Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the derivative using the definition? Calculate the derivate of the given function directly from the definition of derivative, and express the result using differentials
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
when $f(x)= 1/\sqrt{1+x^2}$
any tips/solutions on how to get started on this one? I am able to do more ... | $$\eqalign{
\lim_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\lim_{h \to 0} \frac{\frac{1}{\sqrt{1+\left(x+h\right)^2}}-\frac{1}{\sqrt{1+x^2}}}{h}\\
&=\lim_{h \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+\left(x+h\right)^2}}{h \sqrt{1+x^2} \sqrt{1+\left(x+h\right)^2}}\\
&=\lim_{h \to 0} \frac{\left(\sqrt{1+x^2}-\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer giv... | First note:
$$\frac{z^2 + z + 1}{z^2 - z + 1} = 1 + \frac{2z}{z^2 - z + 1}$$
So the given fraction is real if and only if the fraction $\frac{z}{z^2 - z + 1}$ is real. But a fraction is real if and only if its reciprocal is, so we need:
$$\frac{z^2 - z + 1}{z} = z - 1 + z^{-1}$$
To be a real number. So we get:
$$\boxed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$ Question:
$$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$
Prove that L.H.S.=R.H.S.
My Efforts:
L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\f... | $$\tan^2\theta(\csc^2\theta-1)=(1-\cos^2\theta)\csc^2\theta,$$
$$\frac{\sin^2\theta}{\cos^2\theta}(\frac1{\sin^2\theta}-1)=(1-\cos^2\theta)\frac1{\sin^2\theta},$$
$$\frac{\sin^2\theta}{\cos^2\theta}\frac{\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Differential equation: $y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} $ $$y + t\frac{dy}{dt} = \frac{y}{y^2 - 1} \Rightarrow t\frac{dy}{dt} = \frac{-y^3 + 2y}{y^2 - 1} \Rightarrow \frac{1}{t} dt = \frac{y^2 - 1}{-y^3 + 2y}dy \Rightarrow$$
$$\Rightarrow \ln |t| = - \int \frac{y^2 - 1}{y^3 - 2y}dy$$
How to solve the integral in... | HINT : We have
$$\frac{y^2-1}{y^3-2y}=\frac 12\left(\frac{1}{y}+\frac{y}{y^2-2}\right)=\frac{1}{2y}+\frac{(y^2-2)'}{4(y^2-2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Sum of cubes of roots of a quartic equation $x^4 - 5x^2 + 2x -1= 0$
What is the sum of cube of the roots of equation other than using substitution method? Is there any formula to find the sum of square of roots, sum of cube of roots, and sum of fourth power of roots for quartic equation?
| This is precisely what Newton's Sums are for.
Let $S_n$ be the sum of the $n$-th powers of the roots of $a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0$.
Then, Newton's Sums tells us that $\displaystyle\sum_{k = 0}^{m-1}\left(a_{n-j}S_{m-j}\right)+ma_{n-m} = 0$ for each positive integer $m$.
We can write out the first f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\fr... | $$
\begin{align}
\frac{25}{84} &= \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} \\
&= \frac{bc+c+1}{abc} = \frac{(b+1)c+1}{abc}
\end{align}
$$
Now $ 84 = 2\cdot 2 \cdot 3 \cdot 7$ so $b,c$. Let us try $(b+1)c= c = 24 = 2\cdot 2\cdot 2\cdot 3$ and $abc = 84$ (this is just a guess, it can also be a multiple each time). We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$
My first approach was:
Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sq... | If you replace $x=\sqrt{2} \cosh t$, you end with:
$$\begin{eqnarray*} I = \frac{1}{\sqrt{2}}\int \frac{dt}{\cosh t}=\sqrt{2}\arctan\left(\tanh\frac{t}{2}\right)&=&\sqrt{2}\arctan\left(\frac{\sqrt{x^2-2}}{x+\sqrt{2}}\right)\\&=&\frac{1}{\sqrt{2}}\arctan\sqrt{\frac{x^2-2}{2}}.\end{eqnarray*}$$
Your solution is right, ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to simplify the formula for $n$th Fibonacci number when $n=2$? When n is equal to 2 how do I simplify when the $n=2$ is put into the equation below
(by the way I have to prove this formula by induction that when n= any number it will equal that number)
$$F_n=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^n − \f... | In the first line, $(1+\sqrt 5)^2=1+2\sqrt 5 + 5=6+2\sqrt 5$ Similarly $(1-\sqrt 5)^2=6-2\sqrt 5$. Also the $2$'s in the denominators get squared. It should be $$\begin{align}F_2&=\frac 1{\sqrt 5} \left(\frac{1+\sqrt5}{2}\right)^2 − \frac1{\sqrt 5} \left(\frac{1-\sqrt5}{2}\right)^2\\
&=\frac{6+2\sqrt 5}{4\sqrt 5} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Co... | $$y'=\text{Ax}^2+\text{Bx}+c$$
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d \ \ \ (*) $$
Using the initial condition $y(1)=1$ we get:
$$y(1)=1 \Rightarrow \frac{A }{3}+\frac{B }{2}+c +d=1 \Rightarrow d=1-\frac{A}{3}-\frac{B}{2}-c$$
Replacing this at the relation $(*)$ we get:
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Complex Equations The Equation:
$$ z^{4} -2 z^{3} + 12z^{2} -14z + 35 = 0 $$
has a root with a real part 1, solve the equation.
When it says a real part of 1, does this mean that we could use (z-1) and use polynomial division to extract the other rots? Hence:
$$ (z^{4} -2 z^{3} + 12z^{2} -14z + 35 ) / (z-1) $$
But i ... | If there's a root of that equation of the form $1+ib$ then:
$$(1+ib)^4 - 2(1+ib)^3 -12(1+ib)^2-14(1+ib) + 35 = 0\Longrightarrow \\(1 + 4ib -6b^2 -4ib^3 + b^4) - 2(1 + 3ib - 3b^2 -ib^3) + 12(1 + 2ib - b^2) - 14(1+ib) + 35 =0 \Longrightarrow b^4-12b^2+32 + i(-2b^3+8b) = 0$$
Therefore:
$$\begin{cases}b^4 - 12b^2+32 = 0\\-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ellipse like on sphere Find the locus of all points on a sphere such that the sum of geodesic distances from two fixed points $F_1$ and $F_2$ on it is a constant, less than its diameter. ( When radius of sphere goes to infinity, it would look like an ellipse).
Following is image after Equn ($1$) in achille hui note
| WOLOG, consider the case $\begin{cases}
F_1 &= (+\sin\alpha,0,\cos\alpha),\\
F_2 &= (-\sin\alpha,0,\cos\alpha)
\end{cases}$ with $\alpha \in (0,\frac{\pi}{2})$ and the sphere is the unit sphere.
Let $2\theta \ge 2\alpha$ be that constant for the sum of geodesic distances to $F_1$
and $F_2$. For any point $p = (x,y,z)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{... | $$
\int \ln(z+\sqrt{z^2-1})dz = \int \mathrm{arcosh}(z) dz\,\,\,z>1.
$$
the equality holds because for real x, $z = \sqrt{x^2+1}>1$.
then using the fact
$$
\int \mathrm{arcosh}(z) dz = z\mathrm{arcosh}(z) +\sqrt{z^2-1} + C
$$
then sub back in your substitution.
$\textbf{brief derivation of the arcosh relation}$
$$
y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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To test the convergence of series 2 To test the convergence of series 2
$\displaystyle \frac{a+x}{1!}+\frac{(a+2x)^2}{2!}\frac{(a+3x)^3}{3!}+...\infty$
My Attempt:
$$\begin{align}
\frac{u_n}{u_{n+1}} & = \frac{(a+nx)^n}{n!}\frac{(n+1)!}{(a+(n+1)x)^{n+1}} \\
& = \frac{(a+nx)^n}{1}\frac{n+1}{(a+nx+x)^{n+1}} \\
& = \f... | Follow Did's advice, you should get something like
$$
a^{\frac{1}{n}}_n = \frac{a+nx}{(n!)^{\frac{1}{n}}} \sim \frac{e(a+nx)}{n} (2 \pi n)^{-\frac{1}{2n}}
$$
The last term converges to 1, so the series converges for $x<e^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdot... | \begin{array}{l}
s_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 3} }}{{3k + 3}}} \Rightarrow s'_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 2} } = \frac{{x^2 }}{{1 - x^3 }} \\
s_1 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 4} }}{{3k + 4}}} \Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Calculating multi-variable limit. I am struggling to find a way to approach this limit
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y+x^2y^3)}{x^2+y^2}$$
I would greatly appriciate if You could explain to me how to solve it or at least show how to start.
| By the inequality $|\sin(\theta)| \leq |\theta|$, which is valid for all real $\theta$, we have
$$\left|\frac{\sin(x^2y + x^2 y^3)}{x^2 + y^2}\right| \leq \left|\frac{x^2y + x^2 y^3}{x^2 + y^2}\right|$$
For $x \neq 0$, the right hand side is equal to
$$\left|\frac{y + y^3}{1 + y^2/x^2}\right| \leq \left|y + y^3\right| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\... | You have two planes that go through the same point (4,1,1). Now all you need is having them parallel. You can check this by forming the cross product of the defining vectors, i.e. (0,-1,2) and (0,-3-1) for the first plane, that will give you a normal to the plane. Do the same for the second plane; if your normal vector... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to prove this $\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$ let $n<\dfrac{\pi}{2\arccos{\dfrac{c}{2}}},c\in (0,2),c=2\cos{x}$, show that
$$\dfrac{\sin{(nx)}}{\sin{x}}\ge\dfrac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$
where $0<x<\dfrac{\pi}{2}$
My idea: let $$a_{n}=\dfrac{\sin{(nx)}}{\sin{x}}$$
the... | This inequality $$\frac{\sin{(nx)}}{\sin{x}}\ge\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ with $0<x<\dfrac{\pi}{2}$ and $n>1$ makes a serious problem to me.
If we consider $$f_n(x)=\frac{\sin{(nx)}}{\sin{x}}-\frac{\sqrt{3}}{3}(2n-1)^{\frac{3}{4}}$$ we have $$f_n(0)=n-\frac{(2 n-1)^{3/4}}{\sqrt{3}} > 0$$ but it exists $0<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.
| HINT:
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2 + 2)(3x^2 + 2x + 3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfra... | Related techniques: (I). Here is an approach which enables you to tackle your problems. Let's consider the series
$$ f(x) = \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}. $$
Taking the Laplace transform gives
$$ F(s) = \sum_{n=0}^{\infty}\frac{1}{s^{3n+1}} = \frac{s^2}{s^3-1}. $$
To finish the problem you need to find th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Addition of fractions repetition and convergence Is this a new mathematical concept?
$$
\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} \cdots = \frac{1}{n-1}
$$
If it isn't then what is this called?
I haven't been able to find anything like this anywhere.
| $$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\dots=\sum_{i=1}^{\infty} \frac{1}{n^i}=\sum_{i=1}^{\infty} \left (\frac{1}{n} \right )^i=\sum_{i=0}^{\infty} \left (\frac{1}{n} \right )^i-1=\frac{1}{1-\frac{1}{n}}-1=\frac{n}{n-1}-1=\frac{n-n+1}{n-1}=\frac{1}{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital I stumbled across this silly limit and I am perplexed at how I can arrive to a solution by only relying on the simplest rules of limits.
$$
\lim_{x \to 0}\frac{4-\sqrt{16-7\sin(x)}}{8x}
$$
Any help is appreciated, thanks in advance.
| Multiply the numerator and denominator by $4+\sqrt{16-7\sin x}$ to get
\begin{align}
& \lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \\[8pt]
= {} &\lim_{x \to 0}\dfrac{4-\sqrt{16-7\sin x}}{8x} \cdot \dfrac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}} \\[8pt]
= {} & \lim_{x \to 0}\dfrac{4^2 - (16-7\sin x)}{8x(4+\sqrt{16-7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Ad hoc proof of the convergence of $\sum\limits_n\sin\left(\pi\sqrt{n^2+a^2}\right)$ I use the following method to prove that
$$\sum_{n=0}^\infty \sin(\pi\sqrt{n^2+a^2})$$
converges for any $a \in \mathbb{R}$.
First, we can see that
$$\lim_{n\to\infty} \frac{\sqrt{n^2 + a^2}}{n} = 1,$$
which means precisely that $\fora... | Note that $$\sqrt{n^2+a^2}=n\cdot\sqrt{1+\frac{a^2}{n^2}}=n\cdot\left(1+\frac{a^2}{2n^2}+O\left(\frac1{n^4}\right)\right)=n+u_n,$$
where $$u_n=\frac{a^2}{2n}+O\left(\frac1{n^3}\right).$$ Note also that, when $x\to0$, $$\sin x=x+O(x^3),$$ hence, using this for $x=\pi u_n$, one gets $$\sin\left(\pi\sqrt{n^2+a^2}\right)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How find the least value of the expression: $M = \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A - \cot B)(\cot B - \cot C)(\cot C - \cot A)$? Consider all triangles $ABC$ where $A < B < C \leq \frac{\pi}{2}$. How find the least value of the expression:
$M = \cot^2 A + \cot^2 B + \cot^2 C + 2(\cot A - \cot B)(\cot B - \cot C... | consider $(\cot B - \cot C)(\cot C - \cot A)=-(\cot C-\cot B )(\cot C - \cot A)$
for easy, $x=\cot C,a=\cot A,b=\cot B,ab=-(a+b)x+1 ,f(x)=x^2-(a+b)x+ab=x^2-2(a+b)x+1$ to have max
$x< (a+b) \implies f_{max}=f(0)=1$
$\cot^2 C \ge 0$ so $M$ will get least value when $\cot C=0$
rest is easy.
indeed, without the limitation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/918996",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integrating $\int\frac{\sqrt{16x^2-9}}x\,\mathrm{d}x$? I am trying to differentiate from my previous question, but I am having trouble in the finishing steps. I have the integral $\int\dfrac{\sqrt{16x^2-9}}x\,\mathrm{d}x$.
$$v=4x\implies\mathrm{d}v=4\,\mathrm{d}x$$
$$\int\frac{\sqrt{v^2-9}}{v}\,\mathrm{d}v\implies a=3,... | I would rather like to try to tackle it with rationalization other than substitution. $$
\begin{aligned}
\int \frac{\sqrt{16 x^{2}-9}}{x} d x &=\int \frac{16 x^{2}-9}{x \sqrt{16 x^{2}-9}} d x \\
&=\frac{1}{16} \int \frac{16 x^{2}-9}{x^{2}} d\left(16 x^{2}-9\right) \\
&=\frac{1}{16} \int\left(16-\frac{9}{x^{2}}\right) d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$ Evaluate the following the limit:
$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$
I tried expressing the limit in the form $f(x)g(x)\left[\frac{1}{f(x)} - \frac{1}{g(x)}\right]$ but it did not help... | Write $p(x) = (x-a_{1})(x-a_{2})\cdots (x-a_{n})$. Now
$$
\lim_{x\to\infty}[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}] = \lim_{x \to \infty}[x-\sqrt[n]{p(x)}] \\
= \lim_{x \to \infty}\left((x-\sqrt[n]{p(x)})\frac{x^{n-1}+x^{n-2}\sqrt[n]{p(x)}+\ldots+x(\sqrt[n]{p(x)})^{n-2}+(\sqrt[n]{p(x)})^{n-1}}{x^{n-1}+x^{n-2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the maximum the value $P_{1}\cdot P_{n}$
Suppose that $n \ge 2$ students attend a test of $m \ge 2$ problems. The scoring rule for each problem is: if $x$ students answer a problem incorrectly, then a correct answer is worth $x$ points and an incorrect answer is worth none. The total score of a student is the su... | Let $x_i$ be the score of the $i$-th problem, then
\begin{equation}
\begin{split}
P_1 &\le \sum_{i=1}^{m} x_i \\
P_n &\le \frac{1}{n-1} \Big( \sum_{i=1}^m x_i (n-x_i) - P_1 \Big)
\end{split}
\end{equation}
denote $S = \sum_{i=1}^{m} x_i, T = \sum_{i=1}^m x_i (n-x_i)$, $P_1 P_n \le \frac{1}{n-1} P_1 ( T - P_1 )$.
In ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Find the greatest value of this expression Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 3$. Find the greatest value of the expression
\begin{equation}
P = \sqrt{(x+1)(y^2 + 2)(z^3 + 3)} + \sqrt{(y+1)(z^2 + 2)(x^3 + 3)} + \sqrt{(z+1)(x^2 + 2)(y^3 + 3)}.
\end{equation}
I tried. We have
$$\sqrt{(x+1)(... |
Hint: Above is plot using Mathematica. Looks like maximum is for $(3,0,0),(0,3,0),(0,0,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have the same limit I'm trying to solve the following problem
prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit. In this post https://math.stackexchange.com/a/267499, I do not understand the following steps:
$$\frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} )... | Look at this chain of inequalities in the post you cited:
$$b_{n+1} - a_{n+1} = \frac {1}{2} (\sqrt{b_{n}} -\sqrt{ a_{n}})^2 \leq \frac {1}{2} (\sqrt{b_n} - \sqrt{a_n} ) ( \sqrt{b_n} + \sqrt{a_n} ) = \frac {1}{2} ( b_n - a_n)$$
So for every $n$, we have
$$b_{n+1} - a_{n+1} \leq \frac{1}{2}(b_n - a_n)$$
Therefore,
$$b_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Orbit and stabaliser of $2\times2 * 2\times1$ matrices I have the group action of matrix multiplication, meaning:
$g((x,y))=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ \begin{pmatrix}x\\ y\end{pmatrix}$
$G=\left\{\begin{pmatrix}a&0\\0&b\end{pmatrix}:a,b\in\mathbb{R} - \{0\} \right\} g=\begin{pmatrix}a&0\\0&b\end{pmatrix}\in... | If you write out the group action as $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax \\ by \end{pmatrix}$$
For the stabilizer, we look for elements $g \in G$ such that $$g * \begin{pmatrix} x \\ y \end{pmatrix} = \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$ Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$
$ $
This appears to be an easy problem, but it is consuming a lot of time, I am wondering if an easy way is possible.
WHAT I DID :
Wrote this as $\int \dfrac{1}{4}(\sin^22x)\cos^2x$
And then I wrote $\cos^2x$ in terms of $\c... | $$ 2\sin x \cos x = \sin 2x$$
$$ 2\cos^2 x -1 = \cos 2x \iff \cos 2x = \frac{ \cos 2x+1}{2}$$
$$\sin^2 x \cos^4 x =\sin^2 x \cos^2 x cos^2 x= 1/8 \sin^2 2x ( \cos 2x+1) = 1/8 (\sin^22x\cos2x+\sin^2 2x)$$
$$\int \sin^2 x \cos^4 x dx = 1/8 (\int \sin^22x\cos2xdx+\int \sin^22x dx)$$
denotation:
$$ A = \int \sin^22x\cos2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Taking limits of powers containing 'x'? If we have $lim_{x\rightarrow 0}\sqrt{x^2+sinx-tanx}$ we can write this as
$\sqrt{lim_{x\rightarrow \:0}\left(x^2+sinx-tanx\right)}$
OR If we have $lim_{x\rightarrow \:0}\left(x^3+secx-cosecx\right)^n$. We can write this as
$\left(lim_{x\rightarrow \:\:0}\left(x^3+secx-cosecx\rig... | You can't simply take a product $1/x^2$ times when $1/x^2$ may not be an integer. Regardless, your computation leads to $1^{\infty}$, which is an indeterminate form: it can take on any finite value at least $1$, or $\infty,$ which is why you have to compute the limit in a different way.
In both your introductory exampl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A Definite Integral I Given the definite integral
\begin{align}
\int_{0}^{\pi} \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt = -2\pi
\end{align}
then what is the general value of the indefinite integral
\begin{align}
\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left... | $$\begin{align}
\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt
&= \int\sec^2t\ln\left(1+\sin^2t\right) dt\\
&=\ln(1+\sin^2t)\int\sec^2tdt-\int\frac{2\sin t\cos t}{1+\sin^2t}\int\sec^2tdtdt\\
&=\ln(1+\sin^2t)\tan t-\int\frac{2\sin t\cos t\tan t}{1+\sin^2t}dt\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/934593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer. How do we prove that $\dfrac{\phi^{400}+1}{\phi^{200}}$ is an integer, where $\phi$ is the golden ratio?
This appeared in an answer to a question I asked previously, but I do not see how to prove this..
| We can prove by induction that
if $x+\dfrac1x$ is an integer, $x^n+\dfrac1{x^n}$ will be an integer
as $$\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)=x^{n+1}+\frac1{x^{n+1}}+x^{n-1}+\frac1{x^{n-1}}$$
$$\iff x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)-\left(x^{n-1}+\frac1{x^{n-1}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 2
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How to find these quantities so as to conform to these conditions? Suppose $a \in \mathbb{R}^k$, $b \in \mathbb{R}^k$. Then how to find $c \in \mathbb{R}^k$ and $r > 0$ such that the following holds?
For any $x \in \mathbb{R}^k$, we have $$|x-a| = 2 |x-b|$$ if and only if $$|x-c| = r. $$
| Squaring both sides of the original equation and rewriting both sides as dot products gives
$$(\mathbf{x} - \mathbf{a}) \cdot (\mathbf{x} - \mathbf{a}) = 4 (\mathbf{x} - \mathbf{b}) \cdot (\mathbf{x} - \mathbf{b}).$$ Expanding and rearranging gives
$$3 \mathbf{x} \cdot \mathbf{x} + 2(\mathbf{a} - 4\mathbf{b}) \cdot \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac... | It seems that you already realize that the for $D$, $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x\}$ is the same as $D = \{x^3 \leq y \leq 1\} \cap \{-1 \leq x \leq 1\}$. So, for a function $f(x,y)$, you should have
$$
\iint_D f(x,y) \, dxdy = \int_{-1}^1 \int_{x^3}^1 f(x,y)\,dxdy.
$$
However, you seem to have split up... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/937955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.
| The function $f(x) = \dfrac{6}{x-1}$ is a fractional linear transformation.
Composition of these corresponds to matrix multiplication:
$$\text{if}\ T(x) = \dfrac{a_{11} x + a_{12 }}{a_{21} x + a_{22}} \ \text{and} \ S(x) = \dfrac{b_{11} x + b_{12}}{b_{21} x + b_{22}}$$
$$\text{then}\ T(S(x)) = \dfrac{c_{11} x + c_{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Integral of $\sqrt{x^3 + 8}$? I have issues solving the following integral:
$$\int\sqrt{x^3+8}~dx$$
I tried substitution and integration by parts, but with no use.
I'm guessing I have to use some trigonometric substitution.
Can anybody help solve this integral?
| For any real number $x$:
When $|x|\leq2$ ,
$$\begin{array}\int\sqrt{x^3+8}\,dx
&=\int2\sqrt2\sqrt{\dfrac{x^3}{8}+1}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{8^n4^n(n!)^2(1-2n)}\,dx\\
&=\int\sum\limits_{n=0}^\infty\dfrac{2\sqrt2(-1)^n(2n)!x^{3n}}{32^n(n!)^2(1-2n)}\,dx\\
&=\sum\limits_{n=0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Calculus 2 Integral of$ \frac{1}{\sqrt{x+1} +\sqrt x}$ How would you find $$\int\frac{1}{\sqrt{x+1} + \sqrt x} dx$$
I used $u$-substitution and got this far:
$u = \sqrt{x+1}$ which means $(u^2)-1 = x$
$du = 1/(2\sqrt{x-1}) dx = 1/2u dx$ which means $dx = 2udu$
That means the new integral is $$\int \frac{2u}{u + \sqrt{... | Here is a route.
$$
\begin{align}
\int\frac{1}{\sqrt{x+1} + \sqrt{x}} {\rm d}x &=\int\frac{1}{(\sqrt{x+1} + \sqrt{x})} \times \frac{(\sqrt{x+1} - \sqrt{x})}{(\sqrt{x+1} - \sqrt{x})} {\rm d}x\\\\
&=\int \left(\sqrt{x+1} - \sqrt{x}\right) {\rm d}x \\\\
& = \int(x+1)^{1/2} {\rm d}x-\int x^{1/2} {\rm d}x\\\\
& =\frac{1}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Not understanding how to factor a polynomial completely $$P(x)=16x^4-81$$
I know that this factors out as:
$$P(x)=16(x-\frac { 3 }{ 2 } )^4$$
What I don't understand is the four different zeros of the polynomial...I see one zero which is $\frac { 3 }{ 2 }$ but not the three others.
| Try making the substitutions $a=2x$ and $b=3$. Then we have that
$$
16x^4-81=a^4-b^4.
$$
Recalling that $x^2-y^2=(x+y)(x-y)$, we have
$$
a^4-b^4=[a^2-b^2](a^2+b^2)=[(a-b)(a+b)](a^2+b^2)
$$
Plugging $2x$ in for $a$ and $3$ in for $b$ we have
$$
(a-b)(a+b)(a^2+b^2)=(2x-3)(2x+3)((2x)^2+3^2)=(2x-3)(2x+3)(4x^2+9).
$$
If yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix}
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and ... | Using the Law of Sines, we can write
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $a$, $b$, $c$ are the sides, and $d$ is the circumdiameter, of the triangle. And the Law of Cosines gives us
$$\cos A = \frac{1}{2bc}(-a^2+b^2+c^2) \qquad\text{, etc.}$$
With $\sin 2 x = 2 \sin x \cos x$, we can ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 3
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Computing a limit involving Gammaharmonic series It's a well-known fact that
$$\lim_{n\to\infty} (H_n-\log(n))=\gamma.$$
If I use that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder if it's possible to compute the following limit in a closed-form
$$\lim_{n\t... | From Wolfram Gamma Function equations (35)-(37) provide
\begin{align}\tag{1}
\frac{1}{\Gamma(x)} = x + \gamma x^{2} + \sum_{k=3}^{\infty} a_{k} x^{k}
\end{align}
where, $a_{1}=1$, $a_{2}=\gamma$,
\begin{align}\tag{2}
a_{n} = n a_{1} a_{n-1} - a_{2} a_{n-2} + \sum_{k=2}^{n} (-1)^{k} \zeta(k) \, a_{n-k}.
\end{align}
Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
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$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$
I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.
| Using a trigonometric substitution, let $x+1 = 2\tan u$, so that $dx = 2\sec^2 u\,du$ and $2x+1 = 4\tan u-1$. Then we get
\begin{align*}
\int\frac{2x+1}{(x+1)^2+4}\,dx &= \int\frac{4\tan u-1}{4\tan^2 u+4}\cdot 2\sec^2u\,du \\
&= \frac{1}{4}\int\frac{4\tan u-1}{\sec^2u}\cdot 2\sec^2u\,du \\
&= \frac{1}{2}\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Straight line , gre prolem $170$ score hard What is the distance between the two straight line represented by the equation
$3x+4y=10$ and $6x+8y=10?$ $$A>1$$ $$B>2$$ $$C>\frac43$$ $$D>\frac12$$ $$E>\frac52$$
I try to solve it , firstly find intercept of both line and then find mid point between of them ,and calculate d... | Let me call the lines $l1$ and $l2$, in which:
$l1: y=\frac{-3}{4}x+2.5$
and
$l2: y=\frac{-3}{4}x+1.25$.
With a little notice to the above lines we see that these are the lines mentioned in the question. In another language, we have made the above line equations by altering the shape of the line equations. As we see, ... | {
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"url": "https://math.stackexchange.com/questions/954535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 +... | A quick hint is noticing the symmetry. A rigorous proof is that $$\int\limits_a^b {\frac{{f(\cos x)}}{{g(\sin x\cos x)}}dx} = \int\limits_a^b {\frac{{f\left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}{{g\left( {\cos \left( {\frac{\pi }{2} - x} \right)\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}dx} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$ I want to find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$.
For this, suppose $\pmatrix{e & f\\ g& h}$ such that
$\pmatrix{a & b\\ c & d}\pmatrix{e & f\\ g& h}=\p... | First we have to find out which matrices commute with a Jordan block $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Since
$$ \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}x&y\\z&w\end{pmatrix} = \begin{pmatrix}z&w\\0&0\end{pmatrix}, \begin{pmatrix}x&y\\z&w\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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The relation of angle between two slant faces of a pyramid and the angles between slant vectors Have any of you seen this theorem of relationship of the angles between two slant faces of a pyramid and the angles between slant vectors, provided that two faces of corresponding to $\phi$ and $\eta$ are perpendicular? I at... |
Applying Pythagoras theorem to the base, we have $L^2 = a^2 \sin^2 φ + c^2 \sin^2 η$
Applying cosine law to the slant face, we have $L^2 = a^2 + c^2 – 2ac \cos θ$
∴ $a^2 \sin^2 φ + c^2 \sin ^2 η = a^2 + c^2 – 2ac \cos θ$
$2ac \cos θ = a^2 – a^2 \sin^2 φ + c^2 – c^2 \sin^2 η$
$2ac \cos θ = a^2 \cos^2 φ + c^2 \cos^2 η$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality $\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{3}{2}$ I'm trying to prove this by induction: $$\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}<\frac{3}{2}, \quad n\geq1. $$
I got stuck in the inductive step:
$$\frac{3}{2} +\frac{1}{(n+1)^3}\,\ldots$$
Thank you for your h... | You can use for $n\geq 2$, $$\dfrac{1}{n^3} \leq \dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $ \frac{1}{ a} = \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $ with additional conditions How to solve this equation
$$ \frac{1}{a}= \ \frac{1}{ \ \sqrt{b}} \ +\ \frac{1}{ \ \sqrt{c}} $$
where $$ b = \sqrt{ (x-a/2)^2 + y^2 + z^2 )}$$
& $$ c = \sqrt{ (x+a/2)^2 + y^2 + z^2 )}$$
We have to get an equ... | Partial answer: You need a simplification, not a solution.You already have an implicit equation in x, y, z and a ! The surface is symmetric about x = 0 or y-z plane ( change of sign before x makes no change). One needs to study (x,y) intersection only, as it is formed by rotation about x-axis ( due to appearance of $ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Transforming a cartesian equation to a polar one when it has different x and y denominators? $$\frac{x^2}{9}+\frac{y^2}{16}=1$$
Needs to be replaced with an equivalent polar equation. I'm sure the identity I'll have to use will be $$x^2+y^2=r^2$$
though other options are:
$x=rcos\theta$ and $y=r\sin\theta$.
Originall... | $$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}=1}$$ does not mean $$\frac{x}{3}+\frac{y}{4}=1$$
as $$\sqrt{\frac{x^2}{9}+\frac{y^2}{16}}\ne\frac x3+\frac y4$$
Set $x=r\cos\theta$ and $y=r\sin\theta$ in $16x^2+9y^2=144$
to get $$r^2(16\cos^2\theta+9\sin^2\theta)=144$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this sum $\frac{1}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{3}{4^2}+\frac{2}{5^2}+\frac{4}{6^2}+\cdots+\frac{d(n)}{n^2}+\cdots$ Question:
Find the value
$$\dfrac{1}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{3}{4^2}+\dfrac{2}{5^2}+\dfrac{4}{6^2}+\cdots+\dfrac{d(n)}{n^2}+\cdots$$
where $d(n)$ is The total numb... | Since $d(n) = \sum_{d|n}1$ you can write
$$\sum_{n=1}^{+\infty} \frac{d(n)}{n^2} = \sum_{n=1}^{+\infty} \sum_{d|n} \frac{1}{n^2}$$
Now, the set $\{ (d,n) \in \mathbb{N}^2 : d|n \}$ can be identified with the set $\{ (k, d)\in \mathbb{N}^2 \}$ by the map $(k,d) \mapsto (d, kd)$ (so $n = kd$). This means that
$$\sum_{n=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sketch the set of points determined by $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ Sketch the set of points determined by the following conditions $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ So I started by $(z-i)(\bar{z} -\bar{i}) + (z+i)(\bar{z} + \bar{i}) \le 4$
$z\bar{z} - z\bar{i} - i\bar{z} + i\bar{i... | While the algebraic route is simple, one actually can deduce the result from geometry alone. You have two points, separated by a distance of $c=2$, and you require that their distances $a,b$ to a point $z$ satisfy $a^2+b^2\leq c^2$. For the boundary points, for which this is an equality, we recognize the Pythagorean th... | {
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"source": "stackexchange",
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Tricky sequences and series problem For a positive integer $n$, let $a_{n}=\sum\limits_{i=1}^{n}\frac{1}{2^{i}-1}$. Then are the following true:
$a_{100} > 200$ and
$a_{200} > 100$?
Any help would be thoroughly appreciated. This is a very difficult problem for me. :(
| Have you thought about regrouping some of the terms ?
$\frac{1}{2} \geq \frac{1}{2}$
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$ since $\frac{1}{3} > \frac{1}{4}$
$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{2}$ since $\frac{1}{5} > \frac{1}{8}, \frac{1}{6} > \frac{1}{8}...$
...
$\frac{1}{2^{199}+1} ... | {
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If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have $\gcd(a,b)=1\implies \exis... | If $\gcd(a,b)=1$, then there are $x,y\in\mathbb{Z}$ so that $ax+by=1$.
Then, since
$$
\begin{align}
2
&=\overbrace{[(a+b)+(a-b)]}^{\large2a}\,x+\overbrace{[(a+b)-(a-b)]}^{\large2b}\,y\\[4pt]
&=(a+b)(x+y)+(a-b)(x-y)
\end{align}
$$
we have that
$$
\gcd(a+b,a-b)\mid2
$$
Furthermore
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(a... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $2\cos(x)$ is equal to $2\cos(2x)\sec(x)+\sec(x)\tan(x)\sin(2x)$ This is from the derivative of $\dfrac{\sin(2x)}{\cos x}$
I tried to solve it and arrived with factoring the $\sec(x)$ but I still can't get it to $2\cos(x)$. Could you help me out, please? Thanks
| Another way
$$\begin{align}
2\cos x&=\frac{2\cos^2 x}{\cos x}\\
&=\frac{2(\cos^2 x-\sin^2 x+\sin^2 x)}{\cos x}\\
&=\frac{2(\cos2 x+\sin^2 x)}{\cos x}\\
&=2\left(\frac{\cos2 x}{\cos x}+\frac{\sin x\cdot\sin x}{\cos x}\right)\\
&=2(\cos2 x\sec x+\sin x\tan x)\\
&=2\cos2 x\sec x+2\sin x\cdot\frac{\cos x}{\cos x}\tan x\\
&... | {
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Limit $\lim\limits_{n \to \infty} n \cdot\ln(\sqrt{n^2+2n+5}-n)$ How should this limit be solved ?
$$\lim_{n \to \infty} n \cdot \ln(\sqrt{n^2+2n+5}-n)$$
I've tried to multiply and at the same time divide $\sqrt{n^2+2n+5}-n$ by $\sqrt{n^2+2n+5}+n$, and then make $n$ as the power of $\frac {2n+5}{\sqrt{n^2+2n+5}+n}$. Bu... | Notice that
$$ \lim_{n \to \infty} n \cdot \ln (\sqrt{n^2+2n+5}-n) = \lim_{n \to \infty} n \cdot \ln (\sqrt{(n+1)^2+4}-n) $$
so we change variable: $a= n+1 \to \infty $ to get
$$ \lim_{a \to \infty} (a-1) \cdot \ln (\sqrt{a^2+4}-a+1) $$
But we know that for $a \to \infty $, for a product $ (a-1)$ behaves as $a$ and tha... | {
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"question_score": "6",
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Solving a system of two cubic equations I'm trying to solve a system of two cubic equations with two variables x and y.
The original problem was to solve the equation $z^3=-4i \overline{z}$. I know how to solve it using polar form.
Now I want to solve it using Cartesian form, say $z=x+yi$.
Doing the algebra and simplif... | $$z^3=-4i\bar{z}\Rightarrow (-i)z^4=4|z|^2\Rightarrow |z|^4=4|z|^2$$
A trivial solution would be $|z|=0\Leftrightarrow z=0$. Assume $|z|\neq 0$ then
$$|z|^2=4 \Rightarrow x^2+y^2=4$$
Expressing the main equation in cartesian coordinates:
$$(x+iy)^3=-4i(x-iy)\Rightarrow x^3+3ix^2y-3xy^2-iy^3=-4ix-4y$$
So one gets after... | {
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Logarithmic inequality for a>1 Is $\log_{\sqrt a}(a+1)+\log_{a+1}\sqrt a\ge \sqrt6$ always true for $a>1$?
What is the approach? Do we check the first a's and then form a induction hypothesis?
| rewrite your inequality in the form
$\frac{\ln(a+1)}{\ln(\sqrt{a})}+\frac{\ln(\sqrt{a})}{\ln(a+1)}=$
$\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}\geq 2$
by the theorem that stated
$\frac{x}{y}+\frac{y}{x}\geq 2$ iff $(x-y)^2\geq 0$
for $x,y>0$
if we define $f(a)=\frac{2\ln(a+1)}{\ln(a)}+\frac{\ln(a)}{2\ln(a+1)}$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Integral $\int\limits_{\sqrt{2}}^{2}\frac1{t^3\sqrt{t^2-1}}$ This is my very first trigonometric substitution integral, so please bear with me here. I would mostly like verification that my methods are sound.
$$\int\limits_{\sqrt{2}}^{2}\dfrac{1}{t^3\sqrt{t^2-1}}\text{ d}t\text{.}$$
Because of the $\sqrt{t^2-1} = \sqrt... | There are no issues with your solution.
You can easily find $\sin \frac{2\pi}{3}$ using $\sin x = \sin (\pi - x)$ for $x = \frac{\pi}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all positive integers s.t. $10^m-8^n=2m^2$ Find all pairs of positive integers $(m,n)$ such that $10^m-8^n=2m^2$
| Hint:
$${10}^m - 8^n = {2^m}{5^m} - {2^{3n}}$$
so that
$$m^2 = {2^{m-1}}{5^m} - {2^{3n-1}}.$$
In particular:
$${2^{m-1}}{5^m} - {2^{3n-1}} \geq 1,$$
and
$${2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is a square, which means $m \equiv 1 \pmod 2$.
Suppose that $m > 1$. Then the RHS of
$$m^2 = {2^{m-1}}\left({5^m} - {2^{3... | {
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Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a... | What if there is a condition $a^2+b^2\neq 0?$ How to prove closure under multiplication?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Simplfy trigonometric functions by only considering integer inputs? I have the below function which only takes integer input,
$$ 2 \sqrt{3} \sin \left(\frac{\pi t}{3}\right)+\sqrt{3} \sin \left(\frac{2 \pi t}{3}\right)-\sqrt{3} \sin \left(\frac{4 \pi t}{3}\right)+6 \cos \left(\frac{\pi t}{3}\right)+\cos \left(\frac... | Since $t$ is an integer,
$$ \sin(\tfrac{4\pi t}{3}) = \sin(2\pi t - \tfrac{2\pi t}{3})
= \sin(2\pi t)\cos(\tfrac{2\pi t}{3}) - \cos(2\pi t)\sin(\tfrac{2\pi t}{3})
= -\sin(\tfrac{2\pi t}{3}) $$
and similarly
$$ \cos(\tfrac{4\pi t}{3}) = \cos(\tfrac{2\pi t}{3}) $$
Thus
\begin{align*}
& 2\sqrt3 \sin(\tfrac{\pi t}{3})
+ \s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested w... | Lemma 1: If $|x| < 1$, then we know that, $\frac{1}{1 - x} = \sum_{n = 0}^{\infty}x^n$. Using this, we can further show that, $\frac{1}{(1 - x)^3} = \sum_{n = 0}^{\infty}\frac{(n+1)(n+2)}{2}x^n$.
Proof: The first sum is just the GP. The second sum can be obtained by simply counting the different ways to get $x^n$ in $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Equation: $2\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x^2}=3-x$ $2\sqrt{1-x}-\sqrt{1+x}+\sqrt{1-x^2}=3-x$
Could someone help me solve this problem?
| Assuming that we are restricted to only real square roots:
Substitute $a = \sqrt{1-x}$ and $b = \sqrt{1+x}$ to get $2a-b+ab = 2+a^2$
Solving for $b$ gives us $b = a-1+\dfrac{1}{a-1}$.
Since $b$ must be non-negative, $a-1$ must be positive. Then, by AM-GM, we get that $b \ge 2$.
However, $b = \sqrt{1+x} \ge 2$ iff $x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $f(x) = -3x^3 + 2x^2$, find $f(-1)$ and $f(1/2)$. I hope you can help me out
For number 8 and 9 find each value if $f(x) = -3x^3 + 2x^2$
8) $f(-1)$
9) $f(1/2)$
| If you've given some $f(x)$ and asked to find out $f(a)$
then you have to replace all $x$ by $a$ and then simplify ( If you cannot calculate ) resultant expression.
for example
if $$f(x)=-3x^3+2x^2$$
then $$f(-1)=-3(-1)^3+2(-1)^2=5$$
and $$f\left(\frac{1}{2}\right)=-3\left(\frac{1}{2}\right)^3+2\left(\frac{1}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
| Expanding the answer of Henno Brandsma: a generating function is a way to pack a sequence as the coefficients of the power expansion of a function, by example we can pack the Fibonacci sequence as the coefficients on the power expansion of the function $h(x):=\frac1{1-x-x^2}$.
The important point here is that the algeb... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solve limit using the concept of equivalent functions How to solve this limit $$\lim_{x\rightarrow 0}{\frac{(2+x)^x-2^x}{x^2}}$$ using the concept of equivalent functions?
For example, if $x\rightarrow 0 $ function $\sin x$ is equivalent to $x$, $\ln(1+x)\sim x$, $a^x-1 \sim x \ln a$, etc.
| we have $$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}=\lim_{x\to 0}2^x\lim_{x\to 0}{\frac{(1+\frac{x}{2})^x-1}{x^2}}=\lim_{x\to 0}{\frac{e^{x\ln(1+\frac{x}{2})}-1}{x^2}}$$ therefore we have that$$\lim_{x\to 0}{\frac{(2+x)^x-2^x}{x^2}}\sim\lim_{x\to 0}{\frac{{x\ln(1+\frac{x}{2})}}{x^2}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \... | Hint. Observe that you have
$$
\frac{Bx+C}{(x-4)^2} =\frac{B(x-4)+4B+C}{(x-4)^2}=\frac{B}{(x-4)}+\frac{4B+C}{(x-4)^2}
$$
and
$$
\begin{align}
\frac{Dx^2 + Ex + F}{(x-4)^3} &=\frac{D(x-4)^2+(E+8D)(x-4)+C+4E+32D}{(x-4)^3}\\\\
&=\frac{D}{(x-4)}+\frac{E+8D}{(x-4)^2}+\frac{C+4E+32D}{(x-4)^3}\\\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$ divides $b$, then $a$ divides $3b^3-b^2+5b$.
Prove: Suppose $a$ and $b$ are integers. If $a\mid b$, then $a\mid3b^3-b^2+5b$.
I think I have an idea of how to prove this, but I'm not entirely sure.
I can prove that each individual term in the polynomial $3b^3-b^2+5b$ is a multiple of $a$.
*
*$3b^3 = a(3b^2c... | if $a|b$ then $b=a.t$ for some integer $t$, so $3b^3-b^2+5b=3(at)^3-(at)^2+5(at)=a(3a^2t^3-at^2+5t)$, and thus clearly $a|3b^3-b^2+5b$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ im... | First:$f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $
Then observes that; For $|x|\le1$
$$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le 1$$
and for $|x|\ge1$
$$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$
Hence,
$$|f'(x)|=\frac{2|x|}{(1+x^2)^2} \le 2\implies |f(x)-f(y)|\le 2|x-y|$$
T... | {
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"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
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Find $S=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+...+\frac{2n-1}{2^n}+...$ I'm trying to calculate $S$ where $$S=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+...+\frac{2n-1}{2^n}+...$$
I know that the answer is $3$, and I also know "the idea" of how to get to the desired outcome, but I can't seem ... | $$\sum_{n=1}^\infty\frac{2n-1}{2^n}=\sum_{n=1}^\infty\frac n{2^{n-1}}-\sum_{n=1}^\infty\frac1{2^n}=\frac1{\left(\frac12\right)^2}-\frac{\frac12}{1-\frac12}=4-1=3$$
The above follows from
$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;\left(\frac1{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}$$
and the splitting of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Factoring $x^4 + 4x^2 + 16$ I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.
I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for f... | $$x^4 + 4x^2 + 16$$
Let $y = x^2$, the polynomial is then equal to
$$y^2 + 4y + 16$$
Then, use the quadratic formula:
$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$$
$$y = \frac{-4 \pm \sqrt{-48}}{2}$$
$$y = 2 \pm 2i\sqrt{3}$$
Return to $x$
$$x^2 = 2 \pm 2i\sqrt{3}$$
So, we can now convert $x^4 + 4x^2 + 16$ into fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\gcd(7,abc)=1$ and $a^2+b^2=c^2$, prove that $7$ divides $a^2-b^2$ The only information I have on this problem is that for $a^2+b^2=c^2$ that
$$
a = st, b = \frac{s^2-t^2}{2}, c = \frac{s^2+t^2}{2}
$$
and that $\gcd(7,abc)=1$ gives $7x + abcy = 1$
I have no idea how to proceed, so any help welcomed
| HINT:
$$x\equiv\pm1,\pm2,\pm3\implies x^2\equiv1,4,2\pmod7$$
So, if $a^2\equiv1,b^2$ must be $\equiv1\pmod7$ and so on
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... |
Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$
We have \begin{align*}LHS-RHS&=(x^3+y^3+z^3)\sum x(x-y)(x-z)+(x^2+y^2+z^2-xy-yz-zx)^3\\ &+7(xy+yz+zx)\sum x^2(x-y)(x-z)\\ &+[x^2+y^2+z^2+3(xy+yz+zx)][x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z)]\\ &+ (x+y+z)[4(x^2+y^2+z^2)+9(xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6... | In general, to solve
$$ ax^2+bx+c=0$$
We can use the quadratic formula, which is
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
In your case, we have
$$ 2t^2+t-3=0$$
Which implies that
$$ a=2, b=1, c=-3 $$
Also I'd recommend against your first step of dividing by $2$. This step introduces fractions and the quadratic formula ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral
\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}
\end{equation}
Wolfram Alpha and Mathematica can easily evaluate this integral... | As M.N.C.E. stated in the comments, integrating by parts shows that $$ \begin{align} \int_{0}^{1} \frac{\log(2) - \log(1+x^{2}) }{1-x} \ dx&= - 2 \int_{0}^{1} \frac{x \log(1-x)}{x^{2}+1} \ dx \\ &= - 2 \ \text{Re} \int_{0}^{1} \frac{\log(1-x)}{x+i} \ dx . \end{align}$$
Letting $ \displaystyle t = \frac{x+i}{1+i} $,
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Max. value of $xyz+xy+yz +zx$, If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Maximum value of $xyz+xy+yz +zx$, is
$\bf{My\; Try::}$ First we can write the given expression
$xyz+xy+yz+zx = (x+1)(y+1)(z+1)-(x+y+z)-1 = $
$\displaystyle = (x+1)(y+1)(z+1)-\left\{(x+1)+(y+1)+(z+1)\right\... | Use the fact that $a+b+c=13$ to say you are trying to maximize $abc-14$, which is the same as maximizing $abc$. The AM-GM inequality now tells you that $a=b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How pro... | Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear function.
Hence, $f$ gets a minimal value, when $v^2$ gets an extremal value, which happens when
two numbers from $\{x,y,z\}$ are equal.
Id est, it remains to prove our inequality for $y=x$ and $z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Confirming an answer for finding a limit I have to found the limit of the sequence $(x_n)$, which is equal to:
$((2^3-1)/(2^3+1))$ $((3^3-1)/(3^3+1))$ $...$ $((n^3-1)/(n^3+1))$,
and I used the tricks that $(n^3-1)$=$(n-1)(n^2+n+1)$, and that $(n^3+1)$=$(n+1)(n^2-n+1)$, and after cancellation, I found the limit is 2/3. ... | Yes you are right!
$$\prod_{k=2}^n \dfrac{k^3 - 1}{k^3 + 1} = \dfrac{2(n^2 +n + 1)}{3n(n+1)} \to \frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a non-linear congruence How would I go about solving the following:
$$x^2+1\equiv2\pmod8$$
I know I can subtract, and I get:
$$x^2\equiv1\pmod8$$
I'm not sure if I am allowed to square root both sides, or if I should employ a different technique?
| Rewrite the last relation as $x^2-1\equiv0 \pmod 8$ or equivalently as $$(x-1)(x+1)\equiv0 \pmod 8$$ and assume that $x$ is even, i.e. $x=2n$ for $n \in \mathbb N$. Then $$(x-1)(x+1)=(2n-1)(2n+1)=4n^2-1$$ which cannot be a multiple of $8$.
Now assume that $x$ is odd, i.e. $x=2n+1$ for $n \in \mathbb N$. Then $$(x-1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{3}\cos2\theta+\sin2\theta-1=0$ I tried using the identities $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$. These give
$\sqrt{3}(1-2\sin^2\theta)+2\sin\theta\cos\theta-1=0$
which doesn't seem to lead anywhere. Perhaps I must equate the function to something like $R\sin(2\theta+\alpha... | When you have an equation of the form
$$
a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta+d=0
$$
you can use $1=\cos^2\theta+\sin^2\theta$ and write the equation as
$$
(a+d)\sin^2\theta+b\sin\theta\cos\theta+(c+d)\cos^2\theta=0
$$
If $a+d=0$, this factors; otherwise $\cos\theta=0$ is not a solution and so you can tran... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving a simple first order differential equation I've got the following first order differential equation that is doing something I can't quite figure out. The question is to solve:
$$y' = 2x^2y^2$$
Here is my solution:
$$y' \frac{1}{y^2} = 2x^2$$
$$\int \frac{1}{y^2} y' dx = \int 2x^2 dx$$
$$\int \frac{1}{y^2}dy = ... | Note that $$\int \frac 1{y^2}\,dy = \int y^{-2} \,dy = -y^{-1} + C$$
So you should obtain $$-\frac 1y = \frac{2}{3}x^3 + c \iff y = -\frac 3{2x^3}\; \text{provided } x\neq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor series expansion for $e^{-x}$ could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how
$e^{-i\theta}$=$\cos\theta-i\sin\theta$.
More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.
| $$ e^{-x} = \sum\limits_{k=0}^{\infty}\frac{(-x)^k}{k!} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\dots $$
So plugging in $x=i\theta$ we have that
\begin{align}
e^{-i\theta} &=1-i\theta+\frac{(i\theta)^2}{2!}-\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\dots\\
&=1-i\theta-\frac{\theta^2}{2!}+i\frac{\theta^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{... | This is equal to
$$\frac1{72} \sum_{n=-\infty}^{\infty} \left [\frac1{(n+1/6)^2} + \frac1{(n+5/6)^2}\right ] $$
This may be treated using the following formula from residue theory:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} f(z) \cot{\pi z} $$
where $z_k$ are the poles of $f$. In this c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Maximizing the trinomial coefficient Let $\binom{n} {a,b,c} = \frac{n!}{a!\cdot b! \cdot c!}$
In other words, $\binom{n} {a,b,c}$ is the trinomial coefficient.
I am trying to find the triplets $(a,b,c)$ which maximize this trinomial coefficient.
I have determined, by simply plugging in numbers, that:
When $n\bmod 3 = 0... | Temporarily fix $c$. We want to minimize $x!y!$ given that $x+y=n-c=k$.
Suppose that we have $x\lt y$. Compare $x!y!$ with $(x+1)!(y-1)!$. We have
$$(x+1)!(y-1)!=\frac{x+1}{y}(x!y!).$$
Thus $(x+1)!(y-1)!\lt x!y!$ unless $y=x+1$, in which case we have equality.
So given $a,b,c$ with fixed sum, if two of $a,b,c$ diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
what is the sum of square of all elements in $U(n)$? I know that $\sum\limits_{a\in U(n)} a=\frac{n\varphi(n)}{2}$ where $U(n):=\{1\leq r\leq n: (r, n)=1\}$ is a multiplicative group. And I know how to prove this result.
What I was willing to know was this $\sum\limits_{a\in U(n)} a^2$. is it possible to find in closed... | Lemma 1: For any function $f$ defined on rational values in $[0,1]$, denoting, $\displaystyle F(n) = \sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)$ and $\displaystyle F^{*}(n) = \sum\limits_{1\le k\le n;(k,n)=1} f\left(\frac{k}{n}\right)$.
We have, $\displaystyle 1*F^{*} = F$
Proof: First we observe that each integer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating definite integral I'm having a hard time understanding how to compute this integral.
$$\int_1^4\frac{3x^3-2x^2+4}{x^2}\,\mathrm dx$$
The steps I do is $\dfrac{3x^4}{4} - \dfrac{2x^3}{3} + 4x$ but I don't know how to integrate the $x^2$ in the integral. I know it's suppose to be $\dfrac{x^3}3$.
Is this how th... | $$
\int_1^4 \frac{3x^3 - 2x^2 + 4}{x^2} dx = \int_1^4 (\frac{3x^3}{x^2} - \frac{2x^2}{x^2} + \frac{4}{x^2})dx = \int_1^4 (3x - 2 + 4x^-2)dx
$$
$$ = \frac{3}{2}x^2 - 2x - \frac{4}{x}
$$
Now evaluate this function from 1 to 4. First evaluating the function at 4 and then subtracting off the function evaluated at 1.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
\begin{array}{l}
2x\equiv 3\pmod 5 \\
4x\equiv 2\pmod 6 \\
3x\equiv 2\pmod 7
\end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, gi... | \begin{align}
2x &\equiv 3\pmod 5 \\
4x &\equiv 2\pmod 6 \\
3x &\equiv 2\pmod 7
\end{align}
First, you need to solve each congruence for $x$.
\begin{align}
2x &\equiv 3 \pmod 5 \\
x &\equiv 4 \pmod 5 \\
\hline
4x &\equiv 2 \pmod 6 \\
2x &\equiv 1 \pmod 3 \\
x &\equiv 2 \pmod 3 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \s... | If you're willing to use a symbolic differentiation program like Matlab and have forgotten your trig and calculus formulas like I have, you should be able to recover the answer as $$\frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin x})}{dx^3}}{\frac{d^3(x^3)}{dx^3}} = \frac{\frac{d^3(\sqrt{1 + \tan x} - \sqrt{1 + \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Limit of square root function at $x \to 6$ I'm trying to find the limit of the following function at $x \to 6$:
$$\frac{x^2-36}{\sqrt{x^2-12x+36}}$$
i've simplified it so that it becomes $\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}$, which simplifies to $x+6$.
the problem is that i shouldn't be getting to $x+6$, because then i... | Hint: $$\sqrt{(x-6)^2}=|x-6|$$ which is in turn equal to $$|x-6|=\begin{cases}-(x-6)=-x+6, & x<6\\ \phantom{+}(x-6)=\phantom{-}x-6,&x>6\end{cases}$$ This implies that $$\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}=\lim_{x\to6^-}\dfrac{(x+6)(x-6)}{-(x-6)}=\lim_{x\to6^-}-(x+6)$$ but $$\lim_{x\to6^+}\dfrac{(x+6)(x-6)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$ Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$
| Starting from the representation
$$\pi \cot \pi z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n},$$
we obtain
\begin{align}
\pi \tan \pi z &= -\pi \cot \pi\left(z-\tfrac{1}{2}\right)\\
&= \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{n+\frac{1}{2}-z}\\
&= \lim_{N\to\infty} \left(\frac{1}{N+\frac{1}{2}-z} + \sum_{n=0}^{N-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Limit: $\lim_{x\to 0} \frac{\sqrt {\cos x} - \sqrt[3] {\cos x}}{\sin^2x}$ Compute the following limit:
$$\lim_{x\to 0} \frac{\sqrt {\cos x} - \sqrt[3] {\cos x}}{\sin^2x}$$
How would I go about solving this, I can't used l´Hôpital
| Writing $c$ and $s$ for $\cos x$ and $\sin x$ respectively,
$$\frac{c^{1/2} - c^{1/3}}{s^2} = \frac{c^{1/3}}{s^2} (c^{1/6} - 1) = \frac{c -1}{s^2} \frac{c^{1/3}}{c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1} \ \ \ --(*) $$
as $(c^{1/6} - 1)(c^{5/6} + c^{4/6} + c^{3/6} + c^{2/6} + c^{1/6} +1) = c - 1$
Now $$c - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Logarithmic twin integrals $\int_0^1\frac{\ln x\ln(1+x^2)}{1\pm x}dx$ Here is what I have done
\begin{align}
&\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx\\
=&\int_0^1\frac{(1+x)(1+x^2) \log(x)\log(1-x^4)}{1-x^4} \ dx\\
&-\int_0^1\frac{(1+x)\log(x)\log(1-x^2)}{1-x^2} \ dx
\end{align}
and, then, after letting $x^4\mapsto x$ a... | Denote $K_{\pm} = \int_0^1 \frac{\ln t}{1\pm t}dt$ and integrate the twins as follows
\begin{align}
I_\pm =&\int_0^1\frac{\ln x\ln(1+x^2)}{1\pm x}dx \\
=& \int_0^1 \ln(1+x^2)\> d\left(\int_0^x \frac{\ln t}{1\pm t}dt\right)\\
=& \>\ln2 \int_0^1 \frac{\ln t}{1\pm t}dt -\int_0^1 \frac{2x}{1+x^2}\left(\int_0^x \frac{\ln t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx$ One of the ways to compute the integral
$$\int_0^{\infty} \log(\sin^2(x))\left(1-x\operatorname{arccot}(x)\right) \ dx=\frac{\pi}{4}\left(\operatorname{Li_3}(e^{-2})+2\operatorname{Li_2}(e^{-2})-2\log(2)-\zeta(3)\right)$$
is to m... | First notice that
$$ \begin{align} &\frac{1}{2} \int_{0}^{\infty} \log (4 \sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\log(4)}{2} \int_{0}^{\infty} \Big( 1- x \, \text{arccot}(x) \Big) \, dx + \frac{1}{2} \int_{0}^{\infty} \log (\sin^{2} x) \Big(1 - x \, \text{arccot}(x) \Big) \, dx \\ &= \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$
let $x,y\in (0,1)$, and such
$$(1+x)(1+y)=81(1-x)(1-y)$$
Prove
$$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask $\dfrac{9}{16}$ is best constant?
PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)... | We have $$\frac{81}1=\frac{1+xy+(x+y)}{1+xy-(x+y)}$$
Applying Componendo and dividendo,
$$\frac{1+xy}{x+y}=\frac{41}{40}$$
Let $$\frac{1+xy}{41}=\frac{x+y}{40}=u$$
$$\implies(1-x)(1-y)(10-8x)(10-8y)$$
$$=\{1+xy-(x+y)\}\{36+64(1+xy)-80(x+y)\}$$
$$=(41u-40u)\cdot36(1-16u)=\frac{36}{64}\left[1-\left(8u-1\right)^2\right]\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle... | We have:
$$I=\int_{0}^{+\infty}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}=2\int_{0}^{1}\frac{dx}{\sqrt{1+x+x^2+x^3+x^4}}$$
(just split $[0,+\infty)=[0,1)\cup[1,+\infty)$ and use the substitution $x=1/y$ on the second piece) and since:
$$ \sqrt{1-x}=\sum_{j=0}^{+\infty}\binom{1/2}{j}(-1)^j x^j,\qquad\frac{1}{\sqrt{1-x^5}}=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions How does one write
$$\frac{x^4(1-x)^4}{1+x^2}$$
in terms of partial fractions?
My Attempt
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$
$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$
Multiplying out and comparing coefficient gives:... | You should have tried
$$\frac{x^4(1-x)^4}{1+x^2}=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx +\mathbf{J}+\frac{\mathbf{G + Hx}}{1+x^2}$$
By the way, multiplying out and comparing coefficients gives:
$$\frac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+\mathbf{4}+\frac{\mathbf{-4}}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Binomial identity $\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$ Prove the identity
$$\binom{0}{0}\binom{2n}{n}+\binom{2}{1}\binom{2n-2}{n-1}+\binom{4}{2}\binom{2n-4}{n-2}+\cdots+\binom{2n}{n}\binom{0}{0}=4^n.$$
This is reminiscent of the i... | By way of enrichment here is another algebraic proof using basic
complex variables. As I pointed out in the comment this identity is
very simple using a convolution, so what follows should be considered
a learning exercise.
We seek to compute
$$\sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k}.$$
Introduce the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solve for constants: Derivatives using first principles
*
*Question
Find the values of the constants $a$ and $b$ such that $$\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \frac{5}{12}$$
*
*My approach
*
*Using the definition of the derivative, $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
*I view limit as... | Using
$$ a^3-b^3=(a-b)(a^2+ab+b^2, $$
one has
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{ax + b}-2}{x} = \lim_{x \to 0}\frac{ax+b-8}{x[(\sqrt[3]{ax + b})^2+2\sqrt[3]{ax + b}+4]}.
\end{eqnarray}
Thus if the limit exists, one must have $b=8$, under which it is easy to get
\begin{eqnarray}
\lim_{x \to 0}\frac{\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Using the definition of the limit to show I am stuck an how to proceed in my proof. I am suppose to show, using the definition of the limit, that $\lim_{x \rightarrow 2}{(x^3-2x-4)/(x^2-4)=5/2}$.
I did some algebra with the $|f(x)-5/2|$ and got to the point where I have $|f(x)-5/2| \leq (1/2)|x-2||2x+3|$, but I am unsu... | First $$\begin{align}\left|f(x)-\frac{5}{2}\right|&=\left|\frac{x^2+2x+2}{x+2}-\frac{5}{2}\right|\\&=\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}+\frac{10}{x+2}-\frac{5}{2}\right|\\&\leq\left|\frac{x^2+2x+2}{x+2}-\frac{10}{x+2}\right|+\left|\frac{10}{x+2}-\frac{5}{2}\right|\\&=\frac{|x^2+2x+-8|}{|x+2|}+\frac{|10-5x|}{|2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
My attempt regarding finding critical ponts of $(\cos x)(\cos y)(\cos(x+y))$ Given this problem Restrictions on $x$ any are that $x\in[0,\pi]$ , $y\in[0,\pi]$
I have $f_x=-(\cos y)({\sin(2x+y))}--------*$
$f_y=-(\cos x)(\sin x+2y)-----------**$
So from $*$ I get either $\cos y=0$ or $\sin(2x+y)=0$
From $\sin(2x+y)=0$... | You have $f(x,y)=\cos x\cos y\cos(x+y)$, so
$$
f'_x=-\sin x\cos y\cos(x+y)-\cos x\cos y\sin(x+y)=-\cos y\sin(2x+y)
$$
Since interchanging $x$ and $y$ doesn't change the function, we have
$$
f'_y=-\cos x\sin(x+2y)
$$
We have four cases:
$$
\begin{cases}
\cos y=0\\
\cos x=0
\end{cases}
\qquad
\begin{cases}
\cos y=0\\
\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.