Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Yet another difficult integration question For $r \in (0,1)$ and $k \in \mathbb Z^+$, prove
$$ \frac{1}{\pi} \int_{0}^\pi \ln\left(1 + r \cos(u)\right) \ln \left( 1 + r \cos(3^k u)\right) du = \left(\ln\left(\frac{2(1-\sqrt{1-r^2})}{r^2}\right)\right)^2.$$
Given that for any $n \in \mathbb Z^+$
$$ - \frac{1}{\pi} \int_... |
Lemma 1: Let $n \in \mathbb{Z}\backslash\{0\}$ and let $f$ be a function such
that $f(x+T)=f(x)$ for all $x$. Then $$
\int_0^T f(nx) \mathrm{d}x = \int_0^T f(x) \mathrm{d}x $$
${}$
Lemma 2: Let $f$ be any function such that $\int_0^{\pi} f(x)\,\mathrm{d}x $ converges, then $$\int_0^{\pi} f(\cos x)\,\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/815650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^... | This equation can be rewritten as $$ab=\frac{(a+b)^2}4-\frac{(a-b)^2}4.$$ This is the basis of tables of quarter-squares. Such a table tabulated $\left\lfloor \frac{n^2}4\right\rfloor$ for various integer values $n$. It was used as an aid to multiplying integers. To find $ab$, look up the quarter-squares of $a+b$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Solve a system of linear equations $\newcommand{\Sp}{\phantom{0}}$There is a system of linear equations:
\begin{alignat*}{4}
&x - &&y - 2&&z = &&1, \\
2&x + 3&&y - &&z =-&&2.
\end{alignat*}
I create the matrix of the system:
$$
\left[\begin{array}{rrr|r}
1 & -1 & -2 & 1 \\
2 & 3 & -1 & -2
\end{array}\rig... | Bringing the matrix in reduced row echelon form $R$ (the way other answers have shown) is the first step no matter what you do.
As for "reading off" the solutions off $R$, I think there are several techniques. The one I was taught, and which I find the easiest to use, is the following:
*
*remove all zero rows
In t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\co... | Setting $7\theta=\pi,S=\sin2\theta+\sin4\theta+\sin8\theta=(\sin2\theta+\sin8\theta)+\sin4\theta$
Using Prosthaphaeresis Formula & Double angle formula,
$S=2\sin5\theta\cos3\theta+2\sin2\theta\cos2\theta$
As $5\theta=\pi-2\theta,\sin5\theta=\sin2\theta$
and $3\theta=\pi-4\theta,\cos3\theta=-\cos4\theta$
$S=-2\sin2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Best way to solve $X^3-X^2-X-1=0$ can anyone help me for this cubic equation ?
can be solved without delta method?
$X^3-X^2-X-1=0$
(answer is $\sim 1.8393$)
| @Darksonn 's answer certainly works, but if you only want the positive real solution, there is a "cubic formula" that is reasonably useful in this case.
Theorem: If the cubic equation
$$X^3 + pX + q$$
($p, q$ real) satisfies
$$\frac{p^3}{27} + \frac{q^2}{4} \geq 0,$$
then a solution to the cubic equation is
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2... | $$\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark
}$$
Therefore you can conclude that: $$\eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ satisfying $f(2x+3)+f(2x+5) = 2$ . The period of $f(x)$ If $f:\mathbb{R}\rightarrow \mathbb{{R}}$ and $f(x)$ be a function satisfying $f(2x+3)+f(2x+5) = 2$. Then period
of function $f(x)$ is.
$\bf{My\; Solution::}$ Let $(2x+3) = t\;,$ Then equation is $f(t)+f(t+2) =... | $\displaystyle f(2x+3) + f(2x+5) = 2$
Replace $x$ with $x+1$:
$\displaystyle f(2x+5) + f(2x+7) = 2$
Subtracting and rearranging, $\displaystyle f(2x+3) = f(2x+7)$
Hence $\displaystyle f(2x+3) = f((2x+3)+4)$
At this point, you can conclude that the period is $4$. If this is difficult to see, you can replace $(2x+3)$ wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The 3 Integral $\int_0^\infty {x\,{\rm d}x\over \sqrt[3]{\,\left(e^{3x}-1\right)^2\,}}=\frac{\pi}{3\sqrt 3}\big(\log 3-\frac{\pi}{3\sqrt 3} \big)$ Hi I am trying evaluate this integral and obtain the closed form:$$
I:=\int_0^\infty \frac{x\,dx}{\sqrt[\large 3]{(e^{3x}-1)^2}}=\frac{\pi}{3\sqrt 3}\left(\log 3-\frac{\pi}{... | Following @Pranav Arora's idea, if defining the following integral
$$ I(a)=\int_0^\infty\frac{\ln(1+at^3)}{1+t^3}dt, $$
the calculation will just be basic calculus without using advanced tools. In fact, $I(0)=0$ and
\begin{eqnarray}
I'(a)&=&\int_0^\infty\frac{t^3}{(1+t^3)(1+at^3)}dt\\
&=&\frac{1}{a-1}\int_0^\infty\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
a problem in application of conformal mappings Obtain the complex potential
$F=A(z^
2
+\frac
{1}
{z^
2})$
for a flow in the region $r≥1,0≤θ≤π/2$. Write expressions for $V$ and $ψ$. Note how
the speed $|V|$ varies along the boundary of the region, and verify that $ψ(x, y)=0$ on
the boundary.
this is a problem in applica... | This question is similar to a few other ones in MSE, among these:
How to differentiate Complex Fluid Potential
Applications of conformal mapping
From the first reference it is clear that, in general, the following formulas are valid:
$$F(z) = \phi + i\,\psi \qquad ; \qquad
\frac{dF}{dz} = u - i\,v \quad \Longrightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$?
How to integrate
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$
I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \... | $$
\begin{aligned}
\sin ^{4} x+\cos ^{4} x &=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\
&=1-2 \sin ^{2} x \cos ^{2} x \\
\int \frac{d x}{\sin ^{4} x+\cos ^{4} x} &=\int \frac{d x}{1-2 \sin ^{2} x \cos ^{2} x} \\
&=\int \frac{\sec ^{4} x}{\sec ^{4} x-2 \tan ^{2} x} d x \\
&\stackrel{t=\tan x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/820830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
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Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\t... | \begin{align}
I &= \int\frac{x^2dx}{(x\sin x+\cos x)(x\cos x-\sin x)} \\ &= -\int\frac{x^2dx}{(1+x^2)\left(\frac{x}{\sqrt{1+x^2}}\sin x+\frac{1}{\sqrt{1+x^2}}\cos x\right)\left(\frac{1}{\sqrt{1+x^2}}\sin x-\frac{x}{\sqrt{1+x^2}}\cos x\right)}\\
&=-\int\frac{\left(1-\frac{1}{1+x^2}\right)dx}{\cos (x-\arctan x)\sin (x-\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 2
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Product of Gamma functions I What is the value of the product of Gamma functions
\begin{align}
\prod_{k=1}^{8} \Gamma\left( \frac{k}{8} \right)
\end{align}
and can it be shown that the product
\begin{align}
\prod_{k=1}^{16} \Gamma\left( \frac{k}{8} \right) = \frac{ 3 \Gamma(11) \ \pi}{2^{19}} \zeta(2) \zeta(4) \approx ... | For the first question, you can simply calculate:
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{8}{k}\cdot\frac{k}{8}! = \frac{8^8}{8!}\cdot\prod_{k=1}^{8}\frac{k}{8}!$$
Or alternatively (even simpler):
$$\prod_{k=1}^{8}\Gamma\left(\frac{k}{8}\right) = \prod_{k=1}^{8}\frac{k-8}{8}!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Help in understanding contour integration I would like help in understanding the process of contour integration. As an (hopefully straightforward) example, I have chosen the calculation of Bernoulli number $B_2$.
I should be very grateful if someone could explain how to get from (and describe the steps between) here
$... | So between $$B_n=\dfrac{n!}{2\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{n+1}}$$ and $$\dfrac{1}{\pi i}\oint\dfrac{z}{e^z-1}\dfrac{\text{d}z}{z^{3}}$$ all they did was plug in $n=2$, so now all we need to do is find the residue at $z=0$ of $$\frac{1}{z^{2}(e^{z}-1)}=\frac{1}{z^{2}(1-1+z+\frac{z^{2}}{2!}+\frac{z^{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the Laurent series of $ \exp \! \bigl( - \frac{1}{z} \bigr) $? I’m thinking that I could simply let $ x = - \dfrac{1}{z} $ in the Maclaurin series for $ e^{x} $:
$$
1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots
= 1 - \frac{1}{z} + \frac{1}{2! z^{2}} - \frac{1}{3! z^{3}} + \cdots.
$$
Is that right?
| The Taylor series for $e^x$ is
$$
e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots
$$
Now as you have said, let $x = -\frac{1}{z}$. Then we have
$$
e^{-1/z} = 1 - \frac{1}{z} + \frac{1}{2z^2} - \frac{1}{3!z^3} + \cdots = \sum_{n=0}^{\infty}\frac{1}{(-z)^nn!} = \sum_{n=-\infty}^0\frac{(-z)^n}{(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Basic algebra, isolating the variable So I have the equation
$$\tan30=\frac{4.9t-\frac{10}{t}}{\frac{8.77}{t}}$$
And I want to find t, but my algebra has failed me.
This is my working so far.
$$\frac{8.77}{t}=\frac{4.9t-\frac{10}{t}}{\tan30}$$
$$\frac{8.77}{t}=\frac{4.9t}{\tan30}-\frac{10}{\tan30t}$$
$$8.77=\frac{4.9... | Multiplying the right side of the equation by $\frac{t}{t} = 1$, with the assumption that $t \neq 0$, we have
$$\tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} \,\,.$$
Now, we can isolate $t$ as follows
$$\begin{align} \tan(30^{\circ}) = \frac{4.9t^2 - 10}{8.77} &\implies 8.77\tan(30^{\circ}) = 4.9t^2 - 10 \\&\implies t^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle.
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2... | The gradient vector $\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)$ is normal to the curve $f(x,y)=0$.
Then
$$\left(\frac{2x}{a^2},\frac{2y}{b^2}\right)\cdot\left(\frac{2x}{\alpha^2},-\frac{2y}{\beta^2}\right)=\frac{4x^2}{a^2\alpha^2}-\frac{4y^2}{b^2\beta^2}=0.$$
We can eliminate $x^2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I... | Hint :
Be careful, at the step $k+1$ the sum is
$$ (k+2)^2+\cdots+(2k+1)^2+(2k+2)^2 $$
and the sum at the step $k$ is
$$ (k+1)^2+\cdots+(2k-1)^2+(2k)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Show that $\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50}$ Show that
$$\int_{-\infty}^{\infty}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}=\frac{7\pi}{50} $$
So I figured since it's an improper integral I should change the limits
$$\lim_{m_1\to-\infty}\int_{m_1}^{0}\frac{x^2dx}{(x^2+1)^2(x^2+2x+2)}+ \lim... | Hint
Use partial fractions to write out the integral in the form:
$$\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E+F
x}{x^2+2 x+2}$$
You'll find that $a=-6/25, c = 2/5$, etc. Now look at at what remains, and use the fact that:
$$\int \frac{1}{x^2+1}dx=\arctan x$$
And substitute $x\rightarrow x+1$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Factor the Expression completely$ (a+b)^2 - (a-b)^2$ I don't understand this question. The answer in the book is $4ab$, but how is that term a factor?
I was thinking along the line that this was a difference of squares example. $a^2-b^2 = (a+b)(a-b)$
My answer is $[(a+b)-(a-b)][(a+b)+(a-b)]$
What do I not understand? ... | Note that you don't even need to factor it, just expand the expression.
$$(a+b)^2-(a-b)^2$$
$$=(a^2+b^2+2ab)-(a^2+b^2-2ab)$$
$$=a^2+b^2+2ab-a^2-b^2+2ab$$
$$=4ab$$
If you have to factor it, remember the difference of squares $a^2-b^2=(a+b)(a-b)$. In this case, $a^2$ is $(a+b)^2$, and $b^2$ is $(a-b)^2$.
$$(a+b)^2-(a-b)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Use induction to show that $a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) .$
Let $a_0$ and $a_1$ be distinct real numbers. Define
$a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Prove that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) $$
This is what I have so far:
Let $P(n)$ be ... | Hint: Subtract $a_{n-1}$ from both sides. We get
$$a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$
The "new difference" is the "old difference" times $-\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\left(\dfrac{5}{3}\right)^3\left(-\dfrac{3}{5}\right)^2$ I need help in solving this problem (sorry I didn't know how to write it on here).
| $\begin{align}\left(\dfrac{5}{3}\right)^3\left(\dfrac{-3}{5}\right)^2 & = \left(\dfrac{5^3}{3^3}\right)\cdot\left(\dfrac{(-1)^2 3^2}{5^2}\right) & \text{by commutativity of exponents} \\ ~ & = \dfrac{(-1)^2 5^1}{3^1} & \text{by associativity of exponents} \\ ~ & = \dfrac{5}{3} & \text{by }(-1)^2=1, a^1 = a \end{align}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/836722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$ The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$.
We can consider a similar nested radical where the degree of the radicals increases:
$$\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cd... | As already mentioned by others, there is no closed form known for this expression (like the majority of infinite nested radicals).
I would consider some bounds for this number, which give quite good approximations.
First, instead of truncating the number, it's much better to replace the remainder by $1$ at any step, f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 4,
"answer_id": 0
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solve the equation for a: (matrix) = 10
how did he expand the determinant?
I tried using the method where you take the determinent by the top left element times the det of the bottom right 4 minus the top element times the det of the bottom left 2 and bottom right 2 + the top right times the det of the bottom left 4 ... | $$
\begin{vmatrix} 1 & -1 & 4 \\ -1 & a & 2 \\ 1 & -2 & 3\end{vmatrix} = 1*\begin{vmatrix} a & 2 \\ -2 & 3\end{vmatrix} - (-1)*\begin{vmatrix} -1 & 2 \\ 1 & 3\end{vmatrix} + 4*\begin{vmatrix} -1 & a \\ 1 & -2\end{vmatrix} = 10
$$
can you take it from here?
Also use:
$$
\begin{vmatrix} a & b \\ c & d\end{vmatrix} = ad-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$n \in \mathbb{N} \ 5|\ 2^{2n+1}+3^{2n+1}$ show for all $n \in \mathbb{N}$,
$$5|\ 2^{2n+1}+3^{2n+1}$$
Indeed,
we've to show that : $2^{2n+1}+3^{2n+1}=0[5] $
note that $2^{2n+1}+3^{2n+1}=2.4^n+3.9^n= $
| You can do it by induction. For $n=0$, we have
$$2^{2n+1}+3^{2n+1}=2+3=5$$
which is divisible by $5$. Now suppose that
$2^{2k+1}+3^{2k+1}$ is divisible by $5$, i.e.
$$\tag{1}2^{2k+1}+3^{2k+1}=5N$$
for some integer $N$. Now consider
$$2^{2(k+1)+1}+3^{2(k+1)+1}=4\cdot 2^{k+1}+9\cdot3^{2k+1}=
4\cdot (2^{k+1}+3^{2k+1})... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How do you integrate the reciprocal of square root of cosine? I encountered this integral in physics and got stuck.
$$\int_{0}^{\Large\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}}.$$
| Substituting $y=\cos{\theta}$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}} &= \int_{0}^{1} \, \frac{1}{\sqrt{y\, \left(1-y^2\right)}}\, dy \\
&= \frac{1}{2}\int_{0}^{1} \, t^{-3/4}\left(1-t\right)^{-1/2}\, dt \tag{where $t=y^2$} \\
&= \frac{1}{2}\mathrm{B}\left(\frac{1}{4}, \frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \c... | You could also just do modular exponentiation and take note of the periods.
The powers of $9 \mod 100$ are: 1, 9, 81, 29, 61, 49, 41, 69, 21, 89, 1, 9, 81, 29, 61, 49, ... They have a period of $10$. This means that if $n \equiv 9 \mod 10$, then $9^n \equiv 89 \mod 100$. Since $9^9 = 387,420,489$, this means that $9^{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Parabola and line proof Given are three non-zero numbers $a, b, c \in \mathbb{R}$. The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$.
Prove that the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx-b$.
| The parabola with equation $y=ax^2+bx+c$ lies above the line with equation $y=cx$. So
$ax^2 + bx + c = cx \quad \Rightarrow \quad ax^2 + (b - c)x + c = 0 \quad \Rightarrow \quad \Delta = (b - c)^2 - 4ac$.
On the other hand, the parabola with equation $y=cx^2-bx+a$ lies above the line with equation $y=cx - b$. So
$cx^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for x if $z$ is a complex number such that $z^2+z+1=0$ I was given a task to solve this equation for $x$:
$$\frac{x-1}{x+1}=z\frac{1+i}{1-i}$$
for a complex number $z$ such that $z^2+z+1=0$.
Solving this for $x$ is trivial but simplifying solution using the given condition is what's bothering me. Thanks ;)
| $$\frac{x-1}{x+1}=z\frac{1+i}{1-i}=z\frac{(1+i)^2}{(1+i)(1-i)}=zi$$
Applying componendo and dividend, $$x=\frac{1+zi}{1-zi}=\frac{(1+zi)^2}{(1-zi)(1+zi)}=\frac{1-z^2+2iz}{-z}=-2i+z-z^2$$ as $z^2+z+1=0\implies z^3=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$ Find $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\ldots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)$$
| From the Hint given by @cameron Williams, I did the following.
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{1}-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\ldots+\frac{\sqrt{2n-1}-\sqrt{2n+1}}{-2}\right)$$
Now the second and consectives gets cancelled. So,
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What are the members of the set $A=7^{n}+5^{n}(mod35)$ I have this set
$A=${$\ x \ \in \mathbb{N}|\ \exists \ n \ \in \mathbb{N}:$
$x \equiv 7^{n}+5^{n}$ (mod $35$) $ $, $ 35\gt x\ge 0$}
I want to know how many members has this set?
thanks in advance
| By Little Fermat's Theorem, $7^4\equiv 1\pmod5$ and $5^6\equiv1\pmod7$. In fact, $4$ is the order of $7$ modulo $5$ and $6$ is the order of $5$ modulo $7$. For any integer $k$ the number $12k$ is a multiple of $4$ and $6$. Therefore,
$$5^{n+12k}+7^{n+12k}=5^n\cdot(1+7r)+7^n\cdot(1+5r)\equiv 5^n+7^n\pmod{35}$$
We see th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\s... | Hint: $\frac{t^4-t^2}{2-t^2} \equiv -t^2-\frac{2}{t^2-2}-1,$ which is easy to integrate.
You should find, after a suitable hyperbolic substitution, that $\int \frac{2}{t^2-2}dt= \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2}-t}{\sqrt{2}+t} \right|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
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Find the equation of a circle which is tangent to $y$-axis at a given point and cuts a chord of given length on the $x$-axis How to find the equation of the circle which touches $y$ axis at $(0,3)$ and cuts a chord of length $8$ on the $x$ axis?
It should look like this:
My approach:
Since the circle touches $y$ axi... | $\begin{vmatrix}
x^2+y^2&x&y&1\\
0^2+3^2&0&3&1\\
1^2+0^2&1&0&1\\
9^2+0^2&9&0&1\\
\end{vmatrix}
=
\begin{vmatrix}
x^2+y^2&x&y&1\\
9&0&3&1\\
1&1&0&1\\
81&9&0&1\\
\end{vmatrix}=
24(x^2+y^2)-240x-144y+216=
x^2+y^2-10x-6y+9=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(... | $\displaystyle 2f(x)=2\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$
$\displaystyle=\sum_{r=0}^{n}\left\{\tan \frac x{2^r}-\tan \frac x{2^{r+1}}\right\}$ which is telescopic
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding the remainder when a polynomial is divided by another polynomial. Find the remainder when $x^{100}$ is divided by $x^2 - 3x + 2$.
I tried solving it by first calculating the zeroes of $x^2 - 3x + 2$, which came out to be 1 and 2.
So then, using the Remainder Theorem, I put both their values, and so the remaind... | We write the Euclidean division:
$$x^{100}=(x^2-3x+2)Q(x)+ax+b$$
and notice that $1$ and $2$ are roots of $x^2-3x+1$ so
*
*let $x=1$ we get $1=a+b$
*let $x=2$ we get $2^{100}=2a+b$
so we find $a=2^{100}-1$ and $b=2-2^{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Eliminate $\theta$ from the equations $\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$
Eliminate $\theta$ from the equations
$$\frac{\cos(\alpha-3\theta)}{\cos^3\theta}=\frac{\sin(\alpha-3\theta)}{\sin^3\theta}=m$$
Ans: $m^2+m\cos\alpha-2=0$.
I tried using the following two ide... | Using $\displaystyle\sin(A-B),\cos(A-B)$ and on rearrangement we have $$\cos\alpha\cos3\theta+\sin\alpha\sin3\theta-m\cos^3\theta=0\ \ \ \ (1)$$
$$\cos\alpha\sin3\theta-\sin\alpha\cos3\theta+m\sin^3\theta=0\ \ \ \ (2)$$
Solving for $\displaystyle\cos\alpha,\sin\alpha,$
$\displaystyle\dfrac{\cos\alpha}m=\cos^3\theta\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
| You should know the Taylor series for $e^x$:
$e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots$
Then, the Taylor series for $e^{-x}$ is:
$e^{-x} = 1 + (-x) + \dfrac{(-x)^2}{2} + \dfrac{(-x)^3}{3!} + \dfrac{(-x)^4}{4!} + \cdots$
$e^{-x} = 1 - x + \dfrac{x^2}{2} - \dfrac{x^3}{3!} + \dfrac{x^4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Inte... | In general, with $t=\sqrt[n]{\tan x}$
\begin{align}
&\int\sqrt[n]{\tan x}\>dx=n\int \frac{t^n}{1+t^{2n}}dt\\
=&\>\frac14
\sum_{k=1}^{2n} (-1)^{k-1}\left(\sin a_k \ln(x^2-2x\cos a_k+1)-2\cos a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k}\right) +C
\end{align}
where $a_k=\frac{(2k-1)\pi}{2n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c... | By C-S
$$\left(\sum_{cyc}\sqrt{a^2+bc}\right)^2\leq\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(2a^2+b^2+c^2+bc).$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^2+bc}{2a^2+b^2+c^2+bc}\sum_{cyc}(4a^2+bc)\leq\frac{9(a+b+c)^2}{4}$$ or
$$\sum_{cyc}(2a^8+2a^7b+2a^7c+9a^6b^2+9a^6c^2-23a^5b^3-23a^5c^3+20a^4b^4+11a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Solve The Triangle I am having a tough time trying to solve this problem.
I have utilized the 30, 60, 90 triangle measures for the length of sides. However, I am stuck since the side that would be √3 has 100 as its length. How do I solve then?
This is what I have done so far:
| To find $x$ here, you will need to look at this as 2 separate triangles, and find the base of each separate triangle, then add the bases up to get $x$.
For the triangle on the left hand side, we can use the $\tan(x)$ function to relate the $60^{\circ}$ angle with the opposite side of length 100, and our adjacent side, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Proving $\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1$. By testing in maple I found that
$$
\int_0^1 \frac{\mathrm{d}x}{1-\lfloor \log_2(1-x)\rfloor} = 2 \log 2 - 1
$$
Does there exists a proof for this? I tried rewriting it as an series but no luck there.
$$
\frac{1}{1-\lfloor \log_2(1-x)... | Start by noting that $\lfloor\log_2(1-x)\rfloor = -(n+1)$ for all $x \in (1-2^{-n},1-2^{-(n+1)})$.
Therefore, $\displaystyle\int_0^1 \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor} = \sum_{n = 0}^{\infty}\int_{1-2^{-n}}^{1-2^{-(n+1)}} \dfrac{\,dx}{1-\lfloor \log_2(1-x)\rfloor}$
$= \displaystyle\sum_{n = 0}^{\infty}\int_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$ What tools or ways would you propose for getting the closed form of this integral?
$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
EDIT: It took a while since I made this post. I'll give a little bounty for the solver of th... | Just a few notes for a series development, because a "closed" formula is very unlikely.
$$\int\limits_0^{\pi/4} \frac{\ln(1-x) \tan^2 x}{1-x\tan^2 x} dx = -\sum\limits_{n=0}^\infty \sum\limits_{k=0}^n \frac{1}{n-k+1} \int\limits_0^{\pi/4} x^{n+1} (\tan x)^{2k+2} dx$$
Or using $\int\limits_0^{\pi/4} \frac{\ln(1-x)}{x (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "82",
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"answer_id": 0
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Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
| This is nonhomogeneneous difference equations. First solve the homogeneous equation
$$
a_n - a_{n-1} - 2a_{n-2} = 0 \quad (1)
$$
Let $a_n = r^n$, so that $a_{n-1} = r^{n-1}$ and $r_{n-2} = r^{n-2} = 0$. Replacing in (1), we have
$$
r^n - r^{n-1} - 2r^{n-2} = 0 \quad r^2 - r - 2 = 0 \quad \Rightarrow \ r_1 = -1, \ r_2 =... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Verify algebraically that the equation $\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$ is an identity I am stuck when I get to this point $\frac{\cos^2(x)}{\sin(x)}$.
Am I on the right track?
Verify algebraically that the equation is an identity:
$$\frac{\cos(x)}{\sec(x)\sin(x)}=\csc(x)-\sin(x)$$
My work:
$$\frac{... | General hints:
Agree on only using $\sin$ and $\cos$ by substituting
$$\csc(x) = \frac1{\sin(x)}, \qquad \sec(x) = \frac1{\cos(x)}$$
Then get rid of any fractions by multiplying with the denominator and finally use
$$\sin^2(x) + \cos^2(x) = 1$$
$$\begin{align*}
\frac{\cos x}{\sec x \sin x} & = \frac{\cos x}{\frac1{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$ This problem is from my Math Challenge II Algebra class, and it's really confusing. How can you factor something like this? Here's the question again: Factor $3x^2-11xy+6y^2-xz-4yz-2z^2$.
| First, split up the expression into pieces.
\begin{align*}
F(x,y,z) &= 3x^2-11xy+6y^2-xz-4yz-2z^2 \\
&= 3x^2\underbrace{-11xy-xz}_{G(x,y,z)}+\underbrace{6y^2-4yz-2z^2}_{H(y,z)}
\end{align*}
Start with the easy one : $H(y,z)$.
\begin{align*}
H(y,z) &= 6y^2-4yz-2z^2 \\
&= 6y^2-6yz+2yz-2z^2 \\
&= 6y(y-z)+2z(y-z) \\
&= 2(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that this expression is a perfect square? Show that this expression is a perfect square?
$(b^2 + 3a^2 )^2 - 4 ab*(2b^2 - ab - 6a^2)$
| In fact, note that $(x + y - z)^2 = x^2 + y^2 + z^2 + 2xy - 2yz - 2xz$. Thus,
$$
(3a^2 + b^2)^2 - 8ab^3 + 4a^2b^2 + 24a^3b = 9a^4 + 10a^2b^2 + b^4 - 8ab^3 + 24a^3b
$$
$$
= 9a^4 + (16a^2b^2 - 6a^2b^2) + b^4 + 24a^3b - 8ab^3
$$
$$
= 9a^4 + 16a^2b^2 + b^4 + 24a^3b - 6a^2b^2 - 8ab^3
$$
$$
= (3a^2)^2 + (4ab)^2 + (-b^2)^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $ How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?
| If we abbreviate $w=\sqrt[3]2$, the left hand side is $L=\frac1{\sqrt[3]9}(1-w+w^2)$. From $(1+w)(1-w+w^2)=1+w^3=3$ we see that $L=\frac3{\sqrt[3]9(1+\sqrt[3]2)}$, hence
$$ L^3=\frac{27}{9(1+w)^3}=\frac3{1+3w+3w^2+w^3}=\frac1{1+w+w^2}$$
As above, note that $(1+w+w^2)(w-1)=w^3-1=1$, hence
$$ L^3=w-1=R^3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to evaluate the limit $\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}}$ How to evaluate the limit as it approaches infinity
$$\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}}$$
| Hint:
$\displaystyle \frac{2^x + 1}{2^{x+1}} = \frac{2^x}{2^{x+1}} + \frac{1}{2^{x+1}} = \frac{1}{2} + \frac{1}{2^{x+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Trigonometric relation between sides and angles of a triangle $$a \cdot \sin (B-C) +b \cdot \sin(C-A) +c \cdot \sin(A-B) =0$$
where $a, b, c$ are the sides of a triangle and $A, B, C$ are the angles of a triangle.
No idea how to solve this problem.
| $$
a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\
= a ( \sin B \cos C - \sin C \cos B) + b(\sin C \cos A - \sin A \cos C) + c (\sin A \cos B - \sin B \cos A)\\
= a \sin B \cos C - a \sin C \cos B + b\sin C \cos A - b \sin A \cos C + c \sin A \cos B - c \sin B \cos A\\
= a \sin B \cos C - b \sin A \cos C + b \sin C \cos A -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac {\sqrt5}{\sqrt3+1} - \sqrt\frac{30}{8} + \frac {\sqrt {45}}{2}$ I am trying to find the value of: $$\frac {\sqrt5}{\sqrt3+1} - \sqrt\frac{30}{8} + \frac {\sqrt {45}}{2}$$
I have the key with the answer $\sqrt 5$ but am wondering how I can easily get to that answer?
I realize this is a very basic questi... | $$\begin{align}
\frac{\sqrt{5}}{\sqrt{3}+1} - \sqrt{\frac{30}{8}} + \frac{\sqrt{45}}{2} &= \frac{\sqrt{5}(\sqrt{3}-1)}{(\sqrt{3}+1)((\sqrt{3}-1))} - \sqrt{\frac{15}{4}} + \frac{\sqrt{9·5}}{2}\\
&= \frac{\sqrt{5}(\sqrt{3}-1)}{2} - \frac{\sqrt{15}}{2} + \frac{3\sqrt{5}}{2}\\
&= \frac{\sqrt{15}-\sqrt{5} -\sqrt{15} + 3\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Why does this infinite series equal one? $\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$ Why does
$$\sum_{k=1}^\infty \binom{2k}{k} \frac{1}{4^k(k+1)}=1$$
Is there an intuitive method by which to derive this equality?
| Your series is a telescoping one, since:
$$\begin{eqnarray*}\frac{1}{4^{k+1}}\binom{2k+2}{k+1}-\frac{1}{4^k}\binom{2k}{k}&=&\frac{1}{4^{k+1}}\binom{2k}{k}\left(\frac{(2k+2)(2k+1)}{(k+1)^2}-4\right)\\&=&-\frac{1}{2(k+1)4^{k}}\binom{2k}{k},\end{eqnarray*}$$
hence:
$$\sum_{k=1}^{+\infty}\binom{2k}{k}\frac{1}{4^k(k+1)}=2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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How are these two integrals related? How to express the integral $$\int_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx $$
in terms of the integral $$ \int_{-1}^{1} \sqrt{1-x^2} \ dx?$$ I know that the latter integral is equal to $\pi / 2$.
We can't use substitution. We can only use the following two results:
*
*$$ \int_{a}^{b}... | Using $k=\frac{1}{2}$ in the second rule, we get
$$\begin{align}
\int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx
\\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx
\\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx
\\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/880799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Vectors with given angle and magnitude Give an example of vectors $\mathbf{v}$ and $\mathbf{w}$ such that the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\frac{2\pi}{3}$ and $\|\mathbf{v} \text{ x } \mathbf{w}\|=\sqrt{3}$.
Should I just use something random for $\mathbf{v}$ and then solve for $\mathbf{w}$?
Thank y... | Let $v = (1,0,0)$ and $w = (a,b,0)$.
$$
v\cdot w = \mid\mid v \mid\mid.\mid\mid w\mid\mid\cos \dfrac{2\pi}{3} \quad \Rightarrow \quad -2a = \sqrt{a^2 + b^2} \ \Rightarrow \ 3a^2 = b^2 \quad (1)
$$
But,
$$
\sqrt{3} = \mid\mid v \mid\mid.\mid\mid w\mid \mid \sin \dfrac{2\pi}{3} \quad \Rightarrow \quad \sqrt{3} = \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rare... | Let $x=b+c,y=c+a,z=a+b$, then $a=\frac{y+z-x}{2},\dots$. Your inequality becomes
$$(x+y+z)\left(\frac{y+z-x}{x^2}+\cdots\right)\geq 9.$$
Write $y+z-x=(x+y+z)-2x,\dots$ we need to show
$$(x+y+z)^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) -2(x+y+z)\left(\frac1x+\frac1y+\frac1z\right)\geq 9.$$
Use $3\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$
How ... | Hint:
$$\sin^3(x) = \left( \frac{e^{ix}-e^{-ix}}{2i} \right)^3 = \dots\,. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for... | We may rewrite your series in the following manner:
\begin{align}
&\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\\
&=\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{1}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{-1}{6n+6}\right)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
How to prove $(1-\frac1{36})^{25}\lt\frac12$? How to prove the inequality?
$(1-\frac1{36})^{25}\lt\frac12$
I'm in trouble.
Thank you very much for your help
| We have that $\displaystyle\left(1-\frac{1}{36}\right)^{25}=\sum_{r=0}^{25} \binom{25}{r}(-1)^{r}\frac{1}{36^r}=1-\frac{25}{36}+\frac{25\cdot24}{2\cdot36^2}-\frac{25\cdot24\cdot23}{6\cdot36^3}+\frac{25\cdot24\cdot23\cdot22}{24\cdot36^4}-\cdots$.
Since the sum is alternating and has terms that are decreasing in absolute... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 8
} |
Limit of a definite integral We need to calculate $$\lim_{x \to 0}\int_{\sin x}^{x}\frac{dt}{t^3(1+\sqrt{t})}$$
Integral itself doesn't seem to be the problem here. When making a substitution $\sqrt{t}=u$, we get $$\lim_{x \to 0}2\int_{\sqrt{\sin x}}^{\sqrt{x}}\frac{du}{u^5(1+u)}=2\lim_{x \to 0}\int_{\sqrt{\sin x}}^{\s... | Your proof is nice but there's more simple:
Let consider this simplified integral
$$\int_{\sin x}^x\frac{dt}{t^3}=-\frac12t^{-2}\Bigg|_{\sin x}^x=-\frac12\left(\frac1{x^2}-\frac1{\sin^2x}\right)\sim_0-\frac12\frac{(x-\frac{x^3}{6})^2-x^2}{x^4}\sim_0\frac16$$
and now we prove that the two integrals have the same limit b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximum and minimum of $z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$ Find the maximum and minimum of the function
$$z=\frac{1+x-y}{\sqrt{1+x^2+y^2}}$$
I have calculated
$\frac{\partial z}{\partial x}=\frac{1+y^2+xy-x}{(1+x^2+y^2)^{\frac{3}{2}}}$
$\frac{\partial z}{\partial y}=\frac{-1-x^2-xy-y}{(1+x^2+y^2)^{\frac{3}{2}}}$
I have... | Another solution for solving your two equations $$1+y^2+xy-x=0$$ $$1+x^2+xy+y=0$$ From the first, you can extract $$x=-\frac{y^2+1}{y-1}$$ and put it in the second, which after vsimplification reduces to $$\frac{2 \left(y^3+1\right)}{(y-1)^2}=0$$ So, $y=-1$ and back to the definition of $x$, $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Chinese Remainder Theorem, redundant information I want to solve the following system of congruences:
$ x \equiv 1 \mod 2 $
$ x \equiv 2 \mod 3 $
$ x \equiv 3 \mod 4 $
$ x \equiv 4 \mod 5 $
$ x \equiv 5 \mod 6 $
$ x \equiv 0 \mod 7 $
I know, but do not understand why, that the first two congruences are redundant. Why i... | $x \equiv 1 \mod 2$
$x \equiv 2 \mod 3$
$x \equiv 3 \mod 4 \implies x \equiv 1 \pmod 2$
$x \equiv 4 \mod 5$
$x \equiv 5 \mod 6 \iff
\left.\begin{cases}
x \equiv 1 \mod 2 \\
x \equiv 2 \mod 3
\end{cases} \right\}$
By the CRT.
$x \equiv 0 \mod 7$
So first we replace $x \equiv 5 \mod 6$ with
$\left.\begin{cases}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a_1a_2\cdots a_n=1$, then the sum $\sum_k a_k\prod_{j\le k} (1+a_j)^{-1}$ is bounded below by $1-2^{-n}$ I am having trouble with an inequality. Let $a_1,a_2,\ldots, a_n$ be positive real numbers whose product is $1$. Show that the sum
$$ \frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a... | Note that for that every positive integer $i$ we have
\begin{eqnarray}
\frac{a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} & = & \frac{1 + a_i}{(1+a_1)(1+a_2) \cdots (1+a_i)} - \frac{1}{(1+a_1)(1+a_2) \cdots (1+a_i)} \nonumber \\
& = & \frac{1}{(1+a_1) \cdots (1+a_{i-1})} - \frac{1}{(1+a_1) \cdots (1+a_i)}. \nonumber
\end{eqna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
| By Taylor series $$\cos x \sim_0 1 - \frac{x^2}{2}, \,\sin 2x \sim_0 2x\, \text{ and }\,\Big(1-\frac{x^2}{2}\Big)^3 \sim_0 1 - \frac{3x^2}{2}$$ so $$\dfrac{1-\cos^3 x}{x\sin2x} \sim_0 \frac{1 - \Big(1-\frac{x^2}{2}\Big)^3}{2x^2} \sim_0 \frac{ \frac{3x^2}{2} }{2x^2} = \frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Evaluate the $\displaystyle\lim_{x\to 0}\left(\frac{1}{x\sqrt{1+x}}-\frac 1x\right)$ Evaluate the limit as $x$ approaches $0$ : $$\frac{1}{x\sqrt{1+x}}-\frac 1x$$
I have so far $$\frac{x -x\sqrt{1+x}}{x(x\sqrt{1+x})}$$
Do I multiply the numerator and denominator by the conjugate of $x - x\sqrt{1+x}$? Or do I multiply t... | You can get
$$\frac{1}{\color{red}{x}\sqrt{1+x}}-\frac{1}{\color{red}{x}}=\frac{1}{x\sqrt{1+x}}-\frac{\sqrt{1+x}}{x\sqrt{1+x}}=\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}.$$
Then, multiply it by $$\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Max of $3$-Variable Function I'm trying the find the maximum of the function
$$f(a,b,c)=\frac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$$
for all nonnegative real numbers $a, b, c$ with $ab + bc + ca > 0$.
I tried in vain to prove that $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
| WLOG,let $c=$Min{$a,b,c$},
$f(a,b,c) \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}}=\dfrac{2ab}{\sqrt{ab}(a+b+\sqrt{a^2+b^2})} \le \dfrac{2ab}{\sqrt{ab}(2\sqrt{ab}+\sqrt{2ab})}=2-\sqrt{2}$
$\dfrac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}} \le \dfrac{a+b-\sqrt{a^2+b^2}}{\sqrt{ab}} \iff \sqrt{ab}(a+b+c-\sqrt{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A cotangent series related to the zeta function $$\sin x = x\prod_{n=1}^\infty \left[1-\frac{x^2}{n^2\pi^2}\right]$$
If you apply $\log$ to both sides and derivate:
$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right]$$
I have to make the expansion:
$$\frac{1}{1-... | Actualy (see [1] pg. 75)
$$
\cot(z)=\frac{1}{z}-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}z^{2n-1}\textrm{, }|z|<\pi
$$
and ([1] pg.807)
$$
\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n-1}}{2(2n)!}B_{2n}\textrm{, }n\in\textbf{N}
$$
where $B_{2n}$ are the Bernoulli numbers i.e.
$$
\frac{x}{e^x-1}=\sum^{\infty}_{n=0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
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Convergence of series with root Given
$$\sum_{n=1}^\infty {((-1)^n + \alpha^3) (\sqrt{n+1} - \sqrt{n})}$$
find all values of $\alpha$ such that the series converges.
My try:
By multiplying the series with the expression
$$\frac {\sqrt{n+1} + \sqrt n} {\sqrt{n+1} + \sqrt n}$$
we get
$$\sum_{n=1}^\infty {((-1)^n + \alph... | You may write
$$
\begin{align}
\sum_{n=1}^N {((-1)^n + \alpha^3) (\sqrt{n+1} - \sqrt{n})} & =\sum_{n=1}^N {\frac{(-1)^n}{\sqrt{n+1} + \sqrt{n}}+ \alpha^3 \sum_{n=1}^N (\sqrt{n+1} - \sqrt{n})} \\&= \sum_{n=1}^N {\frac{(-1)^n}{\sqrt{n+1} + \sqrt{n}}+ \alpha^3 (\sqrt{N+1}-1)}
\end{align}
$$
and, as $N \rightarrow \infty$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/892951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove $n|2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)$ Question:
Today, when I solve other problem, I found this follow interesting result
$$n\mid\left(2^{\frac{n(n-1)}{2}}\cdot (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\right),n\ge 1$$
It is clear $n=1$ is true.
$n=2$
$$2\mid2^{\frac{n(n-1)}{2}}\cd... | Let $n=2^st$ where $t$ is odd. If $t>1$ then
$$t\mid 2^{\phi(t)}-1\ ,$$
and $\phi(t)$ is one of the numbers $1,2,\ldots,n-1$, so
$$t\mid (2-1)(2^2-1)(2^3-1)\cdots (2^{n-1}-1)\ ;$$
obviously this is also true when $t=1$. Also, $s<n\le\frac{1}{2}n(n-1)$, so
$$2^s\mid 2^{n(n-1)/2}\ ,$$
and hence
$$n=2^st\mid 2^{n(n-1)/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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For any base-$n$ number system, what is the average length of a number $\leq100$? By this I mean the amount of symbols (bits / digits / etc. ) for any number from $0$ to $100$. I don't know whether this can be answered, I'm just asking out of interest - not homework or anything.
| Any integer between $n^{d-1}$ and $n^d-1$ in base-$n$ requires exactly $d$ digits. Also $0$ needs $1$ digit.
For instance, in base-$2$:
$0 = 0_2$ through $1 = 1_2$ require $1$ digit each
$2 = 10_2$ through $3 = 11_2$ require $2$ digits each
$4 = 100_2$ through $7 = 111_2$ require $3$ digits each
$8 = 1000_2$ through $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Derivatives of trig polynomials do not increase degree? Let $c = \cos x$ and $s = \sin x$,
and consider a trigonometric polynomial $p(x)$
in $c$ and $s$.
The degree of $p(x)$ is the maximum of
$n+m$ in terms $c^n s^m$.
Is it the case that repeated derivatives of $p(x)$,
expressed again in terms of $c$ and $s$,
nev... | To show this is true for any polynomial of degree $n+m$, it's sufficient to show that this is the case for an arbitrary term $c^ns^m$ - since every term must be of this form for some $n,m$ - and to check that adding such terms together won't affect the derivative.
Note that $$\frac{d}{dx}(c^n s^m)=mc^{n+1}s^{m-1}-nc^{n... | {
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"url": "https://math.stackexchange.com/questions/901071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding multivariable limit I would like to find the following limit $$\lim_{(x,y,z)\to(0,0,0)}\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}.$$
It looks like it would be zero since if we put $M=\max\{x,y,z\}$ and $m=\min\{x,y,z\}$, then $$\Big|\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\Big| \leq \Big|\frac{M^5}{m^4}\Big|$$ so the e... | Let $r = \sqrt{x^2+y^2+z^2}$. Then, using the generalized mean inequality, we have:
$\sqrt[4]{\dfrac{x^4+y^4+z^4}{3}} \ge \sqrt[2]{\dfrac{x^2+y^2+z^2}{3}} \ge \sqrt[3]{|xyz|}$.
Therefore, $x^4+y^4+z^4 \ge 3\left(\dfrac{x^2+y^2+z^2}{3}\right)^2 = \dfrac{r^4}{3}$ and $|xyz| \le \left(\dfrac{x^2+y^2+z^2}{3}\right)^{3/2} =... | {
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"url": "https://math.stackexchange.com/questions/901702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proo... | From Chebyshev's sum inequality we have
\begin{align*}
\frac{x^{n+1}+y^{n+1}+z^{n+1}}{3}\geq \frac{x^n+y^n+z^n}{3}\cdot\frac{x+y+z}{3}.
\end{align*}
By AM-GM we have $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}=1$ and that proves the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int\frac{dx}{x\sqrt{x^2+4}}$ While solving this:
$$
\begin{align}
\int\frac{dx}{x\sqrt{x^2+4}}
&=\int\frac{t(-1/t^2)dt}{\sqrt{(1/t)^2+4}}\tag{t=1/x}\\
&=\int\frac{(-1/t)dt}{(1/t)\sqrt{1+4t^2}}\\
&=-\frac12\int\frac{dt}{\sqrt{t^2+1/4}}\\
&=-\frac12\ln(t+\sqrt{t^2+1/4})+C\\
&=-\frac12\ln(1/x+\sqrt{1/x^2+1/4})+... | You can see that your answer is correct and that the given answer is wrong because
$$\left(\frac 12\left(\ln (x)-\ln\left(2+\sqrt{4+x^2}\right)\right)\right)'=\frac{1}{x\sqrt{x^2+4}}$$
$$\left(\ln\left[(x+1/2)+\sqrt{x^2+x+1}\right]\right)'=\frac{1}{\sqrt{x^2+x+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{N\to \infty} \sum_{n=N}^{2N} c_n = 0$ $\Rightarrow \sum_{n=1}^{\infty} c_n$ converges? If $\lim_{N\to \infty} \sum_{n=N}^{2N} c_n = 0$ do we have that $\sum_{n=1}^{\infty} c_n$ converges? At first this did not seem true($\sum_{n=N}^{2N} (-1)^n$ is $0$ when N is odd), but I've failed to find a proper counter-exam... | Let
\begin{align*}
c_1 = c_2 &= \frac{1}{2}\\
& \\
c_3 = c_4 = c_5 = c_6 &= \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}\\
& \\
c_7 = c_8 = c_9 = c_{10} = c_{11} = c_{12} = c_{13} = c_{14} &= \frac{1}{8}\times\frac{1}{3} = \frac{1}{24}
\end{align*}
and so on. More precisely, let $c_n = \dfrac{1}{2^kk}$ where $k = \lfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.
| Consider the picture below.
On the left hand side we have the original picture which we have put in the complex plane, and on the right hand side is its image under the Möbius transformation $f(z) = \frac{z}{2-z}$. The inverse transformation is $g(z) = \frac{2z}{z+1}$. In the new coordinates the green circles have rad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 2,
"answer_id": 0
} |
Test the convergence of a series To test the convergence of a series:
$$
\sum\left[\sqrt[3]{n^3+1}-n\right]
$$
My attempt:
Take out $n$ in common: $\displaystyle\sum\left[n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\right]$.
So this should be divergent.
But, the given answer says its convergent.
| $$ \sqrt[3]{n^3+1}-n = \left({\sqrt[3]{n^3+1}-n)}\right) \frac{ \left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} = \frac {1}{\left({ (n^3 + 1)^{\frac 2 3} + (n^3 + 1)^{\frac 1 3}n + n^{2 } }\right)} \le \frac{1}{(n^3 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/903928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
What is the maximum value of $ \sin x \sin {2x}$ What is the maximum value of $$ \sin x \sin {2x}$$
I have done my work here
$$f (x)=\sin x \sin 2x =\frac{\cos x - \cos3x}2 $$
$$f'(x)= \frac{- \sin x+3 \sin 3x}2 =4\sin x (2-3\sin^2 x)=0$$
$$x=0,\pi; \sin x= \sqrt{\frac {2}{3}}$$
$$f (0)=f (\pi)=0$$
$$f \left(\arcsin \... | We have $$f(x) = \sin (x)\sin (2x) = 2{\sin ^2}(x)\cos (x) = 2(\cos (x) - {\cos ^3}(x))$$for a maximum we should have $f'(x) = 0$, so that $$ - \sin (x) + 3\sin (x){\cos ^2}(x) = 0$$so finally$$\left\{ \begin{array}{l}\sin (x) = 0 \to f(x) = 0\\\cos (x) = \frac{1}{{\sqrt 3 }} \to f(x) = \frac{4}{{3\sqrt 3 }}\\\cos (x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
For how many integral value of $x\le{100}$ is $3^x-x^2$ divisible by $5$? For how many integral value of $x\le{100}$ is $3^x-x^2$ divisible by $5$?
I compared $3^x$ and $x^2$ in $\mod {5}$ i found some cycles but didn't get anything
| HINT : For $n\in\mathbb N$,
$n$ can be divided by $5$ $\iff$ The right-most digit of $n$ is either $0$ or $5$.
You'll find some patterns in the followings from $x=1$ to $x=20$.
The right-most digit of $3^x$ : $3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1.$
The right-most digit of $x^2$ : $1,4,9,6,5,6,9,4,1,0,1,4,9,6,5,6,9,4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find this $\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{x-z}$ minimum of the value let $x,y,z\in R$,and such $x>y>z$,and such
$$(x-y)(y-z)(x-z)=16$$
find this follow minimum of the value
$$I=\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}$$
My idea: since
$$\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{x-z}=\dfrac{x-z}{(x-y)(y-z)}... | Let $\displaystyle a=x-y,b=y-z, c=x-z\implies abc=16$ and $\displaystyle a+b-c=0\iff c=a+b$
$$\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}=\dfrac{\dfrac{16}c+c(a+b)}{16}=\dfrac{\dfrac{16}c+c(c)}{16}$$
Now use Second Derivative Test
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Primitive of $\frac{3x^4-1}{(x^4+x+1)^2}$ How to find primitive of:
$$\frac{3x^4-1}{(x^4+x+1)^2}$$
I am having a faint idea of a type which may or maynot be in the primitve, i.e.:
$$\frac{p(x)}{x^4+x+1}$$
The problem is I am not getting an idea of a substitution to solve this problem.
I might show my work but it is tot... | $$ \int \frac{3x^4-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2}$$
$$ \int \frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2} = \int \frac{3x^4+4x^3}{(x^4+x+1)^2}- \int \frac{4x^3+1}{(x^4+x+1)^2}$$
Consider,
$$ \int \frac{3x^4+4x^3}{(x^4+x+1)^2} = \int \frac{4x^3(x+1)-x^4}{(x^4+x+1)^2} = \int \frac{(4x^3+1)(x+1)+(-1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the sum $1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + \dots + n\cdot 10^n$ How can I calculate
$$1\cdot 10^1 + 2\cdot 10^2 + 3\cdot 10^3 + 4\cdot 10^4+\dots + n\cdot 10^n$$
as a expression, with a proof so I could actually understand it if possible?
| Put
$$ S_n = \sum_{k=1}^{n} k\cdot 10^k.$$
Then:
$$ 9 S_n = (10-1)S_n = \sum_{k=1}^{n}k\cdot 10^{k+1}-\sum_{k=1}^{n}k\cdot 10^k=\sum_{k=2}^{n+1}(k-1)\cdot 10^k-\sum_{k=1}^{n}k\cdot 10^k$$
hence:
$$ 9 S_n = n\cdot 10^n -\sum_{k=1}^n 10^k = n\cdot 10^n-\frac{10^{n+1}-10}{9},$$
so:
$$ S_n = \color{red}{\frac{10}{81}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I w... | $$ (1+\dfrac{1}{16})^{16}
= \sum \frac{1}{16^k}\binom{16}{k}
= \sum \frac{1}{ 16^k}\frac{16!}{(16-k)!} \frac{1}{k!} < 1 + 1 + \frac{1}{2} + \frac{1}{6} + \dots <\dfrac{8}{3} + \frac{2}{4!}$$
How do we know the remaining terms are small enough? Let's try an inequality. In our case, $n=4$.
$$ \sum_{k=n}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
} |
When is a sum of products of positive powers of 2 and 3 divisible by $2^b-3^n$? Here we have a really tough exercise.
Find all natural solution:
$$\frac{\sum\limits_{k=1}^n 2^{a_k} 3^{n-k}}{c}+3^n=2^{b} ,\quad
b\geq a_n; \quad a_k, b, c ,n\in \mathbb N $$
Any ideas, hints?
| $$2^a3^{n-1}+2^{2a}3^{n-2}+\cdots+2^{an}3^0=c(2^b-3^n)$$
$$3^n\left(\left(\frac{2^a}3\right)^1+\left(\frac{2^a}3\right)^2+\left(\frac{2^a}3\right)^3+\cdots+\left(\frac{2^a}3\right)^n\right)=c(2^b-3^n)$$
$$3^n\left(\frac{\frac{2^a}3((2^a/3)^n-1)}{2^a/3-1}\right)=c(2^b-3^n)$$
$$2^a\left(\frac{2^a-3^n}{2^a-3}\right)=c(2^b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| Slightly modifying the answer from @evinda:
$\begin{align}
\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}&=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}} \\
&=\frac{1}{\sqrt{1+ \sqrt{\frac{n+\sqrt{n}}{n^2}}}} \\
&=\frac{1}{\sqrt{1+ \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Short form of few series Is there a short form for summation of following series?
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$$
$$\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}$$
$$\sum\limits_{n=0}^\infty\s... | For $\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$ ,
$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{2k+1}}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^kn!\alpha^{2n}(2y-1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Squeeze Theorem: Finding the limit of a trig function I'm stuck on finding the limit of a complex fraction/trig function. Could someone please assist, or point out where I'm going wrong?
Determine
$$\lim\limits_{x \to 0} \frac{(x+1)\cos(\ln(x^2))}{\sqrt{(x^2+2)}}$$
For all $x$: $$-1 \le \cos(\ln(x^2)) \le 1$$
Multip... | $$f(x)=\frac{(x+1)\cos(\ln(x^2))}{\sqrt{x^2+2}}$$
Take two sequences tending to $0$:
$$\begin{array}{}
a_n=\exp\left(\tfrac{\pi}{4}(1-2n)\right),&\lim\limits_{n\to\infty} a_n=0\\
b_n=\exp(-n\pi),&\lim\limits_{n\to\infty} b_n=0
\end{array}$$
We have
$$f(a_n)=\frac{(a_n+1)\cos\left(\ln\left(\exp\left(\tfrac{\pi}{4}(1-2n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is sum of square of primes a square of prime? I would like to know if it has been proved that :
*
*There are no $a$, $b$ and $c$, all prime numbers, such that $a^2 + b^2 = c^2$
*There are no $a$, $b$, $c$ and $d$, all prime numbers, such that $a^2 + b^2 + c^2 = d^2$
*There are no $a$, $b$, $c$, $d$ and $e$, all pr... | It turns out that the (full) solution is longer than a comment. So here it goes.
Suppose that for $a,b,c,d$ primes,
$$a^2+b^2+c^2=d^2$$
Obviously $d=3$ is not a solution. That implies
$$a^2+b^2+c^2\equiv 1\pmod{3}.$$
So exact two of $a,b,c$ are equal to $3$. We are left to solve
$$d^2-a^2=(d-a)(d+a)=18.$$
But $d-a$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdot... | Your sum is equal to:
\begin{align}
\sum_{i=1}^{\infty} \left ( \frac{1}{3i-1}+\frac{1}{3i+1}-\frac{2}{3i}\right ) &=\sum_{i=1}^{\infty} \frac{(3i+1) \cdot 3i+3i \cdot (3i-1)-2(3i-1) \cdot (3i+1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{9i^2+3i+9i^2-3i-2(9i^2-1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=... | {
"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": 1
} |
Prove $a^3+b^3+c^3\ge a^2+b^2+c^2$ if $ab+bc+ca\le 3abc$
if $a,b,c$ are positive real numbers and $ab+bc+ca\le 3abc$ Prove:$$a^3+b^3+c^3\ge a^2+b^2+c^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: We are given $a... | First
$$3abc\ge ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}.$$
So $abc\ge 1$.
It's then follows that $$a+b+c\ge 3\sqrt[3]{abc}\ge 3.$$
Now
$$3(a^3+b^3+c^3)\ge (a^2+b^2+c^2)(a+b+c)$$
by Rearrangement Inequality or by
$$(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2\ge (a^2+b^2+c^2)\frac{(a+b+c)^2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$
If $x$, $y$ and $z$ are positive numbers,prove:
$$(x+y)(y+z)(z+x)\ge\frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}.$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so... | Well I think i figured it out.by following what I have written in my post:$$LHS\ge 3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$$
For Proving $3(\sum_{cyc}x)\sqrt[3]{x^2y^2z^2} - xyz \ge \frac{8}{3}(x+y+z)\sqrt[3]{x^2y^2z^2}$ :$$\sqrt[3]{x^2y^2z^2}(3(\sum_{cyc}x) - \sqrt[3]{xyz})\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$ it has two singular point $x_0 =1$ and $x_0=1$
I rearrange the equation to
$$y'' - \frac{1}{4x(1-x)}y =0$$
and by $v(x)=\frac{(x-x_0)^2}{4x(1-x)}$ are analytic at $x_0$ for both $0$ and $1$;
that is, $x_0=1,0$ are regular singular points,
so that, I can find ... | $$a_n=\frac{(4*2*1+1)(4*3*2+1)...(4(n-1)(n-2)+1)}{4^{n-1}n!(n-1)!}=\frac{4^{n-2}(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4^{n-1}n!(n-1)!}=\frac{(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4*n!(n-1)!}$$
Thanks, I just tried it again..for this part
$$I=(2*1+\frac{1}{4})(3*2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Two overlapping squares $ABCD$ is a square. $BEFG$ is another square drawn with the common vertex $B$ such that $E,\ F$ fall inside the square $ABCD$. Then prove that $DF^2=2\cdot AE^2$.
|
$(NB)^2+(EN)^2=(EB)^2=y^2$$(NB)^2=y^2-(EN)^2$$(x-AN)^2=y^2-(y\sin\theta)^2=(y\cos\theta)^2$$x-AN=\pm y\cos\theta$$AN=x-y\cos\theta(\because y,\cos\theta \text{ both positive})$$(AE)^2=(AN)^2+(EN)^2$$(AE)^2=(x-y\cos\theta)^2+(y\sin\theta)^2$$(AE)^2=x^2-2xy\cos\theta+y^2\cos^2\theta+y^2\sin^2\theta=x^2-2xy\cos\theta+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Question about initial value problem This seems like it should be something simple.
While solving some HW problems I ran across this:
$dy/dx = ({1-2x})/y$
which I separated and integrated:
$\int{Ydy} = \int{1-2x}dx$
$y^2/2= x-x^2 + c$
$y^2=2(x-x^2+c)$
$y=\sqrt{2(x-x^2+c)}$
$y=\sqrt{2x-2x^2+2c}$
I was given $y(1)=-2$ ... | $$\frac {y^2}{2} = -x^2+x+c \iff y=-\sqrt2 \sqrt{-x^2+x+c}$$ or $$y=\sqrt2 \sqrt{-x^2+x+c}$$
The first one when solved for the initial condition gives $c=2$ while the second one has no solution, thus can't satisfy the initial condition, ergo no solution exists. So you're left with only $$y=-\sqrt2 \sqrt{-x^2+x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/918557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Why does $\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$ Why does $$\lim_{x\rightarrow\infty} x-x^{\frac{1}{x}^{\frac{1}{x}}}-\log^2x=0?$$
Moreover, why is $$x-x^{\frac{1}{x}^{\frac{1}{x}}}\approx\log^2 x?$$
| By a Taylor expansion,
$$
\left( \frac{1}{x} \right)^{1/x} = \exp\left( -\frac{1}{x}\ln x \right) = 1-\frac{\ln x}{x} + O\left(\frac{\ln^2 x}{x^2}\right) .
$$
Thus the second term equals
$$
(1+\delta)\exp\left( \ln x \left( 1 - \frac{\ln x}{x} \right) \right) = x \left( 1-\frac{\ln^2 x}{x} + O\left(\frac{\ln^4 x}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a ... | $$\frac{1}{N+1}+\frac{1}{N(N+1)}=\frac{N}{N(N+1)}+\frac{1}{N(N+1)}=\frac{N+1}{N(N+1)}=\frac{1}{N}.$$
Here's a question for further investigation: is the above decomposition of a unit fraction into a pair of distinct unit fractions unique? After all, there is more than one way to split a unit fraction into a triplet of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seem... | HINT
Multiply by a special kind of $1$ :
$$\lim\limits_{x \to \infty}\left[x(\sqrt {x^2+a} - \sqrt {x^2+b})\right]=\lim\limits_{x \to \infty}\left[ \dfrac{x(\sqrt {x^2+a} - \sqrt {x^2+b})(\sqrt {x^2+a} + \sqrt {x^2+b})}{\sqrt {x^2+a} + \sqrt {x^2+b}}\right] $$
and use difference of squares
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How to integrate $\sqrt{1+(2/3)x}$? How would you solve the following (step by step please!):
$$\int^6_5\sqrt{1+\frac23x}\ dx$$
I started with $u=1+\frac23x$, $du=\frac23\,dx$, now what?
| You have $$\int^6_5\sqrt{1+\frac23x}\ dx$$
Now let $u = 1 + \frac{2}{3}x$, and $du = \frac{2}{3}\ dx$, so that means $dx = \frac{3}{2}du$, so make the subsitution:
$$\int^{6 = x}_{5=x}\sqrt{u}\frac{3}{2}\ du = \frac{3}{2} \int^{6 = x}_{5 = x}u^{\frac{1}{2}}\ du$$
Now we need to find the bounds for $u$, and then we will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $x^2 \cos(a x)\; \mathrm{d}x$ I am trying to solve the following problem:
$\int x^2 \cos(a x)\; \mathrm{d}x$
I thought this would be simple and I am pretty sure this is the answer:
$I =\frac{x^2\sin(ax)}{a}+\frac{2x\cos(ax)}{a^2}+\frac{2\sin(ax)}{a^3}$
I have been told that this isn't the correct answer... | $$\int x^2\cos(ax)dx=|u=x^2\Rightarrow du=2xdx, dv=\cos(ax)dx\Rightarrow v=\frac{1}{a}\sin(ax)|=$$
$$=\frac{x^2}{a}\sin(ax)+\frac{2}{a}\int x\sin(ax)dx=\frac{x^2}{a}\sin(ax)+\frac{2}{a}I_1$$
$$I_1=\int x\sin(ax)dx=|u=x\Rightarrow du=dx, dv=\sin(ax)dx\Rightarrow v=\frac{1}{a}\cos(ax)|=$$
$$=\frac{x}{a}\cos(ax)+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/923310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$ I don't know how to solve this problem. What I know is $\sqrt[3]{2 \sqrt{2}}=\sqrt{2}$. But I don't know how to continue.
| $$\sqrt[3]{2 \sqrt{2}}=\sqrt2$$
$$\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt 2},\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt 2}$$
$$\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)=\sqrt[3]{2^{3/2}}\left(\frac{\sqrt{3}-1}{\sqrt 2}+\frac{\sqrt{3}+1}{\sqrt 2}\right)=$$
$$=\sqrt 2\frac{\sqrt3-1+\sqrt3+1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $ \int \frac{dx}{x\sqrt{1-x^4}}$ Find $\displaystyle \int \dfrac{dx}{x\sqrt{1-x^4}}$
I cannot figure out how start this problem, can anyone explain
| $$
\begin{aligned}
\int\frac{\mathrm{d}x}{x\sqrt{1-x^4}}&=\int\frac{\mathrm{d}x}{x^3\sqrt{1/x^4 - 1}}\\
&=\int\frac{1}{\sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\
&=\int\frac{1/x^2+\sqrt{(1/x^2)^2 - 1}}{\sqrt{(1/x^2)^2 - 1}}\,\frac{1}{1/x^2 + \sqrt{(1/x^2)^2 - 1}}\frac{\mathrm{d}x}{x^3}\\
&=-\frac{1}{2}\int \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding x using the pythagoras theorem $$x^2 = (x+1)^2 + (x-7)^2$$
can someone please find $x$? Also this is a quadratic equation problem solving question.
| As in user$169852$'s answer
$$x^2 = (x+1)^2 + (x-7)^2 \quad
\implies x^2 - 12 x + 50 = 0$$
and the quadratic formula shows that $x$ is complex.
$$x=\dfrac{12\pm\sqrt{144-4(1)50}}{2(1)}
=
\dfrac{12\pm2i\sqrt{14}}{2}
= 6\pm i\sqrt{14}
$$
however by the same logic
$$ x^2 = (x+1)^2 + (x+7)^2\quad \implies
x=-8\pm \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int\frac{dx}{x-3y}$ when $y(x-y)^2=x$?
If y is a function of x such that $y(x-y)^2=x$
Statement-I: $$\int\frac{dx}{x-3y}=\frac12\log[(x-y)^2-1]$$
Because
Statement-II: $$\int\frac{dx}{x-3y}=\log(x-3y)+c$$
Question: Is Statement-I true? Is Statement-II true? Is Statement-II a correct explanation for Statement-I?
I... | Differentiating
$$
x=y(x-y)^2\tag{1}
$$
we get
$$
\begin{align}
1&=y'(x-y)^2+2y(x-y)(1-y')\\
&=y'\left[(x-y)^2-2y(x-y)\right]+2y(x-y)\\
&=y'(x-3y)(x-y)+2y(x-y)\\
1-2y(x-y)&=y'(x-3y)(x-y)\\
(x-y)^2-1&=(1-y')(x-3y)(x-y)\tag{2}
\end{align}
$$
Differentiating Statement-I, we get
$$
\frac{(x-y)(1-y')}{(x-y)^2-1}=\frac1{x-3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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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.