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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. $$\begin{align}
\frac{\sqrt\frac{a+b}c+\sqrt\frac{b+c}a+\sqrt\frac{c+a}b}{3} &\ge \sqrt[3]{\sqrt\frac{a+b}c\sqrt\frac{b+c}a\sqrt\frac{c+a}b} \\
\sqrt\frac{a+b}c+\sqrt\frac{b+c}a+\sqrt\frac{c+a}b &\ge 3 \sqrt[3]{\sqrt\frac{a+b}c\sqrt\frac{b+c}a\sqrt\frac{c+a}b} \\
\left(\sqrt\frac{a+b}c+\sqrt\frac{b+c}a+\sqrt\frac{c+a}b\right)^2 &\ge 9 \sqrt[3]{\frac{a+b}c\frac{b+c}a\frac{c+a}b} \\
\end{align}$$
| 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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"timestamp": "2023-03-29T00:00:00",
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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
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x}
$$
The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).
While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).
Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$?
| 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(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}\Big)+-1+o(\frac{1}{x^2})= \\
&=\frac{-4x^2+5x+2}{6(2x+1)^2x^2}+o(\frac{1}{x^2})
\end{align}
Hence the last quantity is less than $0$ for $x$ sufficiently large, and this implies that your inequality is true when $x$ is big enaugh.
| {
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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
$$\begin{align}
e_n &= \left(1+\displaystyle\frac{1}{n}\right)^n\\
&= 1+1+\frac{1}{2!}(1-\frac{1}{n})+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+\cdots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{n-1}{n})
\end{align}$$
and use above two inequalities, we have $$2<e_n<1+1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}}<3$$
Hence $e_n$ is bounded above.
| {
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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, $$LHS=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}$$
$$=\frac{\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta}+1}$$
$$=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$
$$=\frac{(\sin\theta-\cos\theta+1)((\sin\theta+\cos\theta)+1)}{(\sin\theta+\cos\theta-1)((\sin\theta+\cos\theta)+1)}$$
$$=\frac{\sin^2\theta+2\sin \theta+1-\cos^2\theta}{(\sin\theta+\cos\theta)^2-1}$$
$$=\frac{\sin^2\theta+2\sin \theta+\sin^2\theta}{\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1}$$
$$=\frac{2\sin^2\theta+2\sin \theta}{1+2\sin\theta\cos\theta-1}$$
$$=\frac{2\sin\theta(1+\sin \theta)}{2\sin\theta\cos\theta}$$
$$=\frac{1+\sin \theta}{\cos\theta}=RHS$$
| {
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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?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| $$\lim_{x\to -\infty}x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)$$
$$=\lim_{x\to -\infty}x\frac{\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}$$
$$=\lim_{x\to -\infty}x\frac{\left(1-x\right)}{|x|\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=\lim_{x\to -\infty}\frac{x\left(1-x\right)}{(-x)\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=-\lim_{x\to -\infty}x\frac{\left(\frac 1x-1\right)}{\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=-\lim_{x\to +\infty}x\frac{\left(1+\frac 1x\right)}{\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}\longrightarrow \color{red}{-\infty}$$
| {
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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
$\;\;\color{blue}{(a+3)x+by+(c-1)z=-3a+c-1}\;\;\;$ since $a^2+b^2+c^2=-6a+2c-1$.
Substituting the coordinates of $Q$ gives $6(a+3)+10(c-1)=-3a+c-1$,
so $a+c=-1$ and $\color{blue}{a=-c-1}$.
Substituting the coordinates of $P$ gives $16(a+3)-15b=-3a+c-1$,
so $3b+4c=6$ and $\color{blue}{b=2-\frac{4}{3}c}$.
Substituting into $a^2+b^2+c^2+6a-2c+1=0$ gives
$(c+1)^2+(2-\frac{4}{3}c)^2+c^2+6(-c-1)-2c+1=0$,
so $\frac{34}{9}c^2-\frac{34}{3}c=0$ and thus $c^2-3c=c(c-3)=0$.
If $c=0$, $a=-1$ and $b=2$; so the plane has equation $\color{red}{2x+2y-z=2}$.
If $c=3$, $a=-4$ and $b=-2$; so the plane has equation $\color{red}{-x-2y+2z=14}$.
| {
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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 drew the graph of $4x^2+6x+4$ and found it is a upward parabola with minimum value at $(-\frac{3}{4},\frac{7}{4})$ and I drew the graph of $\frac{28}{4x^2-12x+25}$. I found that it is a bell shaped curve whose maximum value occurs at $(\frac{1}{4},\frac{7}{4})$. So I found that the common abscissa is $y=\frac{7}{4}$ which occurs at $x=-\frac{3}{4}$ for the first curve and at $x=\frac{1}{4}$ for the second curve, so solution should be $(-\tfrac{3}{4},\tfrac{1}{4})$
But I do not know how to solve it algebraically.
| 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 $\delta$:
$$f'_x=(8x+6)(4y^2-12y+25)$$
$$f''_{xx}=8(4y^2-12y+25)$$
$$f'_y=(4x^2+6x+4)(8y-12)$$
$$f''_{yy}=8(4x^2+6x+4)$$
$$f''_{xy}=(8x+6)(8y-12)$$
$$\delta(x,y)=(f''_{xy})^2-f''_{xx}f''_{yy}$$
And solve $$\begin{cases}f'_x=0\\f'_y=0\end{cases}$$
$$\begin{cases}(x+\tfrac68)(4y^2-12y+25)=0\\
(4x^2+6x+4)(y-\tfrac32)=0\end{cases}$$
$$\begin{cases}
x=-\frac34\\y=\frac32
\end{cases}$$
We have $\delta(x,y)<0$ and $f(-\tfrac34,\tfrac32)=0$ so it is an extremum and unique solution to $f(x,y)=0$.
| {
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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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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 $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
| Since you only need the terms up to and including $x^4$, it would be easier just to use the Binomial theorem and get$$\cos^6x=\left[1+\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)\right]^6$$
$$=1+6\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)+15\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)^2+...$$
$$=1-3x^2+4x^4...$$
| {
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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 degree of the field extension $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{6}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] =4$
Does trigonometry help us find the other three conjugate roots?
*
*$+\sqrt{2}-\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_1$
*$-\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_2$
*$-\sqrt{2}-\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot \theta_3$
This problem would be easier if we used $\cos$ instead of $\cot$. If I remember the half-angle identity or... double-angle identity:
$$ \cot \theta = \frac{\cos \theta}{\sin \theta} = \sqrt{\frac{1 - \sin \frac{\theta}{2}}{1 + \sin \frac{\theta}{2}}}$$
Sorry I am forgetting, but I am asking about the relationship between trigonometry and the Galois theory of this number.
| 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
$$
\eqalign{a_4 + a_3 x_3 + a_2 x_2 + a_1 x_1 + a_0 x_0 &= 0\cr
b_4 + b_3 x_3 + b_2 x_2 + b_1 x_1 + b_0 x_0 &= 0\cr
c_4 + c_3 x_3 + c_2 x_2 + c_1 x_1 + c_0 x_0 &= 0\cr
d_4 + d_3 x_3 + d_2 x_2 + d_1 x_1 + d_0 x_0 &= 0\cr
}$$
to determine a polynomial $Z^4 + x_3 Z^3 + x_2 Z^2 + x_1 Z + x_0$ of which your number is a root.
Maple says the answer is $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$.
EDIT: Here is another way, maybe a bit deeper. Your number is $z = 1+(1+\sqrt{2})(1+\sqrt{3})$.
$u = 1+\sqrt{2}$ is an eigenvalue of $A = \pmatrix{0 & 1\cr 1 & 2\cr}$, and
$v = 1+\sqrt{3}$ is an eigenvalue of $B = \pmatrix{0 & 2\cr 1 & 2\cr}$.
So $uv = (1+\sqrt{2})(1+\sqrt{3})$ is an eigenvalue of
$$ A \otimes B = \pmatrix{0 & 0 & 0 & 2\cr 0 & 0 & 1 & 2\cr 0 & 2 & 0 & 4\cr 1 & 2 & 2 & 4}$$
which has characteristic polynomial $X^4 - 4 X^3 - 16 X^2 - 8 X + 4$. Substitute $X = Z - 1$ to obtain $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$.
| {
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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 or wrong?
| 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 comparison.
| {
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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'm sure there is a better way.
My strategy: I calculated $(\sqrt[3]{2}+\omega), (\sqrt[3]{2}+\omega)^2, (\sqrt[3]{2}+\omega)^3, \frac{1}{(\sqrt[3]{2}+\omega)}, \frac{1}{(\sqrt[3]{2}+\omega)^2}$. Then I determined the coefficients I would need to multiply each by to finally get $2^{1/3}$
| 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 $[L:\mathbb{Q}] = 6$.
It will be enough to show that $\mathbb{Q}(\sqrt[3]{2} + \omega) = L$. This will follow if we can show that, of the six automorphisms of $L$, the element $\sqrt[3]{2} + \omega$ is fixed only by the identity. These automorphisms are given by $\sqrt[3]{2} \mapsto \sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^2 \sqrt[3]{2}; \omega \mapsto \omega, \omega^2$.
Thus it is enough to check that
$$\sqrt[3]{2} + \omega \not\in \{ \omega \sqrt[3]{2} + \omega, \omega^2 \sqrt[3]{2} + \omega, \sqrt[3]{2} + \omega^2, \omega \sqrt[3]{2} + \omega^2, \omega^2 \sqrt[3]{2} + \omega^2 \}.$$
For the first three elements, this is obvious because their difference with $\sqrt[3]{2} + \omega$ is obviously nonzero. The fourth element is $\omega(\sqrt[3]{2} + \omega)$. The fifth element has modulus $\sqrt[3]{2} + 1$.
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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 $$2\sin\alpha\cos\alpha=\sin(2\alpha),\quad \sin\beta=\sin(\pi-(11\gamma-\beta))$$
$$\begin{align}2^5PQ&=(2\sin\gamma\cos\gamma)(2\sin 2\gamma\cos 2\gamma)(2\sin 3\gamma\cos 3\gamma)(2\sin 4\gamma \cos 4\gamma)(2\sin 5\gamma \cos 5\gamma)\\&=\sin 2\gamma \sin 4\gamma \sin 6\gamma \sin 8\gamma \sin 10\gamma\\&=\sin 2\gamma \sin 4\gamma \sin (\pi-5\gamma) \sin(\pi-3\gamma)\sin(\pi-\gamma)\\&=\sin 2\gamma \sin 4\gamma \sin 5\gamma \sin 3\gamma \sin \gamma\\&=Q\end{align}$$
| {
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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 $\tan\theta.\tan\phi=2,$ prove that $\beta^2=2\alpha^2-7$
The equation of the pair of tangents from the point $(\alpha,\beta)$ to the hyperbola $3x^2-2y^2=6$ is given by $(3x\alpha-2y\beta-6)^2=(3x^2-2y^2-6)(3\alpha^2-2\beta^2-6)$
When i simplify this expression,i get
$(18+6\beta^2)x^2+(6\alpha^2-12)y^2-12xy\alpha\beta+24y\beta-36x\alpha+18\alpha^2-12\beta^2=0$
This is an equation of pair of lines and the angle between pair of lines $ax^2+2hxy+by^2=0$ is given by the formula $\frac{2\sqrt{h^2-ab}}{a+b}$
So $\tan(\theta-\phi)=\frac{2\sqrt{36\alpha^2\beta^2-(18+6\beta^2)(6\alpha^2-12)}}{18+6\beta^2+6\alpha^2-12}$
$\frac{\tan\theta-\tan\phi}{1+\tan\theta\tan\phi}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$
$\frac{\tan\theta-\tan\phi}{3}=\frac{2\sqrt{\alpha^2\beta^2-(3+\beta^2)(\alpha^2-2)}}{1+\beta^2+\alpha^2}$
I am stuck here.I do not know,how to prove it further.
| 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,
$$(\alpha^2-2)\tan^2\psi-2\alpha\tan\psi+\beta^2-3=0$$
Now if the two values of $\psi$ are $\theta,\phi$
$$\tan\theta\tan\phi=\dfrac{\beta^2-3}{\alpha^2-2}$$ and we are done!
| {
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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 :
$$12x \equiv 2\pmod{11}$$
$$x \equiv 2 \pmod{11}$$
Doing the same for the other leads to the other solution :
$$ x \equiv 5 \pmod{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$\sum_{n=1}^{\infty} (-1)^{n}(\frac{1}{n})(\frac{-7}{2}+O(\frac{1}{n^{2}}))(1+O(\frac{1}{n^{2}}))$
And why we are allowed to split this last expression of the series into a convergent part and an absolutely convergent one?
Still trying to wrap my head around big O notation. Thanks in advance.
| 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
$$
\frac{\frac{7}{12}+O(\frac{1}{n^{2}})}{-\frac{1}{6}+O(\frac{1}{n^{2}})}=\left(\frac{7}{12}+O\left(\frac{1}{n^{2}}\right)\right)\times(-6)\times\left( 1+O\left(\frac{1}{n^2}\right)\right)=-\frac72+O\left(\frac{1}{n^2}\right)
$$ as wanted.
The inital series converges being the sum of two convergent series.
| {
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"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, 12, 17, 13 and 15. The first three are out of 25% so no problem, but the 13 is out of 15% and the 15 is out of 10%. How do I add them up to make a grade out of 25% ?
I know this is beyond easy for some, so thank you for your help.
| 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
$$ \frac{15}{100} \cdot 13 + \frac{10}{100} \cdot 15 = \frac{25}{100} \cdot x \implies x = \frac{15 \cdot 13 + 10 \cdot 15}{25} = 13.8$$
If you double check now, you get that the average of $7,12,17$ and $13.8$ is $12.45$ (so that each grade is worth 25%).
| {
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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 $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\frac{(\sec^4 \theta)^2}{a}+\frac{(\tan^4 \theta)^2}{b}\geq \frac{(\sec^4 \theta+\tan^4\theta)^2}{a+b}$$
and equality hold when $$\frac{\sec^4 \theta}{a}=\frac{\tan^4 \theta}{b}$$
Now I did not understand How can I solve after that
Help me
Thanks
| $$\frac{\sec^2 \theta}{a}+\frac{\tan^2 \theta}{b}=\frac{1}{a+b} = \frac{\sec^2 \theta-\tan^2 \theta}{a+b}$$
$$\Rightarrow \frac{\sec^2 \theta}{a(a+b)}\left[(a+b)\sec^2 \theta-a\right]+\frac{\tan^2 \theta}{(a+b)a}\left[(a+b)\tan^2 \theta+b\right]=0$$
$$\Rightarrow\tan^2 \theta+b\sec^2 \theta = 0\Rightarrow \sin^2 \theta=-\frac{b}{a}\geq 0$$
$$\Rightarrow -\frac{b}{a}\geq 0\Rightarrow -ab\geq 0\Rightarrow ab\leq 0$$
| {
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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 satisfies all conditions given above.
While it is easy enough to guess that $f(x)=x^4+x^2$, proving it proved difficult.
First, putting in $(x,0,0)$ gives us that $f(-x)=f(x)$.
Second, putting in $(0,0,0)$ gives us $f(0)=0$
I also substituted $f(x)=g(x^2)$.
However, I was able to progress no further from here. Any help would be appreciated.
| 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 that $xy+yz+zx=(y+s)y+y(y-t)+(y+s)(y-t)=3y^2+2(s-t)y-st=0$. $b^2-4ac=4(s^2+st+t^2)>0$. Thus such a $y$ always exists.
Notice that $s^2+st+t^2=x^2-2xy+y^2+y^2-2yz+z^2-zx-y^2+xy+yz=x^2+y^2+z^2$
Then, $g(s^2)+g(t^2)+g(s^2+2st+t^2)=2g(s^2+st+t^2)$ gives us that $2g(s^2)+g(4s^2)=2g(3s^2)$. If the degree of $g(x)$ is $n$, comparing the coeffecients give us that $2 \times 1 +4^n=2 \times 3^n$. The only solution for this is $n=1,2$. It is easy to proceed from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$
S(a)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a}.
$$
In particular I find by numerical experimentation
$$
\displaystyle \frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}\overset{?}=-\frac{1}{2}\left(1+\sqrt{2}\right)+\sqrt{2+\sqrt{2}}\tag{2}
$$
(Mathematica wasn't able to find a closed form directly, but then I decided to switch to calculation of ratios of the sums, calculated ratios numerically and then was able to recognize this particular ratio as a root approximant. This was subsequently verified to 1000 decimal places).
I simplified this expression from the previous edition of the question.
Unfortunately for other values of $a$ I couldn't find a closed form. Of course $(2)$ together with $(1)$ would imply a closed form for the sum $S(1/\sqrt{2})$
How one can prove $(2)$?
| 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 evaluate precisely your sum: besides, if your aim is to do computational evaluations, I suggest to simplify the sum, evaluate the integral I'm providing or evaluate the sum from $n = -10^3$ to $n = 10^3$: I've seen no massive difference with $n = -10^6$ to $n = 10^6$ or more.
\begin{align}
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx\\
&= -\frac{2 \sqrt{2} \arctan\left(\frac{(\sqrt{2}-2) \tanh \left(\frac{\pi n}{2}\right)}{\sqrt{2}}\right)}{\pi } \big|_{-\infty}^{+\infty}\\
&= \sqrt{2}
\end{align}
and more in general
\begin{align}
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+a} dx\\
&= -\frac{2 \arctan\left(\frac{(a-1) \tanh \left(\frac{\pi n}{2}\right)}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \big|_{-\infty}^{+\infty}\\
&= \left( -\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right) - \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right)\\
&= -2 \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi \sqrt{1-a^2}} \right)
\end{align}
so in your specific case
\begin{align}
\frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} &\approx \frac{\int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx}{\int_{-\infty}^\infty\frac{1}{\cosh\pi x} dx}\\
&= \frac{-2 \left(\frac{2 \arctan\left(\frac{\frac{1}{\sqrt{2}}-1}{\sqrt{1-\frac{1}{\sqrt{2}^2}}}\right)}{\pi \sqrt{1-\frac{1}{\sqrt{2}^2}}} \right)}{-2 \left(\frac{2 \arctan\left(\frac{0-1}{\sqrt{1-0^2}}\right)}{\pi \sqrt{1-0^2}} \right)}\\
&= \frac{1}{\sqrt{2}}\\
&\approx 0.707107
\end{align}
| {
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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 expanding and matching.) Expanding the r.h.s. and comparing, e.g., the coefficients of the $x^3$ terms gives $B = -2$, and comparing the other like coefficients lets us determine $a$ and $C$, too. Nowo, the solutions are $a$, $-a$, and the roots of $x^2 + B x + C$, which (having already determined $B, C$) we can find as usual with the quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $1 + \cos \theta + \dotsb + \cos 4\theta = 0$ by de Moivre.
This came up while calculating the character of the irreducible two-dimensional representation of the dihedral group of order $10$ (i.e. to prove it is irreducible).
| 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 $\zeta$,
$$ {1 \over 4} \left( \left( \zeta^2 + \zeta^{-2} \right)^2 + \left( \zeta^4 + \zeta^{-4} \right)^2 \right)$$
Now, if you expand those out you get
$$ {1 \over 4} \left( \zeta^4 + 2 + \zeta^{-4} + \zeta^8 + 2 + \zeta^{-8} \right) $$
or, recalling that $\zeta^5 = 1$, this is
$$ {1 \over 4} \left( 4 + \zeta^4 + \zeta + \zeta^2 + \zeta^3 \right). $$
Rearranging gives
$$ {1 \over 4} \left( 3 + (1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4) \right)$$
and by de Moivre we have $1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4$, so this is just $3/4$.
| {
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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 writing it as the sum of two terms but I'm having trouble knowing how to break it down (if this is even the right way to attack this problem?). Any advice would be appreciated. Thanks.
| 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
$$
\frac{1}{(j+1)(j+2)} = \frac{1}{j+1} - \frac{1}{j+2} = \int_{0}^{1}x^jdx - \int_{0}^{1}x^{j+1}dx = \int_{0}^{1}x^j(1 - x)dx\\
\text{and} \ \sum_{j=1}^{\infty}x^jdx = \frac{x}{1 - x}, \quad |x| < 1
$$
Hence,
$$
\sum_{k=1}^{\infty}\frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=1}^{\infty}\int_{0}^{1}x^j(1 - x)dx = \frac{1}{2}+\int_{0}^{1}(1 - x)\sum_{j=1}^{\infty}x^jdx\\
= \frac{1}{2} + \int_{0}^{1}(1 - x)\frac{x}{1 - x}dx = \frac{1}{2} + \int_{0}^{1}xdx = \frac{1}{2} + \frac{1}{2} = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 not know of any known formula to do this. Am I missing something really simple? Thanks!
| 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
$$\begin{align*}
3\left(\frac9{11}\right)+4\left(\frac9{11}\right)^2+5\left(\frac9{11}\right)^3+\ldots&=\sum_{n\ge 1}(n+2)\left(\frac9{11}\right)^n\\
&=\sum_{n\ge 1}n\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\\
&=\sum_{n\ge 1}\sum_{k=1}^n\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\\
&=\sum_{k\ge 1}\sum_{n\ge k}\left(\frac9{11}\right)^n+2\sum_{n\ge 1}\left(\frac9{11}\right)^n\;,
\end{align*}\tag{1}$$
and for each $k$ the summation $\sum_{n\ge k}\left(\frac9{11}\right)^n$ is just a convergent geometric series. In particular,
$$\sum_{n\ge k}\left(\frac9{11}\right)^n=\frac{\left(\frac9{11}\right)^k}{1-\frac9{11}}=\frac{11}2\left(\frac9{11}\right)^k\;,$$
and the desired sum is
$$\sum_{k\ge 1}\frac{11}2\left(\frac9{11}\right)^k+2\sum_{n\ge 1}\left(\frac9{11}\right)^n=\left(\frac{11}2+2\right)\sum_{k\ge 1}\left(\frac9{11}\right)^k=\frac{15}2\cdot\frac{11}2\cdot\frac9{11}=\frac{135}4\;.$$
If you’re not used to working with summations, the calculation in $(1)$ may look rather mysterious. If you write it out longhand in the right way, though, it becomes quite straightforward. For convenience I’ll replace $\frac9{11}$ by $r$. Then you can break up the sum in two dimensions:
$$\begin{array}{ccc}
3r&+&4r^2&+&5r^3&+&6r^4&+&\ldots\\ \hline
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{red}r}{1-r}\\
+&&+&&+&&+\\
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{red}r}{1-r}\\
+&&+&&+&&+\\
r&+&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 1}r^n&=&\frac{\color{blue}r}{1-r}\\
&&+&&+&&+\\
&&r^2&+&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 2}r^n&=&\frac{\color{blue}{r^2}}{1-r}\\
&&&&+&&+\\
&&&&r^3&+&r^4&+&\ldots&=&\sum_{n\ge 3}r^n&=&\frac{\color{blue}{r^3}}{1-r}\\
&&&&&&+\\
&&&&&&r^4&+&\ldots&=&\sum_{n\ge 4}r^n&=&\frac{\color{blue}{r^4}}{1-r}\\ \hline
&&&&&&&&&&&&\frac1{1-r}\left(\color{red}{2r}+\color{blue}{\sum_{n\ge 1}r^n}\right)\\
&&&&&&&&&&&&=\frac1{1-r}\left(2r+\frac{r}{1-r}\right)
\end{array}$$
The original series is organized by summing the columns of this array first and then adding up the column sums; when I reversed the order of summation in the first term at the last step of $(1)$, I was switching to summing the rows first and then adding up the row sums.
| {
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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 $$\lim_{n\rightarrow \infty} \underbrace{\frac{1}{\sqrt{n^2+2n+1}}+\frac{1}{\sqrt{n^2+2n+1}}+.......+\frac{1}{\sqrt{n^2+2n+1}}}_{\bf{(2n+2)\; times}}\leq \lim_{n\rightarrow \infty} S \leq \lim_{n\rightarrow \infty} \underbrace{\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+.......+\frac{1}{\sqrt{n^2}}}_{\bf{(2n+2)\; times}}$$
So $$\frac{2n+2}{\sqrt{n^2+2n+1}}\leq S \leq \lim_{n\rightarrow \infty} \frac{2n+2}{\sqrt{n^2}}$$
So Using Sqeeze Theorem, WE get $$\lim_{n\rightarrow \infty} S = \lim_{n\rightarrow \infty} \sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}} = 2$$
My question is can we solve it any other way, If yes then plz explain here
Thanks
| 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 $m$ $$H_{m}^{\left(\frac{1}{2}\right)}=2 \sqrt{m}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{m}}+O\left(\frac{1}{m^{3/2}}\right)$$ and using again expansions leads to $$A_n=2+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
For $n=10$, the above approximation gives $2.09500$ while the exact computation of the sum would give $\approx 2.09546$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 something like $(5+\frac12)^2 -(3+\frac12)^2 = 2\times9 =18$
| 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 $$2a(2b+1) = \left(a+b\right)\left(a+b+1\right)- \left(\left|a-b-\frac12\right|-\frac12\right)\left(\left|a-b-\frac12\right|+\frac12\right) $$ $$c(c+1)-d(d+1)=(c+d+1)(c-d)$$ where the second right hand side is the product of numbers of different parity. So for example you have
*
*$60 = 30\times 31 -29\times 30 = (30+29+1)(30-29)=60\times 1$
*$60 = 11\times 12 -8\times 9 = (11+8+1)(11-8)=20\times 3$
*$60 = 9\times 10 -5\times 6 = (9+5+1)(9-5)=15\times 4$
*$60 = 8\times 9 -3\times 4 = (8+3+1)(8-3)=12\times 5$
but there is an easier way of finding at least one such factorisation of an even number: keep dividing by $2$ until it stops being even.
Added:
If you prefer to consider squares of numbers which are $\frac12$ more than integers, and their differences, this becomes $$ 2a(2b+1) = \left(a+b+\frac12\right)^2 - \left(a-b-\frac12\right)^2$$ $$\left(c+\dfrac12\right)^2-\left(d+\dfrac12\right)^2=(c+d+1)(c-d)$$ and
*
*$60 = 30.5^2 - 29.5^2 = (30.5+29.5)(30.5-29.5) = 60\times 1$
*$60 = 11.5^2 -8.5^2 = (11.5+8.5)(11.5-8.5)=20\times 3$
*$60 = 9.5^2 -5.5^2 = (9.5+5.5)(9.5-5.5)=15\times 4$
*$60 = 8.5^2 -3.5^2 = (8.5+3.5)(8.5-3.5)=12\times 5$
essentially the same as my initial response. It is still the factorisation of even numbers into an even and an odd number.
| {
"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 many possible groups are there to draw exactly three line segments that connect two midpoints. Example line segments 1-2, 3-4, 5-6 are one group and 1-3,2-4, 5-6 are a separate group. I want the set of all groups that meet the following conditions:
1: Each midpoint is connected by a line segment to exactly one other midpoint
2: Each set is not a rotation of another set (the part I'm really having difficulty with, and I am genuinely curious if this is something calculable.)
So, by the non-mathematical powers of brute force, I compiled the following list of groups:
(1-2, 3-4, 5-6),
(1-2, 3-5, 4-6),
(1-2, 3-6, 4-5),
(1-3, 2-4, 5-6),
(1-3, 2-5, 4-6),
(1-3, 2-6, 4-5),
(1-4, 2-3, 5-6),
(1-4, 2-5, 3-6),
(1-4, 2-6, 3-5),
(1-5, 2-3, 4-6),
(1-5, 2-4, 3-6),
(1-5, 2-6, 3-4),
(1-6, 2-3, 4-5),
(1-6, 2-4, 3-5),
(1-6, 2-5, 3-4)
After that I figured each new group would simply be a rotation of one of the existing groups turned 120 degrees.
So what I need is simply someone to check my work and see if I missed something, and help me root out duplicates. for example (1-2, 3-4, 5-6) and (1-6, 2-3, 4-5) are duplicates if one is rotated 120 degrees.
| I doubt there are more than five up to rotation: three with an adjacent pair and two without
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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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 solve this sequence, thought about doing geometric progression, but can't because of $k^2$
| 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,
$$\sum_{k = 0}^{\infty} kx^{k} = \frac{x}{(1 - x)^2} \;\; |x| < 1$$
Differentiating again we get,
$$\sum_{k = 0}^{\infty} k^2x^{k-1} = \frac{1 + x}{(1 - x)^3} \;\; |x| < 1$$
Multiplying throughout by $x$ we get,
$$\sum_{k = 0}^{\infty} k^2x^{k} = \frac{x(1 + x)}{(1 - x)^3} \;\; |x| < 1$$
Substitute $x = \frac{2}{3} < 1$ to get,
$$\sum_{k = 0}^{\infty} k^2\left(\frac{2}{3}\right)^{k} = 30$$
Divide throughout by $2$ to get,
$$\sum_{k = 0}^{\infty} k^2\frac{2^{k-1}}{3^k} = 15$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
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 tails both greater than $A$.
But since one of those cases is always true. (either more heads($p$) or more tails(=$q$ (let))
So by symmetry both should have equal probabilities.
Since $p+q=1$.
Then $p=0.5$
One more way of doing it is by binomial summation i.e. Lets say $A$ has $x$ heads and $B$ has $y$ heads then we add up
The product $$ \binom{n}{x} * \binom{n+1}{y} $$for all $y>x$.
But I am not able to evaluate that...
So how to do I go about summing this?
| 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
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\frac{1/z^{k}-1}{1/z-1}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^n
\frac{1/z^{k}-1}{1-z}
\; dz$$
The second term vanishes and we get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^k} \frac{1}{1-z} (1+z)^n
\; dz$$
which could have been obtained by inspection. Note that this is zero
when $k=0$ so we may extend the outer sum to include $k=0$, getting
$$\frac{1}{2^{2n+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{1-z} (1+z)^n
\sum_{k=0}^{n+1} {n+1\choose k} \frac{1}{z^k}
\; dz
\\ = \frac{1}{2^{2n+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{1-z} (1+z)^n
\left(1+\frac{1}{z}\right)^{n+1}
\; dz
\\ = \frac{1}{2^{2n+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{1-z} (1+z)^{2n+1}
\; dz.$$
Extracting the residue now yields
$$\frac{1}{2^{2n+1}}
\sum_{q=0}^n {2n+1\choose q}
= \frac{1}{2^{2n+1}} \times \frac{1}{2} 2^{2n+1}
= \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
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 know how to handle double series, especially if I have not proven convergence (which I think is the key point).
How does one calculate these? I have never been taught and have no idea.
This is my best guess:
$A_1 = \sum_{i=1}^\infty (-1 + \sum_{j=2}^\infty a_{i,j}) = \sum_{i=1}^\infty (-1+0) = - \infty$.
And likewise for $A_2$. But when I do this, I am getting that both series diverge to $-\infty$.
I know that they should be different. So, what is going wrong? How does one handle these?
| In order to visualize both sums, see the following diagram:
$$\begin{array}{cccccccc}
\downarrow&-1&0&0&0&\cdots&\longrightarrow&\mbox{Sum}\downarrow\\\
\downarrow&2^{-1}&-1&0&0&\cdots&\longrightarrow&\\
\downarrow&2^{-2}&2^{-1}&-1&0&\cdots&\longrightarrow&\\
\vdots&2^{-3}&2^{-2}&2^{-1}&-1&\cdots&\longrightarrow&\\
\mbox{Sum}\rightarrow&&
\end{array}$$
Thus $A_1$ is sum each row and then sum all these sums:
$\begin{eqnarray}A_1&=&-1+(-1+\frac{1}{2})+(-1+\frac{1}{2}+\frac{1}{2^2})+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3})+...\\
&=&-1-\frac{1}{2}+(-1+\frac{1}{2}(1+\frac{1}{2}))+(-1+\frac{1}{2}(1+\frac{1}{2}+\frac{1}{2^2}))+...\\
&=&-1-\frac{1}{2}+(-1+1-\frac{1}{2^2})+(-1+1-\frac{1}{2^3})+...\\
&=&-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...\\
&=&-2
\end{eqnarray}$
On the other hand, $A_2$ is sum each column and then sum all these sums:
$\begin{eqnarray}
A_2&=&(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+(-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)+...\\
&=&(-1+1)+(-1+1)+(-1+1)+...\\
&=&0
\end{eqnarray}$
Thus $A_1\neq A_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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:
$$\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}$$
$$\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$
Now i got stuck. How to find the limit or simplify this expression?
I get intuition that we have to use $$\lim_{x\to0}\frac{\sin x}{x} = 1$$
but that too is leading no where. I am unable to remove h from denominator.
NOTE
I know the derivative of $\sin\sqrt{x}$ is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ using chain rule, but this exercise was given to us for practice using division quotient.
| 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.
$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
\cdot\lim_{h\to 0} \frac{\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}$$
The first limit is easy.
$$\lim_{h\to 0} \cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)
= \cos\left(\frac{\sqrt{x}+\sqrt{x}}{2}\right) = \cos \sqrt x$$
The second isn't that much harder.
\begin{align}
\lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}
&= \lim_{h\to 0} \frac{\sin\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}
{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)} \cdot
\frac{\left(\dfrac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h} \\
&= \lim_{h\to 0} \left(1 \cdot
\dfrac{\sqrt{x+h}-\sqrt{x}}{2h}\right) \\
&= \lim_{h\to 0} \dfrac{(x+h)-x}{2h(\sqrt{x+h}+\sqrt{x})} \\
&= \dfrac{1}{4\sqrt{x}} \\
\end{align}
Putting it all together, you get $\dfrac{\cos\sqrt{x}}{2\sqrt{x}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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 $$P = \sin \left(\frac{\pi}{2n}\right)\cdot \sin \left(\frac{3\pi}{2n}\right)\cdot \sin \left(\frac{5\pi}{2n}\right)\cdot \cdot \cdot \cdot \cdot \sin\left(\frac{n\pi}{2n}\right)\cdot\cdot \cdot \cdot \cdot \sin\left(\frac{(2n-1)\pi}{2n}\right)$$
Now we can write $\displaystyle \sin\left(\frac{(2n-1)\pi}{2n}\right)=\sin \left(\frac{\pi}{n}\right)$ Similarly $\displaystyle \sin\left(\frac{(2n-3)\pi}{2n}\right)=\sin \left(\frac{3\pi}{2n}\right)$
So we get $$P = \left[\sin \left(\frac{\pi}{2n}\right)\cdot \sin \left(\frac{3\pi}{2n}\right)\cdot \sin \left(\frac{5\pi}{2n}\right)\cdot \cdot \cdot \cdot \cdot \sin\left(\frac{(n-2)\pi}{2n}\right)\right]^2$$
Now How can i Solve after that, Help me
Thanks
| $$\prod^{n}_{r=1}\sin\frac{(2r-1)\pi}{2n}=\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{2v\pi}{2n}} =\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{v\pi}n}=\dfrac{F(2n)}{F(n)}$$
where $F(m)=\prod_{v=1}^{m-1}\sin\dfrac{v\pi}m$
Now from How to prove those "curious identities"?,
$$F(m)=\dfrac m{2^{m-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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
$$\sqrt{\dfrac{bc}{(a+b)(a+c)}}+\sqrt{\dfrac{ca}{(b+a)(b+c)}}+\sqrt{\dfrac{ab}{(c+a)(c+b)}}\geq 1$$
| 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",
"url": "https://math.stackexchange.com/questions/1632956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
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:
$\frac{2ab}{a+b}\leq\sqrt {ab}$
$\frac{2a^2b^2}{(a+b)^2}\leq{ab}$
$\frac{2a^2b^2}{a^2+2ab+b^2} - ab \leq 0$
$\frac{2a^2b^2-ab(a^2+2ab+b^2)}{a^2+2ab+b^2}\leq 0$
$\frac{2a^2b^2-a^3b-2a^2b^2-ab^3}{a^2+2ab+b^2}\leq 0$
$\frac{-a^3b-ab^3}{a^2+2ab+b^2}\leq0$
Because $a$ and $b$ are positive, this guarantees the left side is negative, making the inequality true.
| Let $a,b\in\mathbb{R}$ with $a,b>0$.
$$\begin{align*}\frac{2ab}{a+b}\leq \sqrt{ab}&\Longleftrightarrow \frac{2}{a+b}\leq\frac{\sqrt{ab}}{ab}\\&\Longleftrightarrow \frac{2}{a+b}\leq\frac{1}{\sqrt{ab}}\\&\Longleftrightarrow\sqrt{ab}\leq\frac{a+b}{2}\\&\Longleftrightarrow 2\sqrt{ab}\leq a+b\\&\Longleftrightarrow 0\leq a-2\sqrt{ab}+b\\&\Longleftrightarrow 0\leq (\sqrt{a}-\sqrt{b})^2\end{align*}$$ where the last inequality is always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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:
$$\begin{split}\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{z^2} &\Rightarrow \dfrac{x^2y^2}{x^2+y^2}=z^2 \\ & \Rightarrow x^2y^2-x^2z^2-y^2z^2=0 \\ & \Rightarrow x^2y^2-x^2z^2-y^2z^2+z^4=z^4 \\ & \Rightarrow (z^2-x^2)(z^2-y^2)=z^4 \\ & \Rightarrow (x+z)(x-z)(y+z)(y-z)=z^4\end{split}$$
How should I go on?
| 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,
"answer_id": 0
} |
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 step in terms of number manipulation, a hint that my professor gave me was $f(-n) = -f(n)$. Any further help is appreciated.
| $$\bmod3\\
n^1:0,1,2\\
n^2:0,1,1\\
n^3:0,1,2\\
5n^3+7n:0,0,0$$
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",
"url": "https://math.stackexchange.com/questions/1635114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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 independent?
*How can I find the distribution of $Y$?
Can we use the following method to show that $X$ and $Y$ are independent?
$$Cov(X,Y)=0$$
Or is there any other proper way?
| 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 $$\mathbf{X}=\begin{bmatrix}
X_1 \\
X_2 \\
\vdots\\
X_n
\end{bmatrix}\text{.}$$
Then $$P_\mathbf{1}\mathbf{X} = \mathbf{1}\left(\dfrac{1}{n}\right)\begin{bmatrix}
1 & 1 & \cdots & 1
\end{bmatrix}\begin{bmatrix}
X_1 \\
X_2 \\
\vdots\\
X_n
\end{bmatrix} = \mathbf{1}\left(\dfrac{1}{n}\right)\sum_{i=1}^{n}X_i = \mathbf{1}X = X\mathbf{1}\text{.}$$
Furthermore,
$$\begin{align}
Y &= \sum_{i=1}^{n}(X_i-X)^2 \\
&= (\mathbf{X}-X\mathbf{1})^{T}(\mathbf{X}-X\mathbf{1}) \\
&= (\mathbf{X}-P_\mathbf{1}\mathbf{X})^{T}(\mathbf{x}-P_\mathbf{1}\mathbf{X}) \\
&= [(\mathbf{I}-P_{\mathbf{1}})\mathbf{X}]^{T}(\mathbf{I}-P_{\mathbf{1}})\mathbf{X} \\
&= \mathbf{X}^{T}(\mathbf{I}-P_{\mathbf{1}})^{T}(\mathbf{I}-P_{\mathbf{1}})\mathbf{X}\text{.}
\end{align}$$
$\mathbf{I}$ is obviously symmetric, and notice
$$P_{\mathbf{1}}^{T} = \left[\mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\right]^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T} = P_{\mathbf{1}}$$
so hence $P_{\mathbf{1}}$ is symmetric, so that $\mathbf{I}-P_{\mathbf{1}}$ is symmetric as well, and
$$(\mathbf{I}-P_{\mathbf{1}})^{T} = \mathbf{I}-P_{\mathbf{1}}\text{.}$$
This gives $Y = \mathbf{X}^{T}(\mathbf{I}-P_{\mathbf{1}})^{2}\mathbf{X}$. Observe also that
$$P_\mathbf{1}^2 = \left(\mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\right)^2 = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{n}{n}\right)\left(\dfrac{1}{n}\right)\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T} = P_{\mathbf{1}}$$
hence $P_{\mathbf{1}}$ is idempotent. Notice
$$(\mathbf{I}-P_{\mathbf{1}})^{2} = \mathbf{I}^2-2\mathbf{I}P_{\mathbf{1}}+P_{\mathbf{1}}^2 = \mathbf{I}-2P_{\mathbf{1}}+P_{\mathbf{1}} = \mathbf{I}-P_{\mathbf{1}}$$
since $P_{\mathbf{1}}^2 = P_{\mathbf{1}}$, as shown earlier. Hence $Y = \mathbf{X}^{T}(\mathbf{I}-P_{\mathbf{1}})\mathbf{X}$.
It can be shown that the rank of $\mathbf{I}-P_{\mathbf{1}}$ is $n - 1$. Using the theorem 7 here (p. 29), you can show that $$Y \sim \sigma^2\chi^2_{n-1}(\boldsymbol{\mu}^{\prime}(\mathbf{I}-P_{\mathbf{1}})\boldsymbol{\mu}^{\prime})$$
If
$$Q = \left(\mathbf{1}^{T}\mathbf{1}\right)^{-1}\mathbf{1}^{T} = \left(\dfrac{1}{n}\right)\mathbf{1}^{T}\text{.}$$
then $X = Q\mathbf{X}$.
For the remainder of the proof, see this page, under "Examples."
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
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 suggestions?
| 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 maximum value i.e., the equality occurs if $t+\arccos\dfrac4{\sqrt{4^2+(a^2-4)^2}}=2m\pi+\dfrac\pi2\iff t=2m\pi+\arcsin\dfrac4{\sqrt{4^2+(a^2-4)^2}}$
and $a^2-8$ is maximum which can be made arbitrarily small but will remain negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 second part: due a change of variable $u=x+1$, I find
$$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
| 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",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
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 work out the area, I would use:
${\int_{-3}^{-1} (2x -1) - (x^2 + 6x + 2) dx}$
=> ${-{x^3\over 3} - 2x^2 - 3x}$
=> ${(-{1\over 3} - 2 + 3) - ({-27 \over 3} -18 + 9)}$
=> ${{2\over 3} - (-18)}$
=> ${18{2\over 3}}$
This is not the correct answer, can anyone point out where I have gone wrong?
| 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:
$$2x + 1 = x^{2} + 6x + 2$$
$$x^{2} + 4x + 1 = 0$$
$$x = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}.$$
We need to evaluate the integral
$$\int_{-2 - \sqrt{3}}^{-2 + \sqrt{3}} \left(-x^{2} - 4x - 1\right)dx.$$
We do this as follows:
$$\left[-\frac{1}{3}x^{3} - 2x^{2} - x\right]_{-2 - \sqrt{3}}^{-2 + \sqrt{3}}$$
$$= \left(2\sqrt{3} - \frac{10}{3}\right) - \left(-2\sqrt{3} - \frac{10}{3}\right)$$
$$= \boxed{4\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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
$$\int\frac{4u^{2}du}{(u-1)(u^{2}+u+1)}=\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du$$
by substituting in $u=1$ and ignoring the factor in the denominator $(u-1)$ (Grubby Thumb Rule) $A=\frac{4*1^{2}}{1^{2}+1+1}, A=\frac{4}{3}$
and substituting arbitrarily $u=0$ into the expression
$$4u^{2}=\frac{4}{3}(u^2+u+1)+B(u-1)$$
$$0=\frac{4}{3}(1)-B, B=\frac{4}{3}$$
therefore
$$\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du=\int\bigg(\frac{4}{3(u-1)}+\frac{4}{3(u^2+u+1)}\bigg)du$$
I am able to solve this however I dont think the value for B is correct
I have the following questions:
1: Am I able to use partial fractions even though the denominator of B is $u^{2}+u+1$
2: Did I do the substitution properly?
3: am I able to do what I did to find the value for A?
Thanks Stax
| $$
\int\frac{4u^{2}du}{u^3-1} = \frac 4 3 \int \frac {dw} w, \text{ etc.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1641154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 mind that $\tan 30^\circ=\frac{1}{\sqrt3}$ and $\tan 120^\circ=-\sqrt3$
As Lab suggested, you can go as follows:
$$\frac{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}+\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)}{\cos (\theta+120^\circ)}-\frac{\sin (\theta-30^\circ)}{\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
Then you have, $$\frac{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}{\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\cos (\theta+120^\circ)\cos (\theta-30^\circ)}}=\frac{m+n}{m-n}$$
The denominators cancel out to again give you
$$\frac{\sin (\theta+120^\circ)\cos (\theta-30^\circ)+\cos (\theta+120^\circ)\sin (\theta-30^\circ)}{\sin (\theta+120^\circ)\cos (\theta-30^\circ)-\cos (\theta+120^\circ)\sin (\theta-30^\circ)}=\frac{m+n}{m-n}$$
$$\frac{\sin (2\theta + 90^\circ)}{\sin 150^\circ}=\frac{m+n}{m-n}$$
Can you complete the rest now?
Hope this helps.
| {
"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
} |
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,
$\DeclareMathOperator{id}{id}f \circ g = g \circ f = \id$
which can be constructed as mirroring at the line $y = x$, thus the graph of $\id$.
The distance of the curves is the minimum (perhaps infimum) of all possible distances between a point of the graph of $f$ and a point of the graph of $g$.
\begin{align}
d(f, g)
&= \min_{x, y} d((x, f(x)), (y, g(y)) \\
&= \min_{x, y} \sqrt{(x - y)^2 + (f(x) - g(y))^2}
\end{align}
With $q = d^2$ we have
$$
q = (x - y)^2 + (e^x - \ln(y))^2
$$
and
$$
\DeclareMathOperator{grad}{grad}
\grad q = (q_x, q_y)
$$
with
\begin{align}
q_x &= 2 (x - y) + 2(e^x - \ln(y)) e^x \\
q_y &= 2 (x - y)(-1) + 2(e^x - \ln(y)) (-(1/y))
\end{align}
To find out where the gradient vanishes (from the graph it seems $(0,1)$),
we solve
\begin{align}
0 &= x - y + e^x - \ln(y) \\
0 &= x - y + (e^x - \ln(y))/y
\end{align}
From this we get
$$
0 = (e^x-\ln(y))(1 - 1/y) \iff \\
1 - 1/y = 0 \vee e^x - \ln(y) = 0 \iff \\
y = 1 \vee e^x = \ln(y)
$$
Inserting $y = 1$ we get
$$
0 = x - 1 + e^x \\
1 - x = e^x \\
$$
which means $x = 0$. So $(0,1)$ is a critical point. Going on analytically would mean to look at the second partial derivatives, and check for the positive defniteness at $(0,1)$ to establish the local minimum.
We get
\begin{align}
q_{xx} &= 2 + 2 (e^x)^2 + 2 (e^x - \ln(y)) e^x \\
q_{xy} &= -2 - 2 e^x / y \\
q_{yy} &= 2y + 2 (1/y)^2 + 2(e^x - \ln(y))(1/y^2) \\
\end{align}
so
\begin{align}
q_{xx}(0,1) &= 2 + 2 + 2 = 6 \\
q_{xy}(0,1) &= -4 \\
q_{yy}(0,1) &= 2 + 2 + 2 = 6
\end{align}
and
$$
\left. H \right|_{(0,1)} =
\left.
\begin{pmatrix}
q_{xx} & q_{yx} \\
q_{xy} & q_{yy}
\end{pmatrix}
\right|_{(0,1)}
=
\begin{pmatrix}
6 & -4 \\
-4 & 6
\end{pmatrix}
$$
Using linear algebra or a computer algebra system (me lazy) one finds the eigenvalues $2$ and $10$, both positive, so the Hesse matrix is positive definite at $(0,1)$ and it must be a minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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 $[x] = x - \{x\}$ we get our limit equal to
$$\lim_{n\to\infty}\left\{\sqrt{n^2+n+1}\right\}$$
Moving the limit inside the fractional part function and replacing $n=\frac 1h \; \text {where } h\to0^+$ we get
$$\left\{\lim_{h\to0^+} \frac{\sqrt{h^2+h+1}}h\right\}$$
Applying L'Hospital Rule, we get our limit equal to $\left\{\frac 12\right\}$ which is $0$.
The problem:
The answer in the answer key is $\frac12$. So here, the only problem I seem to find in my solution is that $n\in I$ and simply assuming $n = \frac 1h$ doesn't ensure our $n$ to be an integer.
Can anyone provide a way to either correctly assume a new value for $n$ or any alternate way to solve this?
| 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 $\left\lfloor\sqrt{n^2+n+1}\right\rfloor=n$ eventually.
Thus we can see that \begin{align}\sqrt{n^2+n+1}-\lfloor \sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n\\ &=\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)+\frac{1}{2}\\ &=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}+\frac{1}{2}\\ &\to\frac{1}{2}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
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 spectral density of the process?
For instance, for the Ornstein-Uhlenbeck process, the steps would be
*
*Find autocorrelation function. This is easily done with the Itô isometry and the properties of the Wiener process.
*Compute Fourier transform
And that's it, because the Wiener-Khinchin theorem assures us that the process autocorrelation and the PSD are a Fourier transform pair as long as the process is wide sense stationary (and the Ornstein-Uhlenbeck process satisfies this). However, The Ornstein-Uhlenbeck process is too nice in reality.
What can be done to obtain the PSD in a more general, non stationary case? What can be then said about the autocorrelation function relation to the PSD? Could you provide any references with theory/worked examples?
For instance, I would really appreciate your help in the not-so-nice (but still nice) second order harmonic oscillator with additive noise term
$$
m\ddot{x} + \gamma\dot{x} + \delta x = \sigma \eta(t)
$$
with
$$
\eta(t) = \frac{\mathrm{d} W_t}{\mathrm{d}t}
$$
(forgive my abuse of notation).
| 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
$$
m \ddot{x} + \gamma \dot{x} + \delta x = \sigma \eta(t)
$$
Performing a change of variables, $x = e^{\frac{-\gamma}{2m}t}x_1$, we get
$$
m \ddot{x}_1 + \left(\delta - \frac{\gamma^2}{4m}\right)x_1 = \sigma e^{\frac{\gamma}{2m}t}\eta(t)
$$
thus eliminating $\dot{x}$. Setting $a = \frac{\delta}{m} - \frac{\gamma^2}{4m^2}$, $b = \frac{\sigma}{m}$ and rewriting the equation as a first order linear system with
$$
X =
\begin{pmatrix}
x_2\\
v_2
\end{pmatrix}
$$
where $v_2 = \dot{x_2}$, we get in Ito's notation
\begin{equation}
X =
\begin{pmatrix}\label{eq:complete}
0 & 1 \\
-a & 0
\end{pmatrix}
\cdot X \mathrm{d}t +
\begin{pmatrix}
0\\
b e^{\frac{\gamma}{2 m}t}
\end{pmatrix}
\cdot \mathrm{d} W_t
\end{equation}
The solution of a linear homogeneous SDE is
$$
X_t = e^{\int_0^t A(t)\mathrm{d}t}\cdot X_0 + e^{\int_0^t A(t)\mathrm{d}t}\cdot \int_0^t e^{-\int A(s)\mathrm{d}s}\sigma(s)\mathrm{d}W_s
$$
where $A(t)$ is the (generally vector) coefficient of $X$. For this SDE a fundamental matrix solution of the associated homogeneous noise-free system is
$$
\Phi(t) =
\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a} \\
-\sqrt{a} \sin\sqrt{a}t & \cos \sqrt{a}t
\end{pmatrix}
$$
The determinant of this matrix is 1, so its inverse matrix will be
$$
\Phi^{-1}(t) = e^{-\int A(\tau)\mathrm{d}\tau} = \det \Phi(t)^{-1}\cdot
\begin{pmatrix}
\cos \sqrt{a}t & -\sin \sqrt{a}t/\sqrt{a} \\
\sqrt{a} \sin\sqrt{a}t & \cos \sqrt{a}t
\end{pmatrix}
=
\begin{pmatrix}
\cos \sqrt{a}t & -\sin \sqrt{a}t/\sqrt{a} \\
\sqrt{a} \sin\sqrt{a}t & \cos \sqrt{a}t
\end{pmatrix}
$$
and hence we can solve the complete system. We are interested in the first component of $X$, the position (we will only calculate the PSD of $x$, although this can be done considering the full matrix system)
\begin{equation*}
\begin{split}
x_1(t) = &
\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a}
\end{pmatrix}
\cdot
\begin{pmatrix}
x_1(0)\\
v_1(0)
\end{pmatrix}
+
\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a}
\end{pmatrix}
\cdot
\int_0^{t}
b e^{\frac{\gamma}{2m}r}\cdot
\begin{pmatrix}
-\sin \sqrt{a}r/\sqrt{a}\\
\cos \sqrt{a}r
\end{pmatrix}
\mathrm{d}W_r
\end{split}
\end{equation*}
Finally, $x_1(t) = e^{\frac{\gamma t}{2m}}x(t)$, so
\begin{equation}
\begin{split}
x(t) = &
e^{-\frac{\gamma t}{2m}}\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a}
\end{pmatrix}
\cdot
\begin{pmatrix}
x(0)\\
v(0) + \frac{\gamma}{2m}x(0)
\end{pmatrix}
+\\
&
e^{-\frac{\gamma t}{2m}}
\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a}
\end{pmatrix}
\cdot
\int_0^{t}
b e^{\frac{\gamma}{2m}r}\cdot
\begin{pmatrix}
-\sin \sqrt{a}r/\sqrt{a}\\
\cos \sqrt{a}r
\end{pmatrix}
\mathrm{d} W_r
\end{split}
\end{equation}
We see that, after a transient time, only the term depending on $\mathrm{d} W_r$ remains, so the first moment of the process is zero. Now, applying Ito's isometry to calculate the covariance we get
$$
\mathbb{E} \left[ \int_0^{T} f(u)\,\mathrm{d} W_u \int_0^S f(v) \,\mathrm{d} W_v \right] =
$$
$$
b^2 e^{-\frac{\gamma (t + s)}{2m}}
\begin{pmatrix}
\cos \sqrt{a}t & \sin \sqrt{a}t/\sqrt{a}
\end{pmatrix}
\cdot \mathbb{E} \left[ \int_0^{\min(t,s)}
e^{\frac{\gamma}{m}u}
\begin{pmatrix}
\frac{\sin^2 \sqrt{a}u}{a} & -\frac{\sin \sqrt{a}u \cos \sqrt{a}u}{\sqrt{a}}\\
-\frac{\sin \sqrt{a}u \cos \sqrt{a}u}{\sqrt{a}} & \cos^2 \sqrt{a}u
\end{pmatrix}
\, \mathrm{d} u \right]
\cdot
\begin{pmatrix}
\cos \sqrt{a}s \\
\sin \sqrt{a}s/\sqrt{a}
\end{pmatrix}
$$
This is a quite uninteresting calculation. After simplification one gets
$$
R(\tau) = \frac{b^2 m^2 e^{\frac{-\Gamma |\tau|}{2}}\left( 2 \sqrt{a}m \cos(\sqrt{a}|\tau|) + \gamma \sin(\sqrt{a}|\tau|) \right)}{\sqrt{a}(\gamma^3 + 4a\gamma m^2)}
$$
where we have defined the normalized damping constant $\Gamma = \frac{\gamma}{m}$.
From the expression of the autocorrelation we see that $R(t,\tau) = R(\tau)$: therefore, the process is wide-sense stationary and the conditions to apply the Wiener-Khinchin theorem are satisfied. The Fourier transform of this autocorrelation function is the power spectral density
$$
S(f) = \frac{16b^2}{\Gamma^4 + 8(a + 4\pi^2f^2)\Gamma^2 + 16(a-4\pi^2f^2)^2}
$$
which, after replacing the variables and some rearranging takes the simpler expression
$$
S(\omega) = \frac{\sigma^2/m^2}{(\omega_0^2 - \omega^2)^2 + \Gamma^2 \omega^2}
$$
where we have replaced the unitary ordinary frequency Fourier transform (in terms of $f$) by the non-unitary angular frequency Fourier transform.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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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:
$$\frac{1\cdot2^2+2\cdot3^2+\cdots+k(k+1)^2}{1^2\cdot2+2^2\cdot3+\cdots+k^2(k+1)}=\frac{3k+5}{3k+1}$$
But verification for $n=k+1$ is creating problem.because $k$ is present in denominator too which is hindering use of equation obtained from inductive hypothesis. Please suggest something.
| 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
$$
P(n):(3n+1)\sum_{k=1}^n k(k+1)^2-(3n+5)\sum_{k=1}^n k^2(k+1)=0
$$
I leave it to you to show that $P(1)$ is true (that does not seem to be your problem). Assume that $P(n)$ is true for some $n\in\mathbb N$. The left-hand side of $P(n+1)$ is then given by
$$
(3n+4)\sum_{k=1}^{n+1}k(k+1)^2-(3n+8)\sum_{k=1}^{n+1}k^2(k+1)
$$
which equals (here we split the sums into the first $n$ terms and the last)
$$
(3n+4)\Bigl[\sum_{k=1}^n k(k+1)^2+(n+1)(n+2)^2\Bigr]-(3n+8)\Bigl[\sum_{k=1}^n k^2(k+1)+(n+1)^2(n+2)\Bigr].
$$
Next, since $P(n)$ was assumed to be true, we cancel $3n+1$ times the sum in the first bracket with $3n+5$ times the sum in the second bracket. Thus, the expression above equals
$$
3\sum_{k=1}^n\bigl[k(k+1)^2-k^2(k+1)\bigr]+(n+1)(n+2)\bigl[(3n+4)(n+2)-(3n+8)(n+1)\bigr]
$$
which simplifies to
$$
3\sum_{k=1}^n k(k+1)-n(n+1)(n+2).
$$
Either we know that this happens to be zero, or we don't. If we don't, we prove it by induction. Thus, let $Q(n)$ be the statement that the expression above is zero. It is easy to verify that $Q(1)$ is true. Next, if $Q(n)$ for some $n\in\mathbb N$ is true, i.e if
$$
3\sum_{k=1}^n k(k+1)=n(n+1)(n+2),
$$
then we find that
$$
3\sum_{k=1}^{n+1}k(k+1)=n(n+1)(n+2)+3(n+1)(n+2)=(n+1)(n+2)(n+3),
$$
which means that $Q(n+1)$ is true. We conclude by induction that $Q(n)$ is true for all $n\in\mathbb N$ and consequently (by induction again) that $P(n)$ is true for all $n\in\mathbb N$.
| {
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"url": "https://math.stackexchange.com/questions/1647955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 got was $\ln10 + 1$.
| 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}\end{align}
Can you proceed from it?
| {
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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 $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.
| 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 angle $\arccos\left(\frac{2}{\sqrt 5}\right)$ is the angle
opposite the shorter leg in a right triangle with legs $1$ and $2$,
so
$\arccos\left(\frac{2}{\sqrt 5}\right) = \arctan\left(\frac12\right)$,
and the fact you want to prove is therefore equivalent to
$$
\tan\left(3 \arctan\left(\frac12\right)\right) = \frac{11}{2}.
$$
Using the triple-angle formula
$$
\tan(3x) = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}
$$
with $x = \arctan\left(\frac12\right)$, so $\tan x = \frac12$ and
\begin{align}
\tan\left(3 \arctan\left(\frac12\right)\right)
& = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \\
& = \frac{3 \left(\frac12\right) - \left(\frac12\right)^3}
{1 - 3 \left(\frac12\right)^2} \\
& = \frac{11}{2}
\end{align}
which is what you needed to show.
| {
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"question_score": "9",
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} |
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:
*
*$0^2+1^2=1^2$
*$3^2+4^2=5^2$
*$20^2+21^2=29^2$
*$119^2+120^2=169^2$
*$696^2+697^2=985^2$
*$4059^2+4060^2=5741^2$
*$23660^2+23661^2=33461^2$
*$137903^2+137904^2=195025^2$
*$803760^2+803761^2=1136689^2$
*$4684659^2+4684660^2=6625109^2$
Now what's so interesting? I discovered that any $c$, divided by the previous (for example $5/1$ or $29/5$) limits to $5.828427...=\left(\frac{1}{\sqrt2-1}\right)^2=\sqrt8+3$. My question: why?
|
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,
$$\Big(\frac{b-1}{2}\Big)^2+\Big(\frac{b+1}{2}\Big)^2=c^2\tag1$$
and factoring (WA link), we get the condition,
$$b^2-2c^2=-1$$
This is a Pell equation, and given one integer solution we can find an infinite more (this link, eqn. 35,36). Since this is the negative Pell equation, we use odd powers $n$,
$$\begin{aligned}
b_n &= \frac{S^n+(-S)^{-n}}{2} = 1,\, 7,\, 41,\, 239,\dots\\
c_n &= \frac{S^n-(-S)^{-n}}{2\sqrt{2}} = 1,\, 5,\, 29,\, 169,\dots
\end{aligned}\tag2$$
The $b_n$ is always odd, so $(1)$ are integers.
II. Answer:
Note that the contribution of $(-S)^{-n} \approx (-0.4142)^n$ in $(2)$ rapidly diminishes as $n$ becomes large. Thus,
$$c_n\approx \frac{S^n}{2\sqrt{2}}$$
Taking the ratio of $c_n$ for successive odd powers $n$,
$$
\frac{c_{2m+3}}{c_{2m+1}} = \Big( \frac{S^{2m+3}}{2\sqrt{2}} \Big)\Big(\frac{2\sqrt{2}}{S^{2m+1}}\Big)= S^2=(1+\sqrt{2})^2 = 5.828427\dots
\tag3$$
and we recover your observation.
III. Neat stuff
Just like the golden ratio can be found in the pentagon, the silver ratio $S$ is in the octagon,
$\hskip2.8in$
It is also an infinitely nested radical,
$$S = 1+\sqrt{2} = 2\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{2^2}+\dots}}}$$
and in that nice continued fraction above using $e^{-\pi}$, as well as in a pi formula, etc, etc.
| {
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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 steps but the steps were not to be found. Despite lack of steps, the answer is $(\sqrt {2+\sqrt{3}})/2$
Here are my steps so far:
\begin{align*}
\sin(135^\circ - 30^\circ) & = (\sin 135^\circ\cos 30^\circ)-(\cos 135^\circ \sin 30^\circ)\\
& = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\\
& = \left(\frac{\sqrt{6}}{4}\right)-\left(-\frac{\sqrt{2}}{4}\right)
\end{align*}
[insert final simplification step]
$(\sqrt {2+\sqrt{3}})/2$
What's the missing step here?
| $x=\frac{\sqrt6+\sqrt2}4$
$====================$
$x^2=\frac{6+2\sqrt{12}+2}{16}=\frac{8+2\sqrt{12}}{16}=\frac{8+4\sqrt3}{16}=\frac{2+\sqrt3}4$
$====================$
$x=\frac{\sqrt{2+\sqrt3}}2$
| {
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"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 $35$ condition seems a little hard so since we immediately know it is divisible by $5$ we just have to work on divisibility by $7$. I would use the divisibility test then for the rest of the digits $\overline{abcdef}$. We must have that $\overline{abcdef}-10$ is divisible by $7$. This is equivalent to testing divisibility by $7$ for $100000a+10000b+1000c+100d+10e+f-10$. Taking this all mod $7$ we get $2a+4b+c+2d+3e+f \equiv 3 \mod 7.$ Now how do I find all such pairs $(a,b,c,d,e,f)$?
| 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
$$2 \cdot 10 \equiv -1 \pmod{7} \\
2 \cdot 10^2 \equiv 3 \pmod{7} \\
2 \cdot 10^3 \equiv 2 \pmod{7} \\
2 \cdot 10^4 \equiv -1 \pmod{7} \\
2 \cdot 10^5 \equiv 3 \pmod{7} \\
2 \cdot 10^6 \equiv 2 \pmod{7} $$
and you have to figure out which numbers from this set you need to add to get $2 \pmod{7}$. You can reduce the problem to solving
$$-a+3b+2c \equiv 2 \pmod{7}$$
where $$a,b,c \in \{0 ,1, 2 \}$$
| {
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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 $\mathop{\sum\sum}_{0\le i<j\le 20}(a_j)^2$
$$A=\mathop{\sum\sum}_{0\le i<j\le 20}(a_j)^2=\frac{1}{2}\left(\mathop{\sum\sum}_{0\le i,j\le20}(a_j)^2-\sum_{0\le i\le20}a_i^2\right)=\frac{1}{2}\left(41\binom{40}{20}-\dbinom{40}{20}\right)=20\dbinom{40}{20}$$
Then I calculated the value of $\mathop{\sum\sum}_{0\le i<j\le 20}(a_ia_j)$
$$B=\mathop{\sum\sum}_{0\le i<j\le 20}(a_ia_j)=\frac{1}{2}\left(\mathop{\sum\sum}_{0\le i,j\le20}(a_ia_j)-\sum_{0\le i\le20}a_i^2\right)=\frac{1}{2}\left(2^{40}-\binom{40}{20}\right)$$
The final answer I got was $2A-2B=41\binom{41}{20}-2^{40}$
But given answer is $41\binom{42}{20}-2^{40}$
| 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 same value you have. This suggests a factor of $2$ error in your computation of $A$. I compute: $$\begin{align}
A &= \sum_{i=0}^{20} \sum_{j=i+1}^{20} a_j^2 \\
&= \frac{1}{2}\left( \sum_{i=0}^{20} \sum_{j=0}^{20} a_j^2 - \sum_{i=0}^{20} a_i^2 \right) \\
&= \frac{1}{2}\left( 21 \binom{40}{20} - \binom{40}{20}\right) & &\text{(contra. your $41 \dots$) }\\
&= 10 \binom{40}{20},
\end{align}$$ getting the expected factor of $2$ discrepancy.
With this new $A$ and the old $B$, we have $$\begin{align}
\sum_{i=0}^{20} \sum_{j=i+1}^{20} (a_i - a_j)^2 &= 2A - 2B \\
&= 20 \binom{40}{20} - \left( 2^{40} - \binom{40}{20} \right) \\
&= 21 \binom{40}{20} - 2^{40}.
\end{align}$$
The given answer is more than $10$ times larger than this, so I am dubious about its correctness.
Aside: Checking numerically: $$ \begin{align}
\sum_{i=0}^{20} \sum_{j=i+1}^{20} (a_i - a_j)^2 &= 1\,795\,265\,477\,444. \\
21 \binom{40}{20} - 2^{40} &= 1\,795\,265\,477\,444. \\
41 \binom{41}{20} - 2^{40} &= 9\,934\,774\,798\,244. \\
41 \binom{42}{20} - 2^{40} &= 19\,965\,944\,276\,444.
\end{align}$$
| {
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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: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.
Is the book right or me ?
| I think that the book is wrong.
$$1-\frac{4}{3x}+\frac{2}{x^2}=\frac{3x^2}{3x^2}\cdot \frac{1-\frac{4}{3x}+\frac{2}{x^2}}{1}=\frac{3x^2-4x+6}{3x^2}=\frac{2x^2+(x^2-4x+6)}{3x^2}$$
$$=\frac{2x^2+0}{3x^2}=\frac{2}{3},x\ne 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $x -1$.
In other words we need to find $x \equiv 0 \mod k$ and $x \equiv 1 \mod (j)$ where $m = j*k$ and $\gcd(j,k) = 1$ (Note if $m = \prod p_i^{k_i}$ there will by $2^i$ such sets of equations). According to the chinese remainder theorem there will be one solution to each set of problems. (So $2^i$ solutions total).
Example 1
Solve $x^2 = x \mod 12 \iff x(x-1) \equiv 0 \mod m$. $12 = 3 \times 4$ so there will be the following solutions:
A) $x \equiv 0 \mod 12$ and $x \equiv 1 \mod 1$ which has one solution. via CRT.
B) $x \equiv 0 \mod 1$ and $x \equiv 1 \mod 12$ which has one solution.via CRT.
C) $x \equiv 0 \mod 3$ and $x \equiv 1 \mod 4$ which has one solution.via CRT.
D) $x \equiv 0 \mod 4$ and $x \equiv 1 \mod 3$ which has one solution. via CRT.
The solution to A) is $x= 0$, B) is $x =1$, C) $x = 9$ D)$ x = 4$
Example 2
Solve $\ x^2 = x \mod 360$.
$360 \equiv 8 \times 9 \times 5$ so there will be 8 solutions:
$ (x \equiv 0 \mod 360) \rightarrow x = 0$
$ (x \equiv 1 \mod 360) \rightarrow x = 1$
$ (x \equiv 0 \mod 8) \ AND \ \ (x \equiv 1\mod 45) \rightarrow x = 136$
$ (x \equiv 0 \mod 45) \ AND \ \ (x \equiv 1 \mod 8) \rightarrow x = 225$
$ (x \equiv 0 \mod 9) \ AND \ \ (x \equiv 1 \mod 40) \rightarrow x = 81$
$ (x \equiv 0 \mod 40) \ AND \ \ (x \equiv 1 \mod 9) \rightarrow x = 280$
$ (x \equiv 0 \mod 5) \ AND \ \ (x \equiv 1 \mod 72) \rightarrow x = 145$
$ (x \equiv 0 \mod 72) \ AND \ \ (x \equiv 1 \mod 5) \rightarrow x = 216$.
Which would have been practically impossible by trial and error.
| {
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integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_{2}^{4} \frac{1-\sin^2u}{\sin u}du=\int_{2}^{4} \frac{1}{\sin u}du-\int_{2}^{4} {\sin u}du$$
$$=\ln\left(\tan\left(\frac{u}{2}\right)\right)+\cos u$$
$\frac{x}{4}=\sin u$
$\tan u=\frac{\sqrt{x}}{x^2-16}$
$\ln\left(\frac{{x}}{8(\sqrt{x^2-16})}\right)+\left(\frac{\sqrt{16-x^2}}{4}\right)$ form $2$ to $4$
but I get a $\frac{4}{0}$ the end result according to Wolfram is $1.80$
| Hint:
Multiply Numerator Denomerator by $x$ then substitute $u^2=16-x^2$
| {
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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 tried finding the coefficients of terms after splitting the terms..: it becomes
$$(\frac{1}{1\cdot 6}-\frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}-\frac{1}{6\cdot4}) + (\frac{1}{6\cdot 2} - \frac{1}{3\cdot2} +\frac{1}{4\cdot 2} -\frac{1}{6\cdot5})+\cdots.$$ I tried solving it but am getting nowhere .Someone please help me with this sum.
| 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 only have to subtract $1/(n+1)(n+2)(n+3)$ from $1/(1\cdot2\cdot3)$ and then devide it by 3, because all other terms cancel out. This gives us the result: $$\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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 $-2x-2 - x +1 = -3x - 1 > 3.$ This holds if $x < -4/3$.
Thus the complete solution is $x \in (-\infty,-4/3) \cup (0,\infty)$.
| {
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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 $2x\omega + \omega^2$. Hence, the increment of $x$ itself, which is $\omega$, has the
ratio to the increment of the square, which is $2x\omega+\omega^2$, as $1$ to $2x+\omega$. This
ratio reduces to $1$ to $2x$, at least when $\omega$ vanishes.
$f(x) = x^2$
$f(x+w) = (x+\omega)^2 = x^2 + 2x\omega + \omega^2$
$(x+\omega)^2 - x^2 = 2x\omega+\omega^2$
And then the part I don't understand where $\omega = 2x\omega+\omega^2$ goes to "as $1$ to $2x+\omega$".
| 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
$$\frac {(x + \omega) - x} {f(x + \omega) - f(x)} = \frac {\omega} {2 \omega x + \omega ^2} = \frac 1 {2 x + \omega}$$
which, when $\omega \to 0$, is exactly $\frac 1 {2x}$.
This prepares you mentally for later considering the limit of the reversed fraction:
$$\lim \limits _{\omega \to 0} \frac {f(x + \omega) - f(x)} {(x + \omega) - x} = 2x ,$$
which is an illustration of the concept of "derivative".
| {
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"timestamp": "2023-03-29T00:00:00",
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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 class:
EX: m = 7
$-\frac{7 -1}{2}$ Which gave us a residue system of {- 3, -2 , -1, 0 , 1, 2, 3}
OR {0, 1, 2, 3, 4, 5, 6}
-Based off this example from class I am a bit confused as to how to approach this problem, any help is appreciated.
| \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 one of $\{0,1,2,3,4,5,6\}$ modulo $7$, and thus your set is a complete residue system.
| {
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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 show that $x^2_n \geqslant 2$ by induction:
*
*Basis step: $x_1=2, x_1^2 = 4 >2$.
*Inductive step:
Assume we have some $k$ for which $x^2_K \geqslant 2$, then
\begin{align*}
(x_{k+1})^2 =&~ \bigg(\dfrac{1} {2} \bigg(x_k + \dfrac{2} {x_k}\bigg)\bigg)^2\\
=&~ \dfrac{1} {4} x_k^2 + 1 + \dfrac{1} {x_k^2}\\
\geqslant&~ \dfrac {1} {4} x_k^2 + 1 \geqslant 2\\
&~\dfrac {1} {4} x_k^2 \geqslant 1 \quad \forall~ n \geqslant 2
\end{align*}
How do I use this to show that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$
and hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$?
Thanks!!
| 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{align*}
x_{n}-x_{n+1} &\leq \frac{x_{1}^{2}-2}{2^{n/2}} \\
&= \frac{2}{2^{n/2}}
\end{align*}
Hence $\lim_{n\to \infty} (x_{n}-x_{n+1})=0$ and $\{ x_{n} \}$ is decreasing but bounded below,
$L=\lim_{n\to \infty} x_{n}$ exists.
Therefore $L=\frac{1}{2} \left( L+\frac{2}{L} \right) \implies L=\sqrt{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $x\in[1,4]\Rightarrow y_1\le y_2$.
$$A=\int_0^1 (y_1-y_2)\mathrm dx+\int_1^4 (y_2-y_1)\mathrm dx=\int_0^1 (1-x)\sqrt{4x-x^2}\mathrm dx+\int_1^4 (x-1)\sqrt{4x-x^2}\mathrm dx$$
How to solve integrals $\int \sqrt{4x-x^2}\mathrm dx$ and $\int x\sqrt{4x-x^2}\mathrm dx$?
Substitution $$u=\sqrt{\frac{x}{4-x}}\Rightarrow du=\frac{2}{(x-4)^2\sqrt{\frac{x}{4-x}}}dx$$ doesn't seems to work.
| 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:
$$I_1=\int \sqrt{4x-x^2}dx=-32\int \frac{z^2}{(1+z^2)^3}dz$$
$$I_2=\int x \sqrt{4x-x^2}dx=-128\int \frac{z^2}{(1+z^2)^4}dz$$
$$I_1=32\left(\int \frac{1}{(1+z^2)^3}dz -\int \frac{1}{(1+z^2)^2}dz \right)$$
$$I_2=128\left(\int \frac{1}{(1+z^2)^4}dz -\int \frac{1}{(1+z^2)^3}dz \right)$$
These integrals contain only rational functions. Moreover, we know the integral:
$$\int \frac{1}{1+z^2}dz=\arctan(z)+C$$
Note that:
$$z(0)=\infty$$
$$z(1)=\sqrt{3}$$
$$z(4)=0$$
I don't know what methods you are supposed to use, but I myself would use a parameter. (I will omit the constants of integration).
$$I_0=\int \frac{1}{1+a^2z^2}dz=\frac{\arctan(az)}{a}$$
$$\frac{\partial I_0}{\partial a}= -\int \frac{2az^2}{(1+a^2z^2)^2}dz=\frac{z}{a(1+a^2z^2)}-\frac{\arctan(az)}{a^2}$$
$$a \to 1$$
$$\int \frac{2z^2}{(1+z^2)^2}dz=\arctan(z)-\frac{z}{1+z^2}$$
$$\int \frac{1}{1+z^2}dz-\int \frac{1}{(1+z^2)^2}dz=\frac{1}{2} \left( \arctan(z)-\frac{z}{1+z^2} \right)$$
$$\int \frac{1}{(1+z^2)^2}dz=\frac{1}{2} \left( \arctan(z)+\frac{z}{1+z^2} \right)$$
The same way you can calculate the rest of the itegrals.
| {
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"timestamp": "2023-03-29T00:00:00",
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"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 $\lim_{x\rightarrow1^-} \sum^{\infty}_{n = 0}a_n x^n$. (Abel Summability)
I already did this for another problem. I was given the series $\sum^{\infty}_{n = 0}b_n = \frac{1}{2\cdot 1} - \frac{1}{3\cdot 2} + \frac{1}{4\cdot 3} - \frac{1}{5\cdot 4} + \cdots$
I showed that $\sum^{\infty}_{n = 0}b_n x^n = \frac{\ln(1+x)}{x^2}+\frac{\ln(1+x)}{x}-\frac{1}{x}\rightarrow 2\ln(2)-1$ as $x\rightarrow1^- \Rightarrow \sum^{\infty}_{n = 0}b_n = 2\ln(2)-1$. (This checks out numerically, as well)
However, I'm having trouble finding the functional representation of $ \sum^{\infty}_{n = 0}a_n x^n$
| 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
$$f(y) = \sum_{k=0}^{\infty} \frac{(-1)^k y^{3 k+1}}{3 k+1}$$
$$\implies f'(y) = \sum_{k=0}^{\infty} (-1)^k y^{3 k} = \frac1{1+y^3} $$
$$\implies f(y) = \int \frac{dy}{1+y^3} = -\frac{1}{6} \log \left(y^2-y+1\right)+\frac{1}{3} \log (y+1)+\frac{\arctan{\left(\frac{2 y-1}{\sqrt{3}}\right)}}{\sqrt{3}} +C$$
$$f(0)=0 \implies C=\frac{\pi}{6 \sqrt{3}} $$
The above integral is evaluated using a partial fraction decomposition. Thus, the original sum is equal to
$$\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{3 k+1} = \frac{\pi x^{-1/3}}{6 \sqrt{3}} + \frac{x^{-1/3}}{6} \log{\left [\frac{\left (1+x^{1/3}\right )^2}{1-x^{1/3}+x^{2/3}} \right ] } + \frac{x^{-1/3}}{\sqrt{3}} \arctan{\left(\frac{2 x^{1/3}-1}{\sqrt{3}}\right)}$$
The limit as $x \to 1^-$ of this sum is then equal to
$$\frac{\pi}{3 \sqrt{3}} + \frac13 \log{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \times 1 + 8 \times 2) +1, (8 \times 1 + 8 \times 2 + 8 \times 3 + 8 \times 4) +1 ....$
equals to
$1^2, 3^2, 5^2, 7^2 ...$
and my brain stuck here and cannot find out the relationship between two series.
Can someone give me a hint why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$
| $$S_1=1=(2\cdot1-1)^2,\\
S_n-S_{n-1}=(2n-1)^2-(2n-3)^2=8n-8.$$
You can establish the formula using the well-known triangular numbers,
$$1+2+3+\cdots n=\frac{n(n+1)}2.$$
Then
$$S_n=8\frac{(n-1)n}2+1=4n^2-4n+1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 seems to work. Let's see:
We know that if $P_a(x)$ is to have a root with multiplicity at least two, then $P_a(x)$ and its derivative $P'_a(x)$ must have a common root. So, in the first approach I have computed the resultant of $P_a(x)$ and $P'_a(x)$ with the aid of Mathematica, which gives $512 - 83 a^2 - 8 a^4$. This resultant must be equal to zero, hence we obtain the following four solutions for $a$:
$$
a = -\sqrt{-(83/16) + (37 \sqrt{17})/16}, \\
a = \sqrt{-(83/16) + (37 \sqrt{17})/16}, \\
a = -(1/4) i \sqrt{ 83 + 37 \sqrt{17}}, \\
a = 1/4 i \sqrt{83 + 37 \sqrt{17}}
$$
In the second approach, I have computed the GCD of $P_a(x)$ and $P'_a(x)$, imposing the condition that the GCD must be at least linear. After successive divisions we get the remainder
$$
\frac{1536 + 861 a^2 + 864 a^4}{3 (16 + 12 a^2)^2},
$$
which we want to become zero. Hence we obtain the following four solutions for $a$:
$$
a = -\sqrt{-(287/576) - (i \sqrt{507455})/576} \\
a = \sqrt{-(287/576) - (i \sqrt{507455})/576} \\
a = -\sqrt{-(287/576) + (i \sqrt{507455})/576} \\
a = \sqrt{-(287/576) + (i \sqrt{507455})/576}.
$$
The first obvious problem is that both sets of solutions are different. Moreover, $a = 2$ should be a solution, since $x^4 - x^2 + 2x + 2$ has root $-1$ with multiplicity two, but this solution is not included in any of the above sets. Where is the error?
| 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 equation verified by the abscissas of intersection point(s) of curve $y=x^4-x^2$ and straight line with equation $y=-a x - 2$. The desired values of $a$ correspond to the two tangency cases.
| {
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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-2)} \right) \\
&= n \left(\frac{n-2}{n}\right)^m \left( 1+ O\left(\frac{(\log n)^2}{n}\right) \right)
\end{align*}
I have tried breaking down the initial binomial term, but it gets messy very quickly. Can someone shed some light here?
Edit: I forgot to add what is probably a relevant fact: here $m=\frac{1}{2}n(\log n+w(n))$ and $w(n)=o(\log n)$.
| 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 obtain
\begin{align*}
&\binom{\binom{n-1}{2}}{m}\binom{\binom{n}{2}}{m}^{-1}\\
&\quad=\left(\frac{1}{m!}\binom{n-1}{2}^m\prod_{i=1}^{m-1}\left(1-\frac{i}{\binom{n-1}{2}}\right)\right)
\left(m!\binom{n}{2}^{-m}\prod_{i=1}^{m-1}\left(1-\frac{i}{\binom{n}{2}}\right)^{-1}\right)\tag{1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}\left(1-\frac{2i}{(n-1)(n-2)}\right)
\left(1-\frac{2i}{n(n-1)}\right)^{-1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\left(\frac{n(n-1)(n-2)-2ni}{n(n-1)(n-2)}\right)\left(\frac{n(n-1)(n-2)-2i(n-2)}{n(n-1)(n-2)}\right)^{-1}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\frac{n(n-1)(n-2)-2ni}{n(n-1)(n-2)-2i(n-2)}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\frac{n(n-1)(n-2)-2i(n-2)-4i}{n(n-1)(n-2)-2i(n-2)}\\
&\quad=\left(\frac{n-2}{n}\right)^m\prod_{i=1}^{m-1}
\left(1-\frac{4i}{n(n-1)(n-2)-2i(n-2)}\right)\\
\end{align*}
Comment:
*
*In (1) we use the identity from above for both binomial coefficients.
| {
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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 single round is $\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}$. In order for Alice to win the game, these repeats one must happen on the single rounds. The probability of rolling the first repeats one to on the $k-th$ round is $\dfrac{35}{36}^{k-1}\dfrac{1}{36}$. So the probability of Alice wins the game is $\sum_{n=1}\dfrac{35}{36}^{2(k-1)}\dfrac{1}{36}=\dfrac{1}{36}(1+\dfrac{35^2}{36^2}+(\dfrac{35^2}{36^2})^2+(\dfrac{35^2}{36^2})^3+...=\dfrac{1}{36}\dfrac{1}{1-\dfrac{35^2}{36^2}}=\dfrac{36}{71}$
Can someone check my solution for me please? Thank you.
| 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 $\frac{1}{6}$.
(ii) For Bob to lose in the second round, the first three tosses have to be different, and there must be a match on the fourth toss. The probability the first $3$ tosses are different is $\frac{5}{6}\cdot \frac{4}{6}$. Given that they are different, the probability of a match on the fourth toss is $\frac{3}{6}$. So the probability Bob loses in the second round is $\frac{5}{6}\cdot \frac{4}{6}\cdot \frac{3}{6}$.
(iii) We leave it to you to find the probability Bob loses in the third round. Then add up.
| {
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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 mathematics knowledge. How do I find the positive integer solutions for this equation?
| 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.
$ay^2-xy-2ax=0\Rightarrow \color{red}{a|xy}\qquad (1)$.
$ay^2-xy-2ax=0\Rightarrow 2ay=x\pm \sqrt{x^2+8a^2x}\Rightarrow \color{red}{x^2+8a^2x=z^2}\Rightarrow x^2+8a^2x=w^2x^2\Rightarrow \color{red}{x|8a^2}\qquad (2)$.
So $0<x|2^3\cdot3^2\cdot11^2\cdot61^2$ hence there are at most
$(3+1)(2+1)(2+1)(2+1)=108$ possibilities for solution $x$.
(There are not infinitely many solutions!)
From $(1)$ and $(2)$ one has $xy=2013k$ and $x^2+8a^2x=\square$ (= square)
We try $k\in \{1,2\}$
►$k=1$
The possible solutions are $(x,y)\in \{(1,2013),(3,671),(11,183),(33,61)\}$ and the opposite $(y,x)$ with the four given possibilities. Here we examine, using $(2)$, how squares, with eventual solutions can agree
$$\begin{cases} 1+8(2013^2=\square\\9+24(2013^2=\square\\11^2+8\cdot 11(2013)^2=\square\\33^2+8\cdot 33(2013)^2=\square\\(2013)^2(1+8\cdot2013)=\square\\(671)^2(1+8\cdot 671\cdot 3^2)=\square\\(183)^2(1+8\cdot 183\cdot11^2)= \square\\(61)^2(1+8\cdot61\cdot 33^2=\square\end{cases}$$
(the reason of the factorization in four last possibilities is to simplify the calculation).
None of these eight possibilities is square; then they are discarded.
►$k=2$
The $16$ possibilities of eventual solution for $xy=4026$, like for $k=1$, give not squares so they must be discarded.
The smallest solution, in case it exists, seems to be large. Without computer I doubt
a student with the middle school math knowledge could find out a solution.
| {
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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 steps.
| 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:
$$\frac{d}{dx}\left(\arctan \frac{2x}{1+x^2}\right)$$
$$=\frac{d}{d\theta}\left[\arctan (\sin 2\theta)\right]\cdot \frac{d\theta}{dx}$$
$$=\frac{1}{1+\sin^2 \theta}\cdot \frac{d}{d\theta}\left(\sin 2\theta\right)\cdot \frac{1}{\sec^2 \theta}$$
$$=\frac{2\cos 2\theta}{1+\sin^2 \theta}\cdot \cos^2 \theta$$
$$=\frac{2\cdot \left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)}{\sec^2 \theta+\tan^2 \theta}$$
$$=\frac{2\cdot \left(\frac{1-x^2}{1+x^2}\right)}{1+2x^2}$$
$$=2\cdot\frac{1-x^2}{(1+x^2)(1+2x^2)}$$
Hope this helps.
| {
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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 :
$|a| + |b| = a+b$
$(|a| + |b|)^2 = (a+b)^2$
$\sqrt{(|a| + |b|)^2} = \sqrt{(a+b)^2}$
$a+b =|a+b|$
Thus : |a+b|= |a| + |b|
For the case where a, b < 0, we use the same technique but with minuses before the constants Opinion ?
| 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",
"answer_count": 3,
"answer_id": 2
} |
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:
$$
\pmatrix{4 & 1 \\ 19 & 1}^{-1}
=
-\frac{1}{15}
\pmatrix{1 & -1 \\ -19 & 4}
=
-7
\pmatrix{1 & -1 \\ -19 & 4}
\bmod 26
$$
and so
$$
\pmatrix{a \\ b }
=
-7
\pmatrix{1 & -1 \\ -19 & 4}
\pmatrix{9 \\ 10 }
=
\pmatrix{7 \\ 7 }
\bmod 26
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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 $$0 < |x-a| < \delta \quad \implies \quad \left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon.$$ I don't see an easy way to do this algebraically so I am going to say let $|x-a| < 1$ and then $a - 1 < x < a+1$. We then have $\left | \dfrac{a+1}{(a-1)^2}-\dfrac{a+2}{a^2-5}\right | < \epsilon$. Then I am not sure how to simplify it to prove the statement.
| $$
\begin{align}
\left |\frac{x+2}{x^2-5}-\frac{a+2}{a^2-5} \right| & = \left | \frac{(x + 2)(a^2 - 5) - (a + 2)(x^2 - 5)}{(a^2 - 5)(x^2 - 5)} \right | \\[10pt]
& = \left | \frac{ax+2a+2x-5}{(a^2 - 5)(x^2 - 5)} \right | \cdot \underbrace{ |x-a|}_\text{important}.
\end{align}
$$
Pulling out $|x-a|$ is predictably possible because the whole expression will of course be $0$ when $x=a$.
And $|x-a|$ is the thing that you will make $<\delta$.
Now you need to get a bound on the first factor that does not depend on $x$, so that you have
$$
\left |\frac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| \le (\text{constant}\cdot|x-a|),
$$
where "constant" means "not depending on $x$". You use the prior decision to make $\delta\le 1$ to assure that $x$ is between $a\pm\delta$, and thus to make the "constant" independent of $x$ [but see the PS below].
PS: As "Ian" points out in comments below, rather than just $\delta\le 1$ we need $\delta$ small enough to keep $a\pm\delta$ away from the two square roots of $5$, i.e. $\pm\sqrt 5$, in order to keep the denominator away from $0$.
| {
"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
} |
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:
$$ \begin{array}{} &4 &= y^4 \pmod 7 \\
& y^4 - 4 &\equiv 0 \pmod 7 \\
&(y^2 - 2)(y^2+2) &\equiv 0 \pmod 7 \\
&& \text{giving two factors}\\
&y^2 - 2 &\equiv 0 \pmod 7 \\
\text{ or } & y^2 + 2 &\equiv 0 \pmod 7 \\
& y^2 &\equiv k \cdot 7 +2 & \to k=1,y^2=9 \text{ or } k=2,y^2=16 \text{ or ...}\\
\text{ or } & y^2 &\equiv j \cdot 7- 2 &\to k=?? \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
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 thought $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}\cdots=\frac{S}{2}$$
But answer given is $\frac{\pi^4}{96}$. Whats the mistake in this?
Edit:
I found a way to get the answer.
$$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots+\frac{1}{2^4}\left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots\right)$$
$$S=S_1+\frac{1}{2^4}S$$
| 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:
$$\sum_{n=1}^{\infty}\frac{1}{n^4}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{1}{n^4}=\lim_{m\to\infty}\left[\zeta(4,1)-\zeta(4,m+1)\right]=\zeta(4,1)=\zeta(4)=\frac{\pi^4}{90}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
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-HM inequality we get $t(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2) \geq 2^{2}.$ let $ m =(1-(\frac{x}{2})^2 + 1-(\frac{1}{3x})^2)$ and using AM-GM inequality we get $m \leq 5/3$.
But from this point my inequality signs are getting mixed up. Am I on right track?
| 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 $x = \sqrt{\frac23}, y = -\sqrt{\frac32}$, so that is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
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
$$
\int_Efg\,\mathrm{d}x\le\left(\int_Ef^2\,\mathrm{d}x\right)^{1/2}\left(\int_Eg^2\,\mathrm{d}x\right)^{1/2}
$$
Both sides of the inequality are even functions so its enough to consider
$x \in [0, \frac{\pi}{2})$.
Maybe I can use that $\int_0^x \cos t \:dt = \sin x$.
I would appreciate a hint or a few on how to continue.
edit: The inequality is supposed to be proved using integration methods.
| 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
$$
f'(x)=\cos(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\tan(x)\tag{3}
$$
and applying $(1)$
$$
\begin{align}
f''(x)
&=2-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\sec^2(x)\\
&=4+2\tan^2(x)-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\\[4pt]
&\ge4+2\tan^2(x)-2\tan^2(x)\\[12pt]
&=4\tag{4}
\end{align}
$$
Since $f(0)=f'(0)=0$ and $f''(x)\ge4$, we get that $f(x)\ge2x^2$. Therefore, applying $(1)$ again, we get
$$
2x^2\le\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\le2\tan^2(x)\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
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{2}} + \frac{2b}{\sqrt{2b + 1} + 1} + \frac{3c}{\sqrt{3c + 1} + 1}\\
\ge\, & \frac{-b - c}{1 + \sqrt{2}} + \frac{2b}{\sqrt{3} + 1} + \frac{3c}{\sqrt{4} + 1}\\
=\, & (\sqrt{3} - \sqrt{2})b + (2 - \sqrt{2})c \\
\ge\, &0.
\end{align*}
Also, if $a = 1, b = c = 0$, then
$\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1} = \sqrt{2} + 2$.
Thus, the minimum of $\sqrt{a + 1} + \sqrt{2b + 1} + \sqrt{3c + 1}$ is $\sqrt{2} + 2$.
| {
"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: $x=\sin(y)$
$x^2=\sin^2(y)$
$1-x^2=\cos^2(y)$
$\frac{1}{1-x^2}=\sec^2(x)=1+\tan^2(y)$
$\tan^2(y)=\frac{1}{1-x^2}-1=\frac{x^2}{1-x^2}$
$\tan(y)=\pm\frac{|x|}{\sqrt{1-x^2}}$
$\sin^{-1}x=\tan^{-1}\left(\pm\frac{|x|}{\sqrt{1-x^2}}\right)$
So I end up with $\pm$ and the absolute-value-sign, which are not part of the original equation. How to get rid of them?
By the way, I know that there are other ways to show this statement is true. But the question is what is wrong with this method.
| 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 $\;f(x)=g(x)+K\;,\;\;K\;$ a constant, for all $\;x\in I\;$ , for for $\;x\in (-1,1)\;$
$$\arcsin x=\arctan\frac x{\sqrt{1-x^2}}+K$$
Now just choose $\;x=0\;$ to check $\;K=0\;$ and thus the equality
| {
"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
} |
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?
What did I do wrong?
| $$|\frac{x+9}{x-9}|≤2$$
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",
"url": "https://math.stackexchange.com/questions/1694793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $s^2+4\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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 include $27$, $43$, $3650$ or $4333$.
So I am thinking you do $$8\cdot9\cdot1\cdot1+1\cdot9\cdot1+8\cdot1\cdot9\cdot1+1\cdot9\cdot9\cdot1 = 234$$
But I feel like that is too low.
| Revised answer
Marking the first $3$ and the first $4$ after $3$, there are only $6$ cases:
$$\begin{array}{}
3&4&\bullet&\bullet&\implies& 10\cdot10 = 100\\
3&\bullet&4&\bullet&\implies& 9\cdot 10 = 90\\
3&\bullet&\bullet&4&\implies& 9\cdot9 =81\\
\bullet&3&4&\bullet&\implies& 9\cdot10 = 90\\
\bullet&3&\bullet&4&\implies& 9\cdot 9 = 81\\
\bullet&\bullet&3&4&\implies& 9\cdot9 = 81
\end{array}$$
Adding up, $523$.
Note: The first $4$ after the first $3$ has been marked, but this doesn't preclude a $4$ coming before the first $3$.
| {
"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
} |
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 vector $x=(1,-3,4)$. Using the Euclidean inner product on $\mathbb{R}^3$ verify that $\| T(x) \| = \| x \|$.
For the first part I have
$$
\begin{bmatrix}
w_1 \\
w_2 \\
w_3
\end{bmatrix}=
\begin{bmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\[0.3em]
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\[0.3em]
-\frac{2}{3} & -\frac{1}{3} & \frac{2}{3}
\end{bmatrix}
\begin{bmatrix}
1 \\
-3 \\
4
\end{bmatrix}=
\begin{bmatrix}
1 \\
4 \\
3
\end{bmatrix}$$
But can anyone help with the second part?
| 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 columns, we see that $u_i \cdot u_j = 0$ if $i \ne j$, but $u_i \cdot u_1 = 1$. Hence the right hadn side above simplifies to just
$$
\|T(x)\|^2 = T(x) \cdot T(x) = a^2 u_1 \cdot u_1 + b^2 u_2 \cdot u_2 + c^2 u_3 \cdot u_3 = a^2 + b^2 + c^2 = \| x \|^2
$$
and you're done.
A slightly more sophisticated way of saying this is to note that $$a \cdot b = a^t b,$$
so that
\begin{align}
\|T(x) \|^2
&= (Ax) \cdot (Ax)\\
&= (Ax)^t (Ax)\\
&= x^tA^t Ax\\
&= x^t( A^t A) x.
\end{align}
You now observe that $A^t A = I$, so that this last expression simplifies to $x^t x = \|x \|^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696970",
"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 anyone's hints or good ideas.
| HINT:
$$\int\frac{1}{\left(1-\sqrt{1-\left(\frac{r}{\text{R}}\right)^2}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\sqrt{1-\frac{r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=$$
$$\int\frac{1}{\left(1-\sqrt{\frac{\text{R}^2-r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\frac{\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(\frac{\text{R}-\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=$$
$$\int\frac{1}{\frac{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}{\text{R}^{\frac{3}{2}}}}\space\text{d}r=\text{R}^{\frac{3}{2}}\int\frac{1}{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}\space\text{d}r=$$
Substitute $u=\text{R}-\sqrt{\text{R}^2-r^2}$ and $\text{d}u=\frac{r}{\sqrt{\text{R}^2-r^2}}\space\text{d}r$:
$$\text{R}^{\frac{3}{2}}\int\frac{\text{R}-u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u=\text{R}^{\frac{3}{2}}\int\left[\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}-\frac{u}{u^2\sqrt{2\text{R}-u}}\right]\space\text{d}u=$$
$$\text{R}^{\frac{3}{2}}\left[\int\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u\right]=$$
$$\text{R}^{\frac{3}{2}}\left[\text{R}\int\frac{1}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{1}{u\sqrt{2\text{R}-u}}\space\text{d}u\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
| The telescoping series cancel off itself after two terms, since $k-1$ and $k+1$ are 2 aparr
| {
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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 numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated.
| Using infinitesimals of equal order,
$$\begin{align*} \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)\tan^4 x}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)}
&= \lim_{x\to 0}\frac{\log(1+\sin^3x \cos^2x)\cot(\log(1+x)^3)x^4}{\sin(\sqrt{x^2+2}-\sqrt{2})\log(1+x^2)} \\
&= \lim_{x\to 0}\frac{ \sin^3x \cos^2x \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\
&= \lim_{x\to 0}\frac{ x^3 \cot(\log(1+x)^3)x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2} \\
&= \lim_{x\to 0}\frac{x^3 x^{-3} x^4}{(\sqrt{x^2+2}-\sqrt{2})x^2}
\end{align*}$$
where $\cot(\log(1+x)^3) \sim x^{-3}$ since $\cot \sim x^{-1}, \log(1+x)^3 \sim x^3$. Some of the other infinitesimals I've used:
$$\sin x \sim x, \cos x \sim 1, \log(1+x) \sim x, \tan x \sim x
$$
It is easily checked with l'Hopital that if $f \sim g$,
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = 1
$$
| {
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"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 as a binomial as follows: $$((1)+(ax+bx^{2}))^{10} = 1-30x+410x^{2}+...$$
But I have no idea how to find the coefficient of a specific power aside from expanding fully which would be an epic job in this case.
The answer in the book is $\frac{-15}{256}$.
How is this done?
| 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*}
[x^1](1+ax+bx^2)^{10}&=[x^1]\sum_{k=0}^{10}\binom{10}{k}(ax+bx^2)^k\\
&=[x^1]\binom{10}{1}(ax+bx^2)\\
&=10a
\end{align*}
Since we are looking for $[x^1]$ only $k=1$ provides a contribution, resulting finally in $10a$.
Comparison with (1) gives
\begin{align*}
10a&=-30\\
a&=-3\\
\end{align*}
$$ $$
Coefficient $[x^2]$:
We obtain
\begin{align*}
[x^2](1+ax+bx^2)^{10}&=[x^2]\sum_{k=0}^{10}\binom{10}{k}(ax+bx^2)^k\\
&=[x^2]\binom{10}{1}(ax+bx^2)+[x^2]\binom{10}{2}(ax+bx^2)^2+\\
&=10b+45a^2
\end{align*}
Since we are looking for $[x^2]$ only $k=1$ and $k=2$ provide a contribution, resulting finally in $10b+45a^2$.
Comparison with (1) gives
\begin{align*}
10b+45a^2&=410\\
10b+405&=410\\
b&=\frac{1}{2}
\end{align*}
We conclude the polynomial under consideration is
\begin{align*}
(1-3x+\frac{1}{2}x^2)^{10}
\end{align*}
Coefficient $[x^{19}]$:
We obtain
\begin{align*}
[x^{19}](1-3x+\frac{1}{2}x^2)^{10}&=[x^{19}]\sum_{k=0}^{10}\binom{10}{k}(-3x+\frac{1}{2}x^2)^k\\
&=[x^{19}]\binom{10}{10}(-3x+\frac{1}{2}x^2)^{10}\\
&=[x^{19}]x^{10}(-3+\frac{1}{2}x)^{10}\\
&=[x^{9}](-3+\frac{1}{2}x)^{10}\tag{1}\\
&=[x^{9}]\sum_{j=0}^{10}\binom{10}{j}\left(\frac{1}{2}x\right)^j(-3)^{10-j}\tag{2}\\
&=\binom{10}{9}\left(\frac{1}{2}\right)^9(-3)^1\\
&=-\frac{15}{256}
\end{align*}
Since we are looking for $[x^{19}]$ only $k=10$ provides a contribution.
In (1) we use the rule $[x^{n-k}]p(x)=[x^n]x^{k}p(x)$.
In (2) we observe that only $j=9$ provides a contribution.
| {
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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 appreciated
| 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 both $n-1$ and $n+1$ cannot be multiples of $4$ since consecutive even numbers cannot be both multiples of $4$.
At most one of them will be a multiple of 8.
So either $$n-1 \equiv 0 \pmod 8$$ or $$n+1 \equiv 0 \pmod 8$$
Combining the two, we get the required condition: $$n\equiv \pm 1 \pmod8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 calculated the partial derivative with respect to $x$ as
$$
f_1(x,y) = \frac{x^4-2xy^3+3x^2y^2}{(x^2+y^2)^2} \tag{1}
$$
This is undefined at $(0,0)$, so I then tried to find the limit around accumulation points. Let $S_1$ be the points on the $x$ axis
$$
\lim_{x \to 0} f_1(x,0) = \frac{x^4}{(x^2)^2} = 1 \tag{2}
$$
Let $S_2$ be the points on the line $y = x$
$$
\lim_{x \to 0} f_1(x,x) = \frac{x^4-2x^4+3x^4}{(x^2+x^2)^2} = \frac{2x^4}{(2x^2)^2} = \frac{1}{2} \tag{3}
$$
So, the limits are different around different accumulation points. That's where I'm confused because the answer should be $1$.
| 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",
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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.
$$\frac{7}{-19} + \frac{-17}{-8} = \frac{267}{152}$$
gives us
$$\frac{\frac{7}{4}}{\frac{267}{152}}.$$
To bring this home, rewrite this division of two fractions as a multiplication:
$$\frac{\frac{7}{4}}{\frac{267}{152}} = \frac{7}{4} \times \frac{152}{267} = \frac{1064}{1068} = \frac{532}{534} = \ldots$$
You get the idea.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 write out. I believe it'd be something like
$(3,1,1)$
$(3,2,2)$
$(3,3,3)$
$(3,4,4)$
$(3,5,5)$
$(3,6,6)$
| 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;
$$P(X\geq 1) = 1-P(X = 0) = 1-\binom{3}{0}(1/6)^0(5/6)^3 = \frac{91}{216}.$$
Using a counting argument, we see that there are $6^3$ triplets possible, the number of triplets that do not contain a three is $5^3$, and hence the probability of interest is
\begin{align*}
P(X\geq 1) &= 1-P(X = 0) \\
&= 1-\frac{\text{# of triplets with no threes}}{\text{# of all triplets}}\\
&= 1-\frac{5^3}{6^3} \\
&= \frac{6^3-5^3}{6^3} \\
&= \frac{91}{216}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"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's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$.
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?
| 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",
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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 really grateful. Also, this is only half the proof, but hopefully if I can get this part ill be able to figure the rest out myself.
Define the set $A = \left \{ x\in \mathbb{R}:x> 0\: and\: x^{2}< 2 \right \}$, let $a= SupA$.
Show that $a^{2}= 2$.
First rule out that $a^{2}< 2\; or\; a^{2}> 2$.
Suppose that $a^{2}< 2$
then we can show that $\left ( a+\frac{1}{n} \right )^{2}< 2$ for large enough $n$. We have: $$\left ( a+\frac{1}{n} \right )^{2}-2=\left ( a^{2} -2\right )+\frac{2a}{n}+\frac{1}{n^{2}}\leq \left ( a^{2} -2\right )+\frac{2a+1}{n}$$
( (1) here I'm assuming that in the final term the reason we can drop one of the $n$'s is because of the less than sign?)
By the archimedian property there exists $n\in \mathbb{N}$ such that $$\frac{1}{n}< \frac{2-a^{2}}{2a+1}$$
( (2) ok, where on earth did they get that last fraction from? why is 2a+1 suddenly on the bottom, and how did $\left ( a^{2} -2\right )$ get flipped around and put on top?)
so that $a^{2}-2+\frac{2a+1}{n}< 0$. Therefore $a+\frac{1}{n}\in A$ so a is not an upper bound for A.
| 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 extra step can be seen as follows: Since $n^2\geq n$, we have $\frac{1}{n^2}\leq\frac{1}{n}$. Replacing $\frac{1}{n^2}$ by $\frac{1}{n}$ thus gives us something larger.
Question 2
The expression $\frac{2-a^2}{2a+1}$ has been carefully chosen, such that when we use $\frac{1}{n}<\frac{2-a^2}{2a+1}$ in the last expression $(a^2-2)+\frac{2a+1}{n}$ from your first question, we get exactly that the expression is less than $0$. In detail, whenever $\frac{1}{n}<\frac{2-a^2}{2a+1}$, we have
\begin{align*}
(a^2-2)+\frac{2a+1}{n}&=(a^2-2)+(2a+1)\frac{1}{n}\\
&<(a^2-2)+(2a+1)\frac{2-a^2}{2a+1}\\
&= (a^2-2) + (2-a^2)\\
&= 0.
\end{align*}
This shows that whenever $\frac{1}{n}<\frac{2-a^2}{2a+1}$ (which happens for large enough $n$), we have $(a+\frac{1}{n})^2-2<0$, and thus $a+\frac{1}{n}\in A$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 within $ \int (x^2)(x^3-1)^2 \, dx $ from the beginning, I don't get the same answer. For example: $ \int (x^2)(x^3-1)^2 \, dx $ = $ \int (x^2)(x^6 -2x^3 + 1) \, dx = \int (x^8 -2x^5 + x^2) \, dx$. The anti derivative of this becomes $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} $.
However, these anti-derivatives are not equal. Through u-substitution, when $x=1$, the anti derivative evaluates to zero. In the second route, when I plug in $1$, I get $1/9$.
Where did I go wrong? From what I see, the $u$-substitution version is correct.
| 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 same as what you got but the constant at the end is different (differs by an additive constant factor).
| {
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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 to write $y=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$ in ascending powers of x?
| 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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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 $3|a$.
| {
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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 equal to $\frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$.
My work and thoughts:
If the series is Cesaro summable to $\frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$, this means that if $S_N$ is the $N$th partial sum
then $\sigma_N \rightarrow \frac{1}{2}\cot\frac{x}{2}$, where
$$\sigma_N = \frac{S_1 + \ldots + S_N}{N}.$$
Using trigonometric identities, Euler's formula (and patience) it is not too difficult to show that for the sequence of partial sum we have
$$S_N = \frac{\cos\frac{x}{2} - \cos(N + \frac{1}{2})x}{2\sin{\frac{x}{2}}}.$$
How do I continue from here to find the given Cesàro sum of the series?
| 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 the limit of the sum on the RHS as $N \to \infty$ is zero.
Note that
$$2\sin{\frac{x}{2}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) = \sum_{k=1}^N2\sin{\frac{x}{2}}\cos\left(kx + \frac{x}{2}\right) \\ = \sum_{k=1}^N \left[\sin(kx+x) - \sin(kx) \right] \\= \sin((N+1)x) - \sin(x),$$
and
$$\lim_{N \to \infty} \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) = \lim_{N \to \infty} \frac{\sin((N+1)x) - \sin(x)}{4N\sin^2{\frac{x}{2}}} \\ = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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",
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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 $$\begin{align*} \int x^3 \sqrt{4-x^2} \, dx &= \int u^4 - 4u^2 \, du \\ &= \frac{u^5}{5} - \frac{4u^3}{3} + C \\ &= (4-x^2)^{3/2} \left( \frac{4-x^2}{5} - \frac{4}{3} \right) + C \\ &= -\frac{(4-x^2)^{3/2}(8+3x^2)}{15} + C. \end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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
$$2(2x^2-1)^2-1=-(4x^3-3x).$$
This is a quartic equation. However, it has the obvious root $x=-1$. Divide the polynomial $2(2x^2-1)^2-1+4x^3-3x$ by $x+1$, and we obtain the cubic of the question.
To finish, verify that if $\varphi=3\pi/7$ or $5\pi/7$, then $\cos(4\phi)=-3\cos(3\phi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 by brackets.
$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$
$$1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$$
$$0=\sqrt{2 - \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$
$$\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$$
$$\frac{1}{\phi}=\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \dots}}}}=(-+)$$
In general, the expansion can be found by a very easy algorithm:
*
*take any number in $(0,2)$, square it
*if the result $>2$ write $+$, if the result $<2$ write $-$
*subtract $2$ from the result, square, repeat
If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.
Examples:
$$\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \dots=1.141592653589793 \dots$$
Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it's not so bad.
The convergence plot, and two binary plots for this $50$ terms can be seen below:
$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \dots=1.71828182845905 \dots$$
Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?
Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?
Edit
Now that I think about it, we can use the general expansion for $x \in [0,a]$:
$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$
$$a=2^{\frac{p}{1-p}}$$
For example:
$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$
$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} - \left(\frac{1}{4} - \left(\frac{1}{4} - \dots \right)^2 \right)^2 \right)^2$$
etc.
However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.
Edit
An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344
| 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 $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have
$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}. $$
This can be used to show that, with an appropriate definition of infinite nested radical, the following identity
$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}} $$
is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then
*
*$\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
*$\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,
and likewise. This explains why signs are determined by OP's algorithm.
| {
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"url": "https://math.stackexchange.com/questions/1731070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "121",
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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 $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.
So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
Since $\lfloor x\rfloor\le x$, we can just start from here and prove
$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
| 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",
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Fractions in Questions and Answers
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