Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$ How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$
This equation How prove it? Thank y... | Start from the RHS. We are counting the number of ways to choice $k$ elements among $n+1$, then choice $n-k$ elements among the $k$ elements previously chosen. If we imagine to assign a $+1$ weight in the first step, then increase the weight in the second step, we are counting the number of ways to assign a $+1$ or $+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/664823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| Let $a=\sqrt{2}\cos\theta$, $y=\sqrt{2}\sin\theta$. Then $a^2+b^2=2$, and $a+b=\sqrt{2}(\cos\theta+\sin\theta)$, which is a maximum at $\theta=\frac{\pi}{4}$, at which case $a+b=2$. So $a+b\le 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/666217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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IMO problem 4, $1998$ Determine all pairs $(a, b)$ of positive integers such that $ab^{2} + b + 7$ divides
$a^{2}b + a + b$. I really have no idea where to start with this. This is the first IMO problem that I attempted, because it looked to be the easiest, but I honestly have no idea where to begin. Any help or hints ... | I've got a different solution to the one offered here already.
If $ab^2+b+7 \mid a^2b+a+b$, then $ab^2+b+7 \mid b(a^2b+a+b) - a(ab^2+b+7) = b^2-7a$
Now there are three cases left to check:
*
*$b^2-7a = 0$. This yields the infinite set of solutions $(a,b) = (7k^2, 7k)$, where $k\ge1$
*$b^2-7a > 0$. Hence, we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Max. and Min. value of $f(x,y) = \frac{x-y}{x^4+y^4+6}$ [1] Calculation of Max. and Min. value of $\displaystyle f(x) = \sqrt{x^3-6x^2+21x+18}$, where $\displaystyle -\frac{1}{2}\leq x\leq 1$
[2] Calculation of Max. and Min. value of $\displaystyle f(x,y) = \frac{x-y}{x^4+y^4+6}\;,$ where $x,y\in \mathbb{R}$
$\bf{My\; ... | Hint: Have a look at this for the second problem
Second partial derivative test
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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When is $3^n + n$ a power of 2? For what $n \in \mathbb{N}$ is $3^n + n$ a power of $2$?
| This is not a complete solution, but shows that any nontrival solutions must be mindbogglingly big if it exists at all.
We have the trivial solutions $3^0+0=2^0$ and $3^1+1=2^2$.
Assume $3^n+n=2^m$ with $n\ge 2$. Then clearly $m>n$.
We find $1+n\equiv 0\pmod 2$, hene $n$ is odd. Then $3^n\equiv 3\pmod 8$ and hence $n\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/671143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
|
$z^3=i$ $⟹ z=\sqrt [3] {i}$
We know,
$\sqrt [3] {i} =a+bi$
$⟹ i=(a+bi)^3 = a^3-3ab^2+3a^2bi-b^3i= a^3-3ab^2+(3a^2b-b^3)i$
From this we can say,
$a^3-3ab^2=0$________________(i)
$3a^2b-b^3=1$_______________(ii)
From (i) we get,
$a(a^2-3b^2)=0$
So, $a=0$
Or,
$a^2-3b^2=0$ $⟹ a=±b\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 6
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Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer... | Let's generalise this: you want to write $\dfrac{1}{x+y\sqrt{z}}$ in the form $b+c\sqrt{z}$:
$$\dfrac{1}{x+y\sqrt{z}} = \dfrac{(x-y\sqrt{z})}{(x+y\sqrt{z})(x-y\sqrt{z})} = \dfrac{x-y\sqrt{z}}{x^2-y^2z} = \dfrac{x}{x^2-y^2z} + \dfrac{-y}{x^2-y^2z} \sqrt{z}.$$
So $b= \frac{x}{x^2-y^2z}$ and $c= \frac{-y}{x^2-y^2z}$.
It w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How Find this $T_{20}-S_{20}$
Let $\{a_{n}\},\{b_{n}\}$ be sequences such that
$$\sqrt{\dfrac{a_{n}+1}{b_{n}}}=\dfrac{1}{n}, \quad S_{n}=\sqrt{T_{n-1}} \quad(n>1)$$
where $\displaystyle S_{n}=\sum_{i=1}^{n}a_{i}$, $\displaystyle T_{n}=\sum_{i=1}^{n}b_{i}$, and $b_{1}=1$.
Find $T_{20}-S_{20}$.
(a) $15980$
(b) $359... | We'll prove that $a_n=(n-1)$ and $b_n=n^3$ by complete induction.
For $n=1$, we have $b_1=1$ straight from problem statement and $a_1=0$ follows from $\sqrt{\frac{a_1+1}{b_1}}=1$.
For any higher $n$, we have
$$S_n=\sum_{k=1}^n a_k=\left(0+1+\ldots+(n-2)\right)+a_n=\frac{(n-1)(n-2)}{2}+a_n$$ and
$$T_{n-1}=\sum_{k=1}^{n-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Two-variable limit question $\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$
How to solve this two-variable limit? Thanks :D
| This might also be an approach:
$$0 \le \frac{x^2y^2}{(x^2 + y^4)\sqrt{x^2 + y^2}} \le \frac{x^2y^2}{x^2\sqrt{x^2+y^2}} \le \frac{y^2}{\sqrt{y^2}} \le |y|$$
So, since $(x,y) \to (0,0)$ then $|y| \to 0$ and passing to a limit we obtain that the initial expression goes to 0.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$ Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$.
Can someone explain to me what a closed form is and how to approach this problem?
| A closed form is an expression that doesn't involve any looping/repeatition/recursion (like sums, products, sequences, etc.). It is not exactly clear, for example some consider binomials $\binom{n}{k}$ a closed form expression, while others do not. However, in case of sums of polynomials, it is quite clear: it is anoth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/680860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the following system of equations. $2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$
I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.
| By multiplying the second by $-2$ you get:
$$2x^2-3xy+2y^2=\frac{6}{4}$$
$$-2x^2+8xy-2y^2=1$$
Then adding them you get:
$$5xy=\frac{10}{4} \Rightarrow xy=\frac{1}{2} \Rightarrow x=\frac{1}{2y}$$
Substituting this in the second equation you get:
$$\frac{1}{4y^2}-2+y^2+\frac{1}{2}=0 \Rightarrow \frac{1}{4y^2}+y^2=\frac{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int... | May I confess that I sawer nicer integrals ?
Anyway, being patient and using Mhenni Benghorbal's suggestions plus a series of integrations by parts as well as a few changes of variables, the antiderivative is
$$\frac{1}{24} e^{-\sqrt[6]{-1}} \left(e^{\sqrt{3}} \left(\left(\sqrt{3}-3 i\right)
\left(\text{Ei}\left(i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Factoring a hard polynomial This might seem like a basic question but I want a systematic way to factor the following polynomial:
$$n^4+6n^3+11n^2+6n+1.$$
I know the answer but I am having a difficult time factoring this polynomial properly. (It should be $(n^2 + 3n + 1)^2$). Thank you and have a great day!
| Exploit the symmetry of the palindromic polynomial to factorise as follows,
$$\begin{align*}
n^4 + 6n^3 + 11n^2 + 6n + 1
&= n^2\left( n^2 + \frac{1}{n^2} + 6n + \frac{6}{n} + 11 \right)\\
&= n^2\left( (n + 1/n)^2 + 6(n + 1/n) + 9 \vphantom{\frac{1}{n^2}} \right)\\
&= n^2\left( n + 1/n + 3 \right)^2\\
&= \left( n^2 + 3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 5
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QR-factorization of a tridiagonal matrix super diagonals question I understand it is possible to QR-factorize a tridiagonal matrix A by performing Given's plane rotations:
$$ J(n-1,n)J(n-2,n-1)... J(1,2) A =R$$ where $R$ is upper triangular. I have read that in this case the first two super-diagonals of R will be non-z... | When working with Givens rotations on structured matrices, it is always instructive to draw a picture and of course to notice that pre-multiplying with $J(k,k+1)$ acts only on rows $k$ and $k+1$ (and post-multiplying affects the columns $k$ and $k+1$). In the pictures, $\times$ denotes nonzero entries, $\color{red}\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where
$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$
Here $x_j=j/(n+1)$.
So, workin... | To see that the given expression is equal to $1$, set
$\omega = e^{\frac{ik\pi}{n + 1}}, \tag{1}$
then the expression becomes
$\dfrac{1}{1 - \omega} + \dfrac{1}{1 - \bar{\omega}} = \dfrac{1 - \bar{\omega} + 1 - \omega}{(1 - \omega)(1 - \omega)} = \dfrac{2 - (\omega + \bar{\omega})}{(1 - \omega)(1 - \bar{\omega})}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
| Solution without complex factorization:
Let,
$ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $
$ y= \sqrt{4-\sqrt{4+\sqrt{ 4-\dots}}} $
Clearly,
$x = \sqrt{4 + y} \implies x^2 - y = 4$ (i)
$y = \sqrt{4 - x} \implies y^2 + x = 4$ (ii)
from (i) and (ii) we get,
$x^2 - y = y^2 + x$
$\implies (x+y)(x-y) = x+y$
Since, $x,y \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these sta... | $x = 5n + k$, then $x^2 + x - 3 = (5n + k)^2 + 5n + k - 3 = k^2 + k - 3\pmod 5$ now check $k = 0, 1, 2, 3$, and $4$ you don't get $0\pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Product of inverse matrices $ (AB)^{-1}$ I am unsure how to go about doing this inverse product problem:
The question says to find the value of each matrix expression where A and B are the invertible 3 x 3 matrices such that
$$A^{-1} = \left(\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ 1& 1& -1\end{array}\right)
$$ and
$$B^... | $(A^{-1}B^{-1})$ is not always equal to $(AB)^{-1}$.
Analogous to matrix transpose $(AB)^T = B^TA^T$, we have $(AB)^{-1} = B^{-1}A^{-1}$.
Further, matrix multiplication is not commutative. Here is a proof to show this, but we can see this fact from a simple counterexample involving two square matrices $A$ and $B$.
$A =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $... | Also in the second parenthesis $a$ squared will be $(a-2)$ squared i.e. $(a-2)(a-2) = a^2 -4a + 4$,
and similarly the $b$ squared will be $(a+1)$ squared i.e. $(a+1)(a+1)$ i.e. $a^2 + 2a + 1$.
I think having $a$ in the expression you want to factorize and in the identity you are trying to use is confusing you. Why not... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the arc length of the curve $x = 1/6*(y^2+ 3)^{3/2}$ from $y = 0$ to $y = 1$ I am trying to find the arclength of the curve $$x = 1/6\cdot\left(y^2 + 3\right)^{3/2},\;\; 0\leq y\leq 1$$ I got this far and now I am stuck and don't know what to do next. Any help please?
$$\begin{align} dx & = \left(1/6\right)\cdo... | Simpson's Rule with n=2 already gives a result correct to four decimal places.
$$\frac16\left(1+4\sqrt{1+\left(\frac{1}{2^4}+3\left(\frac{1}{2^2}\right)\right)/4}+\sqrt{2}\right)\approx1.1336\ldots$$
(A CAS says that the integral is $1.13359\ldots$)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 ... | For the third one
$$3\sin^2x-4\sin x\cos x+5\cos^2x=2\\
\Rightarrow\sin^2x-4\sin x\cos x+4\cos^2x=2-2\sin^2x-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=2(1-\sin^2x)-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=2\cos^2x-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=cos^2x\\
\Rightarrow(\sin x-2\cos x)^2-\cos^2x=0\\
\Rightarrow(\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Definite Integral $\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$ $$\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$$
WA gives $\pi^2/9$
| Mhenni has struck first with the approach I have taken, but I would like to elaborate. Again, the integrand may be Taylor expanded:
$$\begin{align}-\int_0^1 dx \frac{\log{[1-(x-x^2)]}}{x-x^2} &= \sum_{n=0}^{\infty} \frac1{n+1} \int_0^1 dx \, x^n (1-x)^n\\ &= \sum_{n=0}^{\infty} \frac1{n+1} \frac{n!^2}{(2 n+1)!}\\ &=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
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Quadratic Congruence for every prime Let $p$ be a prime number. Prove that exists $a,b \in \mathbb{Z}$ such that $p|a^2+b^2+1$. What I've tried:
If $p \equiv 1 \mod 4$, we put $b=0$ and the condition is simply $a^2 \equiv -1 \mod p$ which has a solution by basic quadratic reprocity.
If $p \equiv 3 \mod 4$, I did not se... | The result is obvious for $p=2$, and as you point out it is straightforward for primes of the form $4k+1$. But we give a proof that works for all odd primes.
Let $A$ be the set $\left\{0^2,1^2,2^2,\dots, \left(\frac{p-1}{2}\right)^2\right\}$.
Let $B$ be the set $\left\{-1-0^2, -1-1^2, -1-2^2, \dots, -1-\left(\frac{p-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
| $$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
Let's factor the $(x-3)$ in the numerator and denominator.
$$\dfrac{(x-3)\left[3(x+2)^2(x-3) - 2(x+2)^3\right]}{(x-3)(x-3)^3}$$
Now we can cancel out $(x-3)$ in the numerator and the denominator. That gives us:
$$\dfrac{3(x+2)^2(x-3) - 2(x+2)^3}{(x-3)^3}$$
Let us exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I get from $\frac{-x+1}{-x+2}$ to $1 + \frac{1}{x-2}$ wolframalpha tells me it's the same but I can not follow how to get from one to another.
$$\frac{-x+1}{-x+2} = \frac{1-x}{2-x} = \>? \dots$$
I don't get any further, always end up where I started.
| $$\begin{align} \dfrac{1-x}{2-x} & = \dfrac{-(x - 1)}{-(x-2)} \\ \\ &= \dfrac{x-1}{x-2} \\ \\ &= \dfrac {x-1 \color{blue}{\bf - 1 + 1}}{x-2} \\ \\ & = \dfrac{(x-2) +1}{x - 2}\\ \\ & = \dfrac{x-2}{x-2} + \dfrac 1{x-2}\\ \\ & = 1 + \frac{1}{x - 2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do I solve delta epsilon proofs for quadratic equations? For the $\lim\limits_{x\rightarrow 2} (x^2 + 5x - 2) = 12$ I need to show how to find a $\delta$ such that $|f(x) - L| < \varepsilon$ for all $x$ satisfying $0 < |x - a| < \delta$.
Help is appreciated.
| We have show that for all $\epsilon>0$ there is a $\delta >0$ such that
$$
|x-x_0|<\epsilon \implies | (x^2+5x-2) - 12|<\epsilon .
$$
We will manipulate the expression $| (x^2+5x-2) - 12|<\epsilon$ to find a candidate to $\delta$.
\begin{align}
| (x^2+5x-2) - 12|<\epsilon \implies & -\epsilon <x^2+5x-14<\epsilon \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove an equation about binomial coefficients Could we prove:
$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$
| Here is a similar approach using basic complex variables.
Suppose we seek to evaluate
$$\sum_{k\ge 0} {2k\choose k} {n+k\choose m+2k}
\frac{(-1)^k}{k+1}$$
where $n\ge m.$
Introduce the integral representation
$${n+k\choose m+2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz.$$
This yie... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proof of $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number . I want to Prove $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number . For this purpose , my trying is as the following :
$ e^x-x^2 = \{1+x+\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \}-x^2$
$= 1+\{x-\dfrac{x^2}{2!... | The second derivative of $e^x-x^2-1$ gives you the inflection point at $x = ln(2)$.
The problem is equivalent to $e^{ln(2)}-len(2)^2>1$ or $1>ln(2)$, wich is true
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find complex solutions Find all solutions of the equation $4\sin(z) + 5 = 0$.
So $4\sin(z) + 5 = 4\sin(x+iy) + 5 = 0$. Hence $\sin(x+iy) = \frac{-5}{4}$. So $x+iy = \sin^{-1}(\frac{-5}{4})$ which gives us $x+iy =$ ...
I am not sure what is really being asked here or how to achieve it. Help?
| $\displaystyle\sin z=-\frac54$
$\displaystyle\cos z=\pm\sqrt{1-\frac{25}{16}}=\pm\frac{3i}4$
Using Euler formula, $$e^{iz}=\cos z+i\sin z=\pm\frac{3i}4+i\left(-\frac54\right)=-\frac{(5\mp3)i}4=-2i\text{ or }-\frac i2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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EXERCISE VERIFICATION: Find where $f(x):=|x|+|x+1|$ is differentiable and calculate its derivative Could someone verify my exercise?
a) $f(x):=|x|+|x+1|$
First, analyse the roots of each absolute value, where they go to zero:
$$|x|:=\left\{\begin{matrix}
& x& x>0 \\
& x- & x<0\\
& 0 & x=0
\end{matrix}\right.$$
$... | It is a good idea to draw the function first:
From this we guess that the function is differentiable except at $x=-1$ and $x=0$.
Your formula for $f$ above is correct, and from this we see that for $x \in \mathbb{R} \setminus \{-1,0\}$, the function $f$ is differentiable.
For $x=-1$, we see that ${f(-1+h)-f(-1) \over ... | {
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"source": "stackexchange",
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"answer_id": 0
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$\epsilon$-$\delta$ limit proof that $\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$ I need to prove that $$\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$$ using the epsilon definition.
I'm having specific trouble understanding how to make it less than epsilon once I've simplified the equation.
| $$
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac 1 3\right| = \left|\frac{3(n^2-n+2)}{3(3n^2+2n-4)} - \frac{3n^2+2n-4}{3(3n^2+2n-4)}\right|
$$
$$
= \left|\frac{-5n+10}{3(3n^2+2n-4)}\right| \le \frac{\text{some constant}}{n}
$$
Massage the whole thing a bit to figure out what "constant" can serve in this role, and then the wh... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the maximum and minimum of a sinusoidal function? I understand basic $\sin x$ and $\cos x$ min/max, but I am having a problem solving the minimum and maximum of the following:
$f(x) = \sin^2 x - \sin x$
Oh, and the range is $0 \le x \le \frac{3\pi}{2}$
| $$\begin{align*} f(x) &= \sin^2 x - \sin x \\ &= \sin^2 x - 2 \cdot \tfrac{1}{2} \sin x + (\tfrac{1}{2})^2 - \tfrac{1}{4} \\ &= \left( \sin x - \tfrac{1}{2} \right)^2 - \tfrac{1}{4}\end{align*}.$$ Because the square of a real number is nonnegative, $f$ attains a minimum if $\sin x = \frac{1}{2}$, and the consequences ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ according as $n$ is odd or even.
Since for any $n\ge3$,$2^{2^n}\equiv2^{2^{n-2}}\pmod9$, if $n=2k+3$
then
$$2^{2^n}=2^{2^{2k... | The proof is by induction on $n$. We have $F_1\equiv 5\pmod{9}$.
Note that $F_{n+1}-1=(F_n-1)^2$. This is because $2^{2^{n+1}}=(2^{2^n})^2$.
Rewrite as
$$F_{n+1}=(F_n-1)^2+1.$$
Now we do the induction step.
If $F_k\equiv 5\pmod{9}$, then $(F_k-1)^2\equiv 16\equiv 7\pmod{9}$, and therefore $F_{k+1}\equiv 8\pmod{9}$.
S... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding an algebraic proof for $r{n \choose r} = n{n-1 \choose r-1}$ I can't seem figure this proof out.
How are both sides equal.
$$r{n \choose r} = n{n-1 \choose r-1}$$
| Here are a couple of approaches in addition to the approach used by Dror and TMM.
Generating Functions:
$$
\frac{\mathrm{d}}{\mathrm{d}x}(1+x)^n=\sum_{r=1}^n r\binom{n}{r}x^{r-1}
$$
$$
\begin{align}
n(1+x)^{n-1}
&=\sum_{r=0}^{n-1}n\binom{n-1}{r}x^r\\
&=\sum_{r=1}^nn\binom{n-1}{r-1}x^{r-1}\\
\end{align}
$$
Compare the c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Solving an equality in 2 variables I need to prove that
$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$
if $a+b = 1$ and $a b \le 1/4$
I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get
$a b \le 4$ in the inequality ?
| Apply Cauchy-Schwarz Inequality to argue that,
$(a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$,
And further, $\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab}=\frac{1}{ab}\ge \frac{4}{(a+b)^2}=4$.
So the first expression becomes,
$25=(1+4)^2 \le (a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How can I solve this question? Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D".
$$\int_0^2 \frac{dx}{\sqrt{4-x^2}}$$
Do I make $u= 4-x^2$ then $du= -2x \, dx$
Not exactly sure..
| Consider the substitution $x = 2\sin\theta$. From this we have $dx = 2\cos\theta\cdot d\theta$. Substitute these in and see what happens:
$$\int\frac{2\cos\theta \cdot d\theta}{\sqrt{4 - 4\sin^2\theta}}\\\\
= \int\frac{2\cos\theta \cdot d\theta}{2\cos\theta}\\
= \int d\theta\\
= \theta + C\\
= \sin^{-1}{\frac{x}{2}} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4) \\
\end{cases}
I am stuck at this point. Not sure on how to move forward. ( A small change made)
| $a=(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$a^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2$
substitute $a^2$ and $a$ in the second equation
$b^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2+c_3^2+2c_3c_4(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$b^2c_2^2=c_1^2(c_3^3-b^3)^2+c_3^2c_2^2b^3+2c_3c_4c_1c_2(c_3^3-b^3)$
$0=c_1^2c_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | Let's recall the definition of the Inverse Hyperbolic Tangent:
$$\begin{cases}\operatorname{arctanh}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\\\frac{d}{dx}\operatorname{arctanh}(x)=\frac{1}{1-x^2}\end{cases}; x\in(-1,1)$$
So, the integral can be rewritten as:
$$\require{cancel}\begin{align}\int\cos(2x)\log\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with proof using induction: $1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ I am having trouble with the following proof:
For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$
My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}... | Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$.
Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}. $
So it suffices to show that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.
But this, by simple algebra, is equivalent to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If a prime can be expressed as sum of square of two integers, then prove that the representation is unique. If a prime can be expressed as sum of two squares, then prove that the representation is unique.
My attempt:
If $a^2+b^2=p$, then it is obvious that $a,b$ of different parity.
Now, I assume the contraposition t... | Here's an answer without Algebraic Number Theory. I found it in Shanks, Solved and Unsolved Problems in Number Theory.
Assume $$p=a^2+b^2=c^2+d^2\tag1$$ with all variables positive integers. Then $$p^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ and you can verify by just multiplying everything out that $$p^2=(ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/719700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Why is $ \lim_{x \to \infty} \ x^{2/x} = 1$ Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can't see any manipulation that would suggest this result?
| $\lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1$
Proof using AM-GM and Sandwich Theorem
$\frac{1 + 1 + 1 + \dots + \sqrt{x} + \sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$\frac{x - 2 + 2\sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$1 - \frac{2}{x} + \frac{2}{\sqrt{x}} \geq \sqrt[x]{x} \geq 1$
$\lim \limits_{x \to \infty} 1 - \frac... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Could someone please help me evaluate this integral? The integral is
$$
\int_{0}^{1/\sqrt{\vphantom{\large A}2\,}}
{\arccos\left(x\right) \over \sqrt{\vphantom{\large A}1 -x^2\,}\,}\;{\rm d}x
$$
I was just wondering if I could use substitution to solve this problem or if I had to solve it a different way.
| We have to evaluate the definite integral:
$$\int_0^{1/\sqrt{2}}\frac{\arccos x}{\sqrt{1-x^2}} \ dx$$
Just use u-substitution. Let:
$$u=\arccos x \implies du=-dx$$
$$\int_0^{1/\sqrt{2}}u \ -du=-\int_0^{1/\sqrt{2}}u \ du=\left.-\frac{u^2}{2}\right|_0^{1/\sqrt{2}}$$
Substituting $\arccos x$ back for $u$:
$$\int_0^{1/\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ I'm trying to find the maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ where $n\ge 2$, on a the set $\{(q_1,\ldots , q_n) :q_1^2+q_2^2+\ldots+q_n^2=1 \ q_i\ge 0 \}$ (With the condition $q_i\ge0$ this is just the upper half of the sphere). This appeared to be a simple Lagrange multiplie... | Addendum: I've included the full set of solutions Mathematica finds for the case $n=3$ at the end of this answer.
The comments and answers so far don't show how Lagrange multipliers can solve the problem (because of the non-negativity condition). Here's how that technique can be used.
To maximize a function $f(q_1,\do... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Does $7$ divide $2 x^2 - 4y^2$ for all $x,y$? Does $7$ divide $2 x^2 - 4y^2$ for all $x,y \in \mathbb{Z}$?
| Let $x = 7a + 1$ and $y = 7b + 1$ for integers $a,b$. Then,
$$\begin{align}2a^2 - 4b^2 &\equiv 2\cdot1^2 - 4\cdot1^2\pmod 7\\
&= 5\pmod7\\
&\equiv 5 \pmod 7\end{align}$$
So it is not generally true, because we can always generate integer pairs (though not all of them) that give a remainder of $5$ when divided by $7$ as... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ Prove that for all $x,y>0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.
It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
| Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $x\ln(x)$. $f''(x) = \frac{1}{x} > 0$, therefore $f$ is convex. By Jensen inequality,
$$f\left(\frac{x + y}{2}\right) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$$
That is,
$$\left(\frac{x+y}{2}\right)\ln\left(\frac{x + y}{2}\right) \leq \frac{1}{2} x\ln(x) + \frac{1}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 1
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Interference beats for a more general trigonometric sum Suppose I have three frequencies $\alpha,\beta,\gamma$ that are all close in value, and I consider the sum $\sin(\alpha x) +\sin(\beta x) +\sin(\gamma x)$ If there were only summands I could find a million explanations online that shows me how to rewrite this as a... | Let us expand
$$\sin\left(\dfrac{\alpha x}{3}+\dfrac{(\beta x+\gamma x)}{3}\right)$$
We get
$$\sin \dfrac{\alpha x}{3} \cos \dfrac{(\beta x+\gamma x)}{3} + \cos \dfrac{\alpha x}{3} \sin \dfrac{(\beta x+\gamma x)}{3}$$
Let us now expand
$$\cos \dfrac{(\beta x+\gamma x)}{3}= \cos \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{... | {
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Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$ I tried
$$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$
Simplifying,
$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$
Then I'm stuck!
| Using that $$-2 = -2\sin^2(x) - 2\cos^2(x)$$ from $$6\sin^2(x) + 2\cos(x)\sin(x) - 2 = 0$$ we get $$4\sin^2(x) + 2\cos(x)\sin(x) - 2\cos^2(x) = 0$$ and dividing by $\cos^2(x)$ we have $$4\tan^2(x) + 2\tan(x) - 2 = 0$$
Can you continue from here? Note that when dividing by $\cos^2(x)$ we assume that it is not $0$. What ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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homogeneous coordinates I am to use homogeneous coordinates to calculate a standard matrix for a projection onto the line $4x-2y=6$ from the point $(3,10)$.
I'm not sure what homogeneous coordinates are and neither how to use them to calculate my problem? Please believe me, I've been looking for information on various ... | In 2D points in homogeneous coordinates have the form $P = (x,y,1)$ and lines $L=(a,b,c)$ such that the equation for the line can be found by
$$ L \cdot P = 0 \} a x + b y + c = 0 $$
So in your case, the homogeneous coordinates for the point is $P=(3,10,1)$ and the line $L=(4,-2,-6)$.
The trick with homogeneous coordin... | {
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Integer solutions of $800000007 = x^2+y^2+z^2$ Prove that the equation, $800000007 = x^2+y^2+z^2$ has no solutions in integers.(That is $8$ followed by $7$ zeroes, with a $7$ at the end).
I tried checking modulo $3$, $5$, $7$, and $10$, but couldn't reach any conclusions.
| Hint: have a look at modulo $8$.
details:
If $x^2+y^2+z^2 = A $
then $A\neq 7\mod 8$:
$$
x^2\in \{0,1,4\}\mod 8\\
x^2+y^2+z^2 \in \{0,1,4\}+\{0,1,4\}+\{0,1,4\} =
\{0,1,2,3,4,5,6\}\mod 8.
$$
NB: the general equation $x^2+y^2+z^2 = A $ has integer solutions iff $A$ has not the form
$$
4^N(8k+7).
$$
Let us prove that if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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a manipulation of Fibonacci recurrence Let $F_n$ be the Fibonacci number, and we know
$F_{n+2} = F_{n+1} + F_{n} $ with $F_0 =1,F_1 = 1$
And this can be manipulated to
$F_{n+6} = 4F_{n+3} + F_n$
if we let n be a multiple of 3, we can write $A_r = F_{3r}$, and we get
$A_{r+2} = 4A_{r+1} + A_r$ with $A_0=1, A_1=1$
So I... | Recall the theory of recurrence relations: The characteristic equation of the fibonacci sequence is $X^2 - X - 1 = 0 $.
This tells us that $F_n = a \alpha^n + b \beta ^n $ (for some constants $a$ and $b$), where $ \alpha + \beta = -1 $ and $ \alpha \beta = -1$.
Setting $A_n = F_{rn}$, we get that $$A_n = a(\alpha^r)^n... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequalit... | First note that $x>1$ and simplify like this:
$$\frac{3x^3+125y^3}{x-y}=\frac{3x^6+125x^3y^3}{x^4-x^3y}=\frac{3x^6+125}{x^4-x^2}$$
Set $t=x^2$. Now we want to find $a\ge0$ such that
$$\frac{3t^3+125}{t^2-t}\ge a\Longleftrightarrow3t^3+at+125\ge at^2$$
for $t>1$ where the equality is possible.
Note that we could use AM-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/733770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Equation in rational numbers? Is it true that this equation $6=\frac{x^2}{y^2+1}$ has no solutions in rational numbers?
If so, why?
It is quite evident that it has no solutions in integers (because $y^2+1$ never divides $3$).
| Equivalently, we are looking for a solution of $6(a^2+b^2)=c^2$ in integers not all $0$.
Suppose to the contrary that there is such a solution. Let $3^k$ be the highest power of $3$ that divides both $a$ and $b$. Let $a=3^ks$ and $b=3^k t$.
Then $3$ does not divide $s^2+t^2$, so the highest power of $3$ that divides... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question with Stokes theorem Show with Stokes that $\oint_C (y\mathbf{i}+z\mathbf{j}+x\mathbf{k})\bullet d\mathbf{r}=\sqrt{3}\pi a^2$ when $C$ is intersection of $x^2+y^2+z^2=a^2$ and $x+y+z=0$.
My work:
$$z=g(x,y)=-x-y$$
$$\mathbf{N}=\frac{-\frac{\partial g}{\partial x}\mathbf{i}-\frac{\partial g}{\partial y}\mathbf{j... | Your double integral is over a circular region, but the intersection of the sphere and plane given is actually $x^2+xy+y^2=\frac{a^2}{2}$, an ellipse.
Edited for more detail:
Note that $\iint_R\,dA$ is exactly the area enclosed by this ellipse, so we just need to find this area and multiply it by 3.
One way to do this ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
| We have our equation to solve for:
$$(x+1)^4+(x-1)^4=16$$
First, expand the terms in the left hand side.
$$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$
Simplify the left hand side.
$$2x^4+12x^2+2=16$$
Factor $2$ out from both sides.
$$x^4+6x^2+1=8$$
Move $8$ to the left hand side.
$$x^4+6x^2-7=0$$
Let $x^2=a$. Now our eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Jordan similar matrix I have matrix
$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$.
I found the characteristic polynomial $(1-x)^4$ and was able to get my Jordan Matrix $J = \begin{bmatrix}1 & 1 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$. I... | Note: It is important to learn to type problems using MathJax for many reasons, see this wonderful MathJax Basic Tutorial and Quick Reference.
We have the matrix:
$$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$$
The characteristic polynomial is:
$$(\lambda - 1)^4 = 0 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $AD=BD$, $\angle ADC=3\angle CAB$, $AB=\sqrt{2}$, $BC=\sqrt{17}$, $CD=\sqrt{10}$. Find $AC$ In quadrilateral $ABCD$, we have
$$AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},$$ $$BC=\sqrt{17},CD=\sqrt{10}$$
Find the $AC=?$
My idea: let $$\angle CAB=x.\angle ADC=3x,\angle ADB=y,$$
then we have
$$\angle CAD=90-\dfrac{y}{2}-... | $ABCD$ on a square lattice">
With three distances given, each as the square root of an integer which is the sum of two squares, this problem looks as if the points are meant to have integer $x$ and $y$ coordinates. So treat the problem as one in coordinate geometry.
We are given:
\begin{align}
AD&=BD\tag{1},\\
\angle A... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Riemann's zeta as a continued fraction over prime numbers. Riemann's zeta function is a function with many faces, I mean representations. I recently derived this one, bellow, as a continued fraction over prime numbers.
$$
\zeta(s)=1
+\cfrac{\frac{1}{2^{s}}}{1-\frac{1}{2^{s}}
-\cfrac{\frac{2^{s}-1}{3^{s}}}{1+\frac{2^{s}... | The continued fraction representation above had its origins on another problem I was working on sometime ago.
It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.
And this is as follows
$$
\frac{1}{\zeta(s)}=\left(1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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How find this positive integer $(a,b)$ such $\left(\frac{a}{b}-\left[\frac{a}{b}\right]\right)\left[\frac{a}{b}\right]=2013$ let $a,b$ is positive integer numbers,and such
$$\begin{cases}
\gcd(a,b)=1\\
b\le 100\\
\left(\dfrac{a}{b}-\left[\dfrac{a}{b}\right]\right)\left[\dfrac{a}{b}\right]=2013
\end{cases}$$
Find the p... | Let $a=kb+r$, where $k, r \in \mathbb{Z}$ and $0 \leq r<b$. Since $\gcd(a, b)=1$, we have $\gcd(r, b)=1$.
Now the third equation becomes $\frac{r}{b}k=2013$, so $b \mid rk$. Since $\gcd(r,b)=1$, we get $b \mid k$. Let $k=lb, l \in \mathbb{Z}$, so $rl=2013$.
Now $r \mid 2013$ and $r<b \leq 100$, so $r=1, 3, 11, 33, 61$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all triangles of which perimeter and area are numerically equal Find all triangles of which perimeter and area are numerically equal. I have got solution for right angle triangles but not of others
| The area is $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter. Thus we get
$$(a+b+c)^2 = \frac{a+b+c}{2}(\frac{a+b+c}{2} - a)(\frac{a+b+c}{2} - b)(\frac{a+b+c}{2} - c).$$
We can further simplify this to
$$16(a+b+c) = (-a+b+c)(a-b+c)(a+b-c).$$
Let $u = -a+b+c$, $v = a-b+c$, $w = a+b-c$. Then
$$16(u+v+w) = u ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/749515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{... | Now,I have solution this problem:let $x,y,z\in R$ and such $x+y+z=\pi$,and
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
$$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$
we have
$$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$
we only pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/749758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the limit of a sequence by diagonalising a matrix Consider the sequence described by:
$\frac11 , \frac32 , \frac75 , ... ,\frac {a_{n}}{b_{n}}$
where $ a_{n+1} = a_n +2b_n $ and $b_{n+1} = a_n+b_n$
Find a matrix $A$ such that
$$\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_... | Define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z)$; multiply the recurrences by $z^n$ and sum over $n \ge 0$. Recognize some sums to get:
\begin{align}
\frac{A(z) - a_0}{z} &= A(z) + 2 B(z) \\
\frac{B(z) - b-0}{z} &= A(z) + B(z)
\end{align}
Solve the resulting linear system:
\begin{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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proof of $\sin(x)$ infinite series I can't seem to find a proof for this
$$\sin( x )= x - \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}+\cdots$$
Which euler used to proof the basel problem.
Would be great if it anyone show us how to this equality
Thanks
| You can use the fact that:
$$\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$$ and that:
$$e^{ix} = 1+(ix)+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots$$
$$e^{-ix} = 1+(-ix)+\frac{(-ix)^2}{2!}+\frac{(-ix)^3}{3!}+\cdots$$
So:
$$\begin{align*}\sin(x) &= \dfrac{e^{ix}-e^{-ix}}{2i} \\ &= (1-1)+ \frac{(ix)-(-ix)}{2i}+\frac{1}{2i}\bigg(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Area of a five pointed star A 5 pointed star is inscribed in a circle of radius $r$.
Prove that the area of the star is
$$
\frac{10 \tan\left(\tfrac{\pi}{10}\right)}{3-\tan^2\left(\tfrac{\pi}{10}\right)} r^2
$$
| Consider the following diagram of the 5-pointed star:
Clearly the area of the 5-star is $10$ times the area of the orange-shaded triangle $\triangle OAP$ which, in turn, equals half the height times the base:
$$
A\left(\triangle OAP\right) = \tfrac{1}{2} \bar{AC} \cdot \bar{OP} = \tfrac{1}{2} \bar{AC} \cdot r
$$
To... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/753290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Trig Identity Proofs I'm having a really hard time understanding how to do these. The directions are to verify that each of the following is an identity:
$$\dfrac{\csc x}{\cot x+\tan x}=\cos x$$
I have to get the left side to equal the right.
| $$ \frac{ \csc x }{ \cot x + \tan x} = \frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x} + \frac{ \sin x}{ \cos x}} = \frac{\frac{1}{\sin x}}{\frac{ \cos^2 x + \sin^2 x}{\sin x \cos x}} = \frac{ \frac{1}{\sin x}}{\frac{1}{\cos x \sin x}} = \frac{ \cos x \sin x}{\sin x} = \cos x$$
I have used the identiy $\sin^2 x + \cos^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
| We have:
\begin{equation}
x\left(1+\frac{3}{x^2+y^2}\right)=3+\frac{y}{x^2+y^2} \tag{3}
\end{equation}
\begin{equation}
y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x}{x^2+y^2} \tag{4}
\end{equation}
Multiplying Eqn $3$ with $y$ and Eqn $4$ with $x$ we get
\begin{equation}
x\,y\left(1+\frac{3}{x^2+y^2}\right)=3\,y+\frac{y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help... | There are lots of ways of solving this. One is to note that $a$ and $b$ are the roots of the quadratic equation $$0=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-px+q$$ Where we use $p=a+b$ and $q=ab$ for the unknown coefficients.
Then we have $a^2-pa+q=b^2-pb+q=0$ and adding the equations we obtain $$(a^2+b^2)-p(a+b)+2q=0$$ which beco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from sol... | Set $a=2^x$, $b=2^y$, then the problem is equivalent to finding the range of $a+b$ where $a,b>0$ and
$$a^2+b^2=2a+2b$$
Without loss of generality $a\ge b$ and we can make the substitution $c=a+b$, $d=a-b$, so now $c>d\ge0$, and we require
\begin{align*}\left(\frac{c+d}2\right)^2+\left(\frac{c-d}2\right)^2&=(c+d)+(c-d)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
How many ways move n pies to m distances? A table size $1\times (m+n)$ squares. Give $n$ pies on the $n$ first squares. Now, I want move $n$ pies to the end of table by $m.n$ steps ($m$ steps for each pie), satify conditions one pie only move to an empty-adjacent square. Define $S(n,m)$ is number of all ways move to fi... | Please take a look at the following articles:
Article One
Article Two
| {
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"url": "https://math.stackexchange.com/questions/757477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Higher-Order Approximation of Catalan-Numbers I have a question considering the higher-order approximations of the Catalan-Numbers, following the book Analytic Combinatorics by Flajolet and Sedgewick. First we set
$$
C_n = \frac{1}{n+1} \binom{2n}{n} = \frac{1}{n+1} [z^n] (1-4z)^{-1/2} = \frac{4^n}{n+1} [z^n] (1-z)^{... | Note that $$\dfrac{n}{n+1} = 1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots $$
and
$$ \left(1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots\right)
\left( 1 - \dfrac{1}{8n} + \dfrac{1}{128 n^2} + \ldots\right)
= 1 - \dfrac{9}{8n} + \dfrac{145}{128 n^2} + \ldots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/759532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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counterexamples to $ \det \Big(A^2+B^2\Big)\ge \det(AB-BA) $ $n\geq3$. A and B are two $n\times n$ reals matrices. For $n\times n$, Could one give counterexamples to show that
$$ \det \Big(A^2+B^2\Big)\ge \det(AB-BA) \tag{$*$}$$
is not necessarily true?
Well, I won't do more than $3\times3$: the following $A,B$ show (... | Let \begin{equation}A=\begin{bmatrix}I_{n-2} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & 0 & -1 \\ \mathbf{0} & 1 & 0\end{bmatrix},\end{equation} then \begin{equation}A^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & -1\end{bmatrix}.\end{equation} Now make $B$ a matrix filled wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/760669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$. If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.
I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?
| Since $\sin A = \frac{3}{5} $, then $\cos A = \frac{4}{5} $. Therefore,
$$ \sec (2A) = \frac{1}{\cos(2A)} = \frac{1}{\cos^2 A - \sin^2 A} = \frac{1}{\frac{16}{25} - \frac{9}{25}} = \frac{1}{\frac{7}{25}} = \frac{25}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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No. of real solutions of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $ How many real solutions are there of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $?
Please illustrate.
| Let $y=2^x$, then
$$
\begin{align}
\ln y&=\ln2^x\\
\ln y&=x\ln 2\\
y&=e^{x\ln 2}.
\end{align}
$$
Consequently, $2^{-x}=e^{-x\ln 2}$ and
$$
\begin{align}
2\cos\left(\frac{x^2+x}{6}\right)&=e^{x\ln 2}+e^{-x\ln 2}\\
\cos\left(\frac{x^2+x}{6}\right)&=\frac{e^{x\ln 2}+e^{-x\ln 2}}{2}
\end{align}
$$
Now, let $x=i\theta$, the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Non-Homogenous System answer not matching Find the Genereal Solution of $\vec{x}^{'}=\begin{pmatrix}2&-1\\3&-2 \end{pmatrix}\vec{x}+\begin{pmatrix}1\\-1 \end{pmatrix}e^t$
I found the eigenvalues to be $\lambda=\pm 1$
Therefore the eigenvectors are $e_1=\begin{pmatrix}1\\1 \end{pmatrix}$ and $e_2=\begin{pmatrix} 1\\3 \e... | Hint:
How about that $\dfrac{1}{2}$ term when you multiply by $g(t)$?
Update
I agree with your solution.
| {
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Number theory - rational number Are there any $x, y$ that fit in below
$\sqrt{4y^2-3x^2}$ such that an rational number is yielded.
Appreciate if explanation is given.
| You want $4y^2-3x^2=n^2 \iff (2y-n)(2y+n) = 3x^2$
So for e.g. we can set $2y-n = 3, 2y+n = x^2 \implies y= \dfrac{x^2+3}4, n = \dfrac{x^2-3}2$ and both will be integer if $x \equiv 1 \pmod 2$. Thus $x = 2k+1, y = k^2+k+1$ should work for any $k \in \mathbb{Z}$, and we find
$$4y^2-3x^2 = 4(k^2+k+1)^2-3(2k+1)^2 = (2k^2+... | {
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Guess the closed form on the following sequence? any help would be appreciated, have no idea where to start
$u_1 = 2/3$ and $u_{k+1}$ such that:
$$u_k + \frac{1}{(k+2)(k+3)}$$ for all, k are natural numbers
guess a general formula (i.e the closed form) of the sequence
| Hint: So $u_1=\frac{2}{3}$ and $u_2=\frac{2}{3}+\frac{1}{(3)(4)}$ and $u_3=\frac{2}{3}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}$ and so on.
Note that $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$.
Do a couple more terms and notice the beautiful cancellations (telescoping). In genera... | {
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Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
| I do not understand that. Same for any number can make this combination.
equation: $XY+XZ+YZ=N$
Solutions in integers can be written by expanding the number of factorization: $N=ab$
And vospolzovavschis solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$
$k$ - what some integer number given by us.
Solutions c... | {
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Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any a... | We have, $$∫\frac{dx}{\sqrt{x^2-x+1}}=\text{?}$$
Now, observe the expression
$$x^2-x+1=\left(x^2-x\right)+1=\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^{2}$$
$$∴ ∫\frac{dx}{\sqrt{x^2-x+1}}=∫\frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^... | {
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Finding an explicit formula for $a_n$ defined recursively by The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.
I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$
If we subtract 1 from each term, r... | The characteristic equation for the homogenous part is: $x^2 - 4x = 0$ gives: $x = 0$ and $x = 4$. So the solution to the equation is: $a_n = c\cdot 4^n + b$. Now $a_0 = 0$ and $a_1 = 1$ gives: $0 = c + b$, and $1 = 4c + b$. So $b = -c$, and $c = \dfrac{1}{3}$, and $b = -\dfrac{1}{3}$. So: $a_n = \dfrac{4^n - 1}{3}$
| {
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given $\cosh u = x$ find $\sinh u$ I'm asked to show that:$\newcommand{\arcosh}{\operatorname{arcosh}}$
$\int{x \arcosh x}dx = \frac{1}{4}(2x^2 -1)\arcosh x - \frac{1}{4}x\sqrt{x^2 -1} + C$
If I integrate by parts:
let $u = \arcosh x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{\sqrt{x^2 -1}}$
and let $\dfrac{dv}{dx} = x \R... | $\sinh u$ is negative when $u$ is negative and positive otherwise whereas $\cosh u$ is always positive. Therefore, the same is true for $\cosh u\sinh u$. To transfer this property to $x \sqrt{x^2 -1}$, we take the square root to be positive and let $x$ do the job.
| {
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$p$ is an odd prime number where $p=3k+1\Longleftrightarrow\exists a,b\in\Bbb Z^+$ such that $p=a^2+ab+b^2$ Let $p$ be odd prime number,show that:
$$p=3k+1\Longleftrightarrow \exists a,b\in\Bbb Z^+ \textrm{ such that } p=a^2+ab+b^2$$
I guess this is true because I find
when: $p=7,k=2$,and $$7=2^2+2\cdot 1+1^2$$
(2) wh... | HINT:
For odd prime $p$,
$$a^2+ab+b^2=p\implies(2a+b)^2\equiv-3b^2\mod p$$
$$\iff\left(\frac{2a+b}b\right)^2=-3\mod p$$
Check when $-3$ is a Quadratic Residue $\pmod p$
Now, use this
Alternatively, $$\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\cdot\left(\frac3p\right)$$
$-1$ is a quadratic residue modulo $p$ if an... | {
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had s... | Hint:As @illysial, @Mehdi and others indicated, the substitution $x = t+1$ transforms the expression $x^4 − 4x^3 + 2x^2 + 4x + 4$ into $t^4-4 t^2+7$.
| {
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Show that $\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$ Show that
$$\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$$.
I am thinking of using Fourier series and Parseval's identity to tackle this, I tried $x^4$, $4x^4+1$, how... | It can be shown using Poisson summation formula that
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b^2}=\frac{\pi}{b}\tanh \pi b.$$
Setting $b_{\pm}=e^{\pm i\pi/4}c$ in this formula, we get
\begin{align}
\sum_{n\in\mathbb{Z}}\frac{1}{n^4+c^4}&=\frac{1}{b_+^2-b_-^2}\left(\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b_-^2}-\sum_{n\in\mathbb... | {
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Convergent or divergent? $\sum_{n=1}^{\infty}{\frac{n^n}{(n+1)^{n+1}}}$ Does the series
$$\sum_{n=1}^{\infty}{\frac{n^n}{(n+1)^{n+1}}}$$
converge?
| $$\begin{align*}
\frac{n^n}{(n+1)^{n+1}} &= \frac{1}{n+1}\left(\frac{n}{n+1}\right)^n = \frac{1}{n+1}\left(\frac{1}{1+\frac{1}{n}}\right)^n
= \frac{1}{n+1}e^{-n\ln\left(1+\frac{1}{n}\right)} \\
&= \frac{1}{n+1}e^{-n\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)}
= \frac{1}{n+1}e^{-1+o\left(1\right)} \operatorname*... | {
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Trying to prove M is a manifold Let $M$ be the set of all points $(x, y, z) \in \mathbb{R^3}$ satisfying both of the equations
$x^3 + y^3 + z^3 = 1$ and $x + y + z = 1$. Prove that M is a manifold, except
perhaps near the points $(x; y; z) = (-1; 1; 1) ; (1;-1; 1) ; (1; 1;-1)$. What is
the dimension of $M$?
I have cal... | Consider $\Phi = x^3 + y^3 + z^3 - 1$, and $\Psi = x + y + z - 1$, and $\Lambda = (\Phi, \Psi)$, then $M = \Lambda^{-1}(0,0)$.
Consider the matrix $\begin{pmatrix} \Phi_x & \Phi_y & \Phi_z \\ \Psi_x & \Psi_y & \Psi_z \end{pmatrix} = \begin{pmatrix} 3x^2 & 3y^2 & 3z^2 \\ 1 & 1 & 1 \end{pmatrix}$
Consider the matrices : ... | {
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Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$ I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/short... | Setting $\displaystyle\frac x2+10^\circ=y\iff x+20^\circ=2y$
$$\frac{\tan2y-\sin2y}{\tan2y+\sin2y}=\frac{\sin2y\cdot\dfrac{1-\cos2y}{\cos2y}}{\sin2y\cdot\dfrac{1+\cos2y}{\cos2y}}=\frac{\cos2y\sin2y(1-\cos2y)}{\cos2y\sin2y(1+\cos2y)}=\tan^2y$$ if $\cos2y\sin2y\ne0$
Now $\cos2y\sin2y=0\iff\sin4y=2\sin2y\cos2y=0\implies4y... | {
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Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.
Here is how I find eigenvalues:
$$\begin{align*}
\det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda... | We could put the solution in a cleaner form:
Let $\lambda =-1$ be our eigenvalue. Then we have
$$
\begin{align}
M&=A-\lambda I\\&= \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix}
\end{align}
$$
Let the corresponding eigenvector be $\overrightarrow{v}$:
$$
\overrightarrow{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}
$$... | {
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Find the minimum value of $P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$ Let $x,y,z$ be real numbers such that $x>y>0, z>0$ and $xy+(x+y)z+z^2=1$.
Find the minimum value of $$P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$$
I tried using some ways, but failed. Please give me an idea. Thank you.
| notice that: $x - y = (x+z) - (y+z)$. So let $a = x+z$, and $b = y+z$, then:
$ab = 1$, and minimize: $P(a,b) = \dfrac{1}{4(a-b)^2} + \dfrac{1}{a^2} + \dfrac{1}{b^2} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4(a^2 - 2 + b^2)} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4\left(a^2 + \dfrac{1}{a^2} - 2\right)}= t + \dfrac{1}{4t - 8}$, w... | {
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$(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$ Let $R$ a ring. Show that $(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$.
That's what I have tried:
$$(a+b) \cdot (a-b)=a^2-b^2 \Rightarrow a \cdot a+a(-b)+ba+b(-b)=a^2-b^2 \Rightarrow a^2-ab+ba-b^2= a^2-b^2 \Rig... | If $\;R\;$ is commutative then
$$(a-b)(a+b)=a^2+\overbrace{ab-ba}^{=0\;, \text{ by comm.!}}-b^2=a^2-b^2$$
| {
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How prove $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge 1$
let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$
show that
$$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$
This problem is from Secrets In Inequalities volume 1 page 30,example 1.24. the comment.the author... | By AM-GM (with all sums being cyclic),
$$\sum \frac{a^2}{a+2b^2} = \sum a - \sum \frac{2ab^2}{a+2b^2}\ge \sum a - \sum \frac{2ab^2}{3\sqrt[3]{ab^4}}= \sum a - \frac23\sum (ab)^{2/3}$$
By Power Mean Inequality, $$1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3} \implies \sum (ab)^{2/3} \le 3$$
It remai... | {
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Help with definition of derivative My textbook says the definition is this:
$$\lim\limits_{\textbf{x} \rightarrow \textbf{x}_0} \frac{\|f(\textbf{x}) - f(\textbf{x}_0) - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\|\textbf{x} - \textbf{x}_0\|} = 0$$
I am trying to learn how to use it on, say for example, the function $f(... | Notice taking limit $x \neq 1$, and $y \neq 1$, but they get very close to $(1,1)$, and this is enough to get through the problem...
So starting at your last line: $|| - 1 + y - xy + x|| = |x - 1|\cdot |y - 1|$, so:
$0 \leq \dfrac{||-1 + y - xy + x||}{\sqrt{(x - 1)^2 + (y - 1)^2}} = \dfrac{|x - 1|\cdot |y - 1|}{\sqrt{(... | {
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convergence of a sum of two geometric series I would like to use the ratio and root test on the following series:
s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + .. = a1 + a2 + a3 + ...
where a2 is (1/2)^2 + (1/3)^2 for example
I know we have a sum of two geometric series so the sum will be convergent but I'd like to find the foll... | So I believe you have decided to describe this series by, $$ \sum_{n = 1}^{\infty} \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n $$
Root Test:
Now by the expansion of the Binomial Theorem for each $n \in \Bbb N$,
$$ 0 \lt \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt \sqrt... | {
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Can be rational? Can $z=\sqrt{1-4y^x}$ be a rational number, where $x>2$ is an integer and $y$ is a rational number and $y>0$?
I have tried with $x=2$ and it has rational solutions, for example $y=12/25$ and then $z=7/25$. But I haven't found any rational solution with $x>2$
| Too long for a comment:
$\sqrt{1-4y^x}=z\quad=>\quad1-4y^x=z^2\iff1-4~\bigg(\dfrac mn\bigg)^x=\bigg(\dfrac ab\bigg)^2,\quad\begin{cases}\gcd\big(~a~,b\big)=1.\\\gcd\big(m,n\big)=1.\end{cases}$
$\iff\underbrace{n^x\overbrace{(b^2-a^2)}^{b~>~a}=\overbrace{4b^2m^x}^{>~0}}_{\gcd(a,b){\large=}1,\quad\gcd(m,n){\large=}1.}~=>... | {
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For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple... | Sub $y=n x$ and use a Weierstrass substitution $t=\tan{(y/2)}$ and get that the integral is
$$\frac1{n} \int_0^{\tan{(\pi n/4)}} \frac{dt}{1+t+t^2} $$
Evaluate this by expressing the denominator as $(t+1/2)^2+3/4$ to get an expression for the integral and the stated equation:
$$\frac{2}{\sqrt{3} n} \arctan{ \left [\fra... | {
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Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $
$x\geq0$, then, we have
$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$
The problem is not easy, even $x=1$. Any help will be appreciated
| From a purely algebraic point of view, this problem is quite interesting if we first notice that $$\frac {d}{dx} \Big( \frac {1}{x+n^2} \Big)=-\frac{1}{\left(x+n^2\right)^2}$$ The second point is to recognize that
$$ \sum_{n=1}^{\infty} \frac{1}{x + n^2}=\frac{\pi \sqrt{x} \coth \left(\pi \sqrt{x}\right)-1}{2 x}$$ So... | {
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Number of Rolls of Fair Dice to get '6' and '5' A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$
| This is my Approach, Let me know if i am on right track...
$$E(X|Y=5)=\sum_{x\in \mathbb{X}}xP(X=x|Y=5)=\sum_{x \in \mathbb{X}}\frac{xP(X=x,Y=5)}{P(Y=5)}$$
The set $\mathbb{X}$ is $\left\{1,2,3,4,6,7,8,\cdots \infty \right\}$ So
$$E(X|Y=5)=\sum_{k=1,2,3,\cdots,\infty_{k\ne 5}}\frac{k\,P(X=k,Y=5)}{P(Y=5)}$$ We have
$$P(... | {
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Integral $\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}$ EDIT: Small Typo Fixed now, Thanks to Sir Chen Wang!
Hi I am trying to prove this result without using a series approach
$$
\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}.
$$I know we can just solve it by wri... | $$
\begin{align*}
I&=\int^1_0\log(1+x)\frac{1+x^2}{(1+x)^4}dx\\
&=-\frac13\int^1_0\log(1+x)d\left(\frac{2+3x+3x^2}{(1+x)^3}\right)\\
&=-\frac13\left(\left.\log(1+x)\frac{2+3x+3x^2}{(1+x)^3}\right|^1_0-\int^1_0\frac{2+3x+3x^2}{(1+x)^3}d\,\log(1+x)\right)\\
&=-\frac{\log 2}{3}+\frac13\int^1_0\frac{2+3x+3x^2}{(1+x)^4}dx\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof of basic properties Can anyone provide proof of properties such as:
$$a(b+c) = ab+ac$$
$$(a+b)^2 = a^2+2ab+b^2$$
And exponent rules:
$$a^n \cdot a^m = a^{n+m}$$
$$(a^n)^m = a^{n \cdot m}$$
For $a, b, c \in \mathbb{R}$
I'm quite sure that the proof is the same or very similar for the first two and the last two, wh... | The first equation is an axiom of $\mathbb{R}$
The second equation is an application of the first one and uses another axiom of $\mathbb{R}$, the commutative property :
$$(a+b)^2 = (a+b)\cdot (a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = a^2+2ab+b^2$$
The two next proves are made for integers. We can extand these... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Linear Algebra matrix notation My question is referring to the following $4 \times 6$ matrix:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
Then the book states "It is useful to denote the column numbers in which the leading entri... | I haven't seen this notation before, and I don't see the purpose of it, which is probably why it is so hard to understand. This seems to be some sort of intermediate (and complicated...) way of describing reduced row echelon forms.
The definition, stated a little less haphazardly, is:
*
*(1) $r$ is the number of no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
linear transformation $\Bbb R^3 \to \Bbb R^2$ For linear transform $T ( 1,1,1 ) = (1 , 2),\,\, T ( 1, 1, 0) = (2 , 3)$ and $\,\,T ( 1,0,0 ) = ( 3 , 4)$,from $\mathbb{R}^3$ to $\mathbb{R}^2$, the function used was : Choose one :
a. $T ( x, y, z) = (x - y - 3z , 4x - y - 2z )$
b. $T ( x, y, z) = (3x + y + z, 4x + y + z... | The answer is (c). One can rewrite the given equations in matrix form as follows:
$$ T \cdot\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix}.$$
Hence, we have
$$ T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \cdot\begin{bmatrix} 1 & 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
I want to prove $k(x,t)=\frac{1}{\sqrt{4\pi t} } e^{\frac{-x^2}{4t}} $ I have this integral $$u(x,t)=\int _{-\infty}^{\infty} f(\eta)\left[\frac{1}{2\pi}\int _{-\infty}^{\infty}e^{iw(x-\eta)-w^2t}\ dw\right]\ d\eta=\int _{-\infty}^{\infty}k(x-\eta,t)f(\eta)\ d\eta$$
I want to prove $$k(x,t)=\frac{1}{\sqrt{4\pi t} }\ e^... | Completing square gives
\begin{align*}
&= \int_{-\infty}^\infty \exp ( i\omega (x-\eta ) - \omega^2t)d\omega \\
&= \int_{-\infty}^\infty \exp \left( - \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t}\right )^2+2 \omega \sqrt{t} i (x-\eta ) \cdot \frac{1}{2\sqrt t} - \omega^2t + \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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.