Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluation of $\int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
Evaluation of $\displaystyle \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
My Try:: Let $\displaystyle I = \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$
So $\displaystyle I = -\ln\left(\cos x... | As you have done, we first use integration by parts to get
$$\begin{align}
I &= \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx
\\&= \int \ln(\cos x+\sqrt{\cos 2x})\csc^2 (x)dx
\\&=-\cot (x)\ln(\cos x+\sqrt{\cos 2x})+\int \left(\frac{1}{\cos x+\sqrt{\cos 2x}}\right) \left(-\sin x-\frac{\sin 2x}{\sqrt{\cos 2x}} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Taylor series convergence for sin x a. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!},$
b. $\forall x\in(0,\pi/2),\quad x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots-\frac{x^{4k-1}}{(4k-1)!}<\sin x<x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots+\frac{x^{4k+1}}{(4k+1)!},$
c. $\forall x\in(0,\pi/2)... | For a) and by the Maclaurin formula for the sine function there's $\theta\in(0,1)$ such that
$$\sin x=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin^{(4)}(\theta x)=x-\frac{x^3}{3!}+\frac{x^4}{4!}\sin(\theta x)>x-\frac{x^3}{3!},\; \forall x\in(0,\frac\pi2)$$
and
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^6}{6!}\sin^{(6)}(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True
if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true.
$\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space... | Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$.
So we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$
Assuming that all three inequalities are false, like you did, we obtain $11(x+y+z)<18$. But
$$(x+y+z)^2\geq3(xy+yz+xz)
\Rightarrow x+y+z\geq\sqrt{3}
\Rightarrow 11(x+y+z)\geq11\sqrt{3}>18$$
We have a contradiction.
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions of the factorial equation $(x!+1)(y!+1)=(x+y)!$ The problem is: are there solutions for the next equation?
$$(x!+1)(y!+1)=(x+y)!$$
with $x,y\in\mathbb{N}$.
My solution:
$\left(x!+1\right)\cdot \left(y!+1\right) = \left(x+y\right)!$
$x!y!+x!+y!+1= \left(x+y\right)!$
$\displaystyle\frac{x!y!+x!+y!+1}{x!... | The pairs $(1,2)$ and $(2,1)$ are the only solutions. To see this assume that $x>1$ and $y>1$. Then $x!$, $y!$ and $(x+y)!$ are even. But $(x!+1)(y!+1)$ is odd, which is a contradiction. Thus, $x=1$ or $y=1$, and since the problem is symmetric in $x$ and $y$ we assume that $y=1$. Then $2(x!+1)=(x+1)!$, which is equival... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/900578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
| If $\log $ is 2-base logarithm, you have:
$\;\;\; 3+4 \cdot 3+4^2 \cdot 3+ \cdots + 4^{\log n -1} \cdot 3
\\ = 4-1+4(4-1)+4^2(4-1)+4^3(4-1)+\cdots+4^{\log n -1} \cdot (4-1)
\\ = 4-1+4^2-4+4^3-4^2+\cdots+4^{\log n}-4^{\log n-1}
\\ = 4^{\log n}-1
\\ = 2^{2\log n}-1
\\ = n^2-1$
$\textbf{Edit}$ In sigma nota... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane. I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is:
If we consider 3 non-concurrent, non parallel lines represented by the equa... | equations in the reduced form
$y=ax+b$
$y=cx+d$
$y=ex+f$
solving by the Cramer rule we get the A, B and C points
calculating the area of the ABC triangle
$ΔABC=\frac{1}{2}\left| \begin{array}{} \frac{b-d}{c-a} & \frac{bc-ad}{c-a} & 1 \\ \frac{d-f}{e-c} & \frac{de-cf}{e-c} & 1 \\ \frac{f-b}{a-e} & \frac{af-be}{a-e} & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 3
} |
Minimizing the expression $(1+1/x)(1+m/y)$ over positive reals such that $mx+y=1$ Let $x$ and $y$ be positive real numbers such that $mx+y=1$. Find the positive $m$ such that the minimum of:
$$\left( 1 + \frac{1}{x} \right)\left( 1 + \frac{m}{y} \right).$$
is $81$.
I have tried expanding, using Cauchy-Schwarz inequal... | Since $x=\frac{1}{m}(1-y)$, we have:
$$\left(1+\frac{1}{x}\right)\left(1+\frac{m}{y}\right)=\left(1+\frac{m}{1-y}\right)\left(1+\frac{m}{y}\right),\tag{1}$$
and since $f(x)=1+\frac{m}{x}$ is a log-convex function, the minimum of the RHS of $(1)$ over $(0,1)$ occurs when $y=\frac{1}{2}$, and the value of such a minimum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$ if $abc=1$
if $a,b,c$ are positive real numbers that $abc=1 $,Prove:$$a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$$
Additional info: We should only use AM-GM and Cauchy inequalities.
Things I have done so far: for $a^2b+b^2c+c^2a $ minimum I can say: $$a^4b^2+b^2c^3a + b^... | Expand $(a^2b+b^2c+c^2a)^2$ and use the fact that $abc=1$ to reduce it and then factorise to get,
$$ a^2(a^2b^2+2c) + b^2(b^2c^2+2a) + c^2(c^2a^2+2b) $$
Using the fact $abc=1$ again,
$$ a^2\left(\frac{1}{c^2}+2c\right) + b^2\left(\frac{1}{a^2}+2a\right) + c^2\left(\frac{1}{b^2}+2b\right) $$
and then it's easy to show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the limits $\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$ and $\lim_{(x,y)\to(\infty,8)}(1+\frac{1}{3x})^\frac{x^2}{x+y}$ I got the following problem:
Evaluate the following limits or show that it does not exist:
$$\lim_{(x,y)\to(\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}$$
and
$$\lim_{(x,y)\to(\infty,... | Alternative solutions:
Consider the limit on all lines passing through the origin, i.e.,
$y=ax$. $$\lim_{(x,y)\to (\infty,\infty)}\frac{2x-y}{x^2-xy+y^2}=\lim_{x\to\infty}\frac{x\left[2-a\right]}{x^2\left[1-a+a^2\right]}=\lim_{x\to\infty}\frac{\left[2-a\right]}{x\left[1-a+a^2\right]}=0, $$ independently of the choice ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the points where a circle intersects an axis A circle has the equation:
x²+y²+4x-2y-11 = 0
What would be the coordinates of the points where the circle intersects with the y-axis and how would you calculate it?
| $x^2+y^2+4x-2y-11=0$
The points of intersection with the $y$ axis are when $x=0$. Thus, plug in $x=0$ and solve.
$$0^2+y^2+4(0)-2y-11=0$$
$$y^2-2y-11=0$$
Use the quadratic formula: $\large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1, b=-2, c=-11$
$$\frac{-(-2)\pm \sqrt{4-(-44)}}{2}$$
$$\frac{2\pm \sqrt{48}}{2}$$
$\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?
$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$
I've found that it cannot be denested into $a+b\sqrt{2}+c... | Try factoring $2\sqrt{10}$ as $2\sqrt{5}\sqrt{2}$, then the expression under the radical can be factored.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving these two equations simultaneously I'm having a hard time to solve these two equations simultaneously. I'm arriving to a very long equation..
$$x_0^2+y_0^2=(7\sqrt{2})^2=98$$
$$\sqrt{25+(x_0+2)^2}+\sqrt{4+(y_0-5)^2}=7\sqrt{2}$$
| Drop subscripts of $x$ and $y$.
Put $u=x+2, v=y-5$.
Then the equations become
$$(u-2)^2+(v+5)^2=(7\sqrt{2})^2 \qquad \cdots (1)$$
and
$$\sqrt{25+u^2}+\sqrt{4+v^2}=7\sqrt{2}\qquad \cdots (2)$$
Squaring $(2)$ equals $(1)$, i.e.
$$\begin{align}(25+u^2)+(4+v^2)+2\sqrt{(25+u^2)(4+v^2)}&=(u^2-4u+4)+(v^2+10v+25)\\
\sqrt{(25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with d... | Observe that $\displaystyle\frac{13\pi}{24}-\frac\pi2=\pi\dfrac{(13-12)}{24}=\frac\pi{24}$
and $\displaystyle\frac{19\pi}{24}-\frac\pi2=\pi\dfrac{(19-12)}{24}=\frac{7\pi}{24}$
So, $\displaystyle\sin\frac{13\pi}{24}=\sin\left(\frac\pi2+\frac\pi{24}\right)=\cos\frac\pi{24}$ and $\displaystyle\sin\frac{19\pi}{24}=\sin\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Finding substitution in the integral $\int{\frac{2+3x}{3-2x}}dx$ In a problem sheet I found the integral $$\int{\frac{2+3x}{3-2x}}dx.$$
In the solution the substitution $z=3-2x$ is given which yields $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2}dz$. We have
$$\int{\frac{2+3x}{3-2x}}dx = \int{\frac{2+3\left(\frac{3-z}{2}\right... | You were on the good way $$I =\int{\frac{2+3x}{3-2x}}dx$$ Change variable using $3-2x=z$, so $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2} dz$ (which is what you obtained). Replace and simplify as much as you can; you should easily arrive to $$I=\int \Big(\frac{3}{4}-\frac{13}{4 z}\Big)dz=\frac{3}{4}\int dz-\frac{13}{4}\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find all values of $z$ at which $\sum\limits_{n=1}^{\infty} \frac{1}{n^2} \exp(\frac{nz}{z-2})$ converges? Could anyone advise me on how to find all $z$ such that $\begin{align} \sum^{\infty}_{n=1} \dfrac{1}{n^2} \end{align}\text{exp}\left(\dfrac{nz}{z-2}\right)$ converges ? Does it suffice to find all $z$ such ... | The radius of convergence of $\sum \frac{1}{n^2} x^n$ is $1$. Plug in $x = \exp(\frac{z}{z-2})$, so that the series converge iff
$$|\exp(\frac{z}{z-2})| < 1 \iff \Re\left( \frac{z}{z-2} \right) < 0.$$
By writing down $z = a+bi$, where $a,b \in \mathbb{R}$, you get (let's assume $z \neq 2$ all the time):
$$\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
| Maple says:
$$
\int \!{\frac {\sqrt {1-{k}^{2} \sin^2 \left( x \right)
}}{\sin \left( x \right) }}{dx}
= \frac{1}{2}\left[k\ln \left( 2 \right) +k
\ln \left( k \right) -k\ln \left( -2\, \cos^2 \left( x \right)
{k}^{2}+2\,\cos \left( x \right)k \sqrt { \cos^2
\left( x \right){k}^{2}-{k}^{2}+1}+{k}^{2}-1 \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^{{\pi}/{2}} \log(1+\cos x)\, dx$ Find the value of $\displaystyle \int_0^{{\pi}/{2}} \log(1+\cos x)\ dx$
I tried to put $1+ \cos x = 2 \cos^2 \frac{x}{2} $, but I am unable to proceed further.
I think the following integral can be helpful:
$\displaystyle \int_0^{{\pi}/{2}} \log(\cos x)\ dx =-\frac{\pi... | \begin{align}
\color{red}{\int^{\pi/2}_0\log(1+\cos{x}){\rm d}x}
&=\int^{\pi/2}_0\frac{x\sin{x}}{1+\cos{x}}{\rm d}x\tag1\\
&=\int^{\pi/2}_0 x\tan{\frac{x}{2}} \ {\rm d}x\tag2\\
&=4\int^{\pi/4}_0x\tan{x} \ {\rm d}x\tag3\\
&=8\sum^\infty_{n=1}(-1)^{n-1}\int^{\pi/4}_0x\sin(2nx) \ {\rm d}x\tag4\\
&=2\sum^\infty_{n=1}(-1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Integral of $\ln(x)\operatorname{sech}(x)$ How can I prove that:
$$\int_{0}^{\infty}\ln(x)\,\operatorname{sech}(x)\,dx=\int_{0}^{\infty}\frac{2\ln(x)}{e^x+e^{-x}}\,dx\\=\pi\ln2+\frac{3}{2}\pi\ln(\pi)-2\pi\ln\!\Gamma(1/4)\approx-0.5208856126\!\dots$$
I haven't really tried much of anything worth mentioning; I've had bas... | You can get the value of the integral you're interested in from the integral $$I(a) =\int_{0}^{\infty} \frac{\ln (1+\frac{x^{2}}{a^{2}})}{\cosh x} \, dx, \quad a>0.$$
Notice that $\lim_{a \to \infty} I(a) = 0$.
Differentiating under the integral sign, we get $$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{2a}{(a^{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Cards in box - probability a given type is picked last I came out with a probability question which I find difficult to solve. I hope some kind souls can provide me with some ideas.
There is a box with four different types of cards, namely A, B, C, D. There are 7 A, 4 B, 3 C and 2 D. One starts to pick cards from the b... | The cards in the box are: 7 of type A, 4 of B, 3 of C and 2 of D.
Let, for example, $A=1,B=2,C=3$ represent the event of encountering the type A first, type B second, type C third, and type D last. (We don't have to write the last, it's implicit.) One such example is to draw cards in order $\mathbf A,A,\mathbf B, A, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proof by induction (exponents)
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$
The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
| *
*Show $3^{1-1}=\dfrac{3^1-1}{2}$
*Assume $\sum\limits_{k=0}^{n-1}3^k=\dfrac{3^n-1}{2}$
*Prove $\sum\limits_{k=0}^{n}3^k=\dfrac{3^{n+1}-1}{2}$:
*
*$\sum\limits_{k=0}^{n}3^k=(\sum\limits_{k=0}^{n-1}3^k)+3^n$
*$(\sum\limits_{k=0}^{n-1}3^k)+3^n=\dfrac{3^n-1}{2}+3^n$
*$\dfrac{3^n-1}{2}+3^n=\dfrac{3^n-1+2\cdot3^n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What's the sum of this series? I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?
Thanks!
| It is well known that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
Thus, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \sum_{m=1}^{\infty}\dfrac{1}{(2m)^2} = \dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)$ I am having a tough time with these TYPE of problems
looking forward ideas, All ideas will be appreciated
| Multiply numerator and denominator by the conjugate of the "numerator": $\sqrt{x^2 + x+1} + \sqrt{x^2 -x}$, to get a difference of squares in the numerator.
$$\lim\limits_{x \to \infty} \left(\sqrt{x^2+x+1} - \sqrt{x^2-x} \right)\cdot \frac{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}{\sqrt{x^2 + x+1} + \sqrt{x^2 -x}}$$
$$ = \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\frac{dy}{dx}$ for $x=2\theta+sin2\theta$ and $y=1-cos2\theta$
The parametric equations of a curve are
$$x=2\theta+\sin2\theta,\:y=1-\cos2\theta.$$
Show that $\frac{dy}{dx}=\tan\theta$.
I can use the chain rule to get
$$\frac{dx}{d\theta}=2+2\cos2\theta$$
$$\frac{dy}{d\theta}=2\sin2\theta$$
$$\frac{dy}{dx}=... | You can see that $$\frac{2\sin 2 \theta}{2+2 \cos 2 \theta} = \frac{\sin 2 \theta}{1+ \cos 2 \theta} \\ = \frac{2 \cos \theta \sin \theta}{2 \cos^2 \theta} = \tan \theta$$
I used the identities $\cos 2\theta = 2 \cos^2 \theta - 1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:
let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that
$$a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$$
I should determine whether this is a converg... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\sum\limits_{cyc}a^3b^3=27v^6-27uv^2w^3+3w^6$, we see that our inequality is equivalent to $f(w^3)\geq0$,
where $f$ is a concave function, which says that it's enough to prove our inequality
for an extremal value of $w^3$, which happence in the following cases.
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate Integrals $ \int \sqrt{\sec 2x-1}\;dx$ and $ \int \sqrt{\sec 2x+1}\;dx$ Calculation of Integral of
$$\displaystyle \int \sqrt{\sec 2x-1}\;dx,\>\>\>\>\>\displaystyle \int \sqrt{\sec 2x+1}\;dx$$
$\bf{My\; Solution}::$ For $(a)::$
Let $$\displaystyle I = \int \sqrt{\sec 2x-1}\;dx = \int \sqrt{\frac{1-\cos 2x}{\c... | Why not just perform this on computer?
$\int \sqrt{\sec{2 x}+1} dx = -8 \cos^4\left(\frac{x}{2}\right) \sqrt{\frac{-\left(1+\sqrt{2}\right) \sec
(x)+\sqrt{2}+2}{\sec (x)+1}} \left(\left(1+\sqrt{2}\right) \sec (x)+\sqrt{2}+2\right)
\sqrt{\frac{1}{\sec (x)+1}-\sqrt{2}+1} \sqrt{\frac{3-2 \sqrt{2}}{\sec (x)+1}-5
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why do the $\sec$ & $\cosh$ substitutions give different antiderivatives for $\frac1{\sqrt{x^2-a^2}} \,?$ Suppose we want to find
$$\int \frac{1}{\sqrt{x^2-a^2}}\,\mathrm dx.$$
Trigonometric substitution: $$=\ln \left| x+\sqrt{x^2-a^2} \right|$$
Hyperbolic substitution: $$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln\left(... |
Suppose we want to find
$$\int \frac{1}{\sqrt{x^2-a^2}}\,\mathrm dx.$$
Trigonometric substitution: $$=\ln \left| x+\sqrt{x^2-a^2} \right|$$
Hyperbolic substitution:
$$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln\left({x+\sqrt{x^2-a^2}}\right)$$
the fact that trig substitution gives a more COMPLETE solution than hyperboli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Differentiation help (lots of square roots) help finding the first derivative of this question:
$y=\sqrt{ 1 + \sqrt{ 1+\sqrt{1+8x}}}$
I get confused, is it meant to be done implicitly or is it just a really long chain rule?
| It seems like it is a long chain rule indeed:
\begin{align}
\frac{dy}{dx}&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1+8x}}}} \cdot \frac d{dx} \left(1+\sqrt{1+\sqrt{1+8x}} \right) \\
&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1+8x}}}}\cdot \frac 1{2\sqrt{1+\sqrt{1+8x}}} \cdot \frac d{dx} \left(1+\sqrt{1+8x} \right) \\
&= \cdots
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int (4x+2)\sqrt{x^2+x+1}\,dx$ Trying to solve this for a while now, so far I was able to come up without a proper answer.
Problem : $\displaystyle \int (4x+2)\sqrt{x^2+x+1}\,dx$.
I tried to take two common from $(4x+2)$ and also to take $(x+1)^2 - x$ from the root, but wasn't able to come up with something to ... | I'm extremely new to integration and my methods are sloppy but I'd like to try out this question. The following isn't an answer but a demonstration that any substitution can do the job as long as it is done right (Although, a fundamental zen of calculus dictates that the rightest way is the fastest way, I think any pat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
if $x+y+z=2$ and $xy+yz+zx=1$,Prove $x,y,z \in \left [0,\frac{4}{3} \right ]$
things i have done: first thing to do is to show that $x,y,z$ are non-negative. $$xy+yz+zx=1 \Rightarrow zx=1-yz-xy \Rightarrow zx=1-y(z+x)\Rightarrow zx=1-y(2-y... | You have
$$
1=xy+z(x+y)=xy+(x+y)(2-(x+y))=xy+x(2-x)+2y-y^2,
$$
so that
$$
0=y^2+(x-2)y+x^2-2x+1=\bigg(y+\frac{x-2}{2}\bigg)^2+\frac{3}{4}x^2-x
$$
and hence $x-\frac{3}{4}x^2 =-\bigg(y+\frac{x-2}{2}\bigg)^2 \leq 0$. So we have
$\frac{3x}{4}(\frac{4}{3}-x) \leq 0$, i.e. $x(x-\frac{4}{3}) \geq 0$, hence
$x\in [0,\frac{3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$? What does this number equal if it goes on forever?
$$\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\fr... | SPOILER. Computation of the expression out to $20$ levels suggests strongly that the answer is...
... don't rollover unless you really want to see it...
$\displaystyle\frac{1}{\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/924601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 6,
"answer_id": 2
} |
Find the quadratic equation equation of $x_1, x_2$.
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.
By Vieta's theorem:
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have... | Here's how I would do the "calculate and check" steps:
$$\begin{align}(x-x_1)(x-x_2)&=0\\
x_1&=1+\sqrt 3\\
x_2&=1-\sqrt 3\\
x^2-(x_1+x_2)x+x_1x_2&=0\\
x_1+x_2&=2\\
x_1x_2&=1^2-\sqrt 3^2=-2\\
x^2-2x-2&=0\end{align}$$
Now, taking the form that you created, we have
$$-\frac b2x^2+bx+b = 0$$
Plugging in $x=x_1=1+\sqrt 3$ r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$? What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$?
I know that $\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle$ or equivalently $\nabla f = \frac{\partial f}{\partial x}\ma... | So you're asking about what is $\nabla_{\vec{v}}f$ where $f = \sqrt{x^2 + y^2 + z^2}$. First, lets organize the answer:
$\nabla_{\vec{v}}f =
\begin{aligned}
\begin{bmatrix}
\frac{\partial f}{\partial x} \\
\frac{\partial f}{\partial y} \\
\frac{\partial f}{\partial z}
\end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric equation $\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$ I want to solve a trigonometric equation below:
$$\tan(\frac{\sqrt{3}x}{2})=-\sqrt{3}$$
What is the value of $x$ for $x>0$
Thank you for your help.
| Note $\tan\left(-\frac{\pi}{3}\right)=-\sqrt{3}$, and $\tan$ is periodic with period $\pi$, so this means that for some $n\in\mathbb{Z}$
$$\frac{\sqrt{3}x}{2}=n\pi-\frac{\pi}{3}\implies x=\frac{2\pi}{\sqrt{3}}\left(n-\frac{1}{3}\right).$$
Since we require $x>0$, this is equivalent to saying $n-\frac{1}{3}>0$, and since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Fermat's Little Theorem to find remainders My professor does not want us to use mods.
Use Fermat's Little Theorem to find the remainder of $12^{7641}$ divided by $7$.
I have $7 \mid \left(\,12^{7} - 12\,\right)$ and that $12^{7651}=\left(\,12^{7}\,\right)^{1093}$
but I'm basically stuck because a remainder of $12... | By Fermat's Little Theorem we know that $12^6 = 1 \pmod 7$
*
*Now $7641 =6 \times 1273 + 3$. So that $(12^6)^{1273} = 1^{1273} = 1 \pmod 7$
*hence $12^{7638} = 1 \pmod 7$
*hence $12^{7638} \times 12^3 = 12^3 \pmod 7$
*hence $12^{7641} = 12^3 \pmod 7$
Now $12 = 2^2 3$. Hence $12^3 = 2^6 \times 3^3$
*
*no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Verifying proof of $\lim_{x \to\sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4} = \frac 2 3$ $$\lim_{x \to\sqrt{2}} \dfrac{x^2-2}{x^2+\sqrt{2}x-4} = \lim_{w \to2} \dfrac{w^2-4}{w^2+2w-8} =\lim_{w \to2} \dfrac{(w-2)(w+2)}{(w+4)(w-2)} = \frac 2 3$$
Change of variable:
$$w=\sqrt{2}x \Rightarrow x=\frac{w}{\sqrt{2}}\Rightarrow x^2=\... | That is correct, although you could do the same without the change of variables, and with some courage to manipulate the $\sqrt{2}$
$\lim\limits_{x\to \sqrt{2}} \frac{x^2 - 2 }{ x^2 + \sqrt{2}x - 4}= \lim\limits_{x\to \sqrt{2}} \frac{ ( x- \sqrt{2} ) ( x + \sqrt{2} ) }{ (x - \sqrt{2} ) ( x + 2 \sqrt{2} ) } = \lim\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
| Perhaps a small reduction in the calculations in other answers. First, we use the standard technique for finding the power of a prime which divides a factorial:
$$\Bigl\lfloor\frac{28}{5}\Bigr\rfloor+\Bigl\lfloor\frac{28}{25}\Bigr\rfloor
=5+1=6$$
so $5^6$ is a factor of $28!$ (and $5^7$ is not), and likewise
$$\Bigl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Find $ \int \frac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$ Find $$\int \dfrac{\sin^2x}{1+\sin^2x}\hspace{1mm}\mathrm{d}x$$
I cannot figure out how start this problem, can anyone explain
| $$
\begin{aligned}
\int\frac{\sin^2 x}{1 + \sin^2 x}\,\mathrm{d}x&=\int\frac{1+\sin^2x - 1}{1+\sin^2 x}\,\mathrm{d}x\\
&=x - \int\frac{\mathrm{d}x}{1+\sin^2 x}\\
&=x+\int\frac{-\csc^2 x}{\csc^2 x + 1}\,\mathrm{d}x\\
&=x+\underbrace{\int\frac{-\csc^2 x}{2+\cot^2 x}\,\mathrm{d}x}_{t=\cot x\implies\mathrm{d}t=-\csc^2x\,\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplifying an expression written as the sum of three fractions Specifically, I don't know what to do first given the following expression:
$$
\frac{4x - 2}{6} - \frac{2 - x}{4} + \frac{x + 3}{3}
$$
So I think of it as $\frac 16(4x-2) - \frac 14(2-x) + \frac 13(x+3)$?
That just gives me more fractions.
| Find the common denominator first, noting that the least common multiple of $6, 4, 3$ is $12: $12 = 2\times 6 = 3\times 4 = 4\times 3$. This way, you can write the entire expression as one fraction:
$$\frac{4x - 2}{6} - \frac{2 - x}{4} + \frac{x + 3}{3} = \frac{2(4x-2) - 3(2-x) + 4(x+3)}{12}$$
Now expand the factors in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving an equation over the reals: $ x^3 + 1 = 2\sqrt[3]{{2x - 1}}$ Solve the following equation over the reals:$$
x^3 + 1 = 2\sqrt[3]{{2x - 1}}
$$
I noticed that 1 is a trivial solution, then I tried raising the equation to the 3rd, then dividing the polynomial by $(x-1)$.. But I can't see the solution, how do I go ... | For $x$ - real
$$2\cdot(2x-1)^\frac13=x^3+1$$
Let $y= (2x-1)^\frac13$
Therefore, $$y^3=2x-1$$
$$x=\frac{(y^3+1)}{2}$$
Then you get
$$2\cdot y=\left(\frac{y^3+1}2\right)^3+1$$
Resolve it for $y$, and then replace find the $x$
Resolving for $y$:
*
*multiply both sides by 8 and you get
$$16\cdot y=(y^3+1)^3+8$$
$$16\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Symmetric inequality for a rational function of three variables If $x,y,z$ are positive real numbers such that $xyz \geqslant 1$ prove: $$\dfrac{x^3+y^3}{x^2+xy+y^2}+\dfrac{y^3+z^3}{y^2+yz+z^2}+\dfrac{x^3+z^3}{x^2+xz+z^2} \geqslant 2$$
I have tried with Hölder's inequality, but it is not working. Can you help, please?... | Since:
$$\frac23x^3+\frac13y^3\ge x^y\text{ or }\frac23y^3+\frac13x^3\ge y^2x$$
So:
$$\large x^3+y^3\ge\frac13x^3+\frac23x^2y+\frac23xy^2+\frac13y^3=\frac13(x+y)(x^2+xy+y^2)$$
Hence:
$$\large \sum_{cyc}\frac{x^3+y^3}{x^2+xy+y^2}\ge\sum_{cyc}\frac{x+y}{3}=\frac23(x+y+z)\ge2\sqrt[3]{xyz}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
Prove $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$
Things I have done: after trying many ways and failing, I reached the fact that$\left(\frac{1}{2}\times\frac{3}{4}\time... | Don't you already have the answer?
$\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)^2>\frac{1}{2}\left(\frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}\right)\left(\frac{2}{3}\times\frac{4}{5}\times\cdots\times\frac{98}{99}\right)=\frac{1}{200}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving a recurrence relation of second order I have a pattern, which goes: $x_n =2(x_{n-1}-x_{n-2})+x_{n-1}$ and this pattern holds for all $n \ge 2$. I also know that $x_0 = 1 \ and \ x_1 = 5.$
$x_2 = 2(x_1-x_0)+x_1$
$\begin{align}
x_3 = 2(x_2-x_1) + x_2 &= 2(2(x_1-x_0)+x_1)+x_2\\
&=2^2x_1-2^2x_0+2x_ 1+x_2
\end{align... | Here is another answer which uses a trick special to this problem. Note that the recurrence may be written as $x_n-x_{n-1}=2(x_{n-1}-x_{n-2})$ for $n\geq 2$. So the difference in terms is doubling at each step, and if we iterate this we come to $x_n-x_{n-1}=2^{n-1}(x_1-x_0)=2^{n+1}$ since $x_1-x_0=4$. So we can solve t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi }{2-x}\right)$ How to solve the following question?
If $n$ is an integer, show that
\begin{eqnarray} \left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+... | $$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$
$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right)$$
Apply de Movire's Theorem assuming $1+\sin x\ne0$
as $\sin x+1=0\implies 1+\sin x-i\cos x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$.
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to ... | Hint:
Let $u=x+2$ for
\begin{equation}
\int\frac{2}{x+2}dx
\end{equation}
Split the latter integral into two parts
\begin{equation}
\int\frac{1-2x}{x^2+1}dx=\int\frac{1}{x^2+1}dx-\int\frac{2x}{x^2+1}dx
\end{equation}
then let $x=\tan\theta$ also use identity $\sec^2\theta=1+\tan^2\theta$ for
\begin{equation}
\int\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\frac{y^2-x^2}{(y^2+x^2)^2}$ with respect to $y$? In dealing with the integration,
$$\int\frac{y^2-x^2}{(y^2+x^2)^2}dy$$
I have tried to transform it to polar form, which yields
$$\int\frac{\sin^2\theta-\cos^2\theta}{r^2}d(r\cos\theta)$$
But, what should I do now to continue?
I am sticking on it now.
| $$
\int\frac{y^2-x^2}{(y^2+x^2)^2}dy
$$
\begin{align}
y & = x\tan\theta \\
dy & = x\sec^2\theta\,d\theta \\
y^2+x^2 & = x^2(\tan^2\theta+1) = x^2\sec^2\theta \\
y^2-x^2 & = x^2(\tan^2\theta-1) \\
& \phantom{=}\text{etc.}
\end{align}
Usually with $\displaystyle\int \Big(\cdots\cdots\text{something involving }(y^2+\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of series with cosines I need to prove this:
$$
\sum_{n = 1}^{\infty}{8 \over \left(\,2n - 1\,\right)^{2}\pi^{2}}\,
\sin\left(\,\left[\,2n - 1\,\right]\,{\pi x \over 2}\,\right)
\sin\left(\,\left[\,2n - 1\,\right]\,{\pi z \over 2}\,\right) = \min\left\{x, z\right\}
$$
I got this:
$$ \frac{8}{\pi ^ 2} \sum\limits_{... | Big Hint:
Integrating the negative of
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(nx)}{n}
&=\mathrm{Im}\left(\sum_{k=1}^\infty\frac{e^{inx}}{n}\right)\\
&=-\mathrm{Im}\left(\log(1-e^{ix})\right)\\
&=\frac{x}{2|x|}(\pi-|x|)\tag{1}
\end{align}
$$
we get
$$
\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}=\frac{2\pi^2-6\pi|x|+3x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sums $\sum_{n=1}^{N}\sqrt{4n+1}$ I need to find sum of the first N terms of the sequence whose nth term is as follow :
T(n)= $\sqrt{4*n+1}$
So the sequence is : $\sqrt{5}$,$\sqrt{9}$,$\sqrt{13}$,$\sqrt{17}$,$\sqrt{21}$......
Please help how to find it for a given N.
| There is an answer which is not simple at all since $$\sum_{i=1}^n \sqrt{4i+1}=2 \left(\zeta \left(-\frac{1}{2},\frac{5}{4}\right)-\zeta
\left(-\frac{1}{2},n+\frac{5}{4}\right)\right)$$ in which appears Hurwitz $\zeta$ function.
However, you could notice that $$2\sqrt{i}<\sqrt{4i+1}<2\sqrt{i+1}$$ and so $$2H_{n}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to factor $(x^5+1) (x^5-1) $ I have this:
$ (x^5+1) (x^5-1) $
And I don't know how to continue factor.
Geogebra's Factor says:
$(x+1)(x-1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$
| Since $-1$ is a root of $x^5+1$, we know that it is divisible by $x+1$. Likewise, since $1$ is a root of $x^5-1$, we know that it is divisible by $x-1$. You can use polynomial long division to obtain the other factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$
Integrate $$I=\int\sqrt\frac{\sin(x-a)}{\sin(x+a)}dx$$
Let $$\begin{align}u^2=\frac{\sin(x-a)}{\sin(x+a)}\implies 2udu&=\frac{\sin(x+a)\cos(x-a)-\sin(x-a)\cos(x+a)}{\sin^2(x+a)}dx\\2udu&=\frac{\sin((x+a)-(x-a))}{\sin^2(x+a)}dx\\
2udu&=\frac{\sin(2a)}{\sin^2(x+a)}d... | Too long for a comment, perhaps the following approach would help.
Express the inner term of square root as follows
\begin{align}
\frac{\sin(x-a)}{\sin(x+a)}&=\frac{\sin(x)\cos(a)-\cos(x)\sin(a)}{\sin(x)\cos(a)+\cos(x)\sin(a)}\qquad\Rightarrow\qquad\text{divide by}\,\cos(x)\cos(a)\\
&=\frac{\tan(x)-\tan(a)}{\tan(x)+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 0
} |
Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$
Show this equation holds by squaring both sides and comparing terms up to $x^3$.
I wonder, how can I square the right hand side?
| You can put $t=\frac 12 x-\frac 18 x^2 + \frac 1{16}x^3$ then, you get:
$$\sqrt{\sqrt{1+x}}=\sqrt{1+t}=1+ \frac 12 t--\frac 18 [t^2]_3 + \frac 1{16}[t^3]_3 ... $$
where $[P(x)]_3$ for a polynom $$P(x)=a_0+a_1 x +a_2x^2+a_3x^3+a_4x^4 ... +a_n x^n $$ is:
$$[P(x)]_3=a_0+a_1x+a_2x^2+a_3x^3.$$
For example , since: $t=x(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Integral $\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$ Integrate:
$$
\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}
$$
| $$I=\int_0^\pi \frac{x\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\int_0^\pi \frac{(\pi-x)\,\operatorname dx}{a^2\cos^2x+b^2\sin^2x}$$
$$2I=\pi\int_0^\pi \frac{\operatorname dx}{a^2\cos^2x+b^2\sin^2x}=\pi\int_0^\pi \frac{\sec^2x\,\operatorname dx}{a^2+b^2\tan^2x}$$
$$\frac{2I}{\pi}=\int_0^{\pi}\frac{\operatorname d(\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the integral $ \int{\frac{x}{\sqrt{2x^2 + 3}}}dx $ I am trying to integrate the following:
$$
\int{\frac{x}{\sqrt{2x^2 + 3}}}dx
$$
It seems to me to be a trig substitution; however, I couldn't seem to get it into one of the three forms, i.e.,
$$\sqrt{a^2 - x^2}$$
$$\sqrt{x^2 - a^2}$$
$$\sqrt{a^2 + x^2}$$
I... | $$ \int \frac{x}{\sqrt{2x^2+3}}dx $$
Let $u=\sqrt{2x^2+3}$, then
$$ \frac{d}{dx}u=\frac{2x}{\sqrt{2x^2+3}}\Rightarrow \frac{1}{2}du=\frac{x}{\sqrt{2x^2+3}}dx $$
So now we have
$$ \frac{1}{2}\int du= \frac{1}{2}u+C = \frac{1}{2}\sqrt{2x^2+3}+C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/941435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there any integer solutions of $3x^3+3x+7=y^3$? $3x^3+3x+7=y^3$
$x, y \in \mathbb{N}$
Having thought about it two hours, and I'm still not sure how to show there actually aren't any integer solutions.
EDIT
Another formulation of this problem: Prove that $3x^3+3x+7$ cannot be a perfect cube.
| There are no natural solutions for $x$ and $y$.
A quick bit of important information: If $C_n$ is the $n$th non-negative cube ($0,1,8,...$), then $C_{n+1}-C_n = 3n^2+3n+1$. (This can be grasped visually by a similar method to the visual proof that the consecutive differences of consecutive squares are consecutive odd i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
Integrating $x^3\sqrt{ x^2+4 }$ Trying to integrate $\int x^3 \sqrt{x^2+4 }dx$, I did the following
$u = \sqrt{x^2+4 }$ , $du = \dfrac{x}{\sqrt{x^2+4}} dx$
$dv=x^3$ , $v=\frac{1}{4} x^4$
$\int udv=uv- \int vdu$
$= \frac{1}{4} x^4\sqrt{x^2+4 } - \int \frac{1}{4} x^4\dfrac{x}{\sqrt{x^2+4}} dx$ ---> i'm stuck her... | As a general rule, whenever dealing with integrands containing $\sqrt{x^2+a^2}$ in their expression, the
natural substitutions are $x=a\tan t$ or $x=a\sinh u$, since, on one hand, $\sqrt{\tan^2t+1}=\dfrac1{\cos t},$
and $\tan't=\dfrac1{\cos^2t},$ transforming our integral into $\displaystyle16\int\frac{\sin^3t}{\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
At what point does normal line intersect curve second time?
At what point does the normal line to $y=-5+4x+3x^2$ at $(1,2)$ intersect the parabola a second time?
$y'=6x+4$
$m_{tangent}=6(1)+4=10$
$m_{normal}=-\dfrac{1}{10}$
$y=f'(1)(x-1)+f(1)$
$=(-1/10)(x-1)+2$
$y=-1/10+21/10 \implies$ equation of normal line
Th... | Correcting some algebra, will look into the analytic geometry when I have time.
$$C_{norm}: y= \frac{21}{10}-\frac{x}{10}$$
Then you have $$\frac{21}{10}-\frac{x}{10}=-5+4x+3x^2 \iff -3x^2-\frac{41x}{10}+\frac{71}{10}=0 \iff -\frac{1}{10} ((x-1)(71+30x))=0 \iff (x-1)(71+30x)=0 \iff x=1 \; \mathrm{or} \; x=-\frac{71}{30... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to na... | Note that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$$
$$\frac{ab+ac+bc}{abc}=\frac{1}{a+b+c}$$
$$(ab+ac+bc)(a+b+c)-abc=0$$
$$(a+b)(b+c)(c+a)=0$$
then $a=-b$, $a=-c$ or $b=-c$.
If $a=-b$, how $n$ is odd, then
$$a^n + b^n + c^n=(-b)^n + b^n + c^n=c^n$$ and $$(a+b+c)^n=(-b+b+c)^n=c^n$$
you continue...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/943529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
differentiating a function of a function $w=\sqrt{u^2+v^2}$ I want to find the total differential of $w$ from a given function
$w=\sqrt{u^2+v^2}\:with\:u\:=\:cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)\:and\:v\:=\:sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\right)$
to solving this... | If we denote $$\phi := \ln \tan \left(p + \frac{\pi}{2}\right),$$ then
$$w = \sqrt{\cos^2 \phi + \sin^2 \phi} = 1,$$
and thus $$\frac{dw}{dp} = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $y = a\sin{x} + b\cos{x} +C$ then find maxima and minima for $y$. I was able to solve it till
$$y = \sqrt{(a^2 + b^2)}\sin(\alpha + x) + C.$$
But I don't know how to find maxima and minima from here.
If $C = 0$ then maxima & minima equals the amplitude of the sine curve but when $C$ is non-zero then?
I need help f... | Hint:
*
*$\sin(\alpha +x) \in [-1,1]$
$$ \therefore \quad \sqrt{a^2+b^2}\sin(\alpha+x) \in \left[ - \sqrt{a^2+b^2}, \sqrt{a^2+b^2} \right]$$
$$ \quad \therefore \sqrt{a^2+b^2}\sin(\alpha+x)+C \in \left[C- \sqrt{a^2+b^2}, C+ \sqrt{a^2+b^2}\ \right] \quad ,$$ therefore the maximum of $y$ is $_____$, and the minimum of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
laurent series of $1/(z-5)$ about $z_o=2$ I'm calculating the Laurent Series of $f=1/(z-5)$ about the point $z_{0}=2$.
The expansion I get has no principal part and is analytic within the circle $|z-2|<3$ since there are no singularities within that disk. I write $f=1/(z-5) = 1/(z-2 -3)= 1/3[(z-2)/3-1)]$, and since $(z... | See if this is it.
For $|z-2| < 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = -\frac{1}{3}\frac{1}{1-\frac{z-2}{3}} = -\frac{1}{3}\sum_{n=0}^{\infty} \frac{(z-2)^n}{3^n} = \sum_{n=0}^{\infty} \frac{(z-2)^n}{3^{n+1}}$$
For $|z-2| > 3$
$$\frac{1}{z-5} = \frac{1}{z-2 - 3} = \frac{1}{z-2}\frac{1}{1-\frac{3}{z-2}} = \frac{1}{z-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The maximum possible value of $ (xv - yu)^2 $ over the surface ... The maximum possible value of $ (xv - yu)^2 $ over the surface given by the equations
$ x^2 + y^2 = 4 $ and $ u^2 + v^2 = 9 $ is :
I solved it and my answer comes out to be $9$ but the correct answer is $36$.
Here is what I did :
Put $ x = r_1\cos\thet... | $x^2+y^2=r_1^2(\cos^2\theta+\sin^2\theta)=r_1^2$, so $r_1=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\int \frac {dx}{\sqrt{4x^{2}+1}}$ $\int \dfrac {dx}{\sqrt{4x^{2}+1}}$
I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:
$\dfrac {1}{2}\ln \left| 2\sq... | we have $\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx$.Now let $1+4x^{2}=u^2$, so we have $xdx=\frac{1}{4}udu$ and so $$\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx=\frac{1}{4}\int \dfrac {u}{u\frac{u^2-1}{4}}\,du=\int \dfrac {1}{u^2-1}\,du$$ also $$\int \dfrac {1}{u^2-1}\,du=\frac{1}{2}\int (\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show that $\lim_{n\rightarrow \infty } Var(Y_{n}) = 0$. Given $Var(Y_{n}) = (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2}$.
My work: $$lim_{n\rightarrow \infty } Var(Y_{n}) = \lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = \lim_{n\rightarrow \infty } \frac{(\theta_{2}-\theta_{1})^2}{(1 +... | Hint:
$\lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = (\theta_{2}-\theta_{1})^2\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2} $
However, $\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=0$ (Do you know why?).
ADDED part:
$\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=$\lim_{n\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is $1 \times 3 \times 5 \times \cdots \times (2k-3) = \frac{(2k-2)!}{2^{(k-1)}(k-1)!}$ In order to find out the Catalan numbers from their generating function you have to evaluate the product above.
Here is what I thought:
\begin{align*}
1 \times 3 \times 5 \times...\times (2k-3) &= \frac{1 \times 2 \times 3 \time... | For example, for the denominator $$2\times 4\times 6\times 8=2\cdot 1\times 2\cdot 2\times 2\cdot 3\times 2\cdot 4=2\times 2\times 2\times 2\cdot 1\times 2\times 3\times 4=2^4\cdot 4!$$
I let you adapt this hint to your exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Missing root of equation $\tan(2x)=2\sin(x)$ I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong?
$$
\tan(2x)=2\sin x
$$
$$
... | with $\tan(2x)=\frac{2\cos(x)\sin(x)}{\cos(x)^2-\sin(x)^2}$ we get
$2\sin(x)\left(\frac{\cos(x)}{\cos(x)^2-\sin(x)^2}-1\right)=0$
thus we obtain
$\sin(x)=0$
or
$0=2\cos(x)^2-\cos(x)-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int\frac{x^3}{\sqrt{81x^2-16}}dx$ using Trigonometric Substitution $$\int\frac{x^3}{\sqrt{81x^2-16}}dx$$
I started off doing $$u =9x$$ to get $$\int\frac{\frac{u}{9}^3}{\sqrt{u^2-16}}dx$$
which allows for the trig substitution of $$x = a\sec\theta$$
making the denomonator $$\sqrt{16\sec^2\theta-16}$$
$$\tan^... | I would prefer to use integration by parts as below:
$$
\begin{aligned}
\int \frac{x^3}{\sqrt{81 x^2-16}} d x=&\frac{1}{81} \int x^2 d \sqrt{81 x^2-16} \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{1}{6565} \int \sqrt{81 x^2-16} \,d\left(81 x^2-16\right) \\
= & \frac{x^2 \sqrt{81 x^2-16}}{81}-\frac{2}{19683}\left(81 x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
how to solve equation with cos I have this equation $\cos2x +5 \cos x + 3=0$. To solve it I rewrite $\cos2x$ to $2 \cos^{2} x- 1$ and set $\cos = t$.
I get the following equation $2t^2 - 1 +5t +3 = 0$ with that and then divide the equation with two $t^2 +\frac{5}{2} t +1 = 0$. I solve this equation and get two $t$, $t_... | It is not necessary to use substitutions when what you're working with isn't going to be too long:
$$
\cos 2x +5 \cos x + 3 = 0\\
\implies 2\cos^2 x - 1 + 5\cos x + 3 = 0\\
\implies 2\cos^2 x + 5\cos x + 2 = 0\\
\implies \cos^2 x + \frac{5}{2}\cos x + 1 = 0\\
\implies \cos^2 x + \left(2 + \frac{1}{2}\right)\cos x + 2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Compute $\sum_{n=0}^\infty \frac{(n+1)^2}{n!}$ This is what I've done so far:
$$\sum_{n=0}^\infty\frac{(n+1)^2}{n!} = 1 +4 +4.5 + \frac 83 + \frac{25}{24} +\frac{3}{10} + \frac {49}{720} +\dots$$
I know I need to manipulate $\frac{(n+1)^2}{n!}=\frac{(n+1)(n+1)}{n(n-1)(n-2)}=$
| Hint: $(n+1)^2=n^2+2n+1=n(n-1)+3n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/951107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $f(x)=-6\sin^4 x$
$f(x)=-6\sin^4x$
$f(x)=-6\sin x^4$
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$
What am I doing wrong? Please show the steps.
| $f(x)=-6\sin^4 x=-6(\sin x)^4$
By Chain Rule,
$f'(x)=4\cdot-6(\sin x)^3(\sin x)'=-24(\sin x)^3\cos x=-24\sin^3x\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\frac{ 1 }{ 1010 \times 2016} + \frac{ 1 }{ 1012 \times 2014} + \frac{ 1 }{ 1014 \times 2012} + \cdots + \frac{ 1 }{ 2016 \times 1010} = ?$ $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = ?
$$
My attempt so far :
$... | First, note that each term is duplicated $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = \\\dfrac{2}{ 1010 \times 2016}+ \dfrac{ 2 }{ 1012 \times 2014} + \dfrac{ 2 }{ 1014 \times 2012} + \cdots + \dfrac{ 2 }{ 1512 \times 15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
polynomial with positive integer coefficients divisible by 24? I have to show that $n^4+ 6n^3 + 11n^2+6n$ is divisible by 24 for every natural number, n, so I decided to show that this polynomial is divisible by 8 and 3, but I'm having trouble showing that it is divisible by 8.
Divisible by 3: $$n^4+ 6n^3 + 11n^2+6n \... | We have
$$
n^4+6n^3+11n^2+6n=n(n+1)(n+2)(n+3)=24\binom{n+3}4.
$$
This settles the claim when $n\ge1$. To get it for all $n$ observe that all integers are congruent to one $>1$ modulo $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solve $ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$ Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$
I first tried to use Eulers formula
$$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik} \right)$$
However to use the geometric formula here, I must subtract the $k=0$ term and that term is und... | The summation is as follows:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{1}{2i} \, \sum_{n=1}^{\infty} \frac{ e^{2ai n} - e^{- 2ai n}}{n} \\
&= - \frac{1}{2i} \left( \ln(1 - e^{2ai}) - \ln(1 - e^{-2ai}) \right) \\
&= - \frac{1}{2i} \ln\left( \frac{1 - e^{2ai}}{1 - e^{- 2ai}} \right) = - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
when ${\rm gcd} (a,b)=1$, what is ${\rm gcd} (a+b , a^2+b^2)$? I want to prove above statement
"what is ${\rm gcd} (a+b , a^2+b^2)$ when ${\rm gcd}(a,b) = 1$"
I've seen some proofs of it, but i couldn't find useful one.
here is one of the proof of it.
some proofs
I want prove it using method like 2nd answer. (user 9413... | Suppose $\gcd(a,b)=1$, that is, there are $x,y$ so that $ax+by=1$. Then
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(ax^3+3bx^2y)+b^2(3axy^2+by^3)
\end{align}
$$
Therefore,
$$
\begin{align}
\small2
&\small=\overbrace{(\color{#00A000}{(a^2+b^2)}+\color{#C00000}{(a+b)}(a-b))}^{\large 2a^2}(ax^3+3bx^2y)+\overbrace{(\color{#00A0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to show that $a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2} $ I want show the following $$a,\ b\in {\mathbb Q},\ a^2+b^2=1\Rightarrow a=\frac{s^2-t^2}{s^2+t^2},\ b= \frac{2st}{s^2+t^2},\ s,\ t\in{\mathbb Q} $$
How can we prove this ?
[Add] Someone implies that we mus... | Consider the line through $(-1,0)$ with slope $\frac{t}{s}$. Compute the other point on the circle that the line passes through.
So, points on the line are of the form $(p,q)=(-1+\alpha s,\alpha t)$ and the points on the circle have $p^2+q^2=1$, or $1-2\alpha s + \alpha^2(s^2+t^2)=1$ or $$\alpha\left(\alpha(s^2+t^2)-2s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{1... | Letting $u = \log(1+x)$,
$$ \begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \int_{0}^{\log 2} \frac{u^{3}}{e^{u}-1} e^{u} \ du \\ &= \int_{0}^{\log 2} \frac{u^{3}}{1-e^{-u}} \ du \\ &= \int_{0}^{\log 2} u^{3} \sum_{n=0}^{\infty} e^{-nu} \ du \\ &= \sum_{n=0}^{\infty} \int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/965504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 3
} |
How to find transformation matrix which converts matrix to simple standard form I have a matrix A$$ \left( \begin{array}{ccc}
0 & 1 \\
a^2 & 0\\
\end{array} \right) $$
Using eigen values, I convert it into simple standard form B:
$$\left( \begin{array}{ccc}
a & 0 \\
0 & -a\\
\end{array} \right) $$
How can I find the t... | If $a \ne 0$, the eigenvalues of $A$ are $\pm a$; this is tacitly given in the problem, by stipulating the diagonalized form of $A$, $B$, is diagonal with diagonal entries $a$, $-a$, but it is also easy to see since the characteristic polynomial of $A$ is
$\det (A - \lambda I) = (-\lambda)^2 - a^2 = \lambda^2 - a^2; \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/966257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplification of a series so that it converges to a given function I am trying to rearrange the series
$ \frac{1}{1-z} - \frac{(1-a)z}{(1-z)^2} + \frac{(1-a)^2z^2}{(1-z)^3} - \cdots$
In such a way that I can show it converges to
$\frac{1}{1-az} $
What I have so far
Let $ w = \frac{z}{1-z} $, we can then writ... | The mistake lies in this step:
$$\frac{w}{z}\left(1-(1-a)w+(1-a)^2w^2-\cdots\right)=\frac{w}{z}\cdot\frac{1}{1+(a-1)w}$$
Recall that $1+x+x^2+\cdots=\frac{1}{1-x}$. If we let $x=-(1-a)w$, then your series actually becomes:
$$\frac{w}{z}\left(1+x+x^2+\cdots\right)=\frac{w}{z}\cdot\frac{1}{1-x}=\frac{w}{z}\cdot\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/969471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine whether $A215928(n)=G_n$. Let $G_0=1$ and $G_{n+1}=F_0G_n+F_1G_{n-1}+\cdots+F_nG_0$, where $F_n$ is the $n$th term of the Fibonacci sequence, i.e., $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$.
Let $P_0=P_1=1,\ P_2=2,$ and $P_{n+1}=2P_n+P_{n-1}$ for $n>1$.
Is $P_n=G_n$?
Edit: set $P_2=2$, thanks to Daniel R.
| Let $G(z) = \sum\limits_{n=0}^\infty G_n z^n$
and $F(z) = \sum\limits_{n=0}^\infty F_n z^n$
where $F_n = \text{Fib}_{n+1}$ is the $(n+1)^{th}$ Fibonacci number.
In following set of recurrence relations,
$$G_{n+1} = F_0 G_n + F_1 G_{n-1} + \cdots + F_n G_0,$$
if one multiply the $n^{th}$ term by $z^n$ and sum over $n$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$. If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.
So far, I let
$d=(a^2+b^2,a+b)$
$\implies d|(a^2+b^2-(a+b)^2)$
$\implies d|(a^2+b^2-(a^2+2ab+b^2))$
$\implies d|(-2ab)$
I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Al... | Lemma 1: If $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.
Proof: We will prove the contrapositive. Suppose $\gcd(a^2,b^2)>1$. Then $a^2$ and $b^2$ must have a prime common divisor, say $p$. Since $p|a^2$, Euclid's lemma implies that $p|a$. Similarly, $p|b$. Thus $p$ is a common divisor of $a,b$ and so $\gcd(a,b)>1$.
Lemma 2: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solving the differential equation: $f(x)yy'=(y')^2-0.5$ I am trying to solve this equation:
$f(x)yy'=(y')^2-0.5$
I have already tried traditional methods...
Any ideas?
| We have
$$
\left(\frac{y'}{y}\right)^2-\frac{y'}{y}=\frac{1}{2y^2},
$$
or
$$
\left(\frac{y'}{y}-\frac{1}{2}\right)^2=\frac{1}{4}+\frac{1}{2y^2},
$$
or
$$
\frac{y'}{y}=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{1}{2y^2}}
$$
or
$$
y'=\frac{y}{2}\pm\sqrt{\frac{y^2}{4}+\frac{1}{2}}=\frac{1}{2}\big(y\pm\sqrt{y^2+2}\big).
$$
Usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Can someone walk me through how this expression simplifies to y/x? I am just wondering how this equation comes to be: it is from an economics problem involving marginal utilities. I have my two variables, $x$ and $y$.
Intuitively, how does $$\frac{0.5\times x^{-0.5}\times y^{0.5}}{0.5\times x^{0.5}\times y^{-0.5}}= \f... | Let me show you. It's just the "Multiply by one" trick:
$\frac{0.5 x^{-\frac{1}{2}} y^\frac{1}{2}}{0.5 x^\frac{1}{2} y^{-\frac{1}{2}}} \cdot 1 \cdot 1 = \frac{0.5 x^{-\frac{1}{2}} y^\frac{1}{2}}{0.5 x^\frac{1}{2} y^{-\frac{1}{2}}} \cdot \frac{y^\frac{1}{2}}{y^\frac{1}{2}} \cdot \frac{x^\frac{1}{2}}{x^\frac{1}{2}}$
You ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/974064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$?
How can this be justified?
Thanks.
| $$\sqrt{a^2+4}\sim a+\frac{1}{a}$$
as $a\to\infty$ if
$$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=1$$ But $$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=\lim_{a\to\infty}\frac{a\sqrt{a^2+4}}{a^2+1}=\lim_{a\to\infty}\frac{a^2\sqrt{1+4/a^2}}{a^2(1+1/a^2)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Show that $\, 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 $ How can I prove that, for $a,b \in \mathbb{Z}$ we have $$ 0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor - 2 \left \lfloor{\frac{a}{b}}\right \rfloor \leq 1 \, ? $$ Here, $\left \lfloor\,\right \rfloo... | $0 \leq \left \lfloor{\frac{2a}{b}}\right \rfloor \leq \frac{2*a}{b} < 2*(\left \lfloor{\frac{a}{b}}\right \rfloor + 1)$ = $2*\left \lfloor{\frac{a}{b}}\right \rfloor$ + 2
$\left \lfloor{\frac{2a}{b}}\right \rfloor$ is an integer, so being < than an other integer means being $\leq$ than this integer -1
=> $\left \lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/976665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Proof of ${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4} = -6$ Observe:
\begin{matrix} F(n)|&{F(n)}^{4}& - {4F(n+1)}^{4}& - {19F(n+2)}^{4}&- {4F(n+3)}^{4}&{F(n+4)}^{4}& = -6\\ 1|& 1& -4& -304& -324& 625&=-6\\ 1|& 1& -64& -1539& -2500& 4096&=-6\\ 2|& 16& -324& -11875& -16384& 28561&=-6\\ 3|& 81... | This is not the most elegant solution, but its a very straight forward computation using the standard tools we use to solve recurence-relations.
Let $a_n = F_n^4$ (where $F_n$ represents the terms in your recurence relation which we don't yet know is the Fibonacci sequence) then $$a_{n+4} - 4a_{n+3} - 19a_{n+2} - 4a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Help in finding $\lim_{n\to\infty}\Bigl( \sum_{k=1}^{n} \frac{1}{({n \atop k})} \Bigr)^n$. I am not able to get a solution for this problem . Of finding the limit
$$\lim_{n\to\infty} \left( \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } \right)^n$$
I have tried using Mathematica and that numerically evaluates it to $7.38905609... | $$\left(\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right )^n=e^{n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}}.$$
$$\binom{n}{1}=\binom{n}{n-1}=n,\binom{n}{2}=\binom{n}{n-2}=\frac{n(n-1)}{2},\binom{n}{2}\leq\binom{n}{k}, \ k=2,\cdots,n-2,$$
$$\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}=\frac{2}{n}+\sum_{k=2}^{n-2}\frac{1}{\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
| $$\begin{align}
&1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)\\
&=n\cdot 1+(n-2)\cdot 2+(n-3)\cdot 3+\cdots +1\cdot n\\
&=\sum_{r=1}^n(n+1-r)r\\
&=\sum_{r=1}^n {n+1-r\choose 1}{r\choose 1}\\
&={n+2\choose 1+2}\\
&={n+2\choose 3}\\
&=\frac16 n(n+1)(n+2)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
How to find the derivative of $f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$? Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:
$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$
would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$... | Something to notice is that $$x^4 - 6x^2 + 9 = (x^2 - 3)^2$$ and consequently, $$f(x) = 2 (x^3 - 4x + 6) \log (x^2 - 3).$$ This makes differentiation slightly easier: $$f'(x) = 2\left( (3x^2 - 4)\log(x^2 - 3) + \frac{x^3-4x+6}{x^2-3} \cdot 2x \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/983200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding maximum points by constrain optimization (multivariable calculus) Find the maximum value of the function $f(x,y)=x^2+y^2+2x+y$, on the closed disc (the circle together with the region inside the circle) of radius 2, centred at the origin.
What i tried
I know that i have to maximize the function
$f(x,y)=x^2+y^2+... | You can rewrite the equation you have as follows $$\begin{cases}2x(1+λ)=-2\\2y(1+λ)=-1\\y^2=2-x^2\end{cases}$$ or equivalently (in order to avoid calculations with roots) as $$\begin{cases}4x^2(1+λ)^2=4\\4y^2(1+λ)^2=1\\y^2=2-x^2\end{cases}$$ Now subsituting the 3rd equation into the 2nd yields $$1=4(2-x^2)(1+λ)^2=8(1+λ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$ Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
I am at a loss at how to start this problem and which contour to pick. I have been try... | This isn't quite complex analytic, but first denote your as $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$
Consider the double integral:
$$J=\int_{0}^{\infty} \int_{0}^{\infty} \frac{x}{(x^2+y^3)(1+x^2)}dydx.$$
We intend to evaluate $J$ and relate $J$ to $I.$
To evaluate, $J$ we integrate with respect to $y.$ You ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.
Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$.
| $$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 3\sqrt[3]{abc}\times 3\sqrt[3]{\frac{1}{abc}}=9.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Asymptotic expansion for Fresnel Integrals If you take the fresnel integrals to be $$S(x) = \int_{0}^{x}\sin \left(\frac { \pi \cdot t^2}{2} \right) dt$$ How do you find the asymptotic expansion? I know it begins with a $1/2$ but how?
| I suppose you want an asymptotic expansion as $x\to \infty$. We start with
$$S(x) = \int_0^x \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt = \frac{1}{2} - \int_x^\infty \sin \left(\frac{\pi\cdot t^2}{2}\right)\,dt.$$
Now, to get a handle on that integral, we substitute $u = \frac{\pi t^2}{2}$ and obtain
$$\int_x^\infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How find this value$\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^22^{3n+1}}$ show that:
$$\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\left(\frac{1}{2}\right)^{1/2}?$$
this sum is from other problem,if I solve this,then the other problem is solve it
| Hint: what is $\frac{f^{(n)}(0)}{n!}$ when $f(x) = \sqrt{1-x}$?
Compute the first terms:
\begin{align}
f^{(0)}(x) &= \sqrt{1-x}\\
f^{(1)}(x) &= -\frac 12 (1-x)^{-1/2} \\
f^{(2)}(x) &= -\frac 14 (1-x)^{-3/2} \\
\end{align}
The exponent of $1-x$ in $f^{(n)}(x)$ is obviously $1/2 - n$. So when you derivate,
the front fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the Sum of the Series: $1/(x+1) + 2/(x^2 + 1) + 4/(x^4 +1) +\cdots$ $n$ terms Find the sum of $n$ terms the following series:
$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$
$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$
But after that I'm not sure as to how to proceed
| Just another approach.
Let $x$ be a real number such that $|x|>1$. Observe that
$$
1+x^{2^{n}}=\frac{x^{2^{n+1}}-1}{x^{2^{n}}-1} \quad n=0,1,2,\ldots,
$$ giving
$$
\prod_{n=0}^{N} \left(1+x^{2^{n}}\right)=\frac{x^{2^{N+1}}-1}{x-1}
$$
Applying the logarithmic function gives
$$
\sum_{n=0}^{N}\log \left(1+x^{2^{n}}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Proving the determinant of a tridiagonal matrix with $-1, 2, -1$ on diagonal. Let $A_n$ denote an $n \times n$ tridiagonal matrix.
$$A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2$$
Set $D_n = \det(A_n)$
Pr... | Expand with respect to the first column: you get
$$
D_n
=2 \times
\begin{vmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{vmatrix}_{(n-1)}
- (-1) \times
\begin{vmatrix}-1 & 0 & & & 0 \\
-1 & 2 & -1 & & 0 \\
& \ddots & \ddots & \ddots & \\ & & -1 & 2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers I am trying to find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers:
a) $2$
b) $7$
c) $12$
d) $25$
and I am stuck on how to do this. I know that you have to set $n \geq 1$ for a) and sta... | $$1!+2!+\dots 100!\\ \equiv 1! +2!+3\times 2!+ 4\times 3\times 2!...\\ \equiv 1!+0+3\times 0+ 4\times 3\times 0...\equiv 1!\pmod 2$$
Because $2!$ is congruent to $0$ $\pmod 2$. As André Nicolas pointed out in the comments, $4!\equiv 0\pmod {12}$ which means that $5!$ which is $5\times 4!$ is congruent to $0$ as well.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$? How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$
for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$.
I do not know where to start. I need some idea and advice on this problem.Thanks
| the given inequality is equivalent to
$abc+ab+ac+bc\geq 4$ (I)from the given condition we get
$abc(a+b+c)=ab+ac+bc$ (1)
this gives in our inequality (I)
$abc+abc(a+b+c)\geq 4$
since $a+b+c\geq 3$ we have in the case $abc\geq 1$:
$abc+abc(a+b+c)\geq 1+3\cdot 1=4$
now our second case:
$abc\le 1$
then we can set $a=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Integration of $ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $ How can I integrate this by changing variable?
$$ \int \frac{2\sin x + \cos x}{\sin x + 2\cos x} dx $$
Thanks.
| Let $u(x)=\log(\sin x + 2\cos x)$, then $(\sin x + 2\cos x)'=\cos x-2\sin x$ hence $$u'(x)=\frac{\cos x-2\sin x}{\sin x + 2\cos x}.$$ More generally, for every $(a,b)$, $$(au(x)+bx)'=\frac{(a+2b)\cos x+(b-2a)\sin x}{\sin x + 2\cos x}$$ Solving for $(a+2b,b-2a)=(1,2)$ yields $(a,b)=(-\frac35,\frac45)$ hence $$ \int \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Possible values of infinitely nested square root $n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$ If $$n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$$
Is it possible that $n$ is a integer for any $x=Z( \text{zahlen number})$.If yes .What is the value of $x$??
| So, $n^2=x+n\iff n^2-x-n=0\implies n=\dfrac{1\pm\sqrt{1+4x}}2$
As $n\ge0,n=\dfrac{1+\sqrt{1+4x}}2$ So, we need $1+4x$ to be Perfect Square
As $1+4x$ is odd, $1+4x=(2m+1)^2\iff x=m^2+m$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer:
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?
| Do it in order
$$\frac{\frac{d}{dx}(1+4\cos x)2\sqrt{x+4\sin x} -\frac{d}{dx}(2\sqrt{x+4\sin x})(1+4\cos x)}{4(x+4\sin x)}.$$
Solve the derivatives one by one and then replace in the big fraction. Remember to use the chain rule!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}dx$ Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$
| Put $y=\tan^{-1}x$. Then $\tan y=x$. So
$$1 = \frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( {\tan y} \right) = \frac{d}
{{dy}}\left( {\tan y} \right).\frac{d}{{dx}}\left( y \right) = \frac{1}{{{{\cos }^2}y}}.\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right),$$
which gives
$$\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A geometry problem involving geometric mean $ABCD$ is a quadrilateral inscribed in a circle of center $O$. Let $BD$ bisect $OC$ perpendicularly. $P$ is a point on the diagonal $AC$ such that $PC=OC$. $BP$ cuts $AD$ at $E$ and the circle circumscribing $ABCD$ at $F$.
Prove that $PF$ is the geometric mean of $EF$ and $B... | This is a coordinate-based solution. Without loss of generality, you can choose your coordinate system such that the circle $\bigcirc ABCD$ becomes the unit circle and $C$ has coordinates $(1,0)$. Then, in order to simplify the formulation, we start by choosing rational coordinates for point $P$:
$$P = (2t^2,2t)/(t^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.