Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to calculate this imaginary part of a complex square root? The calculation of the square root of a complex number $a + ib$ involves solving the equation
$$ (x + iy)^2 = a + ib$$
So far so good. One obtains the equations
$$ 4x^4 -4ax^2 - b^2 = 0, y = b/2x$$
and using the quadratic formula for $x^2$ one gets
$$ x = \pm \sqrt{{a + \sqrt{a^2 + b^2} \over 2}}$$
I am supposed to get
$$ y = \pm \sqrt{- a + \sqrt{a^2 + b^2} \over 2} \cdot \text{sgn}(b)$$
but if I substitute $x$ into $y$ I get
$$ y = {b\over 2x} = \pm {b \sqrt{2} \over 2\sqrt{a + \sqrt{a^2 + b^2}}}$$
What am I doing wrong?
| First off, up to a sign (reflected by that $\mbox{sgn }(b)$ factor), the two expressions are equal. You can see this if you multiply the expression you got by
$$
\frac{\sqrt{\sqrt{a^2+b^2}-a}}{\sqrt{\sqrt{a^2+b^2}-a}}
$$
to get
$$
y = \pm \frac{b}{\sqrt{b^2}} \frac{\sqrt{\sqrt{a^2+b^2}-a}}{\sqrt{2}}
$$
And of course, since all use of the square root symbol implies the positive square root,
$$
\frac{b}{\sqrt{b^2}} = \mbox{sgn }(b)
$$
That is, you did everything right, except not trusting your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using calculator.which one is correct?
| Generally:
$$\begin{aligned}
&\phantom{mm}(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\\
&=((a+b)+c)((a+b)-c) \times ( c+(a-b))(c-(a-b)) \\
&=((a+b)^2 -c^2)(c^2-(a-b)^2)\\
&=[2ab+(a^2+b^2 - c^2) ] [ 2ab - (a^2+b^2-c^2)]\\
&= 4a^2b^2 - (a^2+b^2-c^2)^2.
\end{aligned}$$
Here $a^2=5, b^2=6,$ and $c^2=7$, so the answer is $104$ as you said.
| {
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Finding the roots of $z^4+z^3+z^2+z+1$ I need to find the roots of $z^4+z^3+z^2+z+1$. One approach I have is to guess the first linear solution, then use polynomial long division to find a 3rd degree polynomial, guess another from that, long division again, then use the quadratic formula to solve the quadratic.
I think this will work but seems a bit work-heavy. Is there a simpler way?
| In addition to the multiply-by-$\left(z - 1\right)$ trick, there is also the "symmetric" method to solve for the roots directly:
$$z^4 + z^3 + z^2 + z + 1 = z^2 \left(z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2}\right) = 0 $$
Since $z^2 \ne 0$, we can cancel it out, and we recognize the almost-square $z^2 + \frac{1}{z^2}$:
$$z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} = \left(z + \frac{1}{z}\right)^2 - 2 + \left(z + \frac{1}{z}\right) + 1 = 0 $$
Now we can substitute $y = z + \frac{1}{z}$:
$$y^2 + y - 1 = 0 \to y = \frac{1\pm\sqrt{5}}{2}$$
We also have:
$$z + \frac{1}{z} = y \to z^2 + 1 = yz \to z=\frac{y\pm\sqrt{y^2 - 4}}{2}$$
We plug in $ y = \frac{1\pm\sqrt{5}}{2} $ into the equation above, and the reader verifies that the answers are equivalent to:
$$e^{i\frac{2\pi k}{5}}=\cos{\frac{2\pi k}{5}} + i\sin{\frac{2\pi k}{5}}, 0\le k\le 4$$
as obtained from the multiply-by-$\left(z-1\right)$ method.
| {
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How to compute this limit $\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}$ $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$
such that $2\alpha n\le n$
I tried to use Stirling formula and we get
$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\lim_{n\to ∞}\frac{1}{n}\log\frac{n^{\frac{3n}{2}}}{2\pi(n-2\alpha n)^{\frac{3((n-2\alpha n)}{2}{(2\alpha n)}^{3\alpha n}}}=$$
$$=\lim_{n\to ∞}\log{\frac{n^{\frac{3}{2}}}{2\pi(n-2\alpha n)^{\frac{3((1-2\alpha )}{2}{(2\alpha n)}^{3\alpha }}}}$$
but I couldn't continue
| The Stirling formula is the right way. But ignore the logarithm first. I get
$$\ln \frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
It comes from $\displaystyle \ln \left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e})^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}$:
$$\left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e} )^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}=\frac{\frac{n}{e}}{ (\frac{(1-2\alpha)n}{e} )^{1-2\alpha}\,(\frac{2\alpha n}{e})^{2\alpha}}=\frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
| {
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Improper Integral $\int_0^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)dx = \frac{7\zeta(3)}{\pi^2} $ $\newcommand{\sech}{\operatorname{sech}}$
$\displaystyle \int_0^{\infty}{\left(\frac{\tanh(x)}{x^3} - \frac{\sech^2(x)}{x^2} \right)\ dx }= \frac{7\zeta(3)}{\pi^2} $
What I tried
I simplified it to -
$\displaystyle \int_0^{\infty}{\frac{\sinh(2x) - 2x}{x^3 \cosh^2(x)} \ dx}$
Then I don't know how to solve. I tried Feynman's method
$\displaystyle I(a) = \int_0^{\infty}{\frac{\sinh(ax) - ax}{x^3 \cosh^2(x)} \ dx}$
But then too it didn't help much.
I thought of replacing them with trigonometric forms and then complex number real and imaginary part but wasn't helpful much.
Please try to avoid complex analysis.
| Integrating by parts we get $$I=\int_{0}^{\infty}\left(\frac{\tanh\left(x\right)}{x^{3}}-\frac{1}{x^{2}\cosh^{2}\left(x\right)}\right)dx= $$ $$-\int_{0}^{\infty}\frac{1}{x}\left(-\frac{\tanh\left(x\right)}{x^{2}}+\frac{1}{x\cosh^{2}\left(x\right)}+2\frac{\tanh\left(x\right)}{\cosh^{2}\left(x\right)}\right)dx$$ so $$\int_{0}^{\infty}\left(\frac{\tanh\left(x\right)}{x^{3}}-\frac{1}{x^{2}\cosh^{2}\left(x\right)}\right)dx=-\int_{0}^{\infty}\frac{\tanh\left(x\right)}{x\cosh^{2}\left(x\right)}dx
$$ and now taking $x=-\log\left(u\right)
$ we get $$I=4\int_{0}^{1}\frac{\left(u^{2}-1\right)u}{\left(u^{2}+1\right)^{3}\log\left(u\right)}du=4\int_{0}^{1}\frac{u^{3}-u}{\left(u^{2}+1\right)^{3}\log\left(u\right)}du
$$ $$=4\int_{0}^{1}\frac{1}{\left(u^{2}+1\right)^{3}}\int_{1}^{3}u^{z}dzdu=4\int_{1}^{3}\int_{0}^{1}\frac{u^{z}}{\left(u^{2}+1\right)^{3}}dudz
$$ and the last integral can be written in terms of the Gauss hypergeometric function $$\int_{0}^{1}\frac{u^{z}}{\left(u^{2}+1\right)^{3}}du=\frac{1}{2}\int_{0}^{1}\frac{u^{z/2-1/2}}{\left(u+1\right)^{3}}du=\frac{1}{z+1}\,_{2}F_{1}\left(3,\frac{z+1}{2},1+\frac{z+1}{2},-1\right)
$$ and this particular hypergeometric function has a “closed form” in terms of Digamma function, so we have $$ I=\frac{1}{8}\int_{1}^{3}\left(z^{2}-4z+3\right)\left(\psi\left(\frac{z+3}{4}\right)-\psi\left(\frac{z+1}{4}\right)\right)dz-\frac{1}{8}\int_{1}^{3}2z-8dz.
$$ Now note that every single term is in the form $$a\int_{1}^{3}z^{b}\psi\left(\frac{z+c}{4}\right)dz=4a\int_{(1+c)/4}^{(3+c)/4}\left(4v-c\right)^{b}\psi\left(v\right)dv
$$ with $b=0,1,2
$ and $c=1,3
$ so let us consider the case $b=0$. We have $$\int_{(1+c)/4}^{(3+c)/4}\psi\left(v\right)dv=\log\left(\frac{\Gamma\left(\frac{3+c}{4}\right)}{\Gamma\left(\frac{1+c}{4}\right)}\right)
$$ if $b=1
$ we have, integrating by parts, $$\int_{(1+c)/4}^{(3+c)/4}v\psi\left(v\right)dv=\left(v\log\left(\Gamma\left(v\right)\right)-\psi^{\left(-2\right)}\left(v\right)\right)_{(1+c)/4}^{(3+c)/4}$$ and if $b=2$ we have $$\int_{(1+c)/4}^{(3+c)/4}v^{2}\psi\left(v\right)dv=\left(v^{2}\log\left(\Gamma\left(v\right)\right)-2v\psi^{\left(-2\right)}\left(v\right)+3\psi^{\left(-3\right)}\left(v\right)\right)_{(1+c)/4}^{(3+c)/4}$$ so combining this result and the closed form about polygamma at negative orders we obtain $$I=\color{red}{\frac{7\zeta\left(3\right)}{\pi^{2}}}$$ as wanted.
| {
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how to solve $x^2 = 3x + 4$ I am a programmer in the eighth grade taking algebra 1. I am only about a month into the school year, and I need to know how to solve something similar to this equation:
$x^2 = 3x + 4$
However, the problem is that whenever I try to get the square root of both sides of the equation to get rid of the $x^2$, I get something like this:
$x = \sqrt{3x + 4}$
After this, I couldn't figure out what to do, because my only option was to square both sides of the equation, which would get me back where I started. So, how would I solve this? Please make the answer simple enough so that an algebra 1 student would understand. Also, sorry for the tag that probably isn't great, my rep is so low I cant create a new tag (I would probably tag this as algebra 1).
| The standard method is to multiply everything by $4$ and transfer the $x$ term on the left-hand side:
$$
4x^2-12x=16
$$
Now recall $(a+b)^2=a^2+2ab+b^2$ and observe that we can take $a=2x$, so from $2ab=-12x$ we obtain $4bx=-12x$, that's satisfied for $b=-3$. We need the $b^2$ term, so we add it on both sides:
$$
4x^2-12x+9=16+9
$$
The left-hand side can be rewritten $(2x-3)^2$, so we are reduced to
$$
(2x-3)^2=25
$$
that gives us
$$
2x-3=5\qquad\text{or}\qquad 2x-3=-5
$$
A different strategy is to set $x=t+a$ and try to determine $a$ in such a way that some term vanishes:
$$
t^2+2at+a^2=3t+3a+4
$$
If we set $2a=3$, the $t$-term disappears:
$$
t^2+\frac{9}{4}=\frac{9}{2}+4
$$
and so
$$
t^2=\frac{25}{4}
$$
and therefore $t=5/2$ or $t=-5/2$. Thus
$$
x=\frac{5}{2}+\frac{3}{2}
\qquad\text{or}\qquad
x=-\frac{5}{2}+\frac{3}{2}
$$
You'll learn the “quadratic formula” that can be obtained in the same way as the first method above: if $ax^2+bx+c=0$, then
$$
x=\frac{-b+\sqrt{b^2-4ac}}{2a}
\qquad\text{or}\qquad
x=\frac{-b-\sqrt{b^2-4ac}}{2a}
$$
provided $b^2-4ac\ge0$. In your case $a=1$, $b=-3$ and $c=-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.
Why does this hold?
| Note that (by dividing the equations by $16$ and $32$) this is equivalent to showing one of $a,b,c,d$ is $1$ and the others are $0$ if $a^4 + b^4 + c^4 + d^4 = 1$ and $a^5 + b^5 + c^5 + d^5 = 1$. If none of $a,b,c,d$ are $1$ then they must all have absolute value less than one (otherwise the sum of their fourth powers would exceed $1$). It then suffices to note that
$a^5 + b^5 + c^5 + d^5 < a^4 + b^4 + c^4 +d^4 = 1,$
since $a^5 < a^4, b^5 < b^4, c^5<c^4, d^5<d^4$.
Therefore $a^5 + b^5 + c^5 + d^5 \neq 1$ and we can deduce the result.
| {
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Find $x$ given remainder conditions The problem: Find the smallest positive integer $x$ such that
$x$ divided by $4$ has remainder $1$
$x$ divided by $5$ has reminder $2$
$x$ divided by $6$ has remainder $3$
Now, my first idea was to add to each divisor its the remainder and multiply the quantities obtained.
$x=(4+1)(5+2)(6+3)=315$
But $315$ does not satisfy all the conditions above, and I don't know how to get the smallest integer that satisfies the conditions. Any help?
| If $x \equiv 1\pmod{4}$, then $$x \equiv 1,5,9,13,17\pmod{20}$$ Of these, only $17$ is also $2 \pmod{5}$. Therefore
$$x \equiv 17 \pmod{20}$$
Next, the LCM of $20$ and $6$ is $60$.
If $x \equiv 17 \pmod{20}$, then $$x \equiv 17, 37, 57 \pmod{60}$$
Of these, only $57$ is $3\pmod{6}$. This means
$$x \equiv 57\pmod{60}$$
This means $\boxed{57}$ is the smallest $x$ that satisfies your conditions.
| {
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Evaluation of Trigonometric Integral
Evaluation of $\displaystyle \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx$
$\bf{My\; Try::} $ Let $$I= \int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cdot \cos^2 x}{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{(\sin x+\cos x)(2-\sin 2x)}dx$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 2x}{\sin\left(x+\frac{\pi}{4}\right)(2-\sin 2x)}dx$$
Put $\displaystyle x+\frac{\pi}{4} = t,$ Then $dx = dt$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t}{\sin t(2+\cos 2t)}dt=\frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-4+4}{\sin t(2+\cos 2t)}dt$$
So $$I = \frac{1}{2\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{\cos^2 2t-2}{\sin t}dt+\frac{2}{\sqrt{2}}\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{1}{\sin t(2+\cos 2t)}dt$$
How can i solve it after that, Help required, Thanks
| HINT:
If $f(x)=\sin^2x\cos^2x=\dfrac{\sin^22x}4,f\left(\dfrac\pi4+0-x\right)=\dfrac{\cos^22x}4=\dfrac{(1-2\sin^2x)^2}4$
If $g(x)=\sin^3x+\cos^3x,g\left(\dfrac\pi4+0-x\right)=\dfrac{2\cos x(\cos^2x+3\sin^2x)}{2\sqrt2}=\dfrac1{\sqrt2}\cdot\dfrac{(1-\sin^2x)(1+2
\sin^2x)}{\cos x}$
Now use $$\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$ and choose $\sin x=u$
| {
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Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number? Is there a way to show that the addition of the first n terms of the Fibonacci sequence squared gives an answer divisible by a particular number?
| You can prove by induction that the sum of the squares of the first $6n$ terms is divisible by $8$.
First, show that this is true for $n=1$:
$\sum\limits_{m=1}^{6}F_m^2=40$
Second, assume that this is true for $n$:
$\sum\limits_{m=1}^{6n}F_m^2=8k$
Third, prove that this is true for $n+1$:
$\sum\limits_{m=1}^{6(n+1)}F_m^2=$
$\sum\limits_{m=1}^{6n+6}F_m^2=$
$\left(\color\red{\sum\limits_{m=1}^{6n}F_m^2}\right)+F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2=$
$\color\red{8k}+F_{6n+1}^2+F_{6n+2}^2+F_{6n+3}^2+F_{6n+4}^2+F_{6n+5}^2+F_{6n+6}^2=$
$\small8k+F_{6n+1}^2+F_{6n+2}^2+(F_{6n+1}+F_{6n+2})^2+(F_{6n+1}+2F_{6n+2})^2+(2F_{6n+1}+3F_{6n+2})^2+(3F_{6n+1}+5F_{6n+2})^2=$
$8k+16F_{6n+1}^2+40F_{6n+2}^2+48F_{6n+1}F_{6n+2}=$
$8(k+2F_{6n+1}^2+5F_{6n+2}^2+6F_{6n+1}F_{6n+2})$
Please note that the assumption is used only in the part marked red.
| {
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Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$ How do I find this limit?
$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$
I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!
Is there any other way to get the answer besides using L'Hopital's Rule?
| Here we do not use Hopital or derivatives.
Note that
$$x^2+x+\sin({\pi \over 2}x)-3=(x+2)(x-1)+\sin({\pi \over 2}-{\pi \over 2}(1-x))-1\\=(x+2)(x-1)+\cos({\pi \over 2}(1-x))-1\\
=(x+2)(x-1)-2\sin^2(\frac{\pi}{4}(1-x))$$
because $1-\cos(t)=2\sin^2(t/2)$.
Hence, as $x\to 1$,
$$\frac{x^2+x+\sin({\pi \over 2}x)}{(x-1)}=x+2-2\sin(\frac{\pi}{4}(1-x))\cdot \frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}\to 1+2+0=3$$
where we used the fact that
$$\lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}=-\frac{\pi}{4}\cdot \lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{\frac{\pi}{4}(1-x)}=-\frac{\pi}{4}.$$
| {
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limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| The second factor in $$\frac{\sin x-x\cos x}{x^3}\frac{\sin x+x\cos x}{x}$$ clearly tends to $2$.
Then by L'Hospital,
$$\frac{\cos x-\cos x+x\sin x}{3x^2}$$ tends to $1/3$.
| {
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Question on algebra used in induction proof Can someone take a look at this simplification... it is part of a proof by induction solution I'm examining... I'm interested in the result of this simplification as I'm not sure how the third to last step was achieved.
$$\begin{align*}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}&=\\
\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)}\right)+\frac{1}{(k+1)(k+2)}&=\\
\left(1-\frac{1}{k+1}\right)+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}&=\\
1-\frac{k+1}{(k+1)(k+2)}&=\\
1-\frac{1}{k+2}&=\\
1-\frac{1}{(k+1)+1}.
\end{align*}$$
Now to me it seems that the third to last step would result in
$$\begin{align*}
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}&=\\
\bbox[yellow]{
1-\frac{k+3}{(k+1)(k+2)}
}&=
\end{align*}$$
What am I missing?
| That step should be
$$
1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+(-1)\cdot\left(\frac{k+2}{(k+1)(k+2)}\right)+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
=1+\frac{-k-1}{(k+1)(k+2)}\\
=1-\frac{k+1}{(k+1)(k+2)} \\
=1-\frac{1}{k+2}.
$$
| {
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Finding the two planes which contains a given line and forms an angle of $60^\circ$ with a given plane I am given the following question:
Find the equations of the two planes that contain the line $r \{ x=z+1=y+2$ and form a $60^\circ$ angle with the plane $ \pi \{ x+2y-3z+2=0$
My solution:
Lets call the normal vector of the planes wanted $\vec{n}=(a,b,c)$
The vector of the line is $\vec{v} =(1,1,1)$ and since it is perpendicular to the planes we want,
$$
(1,1,1) \cdot (a,b,c) = 0 \therefore a+b+c = 0
$$
Also, since the angles between the plane $\pi$ and our new planes is $60^\circ$ we have
$$
\frac{1}{2}=\frac{\vert (a,b,c) \cdot (1,2,-3) \vert}{\sqrt{a^2+b^2+c^2} \sqrt{14}}
$$
I need another equation to finish the exercise, but I'm not sure what is that third equation.
Textbook's answer
$$
\pi_1 \{ 2x-3y+z-5=0\\
\pi_2 \{ 3x -y-2z-4=0
$$
Thank you.
| Since as has been oft remarked the lenght of $(a,b,c)$ is variable let us chose the length to be $\sqrt{14}$ in order to cancel with the other $\sqrt{14}$
So we have the equations:
$$a+b+c=0$$
$$a+2b-3c=\frac{1}{2}\sqrt{14}\sqrt{a^2+b^2+c^2}=\frac{1}{2}14=7$$
$$a^2+b^2+c^2=14$$
If we take the first two
$$a+b+c=0$$
$$a+2b-3c=7$$
and solve them we get
$$a=-5c-7$$
$$b=4c+7$$
If we now substitute into the third equation and simplify we get
$$c^2+3c+2=0$$ so $c=-1,-2$.
This gives us the two vectors
$$(-2,3,-1) \text{and} (3,-1,-2).$$
And the two equations
$$2x-3y+z=d$$
$$3x-y-2z=d$$
The value of $d$ can then be calculated in each case since the plane passes through $(0,-2,-1)$.
| {
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How to show that the harmonic function $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions? How can I prove that $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions?
| Hint. Prove by induction that
$$1+\frac{n}{2}\leq H_{2^n}\leq 1+n$$
by using the fact that
$$\frac{1}{2}=\frac{2^n}{2^n+2^n}\leq H_{2^{n+1}}-H_{2^n}=\frac{1}{2^n+1}+\frac{1}{2^n+2}+\cdots +\frac{1}{2^{n}+2^n}\leq \frac{2^n}{2^n+1}<1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a trigonometric expression is identical to $2(\operatorname{cosec}^{2}{B}-1).$ I came across this trigonometric identity:
$$\frac{\operatorname{cosec}{B} - \cot{B}}{\operatorname{cosec}{B} + \cot{B}} + \frac{\operatorname{cosec}{B} + \cot{B}}{\operatorname{cosec}{B} - \cot{B}} = 2(\operatorname{cosec}^{2}{B} - 1) = 2(\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}})$$
And as I solved it, the equation came down to:
$2\operatorname{cosec}^{2}{B}+2\cot^{2}{B}.$
This can be written as $2(\operatorname{cosec}^{2}{B} + \cot^{2}{B}),$ which can further be written as $2\frac{1+\cos^{2}{B}}{1 - \cos^{2}{B}}.$
But I can't seem to get my mind around the middle part of the question, that is, $2(\operatorname{cosec}^{2}{B} - 1)$
Is it possible to write it like this, or is this an error in the question paper itself?
| $$(\csc B-\cot B)(\csc B+\cot B)=1\implies\dfrac{\csc B+\cot B}{\csc B-\cot B}=(\csc B+\cot B)^2$$
$$(\csc B+\cot B)^2+(\csc B-\cot B)^2=2(\csc^2B+\cot^2B)$$
Now $\cot^2B=\csc^2B-1$
Finally, $$\dfrac{1+\cos^2B}{1-\cos^2B}=\dfrac{1+\cos^2B}{\sin^2B}=\csc^2B+\cot^2B$$
| {
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Finding a sum as function of $n$ using summation formulas i need to find the sum as a function of $n$ using summation formulas of the following serie:
\begin{equation}
\sum_{k=1}^n\bigg(4+\cfrac{2k}{n}\bigg)^2\bigg(\cfrac{1}{n}\bigg).
\end{equation}
Any idea how to proceed? Any help will be appreciated!
| I would proceed as follow:
$$\frac{1}{n}\sum_{k = 1}^n\ \frac{(4n^2+2k)^2}{n^2} = \frac{4}{n^3}\sum_{k = 1}^n (2n^2+k)^2 = \frac{4}{n^3}\sum_{k = 1}^n\ 4n^4 + k^2 + 4k$$
Now split the sum into three pieces, knowing that those are well known series:
$$\sum_{k = 1}^n 4n^4 = 4n^4\sum_{k = 1}^n = 4\cdot n^5$$
$$\sum_{k = 1}^n 4k = 4\sum_{k = 1}^n k = 4\frac{n(n+1)}{2} = 2n(n+1)$$
$$\sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
Combine the whole terms obtaining the final sum:
$$\frac{4}{n^3}\left(4n^5 + 2n(n+1) + \frac{n(n+1)(2n+1)}{6}\right)$$
Getting finally:
$$\frac{2 \left(38 n^2+15 n+1\right)}{3 n^2}$$
| {
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"source": "stackexchange",
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Convergence test for the following series The series is :-
$$ \frac{1}{1^2} + \frac{1+2}{1^2+2^2} + \frac{1+2+3}{1^2+2^2+3^2} + ... ∞ $$
I am unable to think for its general term.
Can anybody help me solve this.
Thanks in advance.
| hint
$$1+2+3+....n=\frac{n}{2}(n+1)$$
$$1+2^2+3^2+...n^2=\frac{n}{6}(n+1)(2n+1)$$
for the another series, use the ratio test
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n+1)!}\frac{n!}{n^2}=\frac{(n+1)^2}{(n+1)n^2}=\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}$$
the limit is zero when $n\rightarrow \infty$, so the series converges
| {
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Determinant of a symmetric $7\times 7$-matrix Is there any simple method to calculate the determinant of the following symmetric $7\times 7$-matrix
$$M:=\begin{pmatrix}
1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\
0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\
0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\
0 & 0 & 0 & 1 & -a_{3} & a_{2} & -a_{1}\\
a_{2} & -a_{1} & a_{4} & -a_{3} & a_0 & 0 & 0\\
a_{3} & -a_{4} & -a_{1} & a_{2} & 0 & a_0 & 0\\
a_{4} & a_{3} & -a_{2} & -a_{1} & 0 & 0 & a_0\\
\end{pmatrix}$$
where $a_i$ are real numbers.
| Here is a rather simple method.
Consider the block partition of matrix $M$:
$$M=\pmatrix{I_4&B^T\\B&a_0I_3} \ \ \text{of the form} \ \ \pmatrix{R&S\\T&U}.$$
Using a classical determinantal formula (https://en.wikipedia.org/wiki/Schur_complement) valid for any partition where the diagonal blocks are square and the upper left block is invertible:
$$det(A)=det(R)det(U-TR^{-1}S)=det(I_4)det(a_0I_3-BI_4^{-1}B^T)$$
where $U-TR^{-1}S$ is the classical Schur's complement. Thus:
$$\tag{1}det(A)=det(a_0I_3-BB^T)$$
But $BB^T=(a_1^2 + a_2^2 + a_3^2 + a_4^2)I_3$. Then:
$$\tag{2}det(A)=det((a_0-a_1^2 - a_2^2 - a_3^2 - a_4^2)I_3)=(a_0-a_1^2 - a_2^2 - a_3^2 - a_4^2)^3$$
| {
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Functional equation $f(r \cos \varphi)+f(r \sin \varphi)=f(r)$
Find all such monotonous function $f:[0,+\infty)\to \mathbb R$, such that for any real $r\ge0$ and $\varphi \in \left[\frac{\pi}6, \frac{\pi}4\right]$,
$$f(r \cos \varphi)+f(r \sin \varphi)=f(r)\text.$$
My work so far:
*
*If $\varphi \in \left[\frac{\pi}6, \frac{\pi}4\right]$ then $$\frac{\sqrt2}2\le\cos \varphi\le\frac{\sqrt3}2$$
$$\frac12\le\sin \varphi\le\frac{\sqrt2}2$$
*If $r=0$ then $f(0)=0$
*If $\varphi= \frac{\pi}{4}$ then $$f(r)=4f\left(\frac r2\right)$$
*$f(x)=cx^2$ for $x\ge0$ and $c\in\mathbb R$
| Consider a monotonous function $ f : [ 0 , + \infty ) \to \mathbb R $ satisfying the functional equation
$$ f ( r \cos \phi ) + f ( r \sin \phi ) = f ( r ) \tag 0 \label 0 $$
for every $ r \in [ 0 , + \infty ) $ and every $ \phi \in \big[ \frac \pi 6 , \frac \pi 4 \big] $. You can show that there is a $ c \in \mathbb R $ such that
$ f ( x ) = c x ^ 2 $ for all $ x \in [ 0 , + \infty ) $. As you've mentioned, functions of this form satisfy the above criterions, and thus they form the class of all solutions.
The trick is to observe that when you consider $ ( r , \phi ) $ as polar coordinates of a point (noting that we have $ r \ge 0 $ and that's valid), $ ( x , y ) = ( r \cos \phi , r \sin \phi ) $ will be the cartesian coordinates of the same point. In that case, having $ \frac \pi 6 \le \phi \le \frac \pi 4 $ is equivalent to having $ y \le x \le 2 y $.
So we start by considering $ x $ and $ y $ so that $ y \le x \le 2 y $. Then letting $ r = \sqrt { x ^ 2 + y ^ 2 } $, we would have $ r \in [ 0 , + \infty ) $ and there would be a $ \phi \in \big[ \frac \pi 6 , \frac \pi 4 \big] $ so that
$ x = r \cos \phi $ and $ y = r \sin \phi $. Therefore by \eqref{0} we have
$$ f ( x ) + f ( y ) = f \left( \sqrt { x ^ 2 + y ^ 2 } \right) \tag 1 \label 1 $$
for every $ x $ and $ y $ with $ y \le x \le 2 y $. Define $ g : [ 0 , + \infty ) \to \mathbb R $ by $ g ( z ) = f \left( \sqrt z \right) $. Then by \eqref{1} we have
$$ g ( x ) + g ( y ) = g ( x + y ) \tag 2 \label 2 $$
for every $ x $ and $ y $ with $ 0 \le y \le x \le 4 y $. As $ 0 \le 0 \le 0 \le 4 \cdot 0 $, we can let $ x = y = 0 $ in \eqref{2} and get $ g ( 0 ) = 0 $. We can now inductively show that
$$ g ( n x ) = n g ( x ) \tag 3 \label 3 $$
for every nonnegative integer $ n $. This holds for $ n = 0 $ and $ n = 1 $. For $ n \ge 2 $, we either have $ n = 2 m $ or $ n = 2 m + 1 $ for some nonnegative integer $ m $ less than $ n $. If $ n = 2 m $, then since $ 0 \le m x \le m x \le 4 m x $, substituting $ m x $ for both $ x $ and $ y $ in \eqref{2} we get $ 2 g ( m x ) = g ( 2 m x ) $. As $ m < n $, by induction hypothesis we have $ g ( m x ) = m g ( x ) $, which leads to $ g ( n x ) = n g ( x ) $. If $ n = 2 m + 1 $, since $ n \ge 2 $ we have $ 0 \le m x \le ( m + 1 ) x \le 4 m x $, and substituting $ ( m + 1 ) x $ for $ x $ and $ m x $ for $ y $ in \eqref{2} we get $ g \big( ( m + 1 ) x \big) + g ( m x ) = g \big( ( 2 m + 1 ) x \big) $. As $ m + 1 < n $, by induction hypothesis we have $ g \big( ( m + 1 ) x \big) = ( m + 1 ) g ( x ) $ and $ g ( m x ) = m g ( x ) $, which again leads to $ g ( n x ) = n g ( x ) $.
Now, for $ n > 0 $, we can substitute $ \frac x n $ for $ x $ in \eqref{3} and get
$$ g \left( \frac x n \right) = \frac 1 n g ( x ) \text . \tag 4 \label 4 $$
Substituting $ m x $ for $ x $ in \eqref{4} for some nonnegative integer $ m $ and using \eqref{3}, we find out that $ g ( q x ) = q g ( x ) $ for every nonnegative rational number $ q $. In particular, defining $ c = g ( 1 ) $, we have $ g ( q ) = c q $ for every nonnegative rational number $ q $. Since $ f $ is monotonous, $ g $ is monotonous, too, and hence for every nonnegative real number $ x $ between two nonnegative rational numbers $ p $ and $ q $, $ g ( x ) $ is between $ c p $ and $ c q $. As nonnegative rational numbers are dense in nonnegative real numbers, this forces $ g ( x ) $ to be equal to $ c x $, and thus $ f ( x ) = c x ^ 2 $, as desired.
| {
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Solve for $x$ in $\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$ I am taking physics right now and I have gotten my problem down to the following equation:
$$\sqrt{3.5^2+x^2}-\sqrt{3.0^2+x^2}=0.25$$
I am looking for some guidance as to what to do with the square roots in order to solve for $x$. I know I can't take the square of both sides and I tried to factor and failed.
| The idea is to move the equation's terms around such that one and only one square root appears on one side. Squaring then removes that isolated square root; the other side may still have a square root, but there will be one less overall. Repeat until a polynomial is obtained.
Let $y=x^2$, then (using fractions instead of decimals) we have
$$\sqrt{\frac{49}4+y}-\sqrt{9+y}=\frac14$$
$$\sqrt{\frac{49}4+y}=\sqrt{9+y}+\frac14$$
$$\frac{49}4+y=\left(\sqrt{9+y}+\frac14\right)^2=9+y+\frac12\sqrt{9+y}+\frac1{16}$$
$$\frac{49}4-9-\frac1{16}=\frac12\sqrt{9+y}=\frac{51}{16}$$
$$\sqrt{9+y}=\frac{51}{8}$$
$$9+y=\frac{2601}{64}$$
$$y=\frac{2025}{64}$$
$$x=\pm\frac{45}8=\pm5.625$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the set of strings over $\{a, b\}$ that do not contain the substring aaa
I need to identify the set of all possible strings over $\Sigma = \{a, b\}$ that do not contain the substring $aaa$.
I have becoming more familiar with regular-language and languages explored in Automata Theory, and have been defining a particular language through regular-expressions.
I am having a hard time trying to find the set of strings of $\Sigma^*$ (where $\Sigma = \{a, b\}$) that do not contain the substring aaa.
Any advice on how to generate the regular-expression that will define this language would be helpful.
| We can describe the set of valid strings as the strings built from the alphabet $V=\{a,b\}$ which contain at most one or two consecutive $a$'s.
These are strings
*
*starting with zero or more $b$'s:$$b^*$$
*followed by zero or more occurrences of $a$ or $aa$ each followed by one or more $b$'s: $$(ab^+|aab^+)^*$$
*and terminating with zero, one or two $a$'s
$$(\varepsilon|a|aa)$$
We obtain
\begin{align*}b^*(ab^+|aab^+)^*(\varepsilon|a|aa)\tag{1}\end{align*}
$$ $$
Add-on: The regular expression (1) generates all valid words in a unique manner. In such cases we can use it to derive a generating function $$A(z)=\sum_{n=0}^\infty a_n z^n$$ with
$a_n$ giving the number of valid words of length $n$.
In order to do so all we need to know is the geometric series expansion since the $star$ operator
\begin{align*}
a^*=\left(\varepsilon|a|a^2|a^3|\cdots\right)\qquad\text{ translates to }\qquad 1+a+a^2+a^3+\cdots=\frac{1}{1-a}
\end{align*}
Accordingly $a^+=aa^*$ translates to $\frac{a}{1-a}$ and alternatives like $(\varepsilon|a|aa)$ can be written as $1+a+a^2$.
We obtain by translating the regular expression in the language of generating functions (and by mixing up somewhat the symbolic to provide some intermediate steps)
\begin{align*}
b^*\left(ab^+|aab^+\right)^*(\varepsilon|a|aa)
&\longrightarrow \quad \frac{1}{1-z}\left(\left.\frac{z^2}{1-z}\right|\frac{z^3}{1-z}\right)^*\left(1+z+z^2\right)\\
&\longrightarrow \quad \frac{1}{1-z}\left(\frac{z^2+z^3}{1-z}\right)^*(1+z+z^2)\\
&\longrightarrow \quad \frac{1}{1-z}\cdot\frac{1}{1-\frac{z^2+z^3}{1-z}}(1+z+z^2)\\
&\quad\quad=\frac{1+z+z^2}{1-z-z^2-z^3}
\end{align*}
We conclude: The number of valid words is given by the generating function $A(z)$
\begin{align*}
A(z)&=\frac{1+z+z^2}{1-z-z^2-z^3}\\
&=1+2z+4z^2+7z^3+13z^4+\color{blue}{24}z^5+44z^6+81z^7+\cdots
\end{align*}
The expansion was done with the help of Wolfram Alpha. We see that e.g. the number of valid words of length $5$ is $24$.
So, out of $2^5=32$ binary words of length $5$ there are $8$ invalid words marked blue in the table below.
\begin{array}{cccc}
\color{blue}{aaaaa}\qquad&ab\color{blue}{aaa}\qquad&ba\color{blue}{aaa}\qquad&bb\color{blue}{aaa}\\
\color{blue}{aaaa}b\qquad&abaab\qquad&b\color{blue}{aaa}b\qquad&bbaab\\
\color{blue}{aaa}ba\qquad&ababa\qquad&baaba\qquad&bbaba\\
\color{blue}{aaa}bb\qquad&ababb\qquad&baabb\qquad&bbabb\\
aabaa\qquad&abbaa\qquad&babaa\qquad&bbbaa\\
aabab\qquad&abbab\qquad&babab\qquad&bbbab\\
aabba\qquad&abbba\qquad&babba\qquad&bbbba\\
aabbb\qquad&abbbb\qquad&babbb\qquad&bbbbb\\
\end{array}
| {
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Prove: $\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$ Please help me prove the following inequality:
$$\sin 2 \theta \;\ge\; \frac{1}{2}\left(-1-3 \cos^2 \theta\right)$$
I have been working on it for an hour in vain. I have derived $2\sin \theta \cos \theta$ from the left side and $\frac{1}{2}\left(-4+3 \sin^2 \theta\right)$ from the right side, but I don't know where to go from there.
| The extreme values of $A\cos x +B\sin x$ are $\pm\sqrt {A^2+B^2}.$ This is obvious and trivial if $A=B=0.$ If $A,B$ are not both $0,$ there exists $y$ such that $\cos y=A/\sqrt {A^2+B^2}$ and $\sin y=B/\sqrt {A^2+B^2},$ so $$|A\cos x +B\sin x| =|(\sqrt {A^2+B^2}\;)(\cos x \cos y+\sin x \sin y)|$$ $$=(\sqrt {A^2+B^2}\;) |\cos (x-y)|\leq \sqrt {A^2+B^2}.$$
Putting $\cos^2 \theta=(1+\cos 2 \theta)/2,$ the inequality in the Q can be re-arranged as $5\geq -4\sin 2\theta -3\cos 2 \theta .$ Which is true because $|-4\sin 2 \theta -3\cos 2 \theta| \leq \sqrt {(-4)^2+(-3)^2}=5.$
| {
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Proof of rational numbers I am recently starting in the subject of pure mathematics and this problem has eluded me.
Any rational number $\frac{p}{q}$ can be expressed in a simple series form:
$$\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4},...$$
Show that $\frac{p}{q}$ is the $[\frac{1}{2}(p+q-1)(p+q-2)+q]th$ of the series (where $p$ and $q$ are coprime and $p \neq 0$).
| First of all I think that the first number in the sequence should be $\frac 11$.
Now consider the sum of the numerator and the denominator and write it as a sequence and we have:
$$2,3,3,4,4,5,5,5,5,6,6,6,6,6...$$
Therefore if $p+q=r$ we have that the fraction $\frac{p}{q}$ might appear after $1+2+3+...+{r-2} = \frac{(r-1)(r-2)}{2} = \frac{(p+q-1)(p+q-2)}{2}$ elements in the sequence, as the sum of denominator and numerator before that element is less than $r$.
Now note that the first denominator of the fractions whose sum is $p+q$ is $1$, then $2$ and so on. So $q$ will appear in the denominator after $q$ elements, starting from the $\frac{(p+q-1)(p+q-2)}{2}$-th element. Therefore we have that the $\frac{(p+q-1)(p+q-2)}{2} + q$ elements is the fraction with denominator $q$ and sum of denominator and numerator $p+q$. In other words it's the rational number $\frac{p}{q}$
| {
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"url": "https://math.stackexchange.com/questions/1935563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$(2\sin^2\theta)^2+(2\cos^2\theta)^2=(1-\cos2\theta)^2+(1+\cos2\theta)^2=2(1+\cos^22\theta)$$
Now $2\cos^22\theta=1+\cos2(2\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
} |
Finding $x$ such that $2x^2-2x+1=(2m+1)^2$ We need to find the smallest integral value of $x$ such that $2x^2-2x+1=(2m+1)^2$ and $x \ge 10^{15}$.
| HINT: $$2x^2 - 2x + 1 = (2m+1)^2 \iff x^2 + (x-1)^2 = (2m+1)^2$$
So the problem reduces to finding a specific Pythagorean triplet.
UPDATE:
We have that this is a primitive Pythagorean triplet, as $(x,x-1) = 1$. Therefore we either have $x = m^2 - n^2, x-1 = 2mn$ or $x=2mn, x-1 = m^2 - n^2$ for some positive integers $m,n$ which are coprime and one of them is odd.
Now going through the cases we have:
$$m^2-n^2 = 2mn+1 \iff (m-n)^2 = 2n^2 + 1$$
This is a Pell's Equation and has infinitely many solutions. So eventually you would be able to find a solution that will give you $x \ge 10^{15}$
Similarly for the other case we have:
$$m^2 - n^2 = 2mn - 1 \iff (m-n)^2 = 2n^2 - 1$$
Again this is a Pell's Equation, but this one is negative one, which doesn't change much.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$? I need to find the limit of this sequence:
$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$
So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$
$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$
Then divide by n?
And get $\dfrac{9/n-n/n+9/n}{\sqrt{n^2/n+9/n}+\sqrt{n^2/n-n+9/n}}$
$\lim = \dfrac{1}{\infty} = 0$?
| $$\dfrac{n}{(\sqrt{n^2+9}+\sqrt{n^2-n+9})}=\dfrac{n/n}{\frac{1}{n}(\sqrt{n^2+9}+\sqrt{n^2-n+9})}$$
$$=\dfrac{1}{\sqrt{n^2/n^2+9/n^2}+\sqrt{n^2/n^2-n/n^2+9/n^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $ Given the followning limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{xy}{x^2 + y^2}\right)^{x^2} $$
To find limit I have made following steps:
*
*Let $ x = y $ ,then limit equals $0$
*Let $ x > y $ ,then consider the limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{x^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{1 + \frac{y^2}{x^2}}\right)^{x^2} = 0$$
with respect to $$0 < y^2/x^2 < const$$
*Let $ y > x $ ,then consider the limit:
$$ \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{x^2}{x^2 + y^2}\right)^{y^2} = \lim_{x \rightarrow \infty, y \rightarrow \infty} \left( \frac{1}{\frac{x^2}{y^2} + 1}\right)^{x^2} = 0$$
with respect to $$0 < x^2/y^2 < const$$
What could you say about my solution?
| For $x, y >1$, we have the fact that
\begin{align}
2\leq\frac{x}{y}+\frac{y}{x}
\end{align}
which means
\begin{align}
\left(\frac{xy}{x^2+y^2}\right)^{x^2}=\left(\frac{1}{\frac{x}{y}+\frac{y}{x}}\right)^{x^2} \leq \left(\frac{1}{2}\right)^{x^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifcation of a sum I’m unsure if I’m allowed to simplify like this:
$$\begin{align*}
\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} &= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7^{n} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7}{10}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{10}
\end{align*}$$
Please tell me whether this is right or wrong and if possible how else the manipulated part could be simplified. Thanks in advance!
| $$\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!}+\sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!} =$$
$$=\sum_{n=2}^\infty \frac{8}{5^n(n-1)!}+\frac{1}{7}\sum_{n=2}^\infty \left(\frac{7}{10}\right)^n =$$
$$=\frac{8}{5}\sum_{n=2}^\infty \frac{(1/5)^{n-1}}{(n-1)!}+\frac{1}{7}\sum_{n=2}^\infty \left(\frac{7}{10}\right)^n =$$
$$=\frac{8}{5}(e^{1/5}-1)+\frac{1}{7}\left(\frac{1}{1-7/10}-1-\frac{7}{10}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative vanishes at $x=0$ and $x=2\pi$.
| Sorry for being late
By ‘Rationalisation’
$$
\begin{aligned}
\int_0^{2 \pi} \frac{d x}{5-4 \cos x}=&2 \int_0^\pi \frac{d x}{5-4 \cos x} \\
= & 2 \int_0^\pi \frac{5+4 \cos x}{25-16 \cos ^2 x} d x \\
= & 10 \int_0^\pi \frac{d x}{25-16 \cos ^2 x}+8 \int_0^\pi \frac{\cos x}{25-16 \cos ^2 x} d x \\
= & 20 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{25 \sec ^2 x-16} d x+8 \int_0^\pi \frac{d(\sin x)}{9+\sin ^2 x} \\
= & 4 \int_0^{\frac{\pi}{2}} \frac{d(5 \tan x)}{(5 \tan x)^2+3^2}+\frac{8}{3}\left[\tan ^{-1}\left(\frac{\sin x}{3}\right)\right]_0^\pi \\
= & \frac{4}{3}\left[\tan ^{-1}\left(\frac{5 \tan x}{3}\right)\right]_0^{\frac{\pi}{2}} \\
= & \frac{2 \pi}{3}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Finding the number of solutions of an expression Find the number of integers $1\leq x \leq 2010 $ such that the expression
$\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}}-\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}}$
is a rational number
NOTE -> I LITERALLY do not understand where to start !! HELP ME!!
| The number of integers $1≤x≤2010$, (such that the expression above is a rational number), is 26.
The value of $x$ is
$x=3n^{2}-6n+4$.
In this way,
$\sqrt{\frac{x-1}{27}}=\sqrt{\frac{3n^{2}-6n+3}{27}}=\sqrt{\frac{3(n-1)^{2}}{27}}=\frac{n-1}{3}$;
$(x+8)\sqrt{\frac{x-1}{27}}=(n^{2}-2n+4)(n-1)$;
$x+(x+8)\sqrt{\frac{x-1}{27}}=(3n^{2}-6n+4)+(n^{2}-2n+4)(n-1)=n^{3}$;
$x-(x+8)\sqrt{\frac{x-1}{27}}=(3n^{2}-6n+4)-(n^{2}-2n+4)(n-1)=(2-n)^{3}$;
$\sqrt[3]{x+(x+8)\sqrt\frac{x-1}{27}}-
\sqrt[3]{x-(x+8)\sqrt\frac{x-1}{27}}=$
$=n-(2-n)=2(n-1)$.
The number $n$ such as $1≤x≤2010$, is the integer solution of
$1≤3n^{2}-6n+4≤2010$,
that is 26.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the base numeric system Find the numeric base we are using if $ x = 4 $ and $ x = 7 $ are zeros of $ 5x^2 - 50x + 118. $
When I plug in $ x = 4 $ and $ x = 7, $ I receive $ -2 $ and $ 13, $ respectively, but how do I proceed from that?
| Let $b$ be the base in question. Presumably $b>8$, since $8$ appears as a digit in $118$. We also know that $5x^2-50x+118$ is a multiple of $(x-4)(x-7)$, and since the $x^2$ term of $(x-4)(x-7)$ is simply $x^2$, we must have
$$5x^2-50x+118=5(x-4)(x-7)\;.$$
From this we deduce that $55$ in base ten is $50_b$. This clearly implies that $b=11$ (in base ten), and this is confirmed by the fact that $5\cdot4\cdot7=140$ in base ten is $118_b$: with $b=11$ this is $121+11+8=140$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find all primes $p$ for which $\frac{2^{p-1}-1}p$ is a perfect square
Find all primes p for which the quotient $\frac{2^{p-1}-1}p$ is a perfect square.
It is a number theory problem. By guessing I found the value of $p$ is 3. But how do I prove it and how do I found the other primes?
| Put $p = 2q+1$, then the expression is $\dfrac{4^q-1}{2q+1} = \dfrac{(2^q-1)(2^q+1)}{2q+1}$.
Now, since $2^q-1$ and $2^q+1$ are co prime, one of them must be a perfect square. Let $(2^q-1)(2^q+1)=pm^2$. Then two cases arise, namely that :
1) $2^q-1 = px^2,2^q+1 = y^2$
2) $2^q-1 = x^2,2^q+1 = py^2$
For the first case, note that $(y-1)(y+1) = 2^q \implies y+1=2^m,y-1=2^n$, which gives $2^m-2^n=2$, giving $m=2,n=1$. This gives the case $q=3$ and $p=7$.
The second case is solved by the fact that $2^q - 1$ , if $q \geq 2$ is congruent to $3$ modulo $4$, and hence cannot be a perfect square. So this forces $q = 1$ to be checked, which gives $p = 3$ which works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$ $$\lim\limits_{x \to 1+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}$$
I tried this
$\frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1} = 1-2\sqrt{\frac{\arctan(x)-\frac{\pi}{4}}{x-1}}.\frac{1}{\sqrt{x-1}}$
I am stuck here !
| Let's compute
$$
\lim_{x\to1^+}\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}
$$
One way could be to substitute
$$
\sqrt{\arctan x-\frac{\pi}{4}}=t
$$
so
$$
\arctan x=t^2+\frac{\pi}{4}
$$
and therefore
$$
x=\tan(t^2+\tfrac{\pi}{4})=\frac{\tan(t^2)+1}{1-\tan(t^2)},
\qquad
x-1=\frac{2\tan(t^2)}{1-\tan(t^2)}
$$
so we get
$$
\lim_{t\to0^+}\frac{t(1-\tan(t^2))}{2\tan(t^2)}=
\lim_{t\to0^+}\frac{1}{2t}\frac{t^2}{\tan(t^2)}(1-\tan(t^2))
$$
The first factor goes to $\infty$, the other two go to $1$, so the overall limit is $\infty$.
Now
$$
\lim_{x \to 1^+} \frac{x-\sqrt{\arctan(x)-\frac{\pi}{4}}-1}{x-1}=
\lim_{x\to1^+}\biggl(1-\frac{\sqrt{\arctan x-\frac{\pi}{4}}}{x-1}\,\biggr)
=-\infty
$$
A different way is considering that
$$
\lim_{x\to1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1}
$$
is the derivative of $\arctan$ at $1$, so it is finite (precisely $1/2$); then our limit is
$$
\lim_{x\to 1^+}\frac{\arctan x-\frac{\pi}{4}}{x-1}
\frac{1}{\sqrt{\arctan x-\frac{\pi}{4}}}=\infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are real numbers then $(x-y-z)^2 \geqslant 0$. That is $x^2 + y^2 +z^2 -2xy-2xz-2yz \geqslant 0$. But this implies that $\frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$, so because $x^2 + y^2 +z^2 \geqslant \frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$ we get the desired result.
| Based on hagnatural's comment: No, your proof is incorrect since one of your signs is wrong:
$$(x-y-z)^2 = x^2 + y^2 + z^2 - 2xy - 2xz \color{red}{+} 2yz.$$
And unfortunately there is no direct term $(\pm x \pm y \pm z)^2$ which gives you opposite signs on the cross terms, compared to the squares. (But as noted in the other answers, it is possible to use similar ideas by combining the three terms $(x - y)^2$, $(x - z)^2$, and $(y - z)^2$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$
Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.
I tried to do it by comparing the two expressions but I can't proceed.
| Let $f(x)=(x+b)(x+c)$ with $b,c\in\Bbb Z$.
Wlog $b\le c$.
For $-c\le x\le b$, we have $f(x)\le 0$, hence $f(x)-1<0$.
If $x\ge -b+2$, we have $f(x)-1\ge 2\cdot 2-1>0$. By symmetry, $f(x)-1>0$ also for $x\le -c-2$.
Hence the only possibly integer roots of $f(x)-1$ are $-b+1$ and $-c-1$.
Moreover, $f(-b+1)-1=f(-c-1)=c-b$, hence $f(x)-1$ has integer roots iff $b=c$ and in that case these roots are $-b\pm1$ and hence differ by $2$.
So if one of the integer roots is $-1991$, then the other, $-a$, must be $-1989$ or $-1993$, i.e., $$a\in\{1989,1993\}.$$
On the other hand indeed, $(x+1989)(x+1991)+1=(x+1990)(x+1990)$ and $(x+1993)(x+1991)=(x+1992)(x+1992)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| Hint. Is it possible to find $A,B,C$ such that for all $x\in\mathbb{R}$,
$$(x+3)^2=A(x+2)^2+B(x+1)^2+Cx^2?$$
P.S. The answer is yes: $A = 3$, $B = -3$, $C = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 4
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Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
| Multiply the numerator and denominator by $x(1+x)(1-x)$. This clears away all the "inner" denominators, leaving $${xx(1+x)+(1+x)(1+x)(1-x)\over (1-x)(1-x)(1+x)+xx(1-x)}={(1+x)\over (1-x)}{x^2+1-x^2\over x^2+1-x^2}={1+x\over 1-x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Calculation of an infinite sum $\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26$ Could you give me some advice on how to calculate the following sum?
$$\begin{aligned}\sum_{k=1}^{\infty} \frac{k^3}{2^k} = 26\end{aligned}$$
Thank you!
| $$
\begin{aligned}
& S_{0} = \sum_{n=1}^{\infty} \frac{n^{0}}{2^{n}} = \sum_{n=1}^{\infty} \frac{1}{2^{n}} \Rightarrow \color{red}{S_{0} = 1} \\ \\
& S_{1} = \sum_{n=1}^{\infty} \frac{n^{1}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1 - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} - S_{0} \\
& \qquad \Rightarrow S_{1} + S_{0} = \sum_{n=1}^{\infty} \frac{n + 1}{2^{n}} = 2 \sum_{n=1}^{\infty} \frac{n + 1}{2^{n + 1}} = 2 \sum_{n=2}^{\infty} \frac{n}{2^{n}} = 2 \left[ - \frac{1}{2} + \sum_{n=1}^{\infty} \frac{n}{2^{n}} \right] = - 1 + 2 S_{1} \\
& \qquad \Rightarrow \color{red}{S_{1} = 1 (S_{0} + 1) = 2} \\ \\
& S_{2} = \sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}} = \sum_{n=1}^{\infty} \frac{n^{2} + 2 n + 1 - 2 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n}} - 2 S_{1} - S_{0} \\
& \qquad \Rightarrow S_{2} + 2 S_{1} + S_{0} = S_{2} + 3 (S_{0} + 1) - 1 = 2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{2} \\
& \qquad \Rightarrow \color{red}{S_{2} = 3 (S_{0} + 1) = 6} \\ \\
& S_{3} = \sum_{n=1}^{\infty} \frac{n^{3}}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3} - 3 n^{2} - 3 n - 1}{2^{n}} = \sum_{n=1}^{\infty} \frac{(n + 1)^{3}}{2^{n}} - 3 S_{2} - 3 S_{1} - S_{0} \\
& \qquad \Rightarrow S_{3} + 3 S_{2} + 3 S_{1} + S_{0} = S_{3} + 13 (S_{0} + 1) - 1 = 2 \sum_{n=1}^{\infty} \frac{(n + 1)^{2}}{2^{n + 1}} = - 1 + 2 S_{3} \\
& \qquad \Rightarrow \color{red}{S_{3} = 13 (S_{0} + 1) = 26} \\ \\
& \text{... etc}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Hints to find solution of : $\left\lfloor \frac{x}{100} \left\lfloor \frac{x}{100}\right\rfloor \right\rfloor=5$ Find the number of integer solution of $$\left\lfloor \dfrac{x}{100} \left\lfloor \dfrac{x}{100}\right\rfloor \right\rfloor=5$$
$\lfloor . \rfloor$ is Greatest Integer Function or floor function.
| As $y-1<\lfloor y\rfloor\le y $, we find that for $y\ge1$
$$\tag1y^2-y-1<\lfloor y\lfloor y\rfloor\rfloor\le y^2 $$
whereas for $y<0$, we have
$$\tag2y^2-y-1>\lfloor y\lfloor y\rfloor\rfloor\ge y^2 $$
Finally, for $0\le y<1$, we clearly have $\lfloor y\lfloor y\rfloor \rfloor=0$
Thus we cannot have $\lfloor y\lfloor y\rfloor\rfloor=5$ with $y\le -\sqrt 5$ (right hand side of $(2)$), nor with $-2\le y<0$ (left hand side of $(2)$), nor with $0\le y<1$, nor with $1\le y<\sqrt 5$ (right hand side of $(1)$), nor with $y\ge 3$ (left hand side of $(1)$). In other words, $-\sqrt 5< y <-2$ or $\sqrt 5\le y<3$. Then $\lfloor y\rfloor =-3$ or $\lfloor y\rfloor=2$.
In the first case, $$-3y-1<\lfloor -3y\rfloor=\lfloor y\lfloor y\rfloor \rfloor \le -3y,$$
so that we need $y\ge -\frac 53$, which contardicts $y<-2$.
In the second case
$$2y-1<\lfloor 2y\rfloor=\lfloor y\lfloor y\rfloor \rfloor \le 2y,$$
so that we need $y\ge\frac 52$. And indeed, for $\frac25\le y<3$, we verify directly that $\lfloor y\lfloor y \rfloor\rfloor=5 $.
If we impose that $y=\frac x{100}$ with $x\in\Bbb Z$, this is equivalent to $$250\le x<300$$
whic allows exactly $50$ differnet values of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 16 \cdot 21p - 80+5 = 16 \cdot 21p - 75 $
But its not divisible by 21. Whats I doing wrong ?
| **$$ \color{red} {2^{12}=4096=21*195+1}$$
$$2^{12n+4}+5\equiv\\ 2^4 .2^{12n}+5 \equiv\\16.4096^n+5 \equiv\\16 (4095+1)^n+5\equiv\\16.(21.195+1)^n+5 \equiv \\ 16(1)^n+5 \equiv \\21\equiv0$$ so $$21|2^{12n+4}+5\\n=0,1,2,3,4,5,...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simple way to calculate $n! \pmod p$ I have the exercise "Calculate $10! \pmod{13}$".
I have the following two approaches to solve the exercise:
*
*Brute force approach
$$
1! \equiv 1 \pmod{13} \\
2! = 2 \cdot 1! \equiv 2 \cdot 1 \equiv 2 \pmod{13} \\
3! = 3 \cdot 2! \equiv 3 \cdot 2 \equiv 6 \pmod{13} \\
\cdots \\
10! = 10 \cdot 9! \equiv 10 \cdot 11 = 110 = 8 \cdot 13 + 6 \equiv 6 \pmod{13}
$$
*Approach using Wilson's theorem:
Wilson's theorem states that
$$p \in \mathbb{P} \implies (p-1)! \equiv -1 \pmod p$$
For my exercise:
$$13 \in \mathbb{P} \implies \\
(13-1)! = 12! = 10!\cdot 11 \cdot 12 \equiv -1 \pmod{13} \implies \\
10! \equiv -(11 \cdot 12)^{-1} \pmod{13}
$$
Using Fermat's little theorem
$$
a^p \equiv a \pmod p \implies a^{p-2} \cdot a \cdot a \equiv a^{-1} \cdot a \cdot a \pmod p \implies a^{p-2} \equiv a^{-1} \pmod p \\
$$
For my exercise:
$$10! = -(11 \cdot 12)^{-1} \equiv \\
-(11 \cdot 12)^{13-2} = -(11 \cdot 12)^{11} \equiv \\
-(-2 \cdot -1)^{11} = -2^{11} = \\
-2^6 \cdot 2^5 \equiv 1 \cdot 2^5 = \\
32 \equiv 6 \pmod{13} \\
$$
Both approaches look quite bulky.
In the first method I have to make $O(n)$ multiplications.
In the second method I have to make $O(p-n)$ multiplications which is smaller than in the first method, but also can be huge number for big $p$ and $n$.
Is there a way to improve my solution?
Is there an efficient way to calculate $n! \pmod p$ for big $n$ and $p$?
| Start with your Wilson's Theorem approach but finish off differently. Note that $12\equiv -1$ and $11\equiv -2 \pmod{13}$ and that these two numbers have easy inverses $(-2)(-7) \equiv 1$ and $(-1)(-1) \equiv 1 \pmod{13}$, so
$$10! \equiv -(11)^{-1}(12)^{-1} \equiv -(-2)^{-1}(-1)^{-1} \equiv -(-7)(-1) \equiv 6 \pmod{13}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
| This solution may look different. :-)
Note that $f(x)=(x-\frac{a}{2})^2+1-\frac{a^2}{4}$,
$$2*\frac{1}{4} \ge |f(x) -f(x+2)| = |2*(2x+2-a)| \Rightarrow |(x+2-\frac{a}{2})+(x-\frac{a}{2})|\le \frac{1}{4},$$
so $x-\frac{a}{2}$ and $x+2-\frac{a}{2}$ must have different signs as their difference is 2. It must be $x-\frac{a}{2} \le 0 \le x+ 2-\frac{a}{2}$, since $x-\frac{a}{2} < x+2-\frac{a}{2}.$
So $-x-2 + a -\frac{a}{2} \le 0$, and $f(-x-2+a) = f(x+2)$. Since $-1+ \frac{a}{2}$ is the average of $x$ and $-x-2+ a$, and both $x-\frac{a}{2}$ and $-x-2+a-\frac{a}{2}$ are negative, then we know $f(-1+\frac{a}{2})$ is between $f(x)$ and $f(-x-2+a)=f(x+2)$, so $|f(-1+\frac{a}{2})|\le 1/4$, which yields $$7 \le a^2 \le 9.$$
On the other hand, if $7\le a^2 \le 9$, let $x=-1 + a/2$, then $|f(x+2)|=|f(x)|\le 1/4.$ therefore, $7 \le a^2 \le 9$ is the solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is this the shortest proof of Fermat's last theorem for exponent 3? To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be done with the method outlined below).
Here we use the integer representation of a cube $n^3$ as a sequence of consecutive odd numbers. We get the sequence by writing $n^3=n\cdot n^2$. This means that we need $n$ terms with an average value of $n^2$ to represent the cube $n^3$. This representation has a big advantage over other possible representations and it is that consecutive cubes $n^3$, $(n+1)^3$, and $(n+2)^3$ have a representation that ends when the next one starts (that is they do not share any term).
For example $2^3=2\cdot 2^2= 3+5$ and $3^3=3\cdot 3^2=7+9+11$. This is a well known property of cubes when represented as a sequence of consecutive odd integers (https://en.wikipedia.org/wiki/Cube_(algebra)). This means that we can concatenate the left hand side of the equation to form a new sequence. And we need to figure out what integer this concatenated sequence represents.
It was not really difficult to do that.
So we gave an example and show why the left hand side cannot represent a cube whose value is the right hand side of the equation.
$$6^3 + 7^3 = 8^3.$$
We write down the representation of each sequence:
$$6^3 = 6\cdot 6^2 = 31 + 33 + 35 + 37 + 39 + 41,$$ that is 6 terms with average value of $6^2=36$.
We do the same for $7^3$.
$$7^3 = 7\cdot7^2 = 43 + 45 + 47 + 49 + 51 + 53 + 55,$$ that is 7 terms with average value of $7^2=49$.
When we concatenate the two sequences into one we will call lhs, we get:
$$\text{lhs} = 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55.$$
This sequence has $(6+7)=13$ terms with average $(n^2 + (n+1)^2 +1)/2$. In our case, the lhs is therefore a sequence with 13 terms with average 43 which represents the integer $m=13\cdot 43=559=13\cdot43$.
The right hand side is $\text{rhs}=8^3=8\cdot8^2$, so it will have 8 terms with an average of $64$. $$\text{rhs} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71.$$ So looking at the result, we see that $\text{lhs}=13\cdot 43$ cannot be equal to $\text{rhs}=8^3$. That is $(2n+1)\cdot (n^2 + (n+1)^2 +1)/2$ cannot be equal to $(n+2)^3$.
I am not sure if this method can be extended to the general case where $x,y$ and $z$ can take any value (not necessarily consecutive). But I think it's worth looking into it because of the simplicity of the method.
| $n^3 + (n+1)^2 = (n+2)^3 \implies $
$(m-1)^3 +m^3=(m+1)^3; m=n+1$
$2m^3-3m^2+3m-1=m^3+3m^2+3m+1$
$m^3=6m^2+2$
$m=6 + 2/m^2$ so $m^2|2$ and $m > 6$. So $m=1>6$. Clearly impossible.
Of course, the shortest proof would be "this is a single cubic equation with one variable and can easily be confirmed to have no integer solution".
This is simply far too specific a case to be of any significance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Decimal representation of $1/n$ includes the string of digits $777$
Find an integer $n$ with the property that the decimal representation of $1/n$ includes the string of digits $777$.
For $\dfrac{m}{n}$ we can find a string of digits $777$ just by looking at $0.777$ and writing that as a fraction. But how do we do it for $1/n$?
| $$\frac{7}{9}=0.777\ldots,\qquad \frac{7}{90}=0.0777\ldots $$
and if we consider
$$ \frac{7}{90000} = 0.0000777\ldots $$ we have
$$ \frac{7}{90000}+\frac{1}{10000} = 0.0001777\ldots $$
with
$$ \frac{7}{90000}+\frac{1}{10000}=\frac{7+9}{90000}=\frac{16}{90000}=\frac{1}{\color{red}{5625}} $$
since $7+9=16$ is a divisor of $9\cdot 10^4$. Even better: since $7+18$ is a divisor of $9\cdot 10^2$,
$$ 0.02\color{red}{777}\ldots=\frac{7}{900}+\frac{2}{100}=\frac{1}{\color{red}{36}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
prove or disprove $\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}$ let $n\in N^{+}$ and such $n\ge 2$
prove or disprove
$$\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}\tag{1}$$
and I have use induction prove $$\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}<3$$
because $$\sqrt{(k+1)(k-1)}<k$$
and we known this Ramanujan's indentity: Prove $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}=3$
and also use indution prove this
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+(n-1)\sqrt{1+n\sqrt{1}}}}}}<3$$
see:Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $.
But how to prove $(1)$
| $$L=\sqrt{2\sqrt{3\sqrt{4{\cdots\sqrt{n}}}}}$$
$$\log L=\sum_{k=1}^{\infty }\frac{\log(k+1)}{2^k}=\quad{\text{LerchPhi}^{0,1,0}}(\frac{1}{2},0,2)=1.0156678457368767......$$
so
$$L=e^{1.01566784573687........}=2.761206841957498033230454646......$$
hence
$$2.7612.....<3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $30|(a^3b-ab^3) $ Prove that if three distinct integers are chosen at random then there will exists two among them, say $a$ and $b$ such that $30|(a^3 b-ab^3)$
| 1) Divisibility by $5$.
Let $\{a,b,c\}$ the three numbers reduced modulo $5$. If one of them is $0$ or two of them are equal we clearly have
$$ab(a+b)(a-b)\equiv 0\pmod 5$$
So be all the three, $a,b,c$ distinct and non-zero modulo $5$ so we have
$$\{a,b,c\}=\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$$
In the four cases we clearly have two of the (class) numbers have sum $0$ modulo $5$ (because of $2+3=1+4$) so $ab(a+b)(a-b)\equiv 0\pmod 5$.
Thus the expression is for two of the three integers divisible by $5$.
2) Divisibility by $3$ and by $2$.
The same method but easier than for divisibility by $5$ can be used for divisibility by $3$ and by $2$.
Thus the divisibility by $30=2\cdot3\cdot5$ of the expression for two of the three numbers..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 - 1}{2} + i\frac{\sqrt 3 + 1}{2}$.
$$z = r(\cos \phi + i \sin \phi)$$
$$r = |z| = \sqrt 2$$
$$\phi = \arctan {\sqrt 3 + 1}$$
Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?
Thank you in advance!
| $$\left( \frac { -1+i\sqrt { 3 } }{ 1+i } \right) ^{ 2016 }=\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2\pi }{ 3 } \right) +i\sin { \left( \frac { 2\pi }{ 3 } \right) } } } \right) }^{ 1013 } }{ { \left( { \left( 1+i \right) }^{ 2 } \right) ^{ 1013 } } } =\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2016\pi }{ 3 } \right) +i\sin { \left( \frac { 2016\pi }{ 3 } \right) } } } \right) } }{ { 2 }^{ 1008 }{ i }^{ 1008 } } =\\ ={ 2 }^{ 1008 }{ e }^{ 2016i\pi /3 }={ 2 }^{ 1008 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Arithmetic Mean/Geometrix Mean Inequality of Degree 3 For $a,b$ and $c\geq0$;
$$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Is there a simple way to deduce this result from the degree 2 verson?
That is:
$$\frac{a+b}{2}\geq\sqrt{ab}$$
Some sort of substitution involving $a,b,c \ $ for one of the variables was what I had in mind, but I can't seem to figure it out!
| This is a nice method:
We know that $\frac{a+b}{2}\geq\sqrt{ab}$
and similarly $\frac{c+d}{2}\geq\sqrt{cd}$
so $\frac{a+b+c+d}{2}\geq\sqrt{ab}+\sqrt{cd}\geq2\sqrt[4]{abcd}$
$$\Rightarrow\frac{a+b+c+d}{4}\geq\sqrt[4]{abcd}$$
and by making the substitution $d=\frac{a+b+c}{3} \rightarrow$
$$\frac{a+b+c+\frac{a+b+c}{3}}{4}\geq \sqrt[4]{ \frac{a+b+c}{3} abc }$$
$$\frac{a+b+c}{3}\geq\sqrt[4]{ \frac{a+b+c}{3} abc }$$
Raising both sides the the power of $4$
$$(\frac{a+b+c}{3})^4\geq(\frac{a+b+c}{3})abc$$
which gives us:
$$\frac{a+b+c}{3}\geq\sqrt[3]{abc}$$
as required.
I would love to see some more methods if you have any!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate a given limit Calculate the following limit:
$$\lim_{n \rightarrow \infty} \frac{3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} - n}{\ln(n^2 + 1)}$$
I tried to calculate the sum $3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n}$ but got nothing. I currently have no idea what I should do with this limit.
Thank you!
| Observe
\begin{align} \frac{3+3^{1/2}+3^{1/3}+\ldots + 3^{1/n} -n}{\ln(n^2+1)}
= \frac{(e^{\ln 3}-1)+(e^{\frac{1}{2}\ln 3}-1)+\ldots + (e^{\frac{1}{n}\log 3}-1)}{\ln(n^2+1)}. \end{align}
Using the fact that
\begin{align} x\leq e^x-1 \leq x-2x^2 \end{align}
when $0 \leq x\leq 2$, you can show that
\begin{align}
\alpha \ln 3\leq e^{\alpha \ln 3} -1 \leq \alpha \ln 3-2(\alpha \ln 3)^2
\end{align}
for all $0\leq \alpha\leq 1$. Hence it follows
\begin{align} \frac{\ln 3\cdot (1+\frac{1}{2}+\ldots+\frac{1}{n})}{\ln
(n^2+1)}\leq&\ \frac{(e^{\ln 3}-1)+(e^{\frac{1}{2}\ln 3}-1)+\ldots +
(e^{\frac{1}{n}\log 3}-1)}{\ln(n^2+1)}\\ \leq&\ \frac{\ln 3\cdot
(1+\frac{1}{2}+\ldots +\frac{1}{n})-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots + \frac{1}{n^2})}{\ln(n^2+1)}. \end{align}
Using the fact that
\begin{align}
\lim_{n\rightarrow \infty}\left(1+\frac{1}{2}+\ldots +\frac{1}{n} - \ln n\right) = \gamma
\end{align}
where $\gamma$ is the Euler-Mascheroni constant i.e. the limit exists, then it follows
\begin{align}
\lim_{n\rightarrow \infty}\frac{\ln 3\cdot (1+\frac{1}{2}+\ldots+\frac{1}{n})}{\ln (n^2+1)}= \lim_{n\rightarrow \infty}\frac{\ln 3\cdot [(1+\frac{1}{2}+\ldots+\frac{1}{n})-\ln n]+\ln 3\cdot \ln n }{\ln (n^2+1)} = \frac{\ln 3}{2}
\end{align}
and
\begin{align}
&\lim_{n\rightarrow \infty}\frac{\ln 3\cdot (1+\frac{1}{2}+\ldots +\frac{1}{n})-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots + \frac{1}{n^2})}{\ln(n^2+1)}\\
&= \lim_{n\rightarrow \infty} \frac{\ln 3\cdot [(1+\frac{1}{2}+\ldots +\frac{1}{n})-\ln n]+\ln 3\cdot \ln n-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots + \frac{1}{n^2})}{\ln(n^2+1)} = \frac{\ln 3}{2}.
\end{align}
Hence by the squeeze theorem, we have that
\begin{align} \lim_{n\rightarrow \infty}
\frac{3+3^{1/2}+3^{1/3}+\ldots + 3^{1/n} -n}{\ln(n^2+1)}= \frac{\ln
3}{2}. \end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solve for $\alpha$ in $1+\sqrt3\tan\alpha-\sec\alpha=0$
Solve for $\alpha$, where $0^\circ\le\alpha\le360^\circ$, in
$$1+\sqrt3\tan\alpha-\sec\alpha=0$$
I solved this way. But the answer in my book is only $0^\circ,120^\circ,360^\circ$.
Please help me. Which answer is suitable?
| Squaring both sides is false way because in this case you can get false roots $$1+\sqrt { 3 } \tan { \alpha -\sec { \alpha } =0 } \\ 1+\sqrt { 3 } \frac { \sin { \alpha } }{ \cos { \alpha } } -\frac { 1 }{ \cos { \alpha } } =0\\ \cos { \alpha +\sqrt { 3 } \sin { \alpha } =1 } \\ \frac { 1 }{ 2 } \cos { \alpha +\frac { \sqrt { 3 } }{ 2 } \sin { \alpha } =\frac { 1 }{ 2 } } \\ \sin { { 30 }^{ \circ }\cos { \alpha +\cos { { 30 }^{ \circ }\sin { \alpha } =\frac { 1 }{ 2 } } } } \\ \sin { \left( { 30 }^{ \circ }+\alpha \right) =\frac { 1 }{ 2 } } \\ { 30 }^{ \circ }+\alpha ={ 30 }^{ \circ }+{ 360 }^{ \circ }n,\quad \Rightarrow \quad { \alpha }_{ 1 }={ 360 }^{ \circ }n,\quad \quad n=0,\pm 1,\pm 2,...\\ { 30 }^{ \circ }+\alpha ={ 180 }^{ \circ }-{ 30 }^{ \circ }+{ 360 }^{ \circ }n\quad \Rightarrow \quad { \alpha }_{ 2 }={ 120 }^{ \circ }+{ 360 }^{ \circ }n,\quad n=0,\pm 1,\pm 2,.\\ \\ $$
| {
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"timestamp": "2023-03-29T00:00:00",
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When does $\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$ converge? I am required to find the values of $C$ for which the integral
$$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.
I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.
I get stuck when I need to evaluate $$\lim_{x\rightarrow\infty}(\frac{7}{2}\ln(x^2+1)-\frac{7C}{3}ln(3x+1))$$
Any help would be appreciated.
Taking the hint from DonAntonio, I wrote the following solution. Let me know if there is anything wrong with the way I expanded DonAntonio's solution.
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
When $C=3$
$$7\cdot\,\int_0^{\infty}\frac{(3-3)x^2+x-3}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{x-3}{(x^2+1)(3x+1)}dx<\int_0^{\infty}\frac{x}{x^3}dx = \int_0^{\infty}\frac{1}{x^2} dx$$
Which is convergent.
When $C\neq3$
$$7\cdot\,\int_0^{\infty}\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}+7\cdot\,\int_0^{\infty}\frac{x-C}{(x^2+1)(3x+1)}dx$$
$$=|7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}dx+K| > \int_0^{\infty}\frac{x^2}{x^4} dx = \int_0^{\infty}\frac{1}{x^2} $$
For some constant $K$
Which is divergent.
| You know an antiderivative, namely
$$
\frac{7}{6}\bigl(3\ln(x^2+1)-2C\ln(3x+1)\bigr)=
\frac{7}{6}\ln\frac{(x^2+1)^3}{(3x+1)^{2C}}=
\frac{7}{6}\ln
\frac{\bigl(1+\frac{1}{x^2\mathstrut}\bigr)^3x^6}
{(3+\frac{1}{x})^{2C}x^{2C}}
$$
It's quite easy now to see when the limit is finite.
On the other hand, doing an asymptotic comparison with a power of $1/x$ is easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $ Let $x,y,z > 0$ and $xyz=8.$ Prove that
$$
\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12
$$
I have tried with AM–GM inequality but no result.
| Using the Cauchy-Scwarz inequality
$$ (a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}) \geq (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^{2} $$ with $a_{1}=\frac{x+y-1}{\sqrt{z}} $, $ a_{2}=\frac{x+z-1}{\sqrt{y}} $ $ a_{3}=\frac{z+y-1}{\sqrt{x}} $ and $ b_{1}=\sqrt{z} $, $ b_{2}=\sqrt{y} $ and $ b_{3}=\sqrt{x} $, we obtain that $$\Big (\frac{(x+y-1)^{2}}{z}+\frac{(x+z-1)^{2}}{y}+\frac{(z+y-1)^{2}}{x} \Big) \cdot S \geq (2S-3)^{2} $$
where $ S=x+y+z $.
Therefore it suffices to show that $\frac{(2S-3)^{2}}{S} \geq 12 $ which is equivalent to showing that $ S \in (-\infty, \frac{6-3\sqrt{3}}{2}]\cup [ \frac{6+3\sqrt{3}}{2},\infty ) $.
But we have that $ S=x+y+\frac{8}{xy} $ and using the AM-GM, this is greater than or equal to $ 3(xy\cdot 8\frac{1}{xy})^{\frac{1}{3}}=6 $.
since clearly $ 6> \frac{6+3\sqrt{3}}{2} $, the conclusion follows.
Moreover, $ \frac{27}{2} $ is the sharpest lower bound and in the same manner we can prove that the LHS is $ \geq \frac{27}{2} $ since this is equivalent to $ S \in (-\infty, \frac{3}{8}] \cup [6, \infty) $ after computations and this is true since we proved earlier that $ S \geq 6 $ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ If $\frac{a}{b}$ < $\frac{c}{d}$ then
$$\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$$
I have been searching and trying and couldnt find a reliable proof.
One might do this like here which I think it is very wrong:
$\frac{a}{b}$ < $\frac{c}{d}$ then $a.d$ < $c.b$
Because $\frac{2}{-3}$ < $\frac{4}{5}$ doesn't mean $10 < -12$
| Here is a simple intuitive proof for the case when a, b, c and d are positive. Imagine a team that scores a points in its first b games and c points in the remaining d games. Its seasonal points per game is $\frac{a+c}{b+d}$, which can also be expressed as the weighted average of its performance in each part of the season.
$\frac{a+c}{b+d}$ = ($\frac{a}{b}$)($\frac{b}{b+d}$) + ($\frac{c}{d}$)($\frac{d}{b+d}$). This is a weighted average of $\frac{a}{b}$ and $\frac{c}{d}$, and so must between them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Manipulating functions
Let $g(n)$ be a function, defined for all integers $n\ge 0$, as follows:
$$g(n) = \begin{cases} 0, & \text{if } n=0 \\ 1+g(n/2), & \text{if $n\gt 0$ and $n$ is even} \\ 1+g(n-1), & \text{if $n\gt 0$ and $n$ is odd}\end{cases}$$
What is $g(2^l+2^k)$ where $l\gt k\ge 0$ are integers?
I have have solutions which state:
For natural numbers $l\gt k\ge 0$ $$g(2^l + 2^k) = k+g(2^{l-k}+1) = k+1+g(2^{l-k}) = k+1+l-k+1=l+2$$
But can't understand what it is saying. I see that obviously $\dfrac{2^l + 2^k}{2^k} = 2^{l-k} + 1$. But other than that, I am having a hard time trying to grasp the other operations in the solution.
| Using each of the rules for $g$:
$$\begin{align}
\color{red}{g(2^l + 2^k)} &= g\big(2^k(2^{l-k} + 1)\big) \\ &= 1 + g\big(2^{k-1}(2^{l-k} + 1)\big) \\ & \qquad\vdots \\ &= \color{red}{k+g(2^{l-k}+1)} \\ &= k+1+g\big((2^{l-k}+1)-1\big) \\ &= \color{red}{k+1+g(2^{l-k})} \\ &= k+1 + \big(1+ g(2^{l-k-1})\big)\\ &\qquad \vdots \\ &= k+1 + \big(l-k+g(1)\big)\\ &= \color{red}{k+1+(l-k+1)} \\ &=\color{red}{l+2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking $$x=\sqrt{2}y$$ then using integration by parts, we have
$$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help
| We want to calculate
$$
I=\int_0^1 dx\frac{\arcsin(x)\arcsin\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2-x^2}}
$$
Let us start by a substitution $x=\sqrt{2}\sin(y)$ and integrate by parts so we can write $I$ as
$$
\int_0^{\pi/4}dy y \arcsin(\sqrt{2}\sin(y))\underbrace{=
}_{i.b.p}\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^{\pi/4}dy\frac{y^2\cos(y)}{ \sqrt{1-2\sin(y)^2}}}_{J}$$
but since $1-2\sin(y)^2=\cos(2y)$ and by another substitution $2y =q$
$$
J=\frac{1}{8}\int_0^{\pi/2}dq\frac{ q^2 \cos(q/2) }{\sqrt{\cos(q)}}
$$
Now we perform small trick (credit to Mr. Feynman) by rewriting
$$
J=-\frac{1}{8}\left[\frac{d^2}{da^2}\int_0^{\pi/2} dq\cos(q)^{-1/2}\cos(aq)\right]_{a=1/2}
$$
which equals according to Tunk-Fey's answer here
$$
J=-\frac{1}{8}\frac{d^2}{da^2}\left[\frac{2\pi}{\sqrt{2}}\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2}
$$
Or
$$
I=\frac{\pi^3}{64}+\frac{\pi}{8\sqrt{2}}\frac{d^2}{da^2}\left[\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2}
$$
Since i have no CAS at hand today and are way to lazy to do the above derivative by hand that's it for the moment. But everything from here on should be straightforward so i leave that to you...
| {
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"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| Hint: If $x_0=-1$ is a zero of $x^{2016}-1$, then $(x-(-1))$ is a factor of $(x^{2016}-1)$.
In order to find $G(x)=(x^{2016}-1)/(x+1)$ you need to apply polynomial long division (you should see the pattern).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the average rate of change of $S(x) = -2x^2 + 14x - 12$ I have to show that the average rate of change of $S(x) = -2x^2 + 14x - 12$ in the interval $[x,x+h]$ is $-4x - 2h + 14$ and so far I did:
$$A(x) = \frac{S(x+h)-S(x)}{(x+h)-(x)} = \frac{(-2(x+h)^2+14(x+h)-12)-(-2x^2+14x-12)}{(x+h)-x} = \frac{-2(x+h)^2+14(x+h)-12+2x^2-14x+12}{-x+x+h} = \frac{-2(x+h)^2+14(x+h)+2x^2-14x}{h} = \frac{-2h^2+14h}{h} = \frac{-2h^2+14h}{h}\\ = -2^2 +14 = 4+ 14 = 18$$
I know this is totally wrong. What did I do wrong? How do I solve this?
| $$
\frac{-2(x+h)^2+14(x+h)+2x^2-14x}{h} = \frac{-2h^2+14h-4xh}{h} = -2h-4x+14.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$
$\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$
$$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$
$$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$
Now how can i solve it after that , Help required, Thanks
| $\sin(a) + \sin(b) > \sin(a+b)$ if $(a,b)\in (0, \pi)\implies $
$\sin(a + b +c ) <= \sin(a) + \sin(b + c) < \sin(a) + \sin(b) + \sin(c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots \cdots \cdots +\frac{1}{2n}>\frac{13}{24}$
Proving $$\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots \cdots \cdots +\frac{1}{2n}>\frac{13}{24}\;, n>1\;,n\in \mathbb{N}$$
$\bf{My\; Try::}$ Using Limit as a sum $$\frac{1}{n}\left[\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots \cdots \cdots +\frac{1}{1+\frac{n}{n}}\right] = \int^{1}_{0}\frac{1}{1+x}dx = \ln 2$$
But i did not understand how can i prove That $\displaystyle \bf{L.H.S}>\frac{13}{24}$
Help required, Thanks
| Let $$f(n)=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n}=\sum_{r=1}^{2n}\frac1r-\sum_{s=1}^n\frac1s$$
So, $$f(n+1)-f(n)=\frac1{(2n+1)(2n+2)}>0$$ for integer $n\ge0$
$\displaystyle \implies f(n)$ is an increasing function.
Now, $f(2)=\frac13+\frac14=\frac7{12}>\frac{13}{24}$ as $7\cdot24>12\cdot13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2-x^2)}$ How can we solve the equation:
$$\frac{dy}{dx} = \frac{y^3}{2(xy^2-x^2)}$$
I get the idea of dividing by $y^2$, But it doesn't become any more solvable (not homogenous).
$$\frac{dy}{dx} = \frac{y}{2(x-\frac{x^2}{y^2})}$$
Substituting $\frac{x}{y} = t$ causes even more complications.
I get an idea of the question to convert into homogenous, but cant form the equation. Please give me a hint!
| Nice question!
Given:
$$\frac{dx}{dy}=\frac{2x}{y}-\frac{2x^2}{y^3}$$
$$=>\frac{dx}{dy}-\frac{2x}{y}=-\frac{2x^2}{y^3}$$
On dividing by $-x^2$,
$$=>\frac{dx}{dy}\frac{(-1)}{x^2}+\frac{2}{xy}=\frac{2}{y^3}$$
Substituting $v=\frac{1}{x}$; $\frac{dv}{dy}=\frac{dx}{dy}\frac{(-1)}{x^2}$gives:
$$\frac{dv}{dy}+\frac{2v}{y}=\frac{2}{y^3}$$
Now simply use the method for solving linear differential equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$.
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root,
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$.
Since $w$ is not equal to $1$,
$w^3=1$
Which can be rewritten as $(w-1) (1 + w + w^2) = 0$
And since $w \neq 1,$ $(1 + w + w^2) = 0$
On simplifying the given equation,
I get
$w^3 + w + 1
= 2+w$ (Is this right approach to go about it?)
|
since w is a root, I first substituted w and got w^4=1 Which is w * w^3 = 1. Since w is not equal to 1, w^3=1
This step is wrong. Compare to the obviously false: $(-1)(-1)=1$ and since $(-1) \ne 1$ then it must be that $(-1) =1$.
What you can derive from $w^4=1$ however is that $\frac{1}{w} = w^3$ and $\frac{1}{w^2} = w^2$. So the sum reduces to $w+w^2+w^3=-1$ (which holds because $1+w+w^2+w^3=\frac{1-w^4}{1-w}=0$ given $w^4=1, w \ne 1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$ But now, I'm stuck. Solutions are greatly appreciated.
| Hint:$$\left( \sin a+\cos a \right) \left( \sin ^{ 2 } a-\sin a\cos a+\cos ^{ 2 } a \right) =\\ =\left( \sin a+\cos a \right) \left( { \left( \sin a+\cos a \right) }^{ 2 }-\frac { 3 }{ 2 } \left( { \left( \sin { a } +\cos { a } \right) }^{ 2 }-1 \right) \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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A problem related to arithmetico-geometric sequence Question:
Find the sum of the series: $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ... = 1+\sum_{n=1}^\infty \frac{(4n-2)}{3^n}$
My doubt:
I have taken $\frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + ...$ to be in A.G.P. and calculated the sum using the formula $$S_\infty=\frac{a}{1-r}+\frac{d \cdot r}{(1-r)^2},$$ where $a=\frac23, r=\frac13$ and $d=4$.
I get $S_\infty=4$ and then I add $1$ to it to get the answer $5$.
But my book shows a different method with a different answer ($3$).
Please tell what is wrong with this method.
| Let us make the problem more general, replacing $\frac 13$ by
$x$. So, we have $$S=1+2x+6x^2+10x^3+14x^4+\cdots=1+\sum_{n=1}^\infty(4n-2)x^n$$ that is to say $$S=1+4\sum_{n=1}^\infty nx^n-2\sum_{n=1}^\infty x^n=1+4\color{red}x\sum_{n=1}^\infty nx^{n-1}-2\sum_{n=1}^\infty x^n$$ $$S=1+4x\left(\sum_{n=1}^\infty x^n\right)'-2\left(\sum_{n=1}^\infty x^n\right)$$ $$\sum_{n=1}^\infty x^n=\frac{x}{1-x}\implies \left(\sum_{n=1}^\infty x^n\right)'=\frac{1}{(1-x)^2}$$ All of that makes $$S=1+\frac{4x}{(1-x)^2}-\frac{2x}{1-x}\implies S=\frac{3 x^2+1}{(1-x)^2}$$ Now, replace $x$ by $\frac 13$ to get the answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Deriving the equation of a circle I have two points, $A=(-1,3)$ and $B=(2,7)$. There is a third point $P=(x,y)$.
I have found $|AP|$ and $|BP|$ in terms of $x$ and $y$. These are:
$|AP| = \sqrt{(-1-x)^2 +(3-y)^2}$
$|BP| = \sqrt{(2-x)^2 +(7-y)^2}$
I have shown that the set $S = \{P:|AP| = |BP|\}$ is a straight line and determined the slope and y-intercept by expanding out $|AP| = |BP|$ and simplifying to get:
$y= \frac{-3}{4}x+\frac{43}{8}$.
What I can't derive is an equation for the circle $C= \{P: |AP| = \alpha|BP|\}$. I know it should be of the form $(x-a)^2 + (y-b)^2 = \alpha^2$ but when I try to tackle this algebraically I get stuck.
| Change the coordinate system to map $A$ to the origin in order to facilitate algebraic manipulations.
$$\begin{align}
(\;\ x,\;y)&\rightarrow \;\;(x+1,y-3)\\
A(-1,3)&\rightarrow A'(\;\;\;\;\;0,\;\;0)\\
B(\;\;2,7)&\rightarrow B'(\;\;\;\;\;3,\;\;4)\\
P(\;\;x,y)&\rightarrow P'(x+1, y-3)=P(x', y')
\\\\
|A'P'|&=\alpha |B'P'|\\
A'P'^2&=\alpha^2 B'P'^2\\
x'\ ^2+y'\ ^2&=\alpha^2\left[(x'-3)^2+(y'-4)^2\right]\\
(1-\alpha^2)(x'\ ^2+y'\ ^2)+6\alpha^2x'+8\alpha^2y'-25\alpha^2&=0\\
x'\ ^2+y'\ ^2+\frac{6\alpha^2x'+8\alpha^2y'-25\alpha^2}{1-\alpha^2}&=0\\
\left(x'+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y'+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\frac{25\alpha^2}{1-\alpha^2}+\frac{25\alpha^4}{(1-\alpha^2)^2}\\
\left(x'+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y'+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\left(\frac{5\alpha}{1-\alpha^2}\right)^2\\
\text{Converting back to original coordinates,}\qquad\qquad\qquad\qquad&&\\
\left(x+1+\frac {3\alpha^2}{1-\alpha^2}\right)^2+\left(y-3+\frac{4\alpha^2}{1-\alpha^2}\right)^2
&=\left(\frac{5\alpha}{1-\alpha^2}\right)^2\\
\color{red}{\left(x-\left(-\frac {1+2\alpha^2}{1-\alpha^2}\right)\right)^2+\left(y-\frac{3-7\alpha^2}{1-\alpha^2}\right)^2}
&\color{red}{=\left(\frac{5\alpha}{1-\alpha^2}\right)^2}\\
\end{align}$$
which is the equation of the circle $C$, the locus of $P$.
Circle $C$ has its centre at
$$\left(-\frac {1+2\alpha^2}{1-\alpha^2},\frac{3-7\alpha^2}{1-\alpha^2}\right)$$
and radius of
$$\frac{5\alpha}{1-\alpha^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x$ is a real and $0 < x < 4$, then $\frac{4}{x(4-x)} \geq 1$ I got this exercise from the textbook Book of Proof, CH4 E12. I've tackle this problem in the following manner:
Suppose x is a real and $0 < x < 4$, it follows that,
\begin{align*}
&\Rightarrow 0 - 2 < x - 2 < 4 - 2 \\
&\Rightarrow 4 < (x - 2)^2 < 4\\
&\Rightarrow 0 \leq (x - 2)^2 < 4
\end{align*}
Since, $x(4 - x) = 4x - x^2 = 4 - (x - 2)^2$, then
$$\dfrac{4}{x(4 - x)} = \dfrac{4}{4 - (x - 2)^2}.$$
This expression is greater or equal to $1$ for
$0 \leq (x - 2)^2 < 4$. Thus,
$$\dfrac{4}{x(4 - x)} \geq 1.$$
I'm quite new to proof technique and I'm using this book to self-learn logic and proofing writing. My question is: is the solution stated above logically sound? Would my arguments be considered sufficient to prove that $P \Rightarrow Q$?
| We can do this proof by contradiction: We want to prove $P\rightarrow Q$ and we prove it by assuming $P$ and $\neg Q$, and getting a contradiction.
Suppose $0<x<4$ with $x\in\mathbb{R}$, and the inequality $\dfrac{4}{x(x-4)}\geq 1$ does not hold.
Then we have $0<x,0<4-x$ (both follow from $0<x<4$) and $\dfrac{4}{x(x-4)}<1$, and we have:
\begin{align*}
\dfrac{4}{x(4-x)}&<1 & &\\
4&<x(4-x) & &\text{[multiplying both sides by $x(x-4)>0$]}\\
0&<x(4-x)-4\\
0&<-x^2+4x-4\\
x^2-4x+4&<0\\
(x-2)^2&<0
\end{align*}
which is absurd because every squared real number is positive.
This contradiction allows us to conclude that $\dfrac{4}{(x-4)}\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Limit of product with cubes $\lim\limits_{n \to \infty}\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}\dots\frac{n^3-1}{n^3+1}$ I am trying to evaluate
$$\lim\limits_{n \to \infty}{2^3-1 \over 2^3+1}{3^3-1 \over 3^3+1}\dots{n^3-1 \over n^3+1}$$
It seems to be the perfect candidate for a factoring formula, however, I get stuck rewriting the fractions. Do you have any suggestions?
| One may use a telescoping product, as $n \to \infty$,
$$
\prod_{k=2}^n\frac{k^3-1}{k^3+1}=\prod_{k=2}^n\frac{k-1}{k+1} \cdot \prod_{k=2}^n\frac{k(k+1)+1}{k(k-1)+1}=\frac23 \cdot \frac{n(n+1)+1}{n(n+1)}\to \frac23.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \frac{1}{4}a+\frac{3}{4}b\\
(\frac{3}{4}a+\frac{1}{4}b)^2 & (\frac{1}{2}a+\frac{1}{2}b)^2 &
(\frac{1}{4}a+\frac{3}{4}b)^2\\ \end{array} \right] \cdot
\left[ \begin{array}{c} w_1\\ w_2\\ w_3 \end{array} \right] =
\left[ \begin{array}{c} b-a\\ \frac{b^2-a^2}{2}\\
\frac{b^3-a^3}{3} \end{array} \right] $$ where $a, b$ are
constants? (I'm trying to derive the formula for a 3-point open
Newton-Cotes quadrature rule.) Thanks!
Unfortunately the question was deleted (link), but I think it might be interesting in general.
| matrix(
[1,1,1],
[(3/4)*a+(1/4)*b,(1/2)*a+(1/2)*b,(1/4)*a+(3/4)*b],
[((3/4)*a+(1/4)*b)^2,((1/2)*a+(1/2)*b)^2,((1/4)*a+(3/4)*b)^2]
);
$$
\begin{pmatrix}1 & 1 & 1\\
\frac{b}{4}+\frac{3a}{4} & \frac{b}{2}+\frac{a}{2} & \frac{3b}{4}+\frac{a}{4}\\
{{\left( \frac{b}{4}+\frac{3a}{4}\right) }^{2}} & {{\left( \frac{b}{2}+\frac{a}{2}\right) }^{2}} & {{\left( \frac{3b}{4}+\frac{a}{4}\right) }^{2}}\end{pmatrix}
$$
and the result vector like that:
bcol: matrix(
[b-a],
[(b^2-a^2)/2],
[(b^3-a^3)/3]
);
$$
\begin{pmatrix}b-a\\
\frac{{{b}^{2}}-{{a}^{2}}}{2}\\
\frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix}
$$
This command will solve your linear system by $LU$ decomposition:
ls : linsolve_by_lu(A,bcol);
You extract and simplify the solution vector like this:
xycol: ratsimp(first(ls));
$$
\begin{pmatrix}\frac{2b-2a}{3}\\
-\frac{b-a}{3}\\
\frac{2b-2a}{3}\end{pmatrix}
$$
where first picks the first element of the result and ratsimp is one of Maxima's simplification functions.
With this you can verify the result:
ratsimp(A . xycol);
$$
\begin{pmatrix}b-a\\
\frac{{{b}^{2}}-{{a}^{2}}}{2}\\
\frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix}
$$
Note that Maxima uses a dot for the matrix multiplication operator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim \limits_{(x, y)\rightarrow (1, -1)}\frac{(x-y)^2-4}{x^2+y^2-2}$ I want to calculate $\lim\limits_{(x, y)\rightarrow (1, -1)}\dfrac{(x-y)^2-4}{x^2+y^2-2}$.
I already know that $\lim \limits_{x\rightarrow 1}\lim\limits_{y\rightarrow -1}\dfrac{(x-y)^2-4}{x^2+y^2-2} = \lim\limits_{y\rightarrow -1}\lim\limits_{x\rightarrow 1}\dfrac{(x-y)^2-4}{x^2+y^2-2} = 2$, so if the limit exists, it equals $2$. But how do I prove that?
I can't use polar coordinates since $(x, y) \not\to (0,0)$, I've also failed to find a counterexample of a sequence with different limit than $2$.
| Note that for all $x,y$ such that $x^2+y^2\neq 2$ the following holds:$$\frac{(x-y)^2-4}{x^2+y^2-2}=\dfrac{x^2+y^2-2-2xy-2}{x^2+y^2-2}=1-\dfrac{2xy+2}{x^2+y^2-2}.$$
Conjecture: $\lim \limits_{(x,y)\to (1,-1)}\left(\dfrac{2xy+2}{x^2+y^2-2}\right)=-1$.
Note that $$\dfrac{2xy+2}{x^2+y^2-2}+1=\dfrac{2xy+2}{x^2+y^2-2}+\dfrac{x^2+y^2-2}{x^2+y^2-2}=\dfrac{(x+y)^2}{x^2+y^2-2},$$
therefore $$\left|\dfrac{2xy+2}{x^2+y^2-2}+1\right|=\left|\dfrac{(x+y)^2}{x^2+y^2-2}\right|\leq \left|\dfrac{(x+y)^2}{x^2+y^2}\right|_.$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x}$ Here $f(x) = x + \cos x$ and $g(x)= x- \cos x$
We can't apply L'Hopital's Rule since $\forall c > 0\quad \exists x_c \in (c, + \infty): g'(x) = 1 + \sin x = 0$.
But I don't know how to proceed.
My attempt:
Because $\cos x$ is bounded I was taking an arbitrary number $A >0$ that bounds $\cos x$ from above.
From there I have
$$\lim_{x\to \infty}\frac{x+\cos x}{x - \cos x} = \lim_{x\to \infty}\frac{x+A}{x - A} =1$$
But the answer is $0$. What am I doing wrong?
| For $x \gt 1$, you have $\dfrac{x+\cos x}{x - \cos x}=1 + \dfrac{2\cos x}{x - \cos x}$
Since $0 \le | \cos x | \le 1$ you have $-2 \le 2 \cos x \le 2$ and $x-\cos x \ge x-1$
meaning $1 - \dfrac{2}{x - 1} \le \dfrac{x+\cos x}{x - \cos x} \le 1 + \dfrac{2}{x - 1}$ and thus a limit of $1$ as $x$ increases without limit
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation of tan What is $$\sum _1 ^7 \tan^2 (\frac {n\pi}{16}) -1$$ so I used $\tan(x)=\cot(\frac {\pi}{2}-x)$ for last three angles ie $5\pi/16,6\pi/16,7\pi/16$ . Thus it gets converted to some symmetry of tan,cot but after that I am not sure how to proceed or which identity to use.Use of AM-GM seems to give a very rough estimate . Answer given is $34$
| $$S=\sum^{7}_{n=1}\tan^2 \left(\frac{n\pi}{16}\right) = \tan^2 \left(\frac{\pi}{16}\right)+\tan^2 \left(\frac{\pi}{8}\right)+\tan^2 \left(\frac{3\pi}{16}\right)+1+\tan^2 \left(\frac{5\pi}{16}\right)+\tan^2 \left(\frac{3\pi}{8}\right)+\tan^2 \left(\frac{7\pi}{16}\right)$$
And using $\displaystyle \tan^2 \frac{7\pi}{16} = \cot^2 \frac{\pi}{16}$ and $\displaystyle \tan^2 \frac{5\pi}{16} = \cot^2 \frac{3\pi}{16}$
Using $\displaystyle \tan^2 x+\cot^2 x = \frac{\sin^4 x+\cos^4 x}{\sin^2 x \cos ^2 x} = \frac{4-2(\sin 2x)^2}{\sin ^2 (2x)} = \frac{8-2(1-\cos 4x)}{1-\cos 4x}=\frac{6+2\cos 4x}{1-\cos 4x}$ And Using $\displaystyle \tan^2 \alpha = \frac{2}{\cos 2 \alpha + 1}-1$
So we get $$S = 14+8\sqrt{2}+14-8\sqrt{2}+3-\sqrt{2}+3+2\sqrt{2}+1 = 35$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$ Well, I was trying to find $\tan40$ in terms of $\tan25$=x.
So, I expanded 40 as 25+15 and got the value of $\tan40$ in terms of $x$ as $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$.
Now, I solved $\tan40$ alternatively as
$$\frac{\tan155-\tan115}{1+\tan155\tan115}= \frac{\tan(180-25)-tan(90+25)}{1+\tan(180-25)tan(90+25)}= \frac{-\tan25+\cot25}{1+\tan25{\cot25}}=\frac{1}{2}(-x+\frac{1}{x})=\frac{1-{x^2}}{2x}$$
But, we should have same solution from both the approaches.
Therefore, $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$ must be equals to $\frac{1-{x^2}}{2x}$. But I'm unable to prove it by algebra. So, can you kindly help me?
| On rearrange we have
$$2+\sqrt3=\dfrac{3x-x^3}{1-3x^2}=\cdots=\tan(3\cdot25^\circ)$$
Now $$\tan(75^\circ)=\tan(45^\circ+30^\circ)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the reduction formula for the integral $I_n = \int e^{ax}\sin^nxdx$ My attempt:
Let $$u=\sin^nx \rightarrow \text{d}u=n\sin^{n-1}x\cos x\text{d}x $$ and
$$ v = \int{e^{ax}}\text{d}x=\frac{e^{ax}}{a}.$$
Now we have $u\cdot v- \int{v\cdot \text{d}u} $, so
$$I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}$$
Now solving: $\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}$,
Let
$$ u = e^{ax} \rightarrow \text{d}u=ae^{ax}$$
and
$$ v = \int{\sin^{n-1}x \cos x \text{d}x} = \frac{\sin^nx}{n}. $$
So,
$$\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} = \frac{\sin^nx}{n}e^{ax}-\frac{a}{n}I_n.$$
We win
$$I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\left(\frac{\sin^nx}{n}e^{ax}-\frac{a}{n}I_n\right)$$
$$ 0 = 0$$
| Hint. We assume $a \neq0, \pm i$. From the identity you have already obtained by parts,
$$
I_n = \frac{e^{ax}}{a}\sin^nx-\frac{n}{a}\int{e^{ax}\sin^{n-1}x\cos x\text{d}x} \tag1
$$ you may rather perform another integration by parts as follows
$$
\begin{align}
&\int{e^{ax}\sin^{n-1}x\cos x\text{d}x}
\\&= \frac{e^{ax}}{a}\sin^{n-1}x\cos x-\frac{1}{a}\int{e^{ax}\left[(n-1)\sin^{n-2}x\cos^2 x-\sin^{n}x\right]\text{d}x} \tag2
\end{align}
$$ replacing $\cos^2 x$ with $1-\sin^2 x$ then inserting $(2)$ in $(1)$, one gets
$$
I_n=\frac{e^{ax}}{a}\sin^{n}x-\frac{ne^{ax}}{a^2}\sin^{n-1}x\cos x+\frac{n(n-1)}{a^2}I_{n-2}-\frac{n^2}{a^2}I_{n}
$$ or
$$
(n^2+a^2)I_n=ae^{ax}\sin^{n}x-ne^{ax}\sin^{n-1}x\cos x+n(n-1)I_{n-2} \tag3
$$
with $I_0=\dfrac{e^{ax}}a$ and $I_1=\dfrac{-\cos x+a \sin x}{1+a^2}e^{a x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
factrorize $-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3$ factrorize $-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3$
or
$$ -(a^3 + b^3 + c^3) + (a^2 b + a b^2 + b^2 c + b c^2 + c^2 a + c a^2) - 2(abc) $$
It is a so long polynomial I don't have any ideas for starting factoring what should I do?
| $$ ( a +b - c) ( b + c - a ) ( c + a - b) $$
This is one factor short of Heron's formula,
$$ (a + b + c) ( a +b - c) ( b + c - a ) ( c + a - b) = 2 (b^2 c^2 + c^2 a^2 + a^2 b^2) - (a^4 + b^4 + c^4) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$
If the Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$ in $(0,1)\;,$ Then $k\in $
$\bf{My\; Try::}$ Let $3^x=y>0\;,$ Then inequality is
$$y^2-ky-k+3\leq 0\Rightarrow y^2-ky-k+3\leq 0\forall y\in (1,3)$$
So $$y^2-ky+\frac{k^2}{4}-\frac{k^2}{4}-k+3\leq 0$$
So $$\bigg(y-\frac{k}{2}\bigg)^2-\bigg(\frac{k^2+4k-12}{4}\bigg)\leq 0$$
Now above $\displaystyle 1<y<3\Rightarrow 1-\frac{k}{2}<y-\frac{k}{2}<3-\frac{k}{2}\Rightarrow \left(3-\frac{k}{2}\right)^2<\left(y-\frac{k}{2}\right)^2<\left(3-\frac{k}{2}\right)^2$
Now how can i solve it,Help required, Thanks
| Set $t=\mathrm 3^x$. The given equation has at least one root in $(0,1)$ if and only if the equation $\;p(t)=t^2-kt+3-t=0\;$ has at least one root in $(1,3)$.
The first condition is this equation has real roots , i.e.
\begin{align}&\Delta=k^2-4(3-t)=k^2+4t-12=(k+2)^2-16\ge 0\\
&\iff k\ge 2\enspace\text{or}\enspace k\le-6.
\end{align}
Now observe that if $k\le -6$, the equation has real roots with the same sign (their product $3-k$ is positive) and this sign is negative (their sum $k$ is negative. Hence, if there is one root in $(1,3)$, we have $k\ge 2$.
Furthermore, in this case, $1$ separates the roots, since $p(1)=4-2k\le 0$. On the other hand, $p(3)=9-3k$, so:
*
*if $k>3$, $3$ also separates the roots, hence one root is $<1$, the other is $>3$, and there's no root in $(1,3)$.
*if $2<k<3$, $3$ does not separate the roots, but $1$ does, hence there's one root $<1$ and the other one in $(1,3)$.
*if $k=2$, $1$ is a double root. Ik $k=3$, the roots are $0$ and $3$. Hence in both cases, there's no root in $(0,1)$.
As a conclusion, if the given equation has a root in $(0,1)$, necessarily $k\in (2,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}= (ab+bc+ca)^3$$
I tried everything:
$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.
How can I solve this question?
| I put my solution as an alternative (a little bit simpler). Write the original determinant as
$\begin{vmatrix}
-bc & b(b+c) & c(b+c)\\
a(a+c) & -ac & c(a+c) \\
a(a+b) & b(a+b) & -ab
\end{vmatrix}$
Multiply the first column with $b+c$ and the second one with c and bring $\frac{1}{c(b+c)}$ in front of the determinant:
$\frac{1}{c(b+c)}\begin{vmatrix}
-bc(b+c) & bc(b+c) & c(b+c)\\
a(a+c)(b+c) & -ac^2 & c(a+c) \\
a(a+b)(b+c) & bc(a+b) & -ab
\end{vmatrix}$
Multiply the first row by $\frac{1}{c(b+c)}$ in front of the determinant:
$\begin{vmatrix}
-b & b & 1\\
a(a+c)(b+c) & -ac^2 & c(a+c) \\
a(a+b)(b+c) & bc(a+b) & -ab
\end{vmatrix}$
Zero the first row and third column. Add the second column to the first column, multiply the third column with $-b$ and add it to the first column. Take the factors $ab+ac+bc$ out of the determinant:
$(ab+ac+bc)^2\begin{vmatrix}
0 & 0 & 1\\
a & -c & c(a+c) \\
a+b & b & -ab
\end{vmatrix}=
(ab+ac+bc)^2\begin{vmatrix}
0 & 0 & 1\\
a & -c & 0 \\
a+b & b & 0
\end{vmatrix}$
Now you may expand or continue the elimination in the same way.
The cases $c=0$ and $b+c=0$ must be solved separately (but these cases are much simpler).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y \sqrt{x z},\\
x z +y z \geq 2 z \sqrt{xz}.
\end{gather*}
Add and get
$$
xy+xz+yz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.
$$
By condition
$$
xyz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}
$$
Dividing by $\sqrt{xyz}$ we obtain the inequality.
Question. Are there another ways to prove it?
| Here's a proof that avoids (or disguises) the AM-GM inequality.
Let's start by establishing a completely different inequality: If $r,s\ge1$, then
$${1\over s}+r^2s+s\ge1+r+rs$$
The proof comes from first rewriting it in equivalent form as
$$(r-1)rs^2+(s-1)s\ge rs-1$$
and then letting $r=1+a$ and $s=1+b$ with $a,b\ge0$:
$$a(1+a)(1+b)^2+b(1+b)\ge a+b+ab$$
where we've expanded the right hand side but not the left. However, since there are no negative signs on the left, it's easy to see that its expansion consists exclusively of non-negative terms, including an $a$, a $b$, and an $ab$ (in fact a $2ab$), hence cannot be less than $a+b+ab$. This completes the proof.
Now to the problem about $x,y,z$. The statement is symmetric in the three variables, so we can, without loss of generality, assume $0\lt x\le y\le z$. This allows us to write $x=u^2$, $y=r^2u^2$ and $z=r^2s^2u^2$ with $u\gt0$ and $r,s\ge1$. Then
$$\begin{align}
xyz\ge xy+yz+zx&\implies r^4s^2u^6\ge r^2u^4+r^4s^2u^4+r^2s^2u^4\\
&\implies r^2su^3\ge{u\over s}+r^2su+su= \left({1\over s}+r^2s+s\right)u\\
&\implies r^2su^3\ge(1+r+rs)u=u+ru+rsu\\
&\implies\sqrt{xyz}\ge\sqrt x+\sqrt y+\sqrt z
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
| To factor $x^3+4x^2+3x$, we notice that we can factor $x$ out. Therefore, we get$$x^3+4x^2+3x=x(x^2+4x+3)\tag1$$
Now, we need to see if $x^2+4x+3$ can be factored as a product of two linear terms. An easy way to factor a monic polynomial is to find two numbers $r,s$ that sum to the negated value of $b$ and have a product of $c$ in $x^2+bx+c$.
In other words, we have $$\begin{align*} & r+s=-b\\ & rs=c\end{align*}\tag2$$
for $x^2+bx+c$. In your example, we see that $-b=-4$ and $c=3$. Messing around, we see that when $r=-1,s=-3$, the requirements are met. Thus,$$x^3+4x^2+3x=x(x-1)(x-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = 3 \cdot 5^{x^2} \Leftrightarrow ???$$
What do I do next?
Also, this $3^{x^2+1}$ confuses me. Does it expand to $3^{x^2}\cdot 3$ or $3^{x^2}\cdot 3^x$?
| $$3\cdot 5^{ x^{ 2 } }+3^{ x^{ 2 }+1 }-8\cdot 3^{ x^{ 2 } }=0\\ 3\cdot { \left( \frac { 5 }{ 3 } \right) }^{ { x }^{ 2 } }=5\\ { \left( \frac { 5 }{ 3 } \right) }^{ { x }^{ 2 } }=\frac { 5 }{ 3 } \quad \Rightarrow { x }^{ 2 }=1\quad \Rightarrow \quad x=\pm 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side?
| Try to write $2(3+2\sqrt 2)=6+4\sqrt 2$ as a square of $a+b\sqrt 2$ for some rational numbers $a,b$:
$$(a+b\sqrt 2)^2=a^2+2b^2+2ab\sqrt2 \;=\; 6+4\sqrt 2$$
By $\Bbb Q$-linear independence of $1$ and $\sqrt 2$, this implies $$ab=2,a^2+2b^2=6$$
Can you take it from here?
Here are some hints, if needed:
If we try to find integer solutions $a,b$, we easily find $a=2,b=1$ as solution.
Therefore:
$$\sqrt 2 \cdot \sqrt{3+2\sqrt 2}= \sqrt{6+4\sqrt 2} = \sqrt{(2+\sqrt 2)^2}=2+\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.
$\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$
So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$
So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$
Now how can i solve after that, Help required, Thanks
| I will prove $a^a\cdot b^b\cdot c^c \geq (a+b-c)^{c}\cdot (b+c-a)^{b}\cdot (c+a-b)^{b}$
Let $a+b-c=2z,\ b+c-a=2x,\ a+c-b=2y$. Without losing generality, assume $x \geq y \geq z$. Now, we need prove
$$\prod \left(x+y\right)^{(x+y)} \geq \prod \left(2z\right)^{(x+y)}$$
$$\prod \left(\frac{x+y}{2z}\right)^{(x+y)} \geq 1$$
$$\prod \left(\frac{x+y}{y}\right)^x \left(\frac{x+y}{x}\right)^y \geq 4^{x+y+z}$$
$$\prod \left(\frac{y+x}{2y}\right)^x \left(\frac{x+y}{2x}\right)^y \geq 1$$
Applying Cauchy theorem for left side,
$$\prod \left(\frac{y+x}{2y}\right)^x \left(\frac{x+y}{2x}\right)^y \geq \prod \left(\frac{x}{y}\right)^{x/2} \left(\frac{y}{x}\right)^{y/2} = A$$
After simplifying $A$, we get
$$A = \left(\frac{x}{y}\right)^{\frac{x-y}{2}} \left(\frac{y}{z}\right)^{\frac{y-z}{2}} \left(\frac{x}{z}\right)^{\frac{x-z}{2}} \geq 1$$
Hence the given inequality holds iff $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Time complexity of variation on Coupon's collector problem I need to know the complexity of the following algorithm:
Draw a set of n numbers from a larger set of m numbers, one by one, randomly, with replacement. The result may be any set of numbers as long as the size is n and the elements are different.
This is a variation of the Coupon collector's problem: we do not have to draw all m elements, any set of different numbers of size n will do.
Example:
n = 3, m = 5, result = {1,5,2} produced by successive draws 1,1,5,1,5,2 from pool {1,2,3,4,5}
What is the average time complexity of this problem?
| By way of enrichment here is the complexity using Stirling numbers
of the second kind. Using the notation from this MSE
link we have $n$
coupons, and ask about the expected time until a multiset containing
instances of $j$ different coupons has been drawn.
First let us verify that we indeed have a probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times
{n\choose j-1} \times {m-1\brace j-1} \times (j-1)!
\times (n+1-j).$$
What happens here is that for a run of $m$ samples to produce a
multiset containing instances of $j$ different coupons for the first
time on the last sample we have two parts, a prefix of length $m-1$
and a terminal sample that completes the set. Therefore we must choose
the $j-1$ values excluding the one that occurs last for the prefix
from the $n$ possibilities which gives the first binomial coefficient.
Next we partition the first $m-1$ slots into $j-1$ non-empty sets in
an ordered set partition. (Stirling number and factorial). The
smallest value chosen gets the slots listed in the first set, the next
one those in the second set etc. Finally we get $n-(j-1)$
possibilities ($j-1$ values from the prefix have already been used)
for the terminal sample that completes the selection. Combine with
$n^m$ possible choices.
Recall the OGF of the Stirling numbers of the second kind which says
that
$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$
This gives for the sum of the probabilities
$$\sum_{m\ge 1} P[T=m]
= {n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace j-1}.$$
Focusing on the sum we obtain
$$\sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \prod_{q=1}^{j-1} \frac{1/n}{1-q/n}
\\ = \prod_{q=1}^{j-1} \frac{1}{n-q}
= \frac{(n-j)!}{(n-1)!}.$$
Combining this with the outer factor we get
$${n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \frac{(n-j)!}{(n-1)!}
\\ = {n\choose j-1} (j-1)! \frac{(n+1-j)!}{n!} = 1$$
This confirms it being a probability distribution.
We then get for the expectation that
$$\sum_{m\ge 1} m\times P[T=m]
\\ = {n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace j-1}.$$
We once more focus on the sum to get
$$\sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{j-1} \frac{z}{1-qz}
= \sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m}] z \prod_{q=1}^{j-1} \frac{z}{1-qz}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \right)'\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1-pz}{z}
\frac{1}{(1-pz)^2}
\right)\right|_{z=1/n}
\\ = \left.\left( \prod_{q=0}^{j-1}
\frac{z}{1-qz} \sum_{p=0}^{j-1} \frac{1}{z}
\frac{1}{1-pz}
\right)\right|_{z=1/n}
\\ = \prod_{q=0}^{j-1} \frac{1/n}{1-q/n}
\sum_{p=0}^{j-1} \frac{1}{1/n}
\frac{1}{1-p/n}
\\ = \prod_{q=0}^{j-1} \frac{1}{n-q}
\sum_{p=0}^{j-1}
\frac{n^2}{n-p}
= n \prod_{q=1}^{j-1} \frac{1}{n-q}
\sum_{p=0}^{j-1}
\frac{1}{n-p}.$$
Retrieving the outer factor we have
$${n\choose j-1} (j-1)! (n+1-j)
\frac{1}{n} \frac{(n-j)!}{(n-1)!} \times n
\sum_{p=0}^{j-1} \frac{1}{n-p}.$$
The front simplifies to one as before and we are left with
$$n\sum_{p=0}^{j-1} \frac{1}{n-p}
= n \left(\sum_{p=0}^{n-1} \frac{1}{n-p}
- \sum_{p=j}^{n-1} \frac{1}{n-p}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{\Large
n \times \left( H_n - H_{n-j} \right)}$$
This yields $n H_n$ when $j = n$ and $1$ when $j=1$ which are both
correct. Using $H_n \sim \log n + \gamma$ we get for $j = n/2$ the
expectation $n\log 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
For 1/A = 1/B + 1/C, why is A $\frac{1}{a}$ =$\frac{1}{b}$ +$\frac{1}{c}$
I read that $a$ is always < the smaller of $b$ and $c$
In the case of $0$ < $B$ $<1$ and $0$ < $C$ $<1$,
I can understand the rule as:
Firstly,
$$\frac{1}{A}=\frac{1}{B} +\frac{1}{C}=\frac{C+B}{B*C}$$
Now looking at 3 cases:
Case 1
$$\frac{1}{A}=\frac{1}{0.5} +\frac{1}{0.5}=\frac{0.5+0.5}{0.5*0.5}=\frac{1}{0.25}$$
Case 2
$$\frac{1}{A}=\frac{1}{0.2} +\frac{1}{0.9}=\frac{0.2+0.8}{0.2*0.8}=\frac{1}{0.04}$$
Case 3
$$\frac{1}{A}=\frac{1}{0.001} +\frac{1}{0.999}=\frac{0.001+0.999}{0.001*0.999}=\frac{1}{0.0009}$$
So if we look at the $B*C$ term,
$$D=B*C$$
When $0$ < $B$ $<1$ and $0$ < $C$ $<1$
The smaller value (of $B$ or $C$) brings the product $B*C$ down to its level.
But how does the same hold true for when $B >1$ and $C>1$
Case 4
$$\frac{1}{A}=\frac{1}{1} +\frac{1}{1}=\frac{1+1}{1*1}=\frac{2}{1}$$
Hence $$A=1/2$$
Case 5
$$\frac{1}{A}=\frac{1}{5} +\frac{1}{5}=\frac{5+5}{5*5}=\frac{10}{25}$$
Hence $$A=25/10=2.5$$
I can see in both of these cases $A$ is still $<B$ and $A<C$. I know there is something going on here in relation to the A value and the result on the RHS of the equation but I can't quite put my finger on what causes $A$ to be less than $B$ and less than $C$ where $B$ and $C$ $>1$
| As you said, $\frac{1}{A}=\frac{C+B}{C*B}$
which means $A=\frac{B*C}{B+C}$
Now clearly $B > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = B*\frac{C}{B+C}$ and $\frac{C}{B+C} < 1$.
Similarly, $C > \frac{B*C}{B+C}$ because $\frac{B*C}{B+C} = C*\frac{B}{B+C}$ and $\frac{B}{B+C} < 1$.
Thus $B > \frac{B*C}{B+C} = A$ and $C > \frac{B*C}{B+C} = A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$. What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$? Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$ and $k\in \mathbb{R}$.
What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$?
| Hint:
$$
1-a+a^2=\frac{1+a^3}{1+a}
$$
Or use that with $a=x^5$
$$
a+\frac1a=1+\frac1k\implies a^3+3a+3\frac1a+\frac1{a^3}=\left(1+\frac1k\right)^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is that:
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} \tag{$\ast$} $$
is always an integer? On the same page I found the formula:
$$ \left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\
1 & b & b^2 & b^4 \\
1 & c & c^2 & c^4 \\
1 & d & d^2 & d^4 \\ \end{array} \right| = P \times (a+b+c+d)$$
My guess is that $P$ is short-hand for the product of all the different pairs of numbers:
$$ P = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) $$
This is not the right formula I'm afraid. However, I do see:
$$ \left|\begin{array}{cccc} 1 & 2 & 16 \\
1 & 3 & 81 \\
1 & 5 & 625 \end{array} \right| = P \times \stackrel{2}{C}(2,3,5)$$
which could mean anything. The book is in German - which I do not understand, and the man has invented this wonderful symbol $\stackrel{2}{C}$.
Mainly I am wondering if other integer identities can be found this way. Or is it only the case for $(a,b,c) = (2,3,5)$.
| According to Wolfram Alpha the expression
$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(b-a)}+\frac{c^4}{(c-a)(c-b)} \tag{$\ast$}$$ is not always an integer (see https://www.wolframalpha.com/input/?i=%5Cfrac%7Ba%5E4%7D%7B(b-a)(c-a)%7D%2B%5Cfrac%7Bb%5E4%7D%7B(c-b)(b-a)%7D%2B%5Cfrac%7Bc%5E4%7D%7B(c-a)(c-b)%7D,+a%3D3,b%3D5,c%3D7). But the expression you wrote
$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} \tag{$\ast$}$$ is always an integer. (Note the difference with your example. The denominator of the second fraction changes sign).
With respect to the determinant note that:
\begin{align}\left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\
1 & b & b^2 & b^4 \\
1 & c & c^2 & c^4 \\
1 & d & d^2 & d^4 \\ \end{array} \right| & \\ &\underbrace{=}_{(1)}\left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\
0 & b-a & b^2-a^2 & b^4-a^4 \\
0 & c-b & c^2-b^2 & c^4-b^4 \\
0 & d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right| \\ &\underbrace{=}_{(2)}\left|\begin{array}{ccc}
b-a & b^2-a^2 & b^4-a^4 \\
c-b & c^2-b^2 & c^4-b^4 \\
d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right|\\ &\underbrace{=}_{(3)}(b-a)(c-b)(d-c)\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\
1 & c+b & c^3+c^2b+cb^2+b^3 \\
1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| \end{align}
where in $(1)$ we have made $F_i-F_1\to F_i, i=2,3,4,$ in $(2)$ we have used the Laplacian's expansion using the first column and in $(3)$ we get out of the determinand $b-a$ from the first row, $c-b$ from the second one and $d-c$ from the third one.
Now, proceeding in a similar way we have
\begin{align}\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\
1 & c+b & c^3+c^2b+cb^2+b^3 \\
1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| &\\&=\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\
0 & c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\
0 & d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=\left|\begin{array}{ccc}
c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\
d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=(d-b)(d-a)(c-a)(a+b+c+d).\end{align}
So, your guess about $P$ is correct.
| {
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"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Does anybody recognize this array of rational numbers related to (hyper)spherical harmonics? While writing a program to compute bases of spherical harmonics on $S^n$, I discovered that the following array of rational numbers naturally arises when one considers the inclusion $S^n \subset S^{n+1}$:
\begin{array}{ccccccc}
\frac{2}{3} & \frac{8}{15} & \frac{16}{35} & \frac{128}{315} & \frac{256}{693} & \frac{1024}{3003} & \cdots \\
\frac{3}{4} & \frac{5}{8} & \frac{35}{64} & \frac{63}{128} & \frac{231}{512} & \frac{429}{1024} & \cdots \\
\frac{4}{5} & \frac{24}{35} & \frac{64}{105} & \frac{128}{231} & \frac{512}{1001} & \frac{1024}{2145} & \cdots \\
\frac{5}{6} & \frac{35}{48} & \frac{21}{32} & \frac{77}{128} & \frac{143}{256} & \frac{2145}{4096} & \cdots \\
\frac{6}{7} & \frac{16}{21} & \frac{160}{231} & \frac{640}{1001} & \frac{256}{429} & \frac{4096}{7293} & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{array}
Each entry $a_{ij}$ of this array is computed by taking a spherical harmonic on $S^i$ of degree $j$ and computing its $L^2$ norms with respect to the normalized surface area measures on $S^{i}$ and $S^{i+1}$. The ratio of these two norms turns out to be independent of the spherical harmonic one starts with, and $a_{ij}$ is the square of this ratio.
There are some clear and striking patterns here, but I haven't been able to guess a formula for $a_{ij}$ or relate this array to any well-known integer sequences on OEIS. Any ideas?
| By sheer luck, I managed to stumble upon the correct formula shortly after posting this question.
$$ a_{ij} = \frac{i!!}{(i-1)!!} \cdot \frac{(i+2j-1)!!}{(i+2j)!!} $$
Here, $n!!$ denotes the double factorial.
| {
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"url": "https://math.stackexchange.com/questions/2025397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Let $p_n$ denote the $n$th prime number. For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$ Let $p_n$ denote the $n$th prime number. For $n \ge 3$, prove that $p_{n+3}^2 \lt p_np_{n+1}p_{n+2}$.
The book gives a hint as $p^2_{n+3} \lt 4p_{n+2}^2 \lt 8p_{n+1}p_{n+2}$, but I don't understand how to show that the hint is true either.
The only inequality relationship with prime squares that I know is Bonse's inequality but I can't seem to derive either the hint or the main problem from this.
| For $n=3$, show it manually:
$(p_{n+3})^2=(p_{6})^2=13^2=169<385=5\cdot7\cdot11=(p_{3})(p_{4})(p_{5})=(p_{n})(p_{n+1})(p_{n+2})$
For $n=4$, show it manually:
$(p_{n+3})^2=(p_{7})^2=17^2=289<1001=7\cdot11\cdot13=(p_{4})(p_{5})(p_{6})=(p_{n})(p_{n+1})(p_{n+2})$
For $n\geq5$, use the fact that $p_{n+1}<2p_{n}$:
$(p_{n+3})^2<$
$(2p_{n+2})^2=$
$4(p_{n+2})(p_{n+2})<$
$4(2p_{n+1})(p_{n+2})=$
$8(p_{n+1})(p_{n+2})<$
$11(p_{n+1})(p_{n+2})=$
$(p_{5})(p_{n+1})(p_{n+2})\leq$
$(p_{n})(p_{n+1})(p_{n+2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \frac{x^2\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^{2}}}-x^2(1+\frac{1}{x})}{x(1+\frac{1}{x})}$$
and that won't cancel with the x which I have on the bottom so the limit just blows up. Did I make a mistake somewhere?
The limit is apparently $\frac{-1}{2}$ but i'm not sure how that's the case. Thanks.
| The limit cannot be found in your way.
Multiplying $$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
by
$$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}\ (=1)$$
gives
$$\frac{(x^4+x^3\tanh(x)+x^2)-(x^2+x)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\frac{x^3(\tanh(x)-2)}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\frac{\tanh(x)-2}{(1+\frac 1x)\left(\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^2}}+1+\frac 1x\right)}\to -\frac 12$$
as $x\to\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality. Prove that $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+2\sqrt{3abc(a+b+c)}\geq3(a^2+b^2+c^2)$$
I tried SOS, C-S, the $uvw$'s technique and more, but without success.
| Let $p = a+b+c, \ q = ab+bc+ca, \ r = abc$. We have $q^2 \ge 3pr$.
Using AM-GM, we have $\sqrt{3pr} \ge \frac{6pqr}{q^2 + 3pr}$. Thus, it suffices to prove that
$$\frac{a^3}{b} + \frac{b^3}{c} + \frac{c^3}{a} +
\frac{12(a+b+c)(ab+bc+ca)abc}{(ab+bc+ca)^2 + 3(a+b+c)abc} \ge 3(a^2+b^2+c^2).$$
I can solve it by the Buffalo Way. Maybe someone can give nice proofs.
WLOG, assume that $c = \min(a,b,c)$. Let $b = c+ s, \ a = c+t; \ s,t\ge 0$.
We need to prove that $q_7c^7 + q_6c^6 + q_5c^5 + q_4c^4 + q_3c^3 + q_2c^2 + q_1c + q_0 \ge 0$ where
\begin{align}
q_7 &= 12 s^2-12 s t+12 t^2, \\
q_6 &= 32 s^3+48 s^2 t-60 s t^2+32 t^3, \\
q_5 &= 45 s^4+114 s^3 t+3 s^2 t^2-102 s t^3+45 t^4, \\
q_4 &= 31 s^5+113 s^4 t+115 s^3 t^2-113 s^2 t^3-43 s t^4+31 t^5, \\
q_3 &= 7 s^6+67 s^5 t+91 s^4 t^2+17 s^3 t^3-91 s^2 t^4+15 s t^5+7 t^6, \\
q_2 &= 14 s^6 t+44 s^5 t^2+23 s^4 t^3-18 s^3 t^4-13 s^2 t^5+7 s t^6, \\
q_1 &= 8 s^6 t^2+8 s^5 t^3-3 s^3 t^5+s^2 t^6, \\
q_0 &= s^6 t^3.
\end{align}
It is not hard to prove that $q_7, q_6, \cdots, q_0\ge 0$. Omitted. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$ I'm stuck on a trigonometry question. The formulae I have been given don't seem to help so I had to come here.
The question is:
Prove that $\sqrt{3}\cos(x) + \sin(x) = 2\cos\left(x- \frac{\pi}{6}\right)$
If these formulae can be used I would be glad if you could show me how:
*
*$\cos(2x) = 2\cos^2(x) - 1$
*$\cos(2x) = 1 - 2\sin^2(x)$
*$\cos^2(x) = \frac 12(1 + \cos(2x))$
*$\sin^2(x) = \frac 12(1 - \cos(2x))$
| In the right side you can use:
\begin{equation}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\\
\end{equation}
In this case:
\begin{equation}
\cos(x-\frac{\pi}{6})=\cos x\cos\frac{\pi}{6}+\sin x\sin\frac{\pi}{6}\\
\cos(x-\frac{\pi}{6})=\frac{\sqrt 3}{2}\cos x+\frac{1}{2}\sin x
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| $$
\begin{cases}
\text{a}^2-\text{b}^2-\text{a}=0\\
\\
\text{b}-2\text{a}\text{b}=\frac{1}{2}
\end{cases}\space\space\space\Longleftrightarrow\space\space\space
\begin{cases}
\text{a}=\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\\
\\
\text{b}-2\times\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\times\text{b}=\frac{1}{2}
\end{cases}
$$
Now:
$$\text{b}-2\times\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\times\text{b}=\pm\text{b}\sqrt{1+4\text{b}^2}=\frac{1}{2}\space\space\space\Longleftrightarrow\space\space\space\text{b}=\begin{cases}
\pm\frac{1}{2}\sqrt{\frac{\sqrt{5}-1}{2}}\\
\\
\pm\frac{i}{2}\sqrt{\frac{1+\sqrt{5}}{2}}
\end{cases}$$
| {
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"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| Observe that $1$ and $-1$ are roots of your polynomial. You thus find that it is equal to
$$(x-1)^2(x+1)^2(x^4+3x^2+1)$$
and is thus clearly non-negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding S,T in matrix equivalence problem So guys I could really need some help finding the two matrices S,T. I seem to have some problems getting all the elemantary row,colmun opperations done to get the result. Maybe there is also an easier way of finding S,T. Anything can help.
The rank of the matrix A over the field $\mathbb{Q}(x)$ is 2. Find the matrices $S,T \in GL_{4}(\mathbb{Q}(X))$
$$A = \begin{pmatrix}
1 & x & 0 & -x \\
x+1 &0 &-x &x^2 \\
2x+1 &x^2 &-x &0 \\
1& -x^2 &-x &2x^2
\end{pmatrix}$$
so that
$$
SAT = \begin{pmatrix} \
1 & 0 &0 & 0\\
0 & 1 &0 &0 \\
0& 0 & 0 & 0\\
0 & 0 &0 &0
\end{pmatrix}
\in\ GL_{4}(\mathbb{Q}(X))
$$
| This problem comes down to finding appropriate bases for the domain (input) and codomain (output). Recall that the columns of a transformation matrix are the images of the domain basis vectors. The last two columns are zero, so the last two basis vectors for the domain must be elements of the kernel. We can choose the other two freely. The matrix $T$ will have these vectors for its columns so that $AT$ will be the matrix of their images.
Turning now to the output, we see that the first two basis vectors must be the images of the first two input basis vectors. The other two can again be chosen freely. (In fact, this choice is irrelevant to the form of the resulting matrix since the last two columns of $AT$ are zero.) This matrix converts from the standard basis to this basis, so these vectors form the columns of $S^{-1}$.
We begin by row-reducing $A$ to find a basis for its kernel. In order to keep the expressions relatively simple and so reduce the chance of making an error, I recommend waiting to normalize pivots to $1$ as long as possible. After a bit of work, we reach $$\pmatrix{1&x&0&-x\\0&-x^2-x&-x&2x^2+x\\0&0&0&0\\0&0&0&0}.$$ Divide the second row by $-x$ to get $$\pmatrix{1&x&0&-x\\0&x+1&1&-(2x+1)\\0&0&0&0\\0&0&0&0},$$ at which point we can stop. The rank of the matrix is indeed two, and the isolated $1$’s in each non-zero row allow us to determine the kernel: it consists of vectors of the form $((d-b)x,b,(2x+1)d-(x+1)b,d)^T$, with basis $(x,-1,x+1,0)^T$ and $(x,0,2x+1,1)^T$.
We could’ve also found the kernel directly by multiplying $A$ by $(a,b,c,d)^T$ and setting the result to zero, producing the equations $$\begin{align}(b-d)x+a&=0\\dx^2+ax-cx+a&=0\\bx^2+(2a-c)x+a&=0\\(2d-b)x^2-cx+a&=0.\end{align}$$ The first equation gives us $a=(d-b)x$ and substituting this back into the system will give us the other dependency found in the previous paragraph. An obvious and convenient choice for our other two basis vectors is $(1,0,0,0)^T$ and $(0,0,1,0)^T$, so we have $$T=\pmatrix{1&0&x&x\\0&0&-1&0\\0&1&x+1&2x+1\\0&0&0&1}.$$
With this choice of basis vectors we can immediately write down the first two vectors of the output basis, which are also the first two columns of $S^{-1}$: they are simply the first and third columns of $A$. We can take for the last two columns the corresponding columns of the identity matrix. To find $S$, augment the resulting matrix with the identity and row-reduce, leaving pivots unnormalized at first: $$\left(\begin{array}{cccc|cccc}1&0&0&0 & 1&0&0&0\\x+1&-x&0&0 & 0&1&0&0\\2x+1&-x&1&0 & 0&0&1&0\\1&-x&0&1 & 0&0&0&1\end{array}\right)\to\left(\begin{array}{cccc|cccc}1&0&0&0 & 1&0&0&0\\0&-x&0&0 & -(x+1)&1&0&0\\0&0&1&0 & -x&-1&1&0\\0&0&0&1 & x&-1&0&1\end{array}\right).$$ Divide the second row by $-x$ to obtain $$S=\pmatrix{1&0&0&0\\1+\frac1x&-\frac1x&0&0\\-x&-1&1&0\\x&-1&0&1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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determmine image of $f$ Let $f(x)=\dfrac{1}{x+1}\sqrt{x^2-1}$
$D_f={(-\infty,-1)}\cup {[1, +\infty)}$
How can i show algebraically that $$f\left( (1,+\infty) \right)=(0,1)$$
Indeed,
let $y\in \mathbb{R} $
$$
\begin{align}
y\in f\left( ]1,+\infty[ \right) &\iff f(x)=y\\
&\iff \dfrac{1}{x+1}\sqrt{x^2-1}=y\\
&\iff \sqrt{x^2-1}=yx+y\\
&\iff \begin{cases}
|x^{2}-1|=y^{2}(x+1)^{2}& \\
yx+y>0& \\
x> 1 &\\
\end{cases} \\
&\iff \begin{cases}
x-1=y^{2}(x+1)& \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
&\iff \begin{cases}
x-y^{2}x=y^{2}+1& \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
&\iff \begin{cases}
x=\dfrac{y^{2}+1}{1-y^{2}}& \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
&\iff \dfrac{y^{2}+1}{1-y^{2}}>1
\end{align}$$
i'm trying to show that $y\in (0,1)$
I'm stuck here
| for $x>1$
$$f(x)=\sqrt{1-\frac{2}{x+1}}$$
we see that $f$ is continuous and strictly increasing at $(1,+\infty)$.
but
$$\lim_{x\to 1^+}f(x)=0$$
and
$$\lim_{x\to+\infty}f(x)=1$$
thus
$$f((1,+\infty))=(0,1)$$
algebraically
let $x>1$
$$f(x)=y\implies 1-\frac{2}{1+x}=y^2$$
$$\implies x=\frac{2}{1-y^2}-1$$
*
*$$x>1\implies 0 <\frac{2}{1+x}<1\implies 0<y<1$$
*$$0<y<1\iff 0<1-y^2<1\implies \frac{1}{1-y^2}>1\implies x>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculate with Cauchy integral formula Can you please check if i did any mistakes.
a)
$$\int_{\gamma_{B_2(0)}} \frac{z^7+41}{1-z}$$
$\frac{z^7+41}{1-z}$ = $ \frac{-z^7-41}{z-1}$ = $\frac{f(z)}{z-1}$
with the Cauchy integral formula i get $2 \pi i \cdot (-42) = -84\pi i$
b)
$$\int_{\gamma_{B_r(\frac{i}{2})}} \frac{1}{z^2+1}$$
(1) Let $r = \frac{1}{4}$ then $\frac{1}{z^2+1}$ is holomorphic and there exists a holomorphic antiderivative on $\mathbb C\setminus\{i,-i\}$ so therefore the integral is $0$.
(2)Let $ r = 1$ then we have $ \frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)}=\frac{\frac{1}{z+i}}{z-i} = \frac{f(z)}{z-i}$ . With the CIF i get $2\pi i \cdot f(i) = \pi i$
(3)Let $r=2$ then i get $\pi i$ aswell which is probably wrong
| For $r=2$, both poles are enclosed. We can write
$$\frac{1}{z^2+1}=\frac{1/2i}{z-i}-\frac{1/2i}{z+i}$$
and apply Cauchy's Integral Formula as
$$\begin{align}
\oint_{B_{2}(i/2)}\frac{1}{z^2+1}\,dz&=\oint_{B_{2}(i/2)}\left(\frac{1/2i}{z-i}-\frac{1/2i}{z+i}\right)\,dz\\\\
&=\oint_{B_{2}(i/2)}\frac{1/2i}{z-i}\,dz-\oint_{B_{2}(i/2)}\frac{1/2i}{z+i}\,dz\\\\
&=\frac1{2i}-\frac{1}{2i}\\\\
&=0
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Axes of Symmetry for a General Ellipse
Given a general ellipse
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
where $B^2<4AC$, what are the major and minor axes of symmetry in the form $ax+by+c=0$?
It is possible of course to first work out the angle of rotation such that $xy,x,y$ terms disappear, in order to get an upright ellipse of the form $x^2/p^2+y^2/q^2=1$ and proceed from there. This may involve some messy trigonometric manipulations.
Could there be another approach perhaps, considering only quadratic/diophantine and linear equations?
Addendum
Here's a graphical implementation based on the answer by Ng Chung Tak.
Addendum 2
Based on the answers by amd and by Ng Chung Tak, the equations for the axes are
$$\color{red}{\left(y-\frac {2AE-BD}{B^2-4AC}\right)=\frac {C-A\pm \sqrt{(A-C)^2+B^2}}B\left(x-\frac {2CD-BE}{B^2-4AC}\right)}$$
Note that
$$\frac{C-A\pm \sqrt{(A-C)^2+B^2}}B\cdot \color{lightgrey}{\frac {C-A\mp\sqrt{(A-C)^2+B^2}}{C-A\mp\sqrt{(A-C)^2+B^2}}}=-\frac B{C-A\mp\sqrt{(A-C)^2+B^2}}$$
i.e. it is equal to the negative of its own reciprocal. Hence the equations for the axes can also be written as
$$\color{red}{\left(x-\frac {2CD-BE}{B^2-4AC}\right)=-\frac {C-A\mp \sqrt{(A-C)^2+B^2}}B\left(y-\frac {2AE-BD}{B^2-4AC}\right)}$$
hence the two symmetrical anti-symmetrical forms for the axes. Here's the graphical implementation.
| The principal axes are:
$$\left( y-\frac{2AE-BD}{B^2-4AC} \right)=
\frac{C-A \color{red}{\pm} \sqrt{(A-C)^2+B^2}}{B}
\left( x-\frac{2CD-BE}{B^2-4AC} \right)$$
Also refer to another answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the joint Probability distribution of $X$ and $Y$? If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to solve this problem one must draw a chart.
$$\begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & & &\\ \hline
\end {array}$$
How does one fill up the rest of the table and answer questions one and two.
EDIT
In order to find solve $(1)$ one must add all the related $f(x,y)$ relations. Thus
$(1)$ $$ \quad P(U=2) =0, \\ P(U=3) = \frac{1}{7} + \frac{1}{7} = \frac27, \\ P(U=4)= 0+\frac{4}{7} = \frac47, \\ P(U=5)=\frac{1}{7}$$
One must use this notation to solve.
$$P(x=1|U=4)= \frac{P(x=1,U=4)}{P(U=4)} = ? \\P(x=2|U=4) = ? \\ P(x=3|U=4) = ? $$
Knowing this does anyone know how to solve $(2)$ using this notation? What does one substitute for this question to derive the answer?
| Here is a completed table of the discrete probabilities:
$$ \begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & 1 & 1 & 2 & 2 & 3 & 3 \\ \hline
\end {array} $$
The Reader will recognize that the last row's entries are simply the $x$ values of the coordinate pairs in the first row. The intent of forming such a row is likely to help answer the second part of the exercise given in the Question, what is the conditional probability distribution of $X$, given that the sum $U = 4$?
The table makes it clear that while the events $X=2,Y=2$ and $X=3,Y=1$ both lead to $U=4$, the former of these events is assigned a zero probability (and the latter is assigned a positive probability). Therefore when the conditional probabilities are assigned, conditioned on $U=4$, we have:
$$ \mathbf{Pr}(X=2 \mid U=4) = 0 \; ; \; \mathbf{Pr}(X=3 \mid U=4) = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Polar form is
$$
r^4\left(\cos^4(\theta)+\sin^4(\theta)\right)=r^2\tag{1}
$$
which gives
$$
r^2=\frac1{\cos^4(\theta)+\sin^4(\theta)}\tag{2}
$$
The area is
$$
\begin{align}
\frac12\int_0^{2\pi}r^2\,\mathrm{d}\theta
&=\frac12\int_0^{2\pi}\frac{\mathrm{d}\theta}{\cos^4(\theta)+\sin^4(\theta)}\tag{3}\\
&=\frac12\int_\gamma\frac{8}{z^4+6+z^{-4}}\frac{\mathrm{d}z}{iz}\tag{4}\\
&=\frac4i\int_\gamma\frac{z^3\,\mathrm{d}z}{z^8+6z^4+1}\tag{5}\\
&=\frac1i\int_\gamma\frac{\mathrm{d}z^4}{z^8+6z^4+1}\tag{6}\\
&=\frac4i\int_\gamma\frac{\mathrm{d}z}{z^2+6z+1}\tag{7}\\
&=8\pi\operatorname*{Res}_{z=-3+2\sqrt2}\left(\frac1{z^2+6z+1}\right)\tag{8}\\[6pt]
&=\sqrt2\,\pi\tag{9}
\end{align}
$$
Explanation:
$(3)$: apply $(2)$
$(4)$: $z=e^{i\theta}$, so $\mathrm{d}\theta=\frac{\mathrm{d}z}{iz}$, $\cos(\theta)=\frac12\left(z+z^{-1}\right)$, $\sin(\theta)=\frac1{2i}\left(z-z^{-1}\right)$
$(5)$: algebra
$(6)$: $\mathrm{d}z^4=4z^3\,\mathrm{d}z$
$(7)$: if $z$ goes around the origin once, $z^4$ goes around $4$ times
$(8)$: use the Residue Theorem
$(9)$: the residue of $\frac1{z^2+6z+1}$ at $z=-3+2\sqrt2$ is $\frac1{2z+6}=\frac1{4\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Geometry problem about triangle orthocenter Let $ABC$ be a triangle with orthocenter $H$. Let $AB = 84$, $AC = 32\sqrt{3}$ and $\angle{BAC} = 60^\circ$. Let $D, E, F$ be points on $AH, BH, CH$, respectively, such that: $7AD = AH$, $7BE = BH$, $7CF = CH$. Find $AB\cdot CF + BC\cdot AD + AC\cdot BE$.
This problem is taken from a math competition and you should be able to solve it without the aid of a calculator; I managed to find both $CF$ and $BE$, using similar triangles. However I'm stuck with $AD$: the expression that leads to the evaluation of $AD$ is simply too difficult (a lot of square roots), so I was wondering if someone could give me a hint or sketch the solution. Any help would be greatly appreciated.
| On each point $A, B, C$ of the triangle, pass through it a line parallel to its opposite side. Then, these 3 new lines form a new triangle, similar to the original one. the orthocenter of $ABC$ coincide with the circumcenter of the new triangle. Notice that
$$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = AB \cdot CH + BC \cdot AH + AC \cdot BH$$
$$ 7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = \frac{2AB \cdot CH}{2} + \frac{2BC \cdot AH}{2} + \frac{2AC \cdot BH}{2}$$
Now, the sides of the new triangle are given by $2AB$, $2BC$ and $2CA$, so the expression above corresponds to its area. Thus, we have
$$7(AB\cdot CF + BC\cdot AD + AC\cdot BE) = 4 \text{Area}(ABC) = 4 \frac{84 \cdot 32\sqrt{3} \sin 60^\circ}{2} = 2 \cdot 84 \cdot 32 \cdot \frac{3}{2}$$
implying that
$$ AB\cdot CF + BC\cdot AD + AC\cdot BE = 12 \cdot 32 \cdot 3 = 9 \cdot 128 = 1152.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $\sum\limits_{i=1}^6a_i^2=6$, where $a_i>0$, $a_7=a_1$. Prove that $\sum\limits_{i=1}^6\frac{a_i^2}{a_{i+1}}\geq6$ Let $a_i$ be positive numbers such that $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2=6$. Prove that:
$$\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\frac{a_3^2}{a_4}+\frac{a_4^2}{a_5}+\frac{a_5^2}{a_6}+\frac{a_6^2}{a_1}\geq6$$
I tried C-S and SOS, but without success.
| We can write:
$$ \frac{a_1^2}{a_2} + \frac{a_2^2}{a_2} + \ldots + \frac{a_6^2}{a_1} \geq
a_1^2 + \ldots + a_6^2$$
which is equivalent to this inequality.
after dividing both sides by $(a_1^2 + \ldots + a_6^2)$ we get:
$$ \frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} + \ldots + \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1} \geq
1$$
Function $f(x)= \frac{1}{x}$ is convex in real positive domain, hence from Jensen's inequality we have:
$$\frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} + \ldots + \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}
\geq
\frac{1}{ \frac {a_1^2 a_2 } {a_1^2 + \ldots + a_6^2} +
\frac {a_2^2 a_3 } {a_1^2 + \ldots + a_6^2} +
\ldots + \frac {a_6^2 a_1 } {a_1^2 + \ldots + a_6^2} }
$$.
$$\frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} + \ldots + \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1}
= \frac{a_1^2 a_2 + a_2^2 a_3 + \ldots + a_6^2 a_1 } {a_1^2 + \ldots + a_6^2}
$$
but
$$ a_1^2 a_2 + \ldots + a_6^2 a_1 \geq a_1^2 + \ldots + a_6^2 $$
because by dividing both sides by $a_1^2 + \ldots + a_6^2$
we have
$$ \frac{a_1^2}{ a_1^2 + \ldots + a_6^2} a_2 + \ldots + \frac{a_6^2} { a_1^2 + \ldots + a_6^2} a_1 \geq a_1+ \ldots + a_6 \geq 1.$$ (Jensen's inequality)
$$\frac{a_1^2 a_2 + a_2^2 a_3 + \ldots + a_6^2 a_1 } {a_1^2 + \ldots + a_6^2}\geq 1
$$
and finally:
$$ \frac{a_1^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_2} + \ldots + \frac{a_6^2}{a_1^2 + \ldots + a_6^2} \frac{1}{a_1} \geq
1$$ is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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