Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to prove that $42|a^7-a$? Suppose we are given a number $a \in \mathbb{Z}$ prove that $42|a^7-a$.
I'm not too sure how to start any ideas?
| In fact, $42$ is the largest integer that divides $a^7-a$ for all $a\in\Bbb Z$.
$a^7-a=a\left(a^6-1\right)=a\left(a^3-1\right)\left(a^3+1\right)$
$=a(a-1)\left(a^2+a+1\right)(a+1)\left(a^2-a+1\right)$
$a-1,a,a+1$ are three consecutive integers, so $2,3\mid a^7-a$.
Also by Fermat's Little theorem $7\mid a^7-a$. Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Sequence from generating function $\frac{1}{(1 - \frac{x}{3})^2}$ I know that for
$$ \frac{1}{1 - \frac{x}{3}} $$
sequence would be $a_n = \frac{1}{3^n}$ for $n \geq 0$ ($\sum_{n \geq 0} \frac{1}{3^n} x^n$ ).
How should I approach with that power of two?
| $$\frac{1}{(1-\frac{x}{3})^2}=\frac{1}{1-\frac{x}{3}}\cdot\frac{1}{1-\frac{x}{3}}$$
$$=\sum\limits_{n=0}^\infty(\frac{x}{3})^n\cdot\sum\limits_{n=0}^\infty(\frac{x}{3})^n $$
$$=(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots)(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots).$$
Now we want a single sum in the form of $\sum a_nx^n$, and so w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim_{x\to0}\frac{e^x-1-x}{x^2}$ using only rules of algebra of limits. I would like to solve that limit solved using only rules of algebra of limits.
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
All the answers in How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? do... | $$\lim_{x\to0} \dfrac{e^x-1-x}{x^2}= \lim_{x\to0} \dfrac{e^{-x}-1+x}{x^2}= \lim_{x\to0} \dfrac{e^x+e^{-x}-2}{2x^2}=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-e^{\frac{-x}{2}}}{x}\Bigg)^2$$
$$=\displaystyle \dfrac{1}{2} \lim_{x\to0} \Bigg(\dfrac{e^{\frac{x}{2}}-1}{x}-\dfrac{e^{\frac{-x}{2}}-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 5
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Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give ... | All of the other approaches here are based on guessing that the expressions under the cube root signs are perfect cubes.
Here's an approach that doesn't assume that.
Let $$a = \sqrt[3]{7 + \sqrt{50}} = \sqrt[3]{7 + 5\sqrt{2}}, \quad b = \sqrt[3]{7 - 5\sqrt{2}}.$$
We are trying to evaluate $s = a + b$. We have
$$a^3 = 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The number of ways of writing an integer as a sum of two squares Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$.
It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be writte... | I think we can always translate everything to the basics operations but it will take a huge number of pages! Fortunately, your question has an elementary answer. We denote by $S_n$ the set of representation of the integer $n$ as sum of squares, we have $(a,b)\in S_n$ if and only if $0\leq b<a$ and $ a^2+b^2=n$;
Given t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $a_{n+1} - a_n = n^2$ using generating functions The Full Question
Using the method of generating functions, solve $a_{n+1} - a_n = n^2$ where $a_0 = 1$
My Research
Scanned the website for similar answers, reviewed the following links:
Solve the following recurrences using generating functions.
can we use generat... | Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$. Take the recurrence, multiply by $z^n$, sum over $n \ge 0$:
$$
\sum_{n \ge 0} a_{n + 1} z^n - \sum_{n \ge 0} a_n z^n = \sum_{n \ge 0} n^2 z^n
$$
Recognize some sums, note that:
$\begin{align}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to show $\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$ How to show
$$\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$$
| Stahl's approach basically works. A walk-through solution is also possible: Notice that the digamma function $\psi(z)$ satisfies
$$ \psi(z) = -\gamma+\sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z}\right). $$
Consequently
\begin{align*}
\sum_{n=0}^{\infty} \frac{1}{(24n+5)(24n+19)}
&= \frac{1}{14} \sum_{n=0}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$
Here's what ... | If you're only allowed to use $x+y\geq 2\sqrt{xy}$ (for $x,y\geq 0$), then
$$
1=a+b\geq2\sqrt{ab}\implies ab\leq\frac{1}{4}
$$
and
$$
(1+1/a)(1+1/b)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\geq 1+\frac{2}{\sqrt{ab}}+\frac{1}{ab}\geq1+\frac{2}{\sqrt{1/4}}+\frac{1}{1/4}=9.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Strange things happening with definite integration by substitution There are two ways solving the definite integration by substitution. One is to solve the indefinite integration first, then use the evaluation theorem. Another is to use the substitution rule for definite integral. However, when I solved some problems,... | You got the wrong answer.
Use the method similar to partial fractions,
$$\int \frac{dx}{(2+\cos{x})(3+\cos{x})}
=\int \left(\frac{1}{2+\cos{x}}-\frac{1}{3+\cos{x}}\right)dx.$$
Draw the following figure,
we have
$\cos{\frac{x}{2}}=\frac{1}{\sqrt{1+u^2}}$ and
$\sin{\frac{x}{2}}=\frac{u}{\sqrt{1+u^2}}$.
Then $\sin{x}=2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\s... | The $n$ $n$th roots of a nonzero complex number $z=r(\cos \theta+i\sin\theta)$ are given by
$$
w_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right],\tag{1}
$$
where $k=0,1,2,\ldots,n-1$.
Problem: Find the four fourth roots of $z=1+i$.
Solution. We have, in thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: i... | Use the universal substitution $t=\tan\frac{x}{2}$, we have
$$\sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad dx=\frac{2}{1+t^2}dt$$
So
$$\frac{1}{5\cos x+\sin x+7}dx=\frac{1}{\frac{5(1-t^2)+2t}{1+t^2}+7}\frac{2dt}{1+t^2}=\frac{2}{2t^2+2t+12}dt=\frac{1}{t^2+t+6}dt=\frac{1}{\left(t+\frac{1}{2}\right)^2+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all nonnegative integers
Determine all nonnegative integers $x$ and $y$ so that $$3^x + 7^y$$
is a perfect
square and $y$ is even.
Without trial-and-error of course.
$$3^x + 7^y = a^2$$ For some integer $a$.
$$\implies a = \sqrt{3^x + 7^y}$$
But is there really a way?
| Writing $y = 2z$ we have $3^x = (a + 7^z)(a - 7^z )$. Now $a + 7^x$ and $a - 7^x$ must be two powers of $3$, say $a + 7^x = 3^d$, $a - 7^x = 3^e$.
But $3^d - 3^e = 2 \times 7^x \equiv 2 \mod 3$ so we must have $e = 0$. Then
$a = 1 + 7^x$, and $3^d = a + 7^x = 1 + 2 \times 7^x$. One solution is $x = 0$, $d = 1$, cor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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show this $\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$ let $a_{1},b_{1},a_{2},b_{2}$ be no-zero integer,and such $\gcd{(a_{i},b_{i})}=1,|a_{i}|\neq 1,|b_{i}|\neq 1,i=1,2$
show that
$$\gcd{(a_{1}a_{2}-b_{1}b_{2},a_{1}b_{2}+a_{2}b_{1})}=1$$
My attemp
Assmue that prime $p$ such
$$p|(a_{1}a_{2}-b_{1}b_{2}),p|(a_... | This is not true. For counterexample: $(3+4i)\overbrace{(11+2i)}^{(3-4i)(1+2i)}=25+50i$.
That is, take $\gcd(3,4)=1$ and $\gcd(11,2)=1$. Then
$$
\gcd(3\cdot11-4\cdot2,3\cdot2+4\cdot11)=\gcd(25,50)=25
$$
Another counterexample: $(5+2i)\overbrace{(7+3i)}^{(5-2i)(1+i)}=29+29i$.
That is, take $\gcd(5,2)=1$ and $\gcd(7,3)=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number.
A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(... | Since:
$$-\frac{1}{k+1}+\frac{4}{k+2}-\frac{3}{k+3}=\frac{2k}{(k+1)(k+2)(k+3)}$$
we have:
$$S_N=\sum_{n=1}^{N}\frac{k}{(k+1)(k+2)(k+3)}=\sum_{k=1}^{N}\left(\frac{-1/2}{k+1}+\frac{2}{k+2}+\frac{-3/2}{k+3}\right) $$
and since $(-1/2)+(2)+(-3/2)=0$ we have:
$$ \lim_{N\to +\infty}S_N = \frac{-1/2}{1+1}+\frac{2}{1+2}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Are there any two positive integers such that the square root of the sum of their squares give a perfect square. Are there two positive integers $$ \text{{ (x,y) | $ \sqrt{x^2 + y^2} = z $ }}$$
where $ z$ is a perfect square.
| If you're looking for integers such that $\sqrt{x^2+y^2}=z$, then there's plenty, for example, $3,4,5$ or $12,5,13$.
If you're looking for integers such that $\sqrt{a^2+b^2}=c^2$, then take any solutions to the first equation and multiply everything by $z$, e.g. take $3,4,5$ and multiply by $5$.
Then, since
$$\sqrt{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sqrt{5-12i}$ by square root definition I KNOW it can be solved by the trig formula, but I want to solve it by the square root definition, so please don't just post an alternative way to do it.
By the square root definition:
$$z = 5-12i$$
$$\sqrt{z} = w\implies w^2 = z$$
So if I suppose $w = a+bi$ we have:
$$w^2... | An easier way is to solve:
$$\begin{cases}a^2-b^2=5\ \ (1)\\ 2ab=-12\ \ (2)\\a^2+b^2=|5-12i|=13\ \ (3),\end{cases}$$
because you don't have to solve a equation of degree $4$.
Therefore, by $(1)$
$$a^2=b^2+5$$
By $(3)$
$$2b^2=8\implies b^2=4\implies b=\pm2$$
and by $(2)$
$$ab=-6\implies a=\mp3$$
Therefore $$\sqrt{5-12i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determining convergence of $\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$ I have the following infinite series:
$$\sum_{n=1}^{+\infty}(e^{\frac{1}{n}}-(1+\frac{1}{n}+\frac{1}{2n^2}))^{\frac{1}{2}}$$
I want to examine its convergence. First thing that came to my mind was "unfolding"... | In general,
for $m \ge 1$,
let
$d_n
=e^{1/n}-\sum_{k=0}^{m-1} \frac1{n^k k!}
$.
Then
$d_n
=\sum_{k=m}^{\infty} \frac1{n^k k!}
=\frac1{m!}\sum_{k=m}^{\infty} \frac{m!}{n^k k!}
$.
Therefore,
$d_n > \frac1{m!n^m}$
and,
if $n \ge 2$,
$d_n
<\frac1{m!}\sum_{k=m}^{\infty} \frac{1}{n^k }
=\frac1{m!n^m(1-1/n)}
\le\frac{2}{m!n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving a radical equation for real roots I'm attempting to solve the derivative of my function $f(x)$ for real roots.
$$
\\ \begin{align*}
\\ f(x) &= 3x^2 + 3\arcsin{x}
\\ f^{\prime}(x) &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x + \dfrac{3}{\sqrt{1-x^2}}
\\
\\ 0 &= 6x \sqrt{1-x^2} + 3
\\ 0 &= \sqrt{36x^2 - 36x^4} ... | When you find yourself in such a situation, a good trick is to insert the root (any number will work but the root will often make the calculations easy) at each step and see when the value change.
Personally I think you are doing to much at once in the last step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Number of ways to set 3 queens to attack each other
We play chess and want to set 3 queens to attack each other. How many ways we can do it?
I know to solve this problem when I have 2 queens. I see the chess board as 4 squares, from an outer square (the 28 squares on the edges and corners) to the inner square, the 4 ... | $3$ queens on a chessboard attack each other if they are in the vertices of a right and isosceles triangle. There are three possibilities, according to the hypotenuse of such a triangle lying on a diagonal, a row or a column of the chessboard. Counting them is not so difficult:
*
*hypotenuse being on the $a7-b8$ dia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Elements of order $10$ in $S_6$
Find the number of elements of order 10 in $S_6$
I conjecture that there are no elements of order 10 in $S_{6}$.
If there were a $\sigma \in S_6$ s.t. $|\sigma| = 10$, then $\sigma$ is a 10 cycle, or $\sigma$ is the product of two disjoint 2 and 5-cycle. Since there is no 10-cycle or ... | The order of a permutation is the LCM of its cycle lengths. Therefore we are interested in all possible partitions of $6$, and the corresponding LCMs:
$$
\begin{align*}
6 && 6 \\
5+1 && 5 \\
4+2 && 4 \\
4+1+1 && 4 \\
3+3 && 3 \\
3+2+1 && 6 \\
3+1+1+1 && 3 \\
2+2+2 && 2 \\
2+2+1+1 && 2 \\
2+1+1+1+1 && 2 \\
1+1+1+1+1+1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Taylor Series for $\frac{1}{1+e^z}$ and radius of convergence I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:
$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\fr... | I assume you want the Maclaurin series, i.e. the Taylor series about $0$.
Write $$1 + e^z = 2 (1 + Q(z))$$
where
$$Q(z) = \dfrac{z}{2\cdot 1!} + \dfrac{z^2}{2\cdot 2!} + \dfrac{z^3}{2\cdot 3!} + \ldots $$
So $$ \dfrac{1}{1+e^z} = \dfrac{1}{2(1+Q(z))} = \dfrac{1}{2} \left( 1 - Q(z) + Q(z)^2 - Q(z)^3 + \ldots \right) $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Optimisation to solve for trigonometric expression? I have a question that requires the use of optimisation to solve for the following expression:
$$\cos ec{(\cos^{-1}{(-\frac{\sqrt{3}}{2})}+\sin^{-1}{(-\frac{\sqrt{3}}{2})})}$$
I'm a bit baffled, as I'm not sure whether this refers to finding the value of cosec or find... | Since $\left|-\dfrac{\sqrt{3}}{2}\right|\le 1$, we have that $\cos^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)+\sin^{-1}\left( -\dfrac{\sqrt{3}}{2}\right) =\dfrac{\pi}{2} $.
So $\operatorname{cosec} \left( \cos^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)+\sin^{-1}\left( -\dfrac{\sqrt{3}}{2}\right)\right)=\operatorname{cosec} \le... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$
I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significa... | the taylor series of $$y=\frac{1}{x}\log(\frac{1}{1-x})$$ is
$$y=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+...\frac{x^{n-1}}{n}$$
$$\int_{0}^{0.5}ydx=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\frac{x^4}{4^2}+...\frac{x^n}{n^2}=\sum_{n=1}^{\infty }\frac{(.5)^n}{n^2}=\sum_{n=1}^{\infty }\frac{1}{2^nn^2}$$
$$\sum_{n=1}^{\infty }\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $r_k^n/n \le \binom{kn}{n} < r_k^n$ where $r_k = \dfrac{k^k}{(k-1)^{k-1}}$ Show that
for $n \ge 2$,
$\dfrac{r_k^n}{n+1} \le \binom{kn}{n} < r_k^n$ where $r_k = \frac{k^k}{(k-1)^{k-1}}$.
This is a generalization of
How to prove through induction
which asks for a proof that
$\binom{2n}n<4^n$.
I wondered what... | No one else has done anything,
so I thought I'd see what happens
when I follow my own suggestion
of using Stirling's formula.
Definitely easier
and, of course,
more precise.
Here we go.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{kn}{n}
&=\dfrac{(kn)!}{n!(kn-n)!}\\
&\sim \dfrac{\sqrt{2\pi kn}(kn/e)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $1/(x^2+y^2)^a$. I'm curious about the convergence of the series:
\begin{align}
\sum_{x,y=1}^\infty \frac{1}{(x^2+y^2)^\alpha}\ ,\ \alpha \in \mathbb{N}
\end{align}
I'm wondering for what values of $\alpha$ this converges, and if so is there an analytic way to find the limit? I'm not really sure where t... | Let $N,M \in \Bbb N$. On the square $[N,N+1]\times [M,M+1]$,
$$\frac{1}{[(N+1)^2 + (M + 1)^2]^\alpha} \le \frac{1}{(x^2 + y^2)^\alpha} \le \frac{1}{(N^2 + M^2)^\alpha}.$$
Integrating the inequality over the square, we find
$$\frac{1}{[(N+1)^2 + (M+1)^2]^\alpha} \le \int_N^{N+1}\int_M^{M+1} \frac{dx\, dy}{(x^2 + y^2)^{\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$. To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.
$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.
$199x \equiv 1\pmod {10}$ has a solutio... | Yes, it is correct. More simply, by little Fermat $\,2^4\equiv 1\pmod 5\,$ and $\,2^{198}\equiv 1\pmod{199}\,$ so $\,4,198\mid 1980\,\Rightarrow\,2^{1980}\equiv 1\,\Rightarrow\,2^{1990}\equiv 2^{10}\,$ mod $5$ and mod $199$. Finally, combining these we conclude that $\,2^{1990}-2^{10}\,$ is divisible by $\,2,5,199,\,$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help evaluating $ \int \sqrt{{x}^{2} + 3} \; dx $ Can you help me evaluating the following indefinite integral?
$$
\int \sqrt{{x}^{2} + 3} \; dx
$$
Please, don't give a full solution, just some hint on which method to use...
** UPDATE **
Thank you very much to everybody for the useful comments and suggestions. I'm sorr... | You have modified your question and now you want to determine the substitution for $\sec\theta$. So, here we go:
Given:
$$ \tan\theta = \frac{x}{\sqrt{3}} $$
knowing that $\tan\theta$ is positive for positive $x$, squaring both sides:
$$ tan^2\theta = \frac{x^2}{3} $$
$$ 1+ tan^2\theta = 1+ \frac{x^2}{3} $$
$$ 1+ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Simplify $\left(\sqrt{\left(\sqrt{2} - \frac{3}{2}\right)^2} - \sqrt[3]{\left(1 - \sqrt{2}\right)^3}\right)^2$ I was trying to solve this square root problem, but I seem not to understand some basics.
Here is the problem.
$$\Bigg(\sqrt{\bigg(\sqrt{2} - \frac{3}{2}\bigg)^2} - \sqrt[3]{\bigg(1 - \sqrt{2}\bigg)^3}\Bigg)^... | A square root is always a non negative number so $\sqrt{x^2}=|x|$
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Bijection from $\{ (x,y) : 0 \leq x ,y \leq 1\}$ to $\left\lbrace (u,v) : u,v \geq 0, u+v \leq \frac{\pi}{2} \right\rbrace$ Suppose the set $M :=\{ (x,y) : 0 \leq x ,y \leq 1\}$. Now we define
$$ u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \ \ \ \ \ \text{and} \ \ \ \ \ v:= \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}}.$$
Ho... | One can show that
$$
u := \arccos \sqrt{\frac{1-x^2}{1-x^2y^2}} \quad \text{and} \quad
v:= \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}} \quad \tag 1
$$
is a bijective mapping of the open rectangle
$M :=\{ (x,y) : 0 < x ,y < 1\}$ onto the open triangle
$\widetilde{M} := \left\lbrace (u,v) : u,v > 0, u+v < \frac{\pi}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof of identity: cross product of three vectors A book I'm reading contains the following (paraphrased)
\begin{equation}
(a \times b) \times c = (a \cdot c)b - (b \cdot c)a
\end{equation}
This is supposed to follow from:
\begin{equation}
(a \times b) \cdot (c \times d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c... | Chappers' proof is the way that was most likely intended (using the identity given). However, here is another approach (if only to show how much work is saved by using the identity given)
Since $(a\times b)\times a$ is perpendicular to $a$,
$$
((a\times b)\times a)\cdot a=0\tag{1}
$$
Since $|a\times b|=|a|\,|b|\,|\sin... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatri... | You have basically put your matrix $A$ in Jordan normal form.
This means that you can split your matrix $J$ as follows:
$$J = D + N$$
where $D$ is diagonal and $N$ is nilpotent, i.e. there exists an a positive integer $n$ such that $N^n = 0$.
In your case
$$J = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}
=
\underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sq... | $$\begin{align}(x + x^{1/2})^{1/2} &= x^{1/2} + \frac 12 x^{-1/2}x^{1/2} -\frac 1 8 x^{-3/2}x+\cdots\\&=\sqrt x+\frac 12-\frac1{8\sqrt x} +\cdots\\
(x-1)^{1/2} &=\sqrt x -\frac 1{2\sqrt x}+\cdots \end{align}$$
therefore $$(x + x^{1/2})^{1/2} - (x-1)^{1/2}=\frac 1 2 + \frac 3{8\sqrt x} + \cdots \rightarrow \frac 12 \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| ![The method I was speaking of if anyone especially OP is interested.... ][1]
[1]: http://i.stack.imgur.com/r05sp.jpg $$
\text{ Let there be a right triangle with angle A (not with measurement 90 deg) } \\
\text{ whose adjacent has measurement } 3 \text{ and whose opposite has measurement } 2. \\
\text{ Therefore ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
} |
Help me to prove this statement about quadratic equations? (from Gelfand's Algebra). $ x^2+px+q=0 ${p,q are integers; a,b are roots}. Prove $a^n+b^n$(n is any natural number) is an integer.
This is the third part of the problem.I have previously proved that $a^2+b^2$ and $a^3+b^3$ are integers using the binomial theore... | It can be proved by induction on $n$, as follows:
From
$x^2 + px + q = 0 \tag{1}$
it follows that the roots $a$, $b$ satisfy
$a + b = -p, \tag{2}$
$ab = q; \tag{3}$
from (2), (3) we have
$p^2 = (a + b)^2 = a^2 + 2ab + b^2 = a^2 + 2q + b^2, \tag{4}$
or
$a^2 + b^2 = p^2 - 2q \in \Bbb Z, \tag{5}$
since $p, q \in \Bbb Z$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Recovering the optimal primal solution from dual solution I'm having trouble finding the optimal primal solution of a particular problem from its dual solution.
Primal:
$\texttt{Maximize} \ \ 10 x_1 + 24 x_2 + 20 x_3 + 20 x_4 + 25 x_5$
Subject to
$x_1 + x_2 + 2 x_3 + 3 x_4 + 5 x_5 \leq 19$
$2 x_1 + 4 x_2 + 3 x_3 + 2 x... | The following condition must hold:
$(A^T\cdot u^*-c)^T\cdot x^*=0$
Inserting the given values:
$\left[ \left( \begin{array}{} 1&2 \\ 1&4 \\ 2&3 \\ 3&2 \\ 5&1 \end{array} \right) \cdot \left( \begin{array}{} 4 \\ 5 \end{array} \right)-\left( \begin{array}{} 10 \\ 24 \\ 20 \\ 20 \\ 25 \end{array} \right)\right]^T\cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{x\rightarrow0}\frac{e^{x^2} - \cos x}{\sin^2 x}$ Find the limit as $x$ approaches $0$ of
$\dfrac{e^{x^2} - \cos x}{\sin^2 x}$
What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\fr... | As Jean-Claude Arbaut showed, Taylor expansions are extremely useful for this kind of problems.
But, using them with a few more terms, you also see how is the limit approached $$A=\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{\Big(1+x^2+\frac{x^4}{2}+O\left(x^6\right)\Big)-\Big(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
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... | If the general case proves obfuscatory, it often proves enlightening to first examine a few small special cases in order to help grasp the general idea, e.g. let's examine the two smallest cases first $\,p=5,13.\,$ mod $\, p = 5\!:\,\ \color{#0a0}{4\equiv -1},\ 3\equiv \color{#c00}{ -2}\ $ so substituting these into W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Equilateral Triangle equality Let ABC be an equilateral triangle, and P be an arbitrary point within the triangle.
Perpendiculars PD, PE, PF are drawn to the three sides of the triangle. Show
that, no matter where P is chosen,
PD + PE + PF / AB + BC + CA = 1/2√3
| Let $AB=BC=AC=a$.
It follows from the areas equality:
\begin{align}
S_{\triangle APB}+S_{\triangle APC}+S_{\triangle BPC}
&=
S_{\triangle ABC}
\\
\frac{1}{2}PD\cdot a+\frac{1}{2}PE\cdot a+\frac{1}{2}PF\cdot a
&=
a^2\frac{\sqrt{3}}{4}
\\
PD+PE+PF
&=
\frac{3a}{2\sqrt{3}}
\\
\frac{PD+PE+PF}{AB+BC+AC}
&=
\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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|$$
I think you have to multiply by the conjugate. And then make the change equivalent small. Right?
| I am waiting for a simpler more succinct answer but in the meantime we can use Taylor's theorem
$$ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n} = 1 - \frac{x^4}{2} + \frac{x^8}{120} + O(x^{12})$$
$$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2} + \frac{x^4}{120} + O(x^{6})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 5
} |
Integrating $ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $ I am trying to show that
$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \frac{\pi}{a(a+b)} $$ where $ a,b > 0$.
I have tried a few things, but none have worked.
For example, one approach was
$$ \int^{\pi}_{0} \frac{\... | Suppose we seek to evaluate
$$\int_0^\pi \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx
= \frac{1}{2}
\int_0^{2\pi} \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx$$
with $a,b>0.$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2}\int_{|z|=1}
\frac{(z+1/z)^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
4 girls and 8 boys are randomly divided into 3 groups of equal size 4 girls and 8 boys are randomly divided into 3 groups of equal size
What is the probability that there will be at least one girl in each group?
| You can calculate the probability that all girls are in a only group, the probability that four girls are disposed in two groups and then you can use the inverse probability. The probability that all girls are in only group is: $$1\cdot \frac {3}{11}\cdot \frac{2}{10}\cdot {1}{9}$$ while the probability that all girls ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rew... | Suppose that $p>7$. Then $p\ge 11$ so that $p^2+pq+275p+10q\ge 3188$ since $q\ge 2$. Hence $p\le 7$ and we have to check $p=2,3,5,7$. It follows that $(p,q)=(7,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Minimum value of function $f(x)=x+\log_2(2^{x+2}-5+2^{-x+2})$ out of 5 options Minimum value of function $f(x)=x+\log_2(2^{x+2}-5+2^{-x+2})$ out of 5 options
A : $\log_2(1/2)$
B : $\log_2(41/16)$
C : $39/16$
D : $\log_2(4.5)$
E : $\log_2(39/16)$
I just... don't know how to approach this.
| $$x=\log_2(2^x)$$
$$f(x)=\log_2[2^x(2^{x+2}-5+2^{-x+2})]=\log_2[2^{2x+2}-5\cdot2^x+4]$$
Now $4\cdot2^{2x}-5\cdot2^x+4=(2^{x+1})^2-2\cdot2^{x+1}\cdot\dfrac54+\left(\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $=\left(2^{x+1}-\dfrac54\right)^2+4-\left(\dfrac54\right)^2$ $\ge4-\left(\dfrac54\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Last nonzero digit of $2010!$ I have to calculate the last nonzero digit of $2010!$
Till now I couldn't find any pattern.
| $2010!$ ends with $501$ zeroes since:
$$\sum_{n=1}^{5}\left\lfloor\frac{2010}{5^n}\right\rfloor=501.\tag{1}$$
For the same reason,
$$ \nu_2(2010!)=\sum_{n=1}^{10}\left\lfloor\frac{2010}{2^n}\right\rfloor=2002\tag{2}$$
so $\frac{2010!}{10^{501}}$ is an even number, and since:
$$ 1\cdot 2\cdot 3\cdot 4\equiv -1\pmod{5} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}... | your calculations are ok, we are looking at the inequality $$(1-2c)(c^2+4)\geq \frac{65}{2}c^2$$ with $c=ab$ for $c$ we get $$a^2+b^2=1-2ab\geq 2ab$$ from here we get $$\frac{1}{4}\geq ab=c$$ and our inequality is equivalent to $$\frac{-1}{2}(4c-1)(c^2+16c+8)\geq 0$$
ready.
| {
"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": 2
} |
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... | $\bf{My\; Solution::}$ Given $$\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right) = -i\cdot \left[\cos \left(\frac {2k\pi}{11}\right)+i\cdot \sin \left(\frac {2k\pi}{11}\right)\right]$$
Now We know that $$e^{i\phi} = \cos \phi+i\sin \phi$$ and $$e^{-i\phi}=\cos \phi+i\sin \phi$$
So W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How many permutations How many permutations $\pi \in S_{2n} $ for which $\exists a\in [2n] $ such that set $\lbrace a,\pi (a),\pi ^2(a),\pi^3(a),... \rbrace $ has exactly $n$ elements.
I need help to solve this.
| By way of enrichment here is an alternate formulation using
combinatorial species. The species of permutations with cycles of size $n$ marked is
$$\mathfrak{P}(\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathfrak{C}_{=2}(\mathcal{Z}) + \cdots
+ \mathcal{U}\mathfrak{C}_{=n}(\mathcal{Z})
+ \mathfrak{C}_{=n+1}(\mathcal{Z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How do I calculate the limit for this multiplication? $$\lim_{n\to\infty}\left(1-\frac{2}{3}\right)^{\tfrac{3}{n}}\cdot\left(1-\frac{2}{4}\right)^{\tfrac{4}{n}}\cdot\left(1-\frac{2}{5}\right)^{\tfrac{5}{n}}\cdots\left(1-\frac{2}{n+2}\right)^{\tfrac{n+2}{n}}$$
(original image)
I mean,I tried to use Sandwitch rule but it... | The $m^{th}$ term is
\begin{align}
a_m & = \prod_{n=1}^m \left(1-\dfrac2{n+2}\right)^{(n+2)/m} = \prod_{n=1}^m \left(\dfrac{n}{n+2}\right)^{(n+2)/m} = \left(\dfrac{\prod_{n=1}^m n^{(n+2)/m}}{\prod_{n=1}^m (n+2)}\right)^{(n+2)/m}\\
& = \left(\dfrac{\prod_{n=1}^m n^{n/m}}{\prod_{n=3}^{m+2} n^{n/m}}\right) \prod_{n=1}^m n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that if $a\neq 1$, then $\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$
Need to show that if $a\neq 1$, then
$$\sum_{k=0}^{n-1}ka^k = \frac{1-na^{n-1}+(n-1)a^n}{(1-a)^2}$$
Here is my attempt:
$$\begin{aligned}
S & =\sum_{k=0}^{n-1}ka^k \\
&= \sum_{k=0}^{n}(k-1)(a^{k-1}) \\
\end{aligned}$$
from he... | for $1 \le m \le n-1$, set
$$
S_m = a^m + \cdots + a^{n-1} = \frac{a^m -a^n}{1-a}
$$
then
$$
\sum_{k=0}^{n-1} ka^k = \sum_{m=1}^{n-1} S_m =(1-a)^{-1}\sum_{m=1}^{n-1}(a^m-a^n) \\ =(1-a)^{-2}\left( a-a^n-(n-1)a^n(1-a) \right) \\
= \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it... | $b_i:= a_i^2$ Then $$ b_1=2,\ b_{n+1}=2+\sqrt{b_{n}}$$
Show that $b_i$ is an increasing sequence bounded by $4$ :
(1) $b_1< 4$ If $b_n<4$ then $b_{n+1}=2+\sqrt{b_n}< 2+2=4$.
(2) $b_2=2+\sqrt{2} > b_1$ If $b_n>b_{n-1}$ then $$
b_{n+1}=2+\sqrt{b_n} > 2+\sqrt{b_{n-1}}=b_n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
| By setting $a=\cos\theta,b=\sin\theta$ you just have to prove that for any $\theta\in\left(0,\frac{\pi}{2}\right)$ we have:
$$ f(\theta)=(5a+2)(5b+5)=(5\cos\theta+2)(5\sin\theta+5)\leq 48,$$
or, by using Weierstrass substitution $\theta=2\arctan t$,
$$\forall t\in(0,1),\quad g(t) = (7-3t^2)(t+1)^2- \frac{48}{5}(1+t^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Exercise about Matrix diagonalization Well I have to diagonalize this matrix :
$$
\begin{pmatrix} 5 & 0 & -1 \\ 1 & 4 & -1 \\ -1 & 0 & 5 \end{pmatrix}
$$
I find the polynome witch is $P=-(\lambda-4)^2(\lambda-6)$
Now I want to know eignevectors so I solve $AX=4X$ and $AX=6X$ with $X=\begin{pmatrix} x \\ y \\ z \end{pma... | You calculated correctly. The set of all eigenvectors of eigenvalue $4$ is a $2$-dimensional subspace. The condition $x = z$ is equivalent to all matrices of the form
$$
\begin{pmatrix} x \\ y \\ x \end{pmatrix}
= \begin{pmatrix} x \\ 0 \\ x \end{pmatrix}
+ \begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}
= x \begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the type of triangle from equation. In triangle $ABC$, the angle($BAC$) is a root of the equation
$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$
Then the triangle $ABC$ is
a) obtuse angled
b) right angled
c) acute angled but not equilateral
d) equilateral.
Thanks in advance.
| given that
$$\sqrt{3}cosx+sinx=\frac{1}{2}$$ Now, diving the above equation by $\sqrt{(\sqrt{3})^2+(1)^2}=2$ we get
$$\begin{align}
\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=\frac{1}{4}\\
cosxcos\frac{\pi}{6}+sinxsin\frac{\pi}{6}=\frac{1}{4}\\
cos\left(x-\frac{\pi}{6}\right)=\frac{1}{4}\\
x-\frac{\pi}{6}=cos^{-1}\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
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.
| We can actually prove a more general version of what you hope to prove (set $a=0$ for your problem, specifically):
Claim: For every $n\in\mathbb{Z^+}$ and every non-negative real number $a$
$$
\sqrt{a+1+\sqrt{a+2+\cdots+\sqrt{a+n}}}< a+3.
$$
Proof. For $n\geq 1$, let $S(n)$ denote the statement that for any non-negativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can... | Sometimes it might help to take things a little more symbolically. Without worrying about what $x$ is for now, think about $3x + 5x$. Clearly that's $8x$, just as you thought.
For the next problem you have $x - 3x$. The answer is $-2x$. Let's say $x = 1$. Then this boils down to $1 - 3 = -2$. When we put $x = \sqrt{11}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~... | Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}=\lambda$$
we have:
$$ x = e^{\lambda(b-c)},\quad y=e^{\lambda(c-a)},\quad z=e^{\lambda(a-b)}, $$
hence:
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = \exp\left(\lambda\cdot\sum_{cyc}\left(b^2-c^2-a(b-c)\right)\right)=\exp(0)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 0
} |
differential equation power series solution I am trying to solve this equation using power series
$$
(1-x)y"-xy'+y=0
$$
Knowing that $y(0)=-2$ and $y'(0)=6$.
Please I need someone's help, I get a relation between
$c(n)$,$c(n+1)$, and $c(n+2)$.
| taking
$$\begin{align}
y&=\sum_{n=0}^{+\infty}a_n x^n\\
y'&=\sum_{n=0}^{+\infty}na_n x^{n-1}\\
y''&=\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}
\end{align}$$
we have
$$\begin{align}
(1-x)y''-xy'+y&=0\\
(1-x)\sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2}-x\sum_{n=0}^{+\infty}na_n x^{n-1}+\sum_{n=0}^{+\infty}a_n x^n&=0\\
\sum_{n=0}^{+\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the equation of a circle given two points on the circle
11. Find the equation of the circle which touches $x^{2} + y^{2} - 6x + 2y + 5 = 0$ at $(4, -3)$ and passes through $(0, 7)$.
My textbook has a worked example for obtaining the equation of a circle from three points on the circle. It also talks you throu... | I've come back to this question after a while and have found a solution which agrees with that in the textbook. My method is based primarily on the tips given by @Mufasa.
Let the circle which touches $x^{2} + y^{2} − 6x + 2y + 5 = 0$ at $(4,−3)$ and passes through $(0,7)$ be $C_{1}$.
Let the centre of $C_{1}$ be $(p, q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $2^{12^{7}+3}\equiv x \pmod{36}$, then what is the value of $x$?
If $2^{12^{7} + 3} \equiv x \pmod{36}$, then what is the value of $x$?
We have:
$$
\begin{align}
2^5 & \equiv - 4 \pmod{36} \\
2^{10} & \equiv 16 \pmod{36} \\
2^{12} & \equiv - 8 \pmod{36} \\
(2^{12})^{12} & \equiv 8^{12} \pmod{36} \\
2^{12^2} & \equ... | It's probably easier to work modulo $4$ and modulo $9$, and use CRT.
First we have
$$x \equiv 2^{12^{7}+3}\equiv 0 \pmod{4}$$
To compute
$$x \equiv 2^{12^{7}+3} \pmod{9},$$
it suffices to compute $12^{7} + 3 \equiv 3 \pmod{6}$, as $6 = \varphi(9)$. (Or, if you prefer, as $2^{6} \equiv 1 \pmod{9}$.) So
$$2^{12^{7}+3}\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
| Since
$$\left(\frac{x}{1+x^2}\right)'=\frac{2-(1+x^2)}{(1+x^2)^2}=\frac{2}{(1+x^2)^2}-\frac{1}{1+x^2}$$
one has
$$2\int \frac{1}{(1+x^2)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$$
$$\Rightarrow \int\frac{1}{(1+x^2)^2}dx=\frac{x}{2(1+x^2)}+\frac{\tan^{-1} x}{2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How to prove a sum of series How do I prove that for any natural number $n$ we have
$$\sum_{i=0}^n i^4 \neq \left(\sum_{i=0}^n i\right)^3?$$
Any help would be greatly appreciated.
| Recall that $$\sum_{i=1}^n i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
and
$$\sum_{i=1}^n i = \dfrac{n(n+1)}2$$
We hence need
$$\left(\dfrac{n(n+1)}2\right)^3 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
This gives us either $n(n+1)=0$ or
$$15(n(n+1))^2 = 4(2n+1)(3n^2+3n-1) \,\,\,\, (\spadesuit)$$
$(\spadesuit)$ can be simp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Orthogonal circles What is the equation of the circle that is orthogonal to the circles $x^2 + y^2 - 8x +5 =0$ and $x^2 + y^2 +6x +5 = 0$ and passes through the point $(3,4)$? I've spent hours trying to figure this out - help please
| you have $$C_1:(x-4)^2+y^2=11,\quad C_2:(x+3)^2+y^2= 4$$ we will look for a point $(a,0)$ on the $x$-axis so that the tangents to the circle have the same length.
we have $$(4-a)^2 -11=(a+3)^2-4 \to-8a+5=6a+5\to a = 0.$$
we will use the fact that if a circle is orthogonal to both $C_1, C_2,$ then the $x$-cdt of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$\lim_{x\to\infty}(\frac{x^2}{x^2+x})^x = 1 $ or $0$? $$\lim_{x\to\infty}\left(\frac{x^2}{x^2+x}\right)^x$$
as $\lim_{x\to\infty}\frac{x^2}{x^2+x} = 1$ but we all know that $\frac{x^2}{x^2+x} < 1$ and for $a<1$ lim $\lim_{x\to\infty} a^x = 0$
I don't know how to find the limit in this case
| Mathlove privided an efficient way forward. Here is a slightly different approach. Take the logarithm of the function of interest. Then, we have
$$\log\left(\frac{x^2}{x+x^2}\right)^x=x\log\left(\frac{x}{x+1}\right)=\frac{\log\left(\frac{x}{1+x}\right)}{\frac{1}{x}}=\frac{\log x- \log (1+x)}{\frac{1}{x}}$$
Now, app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I need help proving the following: Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I know that the smallest $n$ is 8 by te... | Hint:
$$ x^4+4m^4 = (x^4+4x^2m^2+4m^4)-(4x^2m^2)=(x^2+2m^2)^2-(2mx)^2 = (x^2-2mx+2m^2)(x^2+2mx+2m^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the interval on which $ x^{2} - \lfloor x \rfloor - 3 < 0 $ holds. On what interval does the equation $ x^{2} - \lfloor x \rfloor - 3 < 0 $ hold?
My attempt: I tried sketching the graph, but it’s a bit complicated.
Is there any other approach?
| Probably graphing $y=x^2-3$ and $y=\lfloor x\rfloor$ and seeing where the first graph is below the second is the easiest way to solve this, but here is another method:
$\textbf{1)}$ Since $\lfloor x\rfloor \le x$,
$x^2-3<\lfloor x\rfloor\implies x^2-3<x\implies x^2-x-3<0\implies x^2-x+\frac{1}{4}<\frac{13}{4}\implies$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Residue $\frac{e^z}{z^3\sin(z)}$ I want to find the residue of $$\frac{e^z}{z^3\sin(z)}$$ and get $$ \frac 1 {3!} \lim_{z \to 0} \left( \frac{d^3}{dz^3} \left(\frac{ze^z}{\sin(z)} \right)\right) = \frac{1}{3}$$
Can anyone confirm this? I tried using the Laurent Series, but I didn't know how to compute it.
Oh, I was ab... | $$\begin{align}\frac{e^z}{z^3 \sin{z}} &= \frac1{z^4} \frac{\displaystyle 1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots}{\displaystyle 1-\frac{z^2}{6} + \frac{z^4}{120}+\cdots} \\ &= \frac1{z^4} \left (1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots \right )\left [1-\left (-\frac{z^2}{6} + \frac{z^4}{120}+\cdots\right )+ \left (-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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show $p$ is divisible by $(x^2 +y^2 +1)$ Show that, for any prime $p$, there are integers $x$, and $y$ such that $p$ is divisible by $(x^2+y^2+1)$ Can you show me what to start with? do I prove $p$ is divisible by $x^2$ and $y^2$ separately?
| Take $x=y=0$, then $x^2+y^2+1=1$, then $p$ is divisible by $x^2+y^2+1=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.
| Hint: $3-2\cdot 3^{x+1}>0$
So we'll have: $\:3^{2+2x}+2\cdot 3^{x+1}=3\:\:\rightarrow 9\cdot 3^{2x}+6\cdot 3^x=3\:\:\rightarrow \:3\cdot 3^{2x}+2\cdot 3^x-1=0\:$
let $3^x=a\:\:\rightarrow \:3a^2+2a-1=0$ with $a_1=-1$ and $a_2=\frac{1}{3}$, and only $a_2=\frac{1}{3}$ is good. Therefore, comeback at substitution we'll o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$... | First use the fact that the factorial is divisible by $9$, and therefore the digit sum is divisible by $9$.
Then use the fact that the number with decimal representation $abcdefg\dots$ is divisible by $11$ if and only if $a-b+c-d+e-f+\cdots$ is divisible by $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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If $a+b+c\le 1$ then $3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
Let $a,b,c\ge 0$ such that $a+b+c\le 1$, prove that
$$3(a+b+c)-(a^2+b^2+c^2-ab-bc-ac)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\tag{1}$$
I conjecture:
Let $a_{i}\ge 0$, $i=1,2,\cdots$, $a_{1}+a_{2}+\cdots+a_{n}\le 1$, $n\ge 3$,then
$... | We need to prove that
$$\sum_{cyc}(3a-a^2+ab)\geq\sum_{cyc}(a+2\sqrt{ab})$$ or
$$2\sum_{cyc}(a-\sqrt{ab})\geq\sum_{cyc}(a^2-ab)$$ or
$$2\sum_{cyc}\left(\sqrt{a}-\sqrt{b}\right)^2\geq\sum_{cyc}(a-b)^2$$ or
$$\sum_{cyc}(\sqrt{a}-\sqrt{b})^2\left(2-(\sqrt{a}+\sqrt{b})^2\right)\geq0,$$
for which it's enough to prove that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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A difficult integral: $\int_0^{+\infty} e^{ - x}\left(\frac1{x( e^{ - x} - 1 )} + \frac1{x^2} + \frac1{2x} \right) \, dx$. Could you help me calculate the integral?
$$\int_0^{+\infty} e^{-x} \left(\frac1{x(e^{-x}-1)} + \frac1{x^2} + \frac1{2x} \right) \, dx .$$
| One may recall Binet's formula, for $\Re z >0$,
$$
\log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) +
\int_0^{\infty} \!
\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x. \tag1
$$ Observing that
$$
{\left( {\frac{1}{{x\left( {{e^{ - x}} - 1} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to compute the series $\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$ I have to compute the series
$$\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$$
It's easy to prove that the series converges using comparison test but how to compute it?
| Consider
$$ S(x) = \sum_{n=0}^{\infty} \frac{x^n}{n+2}. $$
Your sum is $S(1/7)$. Then
$$ x S'(x)+2 S(x) = \sum_{n=0}^{\infty} \frac{(n+2)x^{n}}{n+2} = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} $$
An integrating factor for this differential equation is $x$, so we find that
$$ (x^2S)' = \frac{x}{1-x} = \frac{1}{1-x}-1 \\
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Boolean Algebra - proof without associativity? I would like to prove the following:
$(x\cdot y) + (\overline{x} + \overline{y}) = 1$
without the Associativity Property. I can't seem to do this algebraically (without truth tables).
| First note that
\begin{align*}
x + (\overline x + \overline y)
&= 1\cdot (x + (\overline x + \overline y)) \\
&= (x + \overline x)\cdot (x + (\overline x + \overline y)) \\
&= x + (\overline x\cdot (\overline x + \overline y)) \\
&= x + ((\overline x + 0)\cdot (\overline x + \overline y)) \\
&= x + (\overline x + (0\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Application of the Banach fixed point theorem Let $a > 0$. We consider the function: $f: (0, \infty) \to (0, \infty)$, defined by $f(x) = \frac{1}{2}(x + \frac{a}{x})$.
Let $(x_n)_{n \in \mathbb{N}_0}$ be defined by:
$x_0 \in (0, \infty)$, $x_{n+1} := f(x_n)$
What is the smallest $b > 0$ so that f is contracting on $[b... | *
*To answer that question, it is most convenient to solve $|f'(x)| < 1$. Since $f'(x) = \frac{1}{2} \left[ 1 - \frac{a}{x^2} \right],$ the solution for positive $x$ is $x \in \left( \sqrt{ \frac{a}{3} }, \infty \right)$.
So:
*
*for every $b > \sqrt{\frac{a}{3}}$ there is such $L \in (0, 1)$ that $|f'(x)| \leqslant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 =... | Your answer seems to be correct. We see that Excel's answer is wrong by adding just the first 2 terms in the series, $1, \frac 34$ and we see that $1.75>1.66..$, so we can see that Excel's answer is wrong.
| {
"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": 2
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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... | For large $x, x^2 > 2x.$ So it seems like a good idea to write the numerator as $-\int_{2x}^{x^2}t^4e^{t^2}\,dt.$ Notice that the integrand is positive and increasing. Thus the absolute value of the numerator is $\ge (2x)^4 e^{4x^2}(x^2-2x).$ The denominator is $< e^x$ for large $x.$ So in absolute value, it is clear, ... | {
"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": 2
} |
Integrate $\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$using partial fractions How do I integrate the improper integral $$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}$$ using partial fraction decomposition?
I am restricted to use only the principles taught in Calculus 2, which entails partial fraction decomposition. I ... | Decompose the integrand
$$ \frac{6}{1+x^{12}}
=\frac2{x^4+1}-\frac{\sqrt3x^2-2}{x^4-\sqrt3x^2+1}
+ \frac{\sqrt3x^2+2}{x^4+\sqrt3x^2+1} $$
Then
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{12}}
=\frac23 J(0) +\frac{2+\sqrt3}3 J(\sqrt3)+ \frac{2-\sqrt3}3 J(-\sqrt3)\tag1$$
where
\begin{align}
J(a) &= \int_{0}^{\infty} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Integration of $\frac{1}{\sin x+\cos x}$ I'm given this $\int\frac{1}{\sin x+\cos x}dx$.
My attempt,
$\sin x+\cos x=R\cos (x-\alpha)$
$R\cos \alpha=1$ and $R\sin \alpha=1$
$R=\sqrt{1^2+1^2}=\sqrt{2}$,
$\tan\alpha=1$
$\alpha=\frac{\pi}{4}$
So, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}d... | Try using the substitution $t=\tan\frac x2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Is it possible to find the sum of all integer values that $x$ can take? Is it possible to find the sum of all integer values that $x$ can take? In:
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
| hint: $x+3-4\sqrt{x-1}=(\sqrt{x-1}-2)^2, x+8-6\sqrt{x-1}=(\sqrt{x-1}-3)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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... | You have $$(1-2m^2)x^2=-2m^2\\(1-2m^2)x^2=9+8mx\\-2m^2=9+8mx\\x=\frac{-2m^2-9}{8m}\\x^2=\frac{-2m^2}{1-2m^2}\\\frac{4m^4+36m^2+81}{64m^2}=\frac{-2m^2}{1-2m^2}$$Now you can use the quadratic equation in $m^2$
| {
"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": 2
} |
Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $ Cauchy and Ramanujan both gave the formula:
$$ \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} = (2\pi)^{4p+3}\sum_{k=0}^{2p+2} (-1)^{k+1}
\frac{B_{2k}}{(2k)!}\frac{B_{4(p+1)... | There is also a way to obtain this formula using the partial fraction expansion of cotangent:
$$ \frac{\pi}{2x} \coth(\pi x) - \frac{1}{2x^2} = \sum_{k=1}^{\infty} \frac{1}{k^2+x^2} $$
Let $m\geq 0$ be an integer, set $x=n$, divide by $n^{4m+2}$ and then sum from $n=1$ to infinity to get
$$ \sum_{n=1}^{\infty} \frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$
I know that
\begin{equation*}
(1 + x)^n = 1 + nx +\frac {n(n-1)}2!\cdot x^2 +\frac {n(n-1)(n-2)}3! \cdot x^3 +...
\end{equation*}
but how can I use it to solve the above problem>Is there any other ... | $$
(x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x)
$$
So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $(G,*)$ is a group? Let $G=\mathbb{R_0}\times\mathbb{R}$ where $\mathbb{R_0}=\mathbb{R}\setminus\{{0}\}$. Define operation $*$ on $G$ by $(a,b)*(x,y)=(ax,a^2y+b)$.
I'd like to prove that $(G,*)$ is a group. Immediately $(a,b), (x,y)\in G$ implies $(ax,a^2y+b)\in G$.
I need help to prove associativit... | *
*First note that $*$ is well-defined. If $(a,b), (x,y) \in G$, then $a,x \in \Bbb R^*$, so $ax \in \Bbb R^*$, which means that $(ax, a^2 y + b) \in \Bbb R^* \times \Bbb R = G$.
*We show that $(1, 0) \in G$ is the neutral element of $G$:
$$\begin{align*} (1,0) * (x,y) &= (1\cdot x, 1^2 y + 0) = (x,y) \\ (x,y) * (1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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... | You have shown the limit is $0$ along all straight lines through $(0,0).$ That's not enough: There are tons of paths to $(0,0)$ that are not straight lines. For example, the path $(x,x^2)$ as $x\to 0.$
| {
"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": 2
} |
How to find the vertex of a rhombus? I am unable to solve this question.
If the area of a rhombus is 10 sq.unit . It's diagonals intersect at (0,0) if one vertex of the rhombus is (3,4) , then one of the other vertices can be ?
I took a rhombus as ABCD . I took A as (3,4) and took O(0,0) as the point of the intersecti... | As we know that the diagonals of rhombus are equal in length & intersect each other normally. Distance of each vertex from the origin is $\sqrt{3^2+4^2}=5$.
Thus assuming the rhombus ABCD, the vertex C opposite to $A\equiv (3, 4)$ can be easily determined as the moi-point of AC is $(0, 0)$ Hence, $C(-3, -4)$. Now, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0$ I would like to prove that:
\begin{equation*}
\sum\limits_{i=0}^n (-1)^i \binom{n}{i} \binom{n-i}{k}=0;~k\geq0 ; n\geq1.
\end{equation*}
Can any one help me how to do that? Thanks
| Just for fun, here's a generating function answer. Fix $k \geq 0$ and let $a_n = (-1)^n$ and $b_n = \binom{n}{k}$. Letting $a$ and $b$ be the exponential generating functions for these sequences, then
\begin{align*}
a(x) = \sum_{n \geq 0} (-1)^n \frac{x^n}{n!} = \sum_{n \geq 0} \frac{(-x)^n}{n!} = e^{-x}
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
$\lfloor 2x \rfloor \lfloor 3x \rfloor$ From $1$ to $10000$ including both, how many of those integers can be written as:
$$\lfloor 2x \rfloor \lfloor 3x \rfloor$$
Where $x$ is a real number?
| To help you get you started:
If $n \le x < n+\dfrac{1}{3}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n \cdot 3n = 6n^2$.
If $n+\dfrac{1}{3} \le x < n+\dfrac{1}{2}$ for some integer $n$, then $\lfloor 2x \rfloor \lfloor 3x \rfloor = 2n(3n+1) = 6n^2+2n$.
If $n+\dfrac{1}{2} \le x < n+\dfrac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$ Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
This is my try
Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$
The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$
I have no idea how to solve t... | $$
(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1
$$
Let $t=x+1$,
$$2x+3=2t+1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})+\sqrt{2t^2+t}=1$$
$$(t+1)(\sqrt{2t+1}-2\sqrt{t})=1-\sqrt{2t^2+t}$$
Squaring both sides,
$$(t^2+2t+1)(2t+1+4t-4\sqrt{2t^2+t})=1+2t^2+t-2\sqrt{2t^2+t}$$
$$6t^3+12t^2+6t+t^2+2t+1-\sqrt{2t^2+t}(-4t^2-8t-2)=1+2t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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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... | To get rid of the square roots of the denominator, you may use $a^2-b^2=(a+b)(a-b)$.
$$\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{7-4\sqrt{35}+20-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{24-4\sqrt{35}}\\=\dfrac{(\sqrt{7}-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $\sin(a)$ + $\cos(a)\leq\sqrt{2}$ $$\begin{align*}
\sin (a) + \cos(a) &\leq \sqrt{2}\\
(\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\
\sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2}
\end{align*}$$
Am I doing it right? I need help.
| Alternative path:
$$
\sin a+\cos a=
\sqrt{2}\left(\sin a\cos\frac{\pi}{4}+\cos a\sin\frac{\pi}{4}\right)=
\sqrt{2}\sin\left(a+\frac{\pi}{4}\right)\le\sqrt{2}
$$
(and also $\ge-\sqrt{2}$, of course).
However your reasoning is basically correct; only you need to do it backwards:
$$
\sin^2a+2\sin a\cos a+\cos^2a\le2
$$
be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$ Let $a,b,c\in \mathbb{R^+}$.
Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+... | By Holder $\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b^2+c^2}}\right)^3\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3\geq\left(\sum\limits_{cyc}(a^2+ab)\right)^4$.
Thus, it remains to prove that $\left(\sum\limits_{cyc}(a^2+ab)\right)^4(a+b+c)^3\geq729abc\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Limit Problem-To find the value of a and b,when the value of the limit is given. Here,is a limit problem: $\lim \limits _{x \to 0} {x^3 \over {\sqrt {a+x}} (bx - \sin x)} = 1$. Here, $a \in \mathbb R _+$. The question is to find the values of $a$ and $b$.
Here is my workout. Dividing the numerator and denominator by $x... | Suppose that $b \neq 1$.
Concerning $a$, there are two possibilities. Assume first that $a \neq 0$. Then $\sqrt {a+x} \to \sqrt a$. The rest of your expression becomes ${x^3 \over {bx - \sin x}} = {x^2 \over {b - {\sin x \over x}}}$, which will tend to $0$, so your whole limit will be $0$, not $1$. If $a=0$, then your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to show algebraically that $x^3 +3x +1$ is injective?
How to show algebraically that $$x^3 +3x +1$$ is injective?
Working with the usual method of assuming that $f(c)=f(d)$ and then seeing if $c=d$. I've tried several approaches, including factoring by the difference of cubes, followed by use of the quadratic for... | Notice $$\begin{align}
f(x) - f(y) &= (x^3 + 3x + 1) - (y^3 + 3y+1)\\
&= (x-y)(x^2 + xy + y^2 + 3)\\
&= \frac12 (x-y)( x^2 + (x+y)^2 + y^2 + 6)\end{align}$$
Since $x^2 + (x+y)^2 + y^2 + 6 \ne 0$ for any pair of real numbers $x, y$;
we have $f(x) \ne f(y)$ whenever $x \ne y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :
$$\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}$$
where ... | By the conditions, we have:
$$y =\frac{1-xz}{x+z}\ge0\quad\longrightarrow \quad xz\le1, z \le 1/x$$
Now, without loss of generality, assume $z\ge y\ge x$. We therefore have that the expression $f(x,y,z)$ in question satisfies:
$$f(x,y,z) \ge \frac{3}{2} \frac{x^n}{z^m}\ge \frac{3}{2}x^{n-m}$$
Now it remains to be shown... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
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Show that $\sin^7 x + \cos^7 x < 1$ if $0 < x < \frac{\pi}{2}$. Could anybody help me solve this Problem? I don't quite get it.
This is what I got.
$\sin^7x+\cos^7x<1$
$0<x<\frac{\pi}{2}$
$0<\cos x$
$1>\sin x$
What else?
| Using the Pythagorean theorem on the unit circle, we can easily figure out the identity of $\sin^2 x + \cos^2 x = 1$. We also know that if $0<x<\frac{\pi}{2}$, then $0 < \sin x < 1$ and $0 < \cos x < 1$. We know that if $0<\alpha<1$ and $n>1$ are true, then $\alpha ^n < 1$. If $\sin^2 x + \cos^2 x = 1$, then $\boxed{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}$ How to prove
\begin{equation*}
\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}\ ,where \ a,b\ge 0
\end{equation*}
I applied AM-GM inequality but obtained only
\begin{equation*}
\frac{3}{2}(a+b)\ge a\sqrt{b}+b\sqrt{a}+1
\end{equation... | From first principles:
$$(\sqrt{x}-\sqrt{y})^2 \geq 0$$
$$x+y-2\sqrt{xy} \geq 0$$
$$\sqrt{xy} \leq \frac{x+y}{2}$$
And:
$$(\sqrt{x}-\frac12)^2 \geq 0$$
$$x+\frac14-\sqrt{x} \geq 0$$
$$\sqrt{x} \leq x + \frac14$$
Given that:
$$\begin{split}
a\sqrt{b}+b\sqrt{a} &= \sqrt{ab}(\sqrt{a}+\sqrt{b}) \\
&\leq\frac{a+b}2(\sqrt{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$. Hint: $x^2-y^2 = (x+y)(x-y)$. This is the exercise verbatim:
An integer n is a square modulo p if there exists another integer x
such that $n \equiv x^2 \pmod p$. Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equi... | You can use the theorem : p( a prime) divides the product $ab$ if and only if p divides $a$ or $p$ divides $b$. Now $x^{2} \equiv y^{2}$ mod $p$ if and only if $p$ divides $x^{2}-y^{2}=(x-y)(x+y)$, if and only if $p$ divides $x+y$ or $p$ divides $x-y$ if and only if $x \equiv y$ mod $p$ o $x \equiv -y$ mod p.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Part of an induction question I might have done or not realize something stupid, but I can't seem to prove the following...
Inductive hypothesis
Assume $\exists$k$\in$N such that P(k) is true.
P(k): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k)}$ $\leq$ $\frac{1}{\sqrt{k + 1}... | You have given away too much. You want to prove that
$$\frac{2k+1}{2(k+1)\sqrt{k+1}}\lt \frac{1}{\sqrt{k+2}}.$$
Equivalently, you want to prove that $(2k+1)^2(k+2)\lt 4(k+1)^3$. Expand, and the result will drop out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $... | You may try to show that $A^TA+I$ is invertible. Hint: if $(A^TA+I)x=0$, consider $x^T(A^TA+I)x$. Try to rewrite it as a sum of squares and infer that $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$... | I started out by observing that $2+3 = 5$; I thought that might make it easier. Then
$$
x-\sqrt{5} = \sqrt{2}+\sqrt{3}
$$
$$
x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5
$$
$$
x^2-2\sqrt{5}x = 2\sqrt{6}
$$
$$
x^4-4\sqrt{5}x^3+20x^2 = 24
$$
$$
x^4+20x^2-24 = 4\sqrt{5}x^3
$$
$$
x^8+40x^6+352x^4-960x^2+576 = 80x^6
$$
$$
x^8-40x^6+352x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.