Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$ While calculating the aforementioned limits I did the following.
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -4\sqrt2 \left (\frac{1}{\sqrt 2} \cos x +\frac{1}{\sqrt 2} \sin x \right )^5}{1-\sin {2x}}\\=
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin^5 (x + \frac{\pi}{4}) \right )}{1-\sin {2x}}\\
=\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin (x + \frac{\pi}{4}) \right ) \left(1 +\sin (x + \frac{\pi}{4}) +\sin^2 (x + \frac{\pi}{4}) +\sin^3 (x + \frac{\pi}{4}) +\sin^4 (x + \frac{\pi}{4}) \right)}{1-\sin {2x}}\\$$
As $x$ approaches $\frac{\pi}{4}$, the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough. So, I canceled them($1-\sin {2x}$ and $ 1 - \sin (x + \frac{\pi}{4})$ ) and the limit turns out to be $20\sqrt2$.
I made the above claim after comparing the graphs. But the fact that bothers me is L'Hospital's rule yielded the answer $5\sqrt2$ (and it is the correct answer to the problem)
Can anyone please explain?
| Make life easier using $x=y+\frac{\pi}{4}$
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos (x) + \sin (x) \right )^5}{1-\sin ({2x})}$$ Combine
$$\cos (x) + \sin (x)=\sqrt{2} \cos (y)$$ So, in terms of $y$, the expression reduces to
$$\frac{4\sqrt 2 -4 \sqrt{2} \cos ^5(y) )}{2 \sin ^2(y)}=2\sqrt 2\,\frac{1- \cos ^5(y) }{ \sin ^2(y)}$$
Now Taylor series
$$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$
$$\cos^5(y)=1-\frac{5 y^2}{2}+\frac{65 y^4}{24}+O\left(y^6\right)$$
Just finish
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$ How to evaluate:
$$\lim\limits_{n\rightarrow +\infty}\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right)$$
Personally, i always have trouble to evaluate sum of trignometric series, if you know a paper to recommend, or part of a book, about this, please let me know.
Anyway
$$\frac{1}{n^2}\sum_{k=1}^{n} \cot^2\left(\frac{\pi k}{2n+1}\right) = \frac{1}{n^2}\left(\sum_{k=1}^{n} \frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} - \sum_{k=1}^{n} 1\right)$$
What now?
| It is possible to prove, by studying the function $ f: x\mapsto 1-x^{2}\cot^{2}{x}-\frac{2 x^{2}}{3} $ on $ \mathcal{D}=\left]-\frac{\pi}{2},\frac{\pi}{2}\right[\setminus\left\lbrace 0\right\rbrace $, that for any $ x\in\mathcal{D} $, we have : $$ \left|1-x^{2}\cot^{2}{x}\right|\leq\frac{2x^{2}}{3} $$
Let $ n \in\mathbb{N}^{*} $, and $ k\leq n $. Setting $ x\leftarrow \frac{k\pi}{2n+1} $, we get : $$ \left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|\leq\frac{2k^{2}\pi^{2}}{3\left(2n+1\right)^{2}} $$
Using the previous inequality, we can write the following : \begin{aligned} \left|\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}-\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}\right|&\leq\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}\left|1-\frac{k^{2}\pi^{2}}{\left(2n+1\right)^{2}}\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}\right|}\\ &\leq \frac{2n}{3\left(2n+1\right)^{2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Thus, the sequence $ \left(\frac{1}{\left(2n+1\right)^{2}}\sum\limits_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}\right)_{n\in\mathbb{N}^{*}} $ does converge and : $$ \lim_{n\to +\infty}{\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}=\lim_{n\to +\infty}{\frac{1}{\pi^{2}}\sum_{k=1}^{n}{\frac{1}{k^{2}}}}=\frac{1}{6} $$
Hence : \begin{aligned}\lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\lim_{n\to +\infty}{\left(\frac{2n+1}{n}\right)^{2}\times\frac{1}{\left(2n+1\right)^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}\\ &=4\times\frac{1}{6}\\ \lim_{n\to +\infty}{\frac{1}{n^{2}}\sum_{k=1}^{n}{\cot^{2}{\left(\frac{k\pi}{2n+1}\right)}}}&=\frac{2}{3}\end{aligned}
| {
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Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$ Using AM-GM inequality we have:
$$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$
As it turns out it is not the minimum, can you spot the mistake?
| There is no flaw in your work - it is true that $$2x^3 + x + \frac{3}{x^2} \ge 4 \sqrt[4]{\frac{9}{2}}$$ for $x > 0$. However, AM-GM inequality won't be enough to find the minimum, as it only attains equality for $$2x^3 = x = \frac{3}{2x^2} = \frac{3}{2x^2}$$
which never happens.
| {
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Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.
So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.
|
The area of the triangle ABC is
\begin{align}
Area & = \frac12 r(AB + BC +CA)
= \frac12 rAB \frac{\sin A+\sin B +\sin C}{\sin C} \\
&= \frac12 rAB \frac{4\cos \frac A2 \cos \frac B2\cos\frac C2 }{2\sin \frac C2\cos \frac C2}
= rAB \frac{\cos \frac A2 \cos \frac B2 }{\cos \frac{A+B}2}
= \frac{rAB }{1- \tan\frac A2\tan\frac B2 }
\end{align}
Apply the Pythagorean theorem to the right triangle IFO, with $AB = 2R$
$$d^2 = (R-r)^2 - r^2 = R^2 - 2rR$$
$$\tan\frac A2\tan\frac B2= \frac r{R+d} \frac r{R-d}=\frac {r^2}{R^2-d^2} = \frac {r^2}{2rR} = \frac r{AB}
$$
Thus
$$Area = \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$
| {
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| HINT
Since $x > 0$ and $y > 0$, the proposed inequality is equivalent to
\begin{align*}
x^{3} + y^{3} \geq \sqrt{2}xy & \Longleftrightarrow (x^{3} + y^{3})^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (x+y)^{2}(1 - xy)^{2} \geq 2x^{2}y^{2}\\\\
& \Longleftrightarrow (1 + 2xy)(1-xy)^{2} \geq 2x^{2}y^{2}\\\\
\end{align*}
If we make the change of variable $t = xy$, we obtain the following equivalent inequality:
\begin{align*}
(1 + 2t)(1 - t)^{2} \geq 2t^{2} & \Longleftrightarrow (1+2t)(1 - 2t + t^{2})\geq 2t^{2}\\\\
& \Longleftrightarrow 1 - 2t + t^{2} + 2t - 4t^{2} + 2t^{3} \geq 2t^{2}\\\\
& \Longleftrightarrow 2t^{3} - 5t^{2} + 1 \geq 0\\\\\
& \Longleftrightarrow (2t^{3} - t^{2}) - (4t^{2} - 1) \geq 0\\\\
& \Longleftrightarrow t^{2}(2t - 1) - (2t-1)(2t+1) \geq 0\\\\
& \Longleftrightarrow (2t-1)(t^{2} - 2t - 1) \geq 0
\end{align*}
Can you take it from here?
| {
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Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).
How would I work this out?
| Since in a neighbourhood of the origin $\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have $\operatorname{sech}(x)=1-\frac{x^2}{2}+Ax^4+O(x^6)$
and since
$$ \left(1+\frac{x^2}{2}+\frac{x^4}{24}\right)\left(1-\frac{x^2}{2}+Ax^4\right)=1+\left(A-\frac{5}{24}\right)x^4+O(x^6) $$
we must have $A=\frac{5}{24}$. This unique non-trivial coefficient can also be found by considering that
$$\cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right)\tag{1} $$
$$\operatorname{sech}(x)=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{2m}}{\pi^{2m}(2n+1)^{2m}}\tag{2} $$
$$\operatorname{sech}(\sqrt{x})=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{m}}{\pi^{2m}(2n+1)^{2m}}\tag{3} $$
$$ [x^2]\operatorname{sech}(\sqrt{x})=\frac{16}{\pi^4}\sum_{a>b\geq 0}\frac{1}{(2a+1)^2(2b+1)^2}+\frac{16}{\pi^4}\sum_{n\geq 0}\frac{1}{(2n+1)^4}\tag{4} $$
and the RHS of $(4)$ just depends on the values of $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.
| {
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Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.
Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.
What I have tried: Put $x = \frac{1}{9}$. We have, $\sqrt[3]3\big(4x^\frac{1}{3} - 2x^\frac{1}{3} + x^\frac{1}{3})$
$$\rightarrow \sqrt[3]3\big(x^\frac{1}{3}(4-2+1)\big)$$
$$\rightarrow \sqrt[3]3 \times \frac{1}{3\sqrt[3]x}$$
$$\rightarrow \frac{3}{27x} = \frac{1}{9x}$$
Putting back the value of $x$ gives the answer to be $1$.
However, in my text the answer is given to be $\sqrt[3]2 + 1$ , so where did I go wrong?
Can anyone help me?
| : $$\sqrt[3]{4/9}-\sqrt[3]{2/9}+\sqrt[3]{1/9}=\frac{(2^{1/3}+1)(2^{2/3}-2^{1/3}+1)}{(2^{1/3}+1)(3^{2/3})}=\frac{3}{3^{2/3}(2^{1/3}+1)}$$ so:$$\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}=\sqrt[3]{3}\cdot \frac{(2^{1/3}+1)(3^{2/3})}{3}=\sqrt[3]{2}+1$$
| {
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What is Riemann sum of this integral and using it to prove this This is Question 11.19 (e) part of Tom Apostol's Mathematical Analysis.
Let $f$ be a function of period $2\pi$ whose values on $[-\pi ,\pi]$ are $f(x)=1$ if $0<x < \pi$, $f(x)=0$ if $x =\pi$, or $0$ and $f(x)=-1$ if $-\pi <x<0$ and $f(x)= 4/\pi \sum_{n=1}^{\infty} \frac{\sin(2n-1) x} {(2n-1)} dx$, for every $x$.
Consider the function$s_n(x)=\frac{2}{\pi} \int_{0}^{x} \frac{\sin (2nt)} {\sin (t)} dt $ . I proved that this function has local maxima at $x_1, x_3, x_5,\dots, x_{2n-1}$ where $x_m=\frac{m \pi } {2n}$.
The question is Interpret $s_n ( \frac{\pi} {2n}) $ as a Riemann sum and prove that $\lim_{n\to \infty} s_n( \frac{\pi} {2n}) = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin t}{t} dt$ .
I am getting different answer/ answer which is not even close to this representation when I used Riemann sum method on $f(x)= 4/\pi \sum_{n=1}^{\infty} \frac{\sin(2n-1) x} {(2n-1)}\, dx$ so I request you to kindly outline a proof. No need to fully solve it.
I have tried this a lot but couldn't solve it.
Thanks
| The partial sum of the Fourier series for $f$ evaluated at $x = \frac{\pi}{4n}$ is
$$s_n\left(\frac{\pi}{2n}\right) = \frac{4}{\pi}\sum_{k=1}^n\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1}$$
Consider a partition $P_n: 0 < \frac{\pi}{2n}< \frac{3\pi}{2n} < \frac{5\pi}{2n} < \ldots < \frac{(2n-1)\pi}{2n}< \pi$ of the interval $[0,\pi]$ and the function
$$g(x) = \begin{cases}\frac{\sin x }{x}, & 0 < x \leqslant \pi \\1, & x = 0 \end{cases}$$
The partition points are given by $x_0 = 0$, $x_k = \frac{(2k-1)\pi}{2n}$ for $k = 1, \ldots, n$, and $x_{n+1} = \pi$ and the Riemann sum for $g$ using right endpoints of the subintervals as tags is
$$S(P_n,g) = \sum_{k=1}^{n+1}f(x_k) (x_k - x_{k-1}) = \\\frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}}\left( \frac{\pi}{2n}-0\right)+\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{\frac{(2k-1)\pi}{2n}}\left(\frac{(2k-1)\pi}{2n}- \frac{(2k-3)\pi}{2n} \right) +\frac{\sin \pi}{\pi}\left( \pi - \frac{(2n-1)\pi}{2n}\right)\\ = \sin \frac{\pi}{2n}+ \sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{\frac{(2k-1)\pi}{2n}}\frac{\pi}{n} = \sin \frac{\pi}{2n}+ 2\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1}$$
Hence,
$$s_n\left(\frac{\pi}{2n}\right)= \frac{2}{\pi}\cdot 2\sum_{k=1}^{n}\frac{\sin\frac{(2k-1)\pi}{2n} }{2k-1} = \frac{2}{\pi}S(P_n,g) - \frac{2}{\pi}\sin \frac{\pi}{2n}$$
Since the Riemann sum $S(P_n,g)$ converges to $\int_0^\pi g(x) \, dx = \int_0^\pi \frac{\sin x}{x} \,dx$ as $n \to \infty$, we have
$$\lim_{n \to \infty}s_n\left(\frac{\pi}{2n}\right) = \lim_{n \to \infty}\frac{2}{\pi}S(P_n,g) - \lim_{n \to \infty}\frac{2}{\pi}\sin \frac{\pi}{2n} = \frac{2}{\pi} \int_0^\pi \frac{\sin x}{x} \, dx$$
The expression for $s_n(x)$ given in the problem is equal to the partial sum used above.
Differentiating $s_n\left(x\right) = \frac{4}{\pi}\sum_{k=1}^n\frac{\sin(2k-1)x }{2k-1}$ we get
$$s_n'\left(x\right) = \frac{4}{\pi}\sum_{k=1}^n\cos(2k-1)x,$$
which can be summed in closed form and integrated to obtain
$$s_n(x) = \frac{2}{\pi} \int_{0}^{x} \frac{\sin 2nt} {\sin t} dt$$
In this form we have
$$s_n\left(\frac{\pi}{2n}\right) = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2n}} \frac{\sin 2nt} {\sin t}\, dt = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u} {2n \sin \frac{u}{2n}}\, du = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u}{u} \frac{\frac{u}{2n}}{ \sin \frac{u}{2n}}\, du,$$
and by the dominated convergence theorem
$$\lim_{n \to \infty}s_n\left(\frac{\pi}{2n}\right) = \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin u}{u}\, du$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Permutation $\alpha^2$ Show that there is no permutation $\alpha \in S_4$ so that $\alpha^2=\begin{pmatrix}
1 &3 &4 &2
\end{pmatrix}$.
$S_4=\begin{pmatrix}
1 &2 &3 &4
\end{pmatrix}$
If $ \alpha=\begin{pmatrix}
2 &4 &3
\end{pmatrix}$, we get:
$$\alpha^2=\alpha\cdot\alpha=\begin{pmatrix}
1 &2 &3 &4 \\
1&4 &2 &3 \\1
&3 &4 &2
\end{pmatrix}$$
Why can't I do that?
| $\beta = (1 \ 3 \ 4 \ 2)$ is on odd permutation: a cycle of length equal to $4$. As the square of any permutation is even, $\beta$ can't be written as a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974048",
"timestamp": "2023-03-29T00:00:00",
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$\int \sqrt{x^4+1}\ dx$ I have tried to find $\int \sqrt{x^4+1} dx$ but can not find a way to get the answer.
| By the reduction and transformation formulas mentioned in my Elliptic Integrals and Functions monograph
$$\int_0^y\sqrt{x^4+1}\,dx=\int_0^y\frac{x^4+1}{\sqrt{x^4+1}}\,dx=\frac{y\sqrt{y^4+1}}3+\frac23\int_0^y\frac1{\sqrt{x^4+1}}\,dx$$
$$=\frac{y\sqrt{y^4+1}}3+\frac23F\left(\sin^{-1}\sqrt{y^2+1-\sqrt{y^4+1}},m=\frac12\right)=\frac{y\sqrt{y^4+1}+F(2\tan^{-1}y,1/2)}3$$
In hypergeometric terms the integral is
$$y{_2F_1}(-1/2,1/4;5/4;-y^4)$$
| {
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How to find the series $a_{n}$ which has this generating function: $f(x)=\frac{1}{1-x^{2}}$? I'm given a generating function $f(x)$:
$$
f(x)=\frac{1}{1-x^2}
$$
Naturally, I tried to transform it into the form $f(x)=\sum_0^\infty a_n \cdot x^n$.
So far I figured out what follows:
$$
f(x)=\frac{1}{(1-x)(1+x)}=\frac{\frac{1}{2}}{1-x} + \frac{\frac{1}{2}}{1+x}=\frac{1}{2}\cdot\left( \sum_{n=0}^\infty x^n+\frac{1}{1-(-x)} \right)
$$$$
\frac{1}{1-(-x)}=1+(-x)+x^2+(-x^3)+x^4+\cdots
$$$$
\frac{1}{1-(-x)}=(1+x^2+x^4+\cdots)-(x+x^3+x^5+\cdots)=\sum_{n=0}^\infty x^{2n}-x^{2n+1}
$$
Therefore:
$$
f(x)=\frac{1}{2} \left( \sum_{n=0}^\infty x^n+x^{2n}-x^{2n+1} \right)
$$
Now I am lost. Is my train of thought good? Am I anywhere near finding the series which is generated by function $f(x)$?
| To repair the approach you were taking:
$$f(x) = \frac{1}{1-x^2} = \frac{1/2}{1-x} + \frac{1/2}{1+x} = \frac{1}{2} \sum_{n=0}^\infty x^n + \frac{1}{2} \sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{1+(-1)^n}{2} x^n,
$$
so
$$
a_n = \frac{1+(-1)^n}{2} = \begin{cases} 1 &\text{if $2\mid n$} \\0 &\text{otherwise}\\ \end{cases}
$$
| {
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Prove that$ \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}$ Prove the following inequality where positive reals $a$, $b$, $c$ satisfies $ab+bc+ca=1$.
$$
\frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}
$$
I got that $ \frac{a+b}{\sqrt{ab(1-ab)}}=\frac{a+b}{\sqrt{ab(bc+ca)}}=\frac{a+b}{\sqrt{abc(a+b)}}.$
So rationalising, we get $\frac{a+b}{\sqrt{abc(a+b)}}=\frac{(a+b){\sqrt{abc(a+b)}}}{abc(a+b)}=\frac{\sqrt{abc(a+b)}}{{abc}}$
Then enough to show that $\sum_{cyc}\sqrt{abc(a+b)}\le \sqrt2.$
Any help?
| let $x=ab,y=bc,z=ca$ and $x+y+z=1$
after your work we have to show
$$\sum_{cyc} \sqrt{2x(y+z)}\le 2$$ which is true by AM-GM $$\sum_{cyc} \sqrt{2x(y+z)}\le \sum_{cyc}\frac{2x+y+z}{2}=2$$
| {
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Compute the $n$-th power of the difference of the following matrices. Given the matrices $$A=\begin{pmatrix} a & c & b \\ c & b & a \\ b & a & c \end{pmatrix}$$ and $$B=\begin{pmatrix} b & a & c \\ a & c & b \\ c & b & a \end{pmatrix}$$
for $a,b,c \in \mathbb{R}$
find $$(A-B)^n$$ for $n \in \mathbb{N}$.
I can see that the matrices are symmetric and also they have the same determinant since $B$ can be obtained by swapping the rows of $A$ twice.
I attempted to solve this by using the Cayley-Hamilton theorem.
First I find the characteristic polynomial which is $$p(k)=-k^3+(a+b+c)k^2-(ab+bc+ac)k+3abc-b^3-a^3-c^3$$
which is the same for both matrices.
Now by Cayley-Hamilton theorem, I have $$-A^3+(a+b+c)A^2-(ab+bc+ac)A+3abc-b^3-a^3-c^3=0$$
and $$-B^3+(a+b+c)B^2-(ab+bc+ac)B+3abc-b^3-a^3-c^3=0$$
By subtracting those two equations by parts, I get $$-A^3+B^3+(a+b+c)(A^2-B^2)-(ab+bc+ac)(A-B)+3abc-b^3-a^3-c^3$$
and this is where I am stuck, I attempted to use the identities $$a^3+b^3+c^3-3abc=\frac{1}{2} (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
and $$2ab+2bc+2ca=(a+b+c)^2-a^2-b^2-c^2$$ but couldn't end up with something useful.
My question is, first of all, is my approach correct? How should I proceed?
| Note that $M = A - B$ can be written in the form
$$
\pmatrix{
p&r&q\\
r&q&p\\
q&p&r},
$$
where $p+q+r = 0$. Verify that
$$
M \pmatrix{1\\1\\1} = 0,
$$
which means that $\det(A - B)$ must be zero. With that, we find the characteristic polynomial to be
$$
\det(M - kI) = -k^3 - (pq + pr + qr)k.
$$
Let $\alpha = pq + pr + qr$. The Cayley Hamilton theorem tells us that
$$
-M^3 - \alpha M = 0 \implies M^3 = -\alpha M.
$$
With that established, we can conclude that for $n \geq 2$ we have
$$
M^n = \begin{cases}
\alpha^{n/2 - 1} M^2 & n \text{ is even,}\\
\alpha^{(n-1)/2} M & n \text{ is odd.}
\end{cases}
$$
| {
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Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
| mrsamy's method is quite elegant, though it says nothing about how large $x\gg 1$ should be.
If you need a thighter inequality, the best way is to develop
$$f(x)=(5x^2+3x+1)x-10(x^3-x^2-1)$$
Graph it and notice it is negative for some $x$ close to (and below) $3$
Then just calculate $f(u+3)=-5u^3-32u^2-56u-5<0\quad$ for $\ u\ge 0$
Therefore the inequality is valid as soon as $x\ge 3$ (the actual value is $\approx 2.9057$).
| {
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In how many ways can $6$ different candies be divided between three children? I am the freshman in Computer Science at university, and currently struggling with some tasks from the Combinatorics part of the Discrete Math class. Any help would be appreciated.
In how many ways can $6$ different candies be divided between three children?
My vision of solving this problem:
*
*For every candy there is $3$ children to give it to $\implies 3^6$ possible ways of candy distribution
OR
*Each child can get from $0$ to $6$ candies:
$\implies 7$ possible distributions $\implies 7^3$ possible ways of candy distribution between the $3$ children?
But I feel like I am missing something which leads to the duality of my solutions. (Probably that the candies are different?) And I don't understand how to count that in.
| Your first solution is correct. There are three possible recipients for each of the six candies, so there are $3^6$ possible distributions of the candies.
If you wish to consider how many candies each child receives, you need to consider cases based on partitions of $6$ into at most three parts.
\begin{align*}
6 & = 6\\
& = 5 + 1\\
& = 4 + 2\\
& = 4 + 1 + 1\\
& = 3 + 3\\
& = 3 + 2 + 1\\
& = 2 + 2 + 2
\end{align*}
One child receives six candies: There are three ways to select the recipient of all six candies.
One child receives five candies and another child receives one candy: There are three ways to select the child who will receive five candies, $\binom{6}{5}$ ways to select which five of the six candies that child receives, and two ways to choose the child who receives the remaining candy. Hence, there are
$$\binom{3}{1}\binom{6}{5}\binom{2}{1}$$
such distributions.
One child receives four candies and another child receives two candies: There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose the child who receives the remaining two candies. Hence, there are
$$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$
such distributions.
One child receives four candies and each of the other children receive one candy each: There are three ways to select the child who will receive four candies, $\binom{6}{4}$ ways to select which four of the six candies that child receives, and two ways to choose which of the remaining two candies the younger of the two remaining children receives. The other child must receive the remaining candy. Hence, there are
$$\binom{3}{1}\binom{6}{4}\binom{2}{1}$$
such distributions.
Two children each receive three candies: There are $\binom{3}{2}$ ways to select which two children receive three candies each and $\binom{6}{3}$ ways to select which three of the six candies the younger of the two selected children will receive. The other selected child must receive the three remaining candies. Hence, there are
$$\binom{3}{2}\binom{6}{3}$$
such distributions.
One child receives three candies, a second child receives two candies, and the third child receives one candy: There are three ways to select the child who will receive three candies, $\binom{6}{3}$ ways to select which three candies that child will receive, two ways to decide which of the remaining children will receive two candies, and $\binom{3}{2}$ ways to select which two of the remaining three candies that child will receive. The remaining child must receive the remaining candy. Hence, there are
$$\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2}$$
such distributions.
Each of the three children receives two candies: There are $\binom{6}{2}$ ways to select which two candies are received by the youngest child and $\binom{4}{2}$ ways to select which two of the remaining four candies are received by the next youngest child. The oldest child must receive the two remaining candies. Hence, there are
$$\binom{6}{2}\binom{4}{2}$$
such distributions.
Total: Since the above cases are mutually exclusive and exhaustive, the number of ways of distributing six different candies to three children is
$$\binom{3}{1} + \binom{3}{1}\binom{6}{5}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{1}\binom{6}{4}\binom{2}{1} + \binom{3}{2}\binom{6}{3} + \binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{2} + \binom{6}{2}\binom{4}{2} = 3^6$$
| {
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Simplex Algorithm Cycles How can I show that the linear program
\begin{align*}&\min -2x-3y+z+12w\\
&s.t.
\\&-2x-9y+z+9w+s=0\\&1/3x+2y-1/3z-2w+t=0\\&x,y,z,w,s,t\geq 0\end{align*}
can induce cycling in the Simplex Algorithm if we use the original pivot rule of Dantzig? I know that I have to show that the algorithm repeats bases which all correspond to the same vertex but I am not sure how to do that here.
Edit: the linear program above is a conversion of the program
\begin{align*}&\min -2x-3y+z+12w\\
&s.t.
\\&-2x-9y+z+9w\leq0\\&1/3x+2y-1/3z-2w\leq0\\&x,y,z,w\geq 0\end{align*}
into standard form to make it accessible to the Simplex Algorithm.
| To simplify things, let’s denote each basis as $B_0,\ldots,B_n$ for the number of bases we visit until we notice a cycle. Thus, with the given model in the question, the first basis will be:
\begin{array} {|c|c|}
\hline \text{Current Basis} & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline & k & 1 & 2 & 3 & -1 & -12 & 0 & 0 & 0 & - \\ B_0 & s_1 & 0 & -2 & -9 & 1 & 9 & 1 & 0 & 0 & 0 \\ & s_2 & 0 & 1/3 & 2 & -1/3 & -2 & 0 & 1 & 0 & 0 \\ \hline \end{array}
For this to work, we’ll always pivot the first row that is the minimum row in the minimum-ratio test. Thus, we’ll pivot the $y$ column with the $s_1$ row to produce the next basis:
\begin{array} {|c|c|}
\hline \text{Current Basis}
& BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline
& k & 1 & 4/3 & 0 & -2/3 & -9 & 1/3 & 0 & 0 & - \\
B_1 & y & 0 & 2/9 & 1 & -1/9 & -1 & -1/9 & 0 & 0 & 0 \\
& s_2 & 0 & -1/9 & 0 & -1/9 & 0 & 2/9 & 1 & 0 & 0 \\ \hline \end{array}
Then we’ll pivot the $x$ column with the $y$ row to produce:
\begin{array} {|c|c|}
\hline \text{Current Basis}
& BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline
& k & 1 & 0 & -6 & 2/3 & -3 & 5/3 & 0 & 0 & - \\
B_2&x& 0 & 1 & 9/2 & -1/2 & -9/2 & -1/2 & 0 & 0 & 0 \\
&s_2 & 0 & 0 & 1/2 & -1/6 & -1/2 & 1/6 & 1 & 0 & 0 \\ \hline \end{array}
Then we’ll pivot the $s_1$ column with the $x$ row to produce:
\begin{array} {|c|c|}
\hline \text{Current Basis} & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline & k & 1 & 2 & 3 & -1 & -12 & 0 & 0 & 0 & - \\ B_3 & s_1 & 0 & -2 & -9 & 1 & 9 & 1 & 0 & 0 & - \\
& s_2 & 0 & 1/3 & 2 & -1/3 & -2 & 0 & 1 & 0 & - \\ \hline \end{array}
From here, notice that basis $B_0=B_3$ in that every element in the the tableau matches with one another. Thus, if we start pivoting again we’ll be going in the exact same path we did to reach the basis $B_3$. Thus, this is how we prove that cycling is occurring via the Simplex Method in tabular form.
| {
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Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$ I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:
$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right| < \epsilon.$$
This is where I am right now:
\begin{align}
\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|&= \frac{\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{x^3 - a^3}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{(x - a)(x^2 + ax + a^2)}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3 + 1 - a^3}\right|}
\end{align}
Any hint is welcome.
| If $a\not=1$, then you can continue with
$${3\delta\over\sqrt{1-x^3}+\sqrt{1-a^3}}\lt{3\delta\over\sqrt{1-a^3}}$$
so it suffices to let $\delta=\epsilon\sqrt{1-a^3}/3$ to make the epsilon-delta proof work for $a\in[0,1)$. But for $a=1$ you need a separate argument to show that there is a $\delta$ for which $|x-1|\lt\delta\implies|\sqrt{1-x^3}|\lt\epsilon$. I'll let you think this over first; please leave a comment if you'd like more help.
| {
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Find the radius of convergence of the power series representation of $\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$ We are given the following function:
$$\frac{tan(\frac{z}{2})}{(z+2)(z+3)}$$
How do we find the radius of convergence for the power series, representing that function around $z = 0$?
The power series representation for $tan(\frac{z}{2})$ is:
$$\frac{z}{2} + \frac{z^3}{24} + \frac{z^5}{240} + O(z^7)$$
The power series representation for $\frac{1}{(z+2)(z+3)}$ is:
$$\frac{1}{6} - \frac{5z}{36} + \frac{19z^2}{216} + O(z^3)$$
Then I multiply both to obtain:
$$\frac{z}{12} - \frac{5z^2}{72} + \frac{11z^3}{216} - \frac{5z^4}{162} + \frac{697z^5}{38800} + O(z^6)$$
I don't know how to find the radius of convergence from that. Is there a way to convert it to a representation of summation and the nth term in the power series, so that for example, we can use the ratio test to find the radius or is there another way to find it?
| The closest singularities of the function are located at $z=-2,-3,\pm\pi$. The closest one to $0$ is $-2$, thus the radius of convergence of the series centered at $0$ would be $2$.
| {
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Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C,\quad (*)$
but there is one step I can't justify.
When I got to $\int \tan^2(u)\sec(u)\ du = \frac{1}{2}\left(\ \tan(u)\sec(u) - \ln\left|\sec(u)+\tan(u)\right|\ \right) + C,$
I then have to substitute stuff back in in terms of $x$.
Now $x=\sec(u) \implies \tan^2(u) = x^2 - 1$. But I don't see how this implies $\tan(u) = \sqrt{x^2 - 1}$.
Comparing the graphs of $\sec(u)$ and $\tan(u)$, I don't see why not: $\ \tan(u) = -\sqrt{x^2 - 1}$, which would give:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ -x \sqrt{x^2-1} - \ln\left|x - \sqrt{x^2-1}\right|\ \right) + C,\quad (**)$
which is a different answer than $(*)$ ?
Now I noticed that $(*) = -(**)\ $ (ignoring the $C \to -C)$.
I can see from the graph of $\sqrt{x^2-1}$ that for $x>1$, the definite integral $\int^x_1\sqrt{t^2-1}\ dt = (*),$ and for $x<-1,\ \int^{-1}_x\sqrt{t^2-1}\ dt = (**)$
So is the indefinite integral sort of poorly defined, or would you say it is:
$\int\sqrt{x^2-1}\ dx = \pm \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C\ $ ?
| The integration-by-substitution theorem applies here when the domain of the substitution function $\sec(u)$ is restricted† to $(0,\frac\pi2).$
(This substitution works even if the original domain of integration is, say, $[-2,-1]$.)
This is why only the positive output of $\tan(u)$ is taken.
† due to the discontinuity of $\tan$ and $\sec$ at $\frac\pi2$ and the non-integrability of $\frac{\mathrm{d}}{\mathrm{d}x}\sec^{-1}$ on $[1,\infty)$
P.S. If you are acquainted with hyperbolic functions, $x=\cosh(u)$ is a more elegant substitution than $x=\sec(u).$ (As with the $\sec(u)$ substitution, there is an implicit understanding that the substitution function has a restricted domain—in this case $[0,\infty).)$
| {
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How to maximize a function when you cannot solve gradient = 0 I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$
$$ a, b \in{R+*} $$
Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$
I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). But because a,b are positive real numbers, I cannot solve $$ \frac{a}{2\sqrt{x}} = 0 $$
And it's the same for y.
How do I deal with such a situation. Thank you very much !!
| With the help of some slack variables $s_k$ we transform the inequalities into equations and then with $f(x,y) = a\sqrt x+b\sqrt y$ forming the lagrangian
$$
L = f(x,y)+\lambda_1(x-s_1^2)+\lambda_2(y-s_2^2)+\lambda_3(x-2+2y+s_3^2)+\lambda_4(y-1+s_4^2)
$$
The lagrangian stationary points are the solutions for
$$
\nabla L = 0 = \left\{
\begin{array}{l}
\frac{a}{2 \sqrt{x}}+\lambda_1+\lambda_3 \\
\frac{b}{2 \sqrt{y}}+\lambda_2+2 \lambda_3+\lambda_4 \\
x-s_1^2 \\
y-s_2^2 \\
s_3^2+x+2 y-2 \\
s_4^2+y-1 \\
-2 s_1 \lambda_1 \\
-2 s_2 \lambda_2 \\
2 s_3 \lambda_3 \\
2 s_4 \lambda_4 \\
\end{array}
\right.
$$
giving the solution
$$
\left[
\begin{array}{ccccccc}
f(x,y)& x & y & s_1^2 & s_2^2 & s_3^2 & s_4^2\\
\sqrt{2a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2 a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2
a^2+b^2} & 0 & \frac{2 a^2}{2 a^2+b^2} \\
\end{array}
\right]
$$
NOTE
Null $s_k$'s indicates that the corresponding constraint is active.
| {
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Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.
If $p(x)=t-1$ or $t-2$, then $A$ is a scalar multiple of the identity and $A^4+A^3-3A^2-3A=-4I \text{ or }6I$, so $A^4+A^3-3A^2-3A$ is invertible.
If $p(t)=(t-1)^2(t-2)$ or $(t-1)(t-2)$, then $A$ is similar to a upper triangular matrix $J$ with diagonal entries $1,2$. Then $A^4+A^3-3A^2-3A=P(J^4+J^3-3J^2-3J)P^{-1}$ and $\det(A^4+A^3-3A^2-3A)=\det(J^4+J^3-3J^2-3J)$. Since $J^4+J^3-3J^2-3J$ is upper triangular and the diagonal entries do not vanish, $\det(J^4+J^3-3J^2-3J) \ne 0$.
If $p(t)=(t-1)^2$, then by the same reasoning, $A^4+A^3-3A^2-3A$ is invertible.
Is there more efficient way to solve this kind of problems or I have to discuss any possible situations every time?
| Since $A^3-4A^2+5A-2=0$, $A$ must be invertible ($0$ is not an eigenvalue, else we get $-2v=0$.) Since $$A^4+A^3-3A^2-3A=A(A^3+A^2-3A-3),$$ it is enough to show that $A^3+A^2-3A-3$ is invertible. This follows from \begin{align}
A^3+A^2-3A-3&=(A+1)(A+\sqrt3)(A-\sqrt3)\end{align} Since $-1$, $\pm\sqrt3$ are not eigenvalues of $A$ (do not satisfy the polynomial in $A$), this polynomial in $A$ is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$\sqrt a+\sqrt b$ is a root of polynomial, then $\sqrt a -\sqrt b$ is so over finite field Let $a,b$ is not a square element in finite field.
If polynomial over the finite field has a root $\sqrt a+\sqrt b$, then the polynomial has $\sqrt a-\sqrt b$ as a root too?
Thank you for your help.
This question arises from concrete problem, $x^4-10x^2+1$（this is a minimal polynomial of $\sqrt{2}+\sqrt{3}$ over rational field） is reducible over arbitrary finite field.
$\sqrt{2}+\sqrt{3}$ is root over finite field, then $\sqrt{2}-\sqrt{3}$ is also a root（why?）.And $\sqrt2+\sqrt3$ and $\sqrt2-\sqrt3$ exists in $2$-degree extension of the finite field, so the minimal polynomial spilts. This leads minimal polynomial is at most $2$ degree, so $x^4-10x^2+1$ is reducible.
| Let's look at the concrete example, to see what is going on, and assume we are not in characteristic $2$.
Then, using the normal quadratic formula if $$x^4-10x^2+1=0$$ then $$x^2=\frac {10\pm \sqrt{100-4}}{2}=5\pm 2\sqrt 6$$ and if $6$ has a square root in your finite field, then the original equation is reducible.
The roots in a splitting field are $\pm \sqrt 2\pm \sqrt 3$ and the factorisation already obtained corresponds to one pairing of the roots: $$(x^2-5-2\sqrt 6)(x^2-5+2\sqrt 6)=$$$$=(x-\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x-\sqrt 2+\sqrt 3)$$
If we were instead to choose other pairings we'd get $$(x-\sqrt 2+\sqrt 3)(x+\sqrt 2+\sqrt 3)\cdot(x-\sqrt 2-\sqrt 3)(x+\sqrt 2-\sqrt 3)=$$$$=(x^2+2\sqrt 3 x+1)(x^2-2\sqrt 3x+1)$$ or $$(x-\sqrt 2+\sqrt 3)(x-\sqrt 2-\sqrt 3)\cdot(x+\sqrt 2-\sqrt 3)(x+\sqrt 2+\sqrt 3)=$$$$=(x^2-2\sqrt 2 x-1)(x^2+2\sqrt 2x-1)$$ and these give factorisations in the case that $\sqrt 3$ or $\sqrt 2$ exist in your finite field - ie that $3$ or $2$ are squares.
Now, you may also know that the product of two non quadratic residues modulo $p$ is a quadratic residue. From this it is easy to deduce that at least one of $2,3, 6$ has a square root in a field of characteristic $p$. So the polynomial is reducible. If two of the three have roots then the third does and the polynomial splits into linear factors.
Modulo $5$ we have that $6$ is a square while $2$ and $3$ are not. Modulo $7$ we have that $2$ is a square while $3$ and $6$ are not. Modulo $11$ we have that $3$ is a square and $2$ and $6$ are not. Modulo $23$ we have that $2, 3, 6$ are all squares. So all four possibilities occur.
Finally it is easy to show that in characteristic $2$ we have $x^4-10x+1=x^4+1=(x+1)^4$.
So I've worked this in some detail in case it helps you to see what is going on. You can generalise to $a$ and $b$. Obviously there are more efficient expositions, but sometimes longhand helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that a 2-variable function is decreasing with respect to a given variable on an interval For all integers $x\geq 7$ and all real numbers $y$ such that $0.249 \leq y \leq 0.25$ define
$$F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}.$$
I claim that $$F(x,y)\geq F(7,y)$$ for all $x \geq 7$.
I thought of several ways to try and prove this:
*
*Working directly on this inequality is complicated (even Maple couldn't do it for me).
*I plotted $F(x,y)$ for fixed values of $y$ and saw that $F(x,y)$ was increasing with respect to $x$ in all of the plots. However, the derivative of $F(x,y)$ with respect to $x$ is very complicated as well.
*Since the values of $x$ are integers, I thought about an inductive argument, like showing that $F(x,y)\leq F(x+1,y)$ for all $x\geq 7$, but again, this is a difficult inequality to work with.
***If it is too difficult to show that $F(x,y)\geq F(7,y)$, then I would also be very happy to figure out how to at least show that $$F(x,y)\geq 1.05-y$$
for all $x\geq 7$ and $y$ sufficiently close to $0.25$.
Any suggestions are greatly appreciated. Thank you.
| $F(x,y)=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( \frac{x+3}{x}\right)+2\sqrt{\frac{x+3}{x}}+\log\left( \frac{x+1}{x}\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{x}\right)+\frac{1}{2}\log\left(\frac{x^2}{x^2-1} \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( x+3\right)-\frac{1}{2}\log\left(x^2-1 \right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{-\log\left( x+3\right)+2\sqrt{\frac{x+3}{x}}+\log\left( x+1\right)-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1-4y}{2}\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\right)+\frac{1}{2}\log\left( \frac{x+1}{x-1} \right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x^2-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{\sqrt{x+1}\sqrt{x-1}}\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=-\frac{\log\left( \frac{x+3}{x-1}\right)-y\log\left( \frac{x+1}{x-1} \right)-2\sqrt{\frac{x+3}{x}}+3-4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
$=\frac{\log\left( \frac{x-1}{x+3}\right)+y\log\left( \frac{x+1}{x-1} \right)+2\sqrt{\frac{x+3}{x}}-3+4y}{\log\left( \frac{x+1}{x+3}\right)+2\sqrt{\frac{x+3}{x}}-2\sqrt{\frac{x+1}{x}}}$
Now it should be easier to take the derivative or do other simplification steps.
EDIT: As mentioned in the comments, it seems to be decreasing instead of increasing. That could you also show by the last term of above simplification.
| {
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"url": "https://math.stackexchange.com/questions/4009011",
"timestamp": "2023-03-29T00:00:00",
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Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$. For $n\ge 0$, show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.
Over a well ordered set of non-negative integers, induction can be used.
Base case: $n=0$
$\binom{x}{0} \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.
Taking the rhs, get :
$\binom{x+y}{0} = 1$.
Induction hypothesis: it is true for natural number $n$.
So, $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$
Inductive step: need show for next natural $n+1$
$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$
This should lead to :
$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$
get on lhs:
$\binom{x+y}{n} + 1.\binom{x}{n+1}$
$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$
Please help to proceed further, or suggest alternative approach.
| $$S=\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$$
$$S=\sum_{k=0}^{n}[t^k]~ (1+t)^x~ [t^{n-k}]~ (1+t)^y$$
$$S=[t^n] (1+t)^{x+y}={x+y \choose n}.$$
Here $[t^j]f(t)$ means co-efficient of $t^j$ in $f(t).$
| {
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"url": "https://math.stackexchange.com/questions/4010522",
"timestamp": "2023-03-29T00:00:00",
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Resolve $x+\frac{1}{x}$ While reading an old mathematics question paper of the Unified International Mathematics Olympiad, I hit a question with two very interesting patterns. Here I first mention them before presenting my question.
Pattern 1
When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$.
Here, the value of $x$ and $x+\frac{1}{x}$ are in inverse proportion. I tried putting the values and here are the results:
$$0.000001 + \frac{1}{0.000001} = 1000001$$
$$0.999999 + \frac{1}{0.999999} = 2.000001000001$$
The value of $x+\frac{1}{x}$ got near about two, but not equal to it.
Pattern 2
When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$.
This pattern also works the same way as the first one did. Its just that here the value of $x+\frac{1}{x}$ will always evaluate to a number small than $2.5$.
$$1.000001 + \frac{1}{1.000001} = 1.999999000001$$
$$1.999999999999999 + \frac{1}{1.999999999999999} = 2.499999999999999$$
The value of $x+\frac{1}{x}$ got near about $2.5$, but not equal to it.
Question
How can we prove the statements, "When $0 < x < 1$, then the value of $x+\frac{1}{x}$ is greater than $2$" and "When $1 < x < 2$, then the value of $x+\frac{1}{x}$ is smaller than $2.5$." mathematically?
Any help is appreciated. Thank you!
| $x+\frac{1}{x}\geq2$ holds for all $x>0$, with equality at $x=1$ only. As a hint, expand
$$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\geq0.$$
For the second claim, observe that for $1<x<2$, we have
$$(x-2)(x-1)<0\implies x^2+2<3x\implies x+\frac{1}{x}<\frac{x}{2}+\frac{3}{2}<\frac{5}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Arrangement without any monotone section
Here's the question:
"In how many ways can the sequence 45678 be arranged such that there aren't any three consecutive terms increasing or decreasing"
The first thing I thought of is to do complementary counting and then make cases for each number of consecutive terms that there could be (3, 4, 5).
However there's a lot of overlap between the three cases and I'm not sure how to deal with that in an efficient way. If someone could solve it and then tell me how they did it that would be great.
| My simple Python script enumerates them as
(4, 6, 5, 8, 7)
(4, 7, 5, 8, 6)
(4, 7, 6, 8, 5)
(4, 8, 5, 7, 6)
(4, 8, 6, 7, 5)
(5, 4, 7, 6, 8)
(5, 4, 8, 6, 7)
(5, 6, 4, 8, 7)
(5, 7, 4, 8, 6)
(5, 7, 6, 8, 4)
(5, 8, 4, 7, 6)
(5, 8, 6, 7, 4)
(6, 4, 7, 5, 8)
(6, 4, 8, 5, 7)
(6, 5, 7, 4, 8)
(6, 5, 8, 4, 7)
(6, 7, 4, 8, 5)
(6, 7, 5, 8, 4)
(6, 8, 4, 7, 5)
(6, 8, 5, 7, 4)
(7, 4, 6, 5, 8)
(7, 4, 8, 5, 6)
(7, 5, 6, 4, 8)
(7, 5, 8, 4, 6)
(7, 6, 8, 4, 5)
(7, 8, 4, 6, 5)
(7, 8, 5, 6, 4)
(8, 4, 6, 5, 7)
(8, 4, 7, 5, 6)
(8, 5, 6, 4, 7)
(8, 5, 7, 4, 6)
(8, 6, 7, 4, 5)
so there are $32$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I bypass the limit for the tan function on the calculator? So basically, my partner and I are creating problems for a project. She created one where we aren't sure what the correct answer is.
Problem:
$Find \ cos\frac{\theta}{2} \ if \ tan \ \theta = \frac{3}{4}; \pi < \theta<\frac{3\pi}{2}$
I'll show you what we did.
First, we used the pythagorean formula to find $cos \ \theta$ (we need it later). The half-angle formula for cosine is $cos \frac {\theta}{2} = \pm \sqrt{\frac{1 + cos \ \theta}{2}}$.
These are our steps.
*
*$cos \frac {tan^{-1}\frac{3}{4}}{2} = \pm \sqrt{\frac{1 + cos \ (tan^{-1}\frac{3}{4})}{2}}$
*$ = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}}$
*$ = \pm \sqrt{\frac{\frac{9}{5}}{2}}$
*$ = \pm \sqrt{\frac{9}{10}}$
*$ = \pm \frac{3\sqrt{10}}{10}$
To decide whether we have a negative sign or a positive, we have to look at the domain.
$ \pi < \theta<\frac{3\pi}{2} \ changes \ to \ \frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$
We now know that $cos \frac {\theta}{2}$ is located in the second quadrant, which makes our final answer $cos \frac {tan^{-1}\frac{3}{4}}{2} = - \frac{3\sqrt{10}}{10}$
But then, I checked the calculator.
The values are equal except for the negative sign
My partner said that the calculator thinks tangent is in the first quadrant, and if it were, cosine would be positive. But since we are in the third quadrant, cosine is in the second quadrant. I know there are domain restrictions, but I don't entirely understand what she meant. Also, how do I get the calculator to display the same answer?
| The "limit" you're trying to bypass is the fact that $\tan^{-1}$ only produces output values between $-\frac\pi2$ and $\frac\pi2.$
It will never directly give you an angle in the third quadrant.
The way you "get around" this is you remember that
$$ \tan\theta = \tan(\theta + \pi) = \tan(\theta + 2\pi) = \tan(\theta - \pi),$$
that is, no matter what whole multiple of $\pi$ you add or subtract from $\theta,$
the tangent of the resulting angle is always the same as $\tan\theta.$
Since you know $\pi < \theta<\frac{3\pi}{2}$, you know that if you subtract $\pi$ from $\theta$ you will get an angle $\alpha = \theta - \pi$ where
$0 < \alpha < \frac\pi2.$
Since $\tan\alpha = \tan\theta = \frac34,$
and $\alpha$ is an angle that you can get as an answer from $\tan^{-1},$
you know that $\alpha = \tan^{-1}\frac34.$
Therefore $\theta = \alpha + \pi = \pi + \tan^{-1}\frac34.$
To continue along the lines you were trying to follow, you will have to compute
$\cos\frac{\pi + \tan^{-1}(3/4)}{2}.$
One way to continue is to use the cosine half-angle formula,
just as you attempted, but with the correct angle, so you get
$$
\cos\frac\theta2
= \cos\frac{\pi + \tan^{-1}\frac34}{2}
= \pm\sqrt{\frac{1 + \cos\left(\pi + \tan^{-1}\frac34\right)}{2}}.
$$
Now use the fact that $\cos(\pi + \beta) = -\cos(\beta).$
So now you have
$$
\cos\frac\theta2
= \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.
$$
You can get to the correct answer from there using similar methods to what you already used in your attempt for this question,
but you get a different final answer since you have a difference inside the square root rather than a sum.
Alternatively, write
$\cos\frac{\pi + \tan^{-1}(3/4)}{2}
= \cos\left(\frac\pi2 + \frac{\tan^{-1}(3/4)}{2}\right),$
and use the fact that
$\cos\left(\frac\pi2 + \phi\right) = -\sin\phi.$
Therefore
$$
\cos\frac\theta2
= -\sin\left(\frac{\tan^{-1}\frac34}{2}\right)
= \pm\sqrt{\frac{1 - \cos\left(\tan^{-1}\frac34\right)}{2}}.
$$
The same result follows.
Having said all that, I think the easier way to solve this is not to "get around" the $\tan^{-1}$ function at all, but rather draw a triangle as recommended in the other answer. That answer uses straightforward geometric intuition that is much less likely to lead you astray.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:
$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$
Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$. in this problem substitute $x=a+b,y=c,z=d:$
$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd-ab^2)$$
Although I am sure my approach is right, but I should have $3(abc+bcd+cda+dab)$ at the RHS. I don't know how to get it.
| $$(a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3-3ab(c+d)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does this pattern work: $1 \cdot{1} = 1, 11 \cdot{11} = 121, 111 \cdot{111} = 12321\ldots$ I have recently learned about pattern that goes like this:
$$\begin{align}
1^2 &= 1\\
11^2 &= 121\\
111^2 &= 12,321\\
1,111^2 &= 1,234,321\\
11,111^2 &= 123,454,321.
\end{align}$$
It is a very cool pattern, but after a bit it stops:
$$1,111,111,111^2 = 1,234,567,900,987,654,321$$
My main question:
Why does this pattern work?
A side question that's less important:
Is there an algebraic equation to describe this pattern?
| Other answers have already provided some intuition behind the pattern. Here is some terminology:
*
*The numbers being squared, e.g. $111111$, are called repunits.
*The resulting products are called Demlo numbers: $\,1, 121, 12321, 1234321,\, \dots$
Repunits can be written as sums of powers of $10$, and from that perspective, base-$10$ repunits are finite geometric series. For example:
\begin{align}
111111 &= 1 + 10 + 100 + 1000 + 10000 + 100000\\
111111&=1 + 10 + 10^2 + 10^3 + 10^4 + 10^5\\\\
111111&=\displaystyle\frac{10^6-1}{10-1}\\\\
111111&=\displaystyle\frac{10^6-1}{9}
\end{align}
The last two lines above use the formula for the sum of a finite geometric series.
Squaring gives:
\begin{align}
111111^2 &= \left(\displaystyle\frac{10^6-1}{9}\right)^2\\
111111^2 &= \displaystyle\frac{10^{12} - 2\cdot 10^6 +1}{81}
\end{align}
In general:
$$\left(\,\underbrace{\,1111\,\dots\,1111\,}_{ n \text{ ones} } \,\right)^2 \, = \,\, \boxed{\displaystyle\frac{10^{2n} - 2\cdot 10^n +1}{81}\,}$$
Demlo numbers are sequence A002477 in the OEIS.
| {
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Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.
| $$(a-b)^4=(a^2+b^2-2ab)^2=(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)$$
$$a^4+b^4=(a^2+b^2)^2-2a^2b^2$$
On addition, $$2((a^2+b^2)^2+a^2b^2-2ab(a^2+b^2))=2(a^2+b^2-ab)^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$
If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$
If we take $t=\tan \theta$ then $x=2\theta$ and for $y$ ,I am getting two values as follows ,
if we apply identity $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$ then we get $y=2\theta$
and if we take $\tan \theta $ common from $\dfrac{1-\tan^2 \theta}{2\tan\theta}$ and simplify then we get $y= \dfrac{\pi}{4} + \theta$ ?
how we are getting two values of $y$ ?
thankyou
| I'm not sure where you're getting $y=\theta+\pi/4$ but it is definitely incorrect as $$\cot(\theta+\pi/4)=\frac{1-\tan\theta\tan\pi/4}{\tan\theta+\tan\pi/4}=\frac{1-\tan\theta}{1+\tan\theta}$$ is not the same as $$\frac{1-t^2}{2t}=\frac{1-\tan^2\theta}{2\tan\theta}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using $\sec(x)$ for integral.
Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$
$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$
I tried to rearrange the square root and I got:
$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:
$$x+2=2\sec(u) \\
\sqrt{x^2+4x}=2\tan(u) \\
\mathrm{dx}=2\tan(u)\sec(u)\mathrm{du}$$
And the integral turns out like this:
$$\displaystyle \int \dfrac{\tan^2(u)\sec(u)}{(\sec(u)-1)^2}\mathrm{du}$$
And I continued to rearrange the integral:
$$\displaystyle\int \dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}\mathrm{du}$$
I apply partial fractions method, saying $\cos(u)=u$ without integral sign and I ended up with:
$$\displaystyle\int \dfrac{1}{\cos(u)}\mathrm{du}+2\displaystyle\int \dfrac{1}{1-\cos(u)}\mathrm{du}$$
One can easily integrate those integrals se I skip the calculating part. In the end I get:
$$\ln\left|\sec(u)+\tan(u)\right|+2(-\cot(u)-\csc(u))+\mathrm{C}$$
I drew a triangle:
Knowing $\sec(u)=\dfrac{x+2}{2}$ says us $\cos(u)=\dfrac{2}{x+2}$
As I calculate I got:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{2x-8}{\sqrt{x^2+4x}}+\mathrm{C}$$
Bu the answer key is:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{8}{x+\sqrt{x^2+4x}}+\mathrm{C}$$
I have been thinking where my wrong is for five hours but I couldn't find anything. Please if you see any gap tell me. Thanks.
| The difference between the two answers is:
\begin{align}
&\quad\frac {2x\color{red}{+}8}{\sqrt{x^2+4x}}-\frac {8}{x+\sqrt{x^2+4x}}
\\&=\frac {(2x+8)(x+\sqrt{x^2+4x}) - 8\sqrt{x^2+4x}}{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2x^2+8x+2x\sqrt{x^2+4x}}
{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}}
\\&=\frac {2(x^2+4x+x\sqrt{x^2+4x})}
{x\sqrt{x^2+4x}+x^2+4x}
\\&=2
\end{align}
which is a constant. So both answers are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration, as sketched below.
\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}
Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown
$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.
| Let
$$J=\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx.$$
Using
$$ \arctan x+\arctan\frac1x=\frac{\pi}2$$
and under $x\to \frac1x$, one has
\begin{eqnarray}
J&=&\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx\\
&=&\int_0^\infty \frac{\tan^{-1}x^{-2}-\tan^{-1}x^{-1/2} }{1+x^2}dx\\
&=&\int_0^\infty \frac{\tan^{-1}x^{1/2}-\tan^{-1}x^{2} }{1+x^2}dx\\
&=&-J.
\end{eqnarray}
So $J=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving $\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$
Prove that (where $a,b,c>0$) $$\sum_{cyc} \frac{a(a^2+2bc)}{b+c}\ge \frac{{(a+b+c)}^2}{2}$$
I have found a proof for this problem but it is very lengthy and is not nice.(Its not using computer though).I shall post it later.
Background:
Here however is a very similar problem
If $a,b,c>0$ then prove that $$\sum_{cyc} \frac{a(a^2+bc)}{b+c}\ge a^2+b^2+c^2$$
There is a beautiful proof of this by Michael Rozenberg (arqady)
WLOG $a\ge b\ge c$ rewrite the inequality as $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\ge 0$$ but $$\sum_{cyc}\frac{a(a-b)(a-c)}{b+c}\geq\frac{a(a-b)(a-c)}{b+c}+\frac{b(b-a)(b-c)}{a+c}=$$
$$=(a-b)(\frac{a(a-c)}{b+c}-\frac{b(b-c)}{a+c})\geq0$$
However this method doesnt work...
I am looking for a clean and smooth proof (without using BW,uvw or complete expanding)
I am not planning to disclose how I proved the inequality as it may spoil the fun!
| Suppose $c = \min\{a,b,c\},$ then
$$\sum \frac{a^3+abc}{b+c}-(a^2+b^2+c^2) = \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a}+\frac{c(c-a)(c-b)}{a+b}$$
$$\geqslant \frac{a(a-b)(a-c)}{b+c}+\frac{b(b-c)(b-a)}{c+a} = \frac{(a-b)^2(a^2+ab+b^2-c^2)}{(b+c)(c+a)} \geqslant 0.$$
Therefore
$$\sum \frac{a^3+abc}{b+c} \geqslant a^2+b^2+c^2. \qquad (1)$$
Now, we write the inequality as
$$\sum \frac{a^3+abc}{b+c} + abc \sum \frac{1}{b+c} \geqslant \frac{(a+b+c)^2}{2}.$$
By the Cauchy-Schwarz inequality, we have
$$\sum \frac{1}{b+c} \geqslant \frac{9}{2(a+b+c)}. \qquad (2)$$
From $(1)$ and $(2),$ we need to prove
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)} \geqslant \frac{(a+b+c)^2}{2},$$
equivalent to
$$a^2+b^2+c^2+\frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
This is Schur inequality. The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find $F=\left(\frac{B}{2}\right)^{2H}-1$ when $H$ is maximum of $W_{1}$ and $B$ is minimum of $W_{2}$? The problem is as follows:
A pair of wavelenghts are $W_{1}$ and $W_{2}$:
$W_{1}=5\sin\alpha-5+12\cos\alpha$
$W_{2}=3\sin\beta+\sqrt{3}\cos\beta$
Assuming $H$ and $B$ represent the maximum and the minimum value for
$W_{1}$ and $W_{2}$ respectively.
Find:
$F=\left(\frac{B}{2}\right)^{2H}-1$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{6560}\\
2.&\textrm{5860}\\
3.&\textrm{6562}\\
4.&\textrm{5680}\\
\end{array}$
I'm confused exactly how to tackle this problem from my precalculus workbook. What I attempted to do was to use the boundaries of the sine and the cosine function individually to see if I could obtain the maximum and the minimum values:
$-1<\sin\alpha<1$
$-10<5\sin\alpha-5<0$
Then for $\cos\alpha$:
$-1<\cos\alpha<1$
$-12<12\cos\alpha<12$
Then adding both:
$-22<5\sin\alpha-5+12\cos\alpha<12$
Thus the maximum value for the earlier function would be $12$.
Whereas for the other would be:
$-1<\sin\beta<1$
$-3<3\sin\beta<3$
and for the other term:
$-\sqrt{3}<\sqrt{3}\cos\beta<\sqrt{3}$
Then when adding both:
$-3-\sqrt{3}<3\sin\beta+\sqrt{3}\cos\beta<3+\sqrt{3}$
Then the minimum value would be for the function would be $-3-\sqrt{3}$.
Hence:
$B=12$ and $H=-3-\sqrt{3}$
But the problemas has not ended yet, they are asking the result of $F$
Thus:
$F=\left(\frac{-3-\sqrt{3}}{2}\right)^{2(2)}-1$
Then this is where I got stuck. What could possibly be wrong here?. Which part went wrong?. Can someone help me here?. Could it be that the problem lies in the method to get the minimum and the maximum value for those functions?. The sort of approach which I'm looking to get is derivative-free in other words without using derivatives as this problem was intended to be solved relying only using precalculus.
Can someone help me please?. I'd appreciate an answer with an explanation step-by-step. Thanks in advance.
| Let's try to make $a\sin x+ b\cos x$ of the form $\boxed{r\sin(x+\theta)}$ and $r>0$
$\begin{align}&5\sin\alpha+12 \cos\alpha = r(\cos\theta\sin\alpha+\sin\theta\cos\alpha) = r\sin(\alpha+\theta) \\&\Rightarrow r^2 = 5^2+12^2 = 13^2 \\&\Rightarrow r = 13 \text{ and } \theta =\arctan\frac{12}{5} \end{align} $
Now,
$W_1= 5\sin\alpha+12 \cos\alpha - 5 = 13\sin(\alpha+\theta) - 5$
Max. value $13\sin(\alpha+\theta)$ can attain is $13$.
So max. value of $W_1$ is $\boxed{H = W_{1_{\max}} = 13-5 = 8}$
Similarly
$\begin{align}&3\sin\beta+ \sqrt3 \cos\beta = r\sin(\beta+\theta') = r(\sin\beta\cos\theta'+\cos\beta\sin\theta') \\\Rightarrow &r = \sqrt{9+3} = 2\sqrt3\text{ and }\theta' = \arctan\frac1{\sqrt3}\end{align} $
So,
$W_2 = 2\sqrt3\sin(\beta+\theta')$
Min. value of $\sin(\beta+\theta') = -1$ and min. value of $W_2$ is $\boxed{B = W_{2_{\min}} = 2\sqrt3 (-1) =-2\sqrt3}$
So, $$F = \left(\frac{-2\sqrt3}{2}\right)^{2(8)}-1 = 3^8-1 = 6560$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The variation of a Ukrainian Olympiad problem: 10982
Given a recursion $a_{n+ 1}= \dfrac{a_{n}}{n}+ \dfrac{n}{a_{n}}$ with $a_{1}= 1.$ Prove that
$$\lim a_{n}^{2}- n= \frac{1}{2}$$
Source: StachMath/@RiverLi _ The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$
The original problem already has an answer, I'll suggest this way of thinking, which is not mine, but @twelve_sakuya
Let $b_{n}:=a_{n}^{2}- n,$ so
$$a_{n+ 1}= \frac{a_{n}}{n}+ \frac{n}{a_{n}}\Leftrightarrow b_{n+ 1}- \frac{1}{2}= -\frac{n}{b_{n}+ n}\left ( b_{n}- \frac{1}{2} \right )+ \frac{b_{n}}{2\left ( b_{n}+ n \right )}+ \frac{b_{n}}{n^{2}}+ \frac{1}{n}$$
That means $\left | b_{n+ 1}- \dfrac{1}{2} \right |\leq\dfrac{n}{\left | b_{n}+ n \right |}\left | b_{n}- \dfrac{1}{2} \right |+ \dfrac{\left | b_{n} \right |}{2\left | b_{n}+ n \right |}+ \dfrac{\left | b_{n} \right |}{n^{2}}+ \dfrac{1}{n}.$ Therefore, if we can get the evaluations of $\left | b_{n} \right |$ or $\dfrac{\left | b_{n} \right |}{n},$ maybe there exists a number $\beta\in\left ( 0, 1 \right )$ so that
$$\left | b_{n}- \frac{1}{2} \right |\leq\beta^{n}B\left ( n \right )\rightarrow 0\,{\rm as}\,n\rightarrow\infty$$
My friend has no confidence to continue, he also said that is the variation of a Ukrainian Olympiad problem #10982 (I searched and got the result, but I couldn't access it). I need to the help, thanks a real lot !
| Note that, for $n\ge 1$, $$ n^6+10n^5+38n^4+68n^3+57n^2+18n>n^6+10n^5+37n^4+62n^3+46n^2+12n+1$$ and, factorising,
$$n(n+1)^2(n+2)(n+3)^2>(n^3+5n^2+6n+1)^2.$$
Also note that the function $F_n$ such that $$F_n(x)=\frac{n}{x}+\frac{x}{n}$$ is monotonic decreasing for $0<x\le n$.
Now suppose that $$\frac {n+2}{\sqrt {n+1}}\ge a_{n+2}\ge \frac{n+1}{\sqrt n} \tag 1$$ which is true for $n=1$. Then $$F_{n+2}\left(\frac {n+1}{\sqrt {n}}\right)\ge F_{n+2}\left(a_{n+2}\right)\ge F_{n+2}\left(\frac{n+2}{\sqrt {n+1}}\right)$$
$$\frac {n^3+5n^2+6n+1}{\sqrt {n}(n+1)(n+2)}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$
$$\frac {n+3}{\sqrt {n+2}}\ge a_{n+3}\ge \frac{n+2}{\sqrt {n+1}} $$ Therefore, by induction, Equation (1) is true for all positive integers $n$.
Let $a_n^2=n+x_n$ for all $n$. Then, for $n\ge 1$, $$\frac{(n+2)^2}{n+1}\ge {a_{n+2}}^2\ge\frac{(n+1)^2}{n} $$ $$\frac{n+2}{n+1}\ge x_{n+2}\ge \frac{1}{n}.$$
We can now use this bound on $x_n$. $${a_{n+1}}^2=\frac{{a_n^2}}{n^2}+\frac{n^2}{{a_n}^2}+2$$
$$n+1+x_{n+1}=\frac{n+x_n}{n^2}+\frac{n^2}{n+x_n}+2$$
$$x_{n+1}=1+\frac{1}{n}+n\left(1+\frac{x_n}{n}\right)^{-1}-n+O\left(\frac{1}{n^2}\right)$$
$$x_{n+1}=1+\frac{1}{n}-{x_n}+\frac{{x_n}^2}{n}+O\left(\frac{1}{n^2}\right)$$
Then $x_n+x_{n+1}=1+O\left(\frac{1}{n}\right)$ and so $x_n+x_{n+1}$ tends to $1$. We also have
$$x_{n+2}=1+\frac{1}{n}-{x_{n+1}}+\frac{{x_{n+1}}^2}{n}+O\left(\frac{1}{n^2}\right)$$ and therefore $$|x_{n+2}-x_{n+1}|=\frac{n-1}{n}|x_{n+1}-x_{n}|+O\left(\frac{1}{n^2}\right)$$
$$|x_{2n}-x_{2n-1}|=\frac{(n-1)n...(2n-3)}{n(n+1)...(2n-2)}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)=\frac{1}{2}|x_{n+1}-x_{n}|+O\left(\frac{1}{n}\right)$$
Then $x_{n+1}-x_n$ tends to $0$ and so $x_n$ tends to $\frac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Use epsilon-delta definition of limits to show that $\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$ and $\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$ Use epsilon-delta definition of limits to show that
$$\lim_{(x,y) \to (1,2)}(3x+2y-1)=6$$
$$\lim_{(x,y) \to (0,0)}(x^2+y^2)=0$$
We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{\left(x-1\right)^{2}+\left(y-2\right)^{2}}<\delta \implies \left|\left(3x+2y-1\right)-6\right|<\epsilon)))$$
Taking $\left|x-1\right|,\left|y-2\right|<\delta$ follows :
$$\left|\left(3x+2y-1\right)-6\right|\le 3\left|x-2\right|+\left|2y-1\right|<3(\delta+1)+(2\delta+3)$$
$$\left|\left(3x+2y-1\right)-6\right|<5\delta +6$$
Taking $ \delta \le{(\epsilon -6)}/5$ shows the claim.
We need to show that $$\forall \epsilon >0( \exists \delta >0( \forall (x,y) \in \mathbb R^2 (0<\sqrt{x^2+y^2}<\delta \implies \left|x^2+y^2 \right|<\epsilon)))$$
$$\left|x^2+y^2 \right|=x^2+y^2$$
Taking $ \delta \le\sqrt{\epsilon }$ shows the claim.
I want to know if my solution is true,so if you have an alternative solution that would be nice to see that,but please first check mine.
| The latter one is correct, but the former is not. A big hint is that the choice $\delta \le (\varepsilon - 6) / 5$, which is negative(!) when $\varepsilon < 6$. Your choice of $\delta$ should always be positive, so that at least one $(x, y)$ satisfies $0 < \|(x, y) - (x_0, y_0)\| < \delta$.
Instead, consider that
$$|(3x + 2y - 1) - 6| = |3(x - 1) + 2(y - 2)| \le 3|x - 1| + 2|y-2|.$$
So, if we take $\delta = \varepsilon / 5$, then
\begin{align*}
0 < \sqrt{(x - 1)^2 + (y - 2)^2} < \delta &\implies |x - 1|, |y - 2| < \frac{\varepsilon}{5} \\
&\implies |(3x + 2y - 1) - 6| < 3\frac{\varepsilon}{5} + 2\frac{\varepsilon}{5} = \varepsilon.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028984",
"timestamp": "2023-03-29T00:00:00",
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What is the chance to get two pairs from a hand of five cards if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards What is the chance to get two pairs if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!
Is my solution right?
$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$
First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).
Lastly, I divided all on $\binom{55}{5}$
| It seems wise to count hands with pairs of kings or queens separately:
Hands with a pair of kings and not a pair of queens:
$$\binom{5}{2}\binom{11}{1}\binom{4}{2}\binom{46}{1}$$
because we have to pick $2$ kings out of the $5$, then one other type of pair not king or queen (so $11$ ranks to choose from), then $2$ cards out of the $4$ available in that rank, and finally, $1$ card out of the remaining $55 - 5- 4 = 46$ cards (subtracting the ranks (kings, and the other one) from the total number of cards we now have). The latter single remaining card I do not see occur in your count at all, which seems fishy to me.
Hands with a pair of queens but not a pair of kings:
$$\binom{6}{2}\binom{11}{2}\binom{4}{2}\binom{45}{1}$$
Which has been simularly computed as the previous type, but now we have $6$ queens to choose $2$ from and now $55-6 - 4 = 45$ final cards to choose from.
Hands with both a pair of queens and a pair of kings:
$$\binom{6}{2}\binom{5}{2}\binom{55-11}{1}$$
Simpler because there are no extra ranks for pairs to choose and more cards left at the end (just kings and queens, $11$ now, are forbidden).
Finally hands with neither pairs of kings nor queens:
$$\binom{11}{2}\binom{4}{2}^2\binom{55- 8}{1}$$
Where we first choose the two ranks to form pairs from and then twice the $2$ pairs themselves, and finally a term for the final card.
Now add these $4$ numbers to get all hands with two pairs and indeed divide by $\binom{55}{5}$ to get the probability (all possible choices of $5$ out of the total expanded deck).
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?
| If $a$ and $b$ are both even, say $a=2^mA$ and $b=2^nB$ with $A$ and $B$ odd, then without loss of generality $n\geq m$ and so
$$2^c=a^3+b^3=2^{3m}\big(A^3+2^{3(n-m)}B^3\big),$$
from which it follows that
$$2^{c-3m}=A^3+2^{3(n-m)}B^3.$$
Because $A$ and $B$ are positive we see that $2^{c-3m}>1$, and so the left hand side is even. Because $A$ is odd it follows that also $2^{3(n-m)}B^3$ is odd, and so $n=m$. Then for $C=c-3m$ we have
$$A^3+B^3=2^C,$$
thus yielding a new solution with $A$ and $B$ both odd.
If $a$ and $b$ are both odd then you already observed that the factorization
$$2^c=a^3+b^3=(a+b)(a^2-ab+b^2),$$
implies that $a^2-ab+b^2=1$. Viewing this as a quadratic in $a$ shows that
$$a=\frac12\big(b\pm\sqrt{b^2-4(b^2-1)}\big)=\frac12\big(b\pm\sqrt{4-3b^2}\big).$$
For this to be an integer we must have $4-3b^2>0$, so $b=1$ as $b$ is positive by assumption. Then also $a=1$ as $a$ is positive by assumption. This shows that $a=b=1$. It follows that every solution is of the form $a=b=2^m$ for some nonegative integer $m$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $f$ verifies $f(1−x) + f(x) = 1$ I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression
\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}
I would like to show that $f(1-x) + f(x) = 1$.
What I have done so far
With a little bit of algebra I have
\begin{align*}
f(x)
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
&& \text{using }\binom{n}{m} = \frac{n!}{m!(n-m)!}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p, j\neq-k}^{p} (x-j)\right)
&& \text{using the change of indices j=i-k}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p}^{p} (x-j)\right) \frac{1}{x+k}
&& \text{completing the product}
\\
&= \frac{1}{(2p-1)!} \left(\prod_{j=1-p}^{p} (x-j)\right) \left(\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}\right)
&& \text{factoring the product from the sum}
\end{align*}
Then I introduce the functions
\begin{align*}
F(x) &= (2p-1)! f(x) = G(x) H(x)\\
G(x) &= \prod_{j=1-p}^{p} (x-j) \\
H(x) &= \sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\end{align*}
So the initial question reduces to proving that $F(1-x) + F(x) = (2p-1)!$
Next, playing with the indices of the product I have been able to show that $G(1-x) = G(x)$.
So that now I only need to show $G(x) \left( H(1-x) + H(x)\right) = (2p-1)!$
EDIT:
I have just been able to simplify $H(1-x) + H(x)$ as follows :
\begin{align*}
H(x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\\
&=\sum_{s=p}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p+k
\end{align*}
And
\begin{align*}
H(1-x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p+k}
\\
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p-1-k}
&& \text{using } \binom{2p-1}{p+k} = \binom{2p-1}{p-1-k}
\\
&=\sum_{s=0}^{p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p-1-k
\end{align*}
So finally
\begin{align*}
H(1-x) + H(x)
&=\sum_{s=0}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
\end{align*}
EDIT2:
I have just made the same work on $G(x)$ as follows
\begin{align*}
G(x)
&=\prod_{j=1-p}^{p} (x-j) \\
&=\prod_{s=0}^{2p-1} (x-p+s)
&& \text{using a change of index } s=p-i
\end{align*}
Now I think the constant $(2p-1)!$ in the relation $F(1-x)+F(x)=(2p-1)!$ comes from the binomial factor in $H(1-x) + H(x)$ but I don't see how to prove that...
Any lead ?
| Here we have a partial fraction decomposition in disguise.
We obtain
\begin{align*}
\color{blue}{f(x)}&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}
\prod_{\substack{j=k-p+1\\j\neq 0}}^{k+p}(x+k-j)\\
&=\prod_{j=-p+1}^{p}(x-j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}\tag{1}\\
&\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}}\tag{2}\\
\end{align*}
Comment:
*
*In (1) we expand the summands with $(x+k)/(x+k)$ and shift the product by $k$ to start with $j=-p+1$. This way the product can be factored out from the sum.
*In (2) we substitute $j$ with $-j$ in the product.
We start with (2) and consider $f(1-x)$. We obtain
\begin{align*}
\color{blue}{f(1-x)}&=\prod_{j=-p}^{p-1}(1-x+j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{1-x+k}\\
&=\prod_{j=-p}^{p-1}(x-j-1)\sum_{k=0}^{p-1}
\frac{(-1)^{k-1}}{k!(2p-1-k)!}\,\frac{1}{-x+p-k}\tag{3}\\
&=\prod_{j=-p+1}^{p}(x+j-1)\sum_{k=0}^{p-1}
\frac{(-1)^{k}}{k!(2p-1-k)!}\,\frac{1}{x+k-p}\tag{4}\\
&\,\,\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{-1}
\frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{5}\\
\end{align*}
Comment:
*
*In (3) we factor out $(-1)^{2p}=1$ from the product. We also change the order of summation $k\to p-1-k$.
*In (4) we substitute $j$ with $-j$ in the product and we expand numerator and denominator of the summands with $-1$.
*In (5) we shift the index $j$ by one to start with $j=-p$. We also shift the index $k$ by $p$ to start with $k=-p$.
We can now sum up (2) and (5) and get
\begin{align*}
\color{blue}{f(x)+f(1-x)}&\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}
\frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{6}
\end{align*}
The right-hand side of (6) looks like a partial fraction decomposition. Indeed, we consider a function
\begin{align*}
g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)}
\end{align*}
The function $g(x)$ is a rational function with $2p$ pairwise different simple poles. Therefore we can find a representation
\begin{align*}
g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)}=\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\qquad\qquad a_k\in\mathbb{R}
\end{align*}
Multiplication with the denominator of $g(x)$ gives
\begin{align*}
\color{blue}{1=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}}\tag{7}
\end{align*}
We can now evaluate (7) at $x=-k_0, -p\leq k_0\leq p-1$. We obtain
\begin{align*}
1&=\left.\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\right|_{x=-k_0}\\
&=\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}\left(-k_0+j\right)a_{k_0}\\
a_{k_0}&=\frac{1}{\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}(j-k_0)}\\
&=\frac{1}{\prod_{j=-p}^{k_0-1}(j-k_0)}\,\frac{1}{\prod_{j=k_0+1}^{p-1}(j-k_0)}\\
&=\frac{(-1)^{p+k_0}}{(p+k_0)!}\,\frac{1}{\left(p-1-k_0\right)!}
\end{align*}
The coefficients $a_{k_0}$ can now be put into (7) and we obtain expression (6). We conclude
\begin{align*}
\color{blue}{f(x)+f(1-x)=1}
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the approach to find the expression of the partial sum of a series? This question is about computing the limit of a series as $n$ goes to $\infty$.
The series $(x_n)$ is defined as:
$$\sum_{i=0}^n \frac{1}{n^2+i}$$
So the $n$th partial sum will be given by: $\frac{1}{n^2+1} + \frac{1}{n^2+2} + \frac{1}{n^2+3} + ... + \frac{1}{n^2+n}$
So $x_1 = \frac{1}{1^2 + 1}$,
$x_2 = \frac{1}{2^2 + 1} + \frac{1}{2^2 + 2}$,
$x_3 = \frac{1}{3^2 + 1} + \frac{1}{3^2 + 2} + \frac{1}{3^2 + 3}$, and so on.
I have thought about partial fraction decomposition and trying to form a telescoping series, but have been unable to re-write the expression in a way that gives such a series. I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the $n$th partial sum. If the above idea is wrong, can I have some hints as to how to approach this problem?
| For the limit itself, you aleady received a good answer.
We can approximate the partial sums using harmonic numbers since
$$S_n=\sum_{i=0}^n \frac{1}{n^2+i}=H_{n^2+n}-H_{n^2-1}$$ Now, using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
p^4}+O\left(\frac{1}{p^6}\right)$$ apply it twice and continue with Taylor series. This could give
$$S_n=\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right)$$ Compute for $n=10$; the exact value is
$$S_{10}=\frac{61961120721796411}{590910221462160600}\approx 0.1048571$$ while the above truncated series gives
$$S_{10}\sim \frac{12583}{120000}\approx 0.1048583$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Parameterizing both branches of a hyperbola Recently I have been studying parametric equations of surfaces and curves, specifically hyperbolic functions. Given by the equations $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad\frac{(x-\alpha)^2}{a^2}-\frac{(x-\beta)^2}{b^2}=1$$
denoting hyperbolic functions centred at $(0,0)$ and $(\alpha,\beta)$ respectively. I am however, unable to derive a parametrisation that represents the entire function. Help to further understand this topic would be greatly appreciated.
| Here's a geometric way to derive a parametrization. Consider the family of all lines passing through the point $(a,0)$. Such a line is of the form $y = t(x-a)$ for some slope $t$. Since the hyperbola has degree $2$, each line will intersect it in exactly two points: $(a,0)$ and one other point. (This is an example of Bézout's Theorem.) As we vary the slope $t$, this other point will trace out the curve.
Let's use this observation to compute equations for the parametrization. Substituting $y = t(x-a)$ into the equation for the hyperbola, we have
\begin{align*}
0 &= \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1 = \frac{x^2}{a^2} - \frac{(t(x-a))^2}{b^2} - 1 \, .
\end{align*}
Multiplying through by $a^2 b^2$ to clear denominators yields
\begin{align*}
0 &= b^2 x^2 - a^2 (t(x-a))^2 - a^2 b^2 = b^2 x^2 - a^2 t^2(x^2 - 2ax + a^2) - a^2 b^2\\
&= (b^2 - a^2 t^2)x^2 + 2 a^3 t^2 x - a^4t^2 - a^2 b^2
\end{align*}
and dividing through by the leading coefficient, we find
$$
0 = x^2 + \frac{2 a^3 t^2}{b^2 - a^2 t^2} x - \frac{a^2 (a^2 t^2 + b^2)}{b^2 - a^2 t^2} \, .
$$
One of the roots is $x = a$ and since the constant term is the product of the roots, then the other is
\begin{align*}
x = -\frac{a (a^2 t^2 + b^2)}{b^2 - a^2 t^2}
\end{align*}
which, together with
$$
y = t(x-a) = t\left(-\frac{a (a^2 t^2 + b^2)}{b^2 - a^2 t^2} - a\right)
$$
gives a parametrization of the whole hyperbola, except for the point $(a,0)$. Note that since $b^2 - a^2 t^2 = (b - at)(b+at)$, then $x$ and $y$ have poles at $t = b/a$ and $t=-b/a$. Here's a SageMath cell plotting this parametrization, and the resulting plot with the ranges $t < -b/a$, $-b/a < t < b/a$, and $b/a < t$ colored differently.
$\hspace{4cm}$
For more on parametrizing curves, see Ch. $1$, $\S3$ of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms. It's a wonderful book, and they use fun examples like this to motivate later topics in the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinates Context
This is an interim problem related to a Green's function solution for a boundary-value problem in the cylindrical coordinate system.
Question
How do I convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinate system?
My worked solution is given in the answer below.
|
$$(x-x')^2 + (y-y')^2 + (z-z')^2
=r^2 + {r'}^2 -2\,r\,{r}' \,\cos\left(\theta - \theta'\right) + (z-z')^2$$
My attempt
I know that to tranform from Cartesian to cylindrical requires the following three equations
\begin{align}
r &= \sqrt{x^2 + y^2} \\
\theta &= \arctan{\left(\frac{y}{x}\right)} \\
z &= z \quad.
\end{align}
Directly, the question reduces to how to covert the following to cylindrical system.
$$(x-x')^2 + (y-y')^2.$$
Its natural to begin with $r$ and $r'$.
\begin{align}
(x-x')^2 + (y-y')^2
&=
x^2 - 2\,x\,x' +{x'}^2 + y^2 - 2\,y\,{y'} +{y'}^2
\\
&=
x^2 + y^2 +{x'}^2 +{y'}^2 - 2\,x\,x' - 2\,y\,{y'}
\\
&=
r^2 +{r'}^2 - 2\,x\,x' - 2\,y\,{y'}
\end{align}
Now the problem is to convert $ - 2\,x\,x' - 2\,y\,{y'} $ to cylindrical coordinate system.
The next part of the answer required trial and error, and some good fortune.
\begin{align*}
r\,r' \,\cos\left(\theta - \theta'\right)
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{y}{x}} - \arctan{\frac{y'}{x'}}\right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y}{x} + \frac{y'}{x'}}} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y\,y'}{x\,x'} }} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{ y\, x' - y'\,x }{x\,x'+ y\,y' }} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{1}{\sqrt{1+\left(\frac{ y\, x' - y'\,x }{x\,x'+ y\,y' }\right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{(x\,x'+ y\,y' )^2+\left( y\, x' - y'\,x \right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{\left(x\,x'\right)^2+ 2\,x\,x' \, y\,y' + \left( y\,y' \right)^2+\left( y\, x' \right)^2 -2\, y\, x'\,y'\,x + \left( y'\,x \right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{\left(x\,x'\right)^2+ \left( y\,y' \right)^2+\left( y\, x' \right)^2 + \left( y'\,x \right)^2}}
\\
&=
x\,x'+ y\,y'
\end{align*}
At this point the nut is cracked, and I rebuild the solution as follows:
\begin{align}
- 2\,x\,x' - 2\,y\,{y'}
&= -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
r^2 + {r'}^2 - 2\,x\,x' - 2\,y\,{y'}
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
(x-x')^2 + (y-y')^2
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
(x-x')^2 + (y-y')^2 + (z-z')^2
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right) + (z-z')^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $\tan(\alpha/2)=(1-\cos\alpha)/\sin \alpha, \,?$ Is it true that
$$\bbox[5px,border:2px solid #138D75]{\tan\left(\frac {\alpha}2\right)=\frac{1-\cos\alpha}{\sin \alpha}, \quad ?} \tag1$$
My solution:
First, we note that the expression $\tan \left ( \frac{\alpha}{2} \right )$ makes sense when $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$: henceforth, we will assume that this condition is met.
By definition of tangent function, we have: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right )}$$
When $\alpha \neq 2k \pi, k \in \mathbb{Z}$, we can multiply numerator and denominator by $\sin \left ( \frac{\alpha}{2} \right )$, because this quantity is non-zero. We therefore get:
$$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ \sin^2 \left ( \frac{\alpha}{2} \right )}{\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right )}. $$
For what we saw earlier: $$\sin^2 \left ( \frac{\alpha}{2} \right ) = \frac{1 - \cos \alpha}{2}$$and instead, applying the sine duplication formula: $$\cos \left ( \frac{\alpha}{2} \right ) \cdot \sin \left ( \frac{\alpha}{2} \right ) = \frac{1}{2} \sin \alpha$$
In conclusion we get the following formula: $$\tan \left ( \frac{\alpha}{2} \right ) = \frac{ 1 - \cos \alpha}{\sin \alpha}.$$
If in the expression of the tangent we had multiplied numerator and denominator by $\cos \left ( \frac{\alpha}{2} \right )$ then with steps similar to what we have done we get another formula:$$\bbox[5px,border:2px solid #118D45]{\tan \left ( \frac{\alpha}{2} \right ) = \frac{\sin \alpha}{1 + \cos \alpha}}. \tag 2 $$
We note that it always holds $\cos \left ( \frac{\alpha}{2} \right ) \neq 0$ since we have imposed $\alpha \neq \pi + 2k\pi, k \in \mathbb{Z}$ (and this entitles us to multiply numerator and denominator by this term).
Is there a faster and more immediate proof of the $(1)$ or $(2)$?
| Replace $\alpha$ by $2\beta$ so that the right side is
$${{1-\cos 2\beta}\over {\sin 2\beta}}=
{{1-(\cos^2\beta-\sin^2\beta)}\over {2\sin\beta\,\cos\beta}}$$
$$={{2\sin^2\beta}\over{2\sin\beta\cos\beta}}=
{{\sin\beta}\over {\cos\beta}}=\tan\beta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity
$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$
I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.
The answers for $c$ and $d$ are $c=29$, $d=-3$.
| Expanding we get $ax^3 + 3abx^2 + 3ab^2x +ab^3 + c = 4x^2-24x^2 +48x + d$
Technically what you have going on is 4 equations.
*
*$a = 4$
*$3ab=-24$
*$3ab^2 = 48$.
*$ab^3 + c = d$.
They aren't linear as the involve $b$ to powers $2$ and $3$ but $1,2$ are independent linear so $a,b$ can be solved $a= 4, b=-2$. And a equation three is completely redundant (it simplifies to the inane and unhelpful $48 = 48$... gee, no ****)
That leaves a final equation $-32 + c = d$. This is the only equation with $c$ or $d$ so we can express $c$ in terms of $d$ ($c =d +32$) or $d$ in terms of $c$ ($d=c-32$) but as it is one equation with two unknowns that's as for as we can go. We can let $d$ be anything we want and $c$ infinite potential values will follow (or vice versa. $c=29$ and $d = c -32 =-3$ is certainly an answer but it is by no means the only solution.
After all if we had replaced $c,d$ with $M,N$ or $K + 856, J+ 856$ or $c -5, d-5$ nothing would have changed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
The triangle formed by the $x$-axis, the tangent to $y=1/x$ at a 1st-quadrant point, and the line from the origin to that point is isosceles Here is Prob. 45, Sec. 3.2, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:
Let $P$ be a point on the first-quadrant part of the curve $y = 1/x$. Show that the triangle determined by the $x$-axis, the tangent at $P$, and the line from $P$ to the origin is isosceles, and find its area.
My Attempt:
Let the point $P$ be given by $P = \left( a, \frac{1}{a} \right)$, where $a > 0$.
Then the line from $P$ to the origin has equation
$$ y = \frac{ \frac{1}{a} }{a} x = \frac{1}{a^2} x. \tag{1} $$
The slope of the tangent line to the curve $y = 1/x$ at the point $P$ is
$$
\left( \frac{d y}{d x} \right)_{x = a} = -\frac{1}{a^2},
$$
and thus the equation of the tangent line is
$$
y = -\frac{1}{a^2} (x-a) + \frac{1}{a} = -\frac{1}{a^2} x + \frac{2}{a}. \tag{2}
$$
And, this line intersects the $x$-axis at the point $(x, 0)$, where
$$
-\frac{1}{a^2} x + \frac{2}{a} = 0,
$$
that is,
$$
x = \frac{ \frac{2}{a} }{ \frac{1}{a^2} } = 2a.
$$
Thus the tangent line to the curve at the point $(a, 1/a)$ intersects the $x$-axis at the point $A$ given by
$$A = (2a, 0). $$
And, the point of intersection of the tangent line to the curve at point $P$ and the line from $P$ to the origin is of course the point $P = (a, 1/a)$ itself.
Thus the three vertices of our triangle are $O = (0, 0)$, $A = (2a, 0)$, and $P = (a, 1/a)$, where $a > 0$. So the sides of our triangle have lengths
$$
\left| \overline{OA} \right| = 2a,
$$
$$
\left| \overline{AP} \right| = \sqrt{ (2a-a)^2 + (0-1/a)^2 } = \sqrt{ a^2 + 1/a^2},
$$
and
$$
\left| \overline{PO} \right| = \sqrt{ (a-0)^2 + (1/a-0)^2 } = \sqrt{a^2 + 1/a^2}.
$$
Thus the sides $AP$ and $PO$ are congruent, showing that our triangle is indeed isosceles.
Finally, since the base of our triangle is the side $OA$ and the altitude is the vertical line segment from the $x$-axis to the point $P$, therefore the area of our triangle is
$$
\frac{1}{2} \times (2a) \times \frac{1}{a} = 1.
$$
Is what I have done correct and clear in each and every detail? Or, are there any issues of accuracy or clarity?
| It looks good to me. A geometric perspective to consider:
When we find point $A =(2a, 0)$, notice that $P$ is directly above the midpoint $OA$. (This midpoint is $C = (a, 0)$.)
The triangles $OCP$ and $ACP$ are congruent, right-angled triangles that are mirror images of each other ($CP$ as a common edge).
Hence $OPA$ is an isosceles triangle.
Moreover if we rotate $ACP$ around $P$ until it forms a rectangle with $OCP$ (which occurs when $A$ moves to $O$), we get a rectangle with base $a$ and height $1/a$ hence it has area $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the particular solution of a differential equation using at least three different methods. Find the particular solution $(x^2+6y^2)dx-4xydy=0$; when $x=1$, $y=1$ using at least three different methods.
I have done the first two. Can somebody help me with the third method.
Method 1: Homogenous Equation
Let $y=vx; dy=vdx+xdv$
$(x^2+6x^2v^2)dx-4x(vx)(vdx+xdv)=0$
$(x^2+2x^2v^2)dx-4x^3vdv=0$
$(1+2v^2)dx-4xvdv=0$
$\int\frac{dx}{4x}-\int\frac{v}{1+2v^2}dv=0$
$\ln{x}-\ln{(2v^2+1)}=C$
$\ln{(\frac{x}{2v^2+1})}=\ln{C}$
$\frac{x}{2v^2+1}=\frac{1}{C}$
$C=3$
$3x=2v^2+1$
$3x^3=2y^2+x^2$
$2y^2=x^2(3x-1)$
The particular solution by method 1 is $2y^2=x^2(3x-1)$.
Method 2: Bernoulli Equation
$2y\frac{dy}{dx}-\frac{x^2+6y^2}{2x}=0$
$2y\frac{dy}{dx}-\frac{3y^2}{x}=\frac{x}{2}$
Let $v=y^2; dv=2ydy$
$\frac{dv}{dx}-\frac{3v}{x}=\frac{x}{2}$
$P(x)=-3x^{-1}$; I.F.$=e^{-3\int x^{-1}dx}=x^{-3}$
$vx^-3=\frac{1}{2}\int\frac{dx}{x^2}$
$2vx^{-3}=-x^{-1}+C^{-1}$
$2y^2x^{-3}+x^{-1}=C^{-1}$
$C=\frac{1}{3}$
$2y^2+x^2=3x^3$
$2y^2=x^2(3x-1)$
The particular solution by method 2 is also $2y^2=x^2(3x-1)$.
| Compare
$$(x^2+6y^2)dx-4xydy=0~~~~(1)$$
with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial N}{dx}=-4y $$ It's integrating factor is
$$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$
Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE:
$$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$
It's solution is
$$\int (x^{-2}+6x^{-4}y^2) dx~~~ \text{(treat y as constant)}=C$$
$$-x^{-1}-2x^{-3}y^{2}=C$$
By the condition $y(1)=1$, we get $C=-3$
Finally $$2y^2=3x^3-x^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Regularized Incomplete Beta Function I try to solve this property of Regularized Incomplete Beta Function. How can you solve a statement :
$I_x (a,a) = 1 - \frac{1}{2} I_{4x(1-x)}(a,\frac{1}{2}) $ where $1/2 \leq x \leq 1$.
Some definitions :
$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)}$ with B(x;a,b) is incomplete Beta function and B(a,b) is Beta function.
I tried to use substitution rule with z = 1 - x. But it doesn't go any further.
Could someone give me idea more please ?
| Using the definitions you wrote
$$\text{lhs}=\frac{B_x(a,a)}{B(a,a)}$$
$$\text{rhs}=1-\frac{B_{4 (1-x) x}\left(a,\frac{1}{2}\right)}{2 B\left(a,\frac{1}{2}\right)}$$ Now, for $ \frac 12 \leq x \leq 1$
$$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-3} (4 (1-x)-4 x) ((1-x) x)^{a-1}}{\sqrt{1-4 (1-x) x}
B\left(a,\frac{1}{2}\right)}$$
Since $ \frac 12 \leq x \leq 1$, the last term simplify and write
$$\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$
$$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ Divide by $(1-x)^{a-1} x^{a-1}$ to get
$$\frac{\frac d {dx}\text{[lhs-rhs]} } {(1-x)^{a-1} x^{a-1} }=\frac{1}{B(a,a)}-\frac{2^{2 a-1}}{B\left(a,\frac{1}{2}\right)}=0$$ by the properties of the beta function.
So, for $ \frac 12 \leq x \leq 1$, $\text{[lhs-rhs]}$ is a constant. For $x=1$ for example
$$\text{lhs}=\frac{B_1(a,a)}{B(a,a)}=1 \qquad \text{and} \qquad \text{rhs}=1-\frac{\Gamma \left(a+\frac{1}{2}\right) B_0\left(a,\frac{1}{2}\right)}{2
\sqrt{\pi } \Gamma (a)}=1$$ Then, the identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Least squares solution problem
Let $A = \begin{pmatrix}
2 & 0 & 2 \\
1 & 1 & 2 \\
1 & 1 & 2 \\
2 & 0 & 2
\end{pmatrix}$. Determine the least squares solution for $Ax = e_4.$
The least squares solution can be found from $A^TAx = A^Te_4$. This is problematic since $A^TAx = \begin{pmatrix}10&2&12\\ 2&2&4\\ 12&4&16\end{pmatrix}x = \begin{pmatrix}2\\ 0\\ 2\end{pmatrix}$, but $x$ cannot be solved from here since I cannot multiply the RHS by $A^{T^{-1}}$... What can I do here?
| As the matrix
$$B=A^TA=\begin{pmatrix}10&2&12\\ 2&2&4\\ 12&4&16\end{pmatrix}$$ is not invertible, the solution of the least square problem is a linear subspace of dimension greater or equal than $1$. Indeed the kernel of $B$ is a linear line.
You can solve the system
$$Bx = \begin{pmatrix}2\\ 0\\ 2\end{pmatrix}$$ to find this linear line.
| {
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Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$ Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $ so $\sin \frac{C-A}{2} = \frac{1}{4}$
I feel like I am awfully close here, but how do I finish it?
| \begin{align*}
\sin B &= 2 (\sin C - \sin A) \\
&=4 \cos \left(\frac{A+C}{2}\right) \sin \left(\frac{C-A}{2}\right)
\end{align*}
Moreover
\begin{align*}
\sin B &= \sin(\pi-(A+C)) \\
&=2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+C}{2}\right)
\end{align*}
Therefore
\begin{align*}
\sin \left(\frac{A+C}{2}\right) &= 2 \sin \left(\frac{C-A}{2}\right) \\
\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right)+\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)&=2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A}{2}\right)-2\sin \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right) \\
3\tan \left(\frac{A}{2}\right)&=\tan \left(\frac{C}{2}\right)
\end{align*}
Let $c=\tan \left(\frac{C}{2}\right)$ and $a=\tan \left(\frac{A}{2}\right)$. We have both $a$ and $c$ in the interval $(0,\infty)$. Using $\sin(2x)=\frac{2\tan x}{1+\tan^2 x}$, we have
\begin{align*}
\frac{5}{\sin A} + \frac{9}{\sin C}&= 5\left(\frac{1+a^2}{2a}\right)+9\left(\frac{1+c^2}{2c}\right)\\
&= \frac{5}{2a}+\frac{5a}{2}+\frac{9}{2c}+\frac{9c}{2}\\
&=\frac{5}{2a}+\frac{5a}{2}+\frac{3}{2a}+\frac{27a}{2}\\
&=16a+\frac{4}{a}
\end{align*}
By the AM-GM inequality, $a=\frac{1}{2}$ minimizes this function in $(0,\infty)$ and the minimum value is $16$.
| {
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If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$
I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.
| let $f_z=(x+y-7)(z(x+y)+24)$,$f_y,f_x$ defined similarly $$f_x=f_y=f_z=k$$ $$f_z-f_y=0\to -(y - z) (x^2 - 7 x - y z - 24)=0$$ so $$x^2-7x-yz=24$$and similarly for others yield $$p_x=p_y=p_z=24$$ where $p_x=x^2-7x-yz$ and others defined similarly $$p_x-p_y=0\to (x-y)(x+y+z-7)=0$$ so $$x+y+z=7\to z=7-x-y$$ Plug this into orginal condition to get $$z^3-7z^2-24z-k=x^3-7x^2-24x-k=y^3-7y^2-24y-k=0$$ thus $x,y,z$ are the roots of cubic $$p(t)=t^3-7t^2-24t-k$$ Now by vieta you can easily find $x^2+y^2+z^2$
| {
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Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because
Although my answer is correct, I was apparently supposed to use some theory to answer this question.
I started by putting everything on common denominator and factoring out $n$:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$
then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$
$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$
$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$
$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$
$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$
$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$
to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,
and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.
Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?
| You can cut some work by just looking at the remainder mod $3$.
Let $n=3q+r$. Easy calculation gives
$$\frac15(3q+r)^5+\frac13(3q+r)^3+\frac{7}{15}(3q+r) = \text{some integer} + \frac{243q^5+7q}{5} + \frac{r^5}5+\frac{r^3}3+\frac{7r}{15}.$$
By Little Fermat we have $q^5 \equiv q \pmod{5}$ so
$$243q^5+7q \equiv 250q \equiv 0\pmod{5}.$$
The second term
$$\frac{r^5}5+\frac{r^3}3+\frac{7r}{15}$$
is the same as the original expression but you only have to check $r=0,1,2$ which is almost trivial.
| {
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Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$
Calculate the sum of all irrational roots of
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}- xy+ y^{2} \right ),$ but have had no meaningful results.
|
Problem. Half-solve
$$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$
Substitute $y^{3}= 8x- 3$ in the original equation_
$$\left ( 4y- 8x^{3}- 3 \right )+ \left ( y^{3}- 8x+ 3 \right )= \left ( y- 2x \right )\left ( 4x^{2}+ 2xy+ y^{2}+ 4 \right )= 0$$
With $y= 2x,$ we have the equivalent one
$$8x^{3}- x+ 3=\!\left ( 2x- 1 \right )\!\left ( 4x^{2}+ 2x- 3 \right )=\!0\!\Rightarrow\exists_{x_{1}, x_{2}\in\left ( {\mathbb{Q}}' \right )^{2}}x_{1}+ x_{2}=\!-\frac{1}{2}$$
With $4x^{2}+ 2xy+ y^{2}+ 4= 0,$ there are no irrational roots, because
$$4x^{2}+ 2xy+ y^{2}+ 4= y^{6}+ 4y^{4}+ 6y^{3}+ 16y^{2}+ 12y+ 73= 0\;{\rm illegal}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the system of equations: $32y+32x^3=6x+17$, $16z+32y^3=6y+9$, $8x+32z^3=6z+5$ where $x,y,z\in \mathbb{R}$ Solve the system of equations: $$\begin{cases} 32y+32x^3=6x+17 \\16z+32y^3=6y+9 \\8x+32z^3=6z+5 \end{cases}$$ where $x,y,z\in \mathbb{R}$ (Bulgaria 1960)
I attempted to solve this question as follows:
$32y+32x^3=6x+17$
$y=-x^3+\frac{6x}{32}+\frac{17}{32}$
$y-\frac{1}{2}=-x^3+\frac{3x}{16}+\frac{17}{32}-\frac{1}{2}$
Here it started getting very complex, and hence I don't think it can be solved this way. After this I tried doing something similar with the other two equations, but once again it was ending up way too complex. It is obvious that the solution is $x=y=z=\frac{1}{2}$, but I can't manage to prove it. Could you please explain to me how to solve this question?
| Proof by Contradiction:
Let, $32x^3-6x>1$ then:
$$\begin{align} 32x^3-6x>1 &\Longrightarrow x>\frac 12\\ 32x^3-6x=17-32y>1 &\Longrightarrow y<\frac 12 \\32y^3-6y=9-16z <1 &\Longrightarrow z>\frac 13\\ 2z^3-6z=5-8x >1 &\Longrightarrow x<\frac 12\end{align}$$ which gives a contradiction.
The same method also works for $32x^3-6x<1$. We obtain again a contradiction.
So, we deduce that $$\begin{align} 17-32y=32x^3-6x=1
&\Longrightarrow y=\frac 12 \\ 9-16z=32y^3-6y
&\Longrightarrow z=\frac 12\end{align}$$
$$\begin{align} x=\frac {6z+5-32z^3}{8}=\frac{4}{8}=\frac 12\end{align}$$
Finally, our real solutions are only $$x=y=z=\dfrac 12.$$
I used :
*
*If $32x^3-6x>1$, then $x>\frac 12$. Because,
$$\begin{align}32x^3-6x-1&=32\left(x+\frac 14\right)^2\left(x-\frac 12\right)>0.&\end{align}$$
*
*If $y<\frac 12$, then $32y^3-6y<1$. Because,
$$\begin{align}32y^3-6y-1&=32\left(y+\frac 14\right)^2\left(y-\frac 12\right)<0.&\end{align}$$
*
*If $z>\frac 12$, then $32z^3-6z>1$. Because,
$$\begin{align}32z^3-6z-1&=32\left(z+\frac 14\right)^2\left(z-\frac 12\right)>0.&\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the slope of the tangent line to the graph of the given function at the given value of x. Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$
I understand that the slope of the line follows the equation of
$$\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$
But I don't know how to go about the rest
| If you use the h-method you start with
$$\lim_{h \rightarrow 0} \frac{y(a+h)-y(a)}{h}=\lim_{h \rightarrow 0} \frac{(a+h)^4-5\cdot (a+h)^3+2-(a^4-5\cdot a^3+2)}{h}$$
To resolve $(a+h)^4$ you can use the binomial theorem: $(x+y)^n=\sum\limits_{k=0}^n\binom{n}x\cdot x^k\cdot y^{n-k}$
$$=\lim_{h \rightarrow 0}\frac{a^4+4a^3h+6a^2h^2+4ah^3+h^4-5a^3-15a^2h-15ah^2-5h^3+2-a^4+5\cdot a^3-2}{h}$$
Next we eliminate the terms with opposite signs, e.g. $a^4$ and $-a^4$
$$=\lim_{h \rightarrow 0}\frac{4a^3h+6a^2h^2+4ah^3+h^4-15a^2h-15ah^2-5h^3}{h}$$
Factoring out $h$ at the numerator.
$$=\lim_{h \rightarrow 0}\frac{h\cdot \left(4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2\right)}{h}$$
Cancelling $h$
$$=\lim_{h \rightarrow 0}4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2$$
$$=4a^3-15a^2=y'(a)$$
This is the derivative of $y(x)$ at value $x=a$ and therefore the slope. Finally you can insert $a=2$ to obtain $y'(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$ A triangle $ABC$ is given with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ.$ Point $M$ lies inside the triangle and $\measuredangle MAB=\measuredangle MBA=15^\circ.$ Find $\measuredangle BMC$.
I am not sure how to approach the problem. We can find some angles:$$\measuredangle ACB=180^\circ-45^\circ-30^\circ=105^\circ\\ \measuredangle CAM=45^\circ-15^\circ=30^\circ\\ \measuredangle MBC=30^\circ-15^\circ=15^\circ\\ \measuredangle AMB=180^\circ-2\cdot15^\circ=150^\circ.$$
I will be very grateful if we can see a solution without using trigonometry.
| Please apply Trigonometric form of Ceva's theorem.
If $\angle BCM = x$,
$\sin \angle BAM \ \sin \angle ACM \sin CBM = \sin \angle MAC \ \sin \angle MCB \ \sin \angle MBA$
$\sin 15^0 \sin (105^0-x) \sin 15^0 = \sin 30^0 \sin x \sin 15^0$
$\cos (15^0-x) \sin 15^0 = 2 \sin 15^0 \cos 15^0 \sin x$
$\sin 30^0 \cos (15^0-x) = \cos 15^0 \sin x$
$\sin (45^0-x) + \sin (15^0+x) = \sin (15^0+x) + \sin(x-15^0)$
$45^0-x = x-15^0 \implies x = 30^0$
So $\angle BMC = 180^0 - 30^0 - 15^0 = 135^0$.
EDIT: For geometric solution, see the below construction that you can use. $AN$ is angle bisector of $\angle CAM$. You can prove $\angle ANM = \angle ANC = 60^0$ and $CN = NM$. So, $\angle MCB = 30^0$. $\therefore \angle BMC = 135^0$.
| {
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Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| You were going in the correct direction, as you said:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
If we apply the same here:
$^3+12^2+48+64 = ^3+ 4^3 + 12^2+48 = (n+4)(n^2 - 4n + 16) + 12n(n+4)$
If we take (n+4) common, we get:
$(n+4)(n^2 - 4n + 16 + 12n) = (n+4)(n^2 + 8n + 16) = (n+4)(n+4)^2 = (n+4)^3$Cheers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
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Locus of a point of intersecting lines The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$
My attempt
Tangent to an ellipse at $P(\theta)$
$$bx\cos\theta +by\sin\theta=ab\tag{1}$$
slope $m_1$ of a normal line to $(1)$
$$m_1=\frac{a\sin\theta}{b\cos\theta}$$
Equation of the normal passing through origin
\begin{align*}
y=m_1x
&=\frac{a\sin\theta}{b\cos\theta}x\tag{2}
\end{align*}
Equation of a line through $P$ and $S$
$$\begin{align*}
y=m_2(x-ae)&=\frac{b\sin\theta}{\left(a\cos\theta-ae\right)}(x-ae)
\tag{3}
\end{align*}$$
Point $Q(h,k)$ satisfies $(2)$ and $(3)$.
How to eliminate $\sin\theta$ and $\cos\theta$ to obtain the required locus?
| $$x(a\sin t)-y(b\cos t)=0\ \ \ \ (1)\implies\tan t=\dfrac{by}{ax}\ \ \ \ (3)$$
$$x(b\sin t)+ya(e-\cos t)-abe\sin t=0\ \ \ \ (2)$$
$$\dfrac x{ab^2e\sin t\cos t}=\dfrac y{a^2be\sin^2t}=\dfrac1{a^2e\sin t(1-e\cos t)}$$
$$\implies\dfrac{1-e\cos t}{\cos t}=\dfrac{b^2}{ax}\implies\sec t=\dfrac{b^2}{ax}+e\ \ \ \ (4)$$
Use $\sec^2t-\tan^2t=1$ to eliminate $t$
$$\left(\dfrac{b^2}{ax}+e\right)^2-\left(\dfrac{by}{ax}\right)^2=1$$
$$\iff(ax)^2+(by)^2=(b^2+aex)^2$$
$$\iff b^4+2ab^2ex=b^2y^2+a^2x^2(1-e^2)$$
Use $b^2=a^2(1-e^2)$
$$b^2(b^2+2aex)=b^2(y^2+x^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4061022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing the squares. Can you give me a hint?
| Compare $S = (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
with $T = (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$
Adding: $S + T = 2(a^3 + 3ab^2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
The coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$ To find the coefficient of $x^2$ in the expansion of $\left(x^3+2x^2+x+4\right)^{15}$.
I saw a solution to this problem in which the above expression was differentiated 2 times and they put x=0. I really can't understand the significance behind these steps.
| We know that, whatever the expansion of $\left(x^3 + 2x^2 + x +4 \right)^{15}$ is, it will look something like
$$
\left(x^3 + 2x^2 + x +4 \right)^{15} = c_0 + c_1 x + c_2 x^2 +c_3x^3+ ... + c_{45} x^{45}
$$
where $c_i$'s are some constants possibly equal to $0$. Now, in the above equation what you want to know is the value of $c_2$. So now what happens if we differentiate the above equation twice? We get
\begin{align*}
\frac{d^2}{dx^2}\left(x^3 + 2x^2 + x +4 \right)^{15} & = \frac{d^2}{dx^2}\left(c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... + c_{45} x^{45}\right)\\
& = \frac{d}{dx}\left(c_1 + 2c_2 x + 3c_3 x^2 + ... + 45c_{45} x^{44}\right)\\
& = 2c_2 + (3\cdot 2c_3) x + ... + (45 \cdot 44c_{45}) x^{43}
\end{align*}
Plugging in $x=0$ in the R.H.S. of the above makes all the terms after $2c_2$ go to $0$, so we get
$$
\frac{d^2}{dx^2}\left(x^3 + 2x^2 + x +4 \right)^{15} \Bigg\vert_{x=0} = 2 c_2
$$
from where, by using the chain rule on the L.H.S., it's easy to solve for $c_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expected Value of Binomial Distribution What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads when a coin is tossed 5 times?
Here is what I think : Let $X$ is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$
$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$
But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.
| Your random variable $Z=|X-Y|$ can take only the values
$$\{1,3,5\}$$
with probabilities
$$\left\{\frac{20}{32};\frac{10}{32};\frac{2}{32}\right\}$$
This because
$Z=5$ when you get 5 consecutive tails (or heads)
$Z=3$ when you get 4 tails (or heads)
$Z=1$ when you get 2 tails (or heads)
No other situations are possible.
Thus its expectation is
$$\mathbb{E}[Z]=1\times \frac{20}{32}+3\times\frac{10}{32}+5\times\frac{2}{32}=\frac{15}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.
$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :
For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.
For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $ and total number of solutions possible in this case is $4$.
For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.
So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.
$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.
The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?
Thanks in advance !
| The possible values of the product are $2,3,4,5$.
Hence the possible factorizations,
$$1\cdot2,2\cdot1,1\cdot3,3\cdot1,1\cdot4,2\cdot2,4\cdot1,1\cdot5,5\cdot1.$$
But $|x-1|\ge2$, which leaves
$$2\cdot1,3\cdot1,2\cdot2,4\cdot1,5\cdot1,$$
while $|y+3|$ can achieve $1$ and $2$ in two ways.
$$10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$ Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\ f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm} and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$
Can somebody please guide?
| For the minimum, if $x\ge 1$ or $x\le 0$ we have $x^4-x^3+y^4=x^3(x-1)+y^4\ge 0$. If $0<x<1$ we have by AM-GM
$$x^3(1-x)=27(x/3)^3(1-x)\le 27\left(\frac{3(x/3)+(1-x)}{4}\right)^4=27/256$$
so $x^4-x^3\ge -27/256$ and so $x^4-x^3+y^4\ge x^4-x^3\ge -27/256$. The minimum value is reached at $x=3/4$, $y=0$.
For the maximum, clearly it doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Domain of $\frac{x^2-3x-10}{x+2}\div\frac{x^2-25}{x}$ I saw a question asked to simplify
$\dfrac{x^2-3x-10}{x+2}\div\dfrac{x^2-25}{x}$
It is not so hard to do : $$\dfrac{(x-5)(x+2)}{x+2}\times\dfrac{x}{(x-5)(x+5)}=\dfrac x{x+5}$$
But my question is: should we have necessary all the conditions $x\ne-2$ , $x\ne5$ , $x\ne-5$ and also $x\ne0$ (because in former fraction we had $x$ in denominator) ? or we should have some of them?
| Yes, you have to state all of the conditions $x\neq2$, $x\neq5$, $x\neq-5$ and $x\neq0$. This is because when we write
$$
\frac{x^2-3x-10}{x+2} \div \frac{x^2-25}{x}=\frac{x}{x+5} \, ,
$$
we are asserting that for every value of $x$ for which the LHS and RHS are defined, the LHS represents the same number as the RHS. When $x=5$, for instance, the LHS does not make sense, even though the RHS does. Hence, equality does not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $\sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}$ Suppose $x,y,z$ are positive real numbers that satisfy
\begin{align*}
\frac{x}{y} + \frac{y}{x} + \frac{x}{z} + \frac{z}{x} + \frac{y}{z} + \frac{z}{y} &= 2018 \\
\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{z}} + \sqrt{\frac{z}{x}} &= 17.
\end{align*}
Find the value of $\sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}.$
I wasn't sure how to start on this problem as there weren't any good factorizations that I could use.
| Squaring the second equation we get
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 289.$
Subtracting from the first equation
$\frac{x}{z} + \frac{z}{y} + \frac{y}{x} - 2(\sqrt{\frac{x}{z}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{y}{x}}) = 2018-289 = 1729.$
Put $a = \sqrt{\frac{y}{x}} + \sqrt{\frac{z}{y}} + \sqrt{\frac{x}{z}}.$
$$a^2 = \frac{x}{z} + \frac{z}{y} + \frac{y}{x} + 2(\sqrt{\frac{x}{y}}
+ \sqrt{\frac{y}{z}} + \sqrt{\frac{z}{x}}) = 1729+2a+34 = 1763+2a$$
This is a quadratic equation in a and has the positive solution a=43.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots + 2^{\left\lceil\frac12 a_n \right\rceil}}$$
represents all rationals between $\dfrac{1}{2}$ and $1$, where $a_1, a_2, \ldots, a_n \in \mathbb{N}$?
I have tried representing few rationals such as $\dfrac{7}{8}$ and $\dfrac{5}{8}$ and it seems like we can represent all of them. However I'm not completely sure why and don't have a rigorous proof.
For example
$$\dfrac{5}{8} = \dfrac{2^{\left\lfloor\frac12 3\right\rfloor} + 2^{\left\lfloor\frac12 4\right\rfloor} + 2^{\left\lfloor\frac12 5 \right\rfloor}}{2^{\left\lceil\frac12 3\right\rceil} + 2^{\left\lceil\frac12 4\right\rceil} + 2^{\left\lceil\frac12 5 \right\rceil}}$$
$$\dfrac{7}{8} = \dfrac{2^{\left\lfloor\frac12 1\right\rfloor} + 2^{\left\lfloor\frac12 2\right\rfloor} + 2^{\left\lfloor\frac12 4 \right\rfloor}}{2^{\left\lceil\frac12 1\right\rceil} + 2^{\left\lceil\frac12 2\right\rceil} + 2^{\left\lceil\frac12 4 \right\rceil}}$$
Any help would be highly appreciated. Thanks
| Let's group the parameters $a_i$ into two groups, the even and odd numbers (lets call them $e_i$ and $o_i$ resp).
Notice that $2^{\lceil e_i/2 \rceil} = 2^{\lfloor e_i/2 \rfloor} $ while $2^{\lceil o_i/2 \rceil} = 2 \times 2^{\lfloor o_i/2 \rfloor} $
Now, let $\frac12 <Z<1$ with $Z=\frac{X}{Y}$, where $ X, Y \in \mathbb{N}$
Let $D=Y-X$ and $C=2X -Y$, with $C, D>0$. Also, $X=D+C$ and $Y=C+2D$.
Then $$Z=\frac{C+D}{C+2D}$$
Further, any natural number can be expressed as sum of powers of two: $C = 2^{c_1} + 2^{c_2} + \cdots 2^{c_j}$, where $(c_1, \cdots c_j)$ correspond to the positions of the ones in the binary digits of its binary representation. Analogously, $D = 2^{d_1} + 2^{d_2} + \cdots 2^{d_k}$
Now, let $e_i = 2 c_i$ and $o_i = 2 d_i +1$, and we are done.
An example follows: Let $Z = \frac{11}{17}$, then
$$C= 2\times 11 -17=5 = 2^0 +2^2 \implies c_i=(0 , 2) \implies e_i=(0 , 4)$$
$$D=17-11=6 = 2^1 + 2^2\implies d_i=(1 , 2) \implies o_i=(3 , 5) $$
and $a_i = (0,4,3,5)$. Indeed:
$$
\dfrac{
2^{\lfloor 0/2\rfloor} +
2^{\lfloor 4/2\rfloor} +
2^{\lfloor 3/2\rfloor} +
2^{\lfloor 5/2\rfloor}
}
{
2^{\lceil 0/2\rceil} +
2^{\lceil 4/2\rceil} +
2^{\lceil 3/2\rceil} +
2^{\lceil 5/2\rceil}
}
=\dfrac{1+4+2+4}{1+4+4+8}=\dfrac{5+6}{5+2\times 6}=\frac{11}{17}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Let $G$ be a group such that for all $x,y\in G,(xy)^3=x^3y^3$ and $(xy)^8=x^8y^8$. Prove that $G$ is Abelian or not? I tried to prove it like $(xy)^3=xyxyxy=x^3y^3 $ which implies
$$ x^{-1}/x^3y^3=xyxyxy$$
$$x^2y^3=yxyxy/y^{-1}$$
$$ x^2y^2=yxyx $$
and I get $x^2y^2=(yx)^2$.
But I dont know how to prove that $(xy)^3=x^3y^3$
| I prove commutativity. (The comments
reveal that a more general result was proved in 2017.) The proof is for cancellative semigroups.
$$\ (xy)^3=x^3y^3\tag{A}$$
$$\ (xy)^8=x^8y^8\tag{B}$$
Using cancellation on (A), $(yx)^2=x^2y^2 \tag 1$
Therefore, $x^2y^3x=(yx)^3=y^3x^3$ so that $x^2y^3=y^3x^2\tag 2$
Using cancellation on (B), $(yx)^7=x^7y^7 \tag 3$
Therefore, $x^7y^8x=(yx)^8=y^8x^8$ so that $x^7y^8=y^8x^7\tag 4$
Apply (2) to (4) to obtain $xy^8x^6=y^8x^7$ and cancel to get
$xy^8=y^8x \tag 5$
From $y^6x^2y^2=x^2y^8=y^8x^2,$ we obtain
$x^2y^2=y^2x^2 \tag 6$
Apply (6) to (2) to obtain the powerful identity
$x^2y=yx^2 \tag 7$
Commutativity now follows by cancellation on
$$x^3y^3=(xy)^3=x(xy)^2y=x^2yxy^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
| Alternative approach:
If $a,b > 0$ and $a^2 = b^2$, then $a$ must $= b$.
Let $a = \frac{1}{2} \left(\sqrt{6} + \sqrt{14}\right),
~b = \sqrt{5 + \sqrt{21}} \implies 0 < a,b.$
Then $a^2 = \frac{1}{4} \left(6 + 14 + 2\sqrt{84}\right)
= 5 + \sqrt{21} = b^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$? I dont know if this is correct...
$(\frac{21}{-17}) = (\frac{3}{-17}) * (\frac{7}{-17})$
for $(\frac{3}{-17}) = (-1) (\frac{-17}{3})(\text{Quadratic reciprocity})= (\frac{17}{3}) = (\frac{2}{3}) = -1$
for $(\frac{7}{-17}) = (-1) (\frac{-17}{7})(\text{Quadratic reciprocity})= (\frac{17}{7}) = (\frac{3}{7}) = (-1)(\frac{7}{3})(\text{Quadratic reciprocity}) = (\frac{2}{3}) = -1$
insert, we get:
$(-1) * (-1) =1$ and we have integer solutions?
| $x=2, y=-1$ is an integer solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Differentiate $\arcsin(\frac{1}{\sqrt{1 + x^2}})$ This is from Calculus by Michael Spivak, 3rd Edition, Chapter 15, problem 1 (iv):
Differentiate $f(x) = \arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right).$
Here's my attempt:
\begin{align}f'(x) &= \left[\arcsin\left(\frac{1}{\sqrt{1 + x^2}}\right)\right]' \\
&= \arcsin'\left(\frac{1}{\sqrt{1 + x^2}}\right)\cdot \left(\frac{1}{\sqrt{1 + x^2}}\right)' \\
&= \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{1 + x^2}}\right)^2}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{1 - \frac{1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{1}{\sqrt{\frac{1 + x^2 - 1}{1+x^2}}} \cdot \frac{-x}{(1+x^2)^{3/2}} \\
&= \frac{-x}{\frac{\sqrt{x^2}}{\sqrt{1+x^2}} \cdot \sqrt{1 + x^2}(1+x^2)} \\
&=\frac{-x}{|x|} \cdot \frac{1}{1+x^2}.\end{align}
Therefore
$$f'(x) =
\begin{cases}
\dfrac{-1}{1+x^2}, & \text{if $x > 0$} \\[1ex]
\dfrac{1}{1+x^2}, & \text{if $x < 0$}
\end{cases}
$$
In the Answer Book the published solution is $\frac{-1}{1+x^2}$.
It seems Spivak is discounting the possibility that $x < 0$. Why might this be? Is there a reason $x$ cannot be negative here? Did I make a mistake? Perhaps with this part of simplifying the denominator?
$$\sqrt{\frac{x^2}{1+x^2}} \cdot \sqrt{1 + x^2} = \sqrt {x^2} = |x|$$
I suspect there's something simple I'm not seeing. I'm just now encountering these inverse trigonometric functions, so my understanding isn't very solid yet.
Edit: This error could have become a misleading, possibly demoralizing waste of time, but thanks your insightful answers it's instead proven to be more instructive than the first 3 (omitted, correct) parts of the problem.
| Observe that
$$f(\sinh(x))=\arcsin(\frac{1}{\sqrt{1+\sinh^2(x)}})=\arcsin(\frac{1}{\cosh(x)})$$
The differentiation gives
$$f'(\sinh(x))\cosh(x)=$$
$$\frac{1}{\sqrt{1-\frac{1}{\cosh^2(x)}}}\frac{-\sinh(x)}{\cosh^2(x)}=$$
$$\frac{-1}{\cosh(x)}\frac{\sinh(x)}{|\sinh(x)|}$$
$$\implies f'(\sinh(x))=\frac{\pm 1}{\cosh^2(x)}=\frac{-sign(x)}{1+\sinh^2(x)}$$
So,
$$f'(X)=\frac{-sign(X)}{1+X^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $bx(x-b) + ax(a-x) + ab(b-a)$ to $-(a-b)(a-x)(b-x)$ I was doing some matrices problems when I got to this factoring
$$bx(x-b) + ax(a-x) + ab(b-a)$$
I found this was the answer, but it was
$$-(a-b)(a-x)(b-x)$$
Doing some researches, I found this remarkable product
$$(a–b)(a–c)(b–c) = ab(a–c) + bc(b–c) + ca(c–a)$$
but with no explanation at all of how to do this simplification.
| Since the target expression has factor $(a-b)$ showing up at the original expression as $(b-a)$, we have a clue how to get this started $\to$ \begin{align}bx(x-b) + ax(a-x) + ab\color{green}{(b - a)}\end{align}
\begin{align}=bx^2-b^2x + a^2x-ax^2 + ab\color{green}{(b - a)} \end{align}
\begin{align}=x^2\color{green}{(b - a)}-\color{green}{(b - a)}(b+a)x+ ab\color{green}{(b - a)}\end{align}
\begin{align}=\color{green}{(b - a)}(x^2-(b+a)x+ ab)\end{align}
\begin{align}=\color{green}{(b - a)}(x-a)(x-b) = -(a-b)(a-x)(b-x)\end{align}
| {
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"question_score": "2",
"answer_count": 2,
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} |
Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
how can I deduce
$$
\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1}?
$$
I have tried clearing the denominators of the first equation, but that just results in a large mess. I don't know how to continue from there.
| The keys is Vieta's formulas.
Let $u$ be a root of $x^3 = 1 $. Then $u+a$ is a root of $(x-a)^3 = 1$ and $\frac{1}{u+a}$ is a root of $(\frac{1}{x}-a)^3=1 $. Multiplying by $x^3$ and expanding the coefficients we have
$$(1+a^3) x^3 -3a^2x^2 + 3ax - 1 = 0$$
Using Vieta's formulas we get that $\frac{3a^2}{1+a^3}$ is equal to
$$ \frac{3a^2}{1+a^3} = \sum_{u^3 =1} \frac{1}{(u+a)} \tag{1} $$
We can apply this to our equation. We know that
$$\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega}$$ and by conjugating: $$\frac{1}{a + \overline{\omega}} + \frac{1}{b +\overline{\omega}} + \frac{1}{c +\overline{\omega}} + \frac{1}{d + \overline{\omega}} = \frac{2}{\overline{\omega}}$$
Now $\omega$ and $\overline{\omega}$ are two of the three roots of unity (solutions to $x^3 = 1$), the other being $1$. From this we conclude:
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3} - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3} + \frac{3d^2}{1+d^3}) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to integrate definite integral of form $\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$ During WKB approximation maths I typically end up deducing integrals that need to be calculated of the form:
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx$$
where a is a constant.
I know the following standard integral
$$\int_0^{1} \sqrt{1-\sqrt{y}} \space dy = \frac{8}{15}$$
My question is how do I Use the standard integral above to deduce that
$$\int_0^{a^2} \sqrt{a-\sqrt{x}} \space dx = \frac{8 a^{\frac{5}{2}}}{15}$$
| Let $x=a^2y$. Then we get that:
\begin{equation}
\begin{split}
\int_0^{a^2}\sqrt{a-\sqrt{x}}dx & =\int_0^{1}a^2\sqrt{a-\sqrt{a^2y}}dy\\
&=\int_0^{1}a^2\sqrt{a-a\sqrt{y}}dy\\
&=\int_0^1a^2\sqrt{a(1-\sqrt{y})}dy\\
&=\int_0^1a^2a^{\frac{1}{2}}\sqrt{1-\sqrt{y}}dy\\
&=a^{\frac{5}{2}}\int_0^1\sqrt{1-\sqrt{y}}dy\\
&=\frac{8a^{\frac{5}{2}}}{15}
\end{split}
\end{equation}
| {
"language": "en",
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} |
Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$ Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$
My attempt: For all $\epsilon > 0$ there is a $\delta > 0$ such that $\left | \frac{xy(y^2-x^2)}{x^2+y^2} \right| < \epsilon \rightarrow \sqrt{x^2+y^2} < \delta$.
Note important facts: $ (1) \ \sqrt{x^2} < \sqrt{x^2 + y^2}$ and $ (2) \ x^2 < x^2 + y^2$
$$\left | \frac{xy(y^2-x^2)}{x^2+y^2}\right | = \left | \frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} \right| < \left| \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2-x^2)}{x^2+y^2}\right | = |y^2-x^2| $$
From $(2)$ we can show that :
$$|y^2-x^2| < |x^2+y^2-x^2| = |y^2| < |x^2 + y^2| < \delta^2 $$
Hence let $\epsilon = \delta^2$. Does this relationship work?
| You should try with polar coordinates, it gets way easier. Impose:
$$\begin{cases} x=\rho\cos\theta\\y=\rho\sin\theta
\end{cases}
$$
and $(x,y)\longrightarrow(0,0)$ will turn into $\rho\longrightarrow0$. As the numerator goes as $\rho^{4}$ and the denominator goes as $\rho^{2}$ your limit is proven. If $\rho$ simplifies from the function then you can't say anything about the limit.
| {
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} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is divisible by $A,B,C,D,E$.
I tried $$\frac{11}{6}=1+\frac{1}{2}+\frac{1}{3}$$ But fourth fraction is not possible.
Any hint?
| The only choice of $a_1,a_2$ is $a_1=1,a_2=2$ with the reasoning given by Hagen von Eitzen.
So we have $$\frac{1}{a_3}+\frac{1}{a_4}=\frac{11}{6}-\frac{3}{2}=\frac{1}{3}$$
$\implies$
$$(3-a_3)(3-a_4)=9$$
The choice of $a_3=3$ is now ruled out.
So $a_4>a_3>3$ and hence number $9$ can be factored out only in one way viz $-9 \times -1$ which gives $a_3=4,a_4=12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find the matricial representation Let $M_{2}\mathbb{R}$ be the matrix space of $2 \times 2$ with real entries , and $P_{2}[x]$ the space of polinomies with real coeficientes with grade at most 2. Considerate the linear transformation $T: M_{2}(\mathbb{R}) \rightarrow P_{2}[x]$ defined like
$$T \begin{pmatrix} a & b \\ c & d \end{pmatrix}= (a+b)+2dx+bx^{2}$$
Calculate the matricial representation of $T$ respect the canonic base of $M_{2}(\mathbb{R})$ and $P_{2}[x]$ respectively:
$$B_{1}= \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \} $$
$$B_{2}= \{1,x,x^{2} \} $$
I got $T(e_{1})= 1$, $T(e_{2})=1+x^{2}$, $T(e_{3})=0$, $T(e_{4})=2x$, so $$T= \begin{pmatrix} 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 2\\
0 & 1 &0 & 0
\end{pmatrix}$$
Am I right?
| I think it is right:
$$\begin{pmatrix} 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 2\\
0 & 1 &0 & 0
\end{pmatrix}\begin{pmatrix} a \\ b \\ c \\ d \\
\end{pmatrix}\,=\,\begin{pmatrix} a+b \\ 2d \\ b \\
\end{pmatrix}$$
| {
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Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations, $ \bullet $ $ a+b+c=63$ and $ \bullet $ $ab+bc+ac=2021$? Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations,
$ \bullet $ $ a+b+c=63$
$ \bullet $ $ab+bc+ac=2021$ ?
I try to solve this problem but going to the result that this problem has no solutions at all ... My attempt about the solution is that which is best of other ones..
$ \forall a,b,c \in \mathbb{R} $, we have
$(a+b+c)^2 = a^2 + b^2 + c^2 +2ab +2bc + 2ca $
from here we get,
$a^2 + b^2 + c^2= (a + b + c)^2 — 2(ab + bc + ca)$
$a^2 + b^2 + c^2= 63^2 — 2(2021)$
$= 3969 — 4042$
$= -73$
which is impossible since $a^2 + b^2 + c^2 \geq 0$.
Thus there are no such $ a, b, c \in \mathbb{R} $ that satisfy the given equation
Is There any set of Real numbers satisfy the conditions?
| By Vieta's formula we have
$$f(x) = x^3 - 63x^2 + 2021x -abc = 0$$
This function has only one real root since $f'(x) > 0$ for all $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
What does $\sum_{r=1}^{n} r^{k}$ equal in general I know that $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$, $\sum_{r=1}^{n} r^{2} = \frac{n(n+1)(2n+1)}{6}$, $ \sum_{r=1}^{n} r^{3} = \frac{n^2(n+1)^2}{4}$
but what does $\sum_{r=1}^{n} r^{k}$ equal in general ?
Can we express it in some simple predetermined form
| We may also express this in terms of Stirling numbers of the second
kind and falling factorials. We start with
$$\sum_{r=0}^n r^k
= k! [z^k] \sum_{r=0}^n \exp(rz)
= k! [z^k] \frac{\exp((n+1)z)-1}{\exp(z)-1}
\\ = k! [z^k] \frac{1}{\exp(z)-1}
\sum_{q=1}^{n+1} {n+1\choose q} (\exp(z)-1)^q
\\ = k! [z^k]
\sum_{q=1}^{n+1} {n+1\choose q} (\exp(z)-1)^{q-1}
\\ = k! [z^k]
\sum_{q=1}^{n+1} (n+1)^{\underline{q}}
\frac{1}{q} \frac{(\exp(z)-1)^{q-1}}{(q-1)!}
\\ = \sum_{q=1}^{n+1} (n+1)^{\underline{q}}
\frac{1}{q} {k\brace q-1}.$$
Note that we may set the upper limit of the sum to $k+1.$ If $n+1\gt
k+1$ we may lower to $k+1$ because the removed terms from the range
$k+2\le q\le n+1$ produce zero by the Stirling number. If $k+1\gt n+1$
we may raise to $k+1$ because the extra terms from the range $n+2\le q\le
k+1$ produce zero through the falling factorial.
We get
$$\sum_{q=2}^{k+1} (n+1)^{\underline{q}}
\frac{1}{q} {k\brace q-1}
= (n+1) \sum_{q=2}^{k+1} n^{\underline{q-1}}
\frac{1}{q} {k\brace q-1}$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
\sum_{r=0}^n r^k =
(n+1)
\sum_{q=1}^{k} n^{\underline{q}}
\frac{1}{q+1} {k\brace q}.}$$
In this way we get e.g. with $k=4$
$$(n+1) \times
\left[
n^{\underline{1}} \frac{1}{2} {4\brace 1} +
n^{\underline{2}} \frac{1}{3} {4\brace 2} +
n^{\underline{3}} \frac{1}{4} {4\brace 3} +
n^{\underline{4}} \frac{1}{5} {4\brace 4}
\right]$$
The Stirling numbers may be evaluated by inspection:
$$(n+1) \times
\left[
n^{\underline{1}} \frac{1}{2}
\times 1 +
n^{\underline{2}} \frac{1}{3}
\times \left( \frac{1}{2} {4\choose 2} + {4\choose 1} \right) +
n^{\underline{3}} \frac{1}{4}
\times {4\choose 2} +
n^{\underline{4}} \frac{1}{5}
\times 1
\right]$$
We find
$$\sum_{r=0}^n r^4 =
(n+1) \times
\left[
\frac{1}{2} n^{\underline{1}} +
\frac{7}{3} n^{\underline{2}} +
\frac{3}{2} n^{\underline{3}} +
\frac{1}{5} n^{\underline{4}}
\right].$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which is equal to
$a^2b^2y^2 + 1/aby$
Using Vieta's formula I get:
$aby = -d/a$
$aby = 1/3$
Subbing in original:
$(1/9 + 1)/1/3$
which is...
10/27
The answer is supposedly 2/3, want to know what I did wrong.
| Hint:
$$(b+\frac{1}{y})(y+\frac{1}{a})(a+\frac{1}{b})=aby + \frac{1}{aby} + (a + b + y) + (\frac{1}{a} + \frac{1}{b} + \frac{1}{y})$$
(Here is the WolframAlpha verification for this computation.)
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex roots of quartic equation during solution of $z^5=1$ While finding the 5th roots of unity $z^5=1$, I arrived at the following
$$(z-1)(z^4+z^3+z^2+z+1)=0$$
Now, I am well aware that I can arrive at the roots by using the fact that each root is separated by a value of $\alpha=2k\pi$
$$z^5=1 +0i \implies arg(z^5)=0$$
$$z^5=\cos(0 +2k\pi)+i\sin(0+2k\pi)$$
$$z=\cos(\frac{2k\pi}{5})+i\sin(\frac{2k\pi}{5})$$
Then, by letting $k$ take the values $0, ±1,±2$, I arrive at the roots
$$z=1,e^{i \frac{2\pi}{5}},e^{-i \frac{2\pi}{5}}, e^{i \frac{4\pi}{5}}, e^{-i\frac{4\pi}{5}}$$
However, for the sake of knowing if it is possible, is there a purely "algebraic" way of solving this quartic polynomial?
(Clearification: By "algebraic" I mean the classic ways you would deal with a regular polynomial such as the rational root theorem, factorization, or Newton-raphson etc)
| Factorize the equation as follows
\begin{align}
z^5-1= &(z-1)(z^4+z^3+z^2+z+1)\\
=&(z-1)z^2(z^2+z+1+\frac1z+\frac1{z^2})\\
=&(z-1)z^2\left( (z+\frac1z)^2+(z+\frac1z)-1\right)\\
=&(z-1)z^2\left( z+\frac1z -\frac{1+\sqrt5}2\right)\left(z+\frac1z -\frac{1-\sqrt5}2\right)\\
=&(z-1)\left( z^2-\frac{1+\sqrt5}2z+1\right)\left(z^2 -\frac{1-\sqrt5}2 z+1\right)\\
\end{align}
which yields the solutions
$$z=1,\>\frac14\left(1+\sqrt5\pm i \sqrt{10-2\sqrt5}\right),\>
\frac14\left(1-\sqrt5\pm i \sqrt{10+2\sqrt5}\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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What is $\sum_{n=1}^n(-1)^m\frac{m^{n-m+1}}{(n-m)!}$? So consider the sum
$$S=\sum_{m=1}^n(-1)^m\frac{m^{n-m-1}}{(n-m)!},$$
where $n$ is some fixed, positive integer. For specific values of $n$, this gives $-1$, $-\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $-\frac{3}{40}$, $-\frac{1}{120}$, $\frac{31}{1008}$, $-\frac{29}{5040}$, $-\frac{7}{640}$, $\frac{2087}{362880}$,... for $n=1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$ ... Is there a way to evaluate this sum for general $n$?
| The generating function for
$$S_n = \sum_{m=1}^n (-1)^m \frac{m^{n-m-1}}{(n-m)!}$$
may be proved as follows:
$$\sum_{m=0}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
= \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{m} [z^{m-1}]
\exp((n-m)z)
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{n-m} [z^m]
\exp((n-m)z)
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{m} \frac{1}{m} [z^{n-m}]
\exp(mz)
\\ = \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} [z^{n-m}]
\exp(mz)
\\ = [z^n] \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} z^m
\exp(mz).$$
Now here the coefficient extractor enforces the range and we get
$$[z^n] \sum_{m\ge 1} (-1)^{m} \frac{1}{m} z^m \exp(mz)
= [z^n] \log\frac{1}{1+z\exp(z)}.$$
| {
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Prove that $H_n - H_m > \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induction on $n$ and base case $n = m$ Problem:
Let $H_n = \sum_{i=1}^{n} \frac{1}{i} $ for all natural numbers $n \geq 1$.
Prove that $H_n - H_m \geq \frac{n-m}{n}$ for $n, m \in \mathbb{N}$ and $n > m \geq 1$, by doing induction on $n$, letting $m$ be any natural number, and starting with the base case $n = m$.
$--------------------------------------$
I have done the base case, but I am confused on how to start with the inductive step.
We want to show that
$H_{n+1} - H_m \geq \frac{n+1-m}{n+1}$
Proof:
Base Case:
Let $n = m$, then $H_m - H_m \geq \frac{m-m}{m} = 0 \geq 0$, this is true.
Inductive Step:
Let P(n) denote the statement $H_n - H_m \geq \frac{n-m}{n}$.
We need to show that $P(n) \implies P(n+1)$, so we need to prove $H_{n+1} - H_m \geq \frac{n+1-m}{n+1}$
Then,
$H_{n+1} - H_m = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} + \frac{1}{n+1} - (\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{m})$
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq \frac{1}{n+1} + \frac{n-m}{n}$, by the inductive hypothesis
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq 1 + \frac{1}{n} + \frac{n-m}{n}$
$= \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m} \geq 1 + \frac{n+1-m}{n}$
which is $ > \frac{n+1-m}{n+1}$
I think this is a dud, since I proved $>$, and not $\geq$. Does anyone have any recommendations? Am I missing something?
| The problem occurs when you are subtracting the sums
$H_{n+1} - H_m = 1 + \frac{1}{2} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) = $
$ = 1 + \frac{1}{2} + ... + \frac{1}{m} + \frac{1}{m+1} + ... + \frac{1}{n+1} - \Big(1 + \frac{1}{2} + ... + \frac{1}{m}) $
$ = \frac{1}{m+1} + ... + \frac{1}{n+1} $
since the first m terms of $H_{n+1}$ cancel out the terms of $H_m$. But this is not very helpful for induction. You must say the following,
$ H_{n+1} - H_m = \frac{1}{n+1} + H_n - H_m \geq \frac{1}{n+1} + \frac{n-m}{n} $
where the last inequality holds because of the induction hypothesis. Now, by rearranging terms it's easy to prove that
$ \frac{1}{n+1} + \frac{n-m}{n} \geq \frac{n+1-m}{n+1} $
Which implies that
$ H_{n+1} - H_m \geq \frac{n+1-m}{n+1} $
Now about using the equality or not is not a big problem. But, if u want to have only use ">", your first step in induction should be for $n=m+1$, and not $n=m$, since $m+1$ is the first natural number $>m$.
| {
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"url": "https://math.stackexchange.com/questions/4110587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $T$ is surjective
Let $T:S_2 \to \mathbb{R}_3[x]$ such that $$T\left(\begin{bmatrix} a &
b\\ b & c \end{bmatrix}\right)=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2.$$
Prove that $T$ is surjective.
My attempt:
Let $w \in \mathbb{R}_3[x]$ such that $w=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2$.
So $w=a(1-x+2x^2)+b(1+3x^2)+c(1+2x+6x^2)$.
Is that proving that $T$ is surjective? I think that I missed something here.
Thanks!
| A function $f:X\rightarrow Y$ is said to be surjective if:
For every $y\in Y$ there exists and $x\in X$ such that $f(x)=y$.
We need to show that for every $a+bx+cx^2 \in \mathbb{R}_3[x]$ for arbitrary $a,b,c \in \mathbb{R}$ that there exists some symmetric matrix $S\in S_2$ such that $T(S)=a+bx+cx^2$.
The short answer is yes because given $\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\in S_2$ then
$T\left(\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\right)=a+bx+cx^2\in \mathbb{R}_3[x]$
The calculations below show you how I got the matrix $\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}$:
$T\left(\begin{bmatrix} A& B\\B&C \end{bmatrix}\right) = (A+B+C)+(-A+2C)x+(2A+3B+6C)x^2=a+bx+c^2$
and equating the coefficients, we can translate into matrix notation:
$$\begin{align}A+B+C & =a\\
-A+2C &=b\\
2A+3B+6C &=c\end{align} \Longrightarrow \begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$$
so if $M=\begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}$ is invertible, then we could solve for $\begin{pmatrix}A\\B\\C\end{pmatrix}$.
You can find the inverse by creating the matrix $[M|I]$ and bring that matrix into RREF to get $[I|M^{-1}]$ through a series of row operations, you can see this here where the step-by-step process is provided as well. It turns out that $M$ is invertible and its inverse is
$$M^{-1}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}$$
(Note that the column vectors of $M$ are exactly the vectors in your linear combination $a(\color{aqua}{1-x+2x^2})+b(\color{aqua}{1+3x^2})+c(\color{aqua}{1+2x+6x^2})$ and the fact that $M$ is invertible means that $[M|I]\cong[I|M^{-1}]$ are row equivalent, thus the RREF of $M$ is $I$ and the column vectors of $M$ are linearly independent, and so, are the vectors in your sum. This would suffice as a proof of surjectivity as well.)
thus $\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=I\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$
and $\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$, only left to multiply and equate:
$$\begin{align}A&= -6a-3b+2c\\
B &=10a+4b-3c\\
C &=-3a-b+c\end{align}$$
$\require{cancel} T\left(\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\right)=a+bx+cx^2\\ \text{Check:}\\
\begin{align}a &=-6a-\cancel{3b}+\cancel{2c}+10a+\cancel{4b}-\cancel{3c}-3a-\cancel{b}+\cancel{c} \\
b &= -(-6a-3b+2c)+2(-3a-b+c)=\cancel{6a}+3b-\cancel{2c}-\cancel{6a}-2b+\cancel{2c} \\
c &=2(-6a-3b+2c)+3(10a+4b-3c)+6(-3a-b+c)=-\cancel{12a}-\cancel{6b}+4c+\cancel{30a}+\cancel{12b}-9c-\cancel{18a}-\cancel{6b}+6c \end{align} $
And we've found a matrix $S$ s.t. $T(S)=a+bx+cx^2$, for every $a+bx+cx^2 \in \mathbb{R}_3[x]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx$ have closed form? I have been trying to find the closed form for integral below $$\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx ,\; \; a>0 $$
My progress to this integral $$\cong\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)-\frac{\pi}{6a\sqrt 2}+\frac{1}{2a^3}\left(\frac{\log(3)}{4} -\frac{\pi}{12\sqrt 3}\right)-\frac{1}{5a^5}\left(\frac{1}{2}-\frac{\pi}{12\sqrt 2}-\frac{\log(3)}{4}\right)+\frac{1}{7a^7}\left(\frac{\pi}{6\sqrt 3}-\frac{1}{2}\right)+\frac{1}{108a^9} +\frac{1}{9a^9}\left(\frac{\log(3)}{9}-\frac{\pi}{12\sqrt 3}\right)-\frac{1}{24a^{11}}+\cdots $$ Using the series of $\tan(ax)$ the above form is obtained. However, I dont find closed form for it. If $a\to\infty+$,then it is equal to $\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)$.
| You could simplify the problem using
$$\frac{x^2}{x^4+x^2+1}=\frac {x^2}{(x^2-r)(x^2-s)}=\frac 1{r-s} \left(\frac{r}{x^2-r}-\frac{s}{x^2-s}\right)$$ where
$$r=-\frac{1+i \sqrt{3}}{2} \qquad \text{and} \qquad s=-\frac{1-i \sqrt{3}}{2}$$ So, the problem boils down to the computation of
$$I(t)=\int_1^\infty \frac{\tan ^{-1}(a x)}{x^2-t} \,dx$$ where $t$ is a complex number.
Surprising or not, there is an antiderivative (a monster found by a CAS).
So, there is a closed form. I saw it !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is? If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is ?
My approach:
The given expression can be written as $(y-1)(y)(y+1)(y+k-2) + 1$ where $y = x + 2$
This gives me $(y)(y^2 - 1)(y+k-2) = n^2 - 1$ something seems similar in LHS and RHS. But not sure how to proceed next? Should I put $(y)(y+k-2) = 1$ ? although it doesn't seem right for some reason also I dont think it leads anywhere.
Any hints/explanations are appreciated!
| $$(x+1)(x+2)(x+3)(x+k) + 1 = f(x)^2$$
for all $x$ then $f(x)=x^2+ax+b$ for all $x$ and some $a,b$.
Since $f(0)^2 = 6k+1$ and $f(-1)=f(-2)=f(-3)=\pm1$ so two of them must be equal and we have next $6$ options:
*
*If $f(-1)=f(-2)=1$ then $f(x) = (x+1)(x+2)+1 = x^2+3x+3$ so $f(-3) = 3$. Impossible.
*If $f(-1)=f(-2)=-1$ then $f(x) = (x+1)(x+2)-1 = x^2+3x+1$ so $f(-3)=1$ and thus $1=6k+1$ so $\boxed{k = 0}$.
*If $f(-1)=f(-3)=1$ then $f(x) = (x+1)(x+3)+1 = x^2+4x+4$ so $f(-2) = 0$. Impossible.
*If $f(-1)=f(-3)=-1$ then $f(x) = (x+1)(x+3)-1 = x^2+4x+2$ so $f(-2) = -2$. Impossible.
*If $f(-2)=f(-3)=1$ then $f(x) = (x+2)(x+3)+1 = x^2+5x+7$ so $f(-1) = 3$. Impossible.
*If $f(-2)=f(-3)=-1$ then $f(x) = (x+2)(x+3)-1 = x^2+5x+5$ so $f(-1) = 1$. Then $25=6k+1\implies \boxed{k=4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Improper integral $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ If I have to solve $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ it is right to write:
$$-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}$$
and so the integral is divergent? I have use the fact that $-\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \frac{-1}{x^{\frac{1}{3}}}$.
If I have instead $\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$, so without minus sign, what can I do to prove the divergence? Is it true that from the following I can deduce the divergence?
$$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx=-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq -\int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}\, dx $$
My doubt arises since I have seen here that a similar integral here (Solution verification of the improper integral $\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$) is solved in a more complicate way...
| $$\int_{2n\pi}^{2n\pi+\pi}\frac{\sin^4x}{x^{1/3}}dx\geq\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{2n\pi}^{2n\pi+\pi}\sin^4xdx\\
=\frac{1}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\pi}\sin^4xdx\\
=\frac{2}{\sqrt[3]{2n\pi+\pi}}\int_{0}^{\frac\pi2}\sin^4xdx
=\frac{1}{\sqrt[3]{2n\pi+\pi}}\cdot\frac{3\pi}{8}.$$
So
$$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq\frac{3\pi}{8}\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{2n\pi+\pi}}=+\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for the derivative of absolute value, gone wrong The absolute value $|x|$ can be represented as $\sqrt{x^{2}}$, as per this question.
Let $f(x) = |x| = \sqrt{x^{2}}$. Solving for $f'(x)$, let $u = x^{2}$. Then,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(\sqrt{u})\cdot\frac{d}{dx}(x^{2}) \\ &=\frac{1}{2\sqrt{u}}\cdot2x \\ &= \frac{x}{\sqrt{x^{2}}} \\ &= \frac{x}{|x|}\end{align*}
We can also see that $\sqrt{x^{2}} = \left(\sqrt{x}\right)^{2}$. Then, let $u = \sqrt{x}$. Solving for $f'(x)$,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(u^{2})\cdot\frac{d}{dx}(\sqrt{x}) \\ &= 2u\cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sqrt{x}}{\sqrt{x}} \\ &= 1\end{align*}
I think the problem here is by letting $u = \sqrt{x}$. What seems to be the problem?
| Both are fine. What happens is that the second approach only makes sense when $x>0$. And then you do indeed have $(\sqrt{x^2})'=1\left(=\frac x{|x|}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120960",
"timestamp": "2023-03-29T00:00:00",
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Prove $9\mid (x^2 + y^2) \iff 3\mid x \text{ and } 3\mid y$ Prove: Let $x, y$ be two non-negative integers. Then $9|(x^2 + y^2)$ if and only if $3|x$ and $3|y$.
General format of proving iff statement.
Proof.
Part 1, proving $p \implies q$
Part 1, proving $q \implies p$
Therefore, $p \iff q. \qed$
| Here's a (very) slightly different approach, presented more to help you with proof-writing than with the math (which I think you understand).
First, assume $3 \mid x$ and $3 \mid y$. Then $9 \mid x^2$ and $9 \mid y^2$, so $9 \mid (x^2+y^2)$. That proves $\Leftarrow$.
Conversely, assume $3 \nmid x$. Then $x^2 \equiv 1 \pmod 3$. Since $y^2 \neq 2 \pmod 3$ (no matter what $y$ is), this means $x^2+ y^2 \neq 0 \pmod 3$ so $9 \nmid (x^2+y^2)$.
Finally, if $3 \nmid y$, then we've just proved that $9 \nmid (y^2+x^2)$, so also $9 \nmid (x^2+y^2)$. This paragraph and the last paragraph prove $\Rightarrow$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $02$
Let $0<a<b<1$ and $E=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}}$
Prove that $E>2$
I tried to change both log bases to $\frac{2ab}{a+b}$ and I got $E=\frac{1}{\log_{\frac{2ab}{a+b}}{a}} + \frac{1}{\log_{\frac{2ab}{a+b}}{b}}$.
Also I noticed that $\frac{2ab}{a+b} < 1$ because $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$ which is the harmonic mean of $a$ and $b$ and since harmonic mean is always smaller than $max(a,b)$, $\frac{2ab}{a+b}$ will be smaller than 1.
Since both the base and the argument of $\log_{\frac{2ab}{a+b}}{a}$ and $\log_{\frac{2ab}{a+b}}{b}$
are smaller than 1, the logarithms will be positive, but I do not know if they will be greater than 1 or smaller than 1.
Any tips ? Thanks in advance !
| Bringing to $\ln$, we have
$$\begin{align}
E
&=\log_{a}{\frac{2ab}{a+b}} + \log_{b}{\frac{2ab}{a+b}} \\
&= \frac{\ln\frac{2ab}{a+b}}{\ln a} + \frac{\ln\frac{2ab}{a+b}}{\ln b} && \left(1\right) \\
&= \ln\frac{2ab}{a+b}\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) \\
&> \ln\sqrt{ab}\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) && \left(2\right)\\
&= \frac{1}{2}(\ln a + \ln b)\left(\frac{1}{\ln a} + \frac{1}{\ln b}\right) \\
&= \frac12 \left(2 + \frac{\ln a}{\ln b} + \frac{\ln b}{\ln a}\right)\\
&> \frac 12\left(2 + 2\right) && (3)\\
&> 2
\end{align}$$
Explanation:
$(1)$ $\displaystyle\because\log_cd = \frac{\ln d}{\ln c}$
$(2)$ $\because\displaystyle \frac{2ab}{a+b} < \frac{2ab}{2\sqrt{ab}} = \sqrt{ab}$ (Thanks to @Angelo) and $ c_1d > c_2d$ for $c_1 < c_2 < 0,~ d <0$
$(3)$ $\because \displaystyle\frac{\ln a}{\ln b} + \frac{\ln b}{\ln a} \gt 2\sqrt{\frac{\ln a}{\ln b} \cdot \frac{\ln b}{\ln a}} = 2$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $ converge Hi guys I have a problem that i need to prove that the sequence:
$$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $$ converges
I need to show its a monotone sequence with an upper and lower bound.
For lower bound, 0 seems to fit nicely.
I can also see that I always multiply by something bigger than one so its always increasing.
My only issue was finding an upper bound, with my calculator I figured that something around e is the upper bound but I can't prove it.
I attempted by induction:
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)<4 $$
after multiplying both sides I got :
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)\left(1+\frac{1}{2^{n+1}}\right)<4\left(1+\frac{1}{2^{n+1}}\right)<4\cdot2=8 $$
But it seems wrong and Im kind of lost.
Also tried using $$log_2a_n=log(3)-log(2)+log(5)-log(4)+log(9)-log(8)+\dots log(2^n+1)-log(2^n)$$
No matter how much I try, I just cant prove an upper bound, any suggestions?
| Note
\begin{eqnarray}
a_n&=&(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})\\
&=&\frac{(1-\frac12)(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})}{1-\frac12}\\
&=&2(1-\frac{1}{2^2})(1+\frac{1}{2^2})\dots (1+\frac{1}{2^n})\\
&=&\cdots\\
&=&2(1-\frac{1}{2^{2n}})
\end{eqnarray}
and hence $\{a_n\}$ converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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A nonnegative function $f(a,b)\geq 0$ Let $c$ be a fixed positive real umber such that $c\geq 1.$ If $a\geq c^m, b\geq c, $ where $m$ is any positive integer, then is
$$f(a,b)=c(ab+1)(a-c^m)(b-c)+c(ab+1)(a-c^m)(b+1)+c^m(b-c)(ab+1)(a+1)-(ab-c^{m+1})(a+1)(b+1)\geq 0?$$
We can check the validity for the two cases $a=c^m,$ or $b=c$ easily. In fact the equality in the above claim occurs at $(a, b)=(c^m,c).$ Kindly suggest me the way to establish $f(a,b)\geq 0$ if it is true.
| Yes, it's true if all my computation is right.
First, the last term can be divided into two part.
$$(ab-c^{m+1})(a+1)(b+1)=(ab-c^{m+1})(ab+1)+(ab-c^{m+1})(a+b) $$
Use the first term to deal the first part and see what remains
$$ c(ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\geq (ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\\
=-(ab+1)(c(a-c^m)+c^m(b-c))$$
Now divide the second and third term in $f$
$$c(ab+1)(a-c^m)(b+1)=cb(ab+1)(a-c^m)+c(ab+1)(a-c^m)$$
$$c^m(b-c)(ab+1)(a+1)=c^ma(b-c)(ab+1)+c^m(b-c)(ab+1)$$
Their second part can fill the difference the first remains and now it's sufficient to prove
$$cb(ab+1)(a-c^m)+c^ma(b-c)(ab+1)-(ab-c^{m+1})(a+b)\geq0 $$
Note the fact $a\geq c^m>1, \; b\geq c \geq 1$, $(ab+1)\geq(a+b)$, so it's sufficient to prove
$$ cb(a-c^m)+c^ma(b-c)-(ab-c^{m+1})\geq 0 $$
Notice
$$cb(a-c^m)+c^ma(b-c)-(ab-c^{m+1})=(c-1)(a-c^m)(b-1)+c^m(a-1)(b-1)+(c-1)a\geq 0$$
And now we know $f$ is always non-negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$ Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$
Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.
For $x=11$, it's $[363; \mathbf{11}, 544, ...]$. The pattern continues (I have only checked up to $21$, but I assume that it's true) that the second term in the continued fraction expansion is $x$.
I have two questions:
*
*Why does this pattern occur and how do we prove that it occurs?
*Is there a similar pattern with later terms?
I haven't found an answer to 2, and I've had no luck as of yet for 1. All I have done so far is simplify it to $x\sqrt[3]{x^3+1}+\sqrt[3]{(x^3+1)^2}+x^2$.
Thanks in advance.
| Partial answer: the provided numerical examples strongly suggest that the continued fraction expansion would be of the form $[3x^2,x,\ldots]$. So we only need to show that for $x>0$ we have
$$
3x^2+\frac{1}{x+1} < \frac{1}{\sqrt[3]{x^3+1}-x} < 3x^2+\frac{1}{x}
$$
A few steps of algebraic manipulations on the right-side inequality give us
$$ \begin{array}{rl}
\dfrac{1}{\sqrt[3]{x^3+1}-x} &< \ 3x^2+\dfrac{1}{x} \\
x &< \ \left(\sqrt[3]{x^3+1}-x\right)\left(3x^3+1 \right) \\
\dfrac{3x^4+2x}{3x^3+1} &< \ \sqrt[3]{x^3+1} \\
\left(\dfrac{3x^4+2x}{3x^3+1}\right)^3 &< \ x^3+1 \\
x^3+1-\dfrac{2}{3(3x^3+1)^2}-\dfrac{1}{3(3x^3+1)^3} &< \ x^3+1 \\
\end{array}$$
which is evidently true (last step courtesy of WolframAlpha). A completely similar derivation shows that the left-side inequality also holds, and we're done with (1).
I'll admit, I have no idea how to approach (2) though.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does parity of $f(a)$ and $f(a+1)$ are same whenever $a$ is even? Question related to sum of digits Let $D(a, b)$ denotes sum of digits of $a$ in base $b$.
Example: $D(5,2)=2,D(1227,10)=1+2+2+7=12$
Define $f(a)=\sum_{i=2}^aD(a, i)$ where $a\ge2$.
Example $f(5)=D(5,2)+D(5,3)+D(5,4)+D(5,5)=2+3+2+1=8$
Can it be shown that
if $a$ is even then $f(a)\equiv f(a+1)\pmod2$
Or parity of $f(a)$ and $f(a+1)$ are same
Table
$$\begin{array}{c | c | c |c | } a & f(a) & f(a)\pmod2 \\ \hline
2 & 1 & 1 \\ \hline
3 & 3 & 1 \\ \hline
4 &4& 0 \\ \hline
5 &8& 0 \\ \hline
6 &10& 0 \\ \hline
7 &16& 0 \\ \hline
8 &17& 1 \\ \hline
9 &21& 1 \\ \hline
\vdots &\vdots &\vdots \\
\end{array}$$
I checked up to $a\le5000$ holds the claim true.
Note: $D(odd, odd)=odd$ and $D(even, odd)=even$
Source Code, PARI/GP
for(a=2,100,print([a,sum(i=2,a,sumdigits(a, i))%2]))
| Assume $a$ is even. That means we already know the parity with $i=2j+1$ odd: $D(a, 2j+1)=2k_j$ (even) and $D(a+1,2j+1)=2k'_j+1$ (odd) for all $j=1,...,\frac{a}{2} - 1$.
Now take a look for $i=2j$ even: We don't know the concrete value of $D(a,2j)$, but we can draw a connection between $D(a,2j)$ and $D(a+1,2j)$. As $a$ is even the last digit in any even-numbered system must be even as well (otherwise we will get a contradiction $\text{mod }2$)! This means the last digit can never be $i-1$ so if we add $1$ we will only increase the last digit to a value $\leq i-1$ and no overflow to the next digit is generated. So all digits besides the last one stay untouched.
That means we get $D(a+1,2j) = D(a,2j) + 1$. Summing all information up we can conclude:
$$
f(a) = \sum_{j=1}^{a/2} D(a,2j) + \sum_{j=1}^{a/2-1} D(a,2j+1) \\
f(a+1) = \sum_{j=1}^{a/2} D(a+1,2j) + \sum_{j=1}^{a/2-1} D(a+1,2j+1) + D(a+1,a+1) \\
= \sum_{j=1}^{a/2} D(a,2j) + \frac{a}{2} \cdot 1 + \sum_{j=1}^{a/2-1}D(a+1,2j+1) + 1
$$
Considering everything $\text{mod }2$ we can use that $D(a+1,2j+1) \equiv D(a,2j+1)+1$:
$$
f(a+1) \equiv \sum_{j=1}^{a/2} D(a,2j) + \frac{a}{2} \cdot 1 + \sum_{j=1}^{a/2-1}D(a,2j+1) + \left(\frac{a}{2}-1\right) \cdot 1 + 1 \\
\equiv f(a) + \frac{a}{2} + \frac{a}{2} -1 + 1 = f(a) + a \equiv f(a) \mod 2
$$
Here we have it that
$$f(a) \equiv f(a+1) \mod 2$$ if $a$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Redefine the P matrix so that it’s made of eigenvectors with integer components.
I’m trying to redefine the P matrix so that it’s made of eigenvectors with integer components (which will give the same diagonalizing matrix). But I’m stuck because I’m not sure how to redefine all the columns. I think the first column will be redefined as 1,-1,1,-1. But for the rest of the columns, I'm stuck. Please help me out, if possible. Thank you.
(Original Matrix -> H)
| $$A = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\end{pmatrix}$$
$$A = P D P^{-1} =\begin{pmatrix}
-1 & 1 & -1 & 1 \\
1 & 0 & 0 & 1 \\
-1 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
\end{pmatrix} \begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}\begin{pmatrix}
-\frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} \\
\frac{1}{4} & -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \\
-\frac{1}{4} & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\
\frac{1}{4} & \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\
\end{pmatrix}$$
Note: You can also factor our $\frac{1}{2}$ from $A$. This will change $D$ to just be two times what is shown, but $P$ and $P^{-1}$ will be the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$
How many terms are there containing the term $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ such as $xyk^2t^2z^2$ or $xyk^2t^4z^0$ or $xyk^2z^3t$ etc.
I made up this question and calculated it , but i do not know whether my solution is correct or not.
$\color{blue}{Solution:}$
$\color{red}{1-)}$ I turned it out to be binomial expression such that $\color{green}{a=}2x-y+4k$ and $\color{green}{b=}t+3z$ , so I will work over $(a+b)^8$.
$\color{red}{2-)}$ I thought that i should look for $xyk^2$ in $a$ ,so i said that the exponential of $a$ must be $4$ , because $1+1+2=4$ which are the sum of the exponents of $xyk^2$.
$\color{red}{3-)}$ Hence, the exponent of $b$ must be $4$ ,as well. Because the sum of the exponents of $a$ and $b$ must be equal to $8$.
$\color{red}{4-)}$ Therefore, the terms which contains $xyk^2$ in the expansion of $(2x-y+t+3z+4k)^8$ must be equal the numbers of terms in the expansion of $b$ ,i.e, $(3z+t)^4$. It is equal to $5$
So , my answer is $5$ . I checked it with second solution. However , i am not sure about the truth of the first way. Is my first way true?
Thanks for all contributions..
$\color{blue}{Solution 2:}$ Because of the fact that the sum of the exponents must be equal to $8$ , i should find the number of partition of $4$ in two part so as to find the number of combinations of the exponents of $t$ and $z$. Then, $t=4,z=0$ or $t=3,z=1$ or $t=2,z=2$ or $t=1,z=3$ or $t=0,z=4$
| Both solutions are correct. The reasoning to obtain $5$ in the first solution is somewhat more detailed. The essence is in both solutions the same, namely to keep the focus on the number of different terms in the expansion of $(t+3z)^4$.
Here is a crosscheck (kinda overkill). We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](ax+z)^n=\binom{n}{k}a^kz^{n-k}\tag{1}
\end{align*}
We obtain by successively applying (1)
\begin{align*}
\color{blue}{[xyk^2]}&\color{blue}{(2x-y+t+3z+4k)^8}\\
&=\binom{8}{1}2^1[yk^2](-y+t+3z+4k)^7\\
&=\binom{8}{1}2^1\binom{7}{1}(-1)^1[k^2](t+3z+4k)^6\\
&=\binom{8}{1}2^1\binom{7}{1}(-1)^1\binom{6}{2}4^2(t+3z)^4\\
&\,\,\color{blue}{=-26\,880(t+3z)^4}\\
\end{align*}
The number of wanted terms is the number of different terms of $(t+3z)^4$ expanded, which is $5$.
| {
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"url": "https://math.stackexchange.com/questions/4138004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\|x^*\|=1$ and $\|x\|=\|y\|=1$ then $\|x-y\|<\epsilon$ Suppose $X$ is a space with inner product. Let $x^{\ast} \in X^{\ast}, \| x^{\ast} \| = 1$, and $0 < \varepsilon < 1$. If
$x, y \in X, \| x \| = \| y \| = 1, x^{\ast} (x) > 1 -
\frac{\varepsilon^2}{8}$, and $x^{\ast} (y) > 1 - \frac{\varepsilon^2}{8}$,
then $\| x - y \| < \varepsilon$.
I know
$$
| x^{\ast} (z) | \leqslant \| x^{\ast} \| \| z \|
\text{ for all } z \in X $$
Then
$$ | x^{\ast} (x) | \leqslant 1 \text{ and } | x^{\ast} (y) | \leqslant 1 $$
Now by hypothesis
$$ \begin{align}
1 - \frac{\varepsilon^2}{8} &< x^{\ast} (x) \leqslant 1\\
- 1 &\leqslant - x^{\ast} (y) < \frac{\varepsilon^2}{8} - 1\\[1em]
\hline
- \frac{\varepsilon^2}{8} &< x^{\ast} (x) - x^{\ast} (y) <
\frac{\varepsilon^2}{8}\\
| x^{\ast} (x - y) | &< \frac{\varepsilon^2}{8},
\end{align} $$
and also I have that
$$ | x^{\ast} (x - y) | \leqslant \| x - y \| . $$
But I don't know what else can I do to arrive at
$$ \| x - y \| < \varepsilon . $$
Any suggestion will be welcome.
| By the parallelogram law we have
\begin{equation}
\| x + y \|^2 + \| x - y \|^2 = 2 (\| x \|^2 + \| y \|^2) \implies \| x +
y \|^2 = 4 - \| x - y \|^2 \tag{1}
\end{equation}
Also by hypothesis
\begin{align}
x^{\ast} (x) &> 1 - \frac{\varepsilon^2}{8}\\
x^{\ast} (y) &> 1 - \frac{\varepsilon^2}{8}\\
\\ \hline
\| x^{\ast} \| \| x + y \| &\geqslant x^{\ast} (x + y) > 2 -
\frac{\varepsilon^2}{4}\\
\Rightarrow \text{ } \| x + y \| &> 2 - \frac{\varepsilon^2}{4}
\end{align}
Using the last result and (1), we have
\begin{align}
4 - \| x - y \|^2 &> \left( 2 - \frac{\varepsilon^2}{4} \right)^2\\
4 - \| x - y \|^2 &> 4 - \varepsilon^2 + \frac{\varepsilon^4}{16}\\
\| x - y \|^2 &< \varepsilon^2 - \frac{\varepsilon^4}{16} <
\varepsilon^2\\
\| x - y \| &< \varepsilon .
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one find values of $m$ for which the roots of $2x^2-mx-8=0$ differ by $m-1$
Find values of $m$ for which the roots of $2x^2-mx-8$ differ by $m-1$.
When I attempted to solve this I tried to simplify it into something like this: $((m-1)+2)((m-1)-4)$ but when I expanded I got $m^2 -4m -5$ which is nowhere near $2x^2-mx-8$ so my question is how to do solve these types of problems. I thank you in advance for your help. :)
| We may also directly construct the polynomials we seek . The parabola corresponding to $ \ y \ = \ 2x^2 - mx - 8 \ = \ 2·\left(x - \frac{m}{4} \right)^2 \ - \ \left(8 + \frac{m^2}{8} \right) \ $ has its vertex at $ \ x = \frac{m}{4} \ \ , $ which is the midpoint of its $ \ x-$ intercepts. (These always exist, since this is an "upward-opening" parabola and the $ \ y-$coordinate of the vertex is always negative .)
Since it is desired that the difference between the roots of $ \ y \ = \ 2x^2 - mx - 8 \ = \ 0 \ \ $ be $ \ (m - 1) \ \ , $ the roots are $ \ \frac{m}{4} \ \pm \ \frac{(m - 1)}{2} \ = \ \frac{3m}{4} - \frac{1}{2} \ , \ -\frac{m}{4} + \frac{1}{2} \ \ . $ A quadratic polynomial with these as zeroes is then
$$ 2·\left(x \ - \ \frac{3m}{4} \ + \ \frac{1}{2} \right)·\left(x \ + \ \frac{m}{4} \ - \ \frac{1}{2} \right) \ \ = \ \ 2x^2 \ - \ mx \ + \ \left(m \ - \ \frac{3m^2}{8} \ - \ \frac12 \right) \ \ . $$
As $ \ -8 \ \ $ is the given constant term, we therefore have
$$ m \ - \ \frac{3m^2}{8} \ - \ \frac12 \ \ = \ \ -8 \ \ \Rightarrow \ \ 3m^2 \ - \ 8m \ - \ 60 \ \ = \ \ (3m \ + \ 10) \ · \ (m \ - \ 6) \ \ = \ \ 0 \ \ . $$
The two quadratic polynomials with the specified property are then
• $ \ \ \mathbf{m \ = \ 6 \ \ :} \quad 2x^2 \ - \ 6x \ - \ 8 \ \ $ with zeroes $ \ 4 \ \ , \ \ -1 \ \ \ $ [difference: $ \ \ 4 \ - \ (-1) \ = \ 6 \ - \ 1 $ ]
and
• $ \ \ \mathbf{m \ = \ -\frac{10}{3} \ \ :} \quad 2x^2 \ + \ \frac{10}{3}x \ - \ 8 \ \ $ with zeroes $ \ \frac43 \ \ , \ \ -3 \ \ \ $
[difference: $ \ \ (-3) \ - \ \frac43 \ = \ -\frac{13}{3} \ = \ -\frac{10}{3} \ - \ 1 $ ] .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Problem with integration by substitution.
Using the substitution $u=1-x$, compute the integral of
$\int{x(1-x)^2}dx.$
My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so
$$\int{x(1-x)^2}\,\mathrm dx\\
=-\int{(1-u)u^2}\,\mathrm du\\
=\frac{u^4}{4}-\frac{u^3}{3}+C\\
=\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+C\\
=\frac{3x^4-8x^3+6x^2-1}{12}+C.$$
However, the answer gotten by expanding the brackets first then integrating does not seem to match that of the answer above arrived by substitution:
$$\int{x(1-x)^2}\,\mathrm dx\\
=\int{x-2x^2+x^3}\,\mathrm dx\\
=\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+C.$$
I cannot account for the $-\frac{1}{12}$ from the method using substitution.
| Let the domain of $f$ be an interval.
Then every pair of the infinitely many antiderivatives of $f$ differ by a constant. Since the indefinite integral $$\int f(x)\,\mathrm dx$$ gives the general specification of these antiderivatives, it has the form $F(x)+C,$ where the constant of integration $C$ is arbitrary.
Both your answers are correct, and their specified constants of integration differ by $C_2-C_1=\frac1{12}.$ To simplify the first answer, substitute $$C:=D+\frac1{12}$$ so that $$\int{x(1-x)^2}\,\mathrm dx=\frac{3x^4-8x^3+6x^2}{12}+D.$$
| {
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"url": "https://math.stackexchange.com/questions/4144376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$.
Evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}.$$
My work so far and background to the problem.
This question was inspired by the second page of this paper. The author of the paper doesn't mention how he managed to prove the $4$ identities (shown below) at the top of the second page of the paper, so I tried to find my own method of doing so.
\begin{align}
\sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1} &= \frac{1}{3}+\frac{2\pi\sqrt3}{27} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1} &= \frac{\pi\sqrt3}{9} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1} &= \frac{\pi^2}{18} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n^4}\binom{2n}{n}^{-1} &= \frac{17\pi^4}{3240}
\end{align}
The method I used was similar to the method the author used in the rest of his paper: trying to find the generating functions for particular sequences. Using this method, like the author I was successful in proving the first $3$ identities, since I obtained the following:
$$\begin{align}\sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1}x^n&=4\sqrt\frac{x}{(4-x)^3}\arcsin\frac{\sqrt{x}}{2}+\frac{x}{4-x}\\
\sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1}x^n&=2\sqrt{\frac{x}{4-x}}\arcsin\frac{\sqrt x}{2}\\
\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1}x^n&=2\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\end{align}$$
I obtained the second and third power series by dividing the power series above each one by $x$ and then integrating.
I then attempted to find a closed form for $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}x^n$ in order to evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. The way I tried to do this was finding
$$\int\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x$$
but according to WA this has no solution in terms of elementary functions and it involves some polylogarithms which I am not familiar with at all really, especially when they have a complex argument.
I then realized that
$$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}=\int_0^1\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x=\int_0^{1/2}\frac{4}{u}\left(\arcsin u\right)^2\mathrm{d}u$$
(where the second equality is found by using the substitution $u=\frac{\sqrt{x}}{2}$) which could be more helpful since definite integrals can often be evaluated in terms of elementary functions even if the indefinite integral cannot.
From there however I could not think of a way of continuing.
Thank you for your help.
| Hint: We find in Evaluations of Binomial Series by J. M. Borwein and R. Girgensohn the formula (47)
\begin{align*}
\sum_{n\geq 1}\frac{1}{n^3\binom{2n}{n}}=\frac{2}{3}\pi\Im\left(L_2\left(e^{i\pi/3}\right)\right)
-\frac{4}{3}\zeta(3)
\end{align*}
with $L_p(z)=\sum_{n>0}\frac{z^n}{n^p}$ the polylogarithms.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "15",
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On the value of $a+b$ such that $a^2+b^2+3ab=2000$ where $a,b \in \mathbb{Z}^+$
Find the value of $a+b$ such that $a,b \in \mathbb{Z}^+$ such that $a^2+b^2+3ab=2000.$
On the outside, this seems like a very easy problem, but I'm not sure how to do this in a non-messy way. What I have so far is that $$a^2+b^2+3ab=(a+b)^2+ab= 2000\implies a+b = \sqrt{2000-ab}.$$ However, is there any way I can do this without trial and error on these values? Solving the quadratic $(a+b)^2 + ab - 2000=0$ wouldn't help a lot either. How should I proceed from this. Thanks in advance.
| One less messy way of doing this is to notice that
*
*if $a^2 + 3ab + b^2$ is even, then both $a, b$ must be even.
*if $a^2 + 3ab + b^2$ is a multiple of $25$, then both $a, b$ are multiples of $5$.
Thus $a^2 + 3ab + b^2 = 2000$ leads to $a = 20x, b = 20y$ with $x^2 + 3xy + y^2 = 5$.
Proof of 1:
We have $0\equiv a^2 + 3ab + b^2 \equiv ab + a + b \equiv (a + 1)(b + 1) + 1\bmod 2$. Therefore $a, b$ must be both even.
Proof of 2:
We have $(a - b)^2 + 5ab \equiv 0 \bmod 25$ and hence $a - b \equiv 0\mod 5$. Thus $5ab \equiv 0\bmod 25$ and one of $a, b$ is multiple of $5$. Hence both are multiples of $5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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