Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How many solutions has this equation? I've the equation $x - \arctan(2x) = a$ and the question is, how many solution has the equation for different values of $a$, where $a$ is a real number.
I've plotted the graph and found the extreme values at $ x = \pm \frac{1}{2}$.
so naturally I would say:
1 solution for $a < -\fr... | Differentiating $f(x) = x - \arctan(2x)-a$, we get ${f}'(x) = 1-\frac{2}{1+4x^{2}} = 0$ when $x = \frac{1}{2}$ or $x= -\frac{1}{2}$. If $|x| > -\frac{1}{2}$, then function is increasing. Otherwise, it is decreasing. Thus, $f(x)$ has at most 3 roots and we just have to check the intervals $x > \frac{1}{2}$, $x < - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\zeta(2) = \frac{\pi^2}6$ It is well known that $$\int_0^1\frac{\log{x}}{1 - x}\,\mathrm{d}x = -\frac{\pi^2}{6} $$
This is generally proved by expanding the geometric series.
My question is: can this be done in reverse?
Can we evaluate the integral $\int_0^1\frac{\log{x}}{1 - x}dx$ using other method, for exam... | Credits to Daniele Ritelli:
$$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\color{red}{\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy}\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$ Find this follow function $f(x)$ range ,where $x\in R$,
$$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}... | $$\text{Let }y_r=\left(2+\cos\left(x+\frac{r\pi}3\right)\right)\left(2+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right)$$
$$y_r=4+2\left[\cos\left(x+\frac{r\pi}3\right)+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right]+\cos\left(x+\frac{r\pi}3\right)\cos\left(x+\dfrac{(r+1)\pi}3\right)$$
Applying $2\cos A\cos B$ and $\cos C+\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate this integral: $\int \frac{x^4}{(x-1)(x^2-1)}dx\;$? $$\int \frac{x^4}{(x-1)(x^2-1)}dx$$
I tried to decompose the $(x^2-1)$ term into $(x+1)(x-1)$ thus getting $(x-1)^2(x+1)$ as the denominator. I can't use the method of partial fraction because of the $x^4$ term. Should I proceed through normal polynomi... | Hint:
$$\frac{x^4}{(x-1)(x^2-1)}=\frac{x^4-1+1}{(x-1)(x^2-1)}=\frac{x^2+1}{x-1}+\frac{1}{(x-1)(x^2-1)}$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)^2}-\frac{1}{4(x-1)}+\frac{1}{4(x+1)}$$
$$\to \int \frac{x^4}{(x-1)(x^2-1)}dx=\frac{x^2}{2}+x-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y$ is a given positive odd integer Let $k$ be a fixed positive odd integer. Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y=k$
My approaches:
*
*Since $k$ is odd, $x$ and $y$ have different parity.... | Since
$$
x^2+y^2=\frac12\left((x+y)^2+(x-y)^2\right)
$$
the minimum comes when $|x-y|$ is smallest, that is $1$ if $x+y$ is odd. Thus, the minimum is
$$
\frac12\left((x+y)^2+1\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
probability matrix with trace $1$ is square of probability matrix Consider as probability matrix a matrix $M \in [0,1]^{n \times n}$ while every row sums up to $1$.
Statement: Consider a $2\times 2$ probability matrix $M' \in [0,1]^{2 \times 2}$. Show, that the following holds:
$$\exists \text{ probability matrix } M :... | I don't know the nice way to do it, but you can brute force it.
Consider probability matrix $ \left( \begin{array}{cc}
a & 1-a \\
b & 1-b \\ \end{array} \right) $ is a probability matrix $ \left( \begin{array}{cc}
a^2+b(1-a) & a(1-a)+(1-a)(1-b) \\
ab+b(1-b) & b(1-a)+(1-b)^2 \\ \end{array} \right) $. So you need to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Partial sum of a geometric series Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$
and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to
$S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.
I would like to arrive at the same sum by computing
$\lim_{N \to \infty} S_N$ where... | Note that
$$S_1=1=\color{green}{\bf\frac23}\color{red}{\bf+\frac13},\ S_2=\frac12=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{2\cdot3}},\ S_3=\frac34=\color{green}{\bf\frac23}\color{red}{\bf+\frac1{4\cdot3}},\ S_4=\frac58=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{8\cdot3}},\ldots$$
As computed by @Ravi in a c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find recursive forumula for integrals I have to find recursve formulas for solving the following two integrals. The assignment tells one to find an Expression that leads from the calculation of $\dfrac{I_{2n}}{I_{2n+1}}$ to the calculation of $\dfrac{I_{2n-2}}{I_{2n-1}}$
a) $$ I_{2n} = \int_a^b \frac{1}{(1+x^2)^n} dx ... | I could show one way to derive the recursive formulas given in one of the comments:
Write
$$I_{2n}=\int\frac{1+x^2-x^2}{(1+x^2)^n}dx= I_{2n-2}-\int\frac{x^2}{(1+x^2)^n}dx
$$
We have to compute the last integral. Using integration by parts (with $u'=x/(1+x^2)^n$) we have
$$
\int\frac{x^2}{(1+x^2)^n}dx= \frac{x}{(2-2n)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$ The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique... | Binet's second Log Gamma integral formula is
$$
\ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t
$$
Differentiating twice w.r.t. $z$ gives
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 4\int_0^{\infty} \frac{zt}{(z^2+t^2)^2(e^{2\pi t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$ Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$
Let $n^4 + a = (n^2+xn+p)(x^2+yn+q)$. Knowing that the cubic and quadratic coefficients must be zero, we can sho... | For any integer $k>1$ we have
$$n^4+4k^4=(n^2+2kn+2k^2)(n^2-2kn+2k^2)=((n+k)^2+k^2)((n-k)^2+k^2),$$
which is composite for all $n$, being the product of two integers greater than one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{... | first we square both sides
$\dfrac{1}{(a+b)^2}=\dfrac{\sin^8x}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^8x}{b^2}$
$=\dfrac{\sin^6x(1-\cos^2x)}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^6x(1-\sin^2x)}{b^2}$
$=\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}-\dfrac{\sin^6x\cos ^2x}{a^2}-\dfrac{\cos^6x\sin^2x}{b^2}+2\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find all $x,y\in\mathbb{Z}$ s.t $2x^3-7y^3=3$ Find all $$x,y\in\mathbb{Z}$$ such that $$2x^3-7y^3=3$$
Solution:
We consider first $$2x^3-7y^3\equiv3 \pmod 2$$ $$5y^3\equiv 1 \pmod 2$$ $$y^3\equiv 1 \pmod2$$ which has solution $y\equiv 1 \pmod 2$
Consider $$2x^3\equiv 3 \pmod 7$$ $$4\cdot 2x^3\equiv 4\cdot3 \pmod 7$$... | Indeed, the equation $2x^3 - 7y^3 = 3$ has no integer solutions. Modulo $7$, we get the congruence
$$2x^3 \equiv 3 \pmod{7}$$
which every solution of the original equation must satisfy. But the group of nonzero remainders modulo $7$ has order $6$, so $x^3 \equiv \pm 1 \pmod{7}$ for all $x\not\equiv 0 \pmod{7}$. Thus, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Changing variables for multivariable functions
Let $x=uv$ and $y=\frac{1}{2}(u^2-v^2)$. Substitute $x$ and $y$ with $u$ and $v$ in the expression $z_{x}^2+z_y^2$
My attempt:
$$z_x=u_xz_u+v_xz_v=\frac{1}{v}z_u+\frac{1}{u}z_v$$
Squaring:
$$z_x^2=\frac{1}{v^2}z_u^2+\frac{2}{vu}z_uz_v+\frac{1}{u^2}z_v^2$$
Using the same ... | Tackle it from the other side, express $z_u^2 + z_v^2$ in terms of $z_x,\,z_y$. You have
$$z_u = z_x x_u + z_y y_u = vz_x + uz_y,$$ and $$z_v = z_xx_v + z_y y_v = uz_x - vz_y.$$ That produces $z_u^2 + z_v^2 = (u^2+v^2)(z_x^2+z_y^2)$, agreeing with the book.
What was your mistake? When computing $u_x$ etc., you wrote $u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Moore-Penrose pseudoinverse of a 3×3 matrix Is there a "simple" formula for computing the Moore-Penrose pseudoinverse of a $3\times 3$ matrix? I mean something like the formula for the inverse (for non-singular matrices), which involves the matrix of minors, etc.
I need that for a computer program, and I feel that usin... | You are probably thinking of the formula
$$
\begin{align}
\mathbf{A}^{-1}
&= \frac{\text{adj } \mathbf{A}} {\det \mathbf{A} } \\
&= \frac{\left( \text{cof } \mathbf{A}\right)^{\mathrm{T}}} {\det \mathbf{A} } \\
\end{align}
$$
The matrix $\text{adj } \mathbf{A}$ is the adjugate of $\mathbf{A}$ and is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sign problem in proof that $\Gamma'(1) = -\gamma$ Convergence of $(u_n)_{n \in \mathbb{N}} := \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ was proven before
The egality $\Gamma'(1) = \lim_{n \to \infty}\int_0^n \log(t) (1-\frac{t}{n})^n dt$ was also proven before
Let $\gamma$ denote the limit of $(u_n)_{n \in \mathbb{Z... | When you substitute $u = 1-x$ in
$$\int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du,$$
you apparently forgot to adapt the integration limits (or maybe the sign from $du = -dx$),
$$\begin{align}
\int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du &= \int_{1-0}^{1-1} \frac{(1-(1-x))^{n+1}-1}{1-x}\,d(1-x)\\
&= \int_1^0\frac{1-x^{n+1}}{1-x}\,dx\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be app... | Let denote
$$\frac{(-1)^nn^2+n}{n^3+1}=\underbrace{\frac{(-1)^nn^2}{n^3+1}}_{=u_n}+\underbrace{\frac{n}{n^3+1}}_{=v_n}$$
The series $\displaystyle \sum_n v_n$ is convergent since $v_n\sim_\infty \frac 1 {n^2}$.
We have
$$u_n=\frac{(-1)^nn^2}{n^3+1}=\frac{(-1)^n}{n}\frac{1}{1+\frac{1}{n^3}}=\frac{(-1)^n}{n}\left(1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Double integral -- tricky?
If $f(x,y) = x^2+y^2$ and $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\geq1, x^2+y^2-2x\leq0 \text{ and } y\geq0\}$, find $\displaystyle\int\displaystyle\int_D f$.
$D$ looks like the intersection between two circumferences, as I draw below:
Polar coordinates seems a obvious choice, but using them I'd... | The integral's basically:
$$\int\limits_{1/2}^1\int\limits_{\sqrt{1-x^2}}^{\sqrt{1-(x-1)^2}}\,f(x,y)dydx+\int\limits_1^2\int\limits_0^{\sqrt{1-(x-1)^2}}f(x,y)\,dydx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/648004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a least common multiple (LCM) My Algebra 2 book explains how to find a least common multiple:
Find the least common multiple of $4x^2 - 16$ and $6x^2 - 24x + 24$.
Solution
Step 1 Factor each polynomial. Write numerical factors as products of primes.
$4x^2 - 16 = 4(x^2 - 4) = (2^2)(x + 2)(x - 2)$
$6x^2 - ... | It works just like with integers. If you want to find the LCM of $84=2^2\cdot 3 \cdot 7$ and $90=2\cdot 3^2 \cdot 5$ you collect the highest power of each prime, getting $2^2 \cdot 3^2 \cdot 5 \cdot 7 = 1260$ The LCM has to incorporate all the polynomial factors to the highest power in any of the things you are takin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the probability no slots contain more than two balls given I am trying to sort 5 balls into 6 slots? I am having a difficult time understanding how to approach this problem. Suppose I have $6$ total slots and $5$ balls. Now, I assign the balls at random to the slots. What is the probability that no slot will c... | Five balls can be distributed in 6 slots following the GF
$(1+b+b^2+b^3+b^4+b^5)(1+b+b^2+b^3+b^4+b^5)...(1+b+b^2+b^3+b^4+b^5)$ (six factors)
The coefficient of $b^5$ is $252$ and represents the ways of placing 5 balls without restrictions in six slots. With the given restriction, the GF is
$(1+b+b^2)(1+b+b^2)...(1+b+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How prove this the equation $\{x^3\}+\{y^3\}=\{z^3\}$has infinitely many rational non-integers solutions, Show that the equation
$$\{x^3\}+\{y^3\}=\{z^3\}$$
has infinitely many rational non-integers solutions,Here,$\{a\}$ denotes the fractional part of $a$
I have solve this follow two problem
the equation 1:
$$\{x\}+\... | For any $n > 2$, pick any $m > 1$ such that $\gcd(m,n) = 1$. Notice
$$\gcd(m,n) = 1 \implies \gcd(m^n,n) = 1$$
We can define a number $\lambda$ by
$$\lambda = \text{mod}( n^{\varphi(m^n)-1}, m^n )$$
where $\varphi(x)$ is the Euler's totient function and $\lambda$ will satisfy $$\lambda n = 1 \pmod{m^n}$$
Let $k \in \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine the largest power of 10 that is a factor of $50!\,$? How would one find the largest power of 10 which is a factor of $50!\,{}$?
| This problem is equivalent to find the number of trailing zeros in $n!=50!$. I adapt this old answer of mine to the question Derive a formula to find the number of trailing zeroes in $n!$.
For every integer $n$ the exponent of the prime $p$ in the prime
factorization of $n!$ is given by de Polignac's formula $$\display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\lef... | Considering the following diagram:
$\hspace{2cm}$
we get that
$$
\begin{align}
\int_0^\pi\log\left(1-2r\cos(\theta)+r^2\right)\,\mathrm{d}\theta
&=\int_0^{2\pi}\log\sqrt{1-2r\cos(\theta)+r^2}\,\mathrm{d}\theta\\
&=\mathrm{Re}\left(\int_\gamma\log(z-1)\frac{\mathrm{d}z}{iz}\right)
\end{align}
$$
along the path $\gamma=r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 4
} |
Find bases for the subspaces $U_1, U_2, U_1 \cap U_2, U_1 + U_2$ Let $$U_1 = \left\{ \left(
\begin{array}{cc}
x_1\\
x_2\\
x_3\\
x_4\\
\end{array}
\right) \in \mathbb{R} :-x_1-x_2+x_3=0 \right\}$$
$$U_2 = span \left( \left( \begin{array}{cc}
-1 \\
1 \\
1 \\
2 \\ \end{array} \right), \left( \begin{array}{cc}
2 \\
-... | I tend to use a bare-hands approach like this:
Suppose $x \in U_1 \cap U_2$. Since $x \in U_2$, this implies that
$$x= s \left(
\begin{array}{cc}
-1\\
1\\
1\\
2\\
\end{array}
\right)
+t \left(
\begin{array}{cc}
2\\
-1\\
-1\\
2\\
\end{array}
\right) =
\left(
\begin{array}{cc}
-s + 2t\\
s-t\\
s-t\\
2s+2t\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/654856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values in... | A different method (though not really easier) is to use the fact that $\sin x$ can be expressed using exponents:
$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$
So that your integral is:
$$\int e^{2x} \sin x\ dx = \int \frac{e^{2x+ix} - e^{2x-ix}}{2i} dx= \frac{e^{2x+ix}}{4ix-2x} - \frac{e^{2x-ix}}{4ix+2x}+C$$
$$=\frac{-1}{10... | {
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"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Simplify the expression where $x>y>0$ Simplify $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Help please, I tried but the answer doesn't match. I did it by multiplying of course and then normally simplifying.
I tried and what came is $\dfrac{-2xy -y^2}{2xy + x^2}$ and t... | $(a+b)(a-b)=a^2-b^2$
So,
$[(\sqrt {x^2+y^2})+x]\cdot [x-(\sqrt {x^2+y^2})]=[x+(\sqrt {x^2+y^2})]\cdot [x-(\sqrt {x^2+y^2})]=x^2-(x^2+y^2)=-y^2$
Can you proceed?
| {
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"answer_count": 2,
"answer_id": 1
} |
Closed-Form Solution to Infinite Sum Does the convergent infinite sum
$$
\sum_{n=0}^{\infty} \frac{1}{2^n + 1}
$$
have a closed form solution? Quickly coding this up, the decimal approximation appears to be $1.26449978\ldots$
| Your series can be re-written in terms of the q-polygamma function $\psi_q(z)$ which is simply the logarithmic derivative of the q-gamma function $\Gamma_q(z)$. Both of which are special functions related to the theory of q-series: $$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{\psi_{1/4}(1)-\psi_{1/2}(1)-\ln(3)}{\ln(2)}-\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Evaluate integral as a logarithm plus an arctangent. Evaluate the integral as a logarithm plus an arctangent.
$$ \int \frac{x}{3x^2-18x+45} \ dx $$
I just completed the square and couldn't continue.
$$ \int \frac{x}{3(x-3)^2+18} \ dx $$
Fixed a typo $18$ changed to $18 x$
| Write the numerator as $x-3+3$ to get
$$
\int \frac{x-3}{3 (x-3)^2 + 18}dx + \int \frac{3}{3 (x-3)^2 + 18} dx$$
This gives
$${{\log \left(3\,\left(x-3\right)^2+18\right)}\over{6}}+{{\arctan \left({{x-3}\over{\sqrt{6}}}\right)}\over{\sqrt{6}}}$$
Here are the details.
In the first integral let $(x-3)^2 = u$. Then $2 (x-3... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For which $x$ does the series $∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n})\, x^n$ converge? Determine for what value of $x$ the series converges
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
Observe that
$∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}) x^n =∑_{n=1}^∞ (∑_{... | Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$,
Then $\frac{H_{n+1}}{H_n}=1+\frac{1}{(n+1)H_n}$ tends to 1, Thus the ratio of convergence is $R=1$ (D'Alembert criterion).
Now note that the Cauchy product of
$$
\sum_{n=0}^\infty x^n, \quad \sum_{n=1}^\infty \frac{x^n}{n}
$$
is exactly
$$
\sum_{n=1}^\infty H_n\, x^n
$$
Th... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Integrating Factor and the Exact Equation $(y-\frac{1}{x})dx+\frac{1}{y}dy=0$.
$$(y - \frac{1}{x})dx + \frac{1}{y}dy = 0 \,\,\, , \,\,\, y(\sqrt[]{2}) = \sqrt[]{2}$$
So I need to find the integrating factor. $\frac{M_y - N_x}{N} = \frac{1 - 0}{0}$ which we cannot have so that leaves us with only $\frac{N_x - M_y}{M} ... | We can get an exact form
$$\left(y - \frac{1}{x} \right)dx + \frac{1}{y}dy = 0 \implies ydx-xdy=xy^2 dx$$
$$\frac{ydx-xdy}{y^2}=xdx \implies d\left(\frac{x}{y} \right)=xdx$$
On integration we get
$$\frac{x}{y} =\frac{x^2}{2} +k$$
$k$ is some constant
Now using $y(\sqrt{2})=\sqrt{2}$ we get $k=0$
Therefore the specific ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $ Does there exist a closed form expression for $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$
| Given $\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$
First we will simplify $\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$
$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Partial fractions: why does $\int dt \implies t + C$ I am working on a partial fraction problem here, I understand everything in the problem except $t+C$, so I'd like to know where did the $t+C$ come from ?
I want to solve this integral
$$
\int \frac{dy}{(y+2)(1-y)} = \int dt
$$
$$\begin{align}
1 &= \frac{A}{y+2} + \... | Think of $\int dt$ as $\int1dt$. The integral of $1$ with respect to any variable is that variable, so in this case, $\int1dt$ is $t+C$. Where $C$ is a constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How can I check the convergence of the sequence? Does it diverge? How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}} $ and above from $\frac{n(n+1)}{2\sqrt{... | Since $n^2\le n^2+k\le\left(n+\frac12\right)^2$ for $0\le k\le n$, we have
$$
\frac1{n+\frac12}\sum_{k=1}^nk\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac1n\sum_{k=1}^nk
$$
Thus,
$$
\frac n2\le\frac{n(n+1)}{2n+1}\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac{n+1}2{}
$$
That is, the sequence diverges.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| Let us prove the equivalent inequality
$$\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} > 0.$$
Since $\frac{n}{1+n} = 1 - \frac{1}{1+n}$, observe that
$$\begin{align*}
\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a}
&= 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+a}\\
&> 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+b ... | {
"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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Expected value for $X, Y \sim \mathcal{U}[0,10]$ vs. expected value of $E(2x)$ $X, Y$ are independent events both $\sim \mathcal{U}[0,10]$.
I know that $$E[X+Y] = E[X] + E[Y] = 5 + 5 = 10$$
and $$E[2X] = 2 E[X] = 10 \Rightarrow E[X+Y] = E[2X].$$
However, is $E\left[(X+Y-10)_+\right] = E\left[(2X-10)_+\right]$ ?
I beli... | The right one is
$$
\frac{1}{10} \int_0^{10} (2x-10)^+dx
= \frac{1}{10} \int_5^{10} (2x-10)dx = 2.5
$$
and the left one is
$$
\begin{split}
\frac{1}{100} \int_0^{10} \int_0^{10} (x+y-10)^+ dydx
&= \frac{1}{100} \int_0^{10} \int_{10-x}^{10} (x+y-10) dydx \\
&= \frac{1}{100} \int_0^{10} \left[ x(x-10) + 50 - (10-x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/673909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Permutations 1-line notation, and inverse Write (15)(286)(479) in 1-line notation. Find the inverse of (15)(286)(479).
Can anyone please help?
Thank you.
| Let me consider a similar permutation: $$\sigma:=(1\:8\:3)(2\:9)(4\:6\:7).$$ Now, since $9$ is the greatest number that appears in the disjoint cycles, then we can assume that $\sigma$ is a permutation on $\{1,2,3,4,5,6,7,8,9\}.$ In particular, $\sigma$ takes $1\mapsto 8\mapsto 3\mapsto 1,$ $2\mapsto 9\mapsto 2,$ $4\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/674046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Ratio and proportion problem
Question:
If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.
My solution:
*
*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$
*... | $$\frac{a+b}6=\frac{b+c}7=\frac{c+a}8=\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$
Similarly, each ratio is equal to $$\frac{a+b+(b+c)+(c+a)}{6+7+8}=\frac{2(a+b+c)}{21}$$
$$\implies \frac{2c}9=\frac{2(a+b+c)}{21}$$
Now, we have $a+b+c=14$
| {
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"source": "stackexchange",
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"answer_id": 1
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If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| Using the factorization identity $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ we obtain
\begin{align}
x^5+\frac{1}{x^5}&=\left(x+\frac{1}{x}\right)\left(x^4-x^2+1-\frac{1}{x^2}+\frac{1}{x^4}\right) \\ &=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^4-5x^2-5-5\frac{1}{x^2}\right) \\&=\left(x+\frac{1}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 4
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Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
| Let $n = 2k$ for some integer $k$. Then,
$$\begin{align}2k((2k^2) + 20) &= 2k(4k^2 + 20)\\
&=8k(k^2 + 5)\end{align}$$
But
$$\begin{align}k(k^2 + 5) &\equiv k(k^2-1)\mod 2\\
&=(k+1)(k)(k-1) \mod 2\end{align}$$
Since $2$ must divide at least one of these three consecutive integers, we have $2 \mid k(k^2 + 5) \implies 16... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 5
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On average, how many times does one need to roll three fair dice to get a sum of 10? Three fair dice are rolled simultaneously. On average, how many times does one need to roll three fair dice to get a sum of $10$?(All rolls are independent of each other, and a roll of the three dice counts as one time.)
So I found the... | The generating function for the 3 die is
$$ \frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Determining which sets are spanning sets for P3? I have a list of sets:
(a): {$1, x^2, x^2 - 2$}
(b): {$2, x^2, x, 2x + 3$}
(c): {$x + 2, x + 1, x^2 - 1$}
(d): {$x + 2, x^2 -1$}
I am supposed to determine which of the following are spanning sets in $P_3$.
My attempted solution:
(A): Ax^2 + bx + c = A(1) + B(x^2) +C... | I found b) and c) are the spanning sets.
For b) You only need to show that you can find reals m, n, p such that:
ax^2 + bx + c = 2m + nx^2 + px. You don't need 2x + 3 since 2x + 3 = 2x + (3/2)*2 is in the span{2,x}. So choose n = a, p = b, and m = c/2.
For c) again ax^2 + bx + c = m(x+2) + n(x+1) + p(x^2 - 1). Choose ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve equation: $5^x = -2x + 7$ How to solve that equation: $$5^x = -2x + 7$$
I already have the answer $x=1$. Can anyone please explain to me?
| The equation can be solved in terms of the Lambert W function.
Writing $5^x$ as ${\rm e}^{x \ln 5}$ the equation becomes
\begin{eqnarray*}
{\rm e}^{x \ln 5} &=& -2x + 7 \\
\Rightarrow (-2x + 7) {\rm e}^{-x \ln 5} &=& 1 \\
\frac{\ln 5}{2} (-2x + 7) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5 + \frac{7}{2} \ln 5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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geometric series used to work out big O notation for resizing an array in a stack It's a geometric series
$$
1 + 2 + 4 + \cdots + 2^k = \frac{1 - 2^{k+1}}{1 - 2}
$$
Here, $2^k$ = N. You get $1 + 2 + 4 + \cdots + N = \frac{1 - 2N}{-1}$. Therefore, $2 + 4 + \cdots + N = 2N−2$. When $N$ is big, you can just drop the $−2$... | $$S=1 + 2 + 4 + \cdots + 2^k
\\2S= 2 + 4 + \cdots + 2^k+2^{k+1}
\\S-2S=1-2^{k+1}
\\S(1-2)= 1-2^{k+1}
\\S= \frac{1 - 2^{k+1}}{1 - 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/686999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\mathbb{E}[B^4(t)]$ with $B$= brownian motion Can anyone help me to find:
$\mathbb{E}[B^4(t)]$ where $B$ is a brownian motion?
I thought using this density function:
$f_{B_t}(x) = \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$,
but I don't know how to apply it.
| Note that
$$\mathbb{E}(B_t^4) = \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx = \frac{2}{\sqrt{2\pi t}} \int_{0}^{\infty} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx.$$
If we set $y := x^2/2t$, then
$$\begin{align*} \mathbb{E}(B_t^4) &= \frac{2}{\sqrt{2\pi t}} \int_0^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Checking multivariable function's differentiability at $(0,0)$ I have a function: $$f(x,y) = \begin{cases} (x^2+y^2)\sin (x^2+y^2)^{-1} \ \ \ &\text{, if } x^2+y^2 \neq 0 \\ 0 &\text{, if } x^2+y^2=0 \end{cases}$$
I calculated the partial derivatives:
$$f_x(x,y)=2x\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2... | If a radial function $f(x,y) = g(x^2+y^2)$ is differentiable, its differential should be zero, by simmetry.
So you should check whether
$$
\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^2+y^2}}{\sin(x^2+y^2)} = 0.
$$
But this is false even for $y=0$ where you get
$$
\lim_{x\to 0}\frac{|x|}{\sin(x^2)} = +\infty.
$$
So your function ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/693217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proof by mathematical induction - Fibonacci numbers and matrices Using mathematical induction I am to prove:
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $ =
$ \left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right) $
where $F_k$ represents the $k^{th}$ Fibonacci number.
my ... | Inductive proof:
For $n=1$ is true, as the OP correctly observed.
Assume that it is true for $n=k$. Then
$$
\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^k=\left(\begin{matrix} F_{k+1}& F_k \\
F_k& F_{k-1}\end{matrix}\right)
$$
Then for $n=k+1$ we have
\begin{align}
\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions? The system is :
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
After Gaussian elimination, I found that
$$
\begin{array}{cccc|cc}
1 & -4 & 6 & a & ... | There's no solution if there's a row of the form $0 0 0 0 | Q$ where $Q$ is not zero. Which row could possibly look like that? What would have to happen (to $a$, $b$, and $c$) for the row to look like that?
If there is a solution, then the variable represented by the 3rd column is arbitrary, so there are infinitely ma... | {
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"url": "https://math.stackexchange.com/questions/694828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt... | $$
\begin{align}
\sum_{n=1}^\infty\left(\left(\sum_{k=0}^{m-1}\frac1{\sqrt{n+k}}\right)-\frac{m}{\sqrt{n+m}}\right)
&=\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=0}^{m-1}\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\sum_{n=1}^N\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false... | If you use proper notation, instead of dots, you get the correct result:
$x^n-y^n=(x-y)\sum\limits_{k=0}^{n-1}x^{n-1-k}y^k$
Then $x^2-y^2=(x-y)(x^{1-0}y^0+x^{1-1}y^1)=(x-y)(x+y).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Express a vector as a combination of a linearly dependent set So I have to express the vector
\begin{pmatrix}
-2 \\
-4 \\
1 \\
0 \\
\end{pmatrix}
as a combination of these vectors:
\begin{pmatrix}
1 \\
0 \\
2 \\
1 \\
\end{pmatrix}
\begin{pmatrix}
1 \\
1 \\
0 \\
2 \\
\end{pmatrix}
\begin{pmatrix}
-1 \\
-3 \\
4 \\
-4 \\
... | I got the following set of equations:
$$\begin{align}
a_1 + 2a_3 &= -1 \\
a_2 - 3a_3 &= 2 \\
a4 &= -3
\end{align}$$
There were infinitely many solutions. I set $a_3 = 0$ and got the following as a valid linear combination:
$$-v_1 + 2v_2 + -3v_4$$
I tried this and got a failing answer before, but it was because I used $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Square root of x plus square root of x plus ... I want to confirm if my answer in this problem is correct:
$$\sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53}) / 2 $$
Solution:
$$x + \sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53})^2 / 4 $$
$$x + (1 + \sqrt{53}) / 2 = (1 + 2\sqrt{53} + 53) / 4 $$
$$ x + (1 + \sqrt{53}) / ... | Yes your answer is correct. Let's write it more readable:
Let
$$y=\sqrt{x+\sqrt{x+\cdots}}$$
hence the equation is
$$y=\frac{1+\sqrt{53}}{2}$$
so squaring the terms of the last equality gives
$$y^2=x+y=\frac{(1+\sqrt{53})^2}{4} $$
hence we have
$$x=y^2-y=\frac{(1+\sqrt{53})^2}{4}-\frac{1+\sqrt{53}}{2}=\frac{1+2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/700210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sin{b}\sin{c}\sin{(b-c)}(\sin^2{b}+\sin^2{c}+\sin^2{(b-c)})+\dots=0$
If $a,b,c\in (0,\pi)$ and $a+b+c=\pi$, show that:
$$\sin b \sin c \sin(b-c) \left(\sin^2 b + \sin^2 c + \sin^2(b-c)\right) \\
+ \sin c \sin a \sin(c-a) \left(\sin^2 c + \sin^2 a + \sin^2(c-a)\right) \\
+ \sin a \sin b \sin(a-b) \le... | Consider the following configuration:
where $ABC$ is an acute triangle and $X'$ is the symmetric of $X$ with respect to the perpendicular bisector of $YZ$. The six depicted points are clearly concyclic, and by assuming that the diameter of the circumcircle is one we have:
$$ z = XY = \sin\widehat{Z},\quad XX'=\left|\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| Hint: Multiply the numerator and denominator by the conjugate of the expression.
That is, you can try working with $$\dfrac{\sqrt{(x+a)(x+b)} - x}1\cdot\frac{\sqrt{(x + a)(x + b)} + x}{\sqrt{(x + a)(x + b)} + x} = \dfrac{(x+a)(x + b) - x^2}{\sqrt{(x + a)(x+b)} + x}$$
Expanding $(x + a)(x + b)$ in the numerator and sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Curve parametrization When plotting following implicit function $x^4 + y^4 = 2xy$, what is the best way to parametrize it? I tried $y=xt$, but then I get $x=\sqrt{2t\over {1+t^4}}$ and $y=\sqrt{2t^3\over {1+t^4}}$, which means I am restricted by $x>0, y>0, t>0$. It is easy to prove that the curve is symmetric about the... | That relation requires $xy \ge 0$, which translates to
$$
\left [
\begin{array}{c}
\left \{
\begin{array}{c}
x \ge 0 \\ y \ge 0
\end{array}
\right . \\
\left \{
\begin{array}{c}
x < 0 \\ y < 0
\end{array}
\right .
\end{array}
\right .
$$
So, the curve defined by the relation doesn't go through second and fourth quadra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$?
Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.
The only reduction I thought about so far would be:
$$n\c... | I think the initial steps towards a formula look a bit simpler than seen so far here.
Beginning with
$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$
we count the number of occurences of the $1$, of the $2$
$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$
and rewrite a bit
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How prove this inequality $\frac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\frac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$ let $a,b,c,d>0$, show that
$$\dfrac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\dfrac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$$
This post three -varible $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{a... | \begin{align}
a& \frac{a+b}2 \frac{a+b+c}3 \frac{a+b+c+d}4 \\
&= \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\tfrac{a+b+c}3+c \right) \left(a+b+c+d \right) \\
&\ge \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\sqrt[3]{abc}+c \right) \left(a+b+c+d\right) \\
&\ge \frac1{4^4} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$
The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0. \end{alig... | Taking logs (as done in OP's post) we can see that $$\lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{f(x)}{x}\right)}{x} = 3$$ or $$\lim_{x \to 0}\log\left(1 + x + \frac{f(x)}{x}\right) = 0$$ or $$\lim_{x \to 0}1 + x + \frac{f(x)}{x} = 1$$ or $$\lim_{x \to 0}\frac{f(x)}{x} = 0 \Rightarrow \lim_{x \to 0}f(x) = 0$$ and hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How find this $\frac{yf_{y}-z}{f_{x}}+\frac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$ solution In plane $R^3$,Find $z=f(x,y)$, such the length of the portion of any tangent line to the astroid
$$z=f(x,y)$$
cut off by the coordinate axes is constant $C$,
This problem is from this post (when I answer it) How find a solut... | my answer is :
firstly,we assume that:
$$X=\dfrac{yf_{y}-z}{f_{x}}+x=P+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y=Q+y,Z=z-xf_{x}-yf_{y}=z-R$$
then,
$C^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$
$C(C-1)=(X+Y+Z)(X+Y+Z-1)=(X+Y+Z)^2-(X+Y+Z)=(x+y+z+P+Q-R)^2-(x+y+z+P+Q-R)=2\sqrt{xy}+2\sqrt{yz}+2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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what is the proof for $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $ Can someone provide a proof for the solution of this series
$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $
| $$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$$
so
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)}&=\dfrac{1}{2}\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)\\
&=\dfrac{1}{2}\lim_{n\to\infty}\left(\dfrac{1}{1\times 2}-\dfrac{1}{(n+1)(n+2)}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Two circles inside a right angled triangle! The other day I was playing with Ms Paint drawing circles here and there - I coincidentally drew a circle inside a right angled triangle which I already drew. Strangely A problem struck to my mind and I tried solving it , but I was unable to do so. I put forward the stateme... |
Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have
$$\begin{align}
|\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt]
\implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
} |
solve partial differential equation
$$y^2u_x + xu_y = \sin(u^2) \\
u(x,0)=x$$
I get the projected characteristic curve on xy plane easily.
However, cannot get the other one.
actually the problem is getting the value of $U_{xx}, U_{xy}, U_{yy}, U_x, U_y$ on x-axis. How can I do this?
| $y^2u_x+xu_y=\sin u^2$
$\dfrac{u_x}{x}+\dfrac{u_y}{y^2}=\dfrac{\sin u^2}{xy^2}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=\dfrac{1}{y^2}$ , letting $y(0)=0$ , we have $y^3=3t$
$\dfrac{dx}{dt}=\dfrac{1}{x}$ , letting $x(0)=x_0$ , we have $x^2=2t+x_0^2$
$\dfrac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the critical numbers of the function. $$\sin^2(x) + \cos(x)$$
$$\{0 < x < 2\pi\}$$
I thought the answer would be $\pi$, but it is not. Can anyone explain why the answer is not $\pi$?
| Taking the derivative:
$$\dfrac{d}{dx}\left(\sin^2 x + \cos x\right)=2\sin x \cos x -\sin x$$
We want $2\sin x\cos x-\sin x$ to equal $0$ (provided that $0 < x < 2\pi$). We immediately see that when $\sin x = 0$, $2\sin x \cos x-\sin x =0$. The only value of $x$ that will make $\sin x = 0$ and is in the interval $0<x<2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What's the easiest way to factor $5^{10} - 1$? What's the easiest way to factor $5^{10} - 1$?
I believe $5 - 1$ is a factor based off the binomial theorem.
From there I do not know.
We are using congruence's in this class.
| Consider $x^{10} - 1.$
The first trivial factorization is $(x^5-1)(x^5+1)$.
Note that $1+x+x^2+x^3+x^4 = \frac{x^5-1}{x-1}$ by the geometric series formula.
Secondly, replace $x$ with $-x$ to see that $1-x+x^2-x^3+x^4 = \frac{x^5+1}{x+1}.$
$$x^{10} - 1 = (1+x+x^2+x^3+x^4)(1-x+x^2-x^3+x^4)(x-1)(x+1).$$
It should be eas... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the summation involving binomials. $\sum _{ i=0 }^{ 100 }{\binom{k}{i}}*{\binom{M-k}{100-i}*\frac{k-i}{M-100}}/{\binom{M}{100}}$
I wrote the first few terms but couldn't find any pattern and how to club the terms. Help.
| Let's generalize a bit by introducing indices:
$$
\sum_i \binom{k}{i}
\cdot \binom{M - k}{r - i}
\cdot \frac{k - i}{M - r}
/ \binom{M}{r}
= \frac{1}{(M - r) \binom{M}{r}}
\cdot \sum_i \binom{k}{i} \binom{M - k}{r - i} \cdot (k - i)
$$
We can split the $k - i$ to simplify.
So we are int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum area of a triangle In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4
My reasonings:
$BC = a$, $AC = b$, $AB = c$
$AD=AF=x$
$FC=CE=y$
$BD=BE=z$
$a=z+y$, $b=x+y$, $c=x+z$
The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac... | You were on the right track.
Using your notation, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(x + 3)xyz}$, but also $A = x + 3$. So
$$\begin{align}
\sqrt{(x + 3)xyz} &= x + 3, \\
xyz &= x + 3, \\
x &= \frac{3}{yz - 1}.
\end{align}$$
Substituting that into $A$ we get
$$A = \sqrt{\left(\frac{3}{yz - 1} + 3\right)\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$ In solving the definite integral $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$$I tried to do $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx=\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx\Longrightarrow$$ If $x=\frac{2}{\sqrt{3}}\sec(u)\Longrightarrow dx=\frac2{\sqrt3... | You are essentially finished. We have $\sec u=\frac{\sqrt{3}x}{2}$ and $\tan u=\frac{\sqrt{3x^2-4}}{2}$. Thus your procedure gives as first term
$$\frac{1}{\sqrt{3}}\log\left|\frac{\sqrt{3}x+\sqrt{3x^2-4}}{2}\right|.$$
This is equal to
$$\frac{1}{\sqrt{3}}\log\left|\sqrt{3}x+\sqrt{3x^2-4}\right|-\frac{1}{\sqrt{3}}\lo... | {
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"url": "https://math.stackexchange.com/questions/725168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help finishing proof via induction for a summation So I have to prove the following equation using induction for n >= 2:
$$
\sum\limits_{i=1}^n 4/5^i < 1
$$
However the question asks me to prove something stronger such as this:
$$
\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}
$$
first to imply the first equation is t... | Hint:
$$\sum_{i=1}^{k+1}4/5^i=\sum_{i=1}^k 4/5^i+4/5^{k+1}$$
Use the induction hypothesis on the sum from $1$ to $k$ and simplify.
| {
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"url": "https://math.stackexchange.com/questions/725335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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A definite integral $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$
I tried putting $-x$ using properties of definite integral, but that doesn't really help.
$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{-\ln(1-b\sin x)}{\sin x}dx$$
I don't think adding these 2 equations would yield anything. And I ... | From the Taylor series expansion of $\ln(1+x)$ we know that
$$\ln(1+b\sin x) = b\sin x - \frac{b^2\sin^2x}{2} +\frac{b^3\sin^3x}{3} - \cdots$$
Therefore
$$\frac{\ln(1+b\sin x)}{\sin x} = b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots$$
Integrating both sides of the equation
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b... | {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci Proof Using Induction
$$f(n) = \left\{\begin{matrix}
0 & n=1\\
1 & n=2\\
f_{n-1} + f_{n-2} &
n\geqslant 2\end{matrix}\right.$$
How can I prove by induction that $$f_{n} \leq \left ( \frac{1+\sqrt{5}}{2} \right )^{n-1}$$ for all$$ n\geq l_{a}$$, I have to find the smallest value for $$l_{a}$$
| $f_1 = 0 \leq \left(\frac{1+\sqrt{5}}{2}\right)^0 = 1$
$f_2 = 1 \leq \frac{1+ \sqrt{5}}{2} \approx 1.618$
Suppose $f_k \leq \left(\frac{1+\sqrt{5}}{2}\right)^{k-1}, f_{k+1} \leq \left( \frac{1 + \sqrt{5}}{2} \right)^{k}$
Then $f_{k+2} = f_k + f_{k+1} \leq \left(\frac{1+\sqrt{5}}{2}\right)^{k-1} + \left( \frac{1 + \sqrt... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group. Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.
Thank you in advance. Any help is appreciated.
| From $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$ it follows that $(ba)^2=a^2b^2$, and $(ba)^4=a^4b^4$. This implies $(ba)^4=(ba)^2(ba)^2=a^2b^2a^2b^2=a^4b^4$, so $b^2a^2=a^2b^2$, hence all squares commute. Thus $(ba)^2=baba=a^2b^2=b^2a^2=bbaa$ and it follows that $ab=ba$.
| {
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"url": "https://math.stackexchange.com/questions/737786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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What is this trigonometric expression equal to? If $\frac{cos{x}}{cos{y}}=\frac{a}{b}$, then $(a\times tan{x}+b\times tan{y})$ equals
(A)$(a+b)cot{\frac{x+y}{2}}$
(B)$(a+b)tan\frac{x+y}{2}$
(C)$(a+b)(tan\frac{x}{2}+tan\frac{y}{2})$
(D)$(a+b)(cot\frac{x}{2}+cot\frac{y}{2})$
I believe know the trigonometric formulas but ... | If $x=y$ then the expression is $(a+b)\tan x$, so the only alternative that could possibly be true is (B): that is,
$$\hbox{if}\quad\frac{\cos x}{\cos y}=\frac{a}{b}\quad\hbox{then}\quad
a\tan x+b\tan y=(a+b)\tan\frac{x+y}{2}\ .$$
It remains to prove it.
We have
$$\frac{a}{a+b}=\frac{\cos x}{\cos x+\cos y}\ ,\quad \f... | {
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How to compute $\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$? How to compute this integral? :
$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$
Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?
| A little roundabout, but here goes. Write
$$\begin{align}I &= \underbrace{\int_0^{\infty} dx \, \sqrt{x} \, e^{-\left (x+\frac1{x} \right )}}_{x=u^2} \\ &= 2 e^2 \underbrace{\int_0^{\infty} du \, u^2 \, e^{-\left (u+\frac1{u} \right )^2}}_{v=u+\frac1{u}}\\ &= e^2 \int_{\infty}^2 dv \, \left (1-\frac{v}{\sqrt{v^2-4}} \... | {
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"answer_id": 0
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Invalid subtraction when solving system of equations? I'm trying to solve these two equations:
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=... | First $x=y$ yields $8x^3=1$
$x=\dfrac{1}{2},\dfrac{-1+\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}$
$\therefore(x,y)=\left(\dfrac{1}{2},\dfrac{1}{2}\right),\left(\dfrac{-1+\sqrt3i}{4},\dfrac{-1+\sqrt3i}{4}\right),\left(\dfrac{-1-\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}\right)$
But $x^2+y^2=0$ yields $1=0$ which is not useful.
Note than ... | {
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"answer_id": 2
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Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
| Well the idea is very simple. We have to use the fundamental limit $$\lim_{y \to 0}\frac{1 - \cos y}{y^{2}} = \frac{1}{2}$$ This is a pretty standard result which can easily be proved by simplifying $(1 - \cos y)$ as $2\sin^{2}(y/2)$ and then applying $\lim\limits_{y \to 0}\dfrac{\sin y}{y} = 1$.
For the current questi... | {
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"source": "stackexchange",
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"answer_id": 1
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)... | Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, our inequality it's just Muirhead.
| {
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How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$ How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$
I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.
I also tried to write all the cosines i... | According to the equation:
$$2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) = -4sin^6(x) - 1$$
this means that:
$$\frac{2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) + 1}{-4} = sin^6(x)$$
And since we are dividing by a negative where $x\in \mathbb{Q}$ I assume, this means that:
$$3(cos^4(x) + sin^4(x)) - 1 > 2(cos^6... | {
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How does Laplace expansion work?
\begin{bmatrix} 1 & 2 & 0 & 0 & a\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 1 & 1 \end{bmatrix}
Caculate its determinant.
Although it has small numbers, it's still fairly long to calculate the determinant of a large matrix. Peeking at the solution to th... | Hint: Expand along the rows that have the most zeros.
Expanding along row 2, we have:
$$\begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{vmatrix}$$
Note, what I mean by expanding along row $2$ using Laplace Expansion (see link below), is:
$$0 \begin{vmatrix} 2 & 0 & 0 & a\\ 0 & 1 & 1... | {
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Suppose $A=\langle a,b\mid ab^2a^{-1}b^{-3},ba^2b^{-1}a^{-3}\rangle.$ Show that $A \cong \{1\}$. Suppose $A=\langle a,b\mid ab^2a^{-1}b^{-3},ba^2b^{-1}a^{-3}\rangle$, where $\langle a_1,\ldots, a_n \mid R\rangle$ is the group generated by $a_1,\ldots, a_n$ with relations in $R$. Show that $A \cong \{1\}$.
After carryin... | $ab^2a^{-1}=b^3 \Rightarrow ab^4a^{-1}=b^6 \Rightarrow a^2b^4a^{-2}=ab^6a^{-1} = b^9$.
Now $a^2 = b^{-1}a^3b$, and substituting for $a^2$ in the equation $a^2b^4a^{-2}=b^9$ gives $a^3b^4a^{-3}=b^9$.
But $a^3b^4a^{-3}=b^9 = a(a^2b^4a^{-2})a^{-1} = ab^9a^{-1}$, so $ab^9a^{-1}=b^9$.
But $a^{-1}b^9a=b^6$, so $b^9=b^6$ and ... | {
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Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\do... | From the OGF of Catalan numbers we have that:
$$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} $$
where the radius of convergence of the LHS is $\frac{1}{4}$ since $\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}$.
By evaluating the previous identity at $x=\frac{1}{6}$ it follows that:
$$ \color{red}{... | {
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Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I di... | If you eliminate, say, $x$ using the second constraint, you are left with a two-variate function and a single constraint: Minimize/Maximize $d^2 = 5z^2 + y^2$ subject to $(y-2)^2 + 5(z-1)^2 = 4$. Now, a Lagrange solution is not bad. From here, you can play also with basic algebra and trig to avoid calculus altogether.
| {
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"answer_count": 4,
"answer_id": 2
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Calculate $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ I am trying to calculate:
$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the follo... | Here is a relatively simple way that only relies on knowing the following Maclaurin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(\sin^{-1} x)^2 = \frac{1}{2} \sum_{n = 1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}{n}}, \qquad |x| \leqslant 1.$$
Note that i... | {
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"question_score": "13",
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"answer_id": 2
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Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$ My try, using $x = \sec(u)$ substitution:
$$
\begin{eqnarray}
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\
&=& \int \tan^2(u) \mathrm{d}u \\
&=& \tan(u) - u + C \\
&=& \tan(arcsec(x)) - arcsec(x) + C... | $$\frac{x}{\sqrt{x^2-1}} - \frac{x}{(x^2-1)^{3/2} \left(1+\frac{1}{x^2-1}\right)} = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1} (x^2-1+1)}$$
$$ = \frac{x}{\sqrt{x^2-1}} - \frac{x}{x^2 \sqrt{x^2-1}} = \frac{x^3 - x}{x^2 \sqrt{x^2-1}} = \frac{x (x^2-1)}{x^2 \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}}{x}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Solving integral $ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x $ there is integral $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x$$
i am trying to separate this :
$$=\int \mathrm{d}x -\int \frac{\mathrm{d}x}{1+x+\sqrt{1+x+x^2}} $$ but have no idea about second
| We have the algebraic form:
$$
\frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}.
$$
Multiplying by $\dfrac{(1+x)-\sqrt{1+x+x^2}}{(1+x)-\sqrt{1+x+x^2}}$ yields
$$
\frac{\sqrt{1+x+x^2}-1}{x}=\frac{\sqrt{1+x+x^2}}{x}-\frac1x.
$$
The integral becomes
$$
\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\ dx=\int \frac{\sqrt{1+x+x... | {
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$\iiint_Dz \;dxdydz$ where D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$ I asked a questions about regions and then tried to compute a tripple integral:
$$\iiint_Dz \;dxdydz$$ D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$
I tried, but now I am stuck: how do I calculate the volume of $D_{1b}$? Since I do... | $$ z>0$$ What implies:$$z\geq\sqrt{2x^2+3y^2-1}$$
Volume I need to calculate is the interior of sphere: $$ x^2+y^2+z^2=3$$ and above (where we have orthogonal projection to the xy plane) the hyperboloid.
Excuse me for lousy picture!
So I have:
$$ U_{tot}= \int\int\int_{D_1} zdV + \int\int\int_{D_2}z dV=U_1+U_2$$
D... | {
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Prove $f(x) = \frac{1}{x^2}$ is uniformly continuous on $[1, \infty]$ I am trying to prove this function is uniformily continuous on $[1, \infty]$, so far i have;
$$|f(x) - f(x)| = |\frac{1}{x^2} - \frac{1}{y^2}| = |\frac{(x-y)(x+y)}{x^2y^2}|$$
and then,
$$|x-y||\frac{x+y}{x^2y^2}|$$
I am not really sure where to go fr... | Since $x, y\ge 1$ then
$$|x-y|\frac{x+y}{x^2y^2}\le |x-y|\frac{x^2+y^2}{x^2y^2}= |x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$$
$$ =|x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\le 2|x-y|$$
| {
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solving $X^2 - 3X - A = 0$ where $A,X \in \mathbb{M_2(\mathbb{R})}$ Given $A = \begin{pmatrix} 7 & 3 \\ 3 & 7 \end{pmatrix}$ find a $2\times 2$ matrix $X$ s.t. $X^2 - 3X - A = 0$, in the previous parts I have diagonalised $A$ and got $P^{-1}AP = \begin
{pmatrix} 4 & 0 \\ 0 & 10 \end{pmatrix}$, but I'm not sure how this... | $$X^2-3X-A=0$$
$$P(X^2-3X-A)P^{-1}=0$$
Which using the distruibitive property and the fact that $PX^2P^{-1}=(PXP^{-1})^2$, yields:
$$(PXP^{-1})^2-3(PXP^{-1})-PAP^{-1}=0$$
Now complete the square:
$$(PXP^{-1})^2-3(PXP^{-1})+\frac{9}{4}I_2=\frac{9}{4}I_2+PAP^{-1}$$
$$(PXP^{-1}-\frac{3}{2}I_2)^2=\frac{9}{4}I_2+PAP^{-1}= \... | {
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Number of integer solutions by generating functions method I'm stuck in the middle of a problem and not sure where to go next. The original problem is:
Find the number of integer solutions to the equation
$$2x + 3y + 4z + w + s + t = n$$ with $$0 \le w \le 2$$ $$2 \le s \le 5$$ $$0 \le t \le 3$$
Now I was able t... | You end up with:
\begin{align}
[x^n] G(x)
&= [x^n] \frac{x^2 + x^4}{(1 - x)^3} \\
&= [x^{n - 2}] (1 - x)^{-3} + [x^{n - 4}] (1 - x)^{-3} \\
&= \binom{-3}{n - 2} (-1)^{n - 2} + \binom{-3}{n - 4} (-1)^{n - 4} \\
&= \binom{n - 2 + 3 - 1}{3 - 1} + \binom{n - 4 + 3 - 1}{3 - 1} \\
&= \frac{n (n - 1)}{2} + \frac{(n... | {
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Analytic Geometry How does one solve:
Find the equation of the circle which has it's center on the line $y= 3-x$ , and which has as tangents the lines $ 2y-x = 22, $ $ 2x+y=11 $ ?
| Let $(a,b)$ be the center of the circle. Since the center lies on the line $y=3-x$, then we have $b=3-a$. The equation of the circle if the center on $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2\tag1
$$
and the equation of its tangent line is
$$
(x_c-a)(x-a)+(y_c-b)(y-b)=r^2,\tag2
$$
where $(x_c,y_c)$ is point of c... | {
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Proving uniform convergence
Prove uniform convergence of function series: $$ \sum_{n=0}^\infty \frac{1}{n^2 + x} \sin \frac{1}{n^2 + x}$$ on $ \Bbb R $
I'm stuck with a problem, because I've proven that is uniformly convergent(proof below), while a friend of mine, gave a counterexample to this problem, $$ x = \frac... | As $x$ approaches $-n^2$, $\frac1{n^2+x}\sin\left(\frac1{n^2+x}\right)$ oscillates wildly between very large numbers. Since uniform convergence only worries about the tail, we can fix this by assuming that $x$ is not the negative of a perfect square and that $x\gt-m^2$.
Let $M,N\gt2m$, then
$$
\begin{align}
\sum_{n=M}^... | {
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"source": "stackexchange",
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"answer_id": 0
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inner product of trivector and bivector in geometric algebra Hestenes's "New Foundations for Classical Mechanics" book (page 47, 1.1c) sets a problem to show:
$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot B=\mathbf{a} \left( \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B \righ... | I figured it out. The key is trying not to be so clever, instead just expanding in successive dot products and then regrouping. With $B = \mathbf{u} \wedge \mathbf{v}$
$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\cdot B&={\left\langle{{ \left( \mathbf{a} \wedge \mathbf{b} \wedge \m... | {
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$99$th derivative of $\sin x$ Can someone help me calculate the $99$th derivative of $\sin(x)$?
Calculate $f^{(99)}(x) $ for the function $f(x) = \sin(x) $
| Notice if $f(x) = \sin x$
$f'(x) = \cos x = \sin( x + \frac{\pi}{2}) $
$f''(x) = - \sin x = \sin( x + \pi) = \sin( x + 2 (\frac{\pi}{2})) $
$f'''(x) = - \cos x = \sin(x + 3( \frac{\pi}{2} ))$
$f''''(x) = \sin x $
Hence, we can say that
$$ f^{(n)} (x) = \sin \left( x + n \cdot\frac{\pi}{2} \right) $$
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Can an odd perfect number be divisible by $5313$? I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $5313$.
| It is quite easy to prove that an odd perfect prime has a factorisation into powers of odd primes, where exactly one power is a prime of the form 4k+1, raised to a power 4n+1 (n can be 0), and all the other powers are even. This implies an odd perfect number which is a multiple of $5313 = 3 * 7 * 11 * 23$ must be a mul... | {
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Use $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} = 6 $ to show $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} i^n = \frac 4 {25} + i \frac {22} {25}$. I've shown that the series $$\sum_{n=0}^{\infty} \frac {2n +1} {2^n}$$ converges to $6$ using elementary series operations.
However how can I use this to show that $$\sum_{n=0}^{\... | Hints:
$$i^n=\begin{cases}\;\;\,i&,\;\;n=1\pmod4\\-1&,\;\;n=2\pmod4\\-i&,\;\;n=3\pmod4\\\;\;\,1&,\;\;n=0\pmod4\end{cases}$$
Added on request: Using the above and the sum of a geometric series with ratio $\;r\;,\;\;|r|<1\;$
$$\sum_{k=0}^\infty\frac{2n+1}{2^n}i^n=2\sum_{n=0}^\infty \frac {i^nn}{2^n}+\sum_{n=0}^\infty\lef... | {
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How to solve this equation $2\cos(\frac {x^2+x}{6})=2^x+2^{-x}$ How do I solve for $x$ from this equation?
$$2\cos\frac {x^2+x}{6}=2^x+2^{-x}$$
| 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",
"url": "https://math.stackexchange.com/questions/770306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integration by parts order question + integration question In integration by parts, does it matter which of your terms is v (or, rather $f(x)$) and which term is du (or, rather $ f'(x) $?
Also, I'm having trouble finding $\int e^{2x} \cos(3x)dx $. I keep getting $$\frac {2e^{2x}(6\sin(3x) - \cos(3x))}{37}$$ when $u=e^... | Yes, it does matter. You can check Wikipedia's page on integration by parts, and in particular the LIATE rule. In general, you want $u$ to be easier to differentiate, whereas $dv$ to be easier to integrate.
In your example, however, if you follow the rule above, both functions should be $u$. You can overcome this by ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/770383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve $f(x) = ax^2 + bx + c$ to find the value of $K$ $f(x)=ax^2+bx+c$, where $a=-9$, $b=12$ and $c=16$. If $$-1<f'(x)<1$$ then $h<x<k$. To $2$ decimal places, what is the value of $k$?
Hi, this is working for solving $f(x) = ax^2 + bx + c$ to find the value of $K$
I used Quadratic Equation to solve for x and got the... | well, suppose $f(x) = ax^2 + bx + c $, then $f'(x) = 2ax + b $. If $-1 < f'(x) < 1$, then
$$ -1 < 2ax + b < 1 \iff -b-1 < 2ax < 1 - b \iff - \frac{b+1}{2a} < x < \frac{1-b}{2a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
using substitution wrongly Solving integral, first way:
$$\int \frac{du}{u^2-9}=-\int \frac{du}{9-u^2}$$
$$u={3\sin v}$$
$$du=3\cos vdv$$
$$-\int \frac{3\cos vdv }{9-9\sin^{2}v}=-\frac 13\int\frac{dv}{\cos v}=-\frac 13\ln\left(\sec v+\tan v\right)=-\frac13 \ln \frac {\frac u3}{\sqrt{1-\frac{u^2}{9}}}$$
$$-\frac13 \ln \... | $-\dfrac 13 \ln|\sec v + \tan v|$ $ = -\dfrac 13 \ln \left| \dfrac{1+\sin v}{\cos v} \right| \\ = -\dfrac 13 \ln \left|\dfrac{3+u}{\sqrt{9-u^2}}\right| \\ = - \dfrac 16 \ln \left|\dfrac{(3+u)^2}{(3-u)(3+u)}\right| \\ = \dfrac 16 \ln\left|\dfrac{3-u}{3+u}\right|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How many different systems are there? A Football team has 10 players besides the Keeper, who are allocated into 3 positions, the defense, the center and the attack. A system is the allocation of the players into these positions.
For example the system 4 4 2 means that there are 4 players at the defense, 4 players at th... | i have another idea. I used a generating function. Here it is $(x+x^2+x^3+x^4+x^5+x^6)*(x+x^2+x^3+x^4+x^5)*(x+x^2+x^3+x^4)$. If you expand this expression you can read off the number of systems can be formed. The Expansion is $x^3+3 x^4+6 x^5+10 x^6+14 x^7+17 x^8+18 x^9+17 x^{10}+14 x^{11}+10 x^{12}+6 x^{13}+3 x^{14}+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.
I made quite a few... | The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (0, -1), (1, -1) with a minimum at (1/2, -5/4), or some reflection of this. Back to 3/2 ...
If $f$ is monotonic in [-1, 1] then trivially, $|f| \le 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/774069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}$ Is my solution correct?
By Faulhaber's formula, $1^4 + 2^4 + \ldots + n^4 = \frac{6n^5+15n^4+10n^3-n}{30} $.
$$\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = \frac{1}{n^5} \left[\sum_{k=1}^{2n} k^4 - \sum_{k=1}^{n} (2k)^4 \right] = \frac{1}{n^5} \le... | Using Stolz:
$$
\lim_{n\to\infty}\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}=
\lim_{n\to\infty}\frac{(2n+1)^4}{(n+1)^5-n^5}=
\lim_{n\to\infty}\frac{16n^4+\cdots}{5n^4+\cdots}=\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral$\int_0^{\pi/4} \log \tan \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}=\pm\frac{\pi^2}{16}$ Hi I am trying to prove $$
\int_0^{\pi/4} \log \tan \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}=\pm\frac{\pi^2}{16}.
$$
What an amazing result and a clever one this is. I tried writing
$$
\int_0^{\pi/4} \lo... | We can write the integral as:
\begin{align*}
\int_0^{\pi/4} \frac{\log{\tan{(\frac{\pi}{4}-x)}}}{\tan{(2x)}}\, dx &= \int_0^{\pi/4} \log{\tan{(x)}}\tan{(2x)} \, dx \\
\end{align*}
Let
\begin{align*}
I(a) &= \int_0^{\pi/4} \tan{(x)}^a\, \tan{(2x)} dx \\
&= \int_0^{\pi/4} \frac{2\, \tan{(x)}^{a+1}}{1-\tan{(x)}^2} dx\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/779262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
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