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Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$
| $$\sum\limits_{j=0}^n j^2(2j-1)=2\sum \limits_{j=0}^n j^3-\sum \limits_{j=0}^n j^2=2(\dfrac{n(n+1)}{2})^2-\dfrac{n(n+1)(2n+1)}{6}.$$ You can prove this formulae for $\sum j^3$ and $\sum j^2$ by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x$ We have to simplify:
$$\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x$$
I came to the conclusion that the answer would be:
$$\frac{e^x}{2\cdot\sqrt{e^x}}$$
But I was wrong and it was:
$$\frac{\sqrt{e^x}}{2}$$
Where did I go wrong and why?
Thanks
| $$\require{cancel}\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x = \frac{1}{2\cdot\sqrt{e^x}}\cdot e^x\cdot\frac{\sqrt{e^x}}{\sqrt{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}\cdot\frac{e^x}{\sqrt{e^x}\cdot\sqrt{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}\cdot\frac{\cancel{e^x}}{\cancel{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Hint: let $y=x^2$, then the given condition writes as $P(y+1)=y^2+5y+3$ and it follows that $P(x^2-1)=P\big((y-2)+1\big)= (y-2)^2+5(y-2)+3 = \ldots$
| {
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What is the integral of $\int\frac{dx}{\sqrt{x^3+a^3}}$?
What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$
I came across this integration in a physics problem. I suspect role of complex numbers here.
'$a$' is a constant
| It's this:
$$\frac{2 \sqrt[6]{-1} \sqrt[3]{a^3} \sqrt{(-1)^{5/6} \left(\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}-1\right)} \sqrt{\frac{(-1)^{2/3} x^2}{\left(a^3\right)^{2/3}}+\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}+1} F\left(\left.\sin ^{-1}\left(\frac{\sqrt{-\frac{(-1)^{5/6} x}{\sqrt[3]{a^3}}-(-1)^{5/6}}}{\sqrt[4]{3}}\right)\right|\sqrt[3]{-1}\right)}{\sqrt[4]{3} \sqrt{a^3+x^3}}$$
where $F$ denotes the Elliptic Integral.
| {
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Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$ Let $a,b,c>0$ reals.
Prove that
$$cyclic \sum\frac {a^2+1}{b+c}\ge 3$$
I proved it using Nesbitt inequality
$$cyclic \sum \frac {a}{b+c} \ge \frac {3}{2} $$
and the fact that
$$a+\frac {1}{a}\ge 2$$
But i would like to know if there is a straight proof without Nesbitt inequality.
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b+c}\sum_{cyc}\frac{1}{b+c}}\geq$$
$$\geq2\sqrt{\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+c)}\sum_{cyc}\frac{1}{b+c}}=\sqrt{2(a+b+c)\sum_{cyc}\frac{1}{b+c}}=$$
$$=\sqrt{\sum\limits_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}}\geq\sqrt{(1+1+1)^2}=3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $? $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=\sum_{n=1}^\infty\left(\frac{2n+1}{n^2}\frac1{{(n+1)}^2}\right)$$
I assume that I should get a telescoping sum in some way, but I'm couldn't find it yet.
| You'll get a telescoping series, but that is likely just only one of the ingredients for a solution. First, I'll separate the numerator in two parts:
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{n+n+1}{n^2(n+1)^2}
$$
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{1}{n(n+1)^2}+\frac{1}{n^2(n+1)}
$$
And of course your telescoping element:
$$
\frac{1}{n(n+1)}=\left( \frac{1}{n}-\frac{1}{n+1}\right)
$$
Therefore:
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}+\frac{1}{n^2}-\frac{1}{n(n+1)}
$$
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=-\frac{1}{(n+1)^2}+\frac{1}{n^2}
$$
And the result is simply 1. As would [WolframAlpha confirm ]( https://www.wolframalpha.com/input/?i=sum((2*n%2B1)%2F((n%5E2(n%2B1)%5E2) )
| {
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.
| You are making a mistake, in that you are assuming that $(-1)^k=1$.
You have
\begin{align}
(-1)^{k-1}\frac{k.(k+1)}{2}+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{(k+1)}{2} [k-(2k+2)]\\ \ \\
&=(-1)^{k-1}\frac{(k+1)}{2} [-(k+2)]\\ \ \\
&=(-1)^{k}\frac{(k+1)(k+2)}{2} \\ \ \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic equation formula help / simplification I have this quadratic equation,
$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $
I use the quadratic formula to solve and simplify
$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$
my book says it should simplify to
$ \frac{1}{3} ( -5 \pm \sqrt{73}) $
but i cant get this simplification can anyone show me if they can? Thank you.
| $$x=\frac{-10 \pm \sqrt{100-(4)(3)(-80)}}{6} =\frac{-2\times 5 \pm \sqrt{4\times25-(4)(3)(-80)}}{6} =\frac{1}{6}\left(-2\times 5 \pm 2\sqrt{25+240} \right) =\frac{2}{6}\left(-5 \pm \sqrt{265} \right)=\frac{1}{3}\left(-5 \pm \sqrt{265} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find value of $A'$ in eliminating cross product terms Quadratic Curve Rotation? I was studying conics and came around the topic of eliminating cross-product terms when rotating coordinates of a quadratic curve of the form $$A x^2 + B x y + C y^2 + D x + E y + F = 0$$ where
$$\begin{align*}
A x^2 &= A\left(\cos(\alpha) x' - \sin(\alpha) y'\right)^2\\
&= A\left(\cos^2(\alpha) x'^2 -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right)\\
B x y &= B\left(\cos(\alpha) x' - \sin(\alpha) y'\right)(\sin(\alpha) x' + \cos(\alpha) y') \\
&= B\left(\sin(\alpha)\cos(\alpha)\left(x'^2 - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\
C y^2 &= C\left(\sin(\alpha) x' + \cos(\alpha) y'\right)^2 \\
&= C\left(\sin^2(\alpha) x'^2 + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\
D x &= D\left(\cos(\alpha) x' - \sin(\alpha) y'\right) \\
E y &= E\left(\sin(\alpha) x' + \cos(\alpha) y'\right)
\end{align*}$$
I am stuck when they converted from the above equation to $$A' x'^2 + B' x' y' + C' y'^2 + D' x' + E' y' + F' = 0$$
I want to know how they found the value of $ A', B', C'$ etc. as shown below:
$$\begin{align*}
A' &= A\cos^2\theta + B\cos\theta\sin\theta+ C\sin^2\theta \\
B' &= B\left(\cos^2\theta - \sin^2\theta\right) + 2\left(C - A\right)\sin\theta\cos\theta\\
C' &= A\sin^2\theta - B\sin\theta\cos\theta + C\cos^2\theta \\
D' &= D\cos\theta + E\sin\theta \\
E' &= -D\sin\theta + E\cos\theta \\
F' &= F
\end{align*}$$
I tried using double angle formulas but wasn't able to derive the value in the image and I need your help. Any of $A', B', C'$ would do. I will do the rest myself. Just need direction.
I believe that I'm missing some information to provide so please let me know.
| Just expand and collect the expressions right above. For example, focusing on the $\,\color{red}{{x'}^2}\,$ terms:
$$
\begin{align}
A x^2 + B x y + C y^2 + D x + E y + F &= \color{red}{A}\left(\color{red}{\cos^2(\alpha) x'^2} -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right) \\
&\;\; + \color{red}{B}\left(\color{red}{\sin(\alpha)\cos(\alpha)}\left(\color{red}{x'^2} - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\
&\;\; + \color{red}{C}\left(\color{red}{\sin^2(\alpha) x'^2} + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\
&\;\;+ \ldots \\
&= \underbrace{\color{red}{\left(A \cos^2(\alpha) + B \sin(\alpha)\cos(\alpha) + C \sin^2(\alpha)\right)}}_{\color{red}{\Large{A'}}}\color{red}{{x'}^2} + \ldots
\end{align}
$$
| {
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Inequality $\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$ Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$$
The full solution (due to @Hanno) see at the very end in $\underline{\textbf{Conclusion}}\,\,$
My attemps:
*
*AM>HM gives $a+b+c\geqslant 6.$ Put $a=a_1t,b=b_1t,c=c_1t, \,t\geqslant 1, \, a_1+b_1+c_1=6$
As $\frac 1{x+1}$ is decreasing function we have $\sum\limits_{cyc}\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}\leqslant \sum\limits_{cyc}\frac{a_1}{\sqrt{b_1c_1}}\cdot\frac1{a_1+1}.$
So is it possible to prove the same inequality with new restriction $a+b+c=6?$
*$a=\tan^2\alpha, b=\tan^2\beta, c=\tan^2\gamma$ gives
$$\sum\limits_{cyc}\sin^2\alpha\cot\beta\cot\gamma\leqslant \sqrt2$$
if $0<\alpha,\beta,\gamma<\frac\pi2$ and $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$
*After hint of Michael Rozenberg. Rewrite the substitution as linear system with determinant $d=\begin{vmatrix}-a&1&1\\
1&-b&1\\ 1&1&-c\\ \end{vmatrix}$
If $d\ne 0$ then $x=y=z=0$ and so the hint doesn't work
The equality $d=0$ gives editional restriction, but I am not sure that this case deserves attention.
In fact, I have an ugly solution with brute force, but I want something beautiful.
*Continue. Agree with $a=\frac{y+z}{x}$, $b$ and $c$ cyclically and $x+y+z=1.$ I'll try not to break the symmetry as in my ugly solution.
Substution gives
$$\sum\limits_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+z)(x+y)}}\leqslant\sqrt2$$
squaring gives
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+2\frac{(z+x)\sqrt{zx}}{\sqrt{(y+x)(y+z)}}\cdot\frac{(x+y)\sqrt{xy}}{\sqrt{(z+y)(z+x)}}\right)\leqslant2$$
or
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2$$
I do not see here where to use AM-GM, or did I miss the moment?
Thanks, @Michael Rozenberg, very cool. I did not miss, I did not get to the point, although I was already near. Why I stubbornly did not want to multiply by the common denominator? Although further my restriction $x+y+z=1$ would not help but would only hinders.
*Yesterday I did not check the last inequality but about this later.
We conside the case $d=0.$ In order for the linear system to have a unique solution, it is necessary to add a new constraint. I put $x+y+z=1$ and violated homogenity. To save it let $x+y+z=I$ where $I$ denotes the sum only.
Then from the system we have $x=\frac I{a+1}$ and $a=\frac Ix-1=\frac{y+z}{x}$ i.e. we simply forget that $I=1$ and save the homogenity.
we amend the previous one:
$$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+
2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2I^2$$
After AM-GM we have
$$\sum\limits_{cyc}\left({yz(y+z)^3}+
x(x+y)(x+z)(xy+xz+2yz)\right)\leqslant2I^2\prod\limits_{cyc}(x+y)$$
It remains to check the last inequality.
For this its sufficent to calculete the all summands in both sides. I do not want to apply brutal forse.
Avidently both sides have can have members only of type $x^\alpha y^\beta z^\gamma$ with $\alpha+\beta+\gamma=5$ and by symmetry we may assume $\alpha\leqslant\beta\leqslant\gamma.$ My quation is: Is there another way?
Start with memmber $yz^4$. It is contained in the LHS but not contained in the right. So the last inequality must contain negative term and the proof must be corrected if it is possibble yet. Or I'm wrong.
$\underline{\textbf{Conclusion}}\,\,$ The inequality turned out to be interesting and the proof gave @Hanno. Allow yourself to make some minor corrections.
The original restriction was
$$\sum_{\text{cyc}}{1\over a+1} =1 \qquad (1)$$
Now $\sum_{\text{cyc}}{a\over a+1} =\sum_{\text{cyc}}{a+1-1\over a+1} =3-1=2.$ So
$$\sum_{\text{cyc}}{a\over a+1} =2 \qquad (2)$$
Apply the Cauchy–Bunyakovsky–Schwarz inequality (CBS):
$$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1}
\:= \:
\sum_{\text{cyc}}\sqrt{{a\over a+1}}\cdot\sqrt{\frac a{(a+1)bc}}\:\leqslant\:
{\sqrt{\sum_{\text{cyc}}{a\over a+1}}}\,\cdot\,
\sqrt{\sum_{\text{cyc}}{a\over (a+1)bc}}$$
Taking into account (2) it remains to proof inequality $\sum\limits_{\text{cyc}}{a\over (a+1)bc}\:{\leqslant}\:1$ which is equivalent to
$$\sum_{\text{cyc}}{a^2\over a+1}\:\leqslant\: abc $$
We have $\sum\limits_{\text{cyc}}{a^2\over a+1}=\sum\limits_{\text{cyc}}{a^2-1+1\over a+1}=\sum\limits_{\text{cyc}}(a-1)+\sum\limits_{\text{cyc}}{1\over a+1}=a+b+c-2<a+b+c+2=abc. $
The last equality $a+b+c+2=abc$ follows from (1) by bringing to a common denominator. DONE!
The inequality is exact. Take small $a$ and $b=c=1+\frac2a.$ Then (1) is true and the LHS of required inequality evaluates to ${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}$ which tends to $\sqrt2$ if $a\to0$.
| The hint.
After substitution $a=\frac{y+z}{x}$, $b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$, where $x$, $y$ and $z$ are positives,
and squaring of the both sides use AM-GM.
Indeed, let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, the condition gives $$\frac{1}{\frac{y+z}{x}+1}+\frac{1}{\frac{x+z}{y}+1}+\frac{1}{c+1}=1$$ or
$$\frac{x+y}{x+y+z}+\frac{1}{c+1}=1$$ or
$$\frac{1}{c+1}=\frac{z}{x+y+z}$$ or
$$\frac{1}{c+1}=\frac{1}{\frac{x+y}{z}+1},$$ which gives $c=\frac{x+y}{z}$ and we need to prove that
$$\sum_{cyc}\frac{\frac{y+z}{x}}{\sqrt{\frac{(x+y)(x+z)}{yz}}\left(\frac{y+z}{x}+1\right)}\leq\sqrt2$$ or
$$\sum_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+y)(x+z)}}\leq\sqrt2(x+y+z)$$ or
$$\sum_{cyc}\sqrt{yz(y+z)^3}\leq(x+y+z)\sqrt{2(x+y)(x+z)(y+z)}$$ or
$$\sum_{cyc}\left(xy(x+y)^3+2x(x+y)(x+z)\sqrt{(x+y)(x+z)yz}\right)\leq2(x+y+z)^2\prod_{cyc}(x+y)$$ and since by AM-GM
$$2\sqrt{(x+y)(x+z)yz}=2\sqrt{(xz+yz)(xy+yz)}\leq xz+yz+xy+yz=xy+xz+2yz,$$ it's enough to prove that
$$\sum_{cyc}\left(xy(x+y)^3+x(x+y)(x+z)(xy+xz+2yz)\right)\leq2(x+y+z)^2\prod_{cyc}(x+y),$$
which is $$\sum_{cyc}(x^3y^2+x^3z^2+4x^3yz+6x^2y^2z)\geq0.$$
Done!
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Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
There is given hint: Suppose that there is a solution. First show that it can be reduced to a solution satisfying $\gcd(x,z)=1$. Then use the fact that $x^3+ xz^4= x(x^2+z^4)$ is a perfect square to show that there are no solutions other than $x=y=0$.
The hint tells to prove by contradiction approach. So, $y\mid x,z\implies y\mid p$, where $p=(x,z)$. Hence, $\exists X,Z\in \mathbb{Z},\,x=Xp, z=Zp\implies y^2= X^3p^3 +XpZ^4p^4$$\implies y^2= Xp^3(X^2+Z^4p^2)$.
Is it valid to conclude from the last line, any of the below two separate conclusions :
1. The set of common divisors of $y^2, x^3, xz^4$ are the same; hence let $gcd(y^2, x^3, xz^4)=p'$
2.$\,\,p^3\mid y^2\implies \exists y=p^2Y, \,\,\,\,p^2\mid y$
| This is not much more than a lengthy remark, but it's worth posting as an answer, I think, because it shows that proving this is at least as hard as proving the strong version of Fermat's Last Theorem for $n=4$. That is, suppose that $a^4+b^4=c^2$ with $abc\not=0$. Then, letting $x=a^2$, $z=b$, and $y=ac$, we first have $x^2+z^4=c^2$, and thus $x(x^2+z^4)=xc^2=(ac)^2=y^2$ with $xyz=a^3cb\not=0$. So if $y^2=x^3+xz^4$ has no solutions with $xyz\not=0$, then $a^4+b^4=c^2$ has no solutions with $abc\not=0$ either.
| {
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How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10 \leq x_i \leq 20$? I am trying to solve the following question. I have worked through a similar question but the question I have worked through the range was all positive, however this one has a negative side of the range so I am not quite sure how to solve it. I will include the question I have worked through as reference below. Here is the question I need help solving, thanks! I need to solve it through the principle of inclusion exclusion
How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10 \leq x_i \leq 20$
Here is the similar problem I mentioned above,
How many integer solutions for the equation $x_1+x_2+x_3+x_4=28$ are there with $0 \leq x_i \leq 10$
For this solution I used the negation $N(a_1',a_2'a_3'a_4') = N -N(a_i)+N(a_i,a_j)-...$
I let $N=$ all possible solutions. I said it was like distrubting 28 balls in 4 bins so, $\binom{28+4=1}{28}$
the next would term would happen after I use $11$ balls, one more than the range of the question as listed above which is $10$, so we have $\binom{17+4-1}{17}$ and there are $\binom{4}{1}$ also, so $\binom{17+4-1}{17} \binom{4}{1}$
next after using 11 more balls we would have $\binom{6+4-1}{6}$ with $\binom{4}{2}$ so putting together we have $\binom{6+4-1}{6} \binom{4}{2}$
the term in the sequence is not possible since $6-11$ makes us run out of balls
so then the solution is, $\binom{28+4=1}{28} - \binom{17+4-1}{17} \binom{4}{1} + \binom{6+4-1}{6} \binom{4}{2}$
So then how can I use that similar process to solve my question
| How many integers solutions of $$x_1 + x_2 + x_3 + x_4 = 28$$ are there that satisfy $-10 \leq x_k \leq 20$ for $1 \leq k \leq 4$?
We can convert this to an equivalent problem in the nonnegative integers.
$$x_1 + x_2 + x_3 + x_4 = 28 \tag{1}$$
Let $y_k = x_k + 10$, where $1 \leq k \leq 4$. Then each $y_k$ is a nonnegative integer satisfying $0 \leq k \leq 30$. Substituting $y_k - 10$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 - 10 + y_2 - 10 + y_3 - 10 + y_4 - 10 & = 28\\
y_1 + y_2 + y_3 + y_4 & = 68 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $3$ addition signs in a row of $68$ ones. The number of solutions of equation 1 is
$$\binom{68 + 3}{3} = \binom{71}{3}$$
since we must choose which $3$ of the $71$ positions required for $68$ ones and $3$ addition signs will be filled with addition signs.
From these, we must subtract those cases that violate one or more of the restrictions that $y_k \leq 30$, $1 \leq k \leq 4$. Notice that at most two of these restrictions can be violated simultaneously since $3 \cdot 31 = 93 > 68$.
Suppose $y_1 > 30$. Then $y_1' = y_1 - 31$ is a nonnegative integer. Substituting $y_1' + 31$ for $y_1$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2 + y_3 + y_4 & = 68\\
y_1' + y_2 + y_3 + y_4 & = 37 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{37 + 3}{3} = \binom{40}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the four variables that could violate the restriction $y_k \leq 30$. Hence, there are
$$\binom{4}{1}\binom{40}{3}$$
cases in which one of the restrictions is violated.
However, if we subtract this number from the total, we will have subtracted too much since we will have subtracted each case in which two of the variables violate the restrictions twice, once for each variable we designated as the variable that violated the restriction. We only want to subtract those cases once, so we must add them back.
Suppose $y_1, y_2 > 30$. Let $y_1' = y_1 - 31$; let $y_2 = y_2 - 31$. Then $y_1'$ and $y_2'$ are nonnegative integers. Substituting $y_1' + 31$ for $y_1$ and $y_2' + 31$ for $y_2$ in equation 2 yields
\begin{align*}
y_1' + 31 + y_2' + 32 + y_3 + y_4 & = 68\\
y_1' + y_2' + y_3 + y_4 & = 6 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions. By symmetry, there are an equal number of solutions for each of the $\binom{4}{2}$ pairs of variables that could violate the restrictions. Hence, there are
$$\binom{4}{2}\binom{9}{3}$$
cases in which two of the variables violate the restrictions.
Thus, by the Inclusion-Exclusion Principle, the number of solutions that satisfy the restrictions is
$$\binom{71}{3} - \binom{4}{1}\binom{40}{3} + \binom{4}{2}\binom{9}{3}$$
| {
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| What you solved is correct but you must notice that $$\sqrt {x^2-3}\ge 0\Rightarrow x-3\ge 0\Rightarrow x\ge 3$$
And hence I'd you get any solution less than $3$ then you have to reject it.
| {
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Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is
Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have
$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.
Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.
Could some help me to solve it , Thanks
| We can treat the equations as quadratic with variables $a,b,c$, and thus we solve for each of their values using the quadratic formula:
$$\begin{align}
a\to& \frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{1}\\
b\to& \frac{1}{2} \left(\sqrt{208-3 c^2}-c\right)\tag{2}\\
c\to& \frac{1}{2} \left(\sqrt{196-3 a^2}-a\right)\tag{3}\\
\end{align}$$
Since $a,b,c \in \mathbb{R}^+$, then we can discard the negative values of $a,b,c$. We can rewrite $(3)$ in terms of $a$, and we get:
$$a\to \frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)\tag{4}$$
Then we substitute $(4)$ in $(1)$ and we get:
$$\frac{1}{2} \left(\sqrt{196-3 c^2}-c\right)=\frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{5}$$
Using $(2)$ and substituting it in $(5)$, we have:
$$2 \sqrt{196-3 c^2}+\sqrt{208-3 c^2}=\sqrt{6} \sqrt{c^2+\sqrt{208-3 c^2} c-80}+3 c$$
For which we get $c=\frac{58}{\sqrt{91}}$, using that we get $a$ and $b$ from $(4)$ and $(2)$, respectively:
$$\begin{align}
a=&\frac{15}{\sqrt{91}}\\
b=&\frac{18}{\sqrt{91}}\\
\end{align}$$
And thus, we can now solve the problem:
$$\begin{align}\displaystyle \frac{49b^2-33bc+9c^2}{a^2}\Rightarrow &\displaystyle \frac{49(\frac{18}{\sqrt{91}})^2-33(\frac{18}{\sqrt{91}})(\frac{58}{\sqrt{91}})+9(\frac{58}{\sqrt{91}})^2}{(\frac{15}{\sqrt{91}})^2}\\
&\Rightarrow 52
\end{align}$$
| {
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| From here: $(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$ we get $b^2-4ac \geq 0$ and not $>$. So if $b^2-4c<0$ there is no real solution. So else from that $>$ your conclusion is correct.
| {
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Calculating pm problem: $5$ man and $5$ women running in a race Let's say there are $5$ man and $5$ women running in a race. Let the variable X that indicates the highest position in ranking for a woman. So, for example, if $X = 1$, that means that a woman took first place(any woman) and if $X = 6$, that means that women took the last places on the ranking.
I guess, obviously there are $10!$ different outcomes for ranking.
How can I calculate $p_X(x)$ with $x = 1, 2, 3, 4, 5, 6$?
Edit : Two or more people cannot take the same place in the ranking. All 10! different outcomes are equally possible.
| The women are assumed to be indistinguishable. There are ${10 \choose 5}$ ways to place them in $10$ empty slots. You're correct that $6$ is highest value $X$ can take on.
If we are assuming every one of the $10$ people are equally fast, then the probability that a woman is in the first position is $\frac{1}{2}$ by symmetry.
The probability that the best position by a woman is $X=2$ is $$\frac{5}{10}\cdot\frac{5}{9}=\frac{{5 \choose 1}}{{10 \choose 1}}\cdot\frac{{5 \choose 1}}{{9 \choose 1}}$$
since a man has to be in the first position and then a woman has to be in the second position
The probability that the best position by a woman is $X=3$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{5}{8}=\frac{{5 \choose 2}}{{10 \choose 2}}\cdot\frac{{5 \choose 1}}{{8 \choose 1}}$$
since a man has to be in the first and second position and then a woman has to be in the third position
$\vdots$
The probability that the best position by a woman is $X=6$ is $$\frac{5}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}\cdot\frac{5}{5}=\frac{{5 \choose 5}}{{10 \choose 5}}\cdot\frac{{5 \choose 1}}{{5 \choose 1}}$$
Can you go from here to obtain $p_X(x)$?
| {
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Minimum of product of three trigonometric function
If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.
I tried Jensen inequality and AM-GM inequality:
$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+z}{3}\bigg)=\frac{3\sqrt{3}}{2},$$
so
$$ \cos x\cos y\cos z\leq \frac{3\sqrt{3}}{8}.$$
Equality holds when $x=y=z=30^\circ$.
Also,$$ \cos x\cos y\cos z=\frac{1}{4}\bigg[\sin(2x)+\sin(2y)+\sin(2x+2y)\bigg].$$
I am unable to find the minimum of $\cos x\cos y\cos z$.
| The least value of function
$$f(x,y) = \cos x\cos y \cos \left(\dfrac\pi2 - x-y\right),$$
or
$$f(x,y) = \cos x\cos y \sin(x+y),$$
can be achieved in the stationary points of $f(x)$ or in the bounds of the domain.
The stationary points can be defined from the system $f'_x = f'_y = 0,$ or
\begin{cases}
-\sin x \cos y \sin(x+y) + \cos x\cos y\cos(x+y)=0,\\
-\cos x \sin y \sin(x+y) + \cos x\cos y\cos(x+y)=0.
\end{cases}
Subtraction leads to equation
\begin{align}
\sin(x-y)\sin(x+y)=0.
\end{align}
Taking in account the issue conditions, this means
$$y=x,\quad -\sin x \sin 2x + \cos x \cos 2x = 0,\quad x \in[15^\circ, 37.5^\circ],$$
$$\cos 3x =0,\quad y=x,\quad x \in[15^\circ, 37.5^\circ],$$
\begin{align}
x=y=z = 30^\circ,\quad f(30^\circ, 30^\circ) = \dfrac{3\sqrt3}8.
\end{align}
The task is symmetric on $x, y, z,$ so WLOG it is sufficient to consider the bounds $z=15^\circ,\ x+y = 75^\circ,$ and $z= 60^\circ,\ x=y = 15^\circ.$
The first bound gives
$$g(x) = f(x,75^\circ-x) = \cos15^\circ\cos x\cos (75^\circ-x),\quad x\in[15^\circ,\ 60^\circ],$$
$$g(x) = \dfrac12\cos15^\circ(\cos 37.5^\circ + \cos(37.5^\circ - 2x)),$$
with the least value
$$g_m = g(15^\circ) = g(60^\circ) = \cos^2 15^\circ\cos 60^\circ = \dfrac14(1+\cos30^\circ),$$
$$g_m = \dfrac{2+\sqrt3}8 < \dfrac{2\sqrt3+\sqrt3}8 = f(30^\circ, 30^\circ).$$
The second bound gives the same value $g_m.$
Thus, in the issue constraints,
$$\boxed{\min \{\cos x\cos y\cos z\} = \dfrac{2-\sqrt3}8\approx 0.46656}$$
at
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in
\left\{
\begin{pmatrix}15^\circ\\15^\circ\\60^\circ\end{pmatrix},
\begin{pmatrix}15^\circ\\60^\circ\\15^\circ\end{pmatrix},
\begin{pmatrix}60^\circ\\15^\circ\\15^\circ\end{pmatrix}\right\}
$$
| {
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Evaluate : $\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$ I am attempting a calculation for an integral and encounter this series:
$$S=\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$$
but I am stuck and find it difficult to evaluate its sum.
Can someone offer suggestions?
My Attempt:
Factorising:
$$4+8k+6k^2+2k^3 = 2(k+1)(k^2+2k+2)$$
$S=0.3359329927...$
$1=A(k^2+2k+2)+(Bk+C)(2k+2)$
$A=1$
| Observe you have
\begin{align}
4+8k+6k^2+2k^3 =2(k+1)(k+1+i)(k+1-i)
\end{align}
which means
\begin{align}
\frac{1}{4+8k+6k^2+2k^3} = \frac{1}{4i(k+1)(k+1-i)}-\frac{1}{4i(k+1).(k+1+i)}
\end{align}
Then we have
\begin{align}
\sum^\infty_{k=0}\frac{1}{4+8k+6k^2+2k^3} =&\ \frac{1}{4}\sum^\infty_{k=1}\left( \frac{-i}{k(k-i)}+\frac{i}{k(k+i)}\right)\\
=&\ \frac{1}{4}\left( \psi_0(1-i)+\psi_0(1+i)+2\gamma\right)
\end{align}
where $\psi_0$ is the digamma function and $\gamma$ is the Euler–Mascheroni constant.
I have used the fact that
\begin{align}
\psi_0(z+1) = -\gamma+\sum^\infty_{k=1}\frac{z}{k(k+z)}.
\end{align}
| {
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How to find the sum of: $a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+\cdots+[a+(n-1)d]^2?$ Given the arithmetic series with a common difference, d, first term, a, and n is the nth terms.
Deriving the formula for $S_n$
$$S_n=a+(a+d)+(a+2d)+(a+3d)+\cdots+[a+(n-1)d]\tag1$$
Reverse the sum
$$S_n=[a+(n-1)d]+\cdots+(a+d)+a\tag2$$
$(1)+(2)$
$$2S_n=n[a+(n-1)d]$$
$$S_n={n\over 2}[a+(n-1)d]$$
How do we go about to find the square sum of the arithmetic series?
$$T_n=a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+\cdots+[a+(n-1)d]^2\tag3$$
| The only method that I can think of to do it is by using the results that
$$
1 + 2 + \dots + n = \frac{n(n + 1)}{2}
$$
and
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.
$$
We then have that
$$ \begin{align}
T_n & = a^2 + (a + d)^2 + \dots + (a + (n - 1)d)^2 \\
& = a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) + \dots + (a^2 + 2(n - 1)ad + (n - 1)^2 d^2) \\
& = na^2 + 2ad(1 + 2 + \dots + (n - 1)) + d^2 (1^2 + 2^2 + \dots + (n - 1)^2) \\
& = na^2 + 2ad \frac{(n-1)n}{2} + d^2 \frac{(n - 1)n(2n - 1)}{6} \\
& = na^2 = n(n - 1) ad + \frac{1}{6} n(n - 1)(2n - 1) d^2.
\end{align} $$
The easiest way to find the formula for the sum
$$
1^2 + 2^2 + \dots + n^2
$$
is by considering
$$
(n + 1)^3 - n^3 = 3n^2 + 3n + 1.
$$
We then have that
$$ \begin{align}
(n + 1)^3 & = \left( (n + 1)^3 - n^3 \right) + \left( n^3 - (n - 1)^3 \right) + \dots + \left( 1^3 - 0^3 \right) \\
& = (3n^2 + 3n + 1) + (3(n - 1)^3 + 3(n - 1) + 1) + \dots + (3 \cdot 0^3 + 3 \cdot 0 + 1) \\
& = 3(0^2 + 1^2 + 2^2 + \dots + n^2) + 3(0 + 1 + 2 + \dots + n) + (1 + 1 + \dots + 1) \\
& = 3(1^2 + 2^2 + \dots + n^2) + 3 \frac{n(n + 1)}{2} + (n + 1).
\end{align} $$
It follows that
$$
1^2 + 2^2 + \dots + n^2 = \frac{1}{3} (n + 1)^3 - \frac{1}{2} n(n + 1) - \frac{1}{3} (n + 1) = \frac{n + 1}{6} \left(2(n + 1)^2 - 3 n - 2 \right) = \frac{(n + 1) (2n^2 + n)}{6} = \frac{n(n + 1)(2n + 1)}{6}.
$$
Alternatively, if someone told you what the formula is, you could prove that it is correct by induction.
| {
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We throw a die $8$ times. What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s? We throw a die $8$ times. What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s?
My work:
The sample space $ S:$ "The set of all solutions of throwing the die [eight times]" and $|S|:6^8$
Then the probability of getting two $3$s, three $1$s, three $6$s is $\frac{8}{6^8}$
Is the reasoning correct?
| To find the number of ways to get two 3s, three 1s and three 6s can also be seen as a simple permutation problem of $8$ digits.
Imagine the result of the $8$ throws is recorded as a sequence of $8$ numbers: $d_1 d_2 \ldots d_8$.
Now you want to know in how many different ways can occur the numbers $3,3,1,1,1,6,6,6$ in such a sequence.
Or, in other words, how many different 8-digit numbers can you create with the digits $3,3,1,1,1,6,6,6$.
*
*there are $8!$ permutations (arrangements) of these $8$ digits but
*each permutation of $3,3$ and $1,1,1$ and $6,6,6$ gives the same 8-digit number, so you get
$$\frac{8!}{2!\cdot 3! \cdot 3!} \mbox{ differnt 8-digit numbers or different throws}$$
So, the probability in question is
$$P = \frac{8!}{2!\cdot 3! \cdot 3!}\cdot \frac{1}{6^8}$$
| {
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Why am I getting a negative sum? $\frac{1}{1\cdot2} - \frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+ \cdots$ The infinite sum I want to find.
$$\frac{1}{1\cdot2} - \frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+ \cdots$$
My attempt at it
$$\begin{align} &\;\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)} - \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)} \\[4pt]
=&\;\sum_{n=1}^{\infty} \left(\frac{1}{2n-1} - \frac{1}{2n}\right) - \sum_{n=1}^{\infty} \left(\frac{1}{2n} - \frac{1}{2n+1}\right) \\[4pt]
=&\;\sum_{n=1}^{\infty} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{\infty} \frac{1}{2n} \\[4pt]
=&\;\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots\right) +
\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right)
- 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots\right) \\[4pt]
=&\;1 + 2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\right) -
\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots\right) \\[4pt]
=\;&-1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \\[4pt]
=\;&\ln 2 - 1
\end{align}$$
The answer I got here is negative, while the series seems to be positive. What is the mistake here?
| Yes, the sum is positive. Note that instead of
$$= \sum_{n=1}^{\infty} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{\infty} \frac{1}{2n}$$
which is indefinite since the series are divergent, you should have the limit as $N$ goes to $\infty$ of
$$\sum_{n=1}^{N} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{N} \frac{1}{2n}\\
=-1+2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2N-1}\right) +\frac{1}{2N+1} - 2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots+\frac{1}{2N}\right)$$
which is a different thing. What is its limit as $N$ goes to $\infty$?
More simply, you may consider the partial sum and then take the limit as $N$ goes to $\infty$
$$S_N=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n(n+1)}=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\sum_{n=1}^{N}\frac{(-1)^{n+2}}{n+1}=
\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\sum_{n=2}^{N+1}\frac{(-1)^{n+1}}{n}\\
=-1+2\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}+\frac{(-1)^{N+2}}{N+1}\to -1+2\ln(2)>0.$$
| {
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"url": "https://math.stackexchange.com/questions/2719878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$
What to do next?
Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.
Is there any trick to such genre of problems?
| We have $x\ge0$, $y\ge0$ and $z\ge0$.
One possible solution is $x=y=z=0$.
If one of $x,y,z$ is non-zero, the other two are also non-zero.
Now suppose that all of them are non-zero.
$$xyz=\frac{8x^2y^2z^2}{(1+x^2)(1+y^2)(1+z^2)}\le\frac{8x^2y^2z^2}{(2x)(2y)(2z)}=xyz$$
The equality holds if and only if $x^2=y^2=z^2=1$ and this would imply that $x=y=z=1$.
There are totally two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| Hint. Note that
$$1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<1+\sum_{n=2}^{2018}\frac{1}{n^3-n}$$
and the sum on the right is easy to be found (it's telescopic).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
2x^2-4y^2-\frac{3}{2}x+y=0 \\
3x^2-6y^2-2x+2y=\frac{1}{2}
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get that $x=\frac{1}{2}$ and that $y=\frac{3}{4}$ but when I put it back into the original systems equation I dont get the right answer. Can somebody explain why?
| If $2x^2-4y^2-\frac32x+y=0$ and $3x^2-6y^2-2x+2y=\frac12$, then$$3\times(2x^2-4y^2-\frac32x+y)-2\times(3x^2-6y^2-2x+2y)=-1;$$in other words, $-\frac x2-y=-1$. So, replace $y$ with $1-\frac x2$ in the first equation, and you'll get a quadratic equation whose roots are $-3$ and $1$. So, the solutions of the system are $\left(-3,\frac52\right)$ and $\left(1,\frac12\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\sin x \cos x -2x}{2x \sin^2 x+ 2x^2\sin x \cos x})$
=$\lim_{x\to 0}(\frac{\sin 2x -2}{2x \sin^2 x+ x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+2x\sin x \cos x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ 2(x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x+x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+x\sin 2x+ x\sin 2x+x^2\cos 2x})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+2x\sin 2x+x^2\cos 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-2\sin 2x}{2\sin x \cos x+2(\sin 2x+ 2x\cos 2x)+(2x\cos 2x-2x^2\sin 2x})$
=$\lim_{x\to 0}(\frac{-2\sin 2x}{3\sin 2x+6x\cos 2x-2x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-4\cos 2x}{6\cos 2x+6(\cos 2x-2x\sin 2x)-2(2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{-4\cos 2x}{12\cos 2x-12x\sin 2x-4x\sin 2x-4x^2\cos 2x)})$
=$\frac{-4\cos 0}{12 \cos 0-0-0-0}$
=$\frac{-4}{12}$
=$\frac{-1}{3}$
My problem: This method is very lengthy and involves a lot of calculations. Is there is any other (better and efficient) method to evaluate this limit?
| $$=\left(\frac{x}{\sin x}\right)^2\left(\frac{\sin x}{x}+1\right)\left(\frac{\sin x -x}{x^3}\right)\to 1^2(1+1)\left(-\frac{1}{6}\right)=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ternary strings with 3 or more consecutive 2's I'm trying to figure out how to find the number of ternary strings of length $n$ that have 3 or more consecutive 2's.
So far I've been able to establish that there is $n(2^{n-1})$ with a single 2.
And I think (but am not certain) that this can be extrapolated to the number of strings with a single group of 2's of length $x$ by:
$$\bigl(n-(x-1)\bigr)(2^{n-x})$$
What I'm getting caught on is the 'or more part', any help would be greatly appreciated.
|
We consider the alphabet $V=\{0,1,2\}$ and we are looking for the number $c_n$ of strings of length $n$ having runs of $2$ less than three. The wanted number is $$a_n=3^n-c_n$$
We derive a generating function $C(z)=\sum_{n=0}^\infty c_nz^n$ from which the number $a_n$ can be obtained easily since
\begin{align*}
a_n=[z^n]\left(\frac{1}{1-3z}-C(z)\right)\tag{1}
\end{align*}
with $[z^n]$ denoting the coefficient of $z^n$ of a series.
Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
A generating function for the number of Smirnov words over a ternary alphabet $V$ is given by
\begin{align*}
\left(1-\frac{3z}{1+z}\right)^{-1}\tag{2}
\end{align*}
Replacing occurrences of $0$ in a Smirnov word by one or more zeros generates words having runs of $0$ with length $\geq 1$.
\begin{align*}
z\longrightarrow z+z^2+z^3+\cdots=\frac{z}{1-z}
\end{align*}
The same can be done with the digit $1$.
Replacing occurrences of $2$ in a Smirnov word by one or two $2$'s generates words with runs of $2$ of length less than three.
\begin{align*}
z\longrightarrow z+z^2=z(1+z)
\end{align*}
The resulting generating function is according to (2)
\begin{align*}
\color{blue}{C(z)}=\left(1- 2\cdot\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}-\frac{z(1+z)}{1+z(1+z)}\right)^{-1}
&\color{blue}{=\frac{1+z+z^3}{1-2z-2z^2-2z^3}}\tag{3}
\end{align*}
We obtain for $n\geq 3$ the number of wanted words of length $n$ according to (1) and (3) as
\begin{align*}
\color{blue}{a_n}&=3^n-c_n\\
&=[z^n]\left(\frac{1}{1-3z}-\frac{1+z+z^3}{1-2z-2z^2-2z^3}\right)\\
&=[z^n]\left(\color{blue}{1}z^3+\color{blue}{5}z^4+\color{blue}{21}z^5+\color{blue}{81}z^6+\color{blue}{295}z^7+\color{blue}{1\,037}z^8+\color{blue}{3\,555}z^9+\cdots\right)
\end{align*}
whereby the last line was obtained with some help of Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $ For $k>0 $
$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $,
the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $
My attempt here was not complying with any of the options given as my attempt was to find the min. value for the series which the question does not ask. What to do?
Options are:
A. 198
B. 197
C. 196
D. 195
I know $ \sum_{k=1}^{n} \frac{1}{k} \approx log n $ But this does not help in this case.
Also $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} < 2(\sqrt{10000}-\sqrt{2}) $
This also does not help.
| It follows from
$$ \frac{1}{2\sqrt{k+1}} \le \sqrt{k+1} - \sqrt{k} \le \frac{1}{2\sqrt{k}}$$
that
\begin{equation}
\sum_{k=1}^{9998}\frac{1}{2\sqrt{k+1}} = \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}
\le \sqrt{9999}-1
\end{equation}
and
\begin{equation}
\sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}.
\end{equation}
Thus we conlcude that
$$ 98.5858 \approx \sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}
\le \sqrt{9999} - 1 \approx 98.9950 $$
which gives you the answer as
$$ 197.1716 \le \sum_{k=2}^{9999}\frac{1}{\sqrt{k}} \le 197.9900.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integral of a piecewise function
Check whether the integral $\int_{-1}^2{f(x)[2x+1]}dx$ exists where
$$f(x)=
\begin{cases}
x+2, &\text{$x<0$}\\
1, & \text{$x=0$}\\
xe^{x^2},& \text{$x>0$}
\end{cases}$$
and $[x]$ denotes the greatest integer function. If it exists then calculate the integral.
I'm clueless about how to do the whole proof by definition stuff. Is there a way to prove that this function is integrable without any epsilon/delta and partition proofs?
| Because of the fact that the discontinuities of $f(x)$ are of measure zero, the function is Riemann integrable. A proof of this exists here. This means we can validly take the integral of $f(x)$ on either side of the discontinuity and sum those:
$$\int_{-1}^{2} f(x)[2x+1]dx = \int_{-1}^{0}(x+2)[2x+1]dx + \int_{0}^{2}xe^{x^2}[2x+1]dx$$
Next, because $[2x+1]$ is the same value on for any x in the interval $[n, n+\frac{1}{2})$ where $n \in \mathbb{Z}$ and these new discontinuities are also of measure zero, we can seperate this into a larger sum with each value of $[2x+1]$ written out.
$$-1 \cdot \int_{-1}^{-\frac{1}{2}} (x+2)dx + 0 \cdot \int_{-\frac{1}{2}}^{0} (x+2)dx + 1 \cdot \int_{0}^{\frac{1}{2}} xe^{x^2}dx + 2 \cdot \int_{\frac{1}{2}}^{1} xe^{x^2}dx + 3 \cdot \int_{1}^{\frac{3}{2}} xe^{x^2}dx + 4 \cdot \int_{\frac{3}{2}}^{2} xe^{x^2}dx$$
Now we remove the term that is multiplied by $0$ and find $\int_{a}^{b} xe^{x^2}dx$ by setting $u=x^2$ and finding that $du=2xdx$.
$$-\int_{-1}^{-\frac{1}{2}} (x+2)dx + \frac{1}{2}\int_{0}^{\frac{1}{4}} e^{u}du + \int_{\frac{1}{4}}^{1} e^{u}du + \frac{3}{2}\int_{1}^{\frac{9}{4}} e^{u}du + 2\int_{\frac{9}{4}}^{4} e^{u}du$$
Finally we take the antiderivatives and evaluate.
$$-\left[\frac{x^2}{2}+2x\right]_{-1}^{-\frac{1}{2}}+\frac{1}{2}[e^u]_{0}^{1/4}+[e^u]_{1/4}^{1}+\frac{3}{2}[e^u]_{1}^{9/4}+2[e^u]_{9/4}^4$$
$$=-\left(\frac{1}{8}-1-\frac{1}{2}+2\right)+\frac{e^{1/4}-1}{2}+\left(e-e^{1/4}\right)+\frac{3e^{9/4}-3e}{2}+\left(2e^4-2e^{9/4}\right)$$
$$=-\frac{9}{8}-\frac{e^{1/4}}{2}-\frac{e}{2}-\frac{e^{9/4}}{2}+2e^4$$
$$=\frac{16e^4-4e^{9/4}-4e-4e^{1/4}-9}{8} \approx 101.326$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part
$$\sum_{n=1}^{\infty} \left ( \frac{1}{2n-1} + \frac{1}{2n+1} \right )$$
diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.
\begin{align*}
\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\
&= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}}
\end{align*}
I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?
Of course alternatives are welcome.
| Notice
$$\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right)
= \sum_{n=1}^p \left[\frac2n - \left( \frac{1}{2n-1} + \frac{2}{2n} + \frac{1}{2n+1} \right)\right]\\
= 2H_p - \left(1 + \sum_{n=2}^{2p} \frac{2}{n} + \frac{1}{2p+1}\right)
= 2(H_p - H_{2p}) + 1 - \frac{1}{2p+1}$$
where $H_p = \sum_{n=1}^p \frac{1}{n}$ is the $p^{th}$ harmonic number.
Since $\lim\limits_{p\to\infty} (H_p - \log p) = \gamma$, we have
$$\begin{align}\lim_{p\to\infty} (H_p - H_{2p})
&= \lim_{p\to\infty} \left((H_p - \log p) - (H_{2p} - \log(2p)) - \log 2\right)\\
&= \gamma - \gamma - \log 2 = -\log 2\end{align}$$
As a result,
$$\sum_{n=1}^\infty \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right) = 2(-\log 2)+1 - 0 = 1 - 2\log 2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Why is $P(X=2.5)=0?$ The discrete random variable X has cdf
$$\left\{
\begin{array}{rcr}
0 & = & \text{for} \ x<1 \\
1/4 & = & \text{for} \ 1\le x<2 \\
3/4 & = & \text{for} \ 2\le x<3 \\
1 & = & \text{for} \ x\ge 3
\end{array}
\right.$$
Find a) $P(X=1)$, b) $P(X=2)$, c) $P(X=2.5)$, d) $P(x\leq2.5).$
I found a), b) and d) easy by drawing out the peacewise cdf. I fail to se why $P(X=2.5)=0$. If $2.5\in 2\le x<3,$ shouldn't the answer be the same as in b), that is $\frac{3}{4}-\frac{1}{4} = \frac{1}{2}?$
| It might help to express this as a probability mass function.
$$P_X(x)=\left\{
\begin{array}{rcr}
1/4 & x=1 \\
2/4 & x=2 \\
1/4 & x=3 \\
0 & \text{otherwise}
\end{array}
\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using Lagrange Multipliers to Find Extrema of $f$ I have been trying to solve this lagrange problem for an hour with no result. I can't seem to isolate any variable.
$$f(x,y) = y^2 -4xy +4x^2$$
With constraint:
$$x^2 + y^2 = 1$$
I have found $x$ to = $-\left(\frac{λ-4}{2}\right)^2 + 1$. Using the constraint $x$ is also equal to $\sqrt{1-y^2}$. This still leaves me with two unknowns. I have tried solving for lambda and setting them equal to each other and I end up with $x^2 = y^2-\frac{3xy}{2}$.
There are still two unknown variables here. I am not sure what I should be isolating to get an answer to work with.
| You are given $f(x,y)= y^2- 4xy+ 4x^2$ with constraint $g(x, y)= x^2+ y^2= 1$. Then $\nabla f= <-4y+ 8x, 2y- 4x>$ and $\nabla g= <2x, 2y>$. The basic idea of the Lagrange multiplier method is that those two vectors are in the same direction- one is a multiple of the other: $<-4y+ 8x, 2y- 4x>= \lambda<2x, 2y>$. Equating the same components, we have the two equations $4x+ 8y= 2\lambda x$ and $2y- 4x= 2\lambda y$. Together with the constraint $x^2+ y^2= 1$ that gives three equations for the three unknowns x, y, and $\lambda$.
But a value for $\lambda$ is not necessary for this problem so a good way to solve these equations is to first eliminate $\lambda$ by dividing one of the first two equations by the other: $\frac{4x+ 8y}{2y- 4x}= \frac{2x+ 4y}{y- 2x}= \frac{x}{y}$. From that $y(2+ 4y)= x(y- 2x)$ so $2y^2+ 4xy= xy- 2x^2$, $2y^2+ 3xy+ 2x^2= 0$. Completing the square, $2(x^2+ (3/2)xy+ (9/16)y^2-$$(9/16)y^2)+ y^2= 2(x+ (3/4)y)^2- (9/8)y^2= 0$. x+ (3/4)y= (3/4)y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When has weighted sum of two specific unimodal functions two peaks? Consider the functions
$f(x) = x e^{-x}$ and $g(x) = xe^{-kx}$ for $x \geq 0$, where $e$ is the Euler constant and $k > 1$. It is easy to check that $f(x)$ is unimodal and has its global maximum at $x_f^* = 1$, while $g(x)$ is also unimodal and has its global maximum at $x_g^* = \dfrac{1}{k} \in (0,1)$. Notice further that both functions are strictly concave up to $\tilde{x}_f = 2$ (for $f(x)$) and $\tilde{x}_g = \dfrac{2}{k}$ (for $g(x)$), after which they are strictly convex.
What I am interested in is the following: Take the weighted sum of the functions
$$h_{\lambda}(x) := (1-\lambda)f(x) + \lambda g(x),$$
where $0 < \lambda < 1$. For which values of $k$ can we guarantee that there exists a range of weights $\lambda$ such that $h_{\lambda}(x)$ has two peaks?
What we can easily deduce is that for every $\lambda$, $h_\lambda(x)$ is strictly increasing on $[0, 1/k]$ and strictly decreasing on $[1, \infty)$. Moreover, due to the concavity property of the individual functions, $h_{\lambda}(x)$ is clearly strictly concave on $[0, 2/k]$. Taken together, this implies that for $k \leq 2$, $h_{\lambda}(x)$ is unimodal and has a single peak somewhere in $(1/k, 1)$.
What about the case where $k > 2$? Using Mathematica, it appears that two peaks can only emerge for $k \gtrapprox 6$. But can we somehow show this analytically? Or at least prove that for $k$ sufficiently large, the function has two peaks when $\lambda$ is set appropriately? Many thanks in advance!
| $\def\e{\mathrm{e}}$Note that between any two adjacent local maximum points, there exists exactly one local minimum point, and $f(x) = (1 - λ)x\e^{-x} + λx\e^{-kx}$ is increasing on $\left[ 0, \dfrac{1}{k} \right]$ and decreasing on $[1, +\infty]$, thus$$
f \text{ has two peaks} \Longleftrightarrow f' \text{ has three zeros}.
$$
Also, the zeros of $f'$ lie on $\left( \dfrac{1}{k}, 1 \right)$, and $f'(x) = (1 - λ)(1 - x) \e^{-x} + λ(1 - kx) \e^{-kx}$, thus\begin{align*}
\mathrel{\phantom{\Longleftrightarrow}}{} f' \text{ has three zeros} &\Longleftrightarrow \frac{kx - 1}{1 - x} \e^{-(k - 1)x} = \frac{1 - λ}{λ} \text{ has three roots}\\
&\Longleftrightarrow \ln(kx - 1) - \ln(1 - x) - (k - 1)x = \ln\frac{1 - λ}{λ} \text{ has three roots}.
\end{align*}
Define $g(x) = \ln(kx - 1) - \ln(1 - x) - (k - 1)x$. Because the range of $\dfrac{1 - λ}{λ}$ is $(0, +\infty)$ given that $λ \in (0, 1)$, then the range of $\ln\dfrac{1 - λ}{λ}$ is $(-\infty, +\infty)$. Therefore,\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} \exists λ \in (0, 1): g(x) = \ln\frac{1 - λ}{λ} \text{ has three roots}\\
&\Longleftrightarrow \exists c \in \mathbb{R}: g(x) = c \text{ has three roots}\\
&\Longleftrightarrow g' \text{ has two zeros}.
\end{align*}
Now, because\begin{align*}
g'(x) &= \frac{k}{kx - 1} + \frac{1}{1 - x} - (k - 1)\\
&= (k - 1) \frac{kx^2 - (k + 1)x + 2}{(kx - 1)(1 - x)}, \quad x \in \left( \frac{1}{k}, 1\right)
\end{align*}
thus\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}}{} g' \text{ has two zeros} \Longleftrightarrow kx^2 - (k + 1)x + 2 = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right).
\end{align*}
Define $h(x) = kx^2 - (k + 1)x + 2$. Note that the symmetry axis of the graph of $y = h(x)$ is $x = \dfrac{k + 1}{2k}$ and $\dfrac{1}{k} < \dfrac{k + 1}{2k} < 1$, also $h\left( \dfrac{1}{k} \right) = h(1) = 1 > 0$, thus\begin{align*}
h(x) = 0 \text{ has two roots on } \left( \frac{1}{k}, 1 \right) &\Longleftrightarrow Δ = (k + 1)^2 - 8k > 0\\
&\stackrel{k > 1}{\Longleftrightarrow} k > 3 + 2\sqrt{2} ≈ 5.828427.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$
| Gibbs has already presented a perfect proof. However, there is another one, that does not involve decomposition of $n^3-n$ and is relatively straightforward.
To say $n^3-n$ is divisible by $6$ is the same as saying $n^3-n \equiv 0 \pmod 6$. Thus, once we compute $n^3-n \pmod 6$ for all possible $n$, we are done. By general algebra, this is the same as computing $n^3-n$ for all elements of the ring $\mathbb Z / 6 \mathbb Z$, and there are exactly $6$ elements in this ring, namely the equivalence classes of $0,1,\dots,5$ modulo $6$.
So, let's compute this $6$ values $\mod 6$:
$$
\begin{array}{ l|c }
n & n^3-n \\ \hline
0 & 0-0 \equiv 0 \pmod 6\\ \hline
1 & 1-1 \equiv 0 \pmod 6 \\ \hline
2 & 8-2 \equiv 6 \equiv 0 \pmod 6 \\ \hline
3 & 27-3 \equiv 24 \equiv 0 \pmod 6 \\ \hline
4 & 64 - 4 \equiv 60 \equiv 0 \pmod 6 \\ \hline
5 & 125 - 5 \equiv 120 \equiv 0 \pmod 6
\end{array}
$$
As you can see, $n^3-n\equiv 0 \pmod 6$ holds for any $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An inequality with condition I have a new inequality this is the following :
Let $x,y,z$ be real strictly positive number such as :
$$-2 = - x y z + x + y + z $$
Then we have :
$$\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\geq \frac{9}{\sqrt{15+((x+1)(y+1)(z+1))^{(\frac{1}{3})}}}$$
My try :
I put the following substitution :
$a^3=x$;
$b^3=y$;
$c^3=z$
And I have tried Holder (a generalized version) to get :
$$((a^3+1)(b^3+1)(c^3+1))^{(\frac{1}{3})}\geq abc+1 $$
The inequality becomes :
$$\sqrt{\frac{1}{a^3}}+\sqrt{\frac{1}{b^3}}+\sqrt{\frac{1}{c^3}}\geq \frac{9}{\sqrt{15+(abc+1)}}$$
And
$$-2=a^3b^3c^3+a^3+b^3+c^3$$
After that I'm stuck...There is someone to achieve this ?
| The condition gives $\sum\limits_{cyc}\frac{1}{x+1}=1.$
Let $\frac{1}{x+1}=\frac{a}{3},$ $\frac{1}{y+1}=\frac{b}{3}$ and $\frac{1}{z+1}=\frac{c}{3}.$
Hence, $a+b+c=3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{15+\frac{3}{\sqrt[3]{abc}}}}$$ or
$$\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq\sqrt{\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}}.$$
Now, by C-S
$$\sum_{cyc}\sqrt{a(a+b)(a+c)}=\sqrt{\sum_{cyc}\left(a(a+b)(a+c)+2\sqrt{a(a+b)(a+c)}\sqrt{b(a+b)(b+c)}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc)\right)}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)\geq\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$27u^3+9w^3\geq\frac{81(9uv^2-w^3)}{15+\frac{3u}{w}},$$
which is a linear inequality of $v^2$.
Hence, it's enough to prove the last inequality for an extreme value of $v^2$,
which happens for an equality case of two variables.
Since the last inequality is homogeneous, we can assume that $b=c=1$.
Also, let $a=t^3$.
Id est, we need to prove that
$$(t^3+15t+2)(t^9+6t^6+21t^3+8)\geq162t(t^3+1)^2$$ or
$$(t-1)^2\left(t^{10}+2t^9+18t^8+42t^7+66t^6+18t^5+3t^4-12t^3-36t^2-10t+16\right)\geq0,$$
which is very strong, but true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding density function of random variable $Y=X^2$
Suppose $X$ is continuous r.v with c.d.f
$$ F(x) = \begin{cases} 1 - \left( \frac{2}{x} \right)^2 \; \; \; \;
\; \; x>2 \\ 0, \; \; \; \; \text{otherwise} \end{cases} $$
Put $Y=X^2$. Please find density function of $Y$.
Attempt
By definition $f_Y(y) = \frac{d}{dy} F_Y(y) $. We know
$$ F_y(y) = P(Y \leq y ) = P(X^2 \leq y ) = P(X \leq \sqrt{y} )$$
And hence
$$ F_y(y) = \int\limits_2^{\sqrt{y}} 1 - \left( \frac{2}{x} \right)^2 = \sqrt{y} + \frac{4}{\sqrt{y}}- 4 $$
It follows that
$$ f_Y(y) = \frac{1}{2 \sqrt{y}} - \frac{4}{2 y^{2/3} } $$
But,
my answer key say $f_Y(y) = \frac{4}{y^2}$.
What am I doing wrong here?
| You were already given the CDF of $X$. Integrating again is not necessary. We simply write $$F_Y(y) = F_X(\sqrt{y}) = 1 - \left(\frac{2}{\sqrt{y}}\right)^2 = 1 - \frac{4}{y}, \quad y > 4,$$ thus $$f_Y(y) = \frac{4}{y^2}$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $P(X + 10/X > 7 )$ where $X$ is uniform
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?
Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \left( X + \frac{10}{X} > 7 \right) = P[(X-5)(X-2)>0] = P[ \{X>2 \} \cup \{ X < 5 \} ] $$
Since we have two cases: either $X>5$ and $X>2$ OR $X<5$ and $X<2$. now, using inclusion-exclusion we get
$$ P(X>2) + P(X<5) - P(2<X<5) = 1 - F(2) + F(5) - F(5) + F(2) = 1$$
What is wrong with my solution? According to my solution key, I should get $\frac{7}{10}$. What am I missing here?
| Note that we get
$$P[(X-2)(X-5)>0] = P[\{X<2\}\cup\{X>5\}]=1-P[2<X<5],$$
because the parabola $(x-2)(x-5)$ is positive for $x<2$ and $x>5$. Taking the complement yields the last step. Since $X$ is uniformly distributed on $[0,10]$, we know $P[2<X<5] = \frac{3}{10}$, so
$$P\left(X+\frac{10}{X}>7\right) = 1-\frac{3}{10} = \frac{7}{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integrate $\int \ln(x^2 +1)\ dx$ $$\int \ln(x^2 +1)\ dx$$
I done it using integration by parts where
$\int u\ dv = uv - \int v\ du$
Let $u$ = $\ln(x^2 +1)$
$du = \frac{2x}{x^2+1} dx $
Let $dv = dx$ so $v=x$
$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $
I integrate $2\int \frac{x^2}{x^2+1} $ separately using substitution.
Let $u$ = $x^2 + 1$ -> $dx= \frac{1}{2x} du $
substitute $u$ back into it I get a simplified $\int \frac{x}{x^2 + 1} dx$
I used substitution again and got $ \frac{1}{2} \int \frac{1}{u} du$
When I sub it back into the original question, this is my final answer,
$x\ln(x^2 +1) + \frac{1}{2} \ln(x^2 +1) + C$
However, this answer is wrong,
The answer is, $x\ln(x^2 +1) - 2x + 2\tan^{-1} (x) + C $
I believe my integration of $2\int \frac{x^2}{x^2+1} $ is wrong. Where did I went wrong ?
| $2\int \frac{x^2}{x^2+1} $
Does not equal the expression with just an x in the numerator after the substitution .
You can add subtract 1 in numerator to integrate $2\int \frac{x^2}{x^2+1} $
This is wrong :- $2\int \frac{x^2}{x^2+1} $
This is what you get after applying by-parts :- $2\int \frac{x^2}{x^2+1} dx $
Since differential of x is $dx$ .
In by parts formula we have differential of $u $ not it's differenciation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| $\dfrac{c^4}x+\dfrac{s^4}y=\dfrac 1{x+y}\iff(c^4y+s^4x)(x+y)=xy\iff s^4x^2+\underbrace{(c^4+s^4-1)}_{-2s^2c^2}xy+c^4y^2=0$
Thus $(s^2x-c^2y)^2=0\iff y=\tan(\alpha)^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} ) $ Let $ a \in \mathbb{R} $ so that the sequence $ (x_{n})_{n\geq 1} $
$ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} ) $ is bounded.
Find $\lim_{n\to\infty} (x_{n}) $
If $ (x_{n}) $ is bounded that means there are $ 2 $ real numbers $m$ and $q$ let's say so that $ m \leq (x_{n}) \leq q $
Is there a more simple way to approach this problem? , other than writing things like :
$ \frac{\sum_{k=1}^{n}\frac{1}{2k-1}-m}{\sum_{k=1}^{n}\frac{1}{k}} \geq a \geq \frac{\sum_{k=1}^{n}\frac{1}{2k-1}-q}{\sum_{k=1}^{n}\frac{1}{k}}$
The correct answer should be $ ln2 $
| Let me give you a sketch of the argument.
When you increase $n$ by $1$, what happens? In net,
$$
x_{n+1}-x_n=\frac{1}{2(n+1)-1}-a\left(\frac{1}{n+1}\right)=\frac{1}{2n+1}-\frac{a}{n+1}.
$$
Using this, you can show that it must be the case that $a=\frac{1}{2}$. Why? If $a=\frac{1}{2}+\delta$ where $\delta>0$, then
$$
x_{n+1}-x_n=\frac{1}{2n+1}-\frac{a}{n+1}=\frac{1}{2n+1}-\frac{1}{2n+2}-\frac{\delta}{n+1}=-\frac{\delta}{n}+O\left(\frac{1}{n^2}\right),
$$
and so the tail of the sum behaves like $-\delta\sum\frac{1}{n}$ (which must diverge to $-\infty$). Similarly, you can show that if $a=\frac{1}{2}-\delta$ where $\delta>0$, then the series must diverge to $+\infty$.
Now, we know that $a=\frac{1}{2}$. Now, note that we can write
$$
\begin{align*}
x_n&=\left(1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\\
&=y_{2n}-y_n,\qquad y_n:=1+\frac{1}{2}+\cdots+\frac{1}{n}.
\end{align*}
$$
Now, we know (or if not, you can show) that as $n\to\infty$, $y_n=\ln n+\gamma+O\left(\frac{1}{n}\right)$, where $\gamma$ is the Euler–Mascheroni constant, so that
$$
x_n=y_{2n}-y_n=(\ln(2n)+\gamma)-(\ln(n)+\gamma)+O\left(\frac{1}{n}\right)\to\ln2\text{ as }n\to\infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem? How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$
using the tensor product and the Clebsch Gordan Theorem?
| We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right) \right) ^{2}-48\, \left(
\cos \left( \theta \right) \right) ^{4}+64\, \left( \cos \left(
\theta \right) \right) ^{6}
$$
expanding the right-Hand side we get
$$-1+12\, \left( \cos \left( \theta \right) \right) ^{2}-48\, \left(
\cos \left( \theta \right) \right) ^{4}+64\, \left( \cos \left(
\theta \right) \right) ^{6}
$$
this is the same!
$$\cos(2x)=2\cos(x)^2-1$$
$$\cos(4x)=8\cos(x)^4-8\cos(x)^2+1$$
$$\cos(6x)=32\cos(x)^6-48\cos(x)^4+18\cos(x)^2-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the $x$ value for when the expression is takes the minimum value
Let $E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}.$ Find the $x$ value for when this expression take it's minimum value.
Using the derivative it's pretty hard... and I think not indicated for this problem... I thought about using the inequality of means. First, rewrite:
$E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}$ as $E(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}$..
But I don't know how to continue...
| Using Minkowski's inequality:
$$\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}=\\
\sqrt{(x-1)^2+2^2}+\sqrt{(4-x)^2+3^2}\ge \\
\sqrt{(x-1+4-x)^2+(2+3)^2}=\sqrt{3^2+5^2}=\sqrt{34}.$$
The equality occurs when $(x-1,2)||(4-x,3)$, that is:
$$\frac{3}{2}(x-1)=4-x \Rightarrow x=\frac{11}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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EDL2 : $(2x+1)y''+(4x-2)y'-8y=0$
Solve this equation :
$(2x+1)y''+(4x-2)y'-8y=0 \qquad $ on $ \left]\dfrac{1}{2},+\infty\right[$ and on $\mathbb{R}$
Attempt :
First step : $\varphi_1$
The first step was to find a solution in that form : $\varphi_1(x)=ax^2+bx+c$
I found $\varphi_1(x)= 4x^2+1$
Second step : $\varphi_2$
We consider now this equation : $y''+\dfrac{4x-2}{2x+1}y'-\dfrac{8}{2x+1}y=0$
Let $W(x)=\begin{array}{|ll|}\varphi_1(x)&\varphi_2(x)\\\varphi_1'(x)&\varphi_2'(x)\end{array}$
As we know that $W'(x)=-\dfrac{4x-2}{2x+1}W(x)$ we have got $\displaystyle W(x) =\exp\left[\int_{x_0}^x -\dfrac{4s-2}{2s+1}\cdot ds\right]=\lambda(2x+1)^2e^{-2x}$
And as $\left(\dfrac{\varphi_2(x)}{\varphi_1(x)}\right)' =\dfrac{W(x)}{(\varphi_1(x))^2}$
So now I must compute $\displaystyle \int \dfrac{W(x)}{(\varphi_1(x))^2}\cdot dx = \int \dfrac{\lambda(2x+1)^2}{(4x^2+1)^2}\cdot e^{-2x}\cdot dx=\gamma(x)$
Thus $\dfrac{\varphi_2(x)}{\varphi_1(x)}=\gamma(x)\iff \varphi_2(x)=\varphi_1(x)\cdot \gamma(x)$
But I can't find $\displaystyle \gamma(x)=\int \dfrac{\lambda(2x+1)^2}{(4x^2+1)^2}\cdot e^{-2x}\cdot dx$
May I have a hint to solve this integral?
| For the integral consider
$$\left(\frac {e^{-2x}}{4x^2+1}\right)'=-2e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}$$
$$\left(\frac {e^{-2x}}{4x^2+1}\right)=-2\int e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}dx$$
$$\phi_2=-\frac 12 e^{-2x} $$
or
$$\phi_2=e^{-2x} $$
Another look at the differential equation
$$(2x+1)y''+(4x-2)y'-8y=0 \qquad$$
$$(2x+1)(y''+2y')-4(y'+2y)=0 \qquad$$
Substitute $z=y'+2y$
$$(2x+1)z'-4z=0 \qquad$$
Which is separable and easy to integrate
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find recurrence for $I_n$. Let $I_n=\int_{0}^{1/2} \frac {x^n}{\sqrt{1-x^2}}dx.$ I must find a recurrence for this so I just started using interation by parts:
Let $$f'(x)=x^n\to f(x)=\frac{x^{n+1}}{n+1}$$ and
$$g(x)=\frac 1{\sqrt{1-x^2}}\to g'(x)=\frac x{\sqrt{(1-x^2)^3}}$$
Therefore:
$$I_n=\frac {x^{n+1}}{(n+1)\sqrt{1-x^2}} - \int_{0}^{\frac 12}\frac{x^{n+2}}{(1-x^2)\sqrt{1-x^2}}dx$$
And I can't continue from here..
I tried rewriting $x^{n+2}=x^2x^n=(1-x^2+1)x^n$ but it's no good.
| $$I_n-I_{n+2}=\int_0^{1/2}\frac{x^n(1-x^2)}{\sqrt{1-x^2}}\,dx=\int_0^{1/2}x^n\sqrt{1-x^2}\,dx\\$$Now do integration by parts on this, by integrating $x^n$ and differentiating the other term.
This gives:
$$I_n-I_{n+2}=\left[\frac{1}{n+1} x^{n+1}\sqrt{1-x^2}\right]_0^{1/2}+\frac{1}{n+1}\int_0^{1/2}\frac{x^{n+2}}{\sqrt{1-x^2}}\,dx\\I_n=\frac{\sqrt3}{2^{n+2}(n+1)}+\frac{n+2}{n+1}I_{n+2}\\I_{n+2}=\frac{n+1}{n+2}I_n-\frac{\sqrt 3}{2^{n+2}(n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate Smith normal form, cyclic group decomposition Can someone please check my working for the following problem?
Let $A$ be the abelian group generated by elements $x,y,z$ with relations $7x+5y+2z=0, 3x+3y=0, 13x+11y+2z=0$. Decompose $A$ as a direct sum of cyclic groups.
$$
M = \begin{pmatrix}
7 & 5 & 2 \\
3 & 3 & 0 \\
13 & 11 & 2 \end{pmatrix}
$$
$$\begin{pmatrix}
7 & 5 & 2 \\
3 & 3 & 0 \\
13 & 11 & 2 \end{pmatrix} \sim \begin{pmatrix}
7 & 5 & 2 \\
3 & 3 & 0 \\
-1 & 1 & -2 \end{pmatrix} \sim \begin{pmatrix}
1 & -1 & 2 \\
3 & 3 & 0 \\
7 & 5 & 2 \end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 0 \\
3 & 6 & -6 \\
7 & 12 & -12 \end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 0 \\
0 & 6 & -6 \\
0 & 12 & -12 \end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 0 \\
0 & 6 & -6 \\
0& 0 & 0 \end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 0 \end{pmatrix} $$
So $A = \mathbb{Z}/1\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z} \oplus \mathbb{Z} / 0\mathbb{Z}$ based on the above Smith normal form which I'm not confident about.
The 6 steps taken are the following:
*
*$R_3=R_3-2R_1$
*Swap $R_3$ and $-R_1$
*$C_2=C_2+C_1$ and $C_3=C_3-2C_1$
*$R_2=R_2-3R_1$ and $R_3=R_3-7R_1$
*$R_3=R_3-2R_2$
*$C_3=C_3+C_2$
| Since you want to check your answer, there is a direct way to obtain Smith normal form; define inductively
$$d_1d_2\cdots d_k=\mbox{gcd of $k\times k$ minors of matrix under consideration}.$$
Then the Smith normal form is
$$\begin{bmatrix} d_1 & & \\ & d_2 & \\ & & d_3\end{bmatrix}.$$
So $d_1=1$ because it is $gcd(7,5,..)=1$; next $d_1d_2=gcd(6,-6,12,-12)=6$, and
$d_1d_2d_3=\det A=0.$ Thus
$$d_1=1, d_2=6, d_3=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show the next inequality Show that:
$$
\frac{3}{8}\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+x}}dx\leq \frac{\sqrt{3}}{4}.
$$
I have
$$m(b-a)\leq\int_{a}^{b}f(x)dx\leq M(b-a),$$
but the function does not present critical points in its derivative.
And how can I show the following?
$$\frac{|a+b|-|a-b|}{ab}=\frac{2}{\max\{|a|,|b|\}}$$
| If you have
written it correctly,
$\frac{3}{8}
\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\leq \frac{\sqrt{3}}{4}
$
is the same as
$\frac{3}{8}\sqrt{3}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{\sqrt{3}}{4}\sqrt{3}
$
or
$\frac{3\sqrt{3}}{8}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{3}{4}
$.
$\sqrt{1-x}$
is monotonic decreasing,
so,
on $[0, \frac12]$,
$1 \ge \sqrt{1-x} \ge \frac{\sqrt{2}}{2}$.
Therefore
$\frac1{2\sqrt{3}}
\approx 0.2886
\geq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\geq \frac{\sqrt{2}}{4\sqrt{3}}
\approx 0.20412
$.
If the denominator
is actually $x+2$,
then
$\sqrt{\frac{1-x}{x+2}}
$
is again monotonic decreasing
so,
on $[0, \frac12]$,
$\sqrt{\frac12}
\ge\sqrt{\frac{1-x}{x+2}}
\ge \sqrt{\frac{\frac12}{\frac52}}
= \sqrt{\frac15}
$
so the integral
is within half of these bounds.
For the second question,
use
$\min(a, b)
=\frac12(|a+b|-|a-b|)
$
and
$\max(a, b)\min(a, b)
=ab
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Generating function to find the sum of digits How many integers between $30,000$ and $100,000$ have a sum of $15$ or less?
I was approaching this problem with a generating function:
$$g(x) = (x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+...+x^9)^4$$
First I'm going to pull out $x^3$ then I would need the coefficients of $x$ to the $15$th, $14$th, $13$th etc correct? If so, can someone help me through finding the coefficients?
Also finding the coefficient of x to the 15th would be the same as finding the coefficient of x to the 12 is I pull out x$^3$
| Assuming you are including $30\,000$ and excluding $100\,000$.
Write:
$$\sum_{k=3}^{9}x^k=\frac{x^3-x^{10}}{1-x}\, ,$$
$$\sum_{k=0}^{9}x^k=\frac{1-x^{10}}{1-x}\, .$$
Then
$$g(x)=\left(\sum_{k=3}^{9}x^k\right)\left(\sum_{k=0}^{9}x^k\right)^4=\frac{x^3-x^{10}}{1-x}\left(\frac{1-x^{10}}{1-x}\right)^4$$
Use the operator $(1-x)^{-1}=1+x+x^2+x^3+\cdots$. Then the coefficient of $x^{15}$ in
$$\frac{1}{1-x}g(x)$$
is the required answer.
Now
$$(x^3-x^{10})(1-x^{10})^4=\sum_{r=0}^{4}(-1)^{r}\binom{4}{r}x^{10r+3}-\sum_{r=0}^{4}(-1)^{r}\binom{4}{r}x^{10r+10}\, ,$$
and therefore, since $(1-x)^{-6}=\sum_{k\ge 0}\binom{5+k}{5}x^k$, we have
$$\begin{align}(1-x)^{-1}g(x)=&\left(\sum_{r=0}^{4}(-1)^r\binom{4}{r}x^{10r+3}\right)\left(\sum_{k\ge 0}\binom{5+k}{5}x^k\right)\\[1ex] &- \left(\sum_{r=0}^{4}(-1)^r\binom{4}{r}x^{10r+10}\right)\left(\sum_{k\ge 0}\binom{5+k}{5}x^k\right)\, ,\\[1ex]
=& \sum_{k\ge 0}\left(\sum_{r\ge 0}(-1)^r\binom{4}{r}\binom{k+2-10r}{5}\right)x^{k}\\[1ex] &-\sum_{k\ge 0}\left(\sum_{r\ge 0}(-1)^r\binom{4}{r}\binom{k-10r-5}{5}\right)x^{k}\, .\end{align}$$
Taking the $x^{15}$ coefficient gives
$$\begin{align}[x^{15}](1-x)^{-1}g(x)&=\sum_{r\ge 0}(-1)^r\binom{4}{r}\binom{17-10r}{5}-\sum_{r\ge 0}(-1)^r\binom{4}{r}\binom{10-10r}{5}\, ,\\[1ex]
&=\binom{4}{0}\binom{17}{5}-\binom{4}{1}\binom{7}{5}-\binom{4}{0}\binom{10}{5}\, ,\\[1ex]
&=\binom{17}{5}-4\binom{7}{5}-\binom{10}{5}\, ,\\[1ex]
&=5852\, .\tag{Answer}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $(x-1)^{-2}$ in Quotient Ring Let $R=\mathbb{Q}[x]$ with $x^3-x^2+x+2=0$. Find $(x-1)^{-2}$. I thought I could use the euclidean algorithm
$$x^3-x^2+x+2=(x-1)(x^2+1)+3$$
but I already have a constant after the first step. Also not sure that finding $(x-1)^{-1}$ would help anyway.
| Well, $(x-1)^{-2}=((x-1)^{-1})^2$, no? As for the rest, you know that $$(x-1)(x^2+1)\equiv -3\pmod{x^3-x^2+x+2}$$ In other words, that $(x-1)\left(-\frac13x^2-\frac13\right)\equiv 1\pmod{x^3-x^2+x+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $\lim_{n\rightarrow \infty} \frac {n^2}{n^3+n +1} + \frac {n^2}{n^3 +n +2} +.....+\frac {n^2}{n^3 + 2n}$?
How to find
$$
\lim_{n\rightarrow \infty} \frac {n^2}{n^3+n +1} + \frac {n^2}{n^3 +n +2} +.....+\frac {n^2}{n^3 + 2n}?
$$
I was thinking about the Riemann sum, but I am not able to do it.
| \begin{align*}
\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\leq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+1}=\dfrac{n^{3}}{n^{3}+n+1}
\end{align*}
and
\begin{align*}
\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\geq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+n}=\dfrac{n^{3}}{n^{3}+2n},
\end{align*}
now use Squeeze Theorem to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=(1,-\frac 52)$ Let $$T:\mathbb R^3\to \mathbb R^2\mid M_{B_1B_2}=\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}$$ a linear transformation and the basis $$B_1=\{(1,0,0),(0,-3,1),(0,0,-2)\},\qquad B_2=\{(2,0),(0,-1)\}.$$
Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=\left(1,-\dfrac 52\right).$
I did a similar exercise and the steps I did are the same:
$$\left(\begin{array}{cc:c}2&0&1\\0&-1&-5/2\end{array}\right)\sim \left(\begin{array}{cc:c}1&0&1/2\\0&1&5/2\end{array}\right).$$ Then
$$\left(\begin{array}{ccc:c}1&0&0&2\\0&-3&0&0\\0&1&-2&1\end{array}\right)\sim \left(\begin{array}{ccc:c}1&0&0&2\\0&1&0&0\\0&0&1&-1/2\end{array}\right)$$
so we need to do
$$\begin{array}{cccccc}
M_{B_1B_2}&\cdot &[v]_{B_1}&=&[T(v)]_{B_2}& \\
\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}&\cdot&\begin{pmatrix}2\\0\\-1/2\end{pmatrix}&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\\
&\begin{pmatrix}2a-1\\-5/2\end{pmatrix}&&=&\begin{pmatrix}\color{red}{1/2}\\5/2\end{pmatrix}&\Rightarrow \begin{cases}2a-1&=&\color{red}{1/2}\\-5/2&=&5/2\end{cases}
\end{array}$$
but $$\boxed{-\dfrac 52\neq \dfrac 52\quad\wedge\quad a=\dfrac 34}$$ and the answer says $\boxed{a=1}\;\text{(apparently without contradictions)}$.
So I came to an absurd and the value of $a$ gives me different, what can be said of the exercise? Is it wrong or am I wrong?
Thanks!
$\color{red}{\text{Edited in red.}}$
| Find the matrix of $T$ with respect to the canonical bases of $\mathbb{R}^{3}$ and $\mathbb{R}^2$, which is
$$
\begin{bmatrix}
2 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
a & 1 & 2 \\
-1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 1 & -2
\end{bmatrix}^{-1}
$$
In order to find the required inverse, one can use Gaussian elimination:
$$
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & -3 & 0 & 0 & 1 & 0 \\
0 & 1 & -2 & 0 & 0 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & -3 & 0 & 0 & 1 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & 0 & -6 & 0 & 1 & 3
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & -1/6 & -1/2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & -1/3 & 0 \\
0 & 0 & 1 & 0 & -1/6 & -1/2
\end{bmatrix}
$$
so the inverse is
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1/3 & 0 \\
0 & -1/6 & -1/2
\end{bmatrix}
$$
and the matrix with respect to the canonical bases is
$$
\begin{bmatrix}
2a & -4/3 & -2 \\
1 & 1/6 & 1/2
\end{bmatrix}
$$
so the equation becomes
$$
\begin{bmatrix}
2a & -4/3 & -2 \\
1 & 1/6 & 1/2
\end{bmatrix}
\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}
=
\begin{bmatrix} 1 \\ -5/2 \end{bmatrix}
$$
or, doing the computation in the left-hand side,
$$
\begin{bmatrix} 4a-2 \\ 5/2 \end{bmatrix} =
\begin{bmatrix} 1 \\ -5/2 \end{bmatrix}
$$
Note that I get essentially the same result as you. There is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bi-nominal expansion of 3 terms Find the coefficient of $x^{17}$ in the expansion of $(3x^7 + 2x^5 -1)^{20}$
I'm stuck in handling this question as I do know how to solve it when it has 2 terms.
But now it has 3.
I have no idea where to begin...
| We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.
We obtain
\begin{align*}
\color{blue}{[x^{17}]}&\color{blue}{(3x^7+2x^5-1)^{20}}\\
&=[x^{17}]\sum_{k=0}^{20}\binom{20}{k}(3x^7)^k(2x^5-1)^{20-k}\tag{1}\\
&=\binom{20}{0}3^0[x^{17}](2x^5-1)^{20}+\binom{20}{1}3^1[x^{10}](2x^5-1)^{19}\\
&\qquad +\binom{20}{2}3^2[x^3](2x^5-1)^{18}\tag{2}\\
&=3\binom{20}{1}[x^{10}]\sum_{j=0}^{19}\binom{19}{j}(2x^5)^j(-1)^{19-j}\tag{3}\\
&=-3\binom{20}{1}\binom{19}{2}2^2\tag{4}\\
&\,\,\color{blue}{=-41\,040}
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem once to $(3x^7+(2x^5-1))^{20}$.
*In (2) we observe that the term $(3x^7)^k$ contributes to $[x^{17}]$ only when $k=0,1,2$. We also apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
*In (3) we note that expanding $(2x^5-1)^n$ gives only exponents of $x$ which are a multiple of $5$. We can therefore stick to the middle summand with $[x^{10}]$ and we apply the binomial theorem again.
*In (4) we select the coefficient of $x^j$ accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _{n=1}^{\infty }\:\frac{1}{n}-\frac{1}{n+3} = \frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots\right)
$$
Wich means some values are zeroed, does that mean it's a telescoping sum?
I also know that $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} \le \sum _{n=1}^{\infty }\:\frac{1}{n^2}\:$ so the series converges by the harmonic p-series.
| Yes, it is telescoping,
\begin{align}
\sum_{n=1}^N \frac{1}{n(n+3)}&= \frac13 \sum_{n=1}^N \left[ \frac1{n}-\frac{1}{n+3}\right]\\
&= \frac13\left[1-\frac{1}{4} +\frac12-\frac1{5}+\frac13-\frac16+\ldots +\frac1{N-2}-\frac1{N+1}+\frac1{N-1}-\frac1{N+2}+\frac1N-\frac1{N+3}\right]\\
&=\frac13\left[1+\frac12+\frac13-\frac1{N+1}-\frac{1}{N+2}-\frac1{N+3} \right]
\end{align}
Now take limit $N \to \infty$,
$$\sum_{n=1}^N \frac{1}{n(n+3)}=\frac13\left[1+\frac12+\frac13 \right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$? Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$?
So what I did was basically simplify the terms on the right $\mod 13$.
$2^n5^{2n+1}+3^{2n+1}7^{n+1} \mod 13$
$= 5\cdot 2^n\cdot 5^n\cdot 5^n+3\cdot 7\cdot 3^{n}\cdot 3^n7^n \mod 13$
$= 5\cdot 50^n+21\cdot 63^n \mod 13$
$= 5\cdot 50^n+21\cdot 63^n \mod 13$
$= 5\cdot 11^n+8\cdot 11^n \mod 13$
$= 13\cdot 11^n \mod 13$
$= 0\cdot 11^n \mod13$
$=0 \mod13$
Thus, for all $n$, $13$ will divide this term, and the statement is true. Can anyone tell me if what I have done is correct/wrong? Would really appreciate it!
| Induction ...
\begin{eqnarray*}
5 \times 50^{n+1}+21 \times 63^{n+1}=50(\color{red}{5 \times 50^{n}+21 \times 63^{n} }) +\color{red}{13} \times 21\times 63^n.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form
Convert complex to trig.
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$
Let us consider
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{3}{3}\biggr) = \arctan (1)$$
$$\theta = \frac{\pi}{4}$$
Thus in trigonometric form we get
$$Z_1 = \biggr [3\sqrt2 \bigg(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\bigg)\bigg]^5$$
$$Z_1 =(3\sqrt 2)^5 \biggr [\bigg(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\bigg)\bigg]$$
Let us consider
$$(-2+2i)^3$$
$a = -2$, $b=2$
$$\sqrt{a^2+b^2}=\sqrt{(-2)^2+2^2} =2\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{2}{-2}\biggr) = \arctan (1)$$
$$\theta = \frac{3\pi}{4}$$
$$Z_2 =(2\sqrt 2)^3 \biggr [\bigg(\cos \frac{9\pi}{4}+i\sin \frac{9\pi}{4}\bigg)\bigg]$$
Multiplying $Z_1 \cdot Z_2$ we get
$$Z_1 \cdot Z_2 = (3\sqrt 2)^5 \cdot (2\sqrt2)^3 \biggr [\bigg(\cos \frac{5\pi}{4}+\cos \frac{9\pi}{4}\bigg)+i\sin \bigg(\frac{5\pi}{4}+i\sin\frac{9\pi}{4}\bigg)\bigg]$$
$$Z_1 \cdot Z_2 = 31104\biggr [\bigg(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\bigg)\bigg]$$
Let us consider
$$(\sqrt 3+i)^{10}$$
Here $a = \sqrt 3$, $b =1$
$$\sqrt{a^2+b^2}=\sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$$
$$\theta = \arctan \biggr(\frac{1}{\sqrt 3}\biggr) = \frac{\pi}{6}$$
$$Z_3 = 2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]$$
Now we have
$$\frac{Z_1 \cdot Z_2}{Z_3} = \frac{31104\biggr [\bigg(cos \biggr(\frac{7\pi}{2}\biggr) +i\sin \biggr(\frac{7\pi}{2}\biggr) \bigg]}{2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]} = \boxed {30.375 \biggr [ \cos \biggr(\frac{11\pi}{6}\biggr)+i\sin \biggr(\frac{11\pi}{6}\biggr)\biggr]}$$
Is my assumption correct?
| As an alternative by exponential form
*
*$(3+3i)^5=(3\sqrt 2e^{i \pi/4})^5$
*$(-2+2i)^3=(2\sqrt 2e^{i 3\pi/4} )^3$
*$(\sqrt 3 +i)^{10}=(2e^{i \pi/6} )^{10}$
then
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}=\frac{3^54\sqrt 2\cdot2^4\sqrt2}{2^{10}}\frac{e^{i 5\pi/4}\cdot e^{i 9\pi/4}}{e^{i 10\pi/6}}=30.375e^{i11\pi/6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\vec{u}=\vec{w_{1}}+\vec{w_{2}}, w_1\in W$ and $w_2\in W^\perp$ Let $S[(x_1,y_1,z_1)(x_2,y_2,z_2)]=3x_1x_2+x_2y_1+x_1y_2+y_1y_2+z_1z_2$ be an inner product,
$W=${$(x,y,z)\in\mathbb{R}^3|x+z=0$} a subspace and $\vec u=(1,0,0)$
Find $w_1\in W,$ $w_2\in W^\perp$ (orthogonal in respect of $S$) such that $\vec u=\vec w_1+\vec w_2$
Here's what I've done so far:
An orthogonal basis of W in respect of the inner product $S$ is $B=${$(1,0,-1),(-\frac{1}{4},1,\frac{1}{4}$)}
If $w=(x,y,z)\in W^\perp$ then $S[(x,y,z),(1,0,-1)]=0$ and $S[(x,y,z),(-\frac{1}{4},1,\frac{1}{4})]=0$
So
$3x+y-z=0$ and $\frac{1}{4}x+\frac{3}{4}y+\frac{1}{4}z=0\Rightarrow x+3y+z=0$
Solving the above system we get that the basis for $W^\perp$ in respect of $S$ is $B'=\left \{(1,-1,0),(0,0,1) \right \}$
I don't know how to proceed from here..
Also, are the $\vec w_1$, $\vec w_2:\vec u=\vec w_1+\vec w_2$ unique?
| With
$$
\begin{array}{rcl}
\vec{u} & = & (1,0,0)\\
\vec{w}_{1} & = & (w_{11},w_{12},w_{13})\\
\vec{w}_{2} & = & (w_{21},w_{22},w_{23})
\end{array}
$$
The inner product is weighted by
$$
M=\left[\begin{array}{ccc}
3 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
$$
The conditions are
$$
\left\{ \begin{array}{rcl}
\vec u & = & \vec{w}_1+\vec{w}_2\\
\left\langle \vec{u},\vec{w}_{1}\right\rangle _{M} & = & \left\Vert \vec{w}_{1}\right\Vert _{M}^{2}\\
\left\langle \vec{u},\vec{w}_{2}\right\rangle _{M} & = & \left\Vert \vec{w}_{2}\right\Vert _{M}^{2}\\
w_{11}+w_{13} & = & 0
\end{array}\right.
$$
Solving for $\vec{w}_1, \vec{w}_2$ we obtain the solutions
$$
\begin{array}{lccccc}
\vec{w}_1 & = & \frac{1}{7} \left(4+\sqrt{2}\right) & \frac{1}{7} \left(2-3 \sqrt{2}\right) & \frac{1}{7} \left(-4-\sqrt{2}\right) \\
\vec{w}_2 & = & \frac{1}{7} \left(3-\sqrt{2}\right) & \frac{1}{7} \left(-2+3 \sqrt{2}\right) & \frac{1}{7} \left(4+\sqrt{2}\right)
\end{array}
$$
and
$$
\begin{array}{lccccc}
\vec{w}_1 & = & \frac{1}{7} \left(4-\sqrt{2}\right) & \frac{2}{7}+\frac{3 \sqrt{2}}{7} & \frac{1}{7} \left(-4+\sqrt{2}\right) \\
\vec{w}_2 & = & \frac{1}{7} \left(3+\sqrt{2}\right) & \frac{1}{7} \left(-2-3 \sqrt{2}\right) & \frac{1}{7} \left(4-\sqrt{2}\right)
\end{array}
$$
NOTE
Here $< \cdot, \cdot >_M$ is the inner product $\vec x_1^{\top}M \vec x_2$ and $||\cdot ||_M$ indicated the norm $\sqrt{\vec x^{\top}M\vec x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}
$$
Following that log suggestion in the comments below:
\begin{align}
&\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\
&= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\
&= \int_1^2 \ln x \, dx \\
&= (x \ln x - x)\vert_1^2 \\
&= (2 \ln 2 - 2)-(1 \ln 1-1) \\
&= (\ln 4-2)-(0-1) \\
&= \ln 4-1 \\
&= \ln 4 - \ln e \\
&= \ln \left( \frac 4e \right)
\end{align}
but I read from somewhere that the answer should be $\frac 4e$.
| If you accept using the following limit that has been asked and proved (for example here) many times here on MSE
*
*$\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,
then your limit is easily derived:
$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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probability that exactly three of them shows the same face and remaining three shows different faces Question
$6$ dice are thrown simultaneously ,find the probability that exactly three of them shows the same face and remaining three shows different faces.
My Approach
Total outcome=$$6^{6}$$
Number of ways to select an outcome which will be same$$=\binom{6}{1}$$
so selected outcome will be $$=\frac{ \binom{6}{1}\times 1 \times 1 \times 1 \times 5 \times4 \times 3}{6^{6}}=\frac{6\times5 \times4\times 3 }{6^6}$$
Am i correct?
| One may see it as follows:
*
*There are 6 options for the equal faces
*$\binom{6}{3}$ choices, which 3 dice show the same face
*$5\cdot 4 \cdot 3$ choices for the mutually different faces of the other three
*all together there are $6^6$ different throws
$$\frac{6\cdot \binom{6}{3}\cdot 5\cdot 4\cdot 3}{6^6}= \frac{25}{162}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$$\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1).$$
| Because$$
a^{2m} - 1 = (a^2)^m - 1 = (a^2 - 1)((a^2)^{m - 1} + (a^2)^{m - 2} + \cdots + 1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you
| from question,
$$\sqrt {11} -1 = \frac {a}{b}$$ where $(a,b)=1$.
From last step,
$10b^2 = a^2 + 2ab$
$LHS $ is multiple of $5$. then $RHS$ is also multiple of $5$.
in $RHS,a^2 + 2ab$ exactly one of $a$ and $b$ is multiple.
if both $a$ and $b$ are multiple of $5$ , it violated $(a,b)=1$.
if $b$ is multiple of $5$. let $b=5b_1 \implies 250b_{1} ^2 =a^2+50ab_1 \implies 5 \mid a $.
but, it violated $(a,b)=1$.
therefore, $a$ is multiple of $5$. let $a=5a_1$.
$10b^2= 25 a_1^2+2ab$
$10b^2-25 a_1^2= 10a_1b$
$/5 \implies 2b^2-5a_1^2=2a_1b$
Therefore $2 \ mid a_1 , \exists a_2$ such that $a_1=2a_2$
$ 2b^2-10a_2^2 =4a_2b$
$\implies b^ 2 -2a_2b= 5a_2^2 $
$\implies 5\mid a_2 , \ exists a_3$ such that a_2=5a_3
$ b^2 -10a_3b =25a_3^2$
$ b^2 =25a_3^2 +10a_3b$
Therefore $ 5 \mid b$
Which contradict our condition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$ Problem
Evaluate $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$
My solution
Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1)}{(n-1)n^n}=\lim_{n \to \infty}\frac{1-\dfrac{1}{n^n}}{1-\dfrac{1}{n}}=1,$$and
$$\lim_{n \to \infty}\frac{1+2^n+3^n+\cdots+n^n}{n^n}=\frac{e}{e-1}.$$
Hence,\begin{align*}\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}&=\lim_{n \to \infty}\frac{\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=\frac{\lim\limits_{n \to \infty}\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\lim\limits_{n \to \infty}\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=1-\frac{1}{e}.\end{align*}
The solution posted above need to quote an uncommon limit. Is there another more simple and more direct solution?
| A Query to @Teddy38's Answer
Yes, we can agree that $$n+n^2+\cdots+n^n=\frac{n}{n-1}(n^n-1)$$ and $$\frac{n^{n+1}}{n+1}=\int_{0}^{n}x^n\ dx<1^n+2^n+\cdots+n^n<\int_{1}^{n+1}x^n\ dx=\frac{(n+1)^{n+1}-1}{n+1}.$$ Thus,we obtain $$\frac{n+1}{n-1}\cdot \frac{n^{n+1}-n}{(n+1)^{n+1}-1}<\dfrac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}<\frac{n+1}{n-1}\cdot\frac{n^n-1}{n^n}.$$Now, let's evaluate the limits of the both sides as $n \to \infty.$ For the left side, we have $$\lim_{n \to \infty} \dfrac{n+1}{n-1} \cdot \lim_{n \to \infty}\dfrac{1-\dfrac{1}{n^n}}{\left(1+\dfrac{1}{n}\right)^n\cdot\left(1+\dfrac{1}{n}\right)-\dfrac{1}{n^{n+1}}}=\frac{1}{e}.$$
As for the right side, we have $$\lim_{n \to \infty} \dfrac{n+1}{n-1} \cdot \lim_{n \to \infty}\left(1-\dfrac{1}{n^n}\right)=1.$$
From the two aspects, we may only conclude that, $$\varliminf_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}\geq\frac{1}{e},$$and $$\varlimsup_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}\leq 1.$$ That's to say, if the limit we want really exists, then $$\frac{1}{e}\leq\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n} \leq 1.$$This is true, but can't give the accurate value of the limit we want.
| {
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"url": "https://math.stackexchange.com/questions/2792441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.
My attempt
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
This is undefined for $x=-2$, because the right hand side becomes: $\frac{1}{2^{-2+2}-1}=\frac{1}{2^0-1}=\frac{1}{1-1}=\frac{1}{0}$. Therefore,
$$x \neq -2$$
Now, let's set $t=2^x$.
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
Multiply both sides with $(t+3)(4t-1)$.
We must split the inequality because we don't know is $(t+3)(4t-1)$ positive or negative. Question: what if it's zero? Should we consider that possibility too?
For $(t+3)(4t-1) > 0$:
$$4t-1 \geq t+3$$
$$3t \geq 4$$
$$t \geq \frac{4}{3}$$
$$2^x \geq \frac{4}{3}$$
$$x \geq \log_2 \left(\frac{4}{3}\right)$$
For $(t+3)(4t-1) < 0$:
$$4t-1 \leq t+3$$
$$3t \leq 4$$
$$t \leq \frac{4}{3}$$
$$2^x \leq \frac{4}{3}$$
$$x \leq \log_2 \left(\frac{4}{3}\right)$$
I think it's already obvious where and how I'm wrong, so I think I don't need to continue with this attempt. If I do, then please ask in the comment.
| The given solution is clearly wrong on the positive side. For instance, take $x=2$, and
$$
\frac17=\frac{1}{2^x+3}\geq\frac{1}{2^{x+2}-1}=\frac1{15}.
$$
Since $t=2^x$ you know that $t>0$. So $t+3>0$. Now if $4t-1>0$, everything is positive and you can multiply to get
$$
4t-1\geq t+3,
$$
which simplifies to $3t\geq4$, or $t\geq4/3$.
If, on the other hand, $4t-1<0$, the inequality will be reversed if we multiply by it. So we get
$$
4t-1\leq t+3,
$$
which is $t\leq 4/3$. This was under the hypothesis $t<1/4$, so we get $t<1/4$.
In summary, the inequality holds when $t<1/4$ and when $t\geq 4/3$.
Now we translate to $x$. The equality $2^x=1/4$ gives $x=-2$. The equality $2^x=4/3$, gives $x=\log_24-\log_23=2-\log_23$. So the inequality holds for
$$
x\in(-\infty,-2)\cup[2-\log_23,\infty)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?
| You are missing the fact that a single even value of $z$ represents several different soluitons, not one.
With $z$ equal to 10, there is only one possible solution (0+0+10).
With $z$ equal to 8, there are 3 possible solutions (0+2+8, 1+1+8, 2+0+8).
With $z$ equal to 6, there are 5 possible solutions.
With $z$ equal to 4, there are 7 possible solutions.
With $z$ equal to 2, there are 9 possible solutions.
With $z$ equal to 0, there are 11 possible solutions.
In total you have 36 solutions with $z$ being even and 66 solutions in total. So the probability is 36/66 or 6/11.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatrix}
-1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$c) \begin{pmatrix}
1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$d) \begin{pmatrix}
-1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
My attempt : I know that Determinant of A = product of eigenvalues of A, as option c and d is not correct because Here Determinant of A = 4
that is $ \det A = -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm in confusion about option a) and b).......how can I find the Jordan canonical form of A ?
PLiz help me.
Any hints/solution will be appreciated.
Thanks in advance
| All the other answers to this question so far overlook the fact that the matrix in option (b) is not in Jordan canonical form in the first place, so you can eliminate that option without doing any work at all. After eliminating (c) and (d) as you’ve done, that leaves only (a).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$
| Using the series expansion for $\log(1+x)$, we get
$$
\frac1x\,\log\left(1+mx+nx^2+O\!\left(x^3\right)\right)
=m+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right)
$$
Using the series expansion for $e^x$, we get
$$
\left(1+mx+nx^2+O\!\left(x^3\right)\right)^{1/x}
=e^m\left(1+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right)\right)
$$
Therefore, writing $a^x=e^{x\log(a)}$ and $b^x=e^{x\log(b)}$ and using the series expansion for $e^x$, we get
$$
\begin{align}
\frac{\left(\frac{a^x+b^x}2\right)^{1/x}-\sqrt{ab}}x
&=\frac{\scriptsize\left(1+\frac x2(\log(a)+\log(b))+\frac{x^2}4\left(\log(a)^2+\log(b)^2\right)+O\!\left(x^3\right)\right)^{1/x}-\sqrt{ab}}x\\
&=\frac{\sqrt{ab}\left(1+\frac{\log(a)^2+\log(b)^2-2\log(a)\log(b)}8x+O\!\left(x^2\right)\right)-\sqrt{ab}}x\\[9pt]
&=\frac18\sqrt{ab}\,(\log(a)-\log(b))^2+O(x)
\end{align}
$$
Thus,
$$
\lim_{x\to0}\frac{\left(\frac{a^x+b^x}2\right)^{1/x}-\sqrt{ab}}x
=\frac18\sqrt{ab}\,(\log(a)-\log(b))^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral results in difference of means $\pi(\frac{a+b}{2} - \sqrt{ab})$ $$\int_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{1/2}dr = \pi\left(\frac{a+b}{2} - \sqrt{ab}\right)$$
What an interesting integral! What strikes me is that the result involves the difference of the arithmetic and geometric mean. Is there an innate geometric explanation that corresponds to this result? And can we generalize this integral to, say, the mean of three or more items?
Some background on where I saw it and how to solve it. This arises in calculating the action variable $I$ for the Kepler problem (Hamiltonian $H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr}-\frac{k}{r}$). The $a,b$ are the minimal and maximal $r$ from the origin set at the focus of an ellipse (i.e. the perihelion/aphelion). See David Tong's Classical Dynamics notes $\S$4.5.4.
I was able to solve the integral by using the third Euler substitution, letting
$$\left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{1/2}=\frac{1}{r}\sqrt{-(r-a)(r-b)}=\frac{1}{r}(r-a)t$$
giving $r=\frac{b+at^2}{1+t^2}$ and
$$2(b-a)\left\{\int_{t(r_2)=0}^{t(r_1)=\infty}\frac{t^2}{(1+t^2)^2}dt - \int_{t(r_2)=0}^{t(r_1)=\infty}\frac{\frac{a}{b}t^2}{(1+t^2)(1+\frac{a}{b}t^2)}dt\right\}$$
$$=\pi\left(\frac{b-a}{2}\right) + \pi\left(a-\sqrt{ab}\right)=\pi\left(\frac{a+b}{2} - \sqrt{ab}\right).$$
| Let
$0\le a\le b,$
then
\begin{align}
&I_1 = \int\limits_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{-1/2}\,\mathrm dr = \int\limits_a^b \dfrac{r}{\sqrt{(r-a)(b-r)}}\,\mathrm dr\\
&=\dfrac12\int\limits_a^b \dfrac{2r-(a+b)}{\sqrt{(r-a)(b-r)}}\,\mathrm dr+\dfrac{a+b}2\int\limits_a^b \dfrac{a+b}{\sqrt{(r-a)(b-r)}}\,\mathrm dr\\
&=-\dfrac12\int\limits_a^b \dfrac{\mathrm d((r-a)(b-r))}{\sqrt{(r-a)(b-r)}} + \dfrac{a+b}2\int\limits_a^b \dfrac{\mathrm dr}{\sqrt{\left(\dfrac{b-a}2\right)^2-\left(r-\dfrac{b+a}2\right)^2}}\\
&=-\sqrt{(r-a)(b-r)}\Bigg|_a^b + \dfrac{a+b}2\arcsin\dfrac{2r-(b+a)}{b-a}\Bigg|_a^b = \pi\dfrac{a+b}2,\\
&I_2 = \int\limits_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{-1/2}\left(1-\left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}\right)\,\mathrm dr\\
& = \int\limits_a^b \dfrac{2r^2-(a+b)r+ab}{\sqrt{(r-a)(b-r)}}\,\dfrac{\mathrm dr}r = \int\limits_a^b \dfrac{2r-(a+b)}{\sqrt{(r-a)(b-r)}}\,\mathrm dr - ab\int\limits_a^b \dfrac{1}{\sqrt{(1-\frac ar)(\frac br-1)}}\,\mathrm d\left(\dfrac1r\right)\\
&=-\sqrt{(r-a)(b-r)}\Bigg|_a^b + ab\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{(1-at)(bt-1)}} = \sqrt{ab}\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{(\frac1a-t)(\frac1b-t)}}\\
&= \sqrt{ab}\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{\left(\frac{\frac1a-\frac1b}2\right)^2 - \left(t-\frac{\frac1a+\frac1b}2\right)^2}} = \sqrt{ab}\arcsin\dfrac{{2t-\left(\frac1a+\frac1b\right)}}{{\frac1a-\frac1b}}\Biggr|_{1/b}^{1/a} = \pi ab.\\
\end{align}
This allows to present the issue integral (or the square under the graph of the according function) as the square between two other similar functions, which present AM and GM of $a$ and $b$ (see also Wolphram Alpha plot).
| {
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If the last 3 digits of $2012^m$ and $2012^n$ are identical, find the smallest possible value of $m+n$.
Let $m$ and $n$ be positive integers such that $m>n$. If the last 3 digits of $2012^m$ and $2012^n$ are identical, find the smallest possible value of $m+n$.
Since the 100's digit is 0 in both cases, I just did $2012^m \equiv 2012^n \mod 1000$, and got $12^m \equiv 12^n\mod 1000$ but I'm not sure where to go from there. Trying to compute the first few powers of $12$ will only get the numbers larger and the pattern doesn't seem to emerge that soon.
| You have $12^m\equiv 12^n\pmod{10^3}$ hence $12^m\equiv 12^n\pmod{2^3}$ and $12^m\equiv 12^n\pmod{5^3}$.
For the latter, we have $12^{m-n}\equiv 1\pmod{5^3}$, hence $m\equiv n\pmod{100}$ because the multiplicative order of $12$ modulo $5^3$ is $100$, as computed here.
On the other hand, $12^n(12^{m-n}-1)\equiv 0\pmod{2^3}$ from which $12^n\equiv 0\pmod{2^3}$ which holds for $n\geq 2$.
Thus $n=2$ and $m=102$ is the smallest solution with sum $m+n=104$.
| {
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The quadratic equation is giving me error can you please help me locate where I am wrong Question. Solve $$\log(x-3) + \log (x-4) - \log(x-5)=0.$$
Attempt. I got $$x^2-8x+17=0.$$
$$\log(x-3)(x-4)/(x-5)=0$$
$$\log(x^2-4x-3x+12)/x-5=0$$
$$x^2-7x+12= 10^0 (x-5)$$
$$x^2-7x-x+12+5=0$$
$$x^2-8x+17=0$$
Hi guys update: apparently the answer was equation is undefined♀️
| What's the problem? We have\begin{align}\log(x-3)+\log(x-4)-\log(x-5)=0&\iff\log\bigl((x-3)(x-4)\bigr)=\log(x-5)\\&\iff\log(x^2-7x+12)=\log(x-5)\\&\iff x^2-7x+12=x-5\\&\iff x^2-8x+17=0.\end{align}
| {
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Recurrence problem: find $a_{1000}$ from $a_{0}$ Hi I'm stuck at yet another question.
$a_{0}=5$. Given $a_{n+1}a_{n} = a_{n}^{2} + 1$ for all $n \ge 0$, determine $\left \lfloor{a_{1000}}\right \rfloor$.
So I got $a_{n+1}=a_{n} + \frac{1}{a_{n}}$ and then:
$a_{1000}=a_{0}+ \frac{1}{a_{0}} + \frac{1}{a_{1}} + \frac{1}{a_{2}} + ... + \frac{1}{a_{999}}$
and I even tried writing the relation as $a_{n}^{2} - a_{n+1}a_{n} + 1 = 0$ but I'm still not getting anywhere. Can someone just tell me how to deal with this recurrence relation, to start off? Thanks.
| Squaring the given relation $a_{n+1} = a_{n} + \frac{1}{a_n}$, we get
$$ a_{n+1}^2 = a_n^2 + \frac{1}{a_n^2} + 2 \Rightarrow a_{n+1}^2 > a_n^2 +2 $$
and iterating we get (for $n = 999$ we get the right side)
$$ a_{n+1}^2 > a_0^2 + 2n + 2 \Rightarrow a_{1000} > 2025 = 45^2 $$
Also, using the terms ($\frac{1}{a_i^2} $ terms) we neglected for relation above, and the fact that $a_{n+1}^2 > 2n + 27 > 2n + 2$ or just $ a_n^2 > 2n$,
$$ a_{n+1}^2 = 2025 + \sum_{i=1}^{n} \frac{1}{a_i}^2 < 2025 + \frac{1}{2} \sum_{i=1}^{n} \frac{1}{i}$$ $$ \Rightarrow a_{1000}^2 < 2025 + \frac{1}{2} \sum_{i=1}^{999} \frac{1}{i} = 2025 + H_{999} < 2025 + \ln(1000) < 46^2 $$
where $H_n$ is nth harmonic number. Try to prove by yourself that $H_n < \ln(n+1)$. (Hint Riemann sums).
It follows that $\lfloor a_{1000} \rfloor = 45$.
| {
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Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$ The title is the statement of the problem.
I did the following:
$(1-w+w^2)(1+w-w^2)=$
$1+w-w^2-w-w^2+w^3+w^2+w^3-w^4=$
$1-w^2+w^3+w^3-w^4=$
$1-w^2+1+1-w^4=4$, * then,by definition of primitive root of order 3*
$w^4+w^2-1=0$ and the problem is that from here I do not know how to continue. I tried to make the following change of variables: $z=w^2$ then $w^4+w^2-1=0$ becomes $z^2+z-1=0$ but it is not right because this equation has only two solutions and the other has four. So I do not know how to solve this problem.
| $\,4 - (1 - w + w^2) (1 + w - w^2) = (1 + w + w^2) (3 - 3 w + w^2)\,$
where $\,1 + w + w^2\,$ is the irreducible polynomial for a primitive cube root of $1$.
Another way is to multiply the product by $\,1-w\,$ and expand
$\,(1 - w + w^2) (1 + w - w^2)(1 - w) = 1 - w - w^2 + 3w^3 - 3w^4 + w^5.\,$ Now use the cube root property to get
$\,w^3 = 1,\, w^4 = w,\, w^5 = w^2\,$ and
$\,1 - w - w^2 + 3 - 3w + w^2 = 4(1 - w).\,$ The original product must be $4.$
| {
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Integrating $\frac{\arctan x}{x\sqrt{\smash[b]{1-x^2}}}$ How to evaluate
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx\text{?}$$
The steps I can think of is integration by parts as
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx=\int_0^1\frac{\arctan x}{x}\,\mathrm d(\arcsin x)$$
or integration by substitution using $x=\sin t$ as
$$\int_0^1 \frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx = \int_0^{\frac{\pi}{2}}\frac{\arctan(\sin t)}{\sin t}\,\mathrm dt,$$
but both seems to make the problem more complicated.
Thanks a lot.
| Let $\displaystyle I(a)=\int^{1}_{0}\frac{\tan^{-1}(ax)}{x\sqrt{1-x^2}}dx$
Now $\displaystyle I'(a)=\int^{1}_{0}\frac{1}{(1+a^2x^2)\sqrt{1-x^2}}dx$
Put $x=\tan t$. Then $dx=dt$ and changing limits
So $$I'(a)=\int^{\frac{\pi}{2}}_{0}\frac{1}{1+a^2\sin^2 t}dt=\frac{1}{1+a^2}\int^{\frac{\pi}{2}}_{0}\frac{\sec^2 t}{k^2+\tan^2 t}dt$$
Where $\displaystyle k^2=\frac{1}{1+a^2}$
So $\displaystyle I'(a)=\frac{1}{(1+a^2)\cdot k}\tan^{-1}\bigg(\frac{t}{k}\bigg)\bigg|^{\infty}_{0}=\frac{\pi}{2}\cdot \frac{1}{\sqrt{1+a^2}}$
So $$I(a)=\frac{\pi}{2}\ln\bigg|a+\sqrt{1+a^2}\bigg|$$
put $a=1$. Then $$I(1)=\frac{\pi}{2}\ln\bigg|1+\sqrt{2}\bigg|.$$
| {
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Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
| We have
$$
\begin{eqnarray}
\sqrt{x}
&=& 3^{\color{blue}{\log_4(x)} \color{brown}{+ \frac{1}{2}}} + 3^{\color{blue}{\log_4(x)} \color{brown}{- \frac{1}{2}}} \\
&=& \left( \color{brown}{3^{\frac{1}{2}}} + \color{brown}{3^{-\frac{1}{2}}} \right) \cdot \color{blue}{3^{\log_4(x)}} \tag{factoring out $3^{\log_4(x)}$} \\
&=& \left( \color{brown}{3^{\frac{1}{2}}} + \color{brown}{3^{-\frac{1}{2}}} \right) \cdot \color{blue}{x^{\log_4(3)}} \tag{using $a^{\log_b(c)} = c^{\log_b(a)}$} \\
&=& \left( \color{brown}{\sqrt{3}} + \color{brown}{\dfrac{1}{\sqrt{3}}} \right) \cdot \color{blue}{x^{\log_4(3)}} \tag{using $a^\frac{1}{2} = \sqrt{a}$}
\end{eqnarray}
$$
If we square both sides, we get
$$
x
= \left( \color{brown}{\sqrt{3}} + \color{brown}{\dfrac{1}{\sqrt{3}}} \right)^2 \cdot \color{blue}{x}^{2\color{blue}{\log_4(3)}}
= \dfrac{16}{3} \cdot x^{\log_4(9)} \;.
$$
Since $x \neq 0$, we are allowed to divide by $x^{\log_4(9)}$. If we do this, we obtain
$$
\dfrac{16}{3} = x^{1 - \log_4(9)} = x^{\log_4\left( \frac{4}{9} \right)} \;.
$$
Now, we raise both sides to the power of $\log_{\frac{4}{9}}(4)$ and, using the identity $\log_{4}\left( \frac{4}{9} \right) \cdot \log_{\frac{4}{9}}(4) = 1$, obtain
$$
x
= \left( \dfrac{16}{3} \right)^{\log_{\frac{4}{9}}(4)}
\approx 0.0572 \;.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given the area, perimeter and an altitude of a triangle, find some products The triangle $ABC$ has perimeter $360$ and area $2100$. The altitudes are named $AD$, $BE$, $CF$.
The altitudes meet at $H$. If $CF=24$, find the values of $HA \times HD$, $HB \times HE$, $HC \times HF$.
I have observed that all of those products are equal (because of some similar triangles), but I don't know how to find them. Can you help me? Thanks!
| Draw circles $(ABDE), (BECF), (CFAD)$. It is clear that $H$ is the radical center of all three circles. Thus, $HA \cdot HD = HB \cdot HE = HC \cdot HF = x$.
Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\cdot CF = 4200$. It follows that
$$
\begin{align}
\sqrt{a^2 - 24^2} + \sqrt{b^2 - 24} = AB &= 175 \\
a+b+175 &= 360
\end{align}
$$
Thus, examining Pythagorean triples, $a = 40$, $b = 145$. By the Pythagorean Theorem, we can establish that $AF = 143$, and from the given area, $AD = 105$, so $DC = 100$ and $BD = 140$.
Finally, $\triangle AFH \sim \triangle CDH \sim ABD$, so
$$x = HC \cdot HF = \frac{BD}{AD} \cdot \frac{AB}{AD} \cdot AF \cdot DC = \frac{140}{105} \cdot \frac{175}{105} \cdot 143 \cdot 100 = \boxed{\frac{286000}{9}}$$
| {
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The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$. The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain complex values
Yet the reasoning for the implication itself is
\begin{align}
x+\frac {1}{x}&=1 \\
\color {green}{x^2}&=\color {green}{x-1 }\\
x^3&=\color {green}{x^2}-x \\
&=(\color {green}{x-1})-x \\
&=-1.
\end{align}
Hence $x^6=1$, which implies $x^7=x$.
I had just started reading this book when it was presented, asking you to prove the implication without finding the roots first.
So my question: is this implication really justified? Should the author have said to only prove it for an implicit (complex) specific value of $x$? If I didn't know any better, I'd say he presented it like it was true for some range of values...he couldn't of...
| Define two propositions: $p:= x+\frac{1}{x}−1=0$ and $q:=x^7+\frac{1}{x^7}−1=0$. You are trying to find the values of $x$ for which $p \wedge q$ is true. The actual question is 'prove that $p \Rightarrow q$ is true'. These two propositions are different.
If you try to solve $x+\frac{1}{x}=1$ you end up with $x = e^{i\frac{\pi}{3}}$ and $x = e^{-i\frac{\pi}{3}}$. Now, in both cases
$$x^7+\frac{1}{x^7} = 2\cos \frac{7\pi}{3} = 2\cos \frac{\pi}{3} = 1.$$
| {
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Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$. Could someone please explain a step in the following proof?
Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.
The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$.
I understand that $|x-1|$ can be made small, but for $|x+1|$, why does the solution say the condition "if $|x-1| < 1$, $x+1\in (1,3)$"? Why and how is the condition $|x-1| <1$ made?
| The condition for $|x-1|<1$ is just a trick, you can set any number for convenience since we want to show that there exists delta.
First, let us observe the following definition of limit
$$\forall \varepsilon > 0, \exists \delta > 0 \ni 0 < |x-1| < \delta \implies \bigg|\frac{1}{x+1}-\frac{1}{2}\bigg|< \varepsilon$$
So, our goal is to show that $|x-1|<\delta \implies \bigg|\frac{1}{x+1}-\frac{1}{2}\bigg|< \varepsilon$
To do this, we have to realise that limit is a concept of getting closer and closer to a certain value. Hence, we want to "set a limit" for the value of $x$ which we will observe. In this case, we want $x$ to get closer and closer to $1$. Hence, we just simply restrict $|x-1|<1$. This can be any convenient number. You can even set to $1/2$ if you want to.
Now, moving on observe that
$\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg| = \bigg|\frac{2-x-1}{2(x+1)}\bigg|= \frac{1}{2}\bigg|\frac{x-1}{x+1}\bigg|$. Since we restrict $|x-1|<1$, we have $1<x+1<2$ and thus $1<|x+1|<2$. As a consequence we have $\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg|<\frac{1}{2}\frac{|x-1|}{1}=\frac{1}{2}|x-1|$.
Finally, we can prove the limit easily using the reasoning above by fixing $\varepsilon > 0$ and set $\delta = \min\{\frac{\varepsilon}{2},1\}$. Observe that :
$$\bigg|\frac{1}{x+1} - \frac{1}{2}\bigg| = \frac{1}{2}\bigg|\frac{x-1}{x+1}\bigg|< \frac{1}{2}|x-1|<\frac{1}{2}\frac{\varepsilon}{2}<\varepsilon$$
Thus, we complete the proof.
| {
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For any prime number $n >5$, prove the final digit of $n^4$ is $1$ So I am struggling a bit with this question
$n$ is prime
we can ignore $2$ and $5$ as $n>5$
now if $n$ is prime
for the digits: $\{0,1,2,3,4,5,6,7,8,9\}$
$\{0,2,4,6,8\}$ can be discounted as $n$ cannot be even that
$5$ can be discounted as $n$ is not a multiple of $5$ either
therefore any prime must have the last digit $q$ such that $q\subset\{1,3,7,9\}$
For $n^4$, if $n=10p+q$
for a natural number $p$
I don't quite know how I can get to the desired result.
| On the claim I've done in my comment above: for a prime $n > 5$ we have
$$n \equiv 1 \pmod 2 \Rightarrow n^2 \equiv 1 \pmod 8 \Rightarrow n^4 \equiv 1 \pmod{16};$$
$$n^2 \equiv 1 \pmod 3 \Rightarrow n^4 \equiv 1 \pmod 3;\quad n^4 \equiv 1 \pmod 5$$
(the last two are implied by FLT, or may be verified by hand, considering each possible residue $\bmod {3, 5}$), so $n^4 \equiv 1 \pmod{240}$, and $240 = \gcd(7^4 - 1, 11^4 - 1)$ is the greatest possible.
| {
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a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$
Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.
Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the following —
$$P(1) = a+b+c+d+e = 120$$
$$P(2) = 16a + 8b + 4c + 2d + e = 60$$
$$P(3) = 81a + 27b + 9c + 3d + e = 40$$
$$...$$
And what the problem asks for is $$P(6) = 1296a + 216b + 36c + 6d + e .$$
However, I'm not sure if all this is helping me very much. So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$) From solving simultaneous equations I got that $31a + 15b + 7c + 3d + e=0$ and similarly $211a + 65b + 19c + 5d + e=0$, but they seem rather useless at this point.
| The previous solutions calculate the polynomial, and then deduce the value of $P(6)$.
This was done correctly of course (the solution by @Martin R is particularly elegant).
However, it is possible to calculate the value of $P(6)$ directly, by calculating the finite differences and by using the fact that the 5th-order difference is equal to $0$ for polynomials of degree $4$.
$$
\begin{array}{c|rrrrrr}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
P(x) & 120 & 60 & 40 & 30 & 24 & \color{red}{40} \\
\Delta_1 & & -60 & -20 & -10 & -6 & \color{red}{16} \\
\Delta_2 & & & 40 & 10 & 4 & \color{red}{22} \\
\Delta_3 & & & & -30 & 6 & \color{red}{18} \\
\Delta_4 & & & & & 24 & \color{red}{24} \\
\Delta_5 & & & & & & \color{red}{0}
\end{array}
$$
In the above array, the values in black are derived from the input data, and the values in red are calculated, starting from the value $\Delta_5=0$.
And the solution is $P(6) = 40$.
| {
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Is there a special case for equations of curves that have parallel oblique asymtotes? I am given the function $f(x,y)=(x-y)^2(x^2+y^2)-10(x-y)x^2+12y^2+2x+y$ and asked whether the curve $f(x,y)=0$ has two parallel oblique asymptotes, two unique asymptotes, or if there is no oblique asymptote at all. Do I manually find the asymptote equations or there is some shortcut to this?
| The algebraic approach here is not too easy, so here's how to blend in some calculus (and some nontrivial computation, perhaps done with the help of a CAS). Of course, using a graphical CAS, you can easily see that there are two asymptotes with slope $1$. My original guess was that $y=x$ was an asymptote because of the leading factors of $x-y$. I was wrong.
Let $f(x,y) = (x-y)^2(x^2+y^2)-10(x-y)x^2+12y^2+2x+y$.
Using implicit differentiation, we can examine the limiting slope of the level curve $f(x,y)=0$ when $y=mx+b$ as $x\to\infty$. The leading behavior is
$$\frac{dy}{dx} \approx \frac{m^3-2m^2+3m-2}{2m^3-3m^2+2m-1} = \frac{(m^2-m+2)(m-1)}{(2m^2-m+1)(m-1)},$$
and this has limit $m$ precisely when $m=1$. So we look for slant asymptotes $y=x+b$.
Let's review what it means for $y=x+b$ to be an asymptote of the curve $y=g(x)$. This should mean that the vertical error between the curve and the line approaches $0$, i.e.,
$$\lim_{x\to\infty} \big(g(x)-(x+b)\big) = 0.$$
Now, consider our level curve $f(x,y)=0$. Using linear approximation, we have
$$0 = f(x,y) = f(x,x+b) + \frac{\partial f}{\partial y}(x,x+b)\big(y-(x+b)\big),$$
and so our vertical error is given by
$$y-(x+b) = -\frac{f(x,x+b)}{\frac{\partial f}{\partial y}(x,x+b)}. \tag{$\star$}$$
Start plugging away, substituting $y=x+b$ (so $x-y=-b$):
\begin{align*}
f(x,x+b) &= b^2(x^2+(x+b)^2)+10bx^2+12(x+b)^2 + 2x+x+b \\ &= 2(b^2+5b+6)x^2+2(b^3+12b)x+\dots.
\end{align*}
This will be smallest (for large $x$) when the leading coefficient vanishes, i.e., when $b=-2$ or $b=-3$. In these cases, $f(x,x+b)$ will be given by a linear polynomial. On the other hand, considering ($\star$),
$$\frac{\partial f}{\partial y}(x,x+b) = 2(2b+5)x^2 + \dots$$
is quadratic for both $b=-2$ and $b=-3$, so the quotient giving $y-(x+b)$ is the quotient of a linear polynomial by a quadratic polynomial, and hence goes to $0$ as $x\to\infty$.
Whew! In sum, the asymptotes are $y=x-2$ and $y=x-3$, as, indeed, @quasi commented.
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
| $4(x^3+2x^2+x)^3 . (3x^2+4x+1)$
$4x^3 (x^2+2x+1)^3 (3x^2+4x+1)$
$4x^3 (x+1)^6 (3x^2+4x+1)$
$4x^3 (x+1)^6 (3x+1)(x+1)$
$4x^3 (x+1)^7 (3x+1)$
So the book resolved your result in more factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Self-intersections of a Lissajous curve Hei,
I want to find the self intersections of a Lissajous curve, for instance:
$$x(t)=\sin 2t$$
$$y(t)=\sin 3t$$
I have been trying for a couple of hours but I really don't get how to compute all the solutions. Basically I was trying to write the relation such that
$$\sin 2u=\sin 2v \text{ and }\sin 3u=\sin 3v.$$
Thanks for the support!
| Start with a rational parametrization, using the tangent half angle formula:
$$c=\cos u=\frac{1-a^2}{1+a^2}\qquad s=\sin u=\frac{2a}{1+a^2}$$
Use Wikipedia to find formulas for double and triple angle.
$$\sin 2u=2sc=\frac{(1-a^2)2a}{(1+a^2)^2}
\qquad
\sin3u=3s-4s^3=\frac{6a(1+a^2)^2-32a^3}{(1+a^2)^3}$$
Now assume a second parameter $b$ instead of $a$ and write your equations, cross-multiplying in the process:
$$(1-a^2)2a(1+b^2)^2=(1-b^2)2b(1+a^2)^2\\
(6a(1+a^2)^2-32a^3)(1+b^2)^3=(6b(1+b^2)^2-32b^3)(1+a^2)^3$$
Now you have a polynomial system of two equations in two variables. You can cancel the trivial solution $a=b$ from both of them (e.g. by writing them as LHS − RHS = $0$, then factorizing and removing the factor $a-b$). Then using a computer algebra system of your choice, or some elimination technique like resultants, you find 6 algebraic solutions ignoring duplicates. Note that the tangent half-angle substitution omits one point, namely $a=\infty$ corresponding to $t=\pi$. You need to check that case separately, and will indeed find one more solution there. Turning parameters back into points you get these 7 points of intersection:
\begin{align*}
x &= \tfrac{1}{2} \, \sqrt{3} & y &= 0 &
\{a,b\}&=\{
-\sqrt{3},
\sqrt{\tfrac{1}{3}}
\}\\
x &= -\tfrac{1}{2} \, \sqrt{3} & y &= 0 &
\{a,b\}&=\{
\sqrt{3},
-\sqrt{\tfrac{1}{3}}
\}\\
x &= \tfrac12 & y &= \tfrac12\sqrt2 &
\{a,b\}&=\{
-\sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2,
\sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2
\}\\
x &= \tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} - 2,
\sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} - 2
\}\\
x &= -\tfrac{1}{2} & y &= -\tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} + \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2,
\sqrt{6} - \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2
\}\\
x &= -\tfrac{1}{2} & y &= \tfrac{1}{2} \, \sqrt{2} &
\{a,b\}&=\{
-\sqrt{6} - \tfrac{1}{2} \, \sqrt{-8 \, \sqrt{6} + 20} + 2,
\sqrt{6} + \tfrac{1}{2} \, \sqrt{8 \, \sqrt{6} + 20} + 2
\}\\
x &= 0 & y &= 0 & \{a,b\}&=\{0,\infty\}
\end{align*}
There are also complex solutions for $\{a,b\}=\{-i,+i\}$ but I assume these are not interesting to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}(\cos(x/2)+\cos (x/4))$
$f' (x)=\cos(3x/4)\cos (x/4))$
$f'(x)=0$ $\implies$ $\cos(3x/4)\cos (x/4))=0$
$\cos(3x/4)=0\:$ or $\:\cos (x/4)=0$
$3x/4= \pi/2\:$ and so $\:x= 2\pi/3$ or $ x=2\pi$
$x/4= \pi/2\:$ and so $\:x= 2\pi $
Is my work ok ? What should I do now?
| Hint:
Set $x=8y$ to find $$2\sin2y(1+\cos2y)=8\sin y\cos^3y$$
Now $\sqrt[4]{3\sin^2y\cos^6y}\le\dfrac{3\sin^2y+\cos^2y+\cos^2y+\cos^2y}4=\dfrac34$
Squaring we get $$16\sqrt3\sin y\cos^3y\le9$$
The equality occurs if $3\sin^2y=\cos^2y\iff\dfrac{\sin^2y}1=\dfrac{\cos^2y}3=\dfrac1{1+3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Show that the Huber-loss based optimization is equivalent to $\ell_1$ norm based. Dear optimization experts,
My apologies for asking probably the well-known relation between the Huber-loss based optimization and $\ell_1$ based optimization. However, I am stuck with a 'first-principles' based proof (without using Moreau-envelope, e.g., here) to show that they are equivalent.
Problem formulation
The observation vector is
\begin{align*}
\mathbf{y}
&= \mathbf{A}\mathbf{x} + \mathbf{z} + \mathbf{\epsilon} \\
\begin{bmatrix} y_1 \\ \vdots \\ y_N \end{bmatrix} &=
\begin{bmatrix}
\mathbf{a}_1^T\mathbf{x} + z_1 + \epsilon_1 \\
\vdots \\
\mathbf{a}_N^T\mathbf{x} + z_N + \epsilon_N
\end{bmatrix}
\end{align*}
where
*
*$\mathbf{A} = \begin{bmatrix} \mathbf{a}_1^T \\ \vdots \\ \mathbf{a}_N^T \end{bmatrix} \in \mathbb{R}^{N \times M}$ is a known matrix
*$\mathbf{x} \in \mathbb{R}^{M \times 1}$ is an unknown vector
*$\mathbf{z} = \begin{bmatrix} z_1 \\ \vdots \\ z_N \end{bmatrix} \in \mathbb{R}^{N \times 1}$ is also unknown but sparse in nature, e.g., it can be seen as an outlier
*$\mathbf{\epsilon} \in \mathbb{R}^{N \times 1}$ is a measurement noise say with standard Gaussian distribution having zero mean and unit variance normal, i.e. $\mathcal{N}(0,1)$.
We need to prove that the following two optimization problems P$1$ and P$2$ are equivalent.
P$1$:
\begin{align*}
\text{minimize}_{\mathbf{x},\mathbf{z}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1
\end{align*}
and
P$2$:
\begin{align*}
\text{minimize}_{\mathbf{x}} \quad & \sum_{i=1}^{N} \mathcal{H} \left( y_i - \mathbf{a}_i^T\mathbf{x} \right),
\end{align*}
where the Huber-function $\mathcal{H}(u)$ is given as
$$\mathcal{H}(u) =
\begin{cases}
|u|^2 & |u| \leq \frac{\lambda}{2} \\
\lambda |u| - \frac{\lambda^2}{4} & |u| > \frac{\lambda}{2}
\end{cases} .
$$
My partial attempt following the suggestion in the answer below
We attempt to convert the problem P$1$ into an equivalent form by plugging the optimal solution of $\mathbf{z}$, i.e.,
\begin{align*}
\text{minimize}_{\mathbf{x},\mathbf{z}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \\
\equiv
\\
\text{minimize}_{\mathbf{x}} \left\{ \text{minimize}_{\mathbf{z}} \right. \quad & \left. \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right\}
\end{align*}
Taking derivative with respect to $\mathbf{z}$,
\begin{align}
0 & \in \frac{\partial}{\partial \mathbf{z}} \left( \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right) \\
\Leftrightarrow & -2 \left( \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \right) + \lambda \partial \lVert \mathbf{z} \rVert_1 = 0 \\
\Leftrightarrow & \quad \left( \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \right) = \lambda \mathbf{v} \ .
\end{align}
for some $ \mathbf{v} \in \partial \lVert \mathbf{z} \rVert_1 $ following Ryan Tibshirani's lecture notes (slide#18-20), i.e.,
\begin{align}
v_i \in
\begin{cases}
1 & \text{if } z_i > 0 \\
-1 & \text{if } z_i < 0 \\
[-1,1] & \text{if } z_i = 0 \\
\end{cases}.
\end{align}
Then, the subgradient optimality reads:
\begin{align}
\begin{cases}
\left( y_i - \mathbf{a}_i^T\mathbf{x} - z_i \right) = \lambda \ {\rm sign}\left(z_i\right) & \text{if } z_i \neq 0 \\
\left| y_i - \mathbf{a}_i^T\mathbf{x} - z_i\right| \leq \lambda & \text{if } z_i = 0
\end{cases}
\end{align}
Also, following, Ryan Tibsharani's notes the solution should be 'soft thresholding' $$\mathbf{z} = S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right),$$
where
\begin{align}
S_{\lambda}\left( y_i - \mathbf{a}_i^T\mathbf{x} \right) =
\begin{cases}
\left( y_i - \mathbf{a}_i^T\mathbf{x} - \lambda \right) & \text{if } \left(y_i - \mathbf{a}_i^T\mathbf{x}\right) > \lambda \\
0 & \text{if } -\lambda \leq \left(y_i - \mathbf{a}_i^T\mathbf{x}\right) \leq \lambda \\
\left( y_i - \mathbf{a}_i^T\mathbf{x} + \lambda \right) & \text{if } \left( y_i - \mathbf{a}_i^T\mathbf{x}\right) < -\lambda \\
\end{cases} .
\end{align}
Now, we turn to the optimization problem P$1$ such that
\begin{align*}
\text{minimize}_{\mathbf{x}} \left\{ \text{minimize}_{\mathbf{z}} \right. \quad & \left. \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - \mathbf{z} \rVert_2^2 + \lambda\lVert \mathbf{z} \rVert_1 \right\} \\
\equiv
\end{align*}
\begin{align*}
\text{minimize}_{\mathbf{x}} \quad & \lVert \mathbf{y} - \mathbf{A}\mathbf{x} - S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right) \rVert_2^2 + \lambda\lVert S_{\lambda}\left( \mathbf{y} - \mathbf{A}\mathbf{x} \right) \rVert_1
\end{align*}
*
*if $\lvert\left(y_i - \mathbf{a}_i^T\mathbf{x}\right)\rvert \leq \lambda$, then So, $\left[S_{\lambda}\left( y_i - \mathbf{a}_i^T\mathbf{x} \right)\right] = 0$.
the objective would read as $$\text{minimize}_{\mathbf{x}} \sum_i \lvert y_i - \mathbf{a}_i^T\mathbf{x} \rvert^2, $$ which is easy to see that this matches with the Huber penalty function for this condition. Agree?
*
*if $\lvert\left(y_i - \mathbf{a}_i^T\mathbf{x}\right)\rvert \geq \lambda$, then $\left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right)$.
the objective would read as $$\text{minimize}_{\mathbf{x}} \sum_i \lambda^2 + \lambda \lvert \left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right) \rvert, $$ which almost matches with the Huber function, but I am not sure how to interpret the last part, i.e., $\lvert \left( y_i - \mathbf{a}_i^T\mathbf{x} \mp \lambda \right) \rvert$. Please suggest...
| Consider the proximal operator of the $\ell_1$ norm
$$
z^*(\mathbf{u})
=
\mathrm{argmin}_\mathbf{z}
\
\left[
\frac{1}{2}
\| \mathbf{u}-\mathbf{z} \|^2_2
+
\lambda \| \mathbf{z} \|_1
\right]
=
\mathrm{soft}(\mathbf{u};\lambda)
$$
In your case, (P1) is thus equivalent to
minimize
\begin{eqnarray*}
\phi(\mathbf{x})
&=&
\lVert \mathbf{r} - \mathbf{r}^* \rVert_2^2 + \lambda\lVert \mathbf{r}^* \rVert_1
\\
&=&
\sum_n |r_n-r^*_n|^2+\lambda |r^*_n|
\end{eqnarray*}
with the residual vector
$\mathbf{r}=\mathbf{A-yx}$ and its
soft-thresholded version
$\mathbf{r}^*=
\mathrm{soft}(\mathbf{r};\lambda/2)
$.
Note further that
$$
r^*_n
=
\left\lbrace
\begin{array}{ccc}
r_n-\frac{\lambda}{2} & \text{if} &
r_n>\lambda/2 \\
0 & \text{if} & |r_n|<\lambda/2 \\
r_n+\frac{\lambda}{2} & \text{if} &
r_n<-\lambda/2 \\
\end{array}
\right.
$$
\noindent
In the case $r_n>\lambda/2>0$,
the summand writes
$\lambda^2/4+\lambda(r_n-\frac{\lambda}{2})
=
\lambda r_n - \lambda^2/4
$
\
In the case $|r_n|<\lambda/2$,
the summand writes
$|r_n|^2
$
\
In the case $r_n<-\lambda/2<0$,
the summand writes
$\lambda^2/4 - \lambda(r_n+\frac{\lambda}{2})
=
-\lambda r_n - \lambda^2/4
$
Finally, we obtain the equivalent
minimization problem
$$
\phi(\mathbf{x})
=\sum_n \mathcal{H}(r_n)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| Well, obviously $6n^2$ is divisible by $3$.
And $4n^3-4n=4(n-1)n(n+1)$. Since $(n-1)n(n+1)$ is the product of three consecutive integers, one of them is divisible by $3$ and therefore their product has that property too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$ Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$
It seemed easy for me, but I couldn't do it. By Wolfram Alpha, I got that $x=1,3$, but I don't know how to get there.
Any hints?
Thanks.
| $$2^{x-1}-3 \sqrt{2^{x-1}}+2=0$$
$$2^{x-1}+2=3 \sqrt{2^{x-1}}$$
$$9\cdot 2^{x-1}=2^{(x-1)*2}+2^{x+1}+4$$
$$9\cdot2^x\cdot2^{-1}=2^{2x-2}+2^x\cdot2+4$$
$$9\cdot2^x\cdot\frac{1}{2}=2^{2x}\cdot2^{-2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{2^2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{4}+2^x\cdot2+4$$
Set $ n=2^x$
$$\frac{9}{2}n=n^2\cdot\frac{1}{4}+n\cdot2+4$$
$$\frac{9}{2}n=\frac{n^2}{4}+2n+4$$
$$0=\frac{n^2}{4}-\frac{5}{2}n+4$$
$$0=n^2-10n+16$$
Factor:$$(n-8)(n-2)=0$$
$$n=2, 8$$
Since $n=2^x$, we get
$$2^x=2$$
$$2^x=8$$
Solving these, we get
$$x=1$$
$$x=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$
Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.
Let
$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$
Put $x=2\sin^2 \theta$ and $dx = 4\sin \theta \cos \theta \,\mathrm d \theta$ and changing limits, We get
$$\displaystyle I_{m,n} = 2^{m+n+2}\int^{\frac{\pi}{2}}_{0}\cos^{2m+1}\theta\cdot \sin^{2n+1}\theta \,\mathrm d \theta$$
How can I form a recursive relation? Could someone help me? Thanks.
| We first express the integrand as a product of $\sin \theta$ and $\cos \theta $ by
letting $x=\cos 2\theta $, yields
$$I = \int_{\frac{\pi}{4}}^{0}\left(2 \cos ^{2} \theta\right)^{m}\left(2 \sin ^{2} \theta\right)^{n}\left(-2 \sin ^{2} \theta\right) d \theta
=2^{m+n+2} \int_{0}^{\frac{\pi}{4}} \cos ^{2 m+1} \theta \sin^{2n+1} \theta d\theta
$$
Putting $s=\sin \theta$ gives $$
I=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}}\left(1-s^{2}\right)^{m} s^{2 n+1} d s .
$$
Using Binomial Expansion gives $$
\begin{aligned}
I&=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}\left(\begin{array}{c}
m \\
k
\end{array}\right)(-1)^{k} s^{2 k} s^{2 n+1} d s\\
&=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) s^{2 k+2 n+1} d s\\
&=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \int_{0}^{\frac{1}{\sqrt{2}}} s^{2 k+2 n+1} d s\\
&=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{l}
m \\
k
\end{array}\right)\left[\frac{s^{2 k+2 n+2}}{2 k+2 n+2}\right]_{0}^{\frac{1}{\sqrt{2}}}\\
&=2^{m+n+1} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \frac{1}{(k+n+1) 2^{k+n+1}}\\
&=2^{m} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \frac{1}{(k+n+1) 2^{k}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Algebra and matrices. Q. Solve for $x$, $y$ and $z$ if
$$(x+y)(x+z)=15, $$
$$(y+z)(y+x)=18, $$
$$(z+x)(z+y)=30. $$
Solution:
I expanded each equation above as :
$$x^2+xz+yx+yz=15, \tag{1}$$
$$y^2+xz+yx+yz=18, \tag{2}$$
$$z^2+xz+yx+yz=30. \tag{3}$$
Then I subtracted $(1)-(3)$, $(2)-(1)$ and $(3)-(2)$; so I got the equations as below:
$$x^2-z^2=-15 \tag{4}$$
$$y^2-x^2=3 \tag{5}$$
$$z^2-y^2=12 \tag{6}$$
I then tried solving $(4)$,$(5)$,$(6)$ by using matrices, but I couldn't reach any solution.
Please advise. Thank You.
| Take $x+y=a,y+z=b,x+z=c$. Then you get
$ac=15\quad[1]$
$ab=18\quad[2]$
$bc=30\quad[3]$
Multiply to get $(abc)^2=8100$, thus getting $abc=90$.
Divide by $[1],[2],[3]$ separately to get
$a=3$
$b=6$
$c=5$
Now, substitute to get
$x+y=3\quad[4]$
$y+z=6\quad[5]$
$x+z=5\quad[6]$
Add to get $x+y+z=7$
Subtract separately to get
$x=1,y=2,z=4$.
EDIT
As @hkBst pointed out, we could also get $abc=-90$, thus giving
$x=-1,y=-2,z=-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to get rid of the absolute value function when solving an ODE? Problem: $y'+yx=0, \quad y(0)=-1$
We separate it:
$\frac{dy}{dx}=-yx \Rightarrow \int\frac{-1}{y}dy=\int x dx = \frac{1}{2}x^2 + C_1$
With $\int\frac{-1}{y}dy=\log(\frac{1}{|y|})+C_2$
we get
$\log(\frac{1}{|y|})=\frac{1}{2}x^2 + C$
We solve for $y$:
$|y|=e^{-\frac{1}{2}x^2}\cdot e^C=e^{-\frac{1}{2}x^2}\cdot \hat{C}$
Now, what is the best argumentation to get "rid" of $|\cdot |$?
Like I know that e.g. $|a|=b \Leftrightarrow a=\pm b$ but then I still have $\pm$. I "know" that some wil ltell me that I can "put it into $C$" but that not really an arugmentation. Maybe I just don't get how $C$ can determine if we have the positive or negative solution or why we can't have both at the same time. It just feels like I lack proper understanding to properly argument here.
|
$y'+yx=0$
$$
\frac{dy}{y} = -xdx \Rightarrow \int \frac{dy}{y} = \int -xdx \Rightarrow \ln |y| = -\frac{1}{2}x^2 + C_1.
$$
*
*$y > 0$. we have $\ln y = -\frac{1}{2}x^2 + C_1$, so $y = C_1e^{-\frac{1}{2}x^2}$.
*$y < 0$. we have $\ln -y = -\frac{1}{2}x^2 + C_1$, so $y = C_2e^{-\frac{1}{2}x^2}$.
*$y = 0$. we have $y = 0$, so $y = 0 \cdot e^{-\frac{1}{2}x^2}$.
Thus we can use constant $C$ replace absolute value sign, $y= Ce^{-\frac{1}{2}x^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int_{\frac{-1}{2}}^{\frac{1}{2}} \int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{x^2}{(x^2+y^2)^2 \log^2(\frac{2}{\sqrt{x^2+y^2}})} \,dx\,dy.$
$$\int_{\frac{-1}{2}}^{\frac{1}{2}} \int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{x^2}{(x^2+y^2)^2 \log^2(\frac{2}{\sqrt{x^2+y^2}})} \,dx\,dy.$$
I encountered this integral while evaluating norm of a function. The first attempt was to change it into polar coordinates, which didn't work well, since the region of integration is rectangular.
I found a similar integral here, but even WolframAlpha couldn't calculate this.
Does anybody know how to evaluate this?
| We may tackle the equivalent integral
$$ \mathcal{J}\stackrel{\text{def}}{=}\iint_{(0,1)^2}\frac{dx\,dy}{(x^2+y^2)\log^2\left(\frac{1}{4}\sqrt{x^2+y^2}\right)} $$
by writing it as
$$ \int_{0}^{1}\frac{\frac{\pi}{2}\rho}{\rho^2\log^2\left(\rho/4\right)}\,d\rho +\int_{1}^{\sqrt{2}}\frac{\left(\frac{\pi}{2}-2\arccos\frac{1}{\rho}\right)\rho}{\rho^2\log^2(\rho/4)}\,d\rho$$
which equals
$$ \frac{\pi}{3\log 2}-2\int_{1}^{\sqrt{2}}\frac{\arccos\frac{1}{\rho}}{\rho \log^2(\rho/4)}\,d\rho=\frac{\pi}{3\log 2}-2\int_{1/\sqrt{2}}^{1}\frac{\arccos\rho}{\rho\log^2(4\rho)}\,d\rho $$
or
$$ \frac{\pi}{6\log 2}+2\int_{1/\sqrt{2}}^{1}\frac{\arcsin\rho}{\rho\log^2(4\rho)}\,d\rho = \frac{\pi}{3\log 2}-\int_{0}^{\pi/4}\frac{\theta\tan\theta}{\log^2(4\cos\theta)}\,d\theta.$$
The last integral is very far from being elementary or expressible through a simple hypergeometric series, but $\frac{\theta\tan\theta}{\log^2(4\cos\theta)}$ approximately behaves like $C\theta^4$ on $(0,\pi/4)$, hence the composite Boole's rule provides excellent numerical approximations with few computations. We get
$$ \mathcal{J}\approx 1.239117616448\approx \frac{35725}{28831}. $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable
$A=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$
$A^2=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}^2P=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}^2P$
So $P_{A^2}(x)=x(x-1)^2$ but $A^2(A^2-1)=0$ so $M_{A^2}(x)=x(x-1)=x^2-x$
Is there another way to conclude it?
| First you can plug in $A^2$ to see whether it satisfies the polynomial. Indeed:
$$(A^2)^2 - A^2 = A^4 - A^2 = A A^3 - A^2 = AA - A^2=0$$
Now check that $A^2$ doesn't satisfy the linear factors $x$ and $x-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question about Riemann-Stieltjes problem with Euler Use Euler summation formula and integration by parts to show that
$$
\int_1^3 \left(\frac{[[x]]}{x^3}\right)dx = \frac{37}{72}
$$
Sorry I'm still learning how to use MathJax or how to put math formulas into the text...
Currently, I'm an undergraduate studying Math Analysis. Any help would be greatly appreaciated! Thanks for your time
| Use the real function defined by $f(x)=\frac{-1}{2x^2}\Rightarrow f'(x)=\frac{1}{x^3}$
Euler summation formula implies:
$$\sum_{n=1}^3f(n)=\int_1^3f(x)dx+\int_1^3f'(x)(x-[|x|]-\frac{1}{2})dx+\frac{f(1)+f(3)}{2}$$
$$\sum_{n=1}^3(\frac{-1}{2n^2})=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^3}(x-[|x|]-\frac{1}{2})dx-\frac{5}{18}$$
$$\sum_{n=1}^3(\frac{-1}{2n^2})=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^2}dx-\int_1^3\frac{[|x|]}{x^3}dx-\frac{1}{2}\int_1^3\frac{1}{x^3}dx-\frac{5}{18}$$
Thus, you will get:
$$\int_1^3\frac{[|x|]}{x^3}dx=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^2}dx-\frac{1}{2}\int_1^3\frac{1}{x^3}dx-\frac{5}{18}-\sum_{n=1}^3(\frac{-1}{2n^2})=\frac{37}{72}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Rotation of ellipsoid(quadric) Consider $$φ(x, y , z) = x^2 + 2y^2 + 4z^2 −xy −2xz −3yz$$
find the coordinate transformation (translation or rotation) to eliminate $xy$, $xz$ and $yz$.
In $\mathbb R²$, with conic sections, I would do this with
$$\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\cos\alpha &-\sin\alpha\\\sin\alpha & \cos\alpha\end{pmatrix}.\begin{pmatrix}x' \\ y'\end{pmatrix}$$
to find the rotation angle. But how is it done with quadrics?
| Perhaps this will help:
\begin{align}
\phi(x,y,z) &= x^2 -x(y+2z)+{(y+2z)^2\over 4} -{(y+2z)^2\over 4}+ 2y^2 + 4z^2 −3yz\\
& = (x-{y+2z\over 2})^2+{7\over 4}y^2+3z^2-4yz\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}(36z^2-48yz+21y^2)\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}((6z)^2-2 \cdot4y\cdot 6z+16y^2+5y^2)\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}(6z-4y)^2+{5\over 12}y^2\\
& = {1\over 4}(2x-y-2z)^2+{1\over 3}(3z-2y)^2+{5\over 12}y^2
\end{align}
$$\begin{pmatrix}x'\\ y'\\z' \end{pmatrix}=
\begin{pmatrix}
1 &-1/2&-1\\
0 &\sqrt{5\over 12}&0\\
0 &-2\over \sqrt{3}& 3
\end{pmatrix}
\begin{pmatrix}x \\ y\\z \end{pmatrix}$$
| {
"language": "en",
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"source": "stackexchange",
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What is going wrong in this log expansion? I am getting a weird result here:
Let $p_1 = q_1 + \Delta$ and Let $p_2 = q_2 - \Delta$
I use the expansion $\log(1-x) = -x -x^2/2 -x^3/3 - ...$ in the third step. This expansion is valid for $x<1$ so there should be no problem with that.
\begin{align*}
D(\vec{p}||\vec{q}) &= p_1\log{\frac{p_1}{q_1}} + p_2\log{\frac{p_2}{q_2}}\\
&= p_1\log{\frac{q_1 + \Delta}{q_1}} + p_2\log{\frac{q_2 - \Delta}{q_2}}\\
&= p_1(\frac{\Delta}{q_1} - \frac{\Delta^2}{q_1^2} + \frac{\Delta^3}{q_1^3} - ...) + p_2(-\frac{\Delta}{q_2} - \frac{\Delta^2}{q_2^2} - \frac{\Delta^3}{q_2^3} - ...)\\
&= (q_1 + \Delta)(\frac{\Delta}{q_1} - \frac{\Delta^2}{q_1^2} + \frac{\Delta^3}{q_1^3} - ...) + (q_2 - \Delta)(-\frac{\Delta}{q_2} - \frac{\Delta^2}{q_2^2} - \frac{\Delta^3}{q_2^3} - ...)\\
&=0
\end{align*}
The last equality follows from comparing terms of powers of $\Delta$ (am I not allowed to do that?).
| As you wrote, $$\log(1-x)\sim -x-\frac12x^2-\frac13x^3-\dots\ne-x-x^2-x^3-\dots$$
| {
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Ways of $30$ people ordering from $5$ spicy dishes and $45$ normal dishes
A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of 30 people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if
a) each dish has to be different?
b) a dish can be ordered multiple times (but still at most $2$ spicy dishes)?
I know my original answers were wrong. While going over the problem, I think I realized what my mistake was, but want to check. Originally, I thought:
a) Let $A_0$ be the case that $0$ spicy dishes are ordered, $A_1$ that $1$ spicy dish is ordered, and $A_2$ that $2$ spicy dishes are ordered.
Case $A_0$: Choose $1$ person from $30$ to order $1$ from $45$ available dishes. Choose $2$nd person from remaining $29$ to choose one of $44$ dishes. Continuing, we get
$$\binom{30}{1}\binom{45}{1}\binom{29}{1}\binom{44}{1} \ldots = 30! \frac{45!}{(45-30)!}$$
Since we only care about final arrangement (not who orders first, second,..), answer simplifies to ${45}\choose{30}$
case $A_1$: Just have to choose $1$ person from $30$ to order one of $5$ spicy dishes. Remaining part is as in part a, except with $29$ people: $$30\cdot 5\cdot \binom{45}{29}$$
case $A_2$: choose $2$ people to order spicy dishes, choose $2$ of $5$ spicy dishes, and as before, we get: $$30\cdot 29 \cdot 5\cdot 4\cdot \binom{45}{28}$$
Was my mistake considering who will order the spicy dishes in $A_1$ and $A_2$; Should the answer have just been
$$A_0 + A_1 + A_2 = \binom{45}{30} + \binom{5}{1}\binom{45}{29} +\binom{5}{2}\binom{45}{28}?$$
B)
Let $B_0$ be $0$ spicy dishes ordered, each dish can be ordered as many times as possible (except at most $2$ spicy dishes). Similar for $B_1$ and $B_2$.
-case $B_0$: $45$ options for each of $30$ people, so $|B_0| = 45^{30}$
-case $B_1$: Someone chooses $1$ spicy dish, $45$ options for each of $29$ people, so $|B_1|=5 \cdot 45^{30}$ (I originally put $30 \cdot 5 \cdot 45^{29}$ but I know this was wrong)
-case $B_2$: as before, $|B_0| = 5^2 \cdot 45^{28}$ (again, originally had $30 \cdot 29 \cdot 5^2 \cdot 45^{28}$)
|
A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of $30$ people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if each dish is different?
The number of ways a subset of exactly $k$ of the $5$ spicy dishes and exactly $30 - k$ of the $45$ normal dishes can be selected is
$$\binom{5}{k}\binom{45}{30 - k}$$
Thus, the number of selections of $30$ different dishes that contain at most two spicy dishes is
$$\sum_{k = 0}^{2} \binom{5}{k}\binom{45}{30 - k} = \binom{5}{0}\binom{45}{30} + \binom{5}{1}\binom{45}{29} + \binom{5}{2}\binom{45}{28}$$
However, it matters which customer receives which dish. Therefore, we must multiply by the $30!$ ways of assigning the selected dishes to customers. Hence, the number of possible orders that can be placed is
$$30!\sum_{k = 0}^{2} \binom{5}{k}\binom{45}{30 - k} = 30!\left[\binom{5}{0}\binom{45}{30} + \binom{5}{1}\binom{45}{29} + \binom{5}{2}\binom{45}{28}\right]$$
A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of $30$ people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if a dish can be ordered multiple times?
There are $\binom{30}{k}$ ways for exactly $k$ of the customers to order a spicy dish. Each of those $k$ customers has $5$ choices. Each of the remaining $30 - k$ customers has $45$ choices. Hence, there are
$$\binom{30}{k}5^k45^{30 - k}$$
orders in which exactly $k$ of the customers orders a spicy dish. Since at most two customers order a spicy dish, the number of possible orders is
$$\sum_{k = 0}^{2} \binom{30}{k}5^k45^{30 - k} = \binom{30}{0}5^045^{30} + \binom{30}{1}5^145^{29} + \binom{30}{2}5^245^{28}$$
| {
"language": "en",
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find: $\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$ Find: $$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ as $x\in\mathbb{R}$
My progress:
$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\lim_{n\longrightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)}{\sin\left(x+\frac{1}{n}\right)}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=$$
$$\lim_{n\longrightarrow\infty} \ \ {1}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=1-\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ at this point I got stuck.
I can't evaluate the Taylor series of $\sin(x+\frac{1}{n})$ because $n$ is not fixed
(even if we'll suppose that there exist some $\epsilon>0$ and there exists $ N\in\mathbb{N}:\forall n\geq N$ s.t:
$$-\epsilon<\frac{1}{n}<\epsilon$$
it doesn't seem like a formal argument to me)
(I might be very wrong - it's only my intuition).
Also trying to apply L'Hopital's rule for this expression isn't much helpful.
| use $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
$$\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
= $$\frac{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n}) - \sin(x)}{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n})}$$
for sin(x) not zero
$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
$$=\lim_{n\rightarrow\infty}\frac{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n}) - \sin(x)}{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n})}$$
$$=\frac{\sin(x)\cos(0) + \cos(x)\sin(0) - \sin(x)}{\sin(x)\cos(0) + \cos(x)\sin(0)}$$
$$=\frac{\sin(x).1 + \cos(x).0 - \sin(x)}{\sin(x).1 + \cos(x).0}$$
$$=\frac{\sin(x) - \sin(x)}{\sin(x)}$$
$$ = 0$$
for sin(x) = 0, you can again plug in values and simplify to
$$\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841804",
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"source": "stackexchange",
"question_score": "3",
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Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks
| By Cauchy inequality we have:
$$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$$
and
$$(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$$
Since $(a_1+a_3)^2=(a_2+a_4)^2$ we have $$2(a_1+a_3)^2 \leq 2(a_1^2+a_3^2) +2(a_2^2+a_4^2) =2$$
So $$ E \leq 4$$
| {
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"source": "stackexchange",
"question_score": "2",
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Explain why would someone factorise $-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$ to get $\frac{1}{6}(3+2)^3$ I am confused by the workflow of my book.
It went from $$-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$$ to $$-\int_{-2}^{3}(x+2)(x-3)\,\mathrm dx$$ and, finally $$\frac{1}{6}(3+2)^3$$
| We have that more generally
$$\begin{align}
-\int_{-a}^{b}(x+a)(x-b)dx&=-\int_{-a}^{b}(x+a)(x+a-(a+b))dx\\&=-\left[\frac{(x+a)^3}{3}-(a+b)\frac{(x+a)^2}{2}\right]_{-a}^{b}\\&=-\frac{(a+b)^3}{3}+\frac{(a+b)(a+b)^2}{2}=\frac{(a+b)^3}{6}.
\end{align}$$
This is also a particular the area of a parabolic segment where $w=a+b$, $h=\frac{a+b}{4}$ and the area is
$$\frac{4}{3}\cdot\frac{w\cdot h}{2}=\frac{(a+b)^3}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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how to prove this limit is zero? $$\lim_{n\to\infty}(\frac{\sqrt{1\times2}}{n^2+1}+\frac{\sqrt{2\times3}}{n^2+2}+\cdots+\frac{\sqrt{n\times(n+1)}}{n^2+n})=0$$
I tried to use $\sqrt{k(k+1)}\le\frac{2k+1}{2}$ or $k(k+1)=\sqrt{k}\times\sqrt{k}\times(k+1)\le\left(\frac{2\sqrt{k}+k+1}{3}\right)^3$ but nothing worked out. Can you show me some hint to deal with $\frac{\sqrt{k(k+1)}}{n^2+k}$ ? Thanks!
| If you enjoy harmonic numbers, starting with
$$\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i} \lt \sum \limits_{i=1}^{n}\dfrac{\sqrt{i(i+1)}}{n^2+i}\lt \sum \limits_{i=1}^{n}\dfrac{i+1}{n^2+i}$$ you have
$$S_1=\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i}=n+n^2\left(H_{n^2}- H_{n^2+n}\right)$$
$$S_2=\sum \limits_{i=1}^{n}\dfrac{i+1}{n^2+i}=n+(n^2-1)\left(H_{n^2}- H_{n^2+n}\right)$$ Now, using the asymptotics and continuing with Taylor series
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^3}\right)$$ for large values of $n$, you should have
$$S_1=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
$$S_2=\frac{1}{2}+\frac{7}{6 n}-\frac{3}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
| {
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Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but it is hard to guess here.
So I set $z=x+iy$ and got
$$(x^3-y^2x+2xy^2-2x^2+2y^2+\frac{1}{4})+i(-2x^2y+yx^2-y^3+4xy)=0$$
Looking at the imaginary part we get
$y(-x^2+4x-y^2)=0$ so or $y=0$ or $-x^2+4x-y^2=0\iff (x-2)^2+y^2=4$
But how I continue from here? and what does
I read and now I am given it a try:
for a. we look at $|z|<1$ and $\frac{1}{4}<|z|$ we look at the boundary
so fo $|z|=1$ $|z^3+\frac{1}{4}|<|-2z^2|$ so we choose $g(z)=z^3+\frac{1}{4}$ and $f(z)=-2z^2$
So $|z^3+\frac{1}{4}|\leq 1+\frac{1}{4}\leq 2=|-2z^2|$
So we can conclude that there are $2$ zeros in $|z|<1$?
| Writing $z = (4 \sin(s) +2)/3$, the equation becomes
$$ 0 = 4 \sin(s)^3 - 3 \sin(s) - \frac{37}{64} = - \sin(3s) - \frac{37}{64}$$
This is $0$ for $s = -\arcsin(37/64)/3+2 \pi k/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving this quadratic equation by completing the square? $-\frac{2}{3}x^{2} - x +2 = 0$
Here's what I did:
However, the textbook answer is $x = -2.6, x = 1.1$. What did I do wrong?
| \begin{align*}
-\frac{2}{3}x^2 - x + 2 & = 0\\
-\frac{2}{3}\left(x^2 + \frac{3}{2}x\right) + 2 & = 0\\
-\frac{2}{3}\left[x^2 + \frac{3}{2}x + \left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right] \color{red}{- \left[-\frac{2}{3}\left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right]} + 2 & = 0 \tag{1}\\
-\frac{2}{3}\left(x^2 + \frac{3}{2}x + \frac{9}{16}\right) + \frac{3}{8} + 2 & = 0\\
-\frac{2}{3}\left(x + \color{red}{\frac{3}{4}}\right)^2 + \frac{19}{8} & = 0 \tag{2}\\
-\frac{2}{3}\left(x + \frac{3}{4}\right)^2 & = -\frac{19}{8}\\
\left(x + \frac{3}{4}\right)^2 & = \frac{57}{16}\\
\left|x + \frac{3}{4}\right| & = \sqrt{\frac{57}{16}}\\
x + \frac{3}{4} & = \pm \frac{\sqrt{57}}{4} \tag{3}\\
x & = -\frac{3}{4} \pm \frac{\sqrt{57}}{4}\\
x & = \frac{-3 \pm \sqrt{57}}{4}
\end{align*}
(1): If you add a term to one side of the equation, you must either add it to the other side of the equation or subtract it from the same side of the equation to balance the equation. You omitted the term in red.
(2): Note that
$$\frac{9}{16} = \left(\frac{3}{4}\right)^2$$
You made the mistake of treating $9/16$ as the square of $3/2$.
(3): Since $16$ is a perfect square, you do not need to rationalize the denominator in this step. What you did was correct, but introducing extra steps introduces extra opportunities to make an error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Compute $\int_{0}^{\pi/2}\csc^2(x)\ln\left(\frac{a+b\sin^2 x}{a-b\sin^2 x}\right)dx$
I would like to compute $I$, using a different method, other than below
$$I=\large\int_{0}^{\pi/2}\csc^2(x)\ln\left(\frac{a+b\sin^2 x}{a-b\sin^2 x}\right)\mathrm dx$$
$a\ge b$
Applying integration by parts:
$\large u=\ln\left(\frac{a+b\sin^2 x}{a-b\sin^2 x}\right)$
$\large dv=\csc^2(x)dx$
$$I=2a\int_{0}^{\pi/2}\frac{\cot^2 x}{a\cot^2 x-b+a}\cdot \frac{\csc^2 x}{a\cot^2 x +b+a}\mathrm dx$$
Making a sub of $y=\cot x$ and using partial fractions decomposition and so on to solve for I.
Finally arrived at $$I=\pi\cdot \frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a}}$$
| As suggested in comments by @Dahaka
Let $I(b)$ represent the given integral. Hence we have $I(0)=0$.
Now differentiating $I(b)$ with respect to $b$ we get $$I'(b) =\int_0^{\infty} \csc ^2x \cdot \frac {a-b\sin^2x}{a+b\sin^2x}\cdot\left( \frac {\sin^2x(a-b\sin^2x)+\sin^2x(a+b\sin^2x)}{(a-b\sin^2x)^2}\right) dx$$
Simplifying we get $$I'(b) =\int_0^{\infty} \left(\frac {1}{a-b\sin^2x} -\frac {1}{a+b\sin^2x}\right)dx$$
On dividing both the numerator and denominator by $\cos^2x$ and then letting $u$ substitution $u=\tan x$ we get $$I'(b) =\int_0^{\infty} \left(\frac {1}{a+(a-b)u^2}+ \frac {1}{a+(a+b)u^2}\right) du$$
And both of these simply evaluate using the fact that $$\int \frac {dx}{a^2+x^2}=\frac 1a \arctan \left(\frac xa\right)+ C$$
Hence $$I'(b) =\frac {\pi}{2\sqrt a}\left( \frac {1}{\sqrt {a-b}}+\frac {1}{\sqrt {a+b}}\right) $$
And now integrating again with respect to $b$ yields $$I(b)=\frac {\pi(\sqrt {a+b}-\sqrt {a-b})}{\sqrt a} +C$$
But using the fact that $I(0)=0$ we get $C=0$ and hence $$I(b)=\frac {\pi(\sqrt {a+b}-\sqrt {a-b})}{\sqrt a} $$
Q. E. D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| We will square the whole inequality
$$\frac{(a+b)^2}{(ab)^2}+(a+b)^2+2\frac{(a+b)^2}{ab}\geq 18$$
simplifying and using that $$a^2+b^2=1$$ we get
$$2(ab)^3-13(ab)^2+4ab+1\geq0$$
this is equivalent to $$(2ab-1)((ab)^2-6ab-1)\geq 0$$
Now we have $$a^2+b^2\geq 2ab$$ this is $$ab\le \frac{1}{2}$$
so both factors $$2ab-1,(ab)^2-6ab-1$$ are non posivite, thus their product is non negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 1
} |
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
| From
$$
\frac{b+1}{a+b-2}= u\Rightarrow L\to a = 2-b\frac{b+1}{u}
$$
So $L$ should be tangent to $a^2+b^2=1\;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = \frac{(b+1)^2}{b^2-b\pm\sqrt{(1-b) (b+1)^3}-2}
$$
but tangency implies on $\sqrt{(1-b) (b+1)^3}=0\;$ hence the solutions for tangency are $b = \pm 1$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
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