\pmod {5,13}$. The other pair $\pm (-2,5),\,$ arise by multiplying the first pair by $\,(-1,1)\equiv 14,\,$ i.e. $\,\pm 8\cdot 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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About the series $\sum_{n\geq 0}\frac{1}{(2n+1)^2+k}$ and the digamma function Let we provide a closed form for
$$ S_k = \sum_{n\geq 0}\frac{1}{(2n+1)^2+k} $$
for $k>0$ in terms of elementary functions. It is quite easy to check that $S_k$ can be computed in terms of the digamma function $\psi(x)$, but it is also true... | Your result just confirms already known special values of the digamma function.
One may recall that the digamma function admits the following series representation
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{k=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What's wrong in this integral Where's the mistake in this solution? $$\int \tan^3x\sec^2xdx = \int \frac{\sin^3x}{\cos^5x}dx=\int\frac{\sin x(1-\cos²x)}{\cos^5x}dx=\int\frac{u^2-1}{u^5}du$$$$=\frac{1}{4u^4}-\frac{1}{2u^2}+C=\frac{1}{4\cos^4x}-\frac{1}{2\cos^2x}+C$$ for $ \cos x=u \to du=-\sin xdx$. I also tried doing $... | \begin{align}
\frac{1}{4\cos^4x}-\frac{1}{2\cos^2x} +\frac{1}{4}
&= \frac{1-2\cos^2 x+\cos^4x}{4\cos^4 x}\\
&= \frac{(\cos^2x-1)^2}{4\cos^4x}\\
&= \frac{(-\sin^2x )^2}{4\cos^4x}\\
&= \frac{\tan^4x}{4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all solutions of the following linear congruence $2x+3y≡1\pmod 7$ I've never done one with two variables, I understand that $(2, 3, 7) = 1$ and $1|1$ so there are $1*7$ solutions, and $x=0,1,2,3,4,5,6$. I have no idea how to proceed from here, and any help would be greatly appreciated!
| \begin{align}2x + 3y - 1 = 7k &\Rightarrow 3y = -2x + 1 - 7k = -3(x+2k) + x + 1 - k\\ &\Rightarrow y = -x - 2k + \dfrac{1+x-k}{3}
\end{align}
Since
\begin{align}
y \in \mathbb{Z} &\Rightarrow 3\mid (1+x-k)\\ &\Rightarrow 1+x-k = 3n \\&\Rightarrow x = k-1 +3n\\ &\Rightarrow y = -(k-1+3n) - 2k + n = -3k + 4n + 1$,
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Bounds for $\log(1-x)$ I would like to show the following
$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequa... | $$\frac{d}{dx}\left(\log(1-x)+x+x^2\right) = 1+2x-\frac{1}{1-x}=\frac{x(1-2x)}{1-x} $$
is a non-negative function on $\left[0,\frac{1}{2}\right]$, hence the LHS-inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Putnam 2005 A1 Solution
Show that every positive integer is a sum of one or more numbers of the form $2^r3^s,$ where $r$ and $s$ are nonnegative integers and no summand divides another.
(For example, $23=9+8+6.)$
Suppose for $k = \{1, 2, 3... ,n-1\}$ this equality holds.
Let $n$ be an even integer. Since $\frac{n}... | Your start is right, but odd numbers that are not prime are not all divisible by $3$.
Rather, you need to first show that if $n$ is divisible by $3$ or $2$, then you can proceed inductively. That is what you have (sort of) done already.
If $n$ is not divisible by $3$ or $2$, pick the largest, $k$, such that $n-2^k$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sum_{n=1}^\infty\frac{\cos({2nt})}{n2^n} =$? for any $t\in\Bbb R$ Given that $\sum\limits_{n=1}^\infty\dfrac{z^{2n}}{n2^n} = -\log\left(1 - \frac{z^2}{2}\right)$ ,
could you calculate the sum in the title for every real $t$ ?
| Since $$\cos(2nt) = \frac{e^{2nit} + e^{-2nit}}{2}$$ we have
\begin{align}\sum_{n = 1}^\infty \frac{\cos(2nt)}{n2^n} &= \frac{1}{2}\left(\sum_{n = 1}^\infty \frac{(e^{it})^{2n}}{n2^n} + \frac{(e^{-it})^{2n}}{n2^n}\right)\\
&= -\frac{1}{2}\left(\log\left(1 - \frac{e^{2it}}{2}\right) + \log\left(1 - \frac{e^{-2it}}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Is it generally true that $\arcsin \theta + \arccos \theta = \frac{\pi}{2}$?
Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)
How can one verify (1) when it is not generally true? We can rewrite (1) as
Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.
What about $\sin \frac{2\pi}{3}$ and $... | here is a geometric argument to show that $$\sin^{-1}(b) + \cos^{-1}b = \pi/2 \tag 1$$ using the unit circle. i will make may argument for the case $0 \le b \le 1.$ the same should go through for $-1 \le b \le 0.$
we will interpret $\sin^{-1} b$ as the arc length $t$ between $-\pi/2$ and $\pi/2$ such that $\sin t$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Application of Taylor's theorem Use Taylor's theorem to prove the estimate
$${|\sin(x)-x+x^3/6|} \le C|x|^5$$
for a suitable constant C. Hence prove that $$\lim_{x \to 0} \frac{\sin(x)-x}{x^3}$$ exists and determine its value.
I used Taylor's theorem with remainder in Lagrange form and got $$\left |\sin(x)-x+\frac{x^3}... | do integration by parts quite a few times staring with
$$\begin{align} \sin x &= \int_0^x d \sin t = \int_0^x \cos t\, d(t-x)\\
&= (t-x)\cos t\big|_0^x + \int_0^x (t-x) \sin t \, dt\\
&=x + \frac 12 (t-x)^2 \sin t\big|_0^x - \frac 12 \int_0^x (t-x)^2 \cos t\, dt\\
&=x - \frac 16 (t-x)^3 \cos t\big|_0^x - \frac 16\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?
Theorem. $\forall x\geq8, x$ can be represented by $5a + 3b$ where $a,b \in \mathbb{N}$.
Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot... | I'm attempting to answer the question intuitively:
Every third number is a multiple of 3, and can definitely be written as a sum of 3s and 5s:
$3x$ + $5$ x $0$.
Now we need to deal with numbers $y$ for which $y mod 3 = 1$ and $y mod 3 = 2$
For $y mod 3 = 2$:
We know that $5 mod 3 = 2$, so 5 x 1 + 3$x$ should give us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 12
} |
Integrate $\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$
$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$
What I did was trig substitution:
$$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$
Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}... | Substitute $y=3x$
$$I= \int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{1}{x^5\sqrt{9x^2-1}}dx = 81\int^2_{\sqrt2} \frac{1}{y^5\sqrt{y^2-1}}dy
$$
and then apply the reduction formula
$$\int \frac{1}{y^n\sqrt{y^2-1}}dy=I_n=\frac{\sqrt{y^2-1}}{(n-1)y^{n-1}}+\frac{n-2}{n-1}I_{n-2}
$$
to obtain
$$I=81\left( \frac{7\sqrt3-16}{64} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $(a^2+ab+b^2)/(a+\sqrt{ab}+b)$ How can I simplify as much as possible:
$$\frac{a^2+ab+b^2}{a+\sqrt{ab}+b}$$
Also, first post here, looking forward to sticking around!
| You could multiply by the conjugate of the denominator, then cancel.
\begin{align*}
\frac{a^2 + ab + b^2}{a + \sqrt{ab} + b} & = \frac{a^2 + ab + b^2}{a + b + \sqrt{ab}} \cdot \frac{a + b - \sqrt{ab}}{a + b - \sqrt{ab}}\\
& = \frac{(a^2 + ab + b^2)(a + b - \sqrt{ab})}{(a + b)^2 - ab}\\
& = \frac{(a^2 + ab + b^2)(a + b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190659",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
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Finding the sum of $\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}$
Find the sum of the series and for which values of $x$ does it converge:
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n} $$
My attempt:
$$\begin{align}&S_n=(x+1)^2+(x+1)^3/3+...+(x+1)^{n+1}/3^{n} \\
-(x+1)^2&S_n=-(x+1)^3/3-...-(x+1)^{n+2}/3^{n} \\
&S_... | $$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}=(x+1)^2\sum_{n=0}^{\infty}\Big(\frac {x+1}{3}\Big)^n=(x+1)^2\frac{1}{1-\frac{x+1}{3}}=\frac{3(x+1)^2}{2-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac{\log(1+a/x)}{\log(1+1/x)}$ is increasing, for $a\geq 1$? I need to prove that the sequence
$$
\frac{\log\big(1+\frac{a}{k}\big)}{\log\big(1+\frac{1}{k}\big)}
$$
is increasing for any $a\geq 1$, so that I thought of defining
$$
f(x)=\frac{\log\big(1+\frac{a}{x}\big)}{\log\big(1+\frac{1}{x}\big... | For later use, consider the function $\phi(x) = x \log x$, for which $\phi'(x) = 1 + \log x$ and $\phi''(x) = \frac{1}{x}$; in particular, $\phi$ is convex on $[1, \infty)$.
If
$$
f(x) = \frac{\log\bigl(1 + \frac{a}{x}\bigr)}{\log\bigl(1 + \frac{1}{x}\bigr)},
$$
then
$$
f'(x) = \frac{(x + a) \log\bigl(1 + \frac{a}{x}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
| $$ \lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )=\lim _{x \to \infty } (\sqrt{x^2 +x} - \sqrt{x^2 +1} )\frac{\sqrt{x^2 +x} + \sqrt{x^2 +1}}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=\lim _{x \to \infty } \frac{x-1}{\sqrt{x^2 +x} + \sqrt{x^2 +1}}=...=1/2$$
(if you need more I could finish it)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When a function contains a sequence, and how to find the function's limit? Suppose $\lim \limits_{n \to \infty} a_n = 0$. Find the limit
$$\lim \limits_{n \to \infty} \left(1+a_n \frac{x}{n}\right)^n$$
It's kind intuitive that the answer is 1, but clearly I can't just say that the limits equal $\lim \limits_{n \to \inf... | Hint: Supose $\lim_{n\to \infty}n\cdot\frac{1}{ x\cdot a_n}=0$. Note that $\lim_{n\to \infty} x\cdot a_n=0$,
$$
\left(
1+a_n\frac{x}{n}
\right)^{n}
=
\left(
1+\frac{1}{n\cdot\frac{1}{ x\cdot a_n}}
\right)^n
=
\left[
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Continued Fraction for Root 5 How can I find the continued fraction expansion for the square root of 5. Do this without the use of a calculator and show all the steps.
| We can use the
$$x^2-5=0$$
$$x^2=5$$
$$x^2+x=5+x$$
$$x(x+1)=5+x$$
$$x=\frac{x+5}{x+1}$$
or
$$x=1+\frac{4}{1+x}$$
$$x=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
$$\sqrt{5}=1\frac{4}{2\frac{4}{2\frac{4}{2\frac{4}{2.....}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find conditions on positive integers so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational
Find conditions on positive integers
$a, b, c$
so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational.
My solution:
if $ab$ is not the square of an integer,
then the expression is irrational.
I find it interesting
that $c$ does ... | The answer is that if any of $a, b, c$ is not a square of an integer, then
$\sqrt{a} +\sqrt{b} +\sqrt{c}$ must be irrational.
The proof of the general case is not very easy. The paper
Square roots have no unexpected linear relationships by Qiaochu Yuan at
https://qchu.wordpress.com/2009/07/02/square-roots-have-no-un... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A quick way to prove the inequality $\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$ Can anyone suggest a quick way to prove this inequality?
$$\frac{\sqrt x+\sqrt y}{2}\le \sqrt{\frac{x+y}{2}}$$
| $2ab \leq a^2+b^2$ so $\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}$. Now choose $a = \sqrt{x},b = \sqrt{y}$.
| {
"language": "en",
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Find rational points on $x^2 + y^2 = 3$ and on $x^2 + y^2 = 17$ $(a)$ Find all rational points on the circle $x^2 + y^2 = 3$, if there are any. If there is none, prove so.
$(b)$ Find all rational points on the circle $x^2 + y^2 = 17$, if there are any. If there is none, prove so.
I'm not sure how proceed with finding ... | As shown in this answer, $n$ can be written as the sum of two squares if and only if, in the prime factorization of $n$, each prime that is $\equiv3\pmod4$ appears with even exponent.
If $x^z+y^2=3z^2$, then $3$ appears with odd exponent. Thus, there are no rational solutions of
$$
\left(\frac xz\right)^2+\left(\frac y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^... | $$3k^2 + 3k + 12=3(k^2 + k +4)= 3(k(k+1)+4)$$
Can you see why $k(k+1)$ and $4$ are each divisible by $2$?
At least one of $k, k+1$ is even, as is $4$, hence $2$ divides $(k(k+1)+4)$, and with three as a factor of $\color{blue}{3}(k(k+1)+ 4)$, we have $$2\cdot 3 = 6\mid (3k^2 + 3k + 12).$$
| {
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Differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.
*
*Show that the equation has $0$ as its singular regular point .
*Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (... | You have already done that third part, notice that if $a = n \in \mathbb{N}$ then $a_{i} = 0$ for all $i \ge n+1$. So you have a polynomial of degree $n$ as a solution.
For the last part, just some ideas - try differentiating the Leguerre polynomial and substituting it into the equation. Or, conversely, try to expand ... | {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to fi... | You should have $(x^2+px+1)(\lambda x + D) = \lambda x^3 + (p\lambda + D) x^2 + (pD + \lambda) x + D$. thus $D = c$ and $\lambda = a$, and you need
$p \lambda + D = ap + c = 0$ and $pD + \lambda = cp + a = b$. Eliminating $p$ from these two gives you $0 = c (ap + c) - a(cp + a - b) = \ldots$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots ... | Indeed, if $\delta=\frac 1m$ then
$$ \frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2=\frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.$$
However, there is no such thing as "smalles possible" and from that point on you get some factros wrong.
What you really get in the ... | {
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"timestamp": "2023-03-29T00:00:00",
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Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
| HINT:
$$[(x+1)^2-1]+(x+2)^3+(x+3)^4-1^4$$
$$=x(x+2)+(x+2)^3+[\{(x+3)^2-1\}\{(x+3)^2+1\}]$$
$$=x(x+2)+(x+2)^3+[(x+4)(x+2)\{(x+3)^2+1\}]$$
and
$$[(x+1)^2-3^2]+(x+2)^3+2^3+(x+3)^4-1^4$$
$$=(x+4)(x-2)+\{(x+2)+2\}[(x+2)^2-2(x+2)+2^2]+[(x+4)(x+2)\{(x+3)^2+1\}]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all points on a surface which have a tangent plane parallel to given plane - is my method correct? The question given is to find all points on the surface given by $x^3 - y^3 - 2xy - z = 0$ which have a tangent plane which is parallel to $6x - 6y - z = 0$.
So, I found the two gradient vectors (which we defined as... | Given the surface
$$z = f(x,y) = {x^3} - {y^3} - 2xy$$
we have a parametrization:
$$\phi (x,y) = (x,y,{x^3} - {y^3} - 2xy)$$
with partials:
$$\begin{gathered}
{\partial _x}\phi = \left( {1,0,3{x^2} - 2y} \right) \hfill \\
{\partial _y}\phi = \left( {0,1, - 3{y^2} - 2x} \right) \hfill \\
\end{gathered} $$
and an ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int\frac{dx}{(x^3-1)^2}$ Please help. I do not know what to do. You can just show the direction where to go and I continue. Here it is: $$\int\frac{dx}{(x^3-1)^2}$$
| Let $$\displaystyle I = \int\frac{1}{(x^3-1)^2}dx = \frac{1}{3}\int\frac{1}{x^2}\cdot \frac{3x^2}{(x^3-1)^2}dx$$
Now Using Integration by parts, we get
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{1}{(x^3-1)}-\frac{2}{3}\int \left[\frac{1}{x^3\cdot (x^3-1)}\right]dx$$
$$\displaystyle I = - \frac{1}{3x^2}\cdot \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and ... | We have:
$$0\leq\frac{x^2y^2}{x^2+y^2}\leq \frac{x^2y^2}{y^2}=x^2$$
Since $f(x,y)$ is between $0$ and $x^2\rightarrow 0$, the function is arbitrarily close to zero, when the distance between $(x,y)$ and $(0,0)$ is sufficiently small. It follows that the limit of $f$ is $0$, just to limit definition.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How many $5$ digit natural numbers have at least one $5$ or one $7$? There are $9 \cdot 10^4=90,000$ possible $5$ digit numbers.
Numbers with no $5$ is $8\cdot 9\cdot 9\cdot 9\cdot 9 = 52,488$.
Numbers with no $7$ is $52,488$.
Numbers with at least one $5$ is $90,000 - 52,488$
Similarly numbers with at least one $7$ i... | Hint: Count the number of five digit-numbers that contain NO $5$'s and no $7$'s; that gives us $7\cdot 8^4$ such five-digit numbers.
The remainder ($90000 - 7\cdot 8^4$) will be the number of five digit numbers with at least one $5$ OR one $7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$ $(2^m - 1, 2^n - 1) = 2^d - 1$ where $d = (m,n)$
my work:
I assumed $m = da$ , $n = db$ for $a,b \in \mathbb{Z}$.
Now, $2^m - 1$ = $2^{da} - 1$ = $(2^d)^a - 1$ = $x^a - 1$ where $x = 2^d$.
similarly
$2^n - 1$ = $x^b - 1$
Now,
using $x^a - 1 = (x - 1)(x^{a-1... | One way to deal with your particular step is to show that $\gcd(x^{a} -1, x^{b} - 1) = x - 1$, when $\gcd(a, b) = 1$. (Actually, a slight variation on the same argument would yield a proof of your original statement, but still.)
Assume $a \ge b \ge 0$, and proceed by induction on $b$. The case $b = 0$ being obvious, as... | {
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$x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that $\sin \tfrac{2\pi x_1}{3}+\sin \tfrac{2\pi x_3}{3}=2\sin \tfrac{2\pi x_2}{3}$. Find $a$. $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that
$\sin \dfrac{2\pi x_1}{3}+\sin \dfrac{2\pi x_3}{3}=2\sin \dfrac{2\pi x_2}{3}$.
Find $a$
(Bulgari 1998)
| Using Prosthaphaeresis Formula,
$\sin\dfrac{2\pi x_1}3+\sin \dfrac{2\pi x_3}3=2\sin\dfrac{\pi(x_1+x_3)}3\cos\dfrac{\pi(x_1-x_3)}3$
As $x_1+x_2+x_3=3,$
$\sin\dfrac{\pi(x_1+x_3)}3=\sin\dfrac{\pi(3-x_2)}3=\sin\dfrac{\pi x_2}3$
So, we have $2\sin\dfrac{\pi x_2}3\left[\cos\dfrac{\pi(x_1-x_3)}3-2\cos\dfrac{\pi(x_2)}3\right]=... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving equation with rational exponent I have this equation: $$\mathrm{r} (x-1) = x^{8/9} - x^{1/9}$$
where $\mathrm{r}$ is a constant. Is there a general technique to solve such equations? Raising it to the 9nth power: $$\mathrm{r}^9 (x-1)^9 = (x^{8/9} - x^{1/9})^{9} \\ \mathrm{r}^9 (x-1)^9 = x (x^{7/9} - 1)^9\\$$ s... | First make the substitution $y=x^{1/9}$ the equation becomes
$$r(y^9-1)=y^8-y$$
We can factor out from both sides $y-1$, giving us one solution $y=1$, and therefore the solution $x=1$. What remains is
$$r(y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=y(y^6+y^5+y^4+y^3+y^2+y+1)$$
or
$$ry^8+(r-1)\sum_{k=1}^{7}y^k+r=0$$
Observe that t... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving the diophantine equation $p^2+n-3=6^n+n^6$
What are the pairs ($p,n$) of non-negative integers where $p$ is a prime number, such that $$p^2+n-3=6^n+n^6$$
How can I solve this diophantine equation?
| If $n \equiv 0,1\pmod3$, we have $3$ divides $6^n+n^6 - n+3$, except for $n=0$. Further, for $n=1$, we have $6^n+n^6 - n+3 = 9 = 3^2$. Hence, $n \equiv 2\pmod3$. However, if $n \equiv 2 \pmod3$, we have $$6^n+n^6 - n+3 \equiv 2\pmod3$$ and no square is $2 \pmod 3$. Hence, the solutions are $(n,p) = (0,2)$ and $(n,p) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$z^n=(z+1)^n=1$, show that $n$ is divisible by $6$. we are given $z^n=(z+1)^n=1$, $z$ complex number. we want to prove that $n$ is divisible by $6$. I showed that $|z|=|z+1|=1$. Hence $z$ is on the intersection of two unit circles, one centered at $(0,0)$ and the other one centered at $(-1,0)$. Then $z$ can take two va... | $z^n=1\implies z=\cos\tfrac{2\pi a}n+i\sin\tfrac{2\pi a}n$ where $a\equiv0,1,\cdots,n-1\pmod n$
$(z+1)^n=1\implies z+1=\cos\tfrac{2\pi b}n+i\sin\tfrac{2\pi b}n$ where $b\equiv0,1,\cdots,n-1\pmod n$
Equating the imaginary parts, $\sin\tfrac{2\pi a}n=\sin\tfrac{2\pi b}n$
$\implies\cos\tfrac{2\pi a}n=\pm\cos\tfrac{2\pi b}... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving a Diophantine equation: $p^n+144=m^2$ I found this Diophantine equation:
$$p^n+144=m^2$$
where $m$ and $n$ are integers and $p$ is a prime number. I solved it but I want to know if there exist other proofs through the use of rules of modular arithmetic.
This is my solution:
$$p^n+144=m^2$$
$$p^n=(m+12)(m-12)$$
... | This gives us $m+12=p^a$ and $m-12 = p^b$. This means $p^a-p^b = 24$, i.e., $p^b(p^{a-b}-1) = 24 = 2^3 \cdot 3$. Note that $p^b$ and $p^{a-b}-1$ are of opposite parity.
*
*If $b=0$, we need $p^a-1 = 24 \implies p^a = 25 \implies p = 5,a=2$. Hence, $m=13$.
*If $b>0$, we need $p^b \mid 24$. Hence, $p=2$ or $p=3$.
*
... | {
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Easy inequality going wrong
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$
My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $... | Your only mistake is right at the end. Your polynomial changes sign at the points $-2,-\frac54, -1, 1, 5$, meaning that it is:
*
*Negative on $(-\infty, -2)$
*Positive on $(-2,-\frac54)$
*Negative on $(\frac54, -1)$
*Positive on $(-1,1)$
*Negative on $(1, 5)$
*Positive on $(5,\infty)$
As for the precise value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $2^{105} + 3^{105}$ is divisible by $7$ I know that $$\frac{(ak \pm 1)^n}{a}$$ gives remainder $a - 1$ is n is odd or $1$ is n is even.
So, I wrote $ 2^{105} + 3^{105}$ as $8^{35} + 27^{35}$ and then as $(7\cdot 1+1)^{35} + (7\cdot 4-1)^{35}$, which on division should give remainder of $6$ for each term and t... | You can write $2^{105} = 2^{6 \cdot 17}\cdot 8$ and $3^{105} = 3^{6 \cdot 17} \cdot 3^3$ and you can use Fermat's little theorem.
$2^{6 \cdot 17} \equiv 1 \pmod 7$, $8 \equiv 1 \pmod 7$ therefore $2^{2015} \equiv 1 \pmod 7$ while $3^{6 \cdot 17} \equiv 1 \pmod 7$ and $3^3 \equiv 6 \pmod 7$ therefore $3^{105} \equiv 6 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx ... | Since $x^2+x-2 = (x-1)(x+2)$, we can use partial fraction decomposition:
$$\frac{x-2}{x^2+x-2} = \frac{A}{x-1} + \frac{B}{x+2}.$$
Then $x-2 = A(x+2) + B(x-1)$.
Equating the coefficients of each power of $x$, we have that $1=A+B$, and $-2=2A-B$.
Solving for $A$ and $B$ decomposes our original rational expression into tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Identity in Ramanujan style Is it possible to represent
$$ \sqrt[3] {7\sqrt[3]{20}-1} =\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$ with rational $A,\,B,$ and $C?$
|
Yes.
$$ \sqrt[3] {-1+7\sqrt[3]{20}} =\sqrt[3]{\frac{16}{9}}+\sqrt[3]{\frac{100}{9}}-\sqrt[3]{\frac{5}{9}}\tag0$$
Solution: More generally, given the three roots $x_i$ of any cubic equation,
$$x^3+ax^2+bx+c=0\tag1$$
then sums involving the cube roots of the $x_i$ can be given in the simple form,
$$(u+x_1)^{1/3}+(u+x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality? $(a+b+c=1)$ Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold:
$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$,... | Firstly, multiply the right-hand side by $a+b+c$, so that both sides are now degree-zero. Then you want to prove
$$2a^3+2b^3+2c^3\geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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Fractions in Questions and Answers
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