Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trying to use AM–GM to prove an inequality Given that $a,b,c > 0$. I'm trying to prove $$\left(\sqrt\frac{a+b}c+\sqrt\frac{b+c}a+\sqrt\frac{c+a}b\right)^2\ge\frac{2(a+b+c)^3}{3abc}.$$ I tried directly applying AM–GM on the LHS, but I can't get $abc$ with degree 1 in the denominator. $$\begin{align}
\frac{\sqrt\frac{a... | It's wrong inequality
Let $a=1,b=2,c=3$,then
$$LHS=(1+\sqrt{5}+\sqrt{2})^2=21.625\cdots$$
But
$$RHS=\dfrac{2(1+2+3)^3}{3\cdot 6}=24$$
| {
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"url": "https://math.stackexchange.com/questions/1591456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}... | This does not prove the inequality for every $x$, but only when $x$ is succiently large.
Taking the $\log$ of the inequality:
$$\frac{1}{2}\log \Big(\frac{1}{x+\frac{1}{2}}+1 \Big)+x \log \Big(\frac{1}{x}+1 \Big) -1<0.$$
Using series expansion for $x=\infty$:
\begin{align}
& \frac{1}{2x+1}-\frac{1}{(2x+1)^2} +x\Big(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592345",
"timestamp": "2023-03-29T00:00:00",
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Let $n = {1,2,3, ...}$, prove that $\left(1+\frac{1}{n}\right)^n$ is bounded above. I know that the sequence tends to $e \approx 2.718281828...$, but I am having a hard time proving it. Does anyone have any hints for me?
| If $p=1,2,\cdots, n$, we have $(1-\frac{p}{n})<1$, and $2^{p-1}\le p!$, so $\frac{1}{p!}\le \frac{1}{2^{p-1}}$.
Therefore, if $n>1$, apply binomial theorem to $e_n$, which gives us
$$\begin{align}
e_n &= \left(1+\displaystyle\frac{1}{n}\right)^n\\
&= 1+1+\frac{1}{2!}(1-\frac{1}{n})+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593594",
"timestamp": "2023-03-29T00:00:00",
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Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
| Notice, $$LHS=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}$$
$$=\frac{\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta}+1}$$
$$=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$
$$=\frac{(\sin\theta-\cos\theta+1)((\sin\theta+\cos\theta)+1)}{(\sin... | {
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"url": "https://math.stackexchange.com/questions/1593686",
"timestamp": "2023-03-29T00:00:00",
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How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| $$\lim_{x\to -\infty}x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)$$
$$=\lim_{x\to -\infty}x\frac{\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}$$
$$=\lim_{x\to -\infty}x\frac{\left(1-x\right)}{|x|\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the tangent to the sphere obtain the equations of tangent to sphere
$$x^2+ y^2+z^2+6x-2z+1 = 0$$
which pass through the line
$$3 (16-x) = 3z=2y+30$$
Now I know if the plane is $$lx +my+n z=p$$
then $$-I/3 +m/2+n/3=0$$
also $(16,-15,0)$ is a point on the plane
I know that there is $2$ answers , but how to proceed
| Let $R(a,b,c)$ be the point of tangency, and let $P(16, -15, 0)$ and $Q(6, 0,10)$ be two points on the line.
Since the center of the sphere is $C(-3, 0, 1)$, a normal vector to the plane is given by $\vec{n}=\langle a+3,b,c-1\rangle$.
Therefore the plane has equation $(a+3)(x-a)+b(y-b)+(c-1)(z-c)=0$, which gives
$\;\;... | {
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"url": "https://math.stackexchange.com/questions/1594111",
"timestamp": "2023-03-29T00:00:00",
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Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$ Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.
Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$
I can solve this question graphically.
$$P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28$$
$$4x^2+6x+4=\frac{28}{4y^2-12y+25}$$
I d... | Define function $f:\Bbb R^2\to R$ as following
$$f(x,y)=P(x)Q(y)-28=(4x^2+6x+4)(4y^2-12y+25)-28$$
It is continuous and differentiable as a polynomial. Since solution $f(x,y)=0$ is unique it means that point $(x,y)$ must be an extremum of this function.
Find partial derivatives and $\delta$:
$$f'_x=(8x+6)(4y^2-12y+25)$$... | {
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Square root of negative number I was reading my school's math textbooks for the next semester, and I found that when $a >0,$ $\sqrt{(-a)^2}=a$, but $\sqrt{(-a)^2}=\sqrt{a^2}i^2=-a$. So which one is right?
|
$a >0,$ $\sqrt{(-a)^2}=a$
This is correct, because $(-a)^2 = (-1)^2 a^2 = a^2$ and $\sqrt{a^2} = a$.
$\sqrt{(-a)^2}=\sqrt{a^2}i^2=-a$
This is not correct, because $\sqrt{(-a)^2} = a$ different from $\sqrt{a^2} i^2 = a (-1) = -a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | Since you only need the terms up to and including $x^4$, it would be easier just to use the Binomial theorem and get$$\cos^6x=\left[1+\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)\right]^6$$
$$=1+6\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)+15\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)^2+...$$
$$=1-3x^2+4x^4...$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
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What is the minimum polynomial of $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$? Inspired by a previous question what let $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$. What is the minimal polynomial of $x$ ?
The theory of algebraic extensions says the degree is $4$ since we have the deg... | All you need to do is write your number and its square, cube and fourth powers in the form $z^j = a_j + b_j \sqrt{2} + c_j \sqrt{3} + d_j \sqrt{6}$ with $a_j,b_j,c_j,d_j$ rational and solve a system of four equations in four unknowns
$$
\eqalign{a_4 + a_3 x_3 + a_2 x_2 + a_1 x_1 + a_0 x_0 &= 0\cr
b_4 + b_3 x_3 + b_2... | {
"language": "en",
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Testing the convergence of cube root of some function of n I have to test the convergence of the following series:-
$$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$
My approach is as follows :-
$$n^3+1>1=\sqrt[3]{n^3+1}>1=\sqrt[3]{n^3+1}-n>1-n$$
Now since$\sum 1-n$ diverges, the series under consideration diverges.
Is this right... | Note that
$$\sqrt[3]{n^3+1}-n=n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right).$$
For positive $t$ we have
$$1\lt \sqrt[3]{1+t}\lt 1+\frac{1}{3}t,$$
since $\left(1+\frac{t}{3}\right)^3\gt 1+t$. It follows that
$$0\lt n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)\lt \frac{1}{3n^2}$$
and now convergence of our series follows by co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\sqrt[3]{2}+\omega\in \Bbb{K}$, prove that $\sqrt[3]{2}\in \Bbb{K}$
Let $K=\Bbb{Q}(\sqrt[3]{2}+\omega)$ be a field extension of $\Bbb{Q}$ where $\omega$ is a root of $x^2+x+1$. Prove that $\sqrt[3]{2}\in K$
How do I prove this? I am at a loss. I did it by a very long method, involving solving for coefficients. I'... | Let $L$ be the splitting field of $X^3 - 2$ over $\mathbb{Q}$. The roots of this polynomial are $\sqrt[3]{2}$, $\omega \sqrt[3]{2}$, $\omega^2 \sqrt[3]{2}$. Therefore $L = \mathbb{Q}(\sqrt[3]{2}, \omega)$. Clearly, $[L:\mathbb{Q}(\sqrt[3]{2})] = 2$, and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$. Hence $[L:\mathbb{Q}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$
If $11 \gamma = \pi$ then prove that
$
\sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$
I could not use the relation $11 \gamma = \pi$.
| Let
$$P=\cos\gamma\cos 2\gamma\cos 3\gamma \cos 4\gamma \cos 5\gamma$$
$$Q=\sin\gamma\sin 2\gamma\sin 3\gamma \sin 4\gamma \sin 5\gamma$$
Then, prove that $$2^5PQ=Q$$ using $$2\sin\alpha\cos\alpha=\sin(2\alpha),\quad \sin\beta=\sin(\pi-(11\gamma-\beta))$$
$$\begin{align}2^5PQ&=(2\sin\gamma\cos\gamma)(2\sin 2\gamma\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tangents are drawn from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ and are inclined at angles $\theta$ and $\phi$ to the $x-$axis. Tangents are drawn from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ and are inclined at angles $\theta$ and $\phi$ to the $x-$axis.If $\tan\theta.\tan\phi=2,$ p... | From this,
$$y=x\tan\psi\pm\sqrt{2\tan^2\psi-3}$$ are always tangents of $$\dfrac{x^2}2-\dfrac{y^2}3=1$$
Now if these tangents pass through $(\alpha,\beta)$
$$\beta=\alpha\tan\psi\pm\sqrt{2\tan^2\psi-3}$$
$$\iff\beta-\alpha\tan\psi=\pm\sqrt{2\tan^2\psi-3}$$
On squaring and rearrangement,
$$(\alpha^2-2)\tan^2\psi-2\alp... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the congruence $3x^2+x+8\equiv 0 \pmod{11}$ How to find the solutions of this congruence? $$3x^2+x+8\equiv 0 \pmod{11}$$
I need to find the inverse of $3$, and there I have a problem.
| One method is to form the square on the LHS :
$$12 \cdot(3x^2+x+8) \equiv 0 \pmod{11}$$
$$36x^2+12x+1+95 \equiv 0 \pmod{11}$$
$$(6x+1)^2 \equiv -95 \equiv 4 \pmod{11}$$
This means that $$6x+1 \equiv \pm 2 \pmod{11}$$
When $6x+1 \equiv 2 \pmod{11}$ multiply by $2$ so :
$$12x \equiv 2\pmod{11}$$
$$x \equiv 2 \pmod{11}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series convergence study using big O? Can someone explain why the following expression:
$\sum_{n=1}^{\infty} (-1)^{n}(\frac{1}{n})(\frac{\frac{7}{12}+O(\frac{1}{n^{2}})}{-\frac{1}{6}+O(\frac{1}{n^{2}})})$
can be rewritten as:
$\sum_{n=1}^{\infty} (-1)^{n}(\frac{1}{n})(\frac{-7}{2}+O(\frac{1}{n^{2}}))(1+O(\frac{1}{n^{2}... | One may recall that, as $x \to 0$, by the Taylor series expansion we have
$$
\frac1{1-x}=1+x+O(x^2)
$$ giving, as $n \to \infty$,
$$
\frac{1}{-\frac{1}{6}+O(\frac{1}{n^2})}=(-6)\times\frac{1}{1-O(\frac{1}{n^2})}=(-6)\times\left( 1+O\left(\frac{1}{n^2}\right)\right)
$$ and
$$
\frac{\frac{7}{12}+O(\frac{1}{n^{2}})}{-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606878",
"timestamp": "2023-03-29T00:00:00",
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5 grades into 4 percentages. English teacher in need of help. I have five numbers, three of them each represent 25% of an average and the last two are the remaining 25%. One the last two numbers is 10% and the other 15%, how do I add them up into one number ?
I have 35 students. One of them has the following grades : 7... | In this case the formula for the weighted mean is
$$ \frac{25}{100} \cdot 7 + \frac{25}{100} \cdot 12 + \frac{25}{100} \cdot 17 + \frac{15}{100} \cdot 13 + \frac{10}{100} \cdot 15 = 12.45 $$
If you want to write one grade with weight 25% that combines the last two grades, the formula is
$$ \frac{15}{100} \cdot 13 + \fr... | {
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"url": "https://math.stackexchange.com/questions/1609332",
"timestamp": "2023-03-29T00:00:00",
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If $\frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Then prove that $ab\leq 0$
If $\displaystyle \frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Prove that $ab\leq 0$
$\bf{My\; Try::}$ I am Trying To solve it Using Inequality.
Using $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\f... | $$\frac{\sec^2 \theta}{a}+\frac{\tan^2 \theta}{b}=\frac{1}{a+b} = \frac{\sec^2 \theta-\tan^2 \theta}{a+b}$$
$$\Rightarrow \frac{\sec^2 \theta}{a(a+b)}\left[(a+b)\sec^2 \theta-a\right]+\frac{\tan^2 \theta}{(a+b)a}\left[(a+b)\tan^2 \theta+b\right]=0$$
$$\Rightarrow\tan^2 \theta+b\sec^2 \theta = 0\Rightarrow \sin^2 \theta... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional Equation for $f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$ The following functional equation proved quite difficult.
$1.$ $f(x)$ is a polynominal with real coeffecients.
$2.$ $f(1)=2,f(2)=20$.
$3.$ When for real $x,y,z$ satisfies the condition $xy+yz+zx=0$, $$f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$$
Find $f(x)$ that satisfie... | After some comments from Michael, and upon further examination, I was able to formulate a solution.
$f(x-y)+f(y-z)+f(z-x)=2f(x+y+z)$. If $f(x)=g(x^2)$, this gives us that
$g(x^2-2xy+y^2)+g(y^2-2yz+z^2)+g(z^2-2zx+x^2)=2g(x^2+y^2+z^2)$.
From the fact that $xy+yz+zx=0$. If $x-y=s,y-z=t$, then $x-z=s+t$. This gives us th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Conjectured closed form for $\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}$ I was trying to find closed form generalizations of the following well known hyperbolic secant sum
$$
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}=\frac{\left\{\Gamma\left(\frac{1}{4}\right)\right\}^2}{2\pi^{3/2}},\tag{1}
$$
as
... | Disclaimer: I verified my self that $\frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} \approx 0.640652 \approx \sqrt{\sqrt{2}+2}-\frac{1}{2} \left(\sqrt{2}+1\right)$ so mine is only an approximation but as far as I could do, no closed form exists to eval... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $x^{4}-2x^{3}+4x^{2}+6x-21=0$ Solve the equation $$x^{4}-2x^{3}+4x^{2}+6x-21=0$$ given that two of its roots are equal in magnitude but opposite in sign.
I don't know how to solve it. The roots are given as $\pm\sqrt{3},1\pm i\sqrt{6}. $
| Hint We can write the two special roots as $a, -a$, and so the polynomial contains has a factor of the form $x^2 - a^2$. So, we can write the polynomial as
$$x^4 - 2 x^3 + 4 x^2 + 6 x - 21 = (x^2 - a^2) (x^2 + B x + C)$$
for some $a, B, C$. (We know the leading coefficient of the second term has to be $1$ just by expan... | {
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Why is $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$? Suppose $\theta=2\pi/5$. Apparently it is true that $1+ \cos^2 \theta + \cos^2(2\theta) + \cos^2(3 \theta) + \cos^2 (4\theta) = 5/2$, or equivalently, $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$. What is the easiest way to see this?
I see that $1 + \cos \theta + \dotsb + \cos 4\t... | Let $\zeta = e^{i \theta}$. Then you have
$$\cos^2 (2\theta) + \cos^2 (4\theta) = {1 \over 4} \left( e^{2\theta} + e^{-2\theta} \right)^2 + {1 \over 4} \left( e^{4\theta} + e^{-4\theta} \right)^2 $$
or, recalling the definition of $\zeta$,
$$ {1 \over 4} \left( \left( \zeta^2 + \zeta^{-2} \right)^2 + \left( \zeta^4 + ... | {
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"timestamp": "2023-03-29T00:00:00",
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proving $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$ I would like to show that $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$
I was wondering if I could get a hint to get me started.
I tried trying to rewrite it in some sort of familiar sum like a Taylor Series for a function I know like the exponential but no luck. I also tried w... | Note that
$$
\sum_{k=1}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=2}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{j=1}^{\infty}\frac{1}{(j+1)(j+2)}, \ \text{with} \ j = k - 1
$$
But
$$
\frac{1}{(j+1)(j+2)} = \frac{1}{j+1} - \frac{1}{j+2} = \int_{0}^{1}x^jdx - \int_{0}^{1}x^{j+1}dx = \int_{0}^{1}x^j(1 - x)dx\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the infinite series: $3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots$? I'm trying to find the infinite sum that is defined by:
$$
3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots
$$
However, I do ... | The ideas in Mario G’s answer are very good ones to learn, but there are also ways to tackle the problem without calculus. You can easily test that the series is absolutely convergent, so we can rearrange terms pretty much at will. Now
$$\begin{align*}
3\left(\frac9{11}\right)+4\left(\frac9{11}\right)^2+5\left(\frac9{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
"answer_id": 1
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Find value of $\lim_{n\rightarrow \infty} \sum^{(n+1)^2}_{n^2}\frac{1}{\sqrt{k}}$
Find value of $$\lim_{n\rightarrow \infty}\sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}}$$
$\bf{My\; Try::}$ Let $$\lim_{n\rightarrow \infty} S = \sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}} $$
So $$\lim_{n\rightarrow \infty} \underbrace{\frac{1}... | Another way using harmonic numbers $$A_n=\sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}}=H_{n^2+2 n+1}^{\left(\frac{1}{2}\right)}-H_{n^2-1}^{\left(\frac{1}{2}\right)}$$ Now, using, for large values of $m$ $$H_{m}^{\left(\frac{1}{2}\right)}=2 \sqrt{m}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{m}}+O\left(\frac{1}{m^{3/2}}... | {
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Diamond Numbers and Fermat's method of Factorization I call numbers which are the product of two consecutive numbers $n(n+1)$ "diamond numbers". Can they be used in factorizing numbers using Fermat's method of the difference of two squares,considering that $(n + \frac 1 2)^2 = n(n+1) + \frac 1 4$? I had in mind somethi... | Such numbers $n(n+1)$ are even, so the differences between them are also even. It turns out that this could be related to factorising even numbers into an even and an odd number, since $$2a(2b+1) = \left(a+b\right)\left(a+b+1\right)- \left(\left|a-b-\frac12\right|-\frac12\right)\left(\left|a-b-\frac12\right|+\frac12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many combinations of connected midpoints for a regular hexagon? Board game designer here looking for some help with tile design for a hex-tile based game. any help with my image example or wording to make this question more clear is greatly appreciated.
Consider the regular hexagon ABCDEF with midpoints 123456
How... | I doubt there are more than five up to rotation: three with an adjacent pair and two without
| {
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"source": "stackexchange",
"question_score": "3",
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Finding $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$ Finding $$\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$$
I think if $$\int_{1}^{\infty}x^2 \frac{2^{x-1}}{3^x}dx$$
exists that this sequence is convergent, but I doubt that this integral is equal to the number to which the sequence converges to. I do not know a way to... | We know that the geometric series converges,
$$\sum_{k = 0}^{\infty} x^k = \frac{1}{1 - x} \;\; |x| < 1$$
Differentiating both sides,
$$\sum_{k = 0}^{\infty} kx^{k-1} = \frac{1}{(1 - x)^2} \;\; |x| < 1$$
Therefore, multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} kx^{k} = \frac{x}{(1 - x)^2} \;\; |x| < 1$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability involving Binomial Summation The problem statement is :
$A$ has $n$ coins and $B$ has $n+1$ coins. They toss their coins simultaneously.
If $p$ be the probability that $B$ will have more heads than $A$ the find $p=?$
One way I did this is by this argument :
Since $B$ cannot have the number of heads and t... | We have by inspection that the desired probability is
$$\frac{1}{2^{2n+1}}
\sum_{k=1}^{n+1} {n+1\choose k}
\sum_{m=0}^{k-1} {n\choose m}.$$
Now introducing
$${n\choose m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z^{m+1}} \; dz$$
we obtain for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z... | {
"language": "en",
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"source": "stackexchange",
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A Certain Double Sum In this problem, let $a_{i,j}$ be defined piecewise: $a_{i,j} = 0$ if $i <j$; $a_{i,j} = -1$ if $i=j$; and $a_{i,j} = 2^{j-i}$ if $i>j$.
I have to find: $A_1 = \sum_{i=1}^\infty \sum_{j=1}^\infty a_{i,j}$ and $A_2 = \sum_{j=1}^\infty \sum_{i=1}^\infty a_{i,j}$. They should be different.
But I don't... | In order to visualize both sums, see the following diagram:
$$\begin{array}{cccccccc}
\downarrow&-1&0&0&0&\cdots&\longrightarrow&\mbox{Sum}\downarrow\\\
\downarrow&2^{-1}&-1&0&0&\cdots&\longrightarrow&\\
\downarrow&2^{-2}&2^{-1}&-1&0&\cdots&\longrightarrow&\\
\vdots&2^{-3}&2^{-2}&2^{-1}&-1&\cdots&\longrightarrow&\\
\mb... | {
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"source": "stackexchange",
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How to find derivative of $\sin\sqrt{x}$ using difference quotient? The definition of derivative of a function $f(x)$ is $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$
Using this definition, the derivative of $\sin\sqrt{x}$ will be:
$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$
$$\lim_{h\to 0} \frac{\cos\left(\frac{\s... | You got stuck here
$$2 \cdot \lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$
You can break that one limit up into the product of two limits.
$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
\cdot\lim_{h\to 0} \frac{\sin\left(\frac{\sqr... | {
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"question_score": "2",
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Evaluation of $\prod^{n}_{r=1}\sin \left(\frac{(2r-1)\pi}{2n}\right)$
Find value of $$\prod^{n}_{r=1}\sin \left(\frac{\left(2r-1\right)\pi}{2n}\right)$$ Where $n\in \mathbb{N}$ and $n>1$
$\bf{My\; Try::}$ Let $$P = \sin \left(\frac{\pi}{2n}\right)\cdot \sin \left(\frac{3\pi}{2n}\right)\cdot \sin \left(\frac{5\pi}{2n}... | $$\prod^{n}_{r=1}\sin\frac{(2r-1)\pi}{2n}=\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{2v\pi}{2n}} =\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{v\pi}n}=\dfrac{F(2n)}{F(n)}$$
where $F(m)=\prod_{v=1}^{m-1}\sin\dfrac{v\pi}m$
Now from How to prove those "curious ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$ Let $a,b,c>0,a+b+c=1$, show that
$$\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}+\sqrt{\dfrac{ab}{c+ab}} \geq 1$$
I tried
$$\dfrac{bc}{a+bc}=\dfrac{bc}{a(a+b+c)+bc}=\dfrac{bc}{(a+b)(a+c)}$$
It suffice to show that
$$\sqrt{\dfrac... | Let $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, we need to prove that$\sum\limits_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\geq1$.
By AM-GM and C-S $\sum\limits_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\geq\sum\limits_{cyc}\frac{2x}{2x+y+z}\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(2x^2+2xy)}\geq1$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ $a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).
Here's my attempt:
$\frac{2ab}{a+b}\leq\sqrt {ab}$
$\frac{2a^... | Let $a,b\in\mathbb{R}$ with $a,b>0$.
$$\begin{align*}\frac{2ab}{a+b}\leq \sqrt{ab}&\Longleftrightarrow \frac{2}{a+b}\leq\frac{\sqrt{ab}}{ab}\\&\Longleftrightarrow \frac{2}{a+b}\leq\frac{1}{\sqrt{ab}}\\&\Longleftrightarrow\sqrt{ab}\leq\frac{a+b}{2}\\&\Longleftrightarrow 2\sqrt{ab}\leq a+b\\&\Longleftrightarrow 0\leq a-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integer solutions for $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$?
Find all triples $(x, y, z)$ where $x, y, z$ are coprime integers such that $$\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{z^2}$$
I did the following:
$$\begin{split}\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{z^2} &\Rightarrow \dfrac{x^2y^2}{x^2+y^2}=z^2 \\ &... | For the equation.
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
Use a Pythagorean triple.
$$a^2+b^2=c^2$$
Obtained solutions.
$$x=ac$$
$$y=bc$$
$$z=ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that $f(n) = 3n^5 + 5n^3 + 7n$ is divisible by 15 for every integer $n$
*
*So far I have only been able to complete the base case for which I got the following:
$$f(n) = 3n^5 + 5n^3 + 7n$$
$$f(n) = 3(1)^5 = 5(1)^3 + 7(1)$$
$$f(n) = 3 + 5 + 7$$
$$15/15 = 1$$
From here I got a bit confused with the inductive st... | $$\bmod3\\
n^1:0,1,2\\
n^2:0,1,1\\
n^3:0,1,2\\
5n^3+7n:0,0,0$$
This show that $3|(3n^5+5n^3+7)$.
$$\bmod5\\
n^1:0,1,2,3,4\\
n^2:0,1,4,4,1\\
n^3:0,1,3,2,4\\
n^5:0,1,2,3,4\\3n^5+7n:0,0,0,0,0
$$
This show that $5|(3n^5+5n^3+7)$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How can I show that $X$ and $Y$ are independent and find the distribution of $Y$? $X_1,X_2,\dots,X_n$ is an i.i.d. sequence of standard Gaussian random variables.
\begin{align}X&=\frac{1}{n}(X_1+X_2+\dots+X_n) \\[0.2cm] Y&=(X_1-X)^2+(X_2-X)^2+\dots+(X_n-X)^2\end{align}
*
*How can I show that $X$ and $Y$ are indepen... | Partial solution here.
$Y$ is a quadratic form. In particular, let $\mathbf{1} \in \mathbb{R}^n$ be a vector of ones, and define $$P_\mathbf{1} = \mathbf{1}\left(\mathbf{1}^{T}\mathbf{1}\right)^{-1}\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\text{.}$$
Let $$\mathbf{X}=\begin{bmatrix}
X_1 \\
X_2 \... | {
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"source": "stackexchange",
"question_score": "10",
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If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4If $(u,v)$ is a point on $4x^2+a^2y^2=4a^2$,where $4<a^2<8$,that is farthest from $(0,-2)$ then $u+v$ is equal to?
My Approach:
I took a parametric point $(t,4-4t^2/a^2)$.And then tried to find the minima of the distance.But that is too lengthy method.Any other sugges... | HINT:
WLOG any point on the ellipse $$(a\cos t,2\sin t)$$
So, if the distance is $d,$
$$d^2=(a\cos t)^2+(2+2\sin t)^2=8+4\sin t+(a^2-4)\cos^2t$$
$$=8+\sqrt{4^2+(a^2-4)^2}\sin\left(t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}\right)$$
$$\le8+\sqrt{4^2+(a^2-4)^2}$$
Now $4<a^2<8,0<a^2-4<4\implies(a^2-4)^2\le4^2$
So, the maximu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the s... | HINT: the second integral is given by $$\int\frac{x+1}{x^2+x+1}dx$$ you must write the denominator as $$(x+1/2)^2+3/4)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the area enclosed by the curve and line Calculate the are enclosed by ${y = 2x - 1}$ and ${y= x^2 + 6x + 2}$
First of all I combine the equations into:
${x^2 + 4x + 3 = 0}$
${(x + 3)(x + 1), x = -3, x = -1}$
They intersect at ${(-3 -7) (-1, -3)}$
I would say that ${y = 2x -1}$ is the top equation so to wor... | First we calculate the intersection points. We know that as long as the line and the parabola intersect, the line will be on top, because the parabola opens upward.
To solve for the intersection points (which will be our limits), we do the following:
$$2x + 1 = x^{2} + 6x + 2$$
$$x^{2} + 4x + 1 = 0$$
$$x = \frac{-4 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Substitution and Partial Fractions (Integration) $$\int\frac{dx}{x-\sqrt[4]{x}}$$
given the substitution $x=u^{4}, dx=4u^{3}du$
$$=\int\frac{4u^{3}du}{u^{4}-u}=\int\frac{4u^{3}du}{u(u^{3}-1)}=\int\frac{4u^{2}du}{(u^{3}-1)}$$
At this point I believe I can factor the denominator and use partial fractions
$$\int\frac{4u^... | $$
\int\frac{4u^{2}du}{u^3-1} = \frac 4 3 \int \frac {dw} w, \text{ etc.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $m\tan (\theta-30°)=n\tan (\theta+120°)$ If $m\tan (\theta-30°)=n\tan (\theta+120°)$ then prove that :
$$\cos 2\theta=\frac{m+n}{2(m-n)}$$
My attempt\
Here,
$$m\tan (\theta-30°)=n\tan (\theta+120)$$
$$\frac{\tan (\theta-30°)}{\tan (\theta+120°)}=\frac{n}{m}$$.
Now, what should I do next?
| $$m\tan (\theta-30^\circ)=n\tan (\theta+120^\circ)$$
$$\frac{\tan (\theta+120^\circ)}{\tan (\theta-30^\circ)}=\frac{m}{n}$$
Applying Componendo-Dividendo, we get
$$\frac{\tan (\theta+120^\circ)+\tan (\theta-30^\circ)}{\tan (\theta+120^\circ)-\tan (\theta-30^\circ)}=\frac{m+n}{m-n}$$
Now expand the L.H.S., keeping in mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Minimum distance between the curves $f(x) =e^x$ and $g(x) =\ln x$ What is the minimum distance between the curves $f(x) =e^x$ and $g(x) = \ln x$?
I didn't understand how to solve the problem. Please help me.
| Have a visualization:
The green curve is $f(x) = e^x$, the red curve is $g(x) = \ln(x)$.
$P$ is a point on the graph of $f$, $Q$ is a point on the graph of $g$.
Shown is the distance between $P$ and $Q$.
The obvious symmetry of the graphs results from the fact that $f$ and $g$ are inverse to each other,
$\DeclareMath... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find the limit $\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)$ $$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)\;=\;?\quad(n\in I) \\ \text{where $\lfloor\cdot\rfloor$ is the greatest integer function.}$$
This is what I did:
Since $[x] = x - ... | The crux of this is that $\sqrt{n^2+n+1}\approx n+\frac{1}{2}$ for large $n$.
Note that $$n^2+n+1=\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\implies\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}$$
One can thus quickly see that for large enough $n$, $n<\sqrt{n^2+n+1}<n+1$, so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
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How to calculate the PSD of a stochastic process Say we have a stochastic process described by a stochastic differential equation (in the Itô sense), and maybe we are able to find an explicit solution of it in terms of deterministic and Itô integrals (and maybe not).
In any of these cases, how can we compute the power... | I'll reply myself with a full derivation of the PSD of the harmonic oscillator. It's only half of the answer to my original question; the other half can be found here, after I asked the same question on mathoverflow
The full equation takes the expression
$$
m \ddot{x} + \gamma \dot{x} + \delta x = \sigma \eta(t)
$$
Per... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$ by Mathematical Induction Prove by Mathematical Induction:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+n(n+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+n^2(n+1)}=\frac{3n+5}{3n+1}$$
Now by inductive hypothesis:
$$\frac{1\cdot2^2+2\cdot3^2+\c... | As mentioned in the comments, this is probably easier to prove by using formulas for sums of powers. Nevertheless, if you want to proceed by induction, then you could do something like this:
Your formula is equivalent to the statement that $P(n)$ is true for all $n\in\mathbb N$, where
$$
P(n):(3n+1)\sum_{k=1}^n k(k+1)^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integral involving greatest integer function
Let $S_n = \sum_{k=1}^n \frac{1}{k}$ and $I_n=\int_1^n \frac{x-[x]}{x^2}dx$. Then, what is $S_{10} + T_{10}$?
The only clue that i can get is break the limits of integration according as the value the greatest integer function takes???please solve this for me. The answer i... | Hint 1:
\begin{align}\int_1^n \frac{x-[x]}{x^2}dx &= \int_1^2 \frac{x-1}{x^2} dx+\int_2^3\frac{x-2}{x^2} dx + \cdots + \int_{n-1}^n\frac{x-n+1}{x^2} dx\end{align}
Hint 2:
\begin{align}\int_{n-1}^n \frac{n-1}{x^2}dx&=\left[\frac{(1-n)}{x}\right]_{n-1}^n\\&=(1-n)\left(\frac{1}{n}-\frac{1}{n-1}\right)\\&=\frac{1}{n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i... | The fact you want to prove is equivalent to
$$
\frac{\arctan\left(\frac{11}{2}\right)}{3} = \arccos\left(\frac{2}{\sqrt 5}\right),
$$
that is,
$$
\arctan\left(\frac{11}{2}\right) = 3 \arccos\left(\frac{2}{\sqrt 5}\right),
$$
that is,
$$
\frac{11}{2} = \tan\left(3 \arccos\left(\frac{2}{\sqrt 5}\right)\right).
$$
The ang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:
*
*$0^2+1^2=1^2$
*$3^2+4^2=5^2$
*$20^2+21^2=29^2$
*$119^2+120^2=169^2$
*$696^2+697^2=985^2$
*$4059^2+4060... |
I. Silver ratio
What you have discovered is the square of the silver ratio,
$$S=1+\sqrt{2} = 2.414213\dots$$
It is a cousin of the golden ratio, $\phi=\frac{1+\sqrt{5}}{2}$, and share similar properties. Your Pythagorean triple can be expressed as,
$$\Big(\frac{b-1}{2}\Big)^2+\Big(\frac{b+1}{2}\Big)^2=c^2\tag1$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 0
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Where did my simplification go wrong? Sum and difference formula simplification I'm struggling with the following: We are to use the sum and difference formulas to find the exact value of the expression. The problem is simplification has been tough. As a last resort I decided to use Symbolab to find the answer and step... | $x=\frac{\sqrt6+\sqrt2}4$
$====================$
$x^2=\frac{6+2\sqrt{12}+2}{16}=\frac{8+2\sqrt{12}}{16}=\frac{8+4\sqrt3}{16}=\frac{2+\sqrt3}4$
$====================$
$x=\frac{\sqrt{2+\sqrt3}}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $7$-digit numbers which use only the digits $5$ and $7$ and are divisible by $35$
Find all $7$-digit numbers which use only the digits $5$ and $7$ and are divisible by $35$.
Attempt
It is easy to see that all numbers of this form must be of the form _ _ _ _ _ _ 5. Working with the divisible by $35$ conditio... | Hint By Fermat Little Theorem
$$555555=5 \cdot \frac{10^6-1}9$$
Is divisible by $7$.
Therefore
$$5555555 \equiv 5 \pmod{7}$$
Now your numbers are of this form plus twice sums of powers of 10.
You have
$$2 \cdot 10 \equiv -1 \pmod{7} \\
2 \cdot 10^2 \equiv 3 \pmod{7} \\
2 \cdot 10^3 \equiv 2 \pmod{7} \\
2 \cdot 10^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(1+x)^{20}=\sum_{r=1}^{20}a_rx^r$ where $a_r=\binom{20}{r}$, find the value $\mathop{\sum\sum}_{0\le i
$(1+x)^{20}=\sum_{r=1}^{20}a_rx^r$ where $a_r=\binom{20}{r}$, find the value $$\mathop{\sum\sum}_{0\le i<j\le 20}(a_i-a_j)^2$$
I first calculated the value of $\mathop{\sum\sum}_{0\le i<j\le 20}(a_j)^2$
$$A=\mathop{\... | From $(a_i - a_j)^2 = a_i^2 - 2 a_i a_j + a_j^2$, I'm pretty sure you want $2A - 2B$. But your $A$ and $B$ give $2A - 2B = 41 \binom{40}{20} - 2^{40}$, which is also not the given answer, so there must be error(s) in your $A$ and/or $B$. When I compute $A$, I get $10 \binom{40}{20}$ and when I compute $B$, I get the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{... | I think that the book is wrong.
$$1-\frac{4}{3x}+\frac{2}{x^2}=\frac{3x^2}{3x^2}\cdot \frac{1-\frac{4}{3x}+\frac{2}{x^2}}{1}=\frac{3x^2-4x+6}{3x^2}=\frac{2x^2+(x^2-4x+6)}{3x^2}$$
$$=\frac{2x^2+0}{3x^2}=\frac{2}{3},x\ne 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Number of solutions for $x^2 \equiv x \pmod m$ What is the number of solutions of $x^2 \equiv x \pmod m$ for any positive integer $m$?
| So we get $x(x-1) \equiv \mod m$ so $m|x(x-1)$. If $m$ is prime then either $m|x$ and $x \equiv 0 \mod m$ or $m|x -1$ and $x \equiv 1 \mod m$.
But what if $m$ is not prime. If $m = \prod p_i^{k_i}$ then as $x$ and $x-1$ are relatively prime (they have only a difference of 1) then each $p_i^{k_i}|x$ or divides $x -1$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_... | Hint:
Multiply Numerator Denomerator by $x$ then substitute $u^2=16-x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the sum to n terms of series :$\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots$ $$
\frac{1}{1\cdot 2\cdot 3\cdot 4} +\frac{1}{2\cdot 3\cdot 4\cdot 5} + \frac{1}{3\cdot 4\cdot 5\cdot 6}+\cdots
$$
up to $n$ terms. I need help in solving this sum. I trie... | We can use the following identity: $$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right).$$
Thanks to this identity, if we want to compute $$\sum_{k=1}^n\frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3}\sum_{k=1}^n\left(\frac{1}{k(k+1)(k+2)}-\frac{1}{(k+1)(k+2)(k+3)}\right),$$
we onl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Can you explain why $x>\frac 23$ is not a solution to this inequality $|2x + 2| + |x - 1| > 3$
why can't $x>\frac 23$ be a part of the solution?
Thanks for your help!
| Suppose $x>2/3$, then if $x \ge 1 $, then $2x+2 + x -1 = 3x + 1 \ge 3 + 1 > 3$.
If $2/3 < x < 1$, then $2x+2 - x +1 = x + 3 > 3$.
This is, however, not a complete solution (probably why it is not a solution!?).
Consider $-1 \le x \le 2/3$, then $2x+2 - x +1 = x + 3 >3 $. This is only true if $x>0$.
If $x<-1$, then $-2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Foundations of Differential Calculus In the preface of Foundations of Differential Calculus there's a section that says:
Thus, if the quantity $x$ is given an increment $\omega$, so that
it becomes $x + \omega$, its square $x^2$ becomes $x^2 + 2x\omega + \omega^2$, and it takes the
increment $2x\omega + \omega^2$.... | The paragraph says that the increment of $x$ (which means $(x + \omega) - x$, i.e. $\omega$) and the corresponding increment of $f$ from $f(x) = x^2$ to $f(x + \omega) = x^2 + 2 \omega x + \omega ^2$ (which means $f(x + \omega) - f(x) = 2 \omega x + \omega ^2$) have the ratio
$$\frac {(x + \omega) - x} {f(x + \omega) -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the numbers $-13, -9, - 4, -1, 9, 18, 21$ form a complete residue system modulo 7 Show that the numbers $-13, -9, - 4, -1, 9, 18, 21$ form a complete residue system modulo 7
We have just started he section on modular arithmetic so I am new to a residue system, we did a similar problem that we as follows in cl... | \begin{array}{ccc}
-13 \equiv \color{red}{1} \mod 7 & -9 \equiv \color{red}5 \mod 7 & -4 \equiv \color{red}3 \mod 7 \\
-1 \equiv \color{red}6 \mod 7 & 9 \equiv \color{red}2 \mod 7 & 18 \equiv \color{red}4 \mod 7 \\
&21 \equiv \color{red}0 \mod 7&
\end{array}
Isn't this enough? Each member in your set is congruent to o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.
$\bullet~$ I can ... | By $AM\geq GM$,
$x_{n+1}=\frac{1}{2} \left( x_{n}+\frac{2}{x_{n}} \right)
\geq \sqrt{2}$ and $x_{1}=2$, therefore
$$x_{n}^{2} \geq 2 \: , \quad \forall n\in \mathbb{N}$$
Now,
\begin{align*}
x_{n}-x_{n+1} &=\frac{x_{n}^{2}-2}{x_{n}} \\
&\geq 0
\end{align*}
So $\{ x_{n} \}$ is decreasing and
\begin{ali... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find area bounded by functions $y_1=\sqrt{4x-x^2}$ and $y_2=x\sqrt{4x-x^2}$. From $y_1=y_2\Rightarrow x=1$. Intersection points of $y_1$ and $y_2$ are $A(0,0),B(1,\sqrt 3),C(4,0)$. Domain of $y_1$ and $y_2$ is $x\in [0,4]$. On the interval $x\in[0,1]\Rightarrow y_1\ge y_2$ and on the interval $x\in[1,4]\Rightarrow y_1\... | If you'd like to get rid of radicals, you can do the following substitution:
$$z=\sqrt{\frac{4}{x}-1}$$
$$x=\frac{4}{1+z^2}$$
$$dx=-\frac{8z}{(1+z^2)^2}dz$$
The integrals will become:
$$I_1=\int \sqrt{4x-x^2}dx=-32\int \frac{z^2}{(1+z^2)^3}dz$$
$$I_2=\int x \sqrt{4x-x^2}dx=-128\int \frac{z^2}{(1+z^2)^4}dz$$
$$I_1=32\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What function does this Taylor Series represent? What is the function $f$ who's Taylor series is $1 - \frac{x}{4} + \frac{x^2}{7} - \frac{x^3}{10} + \cdots$ ?
I need to find the value of the series $ \sum^{\infty}_{n = 0}a_n = 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots$ by finding $\lim_{x\rightarrow1^-} \s... | Write the series as
$$\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3 k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k \left (x^{1/3} \right )^{3 k}}{3 k+1} = x^{-1/3} \sum_{k=0}^{\infty} \frac{(-1)^k \left (x^{1/3} \right )^{3 k+1}}{3 k+1}$$
Consider
$$f(y) = \sum_{k=0}^{\infty} \frac{(-1)^k y^{3 k+1}}{3 k+1}$$
$$\implies f'(y) = \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \... | $$S_1=1=(2\cdot1-1)^2,\\
S_n-S_{n-1}=(2n-1)^2-(2n-3)^2=8n-8.$$
You can establish the formula using the well-known triangular numbers,
$$1+2+3+\cdots n=\frac{n(n+1)}2.$$
Then
$$S_n=8\frac{(n-1)n}2+1=4n^2-4n+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Determine value of parameter, so that a polynomial has a root with multiplicity 2 I have the polynomial $P_a(x) = x^4 - x^2 + ax + 2$, where $a$ is a complex number, and I have to determine the values of $a$ such that $P_a(x)$ has a root with multiplicity at least two. I have tried two approaches, but none of them seem... | Your resultant is false:
Resultant[$x^4 - x^2 + a x + 2$, $4 x^3 - 2 x + a$, $x$ ] = $1568 - 284\,a^2 - 27\,a^4$ whose 2 real roots are $a=2$ and $a=-2$.
Edit: It can be of interest to have a pictorial representation of what is going on ; here is a way:
Equation $x^4 - x^2 + a x + 2=0$ can be considered as the equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Where does this equation for $n\binom{\binom{n-1}{2}}{m}$ come from? I'm reading a proof and am having trouble seeing why the following two lines are true:
\begin{align*}
n\frac{\binom{\binom{n-1}{2}}{m} }{ \binom{\binom{n}{2}}{m}}
&= n \left(\frac{n-2}{n}\right)^m \prod_{i=1}^{m-1} \left(1 - \frac{4i}{n(n-1)(n-2)-2i(n... | Here is the first part:
The following is valid
\begin{align*}
\binom{N}{m}&=\frac{N(N-1)\cdots(N-m+1)}{m!}\\
&=\frac{N^m}{m!}\left(1-\frac{1}{N}\right)\cdots\left(1-\frac{m -1}{N}\right)\\
&=\frac{N^m}{m!}\prod_{i=1}^{m-1}\left(1-\frac{i}{N}\right)
\end{align*}
With $N=\binom{n-1}{2}$, resp. $\binom{n}{2}$ we obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Alice and Bob dice game Alice and Bob are playing a dice game. They take turns rolling a fair $6$-sided die, with Alice going first. The first person whose roll repeats one that has already been seen loses. Before play starts, what’s the probability that Alice will win the game?
The probability to roll repeats one on a... | An infinite sum seems inappropriate, since the game for sure ends on the $7$-th toss or earlier. Alicia wins if (i) Bob loses in the first round, or (ii) the second, or (iii) the third.
(i) The probability that Bob loses in the first round is $\frac{1}{6}$. For whatever Alicia tossed, the probability Bob matches it is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can only the middle school math knowlegde help to find solutions for $2013 y^2 -xy -4026 x=0$? I found the following equation form an answer written for a question.
$$2013 y^2 -xy -4026 x=0$$
But I'm confused that can I really learn how to find the positive integer solutions for $x,y$ with having only the middle school... | An scholar way can be the following:
Put $a=2013=3\cdot 11\cdot 61$; we see there are $(1+1)(1+1)(1+1)=8$ possible factors.
$ay^2-xy-2ax=0\Rightarrow \color{red}{a|xy}\qquad (1)$.
$ay^2-xy-2ax=0\Rightarrow 2ay=x\pm \sqrt{x^2+8a^2x}\Rightarrow \color{red}{x^2+8a^2x=z^2}\Rightarrow x^2+8a^2x=w^2x^2\Rightarrow \color{red... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with ste... | Say $x=\tan \theta$ since both $\tan \theta$ and $x \in \mathbb{R}$.
Also we have that $\frac{dx}{d\theta}=\sec^2 \theta$
Then we get that $\frac{2x}{1+x^2}=\frac{2\tan \theta}{1+\tan^2 \theta}=\sin 2\theta$
So we get that $\arctan \frac{2x}{1+x^2}=\arctan (\sin 2\theta)$
Hence, the problem reduces to:
$$\frac{d}{dx}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that if a and b have the same sign then |a| + |b| = |a + b| We start with a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result :
$|a| + |b| = a+b$
$(|a| + |b|)^2 =... | If you want to use the characteritic that $|a| = \sqrt{a^2}$ you certainly can.
$(|a|+|b|)^2 = a^2 + b^2 + 2 |ab|$
if |ab| = ab then
$(|a|+|b|)^2 = (a+b)^2$
$||a|+|b|| = |a+b|$
$|a|+|b| = |a+b|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find solutions for $a$ and $b$ where $9 \equiv 4a+b \pmod {26} $ and $10 \equiv 19a+b \pmod {26}$? $$9 \equiv 4a+b \pmod {26}$$
$$10 \equiv 19a+b \pmod {26}$$
How can I solve the following system?
| You can write the system in matrix form:
$$
\pmatrix{4 & 1 \\ 19 & 1}
\pmatrix{a \\ b }
=
\pmatrix{9 \\ 10 }
$$
The determinant of the matrix is $-15$, which is invertible mod $26$, and so you can use Cramer's rule. In this $2\times 2$ case, it's easy:
$$
\pmatrix{4 & 1 \\ 19 & 1}^{-1}
=
-\frac{1}{15}
\pmatrix{1 & -1 \... | {
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"timestamp": "2023-03-29T00:00:00",
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Using the $\epsilon$-$\delta$ definition of the limit, evaluate $ \lim_{x \to a} \frac{x+2}{x^2-5}$
Using the $\epsilon$-$\delta$ definition of the limit, evaluate $$ \lim_{x \to a} f(x)$$ where $$f(x) = \frac{x+2}{x^2-5}$$
Attempt
We need to show that $\forall \epsilon, \exists \delta$ such that $$0 < |x-a| < \delta... | $$
\begin{align}
\left |\frac{x+2}{x^2-5}-\frac{a+2}{a^2-5} \right| & = \left | \frac{(x + 2)(a^2 - 5) - (a + 2)(x^2 - 5)}{(a^2 - 5)(x^2 - 5)} \right | \\[10pt]
& = \left | \frac{ax+2a+2x-5}{(a^2 - 5)(x^2 - 5)} \right | \cdot \underbrace{ |x-a|}_\text{important}.
\end{align}
$$
Pulling out $|x-a|$ is predictably possib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Inverting Modular Exponentiation How can I go about solving the equation $4 = y^4 \bmod{7}$? Do I have to try all of the possible $y$'s in between $1$ and $7-2$ or is there a smarter way that can be generalized for larger numbers?
| The following is -in principle-still "searching" but structures the space to be searched into simpler subspaces:
$$ \begin{array}{} &4 &= y^4 \pmod 7 \\
& y^4 - 4 &\equiv 0 \pmod 7 \\
&(y^2 - 2)(y^2+2) &\equiv 0 \pmod 7 \\
&& \text{giving two factors}\\
&y^2 - 2 &\equiv 0 \pmod 7 \\
\text{ or } & y^2 + 2 &\equiv 0 \pm... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$
If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$, then find the value of $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots$
Firstly how is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$?
Secondly, I thoug... | Notice:
*
*$$\sum_{n=a}^{m}\frac{b}{n^c}=b\left[\zeta(c,a)-\zeta(c,m+1)\right]$$
*$$\sum_{n=a}^{\infty}\frac{b}{n^c}=\lim_{m\to\infty}\sum_{n=a}^{m}\frac{b}{n^c}=\lim_{m\to\infty}b\left[\zeta(c,a)-\zeta(c,m+1)\right]=b\zeta(c,a)\space\text{ when }b=0\vee\Re(c)>1$$
So, when we solve your question:
$$\sum_{n=1}^{\i... | {
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"source": "stackexchange",
"question_score": "6",
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Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ Let $x, y ∈ (−2, 2)$ and $xy = −1$. Find the minimum value of $\frac{4}{4-x^2} + \frac{9}{9-y^2} $ ?
My Attempt
let $t=\frac{4}{4-x^2} + \frac{9}{9-y^2} $ , replacing $y$ by $- \frac{1}{x}$ we get $t=\frac{1}{1-(\frac{x}{2})^2} + \frac{1}{1-(\frac{1}{3x})^2... | Using AM-GM repeatedly:
$$\frac4{4-x^2}+\frac9{9-y^2} \ge \frac2{\sqrt{(1-x^2/4)(1-y^2/9)}} \ge \frac4{2-(x^2/4+y^2/9)}$$
Further, again as $\dfrac{x^2}4+\dfrac{y^2}9 \ge \dfrac13|xy|=\dfrac13$, we have
$$\frac4{4-x^2}+\frac9{9-y^2} \ge \frac4{2-\frac13}=\frac{12}5$$
Equality and the constraint is satisfied when $x = \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove: $\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$
Prove: $$\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$$
for $\: -\frac{\pi}{2} < x < \frac{\pi}{2}$
I have thought of three things so far that can be useful.
The Cauchy–Schwarz inequality
$$
\int_Efg\,\mathrm{d}x\le\left(... | For $x^2\lt1$,
$$
\begin{align}
x\log\left(\frac{1+x}{1-x}\right)
&=\sum_{k=0}^\infty\frac{2x^{2k+2}}{2k+1}\\
&\le\sum_{k=0}^\infty2x^{2k+2}\\
&=\frac{2x^2}{1-x^2}\tag{1}
\end{align}
$$
If
$$
f(x)=\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\tag{2}
$$
then
$$
f'(x)=\cos(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\ri... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the minimum value of $\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$ if $a+b+c=1$
Let $a,b,c\ge 0$, and $a+b+c=1$. Find the minimum value of
$$\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$$
I think the minimum value is $\sqrt{2} + 2$? when $a=1,b=c=0$.
Of course, I can't prove it. Can anyone help?
| We have
\begin{align*}
&\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1} - (\sqrt{2} + 2)\\
=\, & (\sqrt{a + 1} - \sqrt{2}) + (\sqrt{2b + 1} - 1) + (\sqrt{3c + 1} - 1) \\
=\, & \frac{a - 1}{\sqrt{a + 1} + \sqrt{2}} + \frac{2b}{\sqrt{2b + 1} + 1} + \frac{3c}{\sqrt{3c + 1} + 1}\\
=\, & \frac{-b - c}{\sqrt{a + 1} + \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ Show that $\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ for $|x|<1$
Question: what is wrong with the following reasoning?
Define: $x=\sin(y)$
$x^2=\sin^2(y)$
$1-x^2=\cos^2(y)$
$\frac{1}{1-x^2}=\sec^2(x)=1+\tan^2(y)$
$\tan^2(y)=\frac{1}{1-x^... | For example:
$$\begin{align*}&\left(\arcsin x\right)'=\frac1{\sqrt{1-x^2}}\\{}\\&\left(\arctan\frac x{\sqrt{1-x^2}}\right)'=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}\cdot\frac1{1-\frac{x^2}{1-x^2}}=\frac1{\sqrt{1-x^2}}\end{align*}$$
and we know that if on some open interval $\;I\;$ we have $\;f'=g'\;$ then $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve modulus inequality $|\frac{x+9}{x-9}|≤2$ In a problem, I've been asked to solve the inequality
$$|\frac{x+9}{x-9}|≤2$$
So I've done $\frac{x+9}{x-9}≤2$ and $\frac{x+9}{x-9}≥-2$, which gave $x≥27$ and $x≥3$, which doesn't seem quite correct because why would it give that $x$ is greater or equal to both 27 and 3? ... | $$|\frac{x+9}{x-9}|≤2$$
Is the same as
$$(\frac{x+9}{x-9})^2≤4$$
$$(x+9)^2≤4(x-9)^2$$
$$0≤(2x-18)^2-(x+9)^2$$
Using $a^2-b^2=(a-b)(a+b)$
$$0≤(2x-18-x-9)(2x-18+x+9)$$
$$0≤(x-27)(3x-9)$$
$$0≤(x-27)(x-3)$$
The inequality holds when $x-27$ and $x-3$ have the same sign
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for $x$ in $x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$ How would I go about solving for $x$ in this equation?
$$x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$$
| $$x+\frac{s}{s^2+4}=\frac{2x}{s^2+4}\Longleftrightarrow x-\frac{2x}{s^2+4}=-\frac{s}{s^2+4}\Longleftrightarrow$$
$$x\left[1-\frac{2}{s^2+4}\right]=-\frac{s}{s^2+4}\Longleftrightarrow x=\frac{-\frac{s}{s^2+4}}{1-\frac{2}{s^2+4}}\Longleftrightarrow \color{red}{x=-\frac{s}{2+s^2}}$$
But notice then that: $s^2+2\ne0$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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In how many of the integer numbers between 0 and 10,000 does the digit 3 appear to the left of 4 In how many of the integer numbers between $0$ and $10\,000$ does the digit $3$ appear some place to the left of the digit $4$? This would include, for example, the numbers $34$, $374$, $4384$ and $3874$, but would not incl... | Revised answer
Marking the first $3$ and the first $4$ after $3$, there are only $6$ cases:
$$\begin{array}{}
3&4&\bullet&\bullet&\implies& 10\cdot10 = 100\\
3&\bullet&4&\bullet&\implies& 9\cdot 10 = 90\\
3&\bullet&\bullet&4&\implies& 9\cdot9 =81\\
\bullet&3&4&\bullet&\implies& 9\cdot10 = 90\\
\bullet&3&\bullet&4&\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 8,
"answer_id": 3
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Verify that $\| T(x) \| = \| x \|$ Let $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ be multiplication by the matrix
$$A=
\begin{bmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\[0.3em]
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\[0.3em]
-\frac{2}{3} & -\frac{1}{3} & \frac{2}{3}
\end{bmatrix}
$$
Find $T(x)$ for the vecto... | Let $x$ be the vector $$\begin{bmatrix} a \\ b \\ c \end{bmatrix}$$, and let $u_1, u_2, u_3$ be the three columns of the matrix $A$. Then
$$
T(x) = au_1 + b u_2 + c u_3
$$
and
$$
\|T(x)\|^2 = T(x) \cdot T(x) = a^2 u_1 \cdot u_1 + 2ab u_1 \cdot u_2 + \ldots + c^2 u_3 \cdot u_3
$$
Computing all pairwise dot products of... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What's a good substitution to solve this integral? Is there a good substitution (or other simple method) to more easily solve this integral?
$$\int \frac{1}{\left(1-\sqrt{1-(\frac{r}{R})^2}\right)^{\frac{3}{2}}}dr$$
Honestly, I was trying $\frac{r}{R}=\sin(\alpha)$, but the result was complicated...
I appreciate anyon... | HINT:
$$\int\frac{1}{\left(1-\sqrt{1-\left(\frac{r}{\text{R}}\right)^2}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\sqrt{1-\frac{r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=$$
$$\int\frac{1}{\left(1-\sqrt{\frac{\text{R}^2-r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$
I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{... | The telescoping series cancel off itself after two terms, since $k-1$ and $k+1$ are 2 aparr
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerat... | Using infinitesimals of equal order,
$$\begin{align*} \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)\tan^4 x}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)}
&= \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)x^4}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)} \\
&= \lim_{x\to 0}\frac{ \sin^3x \cos^2x \cot(\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $(1+ax+bx^{2})^{10} = 1-30x+410x^{2}+...$ find the value of a and b then find the coefficient of $x^{19}$ in this expansion. If: $$(1+ax+bx^{2})^{10} = 1-30x+410x^{2}+...$$ find the value of a and b then find the coefficient of $x^{19}$ in this expansion.
I found $a=-3$ and $b=\frac{1}{2}$ by writing the trinomial a... | Here is a slighty different variation of the theme. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a polynomial.
We consider the polynomial
\begin{align*}
(1+ax+bx^2)^{10}=1-30x+410x^2+\cdots\tag{1}
\end{align*}
$$ $$
Coefficient $[x^1]$:
We obtain
\begin{align*... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove that $\frac{n^2-1}{8}$ is an even integer if and only if $n \equiv 1\pmod 8$ or $n \equiv -1\pmod 8$ prove that $\frac{n^2-1}{8}$ is an even integer if and only if $n \equiv 1\pmod 8$ or $n \equiv -1\pmod 8$
so I know from what i've already solved that $8|n^2-1$
where can I go from here? any help would be greatly... | Given that $8|n^2-1$ and to prove that if $8|n^2-1$ holds, then $16|n^2-1$ iff $n\equiv \pm 1 \pmod8$
PROOF:
Consider that $8|n^2-1 \Rightarrow 8|(n-1)(n+1)$
So $n-1$ and $n+1$ must be consecutive even numbers.
Then either $n-1$ or $n+1$ must be a multiple of $4$.
Now if $16|n^2-1 \Rightarrow 16|(n-1)(n+1)$, then bot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Partial derivative of $f(x,y) = \frac{x^3+y^3}{x^2+y^2}$ at $f_1(0,0)$ To be more precise than the title, the function is actually piecewise
$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$
I checked that the function is continuous at $(0,0)$, so I then calculat... | By definition of $f_1(x_0,y_0)$ you have :
$$f_1(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}$$
So
$$f_1(0,0)=\lim_{h\to 0}\frac{\frac{h^3+0}{h^2+0}}{h}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplifying Fractions involving negative numbers I want to simplify $$\frac{\frac{7}{-10} \times \frac{-15}{6}}{\frac{7}{-19} + \frac{-17}{-8}}$$
I really don't understand how to do this, or even how to start?
Negative numbers make it even harder for me to try and simplify it.
| Just simplify the numerator first and then simplify the denominator.
$$\frac{7}{-10} \times \frac{-15}{6} = \frac{-105}{-60} = \frac{-7}{-4} = \frac{7}{4}$$
leads us to
$$\frac{\frac{7}{4}}{\frac{7}{-19} + \frac{-17}{-8}}.$$
Very similar principle with the denominator.
$$\frac{7}{-19} + \frac{-17}{-8} = \frac{267}{152... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Determine the probability that at least $1$ three turns up when $3$ dice are rolled. The answer is $91/216$, I understand how $216$ is there because of $6\cdot 6\cdot 6$, $91$ though I don't what they got.
I'm so confused on listing the outcomes, I feel like with $3$ die, there are far too many outcomes for me to writ... | Well, using some notation, I will call $X$ the number of ones rolled in 3 rolls.
Then notice that we have $n = 3$ independent trials with probability $p = 1/6$ of success(rolling a one). Hence, $X$ follows a binomial distribution.
$X\sim\text{Binomial}(3, 1/6).$
Using the complement, we derive an easier solution;
$$P(X... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate :
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to :
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let... | Let $a^2-x^2=t$, then $-2xdx=dt$. Thus
$$
\int_0^a \frac{x}{\sqrt{a^2-x^2}}dx=-\frac{1}{2}\int_{a^2}^0 \frac{1}{\sqrt{t}}dt =\left[\sqrt{t}\right]_0^{a^2}=a.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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I'm having trouble understanding this proof regarding Supremum. I am trying to understand this proof, I understand the general form of the argument, and I understand what they are trying to show, Im just having trouble following the steps they've used. if anyone could explain why they've done what they've done I'd be r... | Question 1
Yeah, I believe you have this one right. Let me write it out in detail for completeness, where I have inserted an extra step:
$\left ( a+\frac{1}{n} \right )^{2}-2=\left ( a^{2} -2\right )+\frac{2a}{n}+\frac{1}{n^{2}}\leq (a^2-2) + \frac{2a}{n} + \frac{1}{n}=\left ( a^{2} -2\right )+\frac{2a+1}{n}.$
The extr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integral through $u$-substitution v. multiplying out I have the integral:
$ \int (x^2)(x^3-1)^2 \, dx $ Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $ which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$.
However, if rather than using $u$-substitution, I multiply wit... | Using the binomial expansion:
\begin{align}
(x^3 - 1)^3 =&\ \binom{3}{0}(x^3)^3 + \binom{3}{1}(x^3)^2(-1) + \binom{3}{2}(x^3)^1(-1)^2 + \binom{3}{3}(-1)^3\\
=&\ x^9 - 3x^6 + 3x^3 - 1
\end{align}
Therefore:
$$
\frac{1}{3}\frac{(x^3 - 1)^3}{3} = \frac{x^9}{9} - \frac{x^6}{3} + \frac{x^3}{3} - \frac{1}{9}
$$
Which is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ and $y$ can be expanded in ascending powers of $x$ If $y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ and assuming that $y$ can be expanded in ascending powers of $x$ in the form $a_0+a_1x+a_2x^2+....+a_nx^n$... prove that $(n+1)a_{n+1}=na_{n-1}$
How to proceed in this question? Is it possible t... | HINT:
$$y\sqrt{1-x^2}=\sin^{-1}x$$
Differentiating both sides wrt $x$
$$y_1\sqrt{1-x^2}-y\cdot\dfrac x{\sqrt{1-x^2}}=\dfrac1{\sqrt{1-x^2}}$$
$$\iff(1-x^2)y_1-xy=1$$
As $y=\sum_{r=0}^\infty a_rx^r, y_1=\sum_{r=1}^\infty ra_rx^{r-1}$
Compare the coefficients of $x^{r+2}$
| {
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"url": "https://math.stackexchange.com/questions/1724576",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 0
} |
if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3 Prove that if $(a,b,c)$ is a Pythagorean triple then $a$ or $b$ or $c$ can be divided by 3
| We know that $a^2+b^2=c^2$, and that the perfect squares mod $3$ are $0$ and $1$. Therefore $c^2$ is not $2$ mod $3$, and so it's not the case that both $a^2$ and $b^2$ are $1$ mod $3$. Therefore one of them is $0$ mod $3$. Let it be $a^2$. Then $a$ is also $0$ mod $3$ because $1^2=2^2=1\pmod{3}$. So $3|a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Cesàro sum of the series $\sin x + \sin 2x + \sin 3x + \ldots = \frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$ I'm learning about Fourier series (specifically Cesàro summation) and need help with the following problem:
Show that the Cesàro sum of the series $\sin x + \sin 2x + \sin 3x + \ldots$ is eq... | Using your result, we have
$$\frac1{N} \sum_{k=1}^NS_k = \frac{\cos\frac{x}{2} }{2\sin{\frac{x}{2}}} - \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(k + \frac{1}{2}\right)x \\ = \frac1{2}\cot\frac{x}{2} - \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) .$$
Now you just need to show that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Maclaurin serie of $\int_0^x\frac{sin(t)}{t}$ If $f(x)=\int_0^x\frac{\sin(t)}{t}$. Show that
$$f(x)=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+...$$
Calculate f(1) to three decimal places.
Would you mind showing how to build this Maclaurin serie?
| $$\sin(x) = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots$$
$$\frac{\sin(x)}{x} = 1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots$$
$$\int \frac{\sin(x)}{x}dx = x-\frac{1}{3\cdot 3!}x^3+\frac{1}{5\cdot 5!}x^5-\frac{1}{7\cdot 7!}x^7+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
| The single substitution $$u^2 = 4-x^2$$ will yield the simplest computation. Note this substitution implies $$x^2 = 4 - u^2, \quad x \, dx = -u \, du,$$ hence $$x^3 \sqrt{4-x^2} \, dx = x^2 \sqrt{4-x^2} \cdot x \, dx = (4-u^2)u(-u) \, du = (u^4 - 4u^2) \, du.$$ It immediately follows that $$\begin{align*} \int x^3 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Trigonometric polynom Prove that $$\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}$$ roots of polynomial $8x^3-4x^2-4x+1=0$
I'm confused, what can i do with $\frac{\pi}{7}$
| Let $\theta=\pi/7$. Note that
$$\cos(4\theta)=-\cos(3\theta).\tag{1}$$
Let $x=\cos\theta$. By the double-angle identity for cosine, we have $\cos(2\theta)=2x^2-1$, so $$\cos(4\theta)=2(2x^2-1)^2-1.$$
By the triple-angle identity for cosine we have
$$\cos(3\theta)=4x^3-3x.$$
Now using Equation (1) we obtain
$$2(2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$ I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.
I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted b... | Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $ \epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1 $. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "121",
"answer_count": 3,
"answer_id": 0
} |
How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloo... | Let $[\frac{x}{n}]=q$ then $q\leqslant \frac{x}{n}< q+1$. Then $nq\leqslant x < n(q+1)$. Hence $nq\leqslant [x] < n(q+1)$ and $q\leqslant [\frac{[x]}{n}]< q+1$. Thus $[\frac{[x]}{n}]=q$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"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.