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$a$ and $b$ are solutions of $ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $, $a+b=?$ $a$ and $b$ are solutions of
$$ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $$
What is $a+b=?$
$$ $$
Are there better approaches than the one below?
Solution:
By letting $x^{2} - 10x = y$, then we have
$$ \frac{1}{y-29} + \frac{1}{y-45} - \frac{2}{y-69} = 0, \:\: y \notin \{ 29,45,69 \} $$
and
$$ (y-45)(y-69) + (y-29)(y-69) - 2(y-29)(y -45) = 0 $$
$$ (y- 69)(y-37) = (y-29)(y-45) $$
$$ y^{2} - 106 y + 69 \cdot 37 = y^{2}-74y + 29 \cdot 45 $$
$$ -32y = 3 (29 \cdot 15 - 23 \cdot 37) = -1248 $$
$$ y = x^{2} - 10x = 39$$
Here are the roots:
$$ x^{2} -10x - 39 = 0 \implies (x-13)(x+3) = 0$$
So the answer is $a + b = 13 - 3 = 10$
| Wait so at the start, you say $y=x^2-10x$ then proceed to say $y=x^2-10x-39$
Apart from that, this is probably the best solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^3$.
| Another way to solve the problem is the following, where no Differential Calculus is needed.
By elementary algebraic manipulation of radicals, you get:
$$ \begin{split} \sqrt[3]{n^3 + 3n^2} - \sqrt{n^2 + 2n} &= \underbrace{\sqrt[6]{(n^3 + 3n^2)^2}}_{=:\sqrt[6]{a}} - \underbrace{\sqrt[6]{(n^2 + 2n)^3}}_{=:\sqrt[6]{b}} \\ & = \frac{a - b}{\sqrt[6]{a^5} + \sqrt[6]{a^4b} + \sqrt[6]{a^3b^2} + \sqrt[6]{a^2b^3} + \sqrt[6]{ab^4} + \sqrt[6]{b^5}} \\ &\approx \frac{-3n^4}{6n^5} \\ &= - \frac{1}{2n} \end{split} $$
with the latter member going to $0$ as $n \to \infty$.
| {
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How much ways there are to produce sum = 21 with 4 different natural numbers? 0 isn't natural number , and the sum way is important (e.g. 3+5+6+7 is different from 5+6+7+3).
I got that I have for a+b+c+d = 21 , 2024 options.
I think I need to sub the invalid numbers , so I need to relate for the cases :
a = 0 , b = 0, c = 0, d = 0 ,
a = b , a = c , a = d, b = c, b = d, c = d?
| This is the answer to "How many ways are there to distribute 21 candies to 4 children, where each child gets at least 1 candy?"
To do this, lay out 21 candies in a row:
$$ \underbrace{\bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet}_{21 \,\,\, Candies} $$
To produce the sum, say $3+5+6+7=21$, we can separate the candies with dividers as such:
$$ \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet \bullet | \bullet \bullet \bullet \bullet \bullet \bullet \bullet $$
Notice there are 20 choices for the location of the dividers, and we must pick 3. The question then becomes,
How many ways are there to choose 3 spots out of 20 to place dividers?
The answer is $$ {21 - 1 \choose 4-1 } = {20 \choose 3 } = \frac{20!}{3! 17!}$$
In general, the answer is
$$\mbox{Number of ways to write } N \mbox{ as the sum of } k \mbox{ natural numbers } = {N - 1 \choose k-1 } = \frac{N-1!}{(k-1)! (N-k+1)!} $$
| {
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algorithms - When not to assume $n$ is a power of 2, while solving recurrences? In Udi Manber - Introduction to Algorithms. A Creative Approach, exercise 3.29, p.59
Although in general it is sufficient to evaluate recurrence relations only for powers of 2, that is not always the case. Consider the following relation:
$$T\left(n\right)=\begin{cases}
T\left(\frac{n}{2}\right)+1 & \text{if $n$ is even}\\
2T\left(\frac{n-1}{2}\right) & \text{if $n$ is odd}
\end{cases}$$
a. Prove that the solution of this recurrence for powers of $2$ is $T\left(2^{k}\right)=k+1$ (namely, for powers of $2$, $T\left(n\right)=O\left(\log n\right)$)
b. Show that, for an infinite number of values of $n$, $T\left(n\right)=\Omega\left(n\right)$. Discuss why the usual assumption about the relative behavior for powers of $2$ and nonpowers of $2$ breaks down for this recurrence.
I have no problem solving this problem, but there an important section of the solution provided in the book make me confused, and I can't search for explanations on the Internet or books about that:
We can restrict ourselves to powers of $2$ if the solution of the recurrence relation is a function that behaves "nicely." In particular, the expressions for the running times of most algorithms are monotonically increasing functions. If the difference between the value of $T\left(n\right)$ and $T\left(2n\right)$ is no more than a constant and if $T\left(n\right)$ is monotonically increasing, then $T\left(m\right)$ for $n < m < 2n$ is no more than a constant times $T\left(n\right)$. Therefore, the analysis for powers of $2$ is sufficient in that case.
Here is what I understand about the bold-italic part above:
Consider,
$$\begin{align*}
T\left(n\right)&=cT\left(\frac{n}{2}\right)+O\left(n\right)\\T\left(2n\right)&=cT\left(n\right)+O\left(2n\right)\\&=c^{2}T\left(\frac{n}{2}\right)+O\left(n\right)\\
\end{align*}$$
The difference between $T\left(n\right)$ and $T\left(2n\right)$ is a constant $c$, and it does not affect the analyzing of $O\left(\right)$ of the recurrence if we chose $2n$ as the input value or any $m$, $n<m<2n$.
Now what I confused is what is the case of not to assume that $n$ is a power of $2$. According to the book, it's not applicable for the case, such as $T\left(n\right)=nT\left(\frac{n}{2}\right)$, when $T\left(2n\right)=2n^{2}T\left(\frac{n}{2}\right)$.
But I see some examples were still solved by this way, such as exercise 8.2.53, p.527 K. Rosen's Discrete Mathematics:
$$T\left(n\right)=nT^{2}\left(\frac{n}{2}\right)$$
or this one.
So, my questions are:
*
*Am I wrong?
*Could you provide simple example(s) of monotonic $T\left(n\right)$ where replacing $n$ by a power of 2 is not applicable? (I desire $T\left(n\right)$ have unified formula, without different formulas for specified values of $n$, such as the example provided in Udi Manber's book above).
Thanks!
| For the case of Kenneth Rosen's exercise, I posted an answer's here. As the result of solution 2, I choose $n$ such that $2^{m-1}<n<2^{m}$, in particular, $n=2^{m-1}\cdot2^{\frac{1}{2}}=2^{m-\frac{1}{2}}$.
$$\begin{align*}
T\left(n\right)&=\frac{n^{2^{\left\lfloor \log_{2}n\right\rfloor }-1}\cdot n^{2^{\log_{2}n-\left\lfloor \log_{2}n\right\rfloor }}}{2^{\left\lfloor \log_{2}n-2\right\rfloor 2^{\left\lfloor \log_{2}n\right\rfloor }+2}\cdot2^{\left(\log_{2}n-\left\lfloor \log_{2}n\right\rfloor \right)2^{\log_{2}n-\left\lfloor \log_{2}n\right\rfloor }}}6^{n}\\&=\frac{n^{2^{m-1}-1}\cdot n^{2^{m-\frac{1}{2}-\left(m-1\right)}}}{2^{\left\lfloor m-\frac{1}{2}-2\right\rfloor 2^{m-1}+2}\cdot2^{\left(m-\frac{1}{2}-\left(m-1\right)\right)2^{m-\frac{1}{2}-\left(m-1\right)}}}6^{n}\\&=\frac{n^{2^{m-1}-1}\cdot n^{\sqrt{2}}}{2^{\left(m-3\right)2^{m-1}+2}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&\\&=\frac{n^{2^{m-1}-1}\cdot n^{\sqrt{2}}}{2^{\left(m-\frac{1}{2}-3+\frac{1}{2}\right)2^{m-1}+2}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{2^{m-1}}\cdot n^{\sqrt{2}-1}}{4\cdot\left[n\cdot2^{-\frac{5}{2}}\right]^{2^{m-1}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{2^{m-1}}\cdot n^{\sqrt{2}-1}}{4\cdot n^{2^{m-1}}\cdot2^{-5\cdot2^{m-2}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{n^{\sqrt{2}-1}}{4\cdot2^{-5\cdot n\cdot2^{-1-\frac{1}{2}}}\cdot2^{\left(\frac{1}{2}\right)2^{\frac{1}{2}}}}6^{n}\\&=\frac{2^{\frac{5n}{2\sqrt{2}}}\cdot n^{\sqrt{2}-1}}{4\cdot2^{\frac{1}{\sqrt{2}}}}6^{n}
\end{align*}$$
When, if $n=2^{m}$, $T\left(n\right)=\frac{6^{n}\cdot4^{n-1}}{n}$.
So, I think Udi Manber's statement is right. While I don't exactly know why some recurrences would be solved by replace $n=2^{m}$, but my guess is that it's easier to do, and the results are acceptable (??) (Hope someones would correct it if I was wrong).
| {
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7} + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.
I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so:
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
| Great substitution technique. Notice that $\sf{A=\sqrt{1+x}}$ and $\sf{B=\sqrt{1-x}}$ leads to $$\sf{(A^3+B^3)\sqrt{1+AB}}=2+AB$$ and since $\sf{A^3+B^3=(A+B)(A^2-AB+B^2)}$ and $\sf{A^2+B^2=2\implies (A+B)^2=2(1+AB)}$, we get $$\sf{(A+B)\sqrt{1+AB}=\frac{2+AB}{2-AB}}\implies (1+AB)\sqrt2=1+\frac{2AB}{2-AB}$$ which can be solved for $\sf{AB}$, and thus $\sf{x}$.
| {
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How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| $4(\cos^6 \theta -\sin^6 \theta) = 4(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta + \cos^2 \theta\sin^2 \theta+\sin^4 \theta) = 4(\cos^2 \theta -\sin^2 \theta)( (\cos^2 \theta +\sin^2 \theta)^2 - 2\cos^2 \theta\sin^2 \theta + \cos^2 \theta\sin^2 \theta) = 4(\cos^2 \theta -\sin^2 \theta)( 1 - \cos^2 \theta\sin^2 \theta ) = 4\cos2\theta(1- \dfrac{\sin^2 2\theta}{4}) = 4\cos2\theta - \sin^2 2\theta\cos2\theta = 4\cos2\theta - (1-\cos^22\theta)\cos2\theta = 3\cos2\theta +\cos^3\theta $
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.
| Another solution, this time using Mike Spivey's alternating Euler sum:
\begin{align*}
I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\
&= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\
&= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\
&= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha}\left[ \int_0^1(1-x)^\alpha x^n\,\mathrm dx\right ]\Bigg\vert_{\alpha=0}\\
&= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha} \text{B}\left(\alpha+1,\,n+1 \right )\Bigg\vert_{\alpha=0},\qquad\text{beta function}\\
&= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\left[ \frac{\Gamma(\alpha+1)\Gamma(n+1)\left(\psi(\alpha+1)-\psi(\alpha+n+2) \right )}{\Gamma(\alpha+n+2)}\right ]\Bigg\vert_{\alpha=0}\\
&= -\sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{H_{n+1}}{n+1},\qquad\text{used }\psi(1)=-\gamma\text{ and }\psi(m)=H_{m-1}-\gamma\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-1+2\ln(2)+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\
&= -2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-2+2\ln(2)\\
&= \frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2,\qquad\text{applied Mike's sum.}
\end{align*}
| {
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Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Have two conditions, leading to three intervals;
(i) $x \lt \sqrt{-2}$
(ii) $\sqrt{-2} \le x \lt \sqrt{11}$
(iii) $x \ge \sqrt{11}$
(i) no soln.
(ii) $2x^2 = 9 \implies x = \pm\sqrt{\frac 92}$
(iii) no soln.
Verifying:
First, the two solutions must satisfy that they lie in the given interval,
this means $\sqrt{-2} \le x \lt \sqrt{11}\implies \sqrt{2}i \le x \lt \sqrt{11}$.
Am not clear, as the lower bound is not a real one. So, given the two real values of $x = \pm\sqrt{\frac 92}$ need only check with the upper bound.
For further verification, substitute in values of $x$, for interval (ii):
a) For $x = \sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
b) For $x = -\sqrt{\frac 92}$, $|x^2 + 2| = |x^2 − 11|$ leads to
$\frac 92 + 2 = -\frac 92 + 11\implies 2\frac 92 = 9$.
| Maybe another approach might also be interesting:
$$|x^2 +2| = |x^2-11| \Leftrightarrow (x^2+2)^2 = (x^2-11)^2\Leftrightarrow (x^2+2)^2 - (x^2-11)^2 = 0$$
$$\stackrel{a^2-b^2 = (a-b)(a+b)}{\Leftrightarrow} (x^2+2 - (x^2-11))(x^2 + 2 + x^2-11)= 13(2x^2 - 9)= 0$$
$$\Leftrightarrow x^2= \frac{9}{2}\Leftrightarrow x= \pm\frac{3}{\sqrt{2}}$$
| {
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Evaluate $\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx$
Prove that $$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi \ln 2}{4}-\frac {G}{3}+\frac {2\pi}{3}\ln(2+\sqrt 3)$$ Where $G$ is the Catalan's constant.
Actually I proved this using the Feynman's trick namely by introducing the parameter $a$ such that $$\xi(a)=\int_0^{\infty} \frac {\ln(1+ax^3)}{1+x^2}dx$$
Where it is clear that $\xi(0)=0$, hence we just need $$\int_0^1 \xi'(a)da$$ which I found too. Hence proving the statement, but this method was too much lengthy because it involved heavy partial fraction decomposition and one infinite summation.
Can someone suggest some better method?
Edit: I also tried some trigonometry bashing by using the substitution $x=\tan \theta$ but got stuck midway
| Noticing that
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x =\underbrace{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}}_{\frac{\pi}{4} \ln 2+G} d x+\underbrace{\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}}}_{K} d x
\end{aligned}
$$
where the first integral comes from my post.
Now we going to find the integral $K$ by observing
$$
\begin{aligned}K=\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} = \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^2+x^{4}\right)}{1+x^{2}} d x}_{\pi\ln (2+\sqrt 3)}- \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x}_{
\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3}G}
\end{aligned}
$$
where the first integral comes from my post and @Quanto’s post
.
Now we can conclude that
$$\boxed{
\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x=-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}(2+\sqrt{3})}
$$
| {
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Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\infty}(1+2x^2-2x^2)=1$$
However, if you plot the function in Desmos or you do it with a calculator, you will find that $f(x\rightarrow-\infty)=0$
What am I missing?
| Multiply top and bottom by a conjugate:
$$\begin{align*}f(x) & = \dfrac{1+2x^2+2x\sqrt{1+x^2}}{1}\cdot \dfrac{1+2x^2-2x\sqrt{1+x^2}}{1+2x^2-2x\sqrt{1+x^2}} \\ & = \dfrac{1}{1+2x^2-2x\sqrt{1+x^2}}\end{align*}$$
Now, as $x \to -\infty$, $1+2x^2-2x\sqrt{1+x^2} \to \infty$, so $f(x) \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
| Let $a+b=c$.
Solving in pari/gp:
? lift(Mod(a*x^3-a*x^2+b*x+4,x^2+1))
%9 = (-a + b)*x + (a + 4)
?
? lift(Mod(a*x^3-a*x^2+b*x+4,x-4))
%10 = 48*a + (4*b + 4)
?
? polresultant(%9,a+b-c,a)
%11 = (-2*b + c)*x + (b + (-c - 4))
?
? polresultant(%10-51,a+b-c,a)
%12 = 44*b + (-48*c + 47)
?
? polresultant(%11,%12,b)
%13 = (52*c - 94)*x + (-4*c + 223)
Or in Wolfram.
I.e. $a+b=\dfrac{94x-223}{4(13x-1)}$.
If $x=4$ then $\begin{cases}4(b-a)+a+4=0\\48a+4b+4=51\end{cases} \Longrightarrow \begin{cases}a=1\\b=-1/4\end{cases} \Longrightarrow (a+b)\bigg|_{x=4}=\dfrac{3}{4}$.
Verifying: $(a+b)\bigg|_{x=4}=\dfrac{94\cdot 4-223}{4(13\cdot 4-1)}=\dfrac{3}{4}$.
By modulo $x^2+1$ remainder is $x(b-a)+a+4$ and quotient is $a(x-1)$:
$ax^3-ax^2+bx+4=\bigg(x(b-a)+a+4\bigg)+\bigg(a(x-1)\bigg)\cdot(x^2+1)$.
I think that from remainder$=0$ $\not\Rightarrow ax^3-ax^2+bx+4=a(x-1)(x^2+1)$, but only $ax^3-ax^2+bx+4 \equiv 0 \equiv x(b-a)+a+4 \equiv a(x-1)(x^2+1) \pmod {x^2+1}$. Thus is not correct to match the coefficients of $ax^3-ax^2+bx+4$ and $a(x-1)(x^2+1)$ whitout modulo.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Maclaurin series for $f(x) = \frac{1}{1+x+x^2}$ I need to get the Maclaurin series and its radius of the convergence for $f(x) = \dfrac{1}{1+x+x^2}$. I tried to do it manually, with getting the derivatives, but I gave up after some time, because I thought there had to be a better way to solve this. Could anyone help? Thanks in advance!
| Note that, if $\lvert x\rvert<1$,\begin{align}\frac1{1+x+x^2}&=\frac{1-x}{(1+x+x^2)(1-x)}\\&=\frac{1-x}{1-x^3}\\&=(1-x)\times(1+x^3+x^6+x^9+\cdots)\\&=1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+\cdots\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $ a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$, prove $ S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2 $ Suppose a sequence of positive real numbers with
$$ a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$$
prove that $$ S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2 $$
Solution: I will show two different approaches, one is finished, the other one is still confusing
*
*Finished approach:
By induction, for the base case we have
$$ a_{2} \ge a_{1}/a_{1}^{2} = 1/a_{1}^ \implies a_{1} a_{2} \ge 1 $$
by AM-GM we get $a_{1}+a_{2} \ge 2\sqrt{a_{1}a_{2}} \ge 2 $. (base case proven).
Now assume it is true for $n=k$. We will prove for $n=k+1$.
$$ S_{k+1} = S_{k} + a_{k+1} \ge k + a_{k+1} $$
if $a_{k+1} \ge 1$ then the the proof is finished. Now if $0 < a_{k+1} < 1$, here is another approach in the induction:
notice that the known inequality at the top of the post is equivalent with
$$ a_{k} \ge k/a_{k+1} - (k-1)/a_{k} $$
summing all from $k=1,2,...,m$ we get
$$ S_{m} \ge m/a_{m+1} $$
using this to prove for $n=k+1$ with $0 < a_{k+1} < 1$, we get
$$ S_{k+1} = S_{k} + a_{k+1} \ge k/a_{k+1} + a_{k+1} = (k-1)/a_{k+1} + ( a_{k+1} + 1/a_{k+1} )$$
Now $a_{k+1} + 1/a_{k+1} \ge 2$ this is because $f(x) = x + 1/x \ge 2, \:\: 0 < x < 1$ (function is monotonically decreasing with convergence to 2). So we have
$$ S_{k+1} \ge (k-1)/a_{k+1} + ( a_{k+1} + 1/a_{k+1} ) \ge (k-1) + 2 = k + 1$$
THus we have solved the problem.
*
*Unfinished approach:
Here is a hint of the IMO 2015 shortlisted problem:
"Using AM-GM on $S_{k}$ and $k a_{k+1}$ we may prove:
$S_{k} + k a_{k+1} \ge 2k$
then sum all of them from $k=1,2,....,m$."
the inequality is quite easy to prove, but after the summation idk what else to do:
$$ S_{1} + a_{2} \ge 2 $$
$$ S_{2} + 2 a_{3} \ge 2(2) $$
$$ ...$$
$$ S_{m} + (m) a_{m+1} \ge 2(m) $$
then
$$ S_{1} + ... + S_{m} + a_{2} + ... + (m) a_{m+1} \ge m (m+1) $$
| Remember $S_k=a_1+a_2+\dots+a_k$, so
$$
S_1+S_2+\dots+S_m=ma_1+(m-1)a_2+\dots+a_m.
$$
Hence you have
$$
m(a_1+a_2+\dots+a_m+a_{m+1})\geq m(m+1)
$$
which gives $S_{m+1}\geq m+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Greatest value of shortest distance between axis of x and normal to ellipsoid Show that greatest value of shortest distance between axis of $x$ and a normal to ellipsoid is $$b-c$$
or in other words show that the maximum distance of all the normals to the ellipsoid is $b-c$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
My attempt :
I have tried to find shortest distance between lines $ \vec{a_{1}}+\vec{b_{1}}t $
and $ \vec{a_{2}}+\vec{b_{2}}t $
using
$$ (a1-a2). \frac{\vec{a_{1}}*\vec{a_{2}}}{|\vec{a_{1}}*\vec{a_{2}}|}$$
Assuming that the point at which I am drawing the normal is $(\alpha,\beta,\gamma)$ it comes out to be
$$\gamma\beta \frac{b^2-c^2}{b^2c^2*\sqrt{\frac{\gamma^2}{c^4}+
\frac{\beta^2}{b^4}}}$$
I am not able to go further . Any help/hint would be appreciated
| Observe that the same values of $x,y,z$ which maximize
$$
d(x,y,z)=\frac{|b^2-c^2|}{\sqrt {\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2}}}\tag1
$$subject to the constraint
$$
\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}+\dfrac {z^2}{c^2}=1,\tag2
$$
will minimize the value of
$$
\frac {(b^2-c^2)^2}{d^2}=\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2}$$
subject to the same constraint.
The equation $(2)$ together with the other three Lagrange equations for the latter problem:
$$\begin {align}
\lambda\dfrac {2x}{a^2}&=0;\\
-\dfrac {2b^4}{y^3}+\lambda\dfrac {2y}{b^2}&=0;\\
-\dfrac {2c^4}{z^3}+\lambda\dfrac {2z}{c^2}&=0,\\
\end {align}
$$
can be readily solved with the result:
$$
x=0,\quad y^2=\dfrac{b^3}{b+c},\quad z^2=\dfrac{c^3}{b+c},
\quad\lambda=\dfrac{b^4}{y^2}+\dfrac{c^4}{z^2}=(b+c)^2.
$$
Substituting the last value into $(1)$ one finally obtains:
$$d_\text {max}=|b-c|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? Using the Cauchy-Schwarz Inequality, we have
$$
\begin{align}
1
&=\left(\int_n^{n+1}1\,\mathrm{d}x\right)^2\\
&\le\left(\int_n^{n+1}x\,\mathrm{d}x\right)\left(\int_n^{n+1}\frac1x\,\mathrm{d}x\right)\\
&=\left(n+\frac12\right)\log\left(1+\frac1n\right)
\end{align}
$$
which means that
$$
\left(1+\frac1n\right)^{n+1/2}\ge e
$$
This hints that $\left(1+\frac1n\right)^{n+1/2}$ might be decreasing.
In this answer, it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. The proofs use Bernoulli's Inequality. However, applying Bernoulli to $\left(1+\frac1n\right)^{n+1/2}$ is inconclusive.
Attempt to show decrease:
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}
&=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\
&\ge\left(1+\frac{2n}{n^2-1}\right)\frac{n-1}{n+1}\\[6pt]
&=1-\frac{2}{(n+1)^2}
\end{align}
$$
Attempt to show increase:
$$
\begin{align}
\frac{\left(1+\frac1n\right)^{2n+1}}{\left(1+\frac1{n-1}\right)^{2n-1}}
&=\left(1-\frac1{n^2}\right)^{2n}\frac{n+1}{n-1}\\
&\ge\left(1-\frac2n\right)\frac{n+1}{n-1}\\[6pt]
&=1-\frac{2}{n(n-1)}
\end{align}
$$
Neither works.
Without resorting to derivatives, is there something stronger than Bernoulli, but similarly elementary, that might be used to show that $\left(1+\frac1n\right)^{n+1/2}$ decreases?
| We'll prove that this sequence indeed decreases.
We need to prove that $$\left(1+\frac{1}{n+1}\right)^{n+\frac{3}{2}}<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}.$$Let $$f(x)=\left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)-\left(x+\frac{3}{2}\right)\ln\left(1+\frac{1}{1+x}\right)=$$
$$=(2x+2)\ln(1+x)-\left(x+\frac{1}{2}\right)\ln{x}-\left(x+\frac{3}{2}\right)\ln(x+2),$$ where $x>0$.
But $$f''(x)=\frac{2}{x^2(x+2)^2(x+1)}>0$$ and
$$\lim_{x\rightarrow+\infty}f'(x)=\lim_{x\rightarrow+\infty}\left(\ln\frac{(x+1)^2}{x(x+2)}+2-\frac{x+\frac{1}{2}}{x}-\frac{x+\frac{3}{2}}{x+2}\right)=0,$$
which says that $f'(x)<0$, $f$ decreases and since it's obvious that $$\lim_{x\rightarrow+\infty}f(x)=0,$$ we obtain that $f(x)>0$ and our sequence indeed decreases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{\sin x_1\sin x_2\cdots\sin x_n}{\sin(x_1+x_2)\sin(x_2+x_3)\cdots\sin(x_n+x_1)}\le\frac{\sin^n(\pi/n)}{\sin^n(2\pi/n)}$, for $\sum x_i=\pi$ Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that
$$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$
Show that
$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n
$$
This problem also post MO,Until now No one solve it,I think there might be a solution here, because I 've heard that there are a lot of people here who are good at and like inequality, so the possibility of solving this inequality is very high, and I really look forward to them.
| Update:
I have been thinking about this more.
The idea is to replace $x_i$ with some M such that all the properties holds and the ratio of the $\sin$ product holds and the ratio of the products of sines is smaller than the ratio formed with $\sin(M)$. Initially I thought that selecting the Maximum among $x_i$ is sufficient but since $x_{max} + x_{max} > \pi$ as per comments from @Sangchul. This is insufficient.
Consider this,
We have $\sin(x_i + x_{i+1}) \gt 0$, $\sin(x_i) \gt 0$ and $0 \lt x_i + x_{i+1} < \pi ---->(0)$
$$\left(\frac{x_1 + x_2 + x_3+ ...+x_n}{n}\right)^n \ge x_1x_2...x_n ----> (1)$$
Also, we have for $k > 0$
$$\left(x_1 + x_2 + x_3 + ...+ x_n\right)^k \ge x_1^k + x_2^k + x_3^k + ..+x_n^k --> (2)$$
for all positive reals.
This can be used.
$x_1 + x_2 + x_3 + ...x_n = \pi$
Now from ($0$),($1$) and ($2$) $$\frac{\sin{x_1}\sin{x_2}\sin{x_3}...\sin{x_n}}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \frac{\left(\frac{\sin{x_1} + \sin{x_2}...+\sin{x_n}}{n}\right)^n}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \frac{\left(\sin{x_1} + \sin{x_2}...+\sin{x_n}\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
Now, use MacLaurin's series expansion of $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}...$
$$\le \dfrac{\left(\left((x_1 + x_2 + x_3 + ...+x_n) + \frac{(x_1^5 + x_2^5 + x_3^5+ ...+x_n^5)}{5!} + \frac{(x_1^9 + x_2^9 + x_3^9+ ...+x_n^9)}{9!} + ...\right) - \left(\frac{(x_1^3 + x_2^3 + x_3^3+ ...+x_n^3)}{3!} + \frac{(x_1^7 + x_2^7 + x_3^7+ ...+x_n^7)}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
Now, using ($2$) and keeping the denominator same, we have a ratio that is bigger.
$$\le \dfrac{\left(\left((x_1 + x_2 + x_3 + ...+x_n) + \frac{(x_1 + x_2 + x_3 + ...+x_n)^5}{5!} + \frac{(x_1 + x_2 + x_3+ ...+x_n)^9}{9!} + ...\right) - \left(\frac{(x_1 + x_2 + x_3+ ...+x_n)^3}{3!} + \frac{(x_1 + x_2 + x_3+ ...+x_n)^7}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$.
But then, $x_1 + x_2 + x_3 + ...x_n = \pi$, so we have
$$\le \dfrac{\left(\left(\pi + \frac{\pi^5}{5!} + \frac{\pi^9}{9!} + ...\right) - \left(\frac{(\pi^3}{3!} + \frac{\pi^7}{7!} + ... \right)\right)^n}{n^n\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
$$\le \dfrac{(\sin \pi)^n}{<denominator>}$$
$$\le 0 !!$$
This means the assumption about the positivity of
$$\frac{\sin{x_1}\sin{x_2}\sin{x_3}...\sin{x_n}}{\sin({x_1 + x_2})\sin({x_2 + x_3})\sin({x_3 + x_4})...\sin({x_{n} + x_{1}})}$$
Seems Incorrect?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 7,
"answer_id": 2
} |
Wish to evaluate $\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)} dx$ We want to evaluate this integral,$\newcommand{\cosec}{\operatorname{cosec}}$
$$\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)}\mathrm dx\tag1$$
$$\int_{0}^{1}[x^4\cosec(\pi x)-3x^2\cosec(\pi x)+2x\cosec(\pi x)]\mathrm dx$$
Apply integration by parts, first part:
$$\int x^4\cosec(\pi x)\mathrm dx=-\frac{x^4}{\pi}\ln[\cot(\pi x/2)]-\frac{4}{\pi}\int x^3\ln[\cot(\pi x/2)]\mathrm dx$$ again we apply integration by parts to this remaining integral,
$$\int x^3\ln[\cot(\pi x/2)]\mathrm dx$$
Integrate this part it is very difficult $\int \ln[\cot(\pi x/2)]\mathrm dx$
I am stuck at this point. I guess there must be another method to deal with this integral $(1)$.
I am looking forward to see how you would go about to evaluates $(1)$
| Here is a reasonably nice way:
Claim:
\begin{align*}
\int_0^1 x^{n-2}\frac{x^2-x}{\sin\pi x}\,\mathrm{d}x
&=
\begin{cases}
93\frac{\zeta(5)}{\pi^5}-21\frac{\zeta(3)}{\pi^3} & n=4\\
-\frac72\frac{\zeta(3)}{\pi^3} & n=3\\
-7\frac{\zeta(3)}{\pi^3} & n=2
\end{cases}.
\end{align*}
Proof (sketch):
Define
$$
f_n(a):=\int_0^1 x^n e^{ax}\,\mathrm{d}x=\frac{\mathrm{d}^n}{\mathrm{d}a^n}\left(\frac{e^a-1}{a}\right)
$$
so
$$
\label{eq:fn1}
f_n(a)=
\begin{cases}
\frac{1+e^a(a-1)}{a^2}&n=1\\
\frac{-2+e^a(a^2-2a+2)}{a^3}&n=2\\
\frac{6+e^a(a^3-3a+6a-6)}{a^4}&n=3\\
\frac{-24+e^a(a^4-4a^3+12a^2-24a+24)}{a^5}&n=4
\end{cases}\tag{2}
$$
Hence
\begin{align*}
I_n&:=\int_0^1 x^{n-2}\frac{x^2-x}{\sin\pi x}\,\mathrm{d}x\\
&=2i\int_0^1 x^{n-2}\frac{x^2-x}{e^{i\pi x}-e^{-i\pi x}}\,\mathrm{d}x\\
&=2i\int_0^1 (x^n-x^{n-1})\sum_{k=0}^\infty e^{-i(2k+1)\pi x}\,\mathrm{d}x\\
&=2i\sum_{k=0}^\infty f_n(-i(2k+1)\pi)-f_{n-1}(-i(2k+1)\pi)\\
\end{align*}
and simplify using \eqref{eq:fn1}, noting $e^{i\pi(2k+1)}=-1$ and
$$
\sum_{k=0}^\infty\frac{1}{(2k+1)^n}=\left(1-\frac1{2^n}\right)\zeta(n).\qquad\square
$$
Now take suitable linear combinations to finish off.
The motivation behind using $\frac{x^2-x}{\sin\pi x}$ is to make sure the integrand stays finite at $x=0,1$ so we don't have to worry about convergence issues.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How to compute $\sum_{k=1}^{\infty}{\frac{1}{k^2+2k}}$? To whom this may concern,
i am struggling with partial sum formulas. I don't really get why you would need to perform a partial fraction decomposition or how you know that you have to.
I started by getting trying to get a grasp of the series:$$\sum_{k=1}^{n}{\frac{1}{k^2+2k}}=\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\cdots + \frac{1}{n(n+2)}$$
But the partial sum formula is, according to WolframAlpha, $\sum_{k=1}^{n}{\frac{1}{k^2+2k}}=\frac{3n^2+5n}{4(n+1)(n+2)}$ HOW?????? I beg you to be as detailed as possible, i really want to understand WHY
| We have the partial fractions decomposition
$$\frac{1}{k^2+2k}=\frac{1}{2k}-\frac{1}{2k+4}.$$
If you write out the first few terms, you will notice a lot of cancellation:
$$\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{6}-\frac{1}{10}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+\left(\frac{1}{10}-\frac{1}{14}\right)+\cdots+\left(\frac{1}{2n-4}-\frac{1}{2n}\right)+\left(\frac{1}{2n-2}-\frac{1}{2n+2}\right)+\left(\frac{1}{2n}-\frac{1}{2n+4}\right).$$
When the dust clears, all you are left with is
$$\frac{1}{2}+\frac{1}{4}-\frac{1}{2n+2}-\frac{1}{2n+4}$$
which is the same as your partial sum formula.
This also makes it clear why the limit is $3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The integral $\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi,~ a, b \in \Re ?$ This integral
$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$
looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of proving or disproving this?
| One way:
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2+p^2)(x^2+q^2) \Rightarrow p=(a+c)/2,q=(a-c)/2, c=\sqrt{a^2-4b^2}.$$
Then $$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx=
\frac{2a^2}{p^2-q^2}[p\tan^{-1}(x/p)-q \tan^{-1}(x/q)]_{0}^{\infty}=\frac{a^2\pi}{p+q}=a\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$ Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$ where $r-\frac{s}{2}$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $\frac{\pi r^2}. {2}$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$). In conclusion the resultant equation is (with $r=x$ and $s=y$):
$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$
I plugged this formula into desmos and recieved a straight line with a gradient close to $\frac{360000457}{302000000}$ but I wish to know the exact value.
| Without loss of generality, we can assume $r=1$.
For convenience, position the circles vertically, with centers on the $y$-axis at the points $(0,h)$ and $(0,-h)$.
Then the equation of the lower circle is
$$x^2+(y+h)^2=1$$
Solving for $y$, and noting that
$$h=1-\frac{s}{2}$$
we get that the upper half of the lower circle has the equation
$$y=-1+\frac{s}{2}+\sqrt{1-x^2}$$
Letting $(-a,0)$ and $(a,0)$ be the points where the circles interect, we get
$$a={\small{\frac{1}{2}}}\sqrt{s(4-s)}$$
hence the area of the region where the disks overlap is
\begin{align*}
&
4
\int_0^
{
{\Large{
\frac{1}{2}\sqrt{s(4-s)}
}}
}
\!
\left(
-1+\frac{s}{2}+\sqrt{1-x^2}
\right)
\;dx\\[6pt]
=\;&
2\sin^{-1}\left({\small{\frac{1}{2}}}\sqrt{s(4-s)}\right)-{\small{\frac{1}{2}}}(2-s)\sqrt{s(4-s)}\\[4pt]
\end{align*}
Solving the equation
$$
2\sin^{-1}\left({\small{\frac{1}{2}}}\sqrt{s(4-s)}\right)-{\small{\frac{1}{2}}}(2-s)\sqrt{s(4-s)}
=\pi/2$$
numerically yields
$$s\approx 1.192054493$$
which matches the ratio you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that this multivariable limit exists or not? I was a given a problem to solve (if the limit exists):
$$ \lim_{(x,y)\to(0,0)} \cos\left(\frac{x^3-y^3}{x^2+y^2}\right) .$$
My Approach:
Approaching along the path $y=0$, yield the limit to be $1$. Similarly, going along the path $y=mx$ yields the limit to be $1$. Thus, it seems that if the limit does exist, the limiting value has to be $1$.
To show that it exists, I decided to use the $\epsilon - \delta$ approach (I am unable to think of an algebraic approach).
It will be sufficient to check whether the limit:
$$ \lim_{(x,y)\to(0,0)} \frac{x^3-y^3}{x^2+y^2} $$ exists or not and if it does exist it should be equal to $0$.
We know that for some $\epsilon \in \mathbb{R}$, $$ 0< \left|\frac{x^3-y^3}{x^2+y^2}\right| < \epsilon,$$ we must be able to choose a $\delta$ such that $0<\sqrt{x^2+y^2} < \delta$ for the limit to exist.
Now,
$$ \left|\frac{x^3-y^3}{x^2+y^2}\right| = \left| \frac{(x-y)(x^2+y^2+xy)}{x^2+y^2} \right| < \frac{|(x-y)||xy|}{x^2+y^2} \leq \frac{|(x-y)|}{2} < |x-y| = \sqrt{x^2+y^2-2xy}.$$
We have 2 cases: Case I: $xy>0$ This will result in ,
$$ \left|\frac{x^3-y^3}{x^2+y^2}\right| < \sqrt{x^2+y^2}< \epsilon$$ Here $\delta$ can be chosen as $\epsilon$.
Case II: $xy<0$. This is where I am not able to proceed further.
Is there any alternative approach? Or can the case II part be proven?
| The limit exists and equals $1$. This follows from the squeeze theorem and the inequalities
$$
0\leq\frac{|x^3-y^3|}{x^2+y^2}\leq |x|+|y|.
$$
Indeed, we have that
$$
|x^3-y^3|\leq |x|^3+|y|^3\leq (|x|+|y|)(x^2+y^2),
$$
where the second inequality follows by expanding out the terms on the right and observing that the extra two cross terms are non-negative.
| {
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"url": "https://math.stackexchange.com/questions/3268275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding max and min from polynomial equation
Let $a, b \in \mathbb R$ and $b \geqslant 0$. Given that $x^4 + ax^2 + b = 0$ has $2$ real solutions, find maximum for $|a-b|$ and minimum for $a+2b$.
My solution :
$$x^4+ax^2+b=0 \implies (x^2-c)(x^2-d) = 0$$
there is only 1 case which is possible,that is, both $c$ and $d$ are positive
so $x^2=c$ and $x^2 = d$
if it has 2 real solutions, then $c = d$ ($x=\pm\sqrt c$ and $ x=\pm\sqrt d$). So we get $a=-2c$ , $b=c^2$
$$|a-b|=|-2c-c^2|=c^2+2c+1-1=(c+1)^2-1$$
but I can't find max.
$$a+2b = 2c^2-2c = 2(c-\frac{1}{2})^2-\frac{1}{2}\implies \min = -\frac{1}{2}$$
But I think my method is wrong because I can't solve for $\max$.
| If $b=0$ and $a\geq 0$ then $x^4+ax^2+b=0$ has just one real root, i.e. $x=0$.
If $b=0$ and $a< 0$ then $x^4+ax^2+b=0$ has three real roots, i.e. $x=0, \pm\sqrt{-a}$
If $b>0$ then $x^4+ax^2+b=0$ has two distinct real roots iff $z^2+az+b=0$ has one positive root and a negative root, which is impossible because $b>0$, or two coincident positive roots that is when $\Delta=a^2-4b=0$ and $a<0$.
Hence for $b=a^2/4$ with $a<0$ we have that $|a-b|=|a-a^2/4|$ has supremum $+\infty$, whereas $a+2b=a+a^2/2=\frac{(a+1)^2-1}{2}$ has minimum value $-1/2$, attained for $a=-1$.
| {
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"url": "https://math.stackexchange.com/questions/3270473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the number) and when this polynomial is divided by $x^2 $, it leaves $2x + 4$ (again, not sure about the number). From the given conditions, if this polynomial is divided by $(x-1)x^2$, what would be the remainder?
The solution as far as I figured out is this:
first, from the division of $(x-1)^2$, I got that $f(1) = 3$
in the same way from division of $x^2$, I got $f(0) = 4.$
I can write the polynomial as follows:
$f(x) = (x-1)(x)(x) g(x) + ax^2 +bx +c$
$ax^2 + bx + c$ is the remainder. And to find $a,b,c$, I can use the conditions above, so I got $c = 4$ by substituting $x = 0,$ and I got $a+b+4 = 3$ by substituting $x = 1.$
This leaves $a + b = -1,$ and I can't figure out how to continue; please help.
Edit : I made a mistake $f(1)$ should be equal to $4$ and $a+b+c = 4$
| We have
$$f(x)=(x-1)^2q_1(x)+x+3$$
$$f'(x)=2(x-1)q_1(x)+(x-1)^2q_1'(x)+1$$
Where for $x=1$ we have $f(1)=4$ and $f'(1)=1$
Then from
$$f(x)=x^2q_2(x)+2x+4$$
$$f'(x)=2xq_2(x)+x^2q_2'(x)+2$$
Where for $x=0$ we have $f(0)=4$ and $f'(0)=2$
Now from $$f(x)=x^2(x-1)q_3(x)+ax^2+bx+c$$ and
$$f'(x)=x^2q_3(x)+2x(x-1)q_3(x)+(x-1)x^2q_3'(x)+2ax+b$$
When we substitute the values $x=0$ and $x=1$ in $f$ and $f'$ we get
$f(0)=c=4$ and $f(1)=a+b+4=4$ $$a+b=0$$
$f'(0)=b=2$ from this we have $a=-2$.
Thus the remainder is $r(x)=-2x^2+2x+4$
| {
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"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| Let $x+2=y$, then:
$\sqrt {3y+1}=\sqrt y +1$
squaring both sides we get:
$3y+1=y+1+2\sqrt y$
⇒ $y=\sqrt y $ ⇒ $y^2-y=y(y-1)=0$
that gives:
$y=x+2=0$ ⇒ $x=-2$
$y-1=0$ ⇒ $y=x+2=1$ ⇒ $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 7
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Solving a Fractional Equation Involving a Logarithm I may be being stupid right now, so I've come to Stack to see if this elementary algebra holds up.
Suppose I have the equation $$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$$
My chosen way to solve this would be to cross multiply and expand brackets, solve the quadratic and get the value of $x$.
However, a student I am helping got this by saying $\ln x = 1$ gives $x = \mathrm{e}$ and at $x= \mathrm{e}$, the denominator $(1+ \ln x)^2 = 4$.
Hence $x= \mathrm{e}$.
Is this approach always correct or is it just luck here?
In general if I have $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$, can I solve it by finding the common solutions of $f(x) = m(x)$ and $g(x) = n(x)$?
[Edit: clearly not because if I have $\frac{x}{x+2} = \frac{1}{x+3}$, then $x= 1$ and $x+2 = x + 3$ don't give you anything..., so why does it work in this case?]
| $\frac ab = \frac cd$ does not mean $a = c$ and $b = d$.
Here are two examples:
$\frac 13 = \frac 26$ but $1 \ne 2$ and $3\ne 6$. Likewise if $x =2$ we would have $\frac x{x+4} = \frac 13$. That does not mean $x=1$ and $x+4 =3$.
One (not advised) way of doing this is to notice that $\frac ab = \frac {\beta a}{\beta b}$ for all $\beta$ (excepts $0$) so there must be some $\alpha$ where we have $\frac ab = \frac {\alpha a}{\alpha c} = \frac cd$ and $a = \alpha c$ and $b=\alpha d$. We just don't have any reason to assume that $\alpha$ is equal to $1$.
So when $\frac 13 = \frac 26$ we have $1 = [\frac 12]*2$ and $3 = [\frac 12]*6$. And for $\frac x{x+4} = \frac 13$ we have $x = \alpha$ and $x+ 4 = 3\alpha$. (We can solve this as $\alpha = x$ we have $x +4 = 3x$ so $2x =4$ and $x=2$...)
But the better way is to do:
$\frac ab = \frac cd \implies ad = bc$. For our examples we get $\frac 13 = \frac 26 \implies 1*6 = 2*3$ which it does and $\frac x{x+4} =\frac 13 \implies 3x = x+4\implies x = 2$.
So in your problem you have:
$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$ so that means
$4\frac {\ln x} = (1 + \ln x)^2$. It does not mean $\ln x = 1$ or that $(1+\ln x)^2 = 4$. It does mean that there is some constant $\alpha$ so taht $\ln x = \alpha$ and $1+\ln x)^2 = \alpha *4$ but .... that's just the same thing with a complicated and unnecessary variable thrown in.
To do this I'd just replace $\ln x$ with $w$ and have $4w = (1 + w)^2$ and solve for $w$.
$4w = 1 + 2w + w^2$
$0 = 1-2w + w^2$
$(w-1)^2 = 0$
$w-1 = 0$
$w = 1$
So $\ln x = 1$ so $x = e$.
| {
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"url": "https://math.stackexchange.com/questions/3275438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$ IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$
I only got that smallest side should be $3x%$, how do i solve this?
| The law of cosines gives
$$28^2=(3x)^2+(5x)^2-2(3x)(5x)\cos\frac{2\pi}3=34x^2+15x^2=49x^2$$
Thus $28=7x$, $x=4$ and the smallest side of the triangle is $3x=12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{n\rightarrow \infty} \left[ \frac{1}{(n+1)(n+2)} + \frac{2}{(n+2)(n+4)} + \cdots + \frac{n}{6n^2} \right]$
Evaluate: $$\lim_{n\rightarrow \infty} \left[ \dfrac{1}{(n+1)(n+2)} + \dfrac{2}{(n+2)(n+4)} + \cdots + \dfrac{n}{6n^2} \right]$$
$\text{My Attempt:}$ breaking down the summation series into: $$\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}$$.
Further breaking down into two separate series: $$\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}=\sum_{r=1}^{n} \dfrac{(n+2r)-(n+r)}{(n+r)(n+2r)}$$
This will reduce to give: $$\sum_{r=1}^{n} \dfrac{1}{n+r} - \sum_{r=1}^{n}\dfrac{1}{n+2r}$$
Now, applying limits to the sum: $$\lim_{n\rightarrow\infty}\left[\sum_{r=1}^{n} \dfrac{1}{n+r} - \sum_{r=1}^{n} \dfrac{1}{n+2r}\right]$$
Taking $n$ common in denominator and converting to Definite integral taking $\dfrac{r}{n}=x$ this reduces to: $$\int_{0}^{1}\dfrac{\text{dx}}{1+x}-\int_{0}^{1} \dfrac{\text{dx}}{1+2x}$$
Edit: Solving this we will get the answer as $\ln\left(\dfrac{2}{\sqrt{3}}\right)$.
I have had committed an error in the evaluation of the 2nd integral as Mr. Robert Z has pointed out below.
| After revising my answer, I got: as $n\to \infty$,
$$\begin{align}\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}&=\frac{1}{n}\sum_{r=1}^{n} \dfrac{\frac{r}{n}}{(1+\frac{r}{n})(1+2\frac{r}{n})}\\
&\to \int_0^1 \frac{x}{(1+x)(1+2x)}\,dx\\
&=\int_0^1\left(\frac{1}{1+x}-\frac{1}{1+2x}\right)\,dx\\
&=\left[\ln\left(\frac{1+x}{(1+2x)^{1/2}}\right)\right]_0^1\\
&= \ln(2/\sqrt{3}).
\end{align}$$
So, your approach is correct, and there is just a minor error in the integration of $1/(1+2x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficult Recurrence Problem Find all integers $n\geq 3$ for which there exist real numbers $a_{1}, a_{2},...,a_{n+2}$ satisfying $a_{n+1}=a_{1}$ , $a_{n+2}=a_{2}$ and:
$a_{i}a_{i+1}+1=a_{i+2}$
for $i=1,2,..,n$
| Hint.
Solving backwards we have for
$$
\begin{cases}
n=2\to \{2,-1,-1,2\}\\
n=5\to \{2,-1,-1,2,\frac 12,-1,2\}\\
n=8\to \{2,-1,-1,2,\frac 12,-1,2,\frac 12,-1,2\}\\
\vdots\\
n=2+3k\to \{2,-1,\underbrace{-1,2,\frac 12}_{k},-1,2\}
\end{cases}
$$
NOTE
The real solutions arise from
$$
\left\{
\begin{array}{rcl}
a_1&=&\frac{a_2-1}{a_1} \\
a_2&=&1-\frac{a_1}{a_2-1} \\
\end{array}
\right.
$$
the other possibilities
$$
\left\{
\begin{array}{c}
a_1=1-\frac{a_1}{a_2-1} \\
a_2=\frac{a_1}{a_1-a_2+1} \\
\end{array}
\right.
\ \ \text{and}\ \ \left\{
\begin{array}{c}
a_1=\frac{a_1}{a_1-a_2+1} \\
a_2=\frac{a_2-1}{a_1} \\
\end{array}
\right.
$$
have no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area between $x^3$ and $\sqrt[3]{x}$ I'm having some difficult to find the intersection between $x^3$ and $\sqrt[3]{x}$ for calculate the area between them. Could someone help me?
| \begin{equation}
x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0
\end{equation}
And so we have $x = 0$ or $x^8 - 1 = 0$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus,
\begin{equation}
x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0
\end{equation}
Assuming you are seeking Real Solutions only we see that $x^4 + 1 = 0$ has no solutions. For $x^4 - 1$ employ the same identity:
\begin{equation}
(x^4 + 1)(x^4 - 1) = 0 \rightarrow (x^4 + 1)(x^2 + 1)(x^2 - 1) = 0
\end{equation}
As with $x^4 + 1 = 0$ having no Real Solutions we also observe that $x^2 + 1 = 0$ also have no Real Solutions. Thus the only remaining Real Solutions as those that satisfy:
\begin{equation}
x^2 - 1 = 0 \rightarrow x = \pm 1
\end{equation}
As such, the three intersection points of $x^3$ and $x^{\frac{1}{3}}$ occur at $x = -1, 0, 1$
| {
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"url": "https://math.stackexchange.com/questions/3279437",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solve the equation $|2x^2+x-1|=|x^2+4x+1|$ Find the sum of all the solutions of the equation $|2x^2+x-1|=|x^2+4x+1|$
Though I tried to solve it in desmos.com and getting the requisite answer but while solving it manually it is getting very lengthy.
I tried to construct the two parabola and mirror image the region below y axis but still getting it is getting complicated.
Is there any easy method to solve it and get the sum of all the solutions ?
| $$
|2x^2+x-1|=|x^2+4x+1|\\
(2x^2+x-1)^2=(x^2+4x+1)^2\\
(2x^2+x-1)^2-(x^2+4x+1)^2=0\\
[(2x^2+x-1)+(x^2+4x+1)]\cdot[(2x^2+x-1)-(x^2+4x+1)]=0\\
(3x^2+5x)\cdot(x^2-3x-2)=0\\
x\cdot(3x+5)\cdot(x^2-3x-2)=0\\
Solving\space for\space all\space cases,\space we\space get:\\
x=0\\
x=-\frac{5}{3}\\
x=\frac{3\pm\sqrt{17}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maclaurin series of $\tan(x+x^2)$ to order 3 I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$?
Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.
| Alternatively:
$$\begin{align}y&=\tan{(x+x^2)} \Rightarrow \color{red}{y(0)=0}\\
\arctan y&=x+x^2\\
\frac1{1+y^2}\cdot y'&=1+2x\\
y'&=(1+\color{red}y^2)(1+2x) \Rightarrow \color{blue}{y'(0)=1}\\
y''&=2\color{red}yy'(1+2x)+2(1+\color{red}y^2)\Rightarrow \color{green}{y''(0)=2}\\
y'''&=2\color{blue}{y'}^2(1+2x)+2\color{red}yy''(1+2x)+8\color{red}yy' \Rightarrow \color{brown}{y'''(0)=2}\\
y&=\tan(x+x^2)=\color{red}{y(0)}+\frac{\color{blue}{y'(0)}}{1!}x+\frac{\color{green}{y''(0)}}{2!}x^2+\frac{\color{brown}{y'''(0)}}{3!}x^3+O(x^4)=\\
&=x+x^2+\frac{x^3}{3}+O(x^4).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tanh^{-1}\frac{2a\sin x}{1+a^2}dx$, with $I(0)=0$
$$ J’(a) = \int_0^{\frac\pi{3}}\frac{2(1-a^2)\sin x}{4a^2\cos^2x+(1-a^2)^2}dx
=\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a}
$$
Then
\begin{align}
I
& =J(1) =J(0)+\int_0^1 J’(a)da = \int_0^1\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a} da\\
&=\int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^3\to a}{\left.\frac{\tan^{-1}a^3}{a}\right)}da=\left(1-\frac13\right) \int_0^1\frac{\tan^{-1}a}{a}da=\frac23G
\end{align}
| Note that $2\operatorname{arctanh} x=\ln\left(\frac{1+x}{1-x}\right)$. So by letting $\frac{1-\sin x}{1+\sin x}=y\,$ in the integral we get: $$\int_0^\frac{\pi}{3} \operatorname{arctanh}(\sin x)dx=-\int_0^{(2-\sqrt 3)^2} \frac{\ln y}{\sqrt y(1+y)}dy\overset{\sqrt y=\tan x}=-\int_0^\frac{\pi}{12}\ln(\tan x)=\frac23G$$
See here for the last integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of the infinite series given I've this problem to solve, but no answer key, I tried to do some development, but i didn't succeed :( Could anyone help me, please?
$$\sum_{k=2}^\infty \sum_{l=1}^{k-1} \frac{1}{k^2l^2}=\frac{1}{2^21^2}+\frac{1}{3^21^2}+\frac{1}{3^22^2}+\frac{1}{4^21^2}+\frac{1}{4^22^2}+\frac{1}{4^23^2}+\cdots$$
Hint: Use that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$ and $$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$$
Thanks a lot!
| Denoting the given double sum by $S$, we note that
$$\left(\sum_{n=1}^\infty\frac1{n^2}\right)^2=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2n^2}=\sum_{n=1}^\infty\frac1{n^4}+2S$$
Thus
$$\left(\frac{\pi^2}6\right)^2=\frac{\pi^4}{90}+2S$$
$$S=\frac{\pi^4}{120}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for $(5x-1)(x-3)<0$ The inequality $(5x-1)(x-3)<0$ is true when
$(5x-1)<0$ and $(x-3)>0$
or
$(5x-1)>0$ and $(x-3)<0$.
If I solve for $x $ in the first scenario, $x < \frac{1}{5}$ and $x > 3$ which is wrong. But if I solve for $x$ in the second scenario, $x > \frac{1}{5}$
and $x < 3$ which is correct.
Why is such kind of contradiction occurring in the first scenario?
| $f(x)=(5x-1)(x-3)$ polynomial of degree 2 has 2 zeroes: $5x-1=0$, and $x=3$.
1) $\lim_{x \rightarrow \pm \infty}f(x)= \infty$.
Hence for
$5x-1<0$, or $x <1/5$: $f(x)>0$, and
$x>3$: $f(x) >0$.
2) For $x \in (1/5,3)$:
$f(x)= 5(x-1/5)(x-3) <0$ , since $(x-1/5)>0$ and $(x-3) <0$.
| {
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"url": "https://math.stackexchange.com/questions/3284981",
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"source": "stackexchange",
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$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}+\dots$ Consider the series:
$$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\frac{1}{41}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\dots$$
The numerator of each term is $1$, and the denominators are the natural numbers that contain the digit $1$.
If the denominators, instead, are the natural numbers omitting numbers containing the digit $9$ is called Kempner series, it converges to $22.92067661926415034816...$.
What is the value of $A$?
Does $A$ have an exact form?
| $$A\gt\sum_{k=1}^\infty \frac1{10k+1}\gt\sum_{k=1}^\infty\frac1{11k}=\infty $$
Hence the sum determining $A$ does not converge and $A$ does not exist.
| {
"language": "en",
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"source": "stackexchange",
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Find cumulative distribution function of random variable given probability generating function
Let $f_c(x) := c \cdot \ln\left(1 - \frac{x}{2}\right)$.
Find for what $c \in \mathbb{R}$ $f_c$ is a probability generating function.
I wrote
$$
f_c(x)
= \sum_{k \in \mathbb{N}_0} -c \cdot \frac{x^k}{2^k \cdot k!}.
$$
For $f_c$ to be probability generating function, $p(k) := \frac{-c}{2^k \cdot k!}$ has to be a PMF, i.e
\begin{equation*}
1
\overset{!}{=} \sum_{k \in \mathbb{N}_0} p(k)
= \sum_{k \in \mathbb{N}_0} \frac{-c}{2^k \cdot k!}
= -c \cdot \sqrt{e},
\end{equation*}
yielding $c = - e^{-\frac{1}{2}}$.
Let $X_c$ be a random variable with the distribution obtained from $f_c$.
Calculate $E[X_c]$ and $P(X_c = 1)$.
Similarly to the above I wrote
\begin{align*}
E[X_c]
& = \sum_{k \in \mathbb{N}_0} k p(k)
= e^{-\frac{1}{2}} \sum_{k \in \mathbb{N}_0} \frac{k}{2^k \cdot k!} \\
& = \frac{e^{-\frac{1}{2}}}{2} \sum_{k \in \mathbb{N}_0} \frac{1}{2^{k - 1} \cdot (k - 1)!}
= \frac{e^{-\frac{1}{2}}}{2} e^{\frac{1}{2}}
= \frac{1}{2}.
\end{align*}
and
\begin{equation*}
P(X_c = 1)
= p(1)
= \frac{e^{-\frac{1}{2}}}{2^1 \cdot 1!}
= \frac{1}{2 \sqrt{e}}.
\end{equation*}
Find the cumulative distribution function of $X_c$.
I wrote
\begin{align*}
P(X_c \le x)
& = \sum_{k = 0}^{\lfloor x \rfloor} p(k)
= e^{-\frac{1}{2}} \cdot \sum_{k = 0}^{\lfloor x \rfloor} \frac{1}{2^k \cdot k!}
\end{align*}
Questions
*
*Is there a closed form for the above result (not utilising the incomplete gamma function, which we did not cover in class?
*Are my results correct? Is there an easier way to arrive at them?
| Using standard probability notation you have probability generating function $G_X(z) = c \cdot \ln(1-\tfrac{z}{2})$ and the corresponding random variable is $X$. From the sums in your question, I will assume that $X$ is a non-negative integer random variable. With this condition, one of the properties of the PGF is that:
$$\mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!}.$$
These probability mass values must sum to one, which imposes a constraint on the PGF, which we can use to find the constant $c$. Now, substituting your specified PGF you get:
$$\begin{equation} \begin{aligned}
G_X^{(k)}(z)
= c \cdot \Big( \frac{d}{dz} \Big)^k \ln(1 - z/2)
&= \begin{cases}
c \cdot \ln(1 - z/2) & & & \text{for } k=0, \\[6pt]
- c \cdot (k-1)! \cdot ( 2-z )^{-k} & & & \text{for } k>0, \\[6pt]
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
which gives the mass function:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X=k) = \frac{G_X^{(k)}(0)}{k!}
&= \begin{cases}
0 & & & \text{for } k=0, \\[6pt]
- (c/k) \cdot 2^{-k} & & & \text{for } k>0. \\[6pt]
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
The constraint equation therefore reduces to:
$$\begin{equation} \begin{aligned}
1 = \sum_{k=0}^\infty \mathbb{P}(X=k)
= -c \cdot \sum_{k=1}^\infty \frac{1}{k \cdot 2^k}
= -c \cdot \ln (2). \\[6pt]
\end{aligned} \end{equation}$$
From this constraint we have $c=-1 / \ln (2)$ so your PGF is:
$$G_X(z) = - \frac{\ln(1-\tfrac{z}{2})}{\ln(2)} = 1 - \frac{\ln(2-z)}{\ln(2)},$$
and the corresponding probability mass function is:
$$p_X(k) \equiv \mathbb{P}(X=k) = \frac{1}{\ln (2)} \cdot \frac{1}{k \cdot 2^k} \quad \quad \quad \text{for all } k \in \mathbb{N}.$$
This is the logarithmic series distribution with probability parameter $p=\tfrac{1}{2}$. The CDF, moments, and other properties, can be derived from the mass function.
| {
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"timestamp": "2023-03-29T00:00:00",
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On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
\dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!} &1\\
\dfrac{1}{n!} & \dfrac{1}{(n-1)!} & \dfrac{1}{(n-2)!} & \dfrac{1}{(n-3)!} & \cdots & \dfrac{1}{1!}
\end{pmatrix}
=\dfrac{1}{n!}.
$$
| HINT.-By property of determinants, we lower the order of n to (n-1) as follows
$$\Delta_n=\det\begin{pmatrix}
1 &1 &0 &0 &\cdots &0\\
a_2 &1 &1 &0 &\cdots &0\\
a_3 &a_2 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 &1\\
a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
$$\Delta_n=\det\begin{pmatrix}
1 &0 &0 &0 &\cdots &0\\
a_2 &1-a_2 &1 &0 &\cdots &0\\
a_3 &a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_n &a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
$$\Delta_n=\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\
a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
On the other hand one has $\Delta_n=\dfrac{1}{n!}=\dfrac{1}{n}\dfrac{1}{(n-1)!}=\dfrac 1n\Delta_{n-1}$ so we have to prove that
$$\det\begin{pmatrix}1-a_2 &1 &0 &\cdots &0\\
a_2-a_3 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
a_{n-2}-a_{n-1} &a_{n-3} &\cdots &1 &1\\
a_{n-1}-a_n &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}=\dfrac 1n\Delta_{n-1}$$ $$\dfrac 1n\Delta_{n-1}=\det\begin{pmatrix}\dfrac 1n &1 &0 &\cdots &0\\
\dfrac{1}{n}a_2 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\\
\dfrac 1na_{n-2} &a_{n-3} &\cdots &1 &1\\
\dfrac 1na_{n-1} &a_{n-2} &a_{n-3} &\cdots &1
\end{pmatrix}$$
Note that the columns in these two last determinants are all equal excepting the first.
Can you apply comfortably induction now to prove that both two last determinants are equal?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorization in a proof of induction I have to prove the following:
$1+3^3+ ... + (2n+1)^3=(n+1)^2(2n^2+4n+1)$ by induction.
My try:
Base case, $n=1$:
$1+3^3=(2)^2(2\cdot1^2 + 4\cdot1 + 1)$, which is true.
By inductive hypothesis, assume $n=k$:
$1 + 3^3 + ... + (2k+1)^3=(k+1)^2 (2k^2 + 4k + 1)$
For $n=k+1$
$1 + 3^3 + ... + (2k+1)^3 + (2k+3)^3 = (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
Using the inductive hypothesis we need to prove that
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
and here I have the problem, because I don't know how to manipulate any of the sides of the equation to prove this. I tried in both sides but I can't find the way. One of my last attempts ended here:
$(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)(k+3) + 1 )$
Which is true by wolfram alpha.
PS: I'm aware that there are a lot of papers and information about this proof but I'm not searching another way or something like that, because I saw a lot of posts about proving this statement via induction, but all of them used the last term as (2n-1)^3, and I need to prove it when it is (2n+1)^3, and the final expression is a little bit different. I just need help factoring my last step.
Any corrections on the inductive steps that i followed are appreciated too.
| This may not be the most elegant solution,
but $(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
is true because both sides equal $2k^4+16k^3+47k^2+60k+28$.
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:
Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$
We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so this is only possible when $x^3+y^3=4$ which no $x,y \in \mathbb{Z}$ satisfy.
Now I don't know if that helps at all solve the second bit. We have $z^6 = x^3+y^3-3$. I know that $x^3+y^3 = (x+y)(x^2-xy+y^2)$.
| ►If $z\equiv0\pmod7$ we have $x^3+y^3\equiv 3\pmod7$ but by Fermat's little theorem $w^6\equiv1\pmod7$ if
$w\not \equiv 0\pmod7$ then $w^3\equiv\pm1\pmod7$ so either $\pm1\pm1\not\equiv3\pmod7$ or $0\pm1\not\equiv\pmod7$.
►If $z\not\equiv0\pmod7$ we have the equation $x^3+y^3\equiv 1+3=4\pmod7$ and similar reasoning show the impossibility.
►If $x^3+y^3=z^6+3$ have solution this solution must satisfy the congruence modulo $7$ so there is not solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $100$ the only square number of the form $a^b+b^a$? Conjecture: $100$ is the only square number of the form $a^b+b^a$ for integers $b>a>1$.
In other words, $(a,b)=(2,6)$ is the only solution. Can we prove/disprove this?
Observations:
*
*The solution mentioned should not come as a surprise, since $2^6+6^2=8^2+6^2=10^2$ is a (non-primitive) Pythagorean triple. It is possible to show that $2^b+b^2$ has no other solutions. See Remark 1.
*In the general case where $a$ is a power of $2$; that is, $a=2^d$ for some positive integer $d$, a similar approach can be followed. See Remark 2.
*We can eliminate some values of $b$ when $a=5^r,6^r$, since no matter the value of $r$, we have $a\equiv5,6\pmod{10}$ respectively.
*PARI/GP code: If the conjecture is true it should only ever print 2 6.
sqfun(a,b)={for(i=2,a,for(j=2,b,if(issquare(i^j+j^i)==1,print(i," ",j))));}
Remark 1: Suppose that there is a positive integer $b$ that admits $2^b+b^2=t^2$ for some integer $t$. Then we can write the equation as $$2^b=(t+b)(t-b)\implies\begin{cases}t+b=2^c\\t-b=2^{b-c}\end{cases}$$ for some positive integer $c>\dfrac b2$. Subtracting the two equations yields $$2b=2^c-2^{b-c}\implies b=2^{b-c-1}(2^{2c-b}-1).$$ If $b$ is odd, it cannot have a factor of $2$, forcing $b-c-1=0\implies t=b+2$ and substituting gives $2^b+b^2=(b+2)^2$, or $2^b=4(b+1)$. No solutions exist.
If $b$ is even, then $b=2k$ for some positive integer $k$, so we must have $$\begin{cases}2^k=s(m^2-n^2)\\2k=2mns\end{cases}$$ for some integers $m,n,s$, so that $2^{mns}=s(m^2-n^2)$. Without loss of generality, let $m>n>0$ and $s>0$. [Servaes: If $mns\ge4$ then $2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2)$, so the only solutions with even $b$ are $b=4,6$, and the first case does not yield a square.]
Remark 2: If $b$ is odd, it boils down to the equation $$2^{db}=4\left(b^{2^{d-1}}+1\right)\implies 2^{db-2}-1=b^{2^{d-1}}.$$ [Haran: For $db-2>1$, the LHS is congruent to $3\pmod4$, and since the RHS is a square for $d>1$, we reach a contradiction unless \begin{cases}d=1\implies a=2\quad\text{case covered above}\\db-2=1\implies1=b^{2^{d-1}}\implies 2^{d-1}=0\end{cases} which is impossible.]
If $b$ is even, then $b=2k$ for some positive integer $k$, and the Pythagorean triplet forces $$\begin{cases}2^{dk}=s(m^2-n^2)\\(2k)^{2^{d-1}}=2mns.\end{cases}$$ [Servaes: From the first equation, all three factors on the RHS are powers of two, so $$\begin{cases}m+n=2^u\\m-n=2^v\end{cases}\implies\begin{cases}m=2^{v-1}(2^{u-v}+1)\\n=2^{v-1}(2^{u-v}-1)\end{cases}$$ with $u>v>0$. Since $m$ and $n$ are coprime, we have $v=1$. Plugging this into the first equation yields $$2^{dk}=s(m-n)(m+n)=2^{u+1}s\implies s=2^{dk-u-1}.$$ Substituting this into the second equation yields $$(2k)^{2^{d-1}}=2mns=2(2^{u-1}+1)(2^{u-1}-1)s=(2^{2u-2}-1)2^{dk-u}$$ which is impossible; if we let $k=2^w\ell$ with $\ell$ odd then this implies $\ell^{2^{d-1}}=2^{2u-2}-1$ which by Catalan's conjecture/Mihailescu's theorem is impossible if $d>1$. Note that $u>v$ hence $u\geq2$.]
| $\newcommand{\eps}{\varepsilon}$
$\newcommand{\rad}{\mathrm{rad}}$
At least, under the abc conjecture, there can be only finitely many pairs $(a,b)$ with $b>a>1$ coprime such that $a^b+b^a$ is a square.
As a reminder, the conjecture says that to any $\eps>0$ there corresponds some $K_\eps>0$ such that whenever $u,v$, and $w$ are coprime positive integers with $u+v=w$, one has $\rad(uvw)>K_\eps w^{1-\eps}$. Here $\rad(z)$ is the product of all primes dividing $z$ (thus, for instance, $\rad(8)=2$, $\rad(9)=3$, $\rad(10)=10$, $\rad(11)=11$, and $\rad(12)=6$).
Suppose now that $a^b+b^a=c^2$ with coprime integers $b>a\ge 3$ and $c>0$ (the case $a=2$ is resolved above). Applying the abc conjecture with $u=a^b$, $v=b^a$, $w=c^2$, and $\eps=0.05$, and making the key observation $\rad(a^bb^ac^2)\le abc$, we conclude that
$$ Kc^{2\cdot 0.95} < abc $$
with an absolute constant $K>0$. At the same time, we have $c^2>a^b$ and $c^2>b^a$, implying $a<c^{2/b}$ and $b<c^{2/a}$, respectively. Consequently,
$$ Kc^{1.9} < c^{(2/b)+(2/a)+1}, $$
showing that either $\frac1b+\frac1a>0.4$, or $Kc^{0.1}<1$. Clearly, there are only finitely many values of $c$ satisfying the latter condition, and to each value corresponds finitely many pairs $(a,b)$. On the other hand, since $\frac1b+\frac13\ge\frac1b+\frac1a>0.4$ implies $b<15$, there are only finitely many pairs $(a,b)$ satisfying the former condition. Thus, the total number of exceptional pairs $(a,b)$ is also finite.
| {
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Finding the solution to $xy'' +2y' +xy=0$ around $x_{0}=0$using the method of Frobenius. We know that the solution of this ODE is like:
$$ y=\sum_{n=0}^{\infty}C_nx^{n+r}$$
Them derivative $y$ and $y'$.
$$y'=\sum_{n=0}^{\infty}(n+r)C_nx^{n+r-1}$$
$$y''=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_nx^{n+r-2}$$
Replace $y$ , $y'$ and $y''$ in the ODE.
$$\sum_{n=0}^{\infty}(n+r-1)(n+r)C_nx^{n+r-1}+\sum_{n=0}^{\infty}2(n+r)C_nx^{n+r-1} +\sum_{n=0}^{\infty}C_nx^{n+r+1}=0 $$
Now in the first and second summations if $k=n-1$ and in the third summation if $k=n+1$. We have:
$$\sum_{k=-1}^{\infty}(k+r)(k+r+1)C_{k+1}x^{k+r}+\sum_{n=-1}^{\infty}2(k+r+1)C_{k+1}x^{k+r} +\sum_{n=1}^{\infty}C_{k-1}x^{k+r}=0 $$
Now I separed the $k=-1$ and $k=0$ in the first and second summantions.
$$r(r-1)C_0x^{r-1}+(r+1)(r)C_1x^{r} + 2rC_0x^{r-1} +2(r+1)C_1x^r + \sum_{k=1}^{\infty}[(k+r)(k+r+1)C_{k+1}x^{k+r}+2(k+r+1)C_{k+1}x^{k+r}+C_{k-1}x^{k+r}]=0$$
After that I equalized the right side with the left and I have this 2 equation to find $r$:
$$(r^2 +r)C_0=0$$
and
$$(r^2+3r+2)C_{1}=0$$
If you solve the quadratic equations you have:
$$r_1=0$$
$$r_{2,3}=-1$$
$$r_4=-2$$
And that's all that I did please help me.
| $$xy'' +2y' +xy=0\qquad ......(1)$$
$~x=0~$ is a regular singular point of equation $(1)$.
So the equation admits of a Frobenius series of the form $$y=\sum_{n=0}^{\infty}C_n~x^{n+r},\qquad C_0\neq 0 \qquad ..........(2)$$
which converges for all $~x~$.
From $(2)$,
$$y'(x)=\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1};\qquad \qquad y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}\qquad .....(3)$$
Substituting $(2)$ and $(3)$ in $(1)$ we get,
$$x~\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}+2~\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1}+x~\sum_{n=0}^{\infty}C_n~x^{n+r}=0$$
$$\implies \sum_{n=0}^{\infty}(n+r)~(n+r+1)~C_n~x^{n+r-1}~+~\sum_{n=0}^{\infty}C_n~x^{n+r+1}=0\qquad .....(4)$$
Lowest power of $~x~$ in equation $(4)$ is $~{r-1}~$, so coefficient of $~x^{r-1}~=0$ gives the indicial equation $~r~(r+1)~=0\implies r=0,~-1$
From equation $(4)$ we have the following recursive formula,
$$(n+r+1)~(n+r+2)~C_{n+1}~+~C_{n-1}=0$$
$$\implies C_{n+1}=-\frac{1}{(n+r+1)~(n+r+2)}~C_{n-1}\qquad ........(5)$$
From $(5)$ we have $C_1=C_3=C_5=\cdots =0$
$C_2=-\frac{1}{(r+2)~(r+3)}~C_{0}$
$C_4=-\frac{1}{(r+4)~(r+5)}~C_{2}=\frac{1}{(r+2)~(r+3)~(r+4)~(r+5)}~C_{0}$
$\cdots$
Therefore
$$y(x)=C_0~x^r \left[1-\frac{1}{(r+2)~(r+3)}~x^2+\frac{1}{(r+2)~(r+3)~(r+4)~(r+5)}~x^4-\cdots\right]$$
For $~r=0~$, $$y_1(x)=C_0~ \left[1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots\right]$$
$$\implies y_1(x)=C_0~ \left[1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots\right]$$
Since $~0-(-1)=1,~$ an integer so the other independent solution of equation $(1)$ is $$y_2(x)=\left[\frac{\partial y}{\partial r}\right]_{r=0}$$
$$\implies y_2(x)=y_1(x)~\log x~+~C_0~\left[\frac{5}{36}~x^2+\cdots\right]$$
General solution is $$y(x)=A~y_1(x)~+~B~y_2(x)\qquad \text{where $~A,~B~$are constants.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\prod_{n>1} \frac{1}{1-\frac{1}{n^s}}$ or $-\sum_{n=2}^{\infty} \log(1-\frac{1}{n^s} )$ I know
\begin{align}
\prod_{p~is~ prime} \frac{1}{1-\frac{1}{p^2}} = \zeta(2) = \frac{\pi^2}{6}
\end{align}
which has a convergent number.
actually I can even generalized this to
\begin{align}
\prod_{p ~is~ prime} \frac{1}{1-\frac{1}{p^s}} = \zeta(s)
\end{align}
For $s>1$ [consider $s\in \mathbb{R}$] we know zeta function converges, so this has a convergent number.
How about generalization to arbitrary integers? [i.e., I want to replace $p$ with arbitrary integer $n$.]
For example $s=2$, we have
\begin{align}
\prod_{n>1} \frac{1}{1-\frac{1}{n^2}}
\end{align}
taking log we need to show
\begin{align}
- \sum_{n=2} \log\left(1-\frac{1}{n^2} \right)
\end{align}
is convergent or not.
simply by telescope method I can see this value converges to $\log(2)$, that means $ \prod_{n>1} \frac{1}{1-\frac{1}{n^2}} = 2$.
Now consider $s>1$.
\begin{align}
-\sum_{n>1} \log\left(1-\frac{1}{n^s}\right)
\end{align}
This is convergent from comparison test.
Simply take $a_n = -\log(1-\frac{1}{n^s})$ and $b_n = \frac{1}{n^s}$, then
\begin{align}
\lim_{n\rightarrow \infty} \frac{a_n}{b_n} = \lim_{x\rightarrow 0} \frac{-\log(1-x)}{x} = 1 >0
\end{align}
and since $\sum_{n=1}^{\infty} b_n = \zeta(s)$ is convergent for $s>1$, $\sum_{n=2}^{\infty} a_n$ also converges.
What I want to obtain is the value of such convergent series, first i tried telescope method, but it seems difficult even for $s=3$.
Is there a way to compute exact value of those products?
How and what is the values of those products?
| Let
$$P_s=\prod_{n=2}^\infty \frac{1}{1-n^{-s}}$$
Using a CAS, there are some nice expressions such as
$$\left(
\begin{array}{cc}
s & P_s \\
2 & 2 \\
3 & 3 \pi \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right) \\
4 & 4 \pi \text{csch}(\pi ) \\
6 & 6 \pi ^2 \text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right)
\end{array}
\right)$$
$P_5$ and the other ones are quite ugly.
Have a look here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\sin(x) - \sin(y))(\cos(x) - \cos(y)) = -\frac{1}{6} $$
$$ \sin(x)\cos(x) + \sin(y) \cos(y) - \sin(x+y) = \frac{1}{6}$$
thus
$$ \sin(x+y) =\sin(x)\cos(x) + \sin(y) \cos(y) - \frac{1}{6} $$
Next if I do a sum and organize and squaring we get:
$$ (\sin(x) + \cos(x))^{2} + (\sin(y) + \cos(y))^{2} - 2 (\sin(x) + \cos(x)) (\sin(y) + \cos(y)) = \frac{1}{36} $$
$$ 2 + 2 ( \sin(x) \cos(x) + \sin(y) \cos(y)) - 2 \left( \sin(x+y) + \cos(x-y) \right) = \frac{1}{36} $$
I have no idea after this.
Another method is I suspect that we have to solve for $\sin(x), \cos(x), \sin(y), \cos(y)$ instead of directly finding $\sin(x+y)$ algebraically.
| If you start with the Sum-to-Product Identities, you can rewrite the given equations as
$$2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=-\frac{1}{3} \quad \text{and} \quad -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=\frac{1}{2}.$$
Dividing the second equation by the first one, we'll cancel out $\sin\left(\frac{x-y}{2}\right)$, and as a result we'll find the value of $\tan\left(\frac{x+y}{2}\right)$. Then to finish it up, use one of the half-angle substitutions:
$$\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the constant $k$ from the determinant
Given: $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.
If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.
| It is also simple to compute the posted determinant, $D$ for short, thus ignoring the information on the special form of the result. We are subtracting at the first step the first column from the other two.
$$
\begin{aligned}
D &=
\begin{vmatrix}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{vmatrix}
\\[2mm]
&=
\begin{vmatrix}
(b+c)^2 & a^2- (b+c)^2 & a^2 - (b+c)^2 \\
b^2 & (c+a)^2-b^2 & 0 \\
c^2 & 0 & (a+b)^2-c^2
\end{vmatrix}
\\[2mm]
&=
\begin{vmatrix}
(b+c)^2 & a- b-c & a- b-c \\
b^2 & c+a-b & 0 \\
c^2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
2bc & -2c & - 2b \\
b^2 & c+a-b & 0 \\
c^2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
2bc-bc-bc & -2c & - 2b \\
b^2+\frac b2(c+a-b)+\frac c2\cdot 0 & c+a-b & 0 \\
c^2+\frac c2\cdot 0+\frac c2(a+b-c) & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^2
\\[2mm]
&=
\begin{vmatrix}
0 & -2c & - 2b \\
\frac b2 & c+a-b & 0 \\
\frac c2 & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^3
\\[2mm]
&=
-\begin{vmatrix}
0 & c & b \\
b & c+a-b & 0 \\
c & 0 & a+b-c
\end{vmatrix}
\cdot(a+b+c)^3
\\[2mm]
&=2abc(a+b+c)^3\ .
\end{aligned}
$$
(At the last step, using Sarrus, only two out of six terms survive, we can factor $bc$, and $-(a+b-c)-(c+a-b)=-2a$ leads immediately to the result. No step was omitted, symmetry was used when possible.)
Computer check:
sage: var('a,b,c');
sage: A = matrix(3,3, [(b+c)^2, a^2, a^2, b^2, (a+c)^2, b^2, c^2, c^2, (a+b)^2] )
sage: A.det().factor()
2*(a + b + c)^3*a*b*c
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$
what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2x(1+\sin^2 2x))$
$16\sin^4 (2x)+68\sin^3 2x+34\sin^2 2x+68\sin 2x+68\sin^2 4x+16=0$
How do i solve it Help me please
| Wrong expansion jacky
$$16\sin^42x+68\sin^32x+102\sin^22x+68\sin 2x+16=0$$
$$16(\sin 2x+\csc 2x)^2+68(\sin 2x+\csc 2x)+70=0$$
$$\sin 2x +\csc 2x=-15.0479,-1.9521$$
Rejecting $-1.9521$ (Why)
Thus, $$\sin^22x+(15.0479)\sin 2x+1=0$$
Both the roots of the quadratic are negative and since their product is $1$ only one root will lie on the interval $(-1,0)$
So there should be $4$ solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work because the derivative of $2x^2-4x+3$ is not a multiple of $(x-1)^{-2}$ and the derivative of $x-1$ is not a multiple of $2x^2-4x+3$.
| Note that
$$2x \color{red}{-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}} \color{green}{-2+K} = 2x \color{red}{-\frac{1}{x-1}} \color{green}{+C}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integration of $x^2\cdot\frac{x\sec^2x+\tan x}{(x\tan x+1)^2}$ Integrate
$$\int x^2\cdot\dfrac{x\sec^2x+\tan x}{(x\tan x+1)^2}dx$$
So what is did is integration by parts taking $x^2$ as $u$ and the other part as $v$ . Now I got to use it again which then eventually leads to (integral of $\dfrac1{x\tan x+1}dx $). Can someone help me?
| Let $u:=x\sin x+\cos x$ so $u^\prime=x\cos x$. You want to integrate$$\begin{align}\frac{x^2(x+\sin x\cos x)}{u^2}&=\frac{x^2(x\cos^2x+u\sin x)}{u^2}\\&=\frac{u^\prime x^2\cos x}{u^2}+\frac{x^2\sin x}{u}\\&=\frac{d}{dx}\left(-\frac{x^2\cos x}{u}\right)+\frac{(x^2\cos x)^\prime+x^2\sin x}{u}\\&=\frac{d}{dx}\left(-\frac{x^2\cos x}{u}\right)+\frac{2u^\prime}{u}\\&=\frac{d}{dx}\left(-\frac{x^2\cos x}{u}+2\ln u\right).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| Assume $$\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$$
where $a,b,c,d$ are all rational.
Then
$$a+\sqrt b=c+2\sqrt{cd}+d.$$
By identification
$$\begin{cases}a=c+d,\\b=4cd.\end{cases}$$
From this,
$$4c^2+4cd-4ac=4c^2-4ac+b=0.$$
This is a quadratic equation in $c$, with roots
$$c=\frac{a\pm\sqrt{a^2-b}}2.$$
So for a rational solution to exist, $a^2-b$ must be a perfect rational square.
With $a=2,b=3$,
$$\sqrt{2+\sqrt3}=\sqrt{\frac12}+\sqrt{\frac32}=\frac{\sqrt2+\sqrt6}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Proof Correction: $\cos ' (x) = -2 \sin (x)$ Preliminary Information
Let $$A(y) = \cfrac{y \sqrt{1 - y^2}}{2} + \int_{y}^{1} 1\sqrt{1 - t^2} dt \ \ \ \text{on $[-1, 1]$}$$
Moreover $$ A'(y) = \dfrac{-1}{2\sqrt{1 - y^2}}$$
Define $\cos x$ as $A(\cos x) = \cfrac{x}{2}$ and $\sin x = \sqrt{1 - \cos ^{2} x}$
Problem: Find $\cos' (x)$.
$A(\cos x) = \dfrac{x}{2}$; since $A$ is decreasing it is one-one and $A^{-1}$ a function. Consider that $A^{-1}\left( \dfrac{x}{2} \right) = \cos x$. Let us find then $(A^{-1})' \left (\dfrac{x}{2} \right)$.
$\begin{equation}
\begin{split}
(A^{-1})' \left (\dfrac{x}{2} \right) &= \dfrac{1}{A'(A^{-1} \left (\dfrac{x}{2} \right)} \\
&= -2 \sin x
\end{split}
\end{equation}$
Textbook's Proof
| You forgot to apply the chain rule and multiply by $(x/2)'$ when differentiating $(A^{-1})(x/2)$.
| {
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"url": "https://math.stackexchange.com/questions/3304965",
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"answer_id": 0
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The infinite sum $\sum_{n=1}^\infty (-1)^{n+1} \frac{2n-1}{n^2-n+1}$ My (rather old) version of Mathematica cannot compute
$$\sum_{n=0}^\infty (-1)^{n+1}\frac{2n-1}{n^2-n+1}$$
other than re-writing it as a hypergeometric function as follows:
$$\mbox{HypergeometricPFQ}\left[\left\{\frac{1}{2}, 1, -(-1)^{1/3}, (-1)^{2/3}\right\}, \left\{-\frac{1}{2}, (-1)^{1/3}, -(-1)^{2/3}\right\}, -1\right]$$
But strangely enough, my old Mathematica has no trouble finding
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2n-1}{n^2-n+1}=\frac{2\pi}{e^{\pi \sqrt{3}/2}+e^{-\pi \sqrt{3}/2}}$$
SAGE (with CoCalc) can't compute either version. I can't either. How does the old Mathematica do it?
| Using partial fraction decomposition
$$\frac{2 n-1}{(n-a) (n-b)}=\frac{2 a-1}{(a-b) (n-a)}+\frac{1-2 b}{(a-b) (n-b)}$$ So
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2 a-1}{(a-b) (n-a)}=\frac{(2 a-1) \Phi (-1,1,1-a)}{a-b}$$
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1-2 b}{(a-b) (n-b)}=-\frac{(2 b-1) \Phi (-1,1,1-b)}{a-b}$$ where appears the Hurwitz-Lerch transcendent function.
Now, using $a=(-1)^{1/3}=\frac{1+i \sqrt{3}}{2}$ and $b=-(-1)^{2/3}=\frac{1-i \sqrt{3}}{2}$ gives for the sum
$$S=\sum_{n=1}^\infty (-1)^{n+1} \frac{2n-1}{n^2-n+1}$$
$$S=\frac{1}{2} \left(-\psi \left(\frac{1}{4}-\frac{i \sqrt{3}}{4}\right)+\psi
\left(\frac{3}{4}-\frac{i \sqrt{3}}{4}\right)-\psi
\left(\frac{1}{4}+\frac{i \sqrt{3}}{4}\right)+\psi
\left(\frac{3}{4}+\frac{i \sqrt{3}}{4}\right)\right)$$ which reduces to
$$S=\pi\, \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right)$$
| {
"language": "en",
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Number of ways to chose 6 courses out of 15 (different) courses We got 3 English courses, 6 Chinese courses, 6 Spanish courses.
Each course is different.
In how many was can we choose 6 courses such that we must chose atleast 1 course from each topic (English, Chinese, Spanish)
If possible, I would like to see a solution without the inclusion - exclusion principle.
edit : I got to $\frac{6*6*3}{3!}*\binom{12}{3}\;$ where $\frac{6*6*3}{3!}$ represents picking a course from each topic (we must choose atleast 1 from each topic), then dividing by the number of permutations. then we are left with 3 courses to choose from 12, so I multiplied by $\binom{12}{3}\;$
| The generating function for the number of $k$ courses such that there at least one course from each category is the following:
$$\Bigg( \binom{3}{1}x + \binom{3}{2}x^2 + \binom{3}{3}x^3 \Bigg) \Bigg(\binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 +\binom{6}{5}x^5 + \binom{6}{6}x^6 \Bigg)^2$$
Thus the number of ways to choose $6$ courses with at least one from each category is the coefficient of $x^6$ in the generating function above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding product of trigonometric values (cos/sin) using complex number/roots of unity Using the ninth root of unity of 1, show that $cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$.
Here is my solution.
Let $\omega=cos\theta+isin\theta$,
As $\omega^9=1$,
$$cos9\theta+isin9\theta=1$$
so $$9\theta=2nπ {(n\in Z)}$$
$$\theta=\frac{2nπ}{9} {(n\in Z)}$$
$$\theta=\frac{\pm2π}{9}, \frac{\pm4π}{9}, \frac{\pm6π}{9}, \frac{\pm8π}{9},$$
$$\omega=cis\frac{\pm2π}{9}, cis\frac{\pm4π}{9}, cis\frac{\pm6π}{9}, cis\frac{\pm8π}{9},$$
therefore $$\omega^9-1=0$$
$$(\omega-1)(1+\omega +\omega^2 +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8)=0$$
and because $\omega$ is not real,
therefore $$1+\omega +\omega^2 +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8=0$$
through factorisation,
$$(\omega - cis\frac{2π}{9})(\omega - cis\frac{-2π}{9})...(\omega - cis\frac{8π}{9})(\omega - cis\frac{-8π}{9})=1+\omega +\omega^2 +...+\omega^7+\omega^8$$
$$(\omega^2 - 2cos\frac{2π}{9}\omega +1)...(\omega^2 - 2cos\frac{8π}{9}\omega +1)=1+\omega +\omega^2 +...+\omega^7+\omega^8$$
divide both sides by $\omega^4$,
$$(\omega - 2cos\frac{2π}{9} +\frac{1}{\omega})...(\omega - 2cos\frac{8π}{9} +\frac{1}{\omega})=\frac{1}{\omega^4}+\frac{1}{\omega^3}+... +\omega^3+\omega^4$$
Let $x=\omega+\frac{1}{\omega}$,
$$\omega^2+\frac{1}{\omega^2}=(\omega+\frac{1}{\omega})^2-2=x^2-2$$
similarly,
$$\omega^3+\frac{1}{\omega^3}=x^3-3x$$
$$\omega^4+\frac{1}{\omega^4}=x^4-4(x^2-2)-6=x^4-4x^2+2$$
therefore, substituting in
$$x^4+x^3-3x^2-2x+1=(x - 2cos\frac{2π}{9})...(x - 2cos\frac{8π}{9})$$
$$2^4cos\frac{2π}{9}cos\frac{4π}{9}cos\frac{6π}{9}cos\frac{8π}{9}=1$$
as $cos\frac{8π}{9}=-cos\frac{π}{9}$ and $cos\frac{6π}{9}=-\frac{1}{2}$
$$cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$$
I believe that I might have over-complicated this question. Any other thoughts?
| Root of Unity Approach
Using
$$
\omega=\cos\left(\frac\pi9\right)+i\sin\left(\frac\pi9\right)
$$
we get $\omega^9=-1$
$$
\begin{align}
\overbrace{\left(\omega-\omega^8\right)}^{2\cos\left(\frac\pi9\right)}\overbrace{\left(\omega^2-\omega^7\right)}^{2\cos\left(\frac{2\pi}9\right)}\overbrace{\left(\omega^4-\omega^5\right)}^{2\cos\left(\frac{4\pi}9\right)}
&=\omega^7-\omega^8-\omega^{12}+\omega^{13}-\omega^{14}+\omega^{15}+\omega^{19}-\omega^{20}\\
&=\omega^7-\omega^8+\omega^3-\omega^4+\omega^5-\omega^6+\omega^1-\omega^2\\[3pt]
&=1
\end{align}
$$
Trigonometric Identity Approach
We can also use $\sin(2x)=2\sin(x)\cos(x)$,
$$
\begin{align}
\sin\left(\frac{\pi}9\right)\cos\left(\frac{\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right)
&=\frac12\sin\left(\frac{2\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right)\\
&=\frac14\sin\left(\frac{4\pi}9\right)\cos\left(\frac{4\pi}9\right)\\
&=\frac18\sin\left(\frac{8\pi}9\right)
\end{align}
$$
Now divide by $\sin\left(\frac{\pi}9\right)=\sin\left(\frac{8\pi}9\right)$.
| {
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"url": "https://math.stackexchange.com/questions/3308303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
| You were close. However, you simplified $\frac{2b}b$ into $b$ and $\frac{2a}a$ into $a$ in your final step, which is not right. Correct that mistake, and you should have your solution.
On a less serious note, I would do an extra step here, for clarity:
$$
3^{2a/b} + 4^{2b/a} = (3^a)^{2/b} + (4^b)^{2/a} = 4^{2a/a} + 3^{2b/b}
$$
Also I prefer to not write fractions in exponents unless necessary. I think they're somewhat ugly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$$ holds.
My work. I proved this inequality, but my proof is ugly (it is necessary to check by brute force whether the inequality holds for $n=1,2,3,...,15$). I hope that there is nice proof of this inequality. Michael Rozenberg wrote a very nice solution to a similar problem ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). I think this inequality has a similar proof, but I can’t prove in a similar way. I will write as I proved the inequality. Let $S_n= \sum \limits_{k=1}^n \frac{1}{k} $. Then $$ \sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2n-S_n $$ and $$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3+\frac{3}{n+1}+3S_n$$ We need to prove that $$3S_n^2-S_n \left( 2n^2+8n+3-\frac{3}{n+1}\right)+\frac{n^4}{8}+7n^2+3n-3+\frac{3}{n+1} \ge 0$$ To prove this inequality, I found discriminant of the quadratic polynomial and used the fact that $S_n \le n$. It was possible to prove that the inequality holds for all $n \ge 16$.
| Here is a way, which is motivated by the excellent approach of @MichaelRozenberg's. Let us write $\displaystyle a_k = \frac { k}{k+1}, b_k = 2k-1$. We then have
\begin{align*}
S =& \sum_{i\le n} a_i b_i\sum_{j\le n}\frac{b_j}{a_j} \\
=& \sum_{i,j\le n} \frac{a_i}{a_j}b_ib_j \\
=& \frac 1 2 \sum_{i,j\le n} \left(\frac {a_i}{a_j} + \frac{a_j}{a_i}\right) b_ib_j.
\end{align*} We then observe
\begin{align*}
\frac {a_i}{a_j} + \frac{a_j}{a_i} \le& \begin{cases} 2\quad \text {if}\quad i=j=1\\
\frac{5}{2}\quad\text{if}\quad i=1, j> 1\ \text{ or }\ i>1, j=1\\
\frac{13}{6}\quad \text{if} \quad i > 1, j> 1
\end{cases}
\end{align*} thus
\begin{align*}
S \le &\frac 1 2 \left[2 + 2\cdot\frac {5}{2}(n^2-1) + \frac {13}{6}(n^2-1)^2\right]\\
<& \frac 9 8 n^4
\end{align*} for all $n\ge 1$ except for $n=2$. We can check $n=2$ case manually...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The natural numbers $a, b, c$, formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ satisfy $a^2 + b = c$
Given that the natural numbers $a, b, c$ are formed by the same $n$
digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any
$n \geq 2$, find the digits $x, y, z$ such that $a^2 + b = c$
Greetings, I was doing this question above and I couldn't figure out how to do.
Here's my progress so far:
The relation $a^2 + b = c$ means $$(xxx...x)^2 + (yy..yy) = (zzz...zz)$$ where the $x\text{ and }y$ are $n$ in number and the $z$ is in $2n$ in number. This simplifies to $$x^2(11···1)^2 +y(11···1) = z(11···1)$$
and I couldn't figure out further. :(
By some trial and error, I found $(333)^2 + 222 = 111,111$.
Any help would be appreciated. I tried but couldn't find out even a similar question somewhere else. Please inform if this question has been asked before.
Thank You
| Let
$$f(n) = \sum_{i=0}^{n-1} 10^i \tag{1}\label{eq1}$$
Thus, as you've shown,
$$x^2 f^2(n) + yf(n) = zf(2n) \implies f(n)\left(x^2 f(n) + y\right) = zf(2n) \tag{2}\label{eq2}$$
Also, note that
$$f(2n) = 10^n f(n) + f(n) = f(n)\left(10^n + 1\right) \tag{3}\label{eq3}$$
Thus, substituting that into \eqref{eq2} and dividing by the common factor of $f(n)$ gives
$$x^2f(n) + y = z(10^n + 1) \tag{4}\label{eq4}$$
First, there's the trivial case of $x = y = z = 0$. The rest of this solution will assume that $z \gt 0$. Next, note that
$$f(n) = \frac{10^n - 1}{9} \tag{5}\label{eq5}$$
Thus, substituting \eqref{eq5} into \eqref{eq4} gives
$$\begin{equation}\begin{aligned}
x^2\left(\frac{10^n - 1}{9}\right) + y & = z(10^n + 1) \\
x^2(10^n - 1) + 9y & = 9z(10^n + 1) \\
x^2(10^n) - x^2 + 9y & = 9z(10^n) + 9z \\
(x^2 - 9z)(10^n) & = x^2 - 9y + 9z
\end{aligned}\end{equation}\tag{6}\label{eq6}$$
The RHS is less than $100$ (update: actually, it's less than $1000$, but it can be $100$; see the end for more details) and greater than $-10$, so for $n \ge 2$, this means that $x^2 - 9z = x^2 - 9y + 9z = 0$. Note $x^2 - 9z = 0$ occurs only for $x = 3, z = 1$ and $x = 6, z = 4$. This gives, from $x^2 - 9y + 9z = 0$, that $y = 2$ for the first part and $y = 8$ for the second part.
Another way to see this is that the RHS of \eqref{eq4}, as $z$ is a digit, in base $10$, would be $z$, followed by $n - 1$ zeros, and then $z$ again. On the LHS, $y$ is just a single digit. This means for $n \ge 2$, the digits in $x^2f(n)$ must basically all "disappear" when $y$ is added, i.e., $x^2f(n)$ must be just slightly less (i.e., $\lt 9$) than a power of $10$. Checking the squares of $x$ multiplied by $11$, you can see this only occurs if $x = 3$, so $x^2 = 9$ (with it being $1$ less than a power of $10$), or $x = 6$ so $x^2 = 36$ (with it being $4$ less than a power of $10$). For $x = 3$, this means $z = 1$, which requires that $y = 2$, while for $x = 6$, this means $z = 4$ and $y = 8$.
Note the second solution comes from the first one by multiplying by $4$ in \eqref{eq4}. You can see this changes $x = 3$ to $x = 6$, $y = 2$ to $y = 8$ and $z = 1$ to $z = 4$.
In conclusion, there are $3$ solutions, with the sets of digits $(x,y,z)$ being $(0,0,0)$, $(3,2,1)$ and $(6,8,4)$.
Update: As I saw from the answer by fleablood, I made a mistake in my handling of \eqref{eq6}. The RHS of it is definitely less than $1000$, so what I wrote above is true for $n \ge 3$. However, the RHS can be equal to $100$, which means that for $n = 2$, there's also the solution $(8,3,7)$.
| {
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"url": "https://math.stackexchange.com/questions/3312967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum possible perimeter of a right triangle
The ratio between the perimeter of a right triangle and its area is 2:3. The sides of the triangle are integers. Find the maximum possible perimeter of the triangle.
If the sides of the triangle are $A$, $B$ and $C$ (the hypotenuse), I have deduced that:
$$A+B+\sqrt{A^2+B^2}=\frac{AB}{3}$$
I am stuck here. Any hints? From the above I know that $A^2+B^2$ should be a square and that $AB$ should be divisible by 3.
| I worked out the opposite, i.e. area:perimeter ratio. Michael Rozenberg's answer is correct.
$$ R=\frac{area}{perimeter}=\frac{AB}{2}/P=\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}
\qquad\qquad\qquad\qquad\qquad $$
\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\quad\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1)
\end{equation}
The lower limit insures that $k\in \mathbb{N}$ and the upper limit ensures that $m> k$.
$$R=3/2\implies \lceil\sqrt{8(1.5)}\rceil=4\le m \le (2(1.5)+1)=4 \quad\land\quad m\in\{ 4\}\implies k\in\{ 3,1\}$$
$$F(4,3)=(7,24,25)\quad\land\quad \frac{84}{56}=3/2\qquad\qquad f(4,1)=(15,8,17)\quad\land\quad \frac{60}{40}=3/2$$
We can see that the first triple has the maximum perimeter of $56$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Chord $x+y=1$ Subtends $45^{\circ}$ at the centre If the Chord $x+y=1$ Subtends $45^{\circ}$ at the centre of the circle $x^2+y^2=a^2$, Find $a$
Method $1$:
We have radius=$a$
If $O$ is the centre and chord be $AB$ we have
$$\angle AOB =45$$
If $OM$ is drawn perpendicular to the chord we have:
$$\angle AOM=22.5$$
In $\Delta AOM$ we have
$$cos(22.5)=\frac{OM}{a}$$
$OM$ is perpendicular distance from $(0,0)$ to $x+y=1$ which is
$$OM=\left|\frac{0+0-1}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$$
Hence we get $$a=\sqrt{2-\sqrt{2}}$$
Method $2.$
By Homogenization of line and curve we have:
$$x^+y^2=a^2(x+y)^2$$
$$(a^2-1)x^2+2xya^2+(a^2-1)y^2=0$$
This pair of straight lines subtends angle of $45^{\circ}$
We have formula that:
$$\tan (\theta)=\left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$$
We have
$A=a^2-1=B$ and $H=a^2$
We get
$$1=\left|\frac{2\sqrt{a^4-(a^2-1)^2}}{2(a^2-1)}\right|$$
Simplifying we get
$$a^4-4a^2+2=0$$
Giving
$$a=\sqrt{2\pm \sqrt{2}}$$
Why in first Method, i could not get second answer?
| It's not that you couldn't get the second answer of the second method $\sqrt{2+\sqrt2}$ in the first method. It's that $\sqrt{2+\sqrt2}$ is wrong – drawing the associated circle and line and measuring the subtended angle gives you well over $90^\circ$. There is only one (positive) solution $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$
Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t.
$$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$
Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$
This question is terrifying that I even don't know how to start it. I've found different ways such as multiplying the numerator and denominator by $a$ and substitute, but it is not useful (maybe?). Anyway, please comment or answer if you solve it or having clues that may help solve this question.
| Recently, I find a solution which is quite simple. Check it out!
Claim $1$: $a^2\ge3ab\rightarrow a\ge3b, b\le\dfrac{a}{3}$
Prove:
$$a^2-3ab=\left(x+y+z\right)^2-3\left(xy+yz+zx\right)=x^2+y^2+z^2-xy-yz-zx=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{2}\ge0 \\ \therefore a^2\ge 3ab$$
Claim $2$: $a\ge3\sqrt{3}$
Prove:
By AM-GM Inequality,
$$\left(\dfrac{x+y+z}{3}\right)^3\ge xyz\\\left(\dfrac{a}{3}\right)^3\ge a\\\dfrac{a^3}{27}\ge a\\a^2\ge27\\\therefore a\ge3\sqrt{3}$$
Then comes the solution:
$\dfrac{a^2+6b+1}{a^2+a}\le\dfrac{a^2+2a+1}{a^2+a}=\dfrac{\left(a+1\right)^2}{a\left(a+1\right)}=\dfrac{a+1}{a}=1+\dfrac{1}{a}\le1+\dfrac{1}{3\sqrt{3}}=\boxed{1+\dfrac{\sqrt{3}}{9}}$
Equality holds when $x=y=z=b=\sqrt{3},a=3\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$
Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}}{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}} $$
$$ = \frac{ (\sqrt[3]{16} - 1) [(3 + \sqrt[3]{4}) - \sqrt[3]{2}]}{ (3 + \sqrt[3]{4})^{2} - 2^{2/3}} $$
$$ = \frac{ 3 \sqrt[3]{16} - 3\sqrt[3]{4} + \sqrt[3]{2} + 1}{ (9 + 5 \sqrt[3]{4} + \sqrt[3]{16}) } $$
From here on I don't know how to continue.
I can let $a = \sqrt[3]{2}$, but still cannot do anything.
| With the same notation as in the other answer, i.e. $x = \sqrt[3] 2$, noting that $x^3+1=3,$ you can write your quantity as
\begin{eqnarray}
\frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\
&=&x-1.
\end{eqnarray}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is longer? Length of $2^n$ or number of $0$ in $n!$? What is longer? Length of $2^n$ or number of $0$ at the end of $n!$ ?
Assume that $n$ is really big, big number...
Solution
Look at terms of $2^k$:
$$2,4,8,16,32,64,128,256,512,1024,2048,4096... $$
It can be seen that
$$ \lfloor\frac{n}{3} \rfloor \ge length(2^n)\ge \lfloor\frac{n}{4} \rfloor $$
Now, lets look at $n!$. We know that $0$ is created at the end iif we use $2$ and $5$ or $10$. Number of $2$ is greater than number of $5$ so we can count number of $5$.
$$\mbox{number of 0 in } n! = \sum_{k \ge 0} \lfloor\frac{n}{10^k} \rfloor + \sum_{k \ge 0} \lfloor \frac{n}{5^k} \rfloor $$
$$ (n-1)\left( \sum_{k \ge 0} \frac{1}{10^k} + \sum_{k \ge 0} \frac{1}{5^k} \right)\ge \mbox{number of 0 in } n! \ge n \left( \sum_{k \ge 0} \frac{1}{10^k} + \sum_{k \ge 0} \frac{1}{5^k} \right) $$
$$ \mbox{number of 0 in } n! \approx \frac{13}{36} \cdot n \approx 0.361 \cdot n \ge \frac{1}{3} \cdot n $$
So the longer is number of $0$ in $n!$. Have I done this exercise correctly?
update
$$ (n-1)\left( \sum_{k \ge 0} \frac{1}{5^k} \right)\ge \mbox{number of 0 in } n! \ge n \left( \sum_{k \ge 0} \frac{1}{5^k} \right) $$
$$ \mbox{number of 0 in } n! \le 1/4 $$ so length of 2^n is greater that trailing $0$ in $n!$
| Denote by $l(n)$ the number of digits of $2^n$ and by $t(n)$ the number of trailing zeros of $n!$. We aim to show that $t(n) < l(n)$, at least for large $n$.
Per the update, the number of trailing zeros in $n!$ is the number of factors of $5$ in the prime factorization of $n!$, and it follows from the definition of factorial that this is
$$t(n) = \sum_{k \geq 0} \left\lfloor\frac{n}{5^k}\right\rfloor .$$
The definition of $\lfloor \,\cdot\, \rfloor$ implies that $\lfloor m \rfloor \leq m$, so (as nearly observed in the question statement) $$t(n) \leq \sum_{k \geq 0} \frac{n}{5^k} = n \sum_{k \geq 0} \frac{1}{5^k} = \frac{1}{4} n .$$
To make rigorous the rest of the argument, we should formalize the assertion "It can be seen that..." as an inequality and then use it to show that $t(n) < l(n)$.
Hint To do this, we can use a familiar formula $$l(n) = \lfloor \log_{10} (2^n) \rfloor + 1 = \lfloor n \log_{10} 2 \rfloor + 1 .$$
The definition of $\lfloor \,\cdot\, \rfloor$ implies that $m \leq \lfloor m \rfloor + 1$, immediately giving the lower bound
$$(\log_{10} 2) n \leq l(n) .$$
Comparing our bounds, we can see that if we can show $$\phantom{(\ast)} \qquad \frac{1}{4} < \log_{10} 2 , \qquad (\ast)$$ then we will have $$t(n) \leq \frac{1}{4} n < (\log_{10} 2) n \leq l(n)$$ as desired (in fact, for all $n > 0$, not just large $n$). But some algebraic rearrangement shows that $(\ast)$ is equivalent to $10 < 16$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Explicit description of quotient ring of $\mathbb{Z}[x]$ I am studying for a prelim and I stumbled on this problem:
Describe explicitly the elements in the quotient ring $\dfrac{\mathbb{Z}[x]}{(3,x^3-x+1)}$. First of all I don't see why the ideal $(3,x^3-x+1)$ is a maximal ideal in $\mathbb{Z}[x]$. If there is anyone who can help me with this will be greatly appreciated.
| Let $m$ be a maximal ideal of $\mathbb Z[x]$ with $(3,x^3-x+1)\subseteq m$. Assume tha $(3,x^3-x+1)\not=m$ and let $f\in m\setminus(3,x^3-x+1)\subseteq m$. Since $x^3-x+1$ is monic we have $f=g(x^3-x+1)+h$, where $g,h\in \mathbb Z[x]$ and $1\leqslant deg(h)\leqslant 2$. Now since $x^3-x+1\in m$, we have $h\in m$. We consider the following:
Case 1) $deg(h)=1$: Let $h=ax+b$. Since $3\in m$, we assume that $a, b\in\{1, 2\}$. Thus we have the following subcasis:
Subcase 1a) $a=b=1$: In this case $x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-2)=x^3-2x\in m$. Thus, $x\in m$ or $x^2-2\in m$. Hence, $1\in m$ or $x(x+2)=x^2+2=x^2-2+2(x+1)\in m$.Thus, $1\in m$ or $x\in m$ or $x+2\in m$ and so $1\in m$, a contradiction.
Other subcases are simillar.
Case 2) $deg(h)=2$: Let $h=ax^2+bx+c$. Since $3\in m$, we assume that $a, b, c\in\{1, 2\}$. Thus we have the following subcases:
Subcase 2a) $a=b=c=1$: In this case $x^2+x+1\in m$ and since $x^3-x+1\in m$, we have $x(x^2-x-2)=x^3-x^2-2x\in m$. Thus, $x\in m$ or $x^2-x-2\in m$. Hence, $1\in m$ or $(x^2-x-2)-(x^2+x+1)\in m$, if $1\in m$, a contradiction. Now let $(x^2-x-2)-(x^2+x+1)\in m$, hence a polynominal of degree one is in $m$ and by case 1 we have a contradiction.
Other subcases are simillar.
| {
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"url": "https://math.stackexchange.com/questions/3325549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Problem with $x^{6} - 2 = 0$ compute roots in $\mathbb{C}$
I have problem with simple equation $x^{6} - 2 = $ compute roots in
$\mathbb{C}$
I will try compute roots of $x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=(x-2^{1/6})(x^{2}+2^{1/6}x+2^{1/3})(x^{3}+\sqrt{2})$, but this not looks good.
Maybe is better solution.
Any suggestions?
| With $\sqrt[3]2z=x^2$, we have a cubic equation
$$z^3-1=0$$ which we can factor as (using a remarkable product)
$$(z-1)(z^2+z+1)=0.$$
Besides the obvious root $z=1$, we also have
$$z=\frac{-1\pm i\sqrt 3}{2}.$$
Finally, the six roots are
$$x=\pm\sqrt[6]2,\pm\sqrt[6]2\sqrt{\frac{-1\pm i\sqrt 3}{2}}.$$
Note that $(1\pm i\sqrt3)^2=2(-1\pm i\sqrt3)$ and we can unnest
$$\sqrt{\frac{-1\pm i\sqrt 3}{2}}=\frac{1\pm i\sqrt3}2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of a function in which square roots are involved $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}$$
Could someone please help me solve this problem.
I tried multiplying by a unity factor but I end up stuck.
| $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x\rightarrow0}\frac{((x+2)^2-(2x+4))(3x-1-\sqrt{x+1})}{((3x-1)^2-(x+1))(x+2+\sqrt{2x+4})}=$$
$$=\lim_{x\rightarrow0}\frac{(x^2+2x)(3x-1-\sqrt{x+1})}{(9x^2-7x)(x+2+\sqrt{2x+4})}=\lim_{x\rightarrow0}\frac{(x+2)(3x-1-\sqrt{x+1})}{(9x-7)(x+2+\sqrt{2x+4})}=\frac{2\cdot(-2)}{-7\cdot4}=\frac{1}{7}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use properties of determinant and show
Let $n$ be a positive integer and
\begin{align} M =
\begin{pmatrix}
n! & (n+1)! & (n+2)! \\
(n+1)! &(n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)! \\
\end{pmatrix}
\end{align}
Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\right)\end{align} is divisible by $n$.
I took an $n!$ factor out of the matrix, getting a new matrix $B$ such that $\det A = n^3 \det B$, since $A$ and $B$ are row equivalents. But what is $\det B$ ?
| So, they're looking for the determinant of $M$. We can make that easier to calculate by factoring a common term of $n!$ out of each row, giving us a new matrix:
$$N=\left[\begin{array}{ccc}
1 & n+1 & (n+1)(n+2)\\
n+1 & (n+1)(n+2) & \small{(n+1)(n+2)(n+3)}\\
(n+1)(n+2) & \small{(n+1)(n+2)(n+3)} & \scriptsize{(n+1)(n+2)(n+3)(n+4)}
\end{array}\right]$$
Now, $|N|=\frac{|M|}{(n!)^3}$, so we just need to find $|N|-4$. It looks like we're about to dive into horrible polynomial multiplications. But wait! We only need to show that $|N|-4$ is divisible by $n$. That means that if the constant term of the polynomial $|N|$ is 4, we're done. We can do that by multiplying those terms out and ignoring all but the constant term"
$$\left[\begin{array}{ccc}
1 & ...+1 & ...+2\\
..+1 & ...+2 & ...+6\\
...+2 & ...+6 & ...+24
\end{array}\right]$$
Now the determinant is easy enough to calculate. Using Sarrus' Rule, the value of this determinant is
$$(1\cdot2\cdot24) + (1\cdot6\cdot2) +(2\cdot1\cdot6)- (2\cdot2\cdot2) - (1\cdot6\cdot6) - (1\cdot1\cdot24) $$ $$ = 48+12+12-8-36-24=4$$.
Therefore, $|N|=An^6+Bn^5+Cn^4+Dn^3+En^2+Fn+4$ for some integer values of $A,B,C,D,E,F$. We don't need to find those values, though. The only thing that matters is that every term of $|N|-4$ contains a power of $n$, which is what we were asked to show.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For how many pairs of positive integers n and m is the statement $mn - 8m + 6n =0$ true?
Each interior angle of a regular polygon with $n$ sides is $\frac{3}{4}$
of each interior angle of a second regular polygon with $m$ sides.
How many pairs of positive integers $n$ and $m$ are there for which this
statement is true?
$\frac{(n-2)*180}{n}$ is the value of one interior angle for a polygon with $n$ sides.
Therefore $\frac{(n-2)*180}{n} =\frac{3(m-2)*180}{4m}$ and $mn - 8m + 6n =0$
For how many pairs of positive integers $n$ and $m$ is the statement $mn - 8m + 6n =0$ true?
| $$mn-8m+6n=0\implies n(m+6)=8m\implies n=\frac{8m}{m+6}$$
Testing this formula from $1\le m\le 1000$, the only integer solutions found were
$$\frac{2\times8}{2+6}=2
\quad\frac{6\times8}{6+6}=4
\quad\frac{10\times8}{10+6}=5
\quad\frac{18\times8}{18+6}=6
\quad\frac{42\times8}{42+6}=7$$
If we solve the other way around, $$m=\frac{6n}{8-n}$$ and we can see a limit where $1\le n\le7$ so the pairs are $(2,2), (6,4), (10,5), (18,6), (42,7)$.
| {
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How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
I tried many times, and finally I used Muirhead, but it failed!
$\begin{split}L.H.S.-R.H.S.&=8(1-a)(1-b)(1-c)-abc\\&=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$
But as $(3,0,0)$ majorizes $(2,1,0)$ but $(2,1,0)$ majorizes $(1,1,1)$, so it fails.
Could someone help? Any help is appreciated!
| Denote: $1-a=x,1-b=y,1-c=z$. Then:
$$x+y+z=1 \stackrel{AM-GM}{\Rightarrow} \color{red}1=x+y+z\color{red}{\ge 3\sqrt[3]{xyz}}$$
Hence:
$$8(1-a)(1-b)(1-c)\le abc \Rightarrow 8xyz\le (1-x)(1-y)(1-z)\iff \\
8xyz\le 1-(x+y+z)+(xy+yz+zx)-xyz \iff \\
9xyz\le xy+yz+zx\iff \\
9xyz\le 3\sqrt[3]{(xyz)^2} \stackrel{AM-GM}{\le} xy+yz+zx\iff \\
\color{red}{3\sqrt[3]{xyz}\le 1}.$$
equality occurs for $x=y=z=\frac13$, consequently, for $a=b=c=\frac23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329546",
"timestamp": "2023-03-29T00:00:00",
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Find the minimum $n$ such that $x^2+7=\sum_{k=1}^n f_k(x)^2$ where $f_k(x)\in \mathbb{Q}[x]$ Recently, I have found this problem:
Find the minimum $n \in N$ such that $x^2+7=f_1(x)^2+f_2(x)^2+\cdots+f_n(x)^2$ where $f_1(x),+f_2(x),+\cdots+f_n(x)$ are polynomials with rational coefficients.
I have tried to solve this problem when $n=2$ using $f_1(x)=(a_1x+b_1)^2$ and $f_2(x)=-(a_2x+b_2)^2$, but I can't go on. Any idea?
| You cannot go on this way, because even if you take a function negative, its square will be positive. Observe that degree of all of the polynomials is less than 2, because degree of LHS and RHS should match. It is easy to see that $n=1$ is not a solution. For $n=2$, we observe,
$$(ax+b)^2+(cx+d)^2=x^2+7$$
$$\Longrightarrow (a^2+c^2)x^2+2(ab+cd)x+(b^2+d^2)=x^2+7$$
$$\Longrightarrow a^2+c^2=1, ab+cd=0, b^2+d^2=7$$
Now, apply a trick. From second equation we get,
$$ab=-cd$$
$$\Longrightarrow \frac{a}{d}=\frac{-c}{b}$$
Apply ratio and proportion,
$$\Longrightarrow \frac{a}{d}=\frac{-c}{b}=\pm\sqrt{\frac{a^2+c^2}{d^2+b^2}}$$
The ratio on the RHS is irrational by first and third equations. Hence, $n=2$ is also not a solution.
For, $n=3$, we have the following solution.
$$f_1(x)=x$$
$$f_2(x)=\frac{21}{5}$$
$$f_3(x)=\frac{28}{5}$$
Hence, this is the required solution.
Hope it helps:)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
| Isolate the square root, then square both sides:
\begin{align*}
x-8\sqrt x+7&=0\\
\implies x+7&=8\sqrt x\\
\implies (x+7)^2 &= 64 x\\
\implies x^2 -50x + 49&=0\\
\implies (x-49)(x-1) &=0\\
\implies x = 49 \text{ or } x&=1,
\end{align*}
which are the two solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
| $$4x^2-4x + 1+y^2 + 2y +1 =9\implies (2x-1)^2+(y+1)^2=9$$
So a pair $(x,y)$ is on elipse: $${(x-{1\over 2})^2\over {9\over 4}}+{(y+1)^2\over 9}=1$$ so you can write $x= {3\over 2}\cos t +{1\over 2}$ and $y= 3\sin t -1$
So your expresion is now $${15\over 2}\cos t +18\sin t -{7\over 2}$$
This you can write as $${39\over 2}\sin (t+\varphi) -{7\over 2}$$ for some phase $\varphi$. So the maximum is when $\sin (t+\varphi)=1$ and it is $16$ and minimum is when $\sin (t+\varphi)=-1$ and it is $-23$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.
I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither could think of any other ways to proceed.
How to proceed?
Thank you.
| You can rewrite it as $$(y+1)^3 = x^3 + 5x^2 - 19x + 21$$
Now $y+1$ and $x$ differ for some integer, say $z$, so $(y+1) - x=z$, so we have $$(x+z)^3 = x^3 + 5x^2 - 19x + 21$$ or $$ x^2(3z-5)+x(3z^2+19)+(z^3-21)=0$$ so we have a quadratic equation on $x$ with an integer parameter $z$: It will have an integer solution if it discriminat is a perfect square, so
$$(3z^2+19)^2-4(3z-5)(z^3-21) = d^2$$ or $$-3z^4+20z^3+114z^2+252z-59 =d^2$$
Notice that $$-3z^4+20z^3+114z^2+252z-59\geq 0$$ but that can not happend very often.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which integers $m$ is there a solution to $3^y m - 2^x = 1$? This self-answered question is motivated by a recent question asking whether for every $m$ there is a solution $(x, y)$ of integers to the equation $$3^y m - 2^x = 1$$ for every odd positive integer $m$. As the answers there show there is not---the easiest way to see this is to recall that the only solutions to $3^{y'} - 2 x = 1$ have $y' \leq 2$, so for $m = 3^k$, $k > 2$, the resulting equation $3^{y + k} - 2^x = 1$ has no solutions. But this raises the natural question:
For which integers $m$ is there a solution $(x, y)$ in positive integers to the equation $$3^y m - 2^x = 1 \,?$$
| Reducing modulo $3^y$ and rearranging leaves $$2^x = -1 \pmod {3^y} .$$ On the other hand, since $2$ is a primitive root modulo $3^1, 3^2$, it is a primitive root modulo $3^y$ for all $y \geq 1$. So, since $(\Bbb Z / (3^y \Bbb Z))^\times$ has order $\phi(3^y) = 2 \cdot 3^{y - 1}$, for fixed $y$ the $x$ satisfying the above congruence are exactly those of the form $$x = \frac{\phi(3^y)}{2} (2 k + 1) = 3^{y - 1} (2 k + 1) , \qquad k \in \Bbb Z_{\geq 0} .$$
Thus, substituting in the original equation and solving for $m$ gives that the $m$ that occur are exactly those of the form
$$\color{#df0000}{\boxed{m = \frac{2^{3^{y - 1} (2 k + 1)} + 1}{3^y}, \qquad k \geq 0, y \geq 1}} .$$
The solutions with $\color{#df0000}{m} < 1000$ give the equations
\begin{alignat}{4}
3^1 \cdot \color{#df0000}{1} - 2^1 &= 1 &
3^2 \cdot \color{#df0000}{1} - 2^3 &= 1 &
3^1 \cdot \color{#df0000}{3} - 2^{3\phantom{1}} &= 1 \\
3^1 \cdot \color{#df0000}{11} - 2^5 &= 1 &
3^3 \cdot \color{#df0000}{19} - 2^9 &= 1 &
3^1 \cdot \color{#df0000}{43} - 2^{7\phantom{1}} &= 1 \\
3^2 \cdot \color{#df0000}{57} - 2^9 &= 1 &\qquad
3^1 \cdot \color{#df0000}{171} - 2^9 &= 1 &\qquad
3^1 \cdot \color{#df0000}{683} - 2^{11} &= 1 .
\end{alignat}
The smallest $\color{#df0000}{m}$ giving a solution of the form $(x, 4)$ yields
$$3^4 \cdot \color{#df0000}{1\,657\,009} - 2^{27} = 1 .$$
Incidentally, the sequence $1, 3, 11, 19, 43, \ldots$ of $\color{#df0000}{m}$ that admit solutions does not appear in the OEIS.
| {
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Impossible hyperbolic integral I'm quite good with integral calculation, very rarely happens that I'm not able to solve an indefinite integral. I've tried for 7 days to solve this monster but in the end, I surrendered to this beast.
I'm sure that it has a solution because it was a challenge from my calculus professor. Unfortunately, he didn't give the solution of this integral. It has been unsolved for years, since 7 days ago when I found it in my old notebook.
Can some expert help me?
$$\int { \frac { \sinh { x } }{ \sqrt { 2\cosh { x+2 } } }\frac { 1+\sqrt { \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 } }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 } } } }{ 1-\sqrt [ 3 ]{ \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 } }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 } } } } } dx$$
| As suggested in comments, let $x=2 \cosh ^{-1}(t)$ to end with
$$I=-2\int\frac{
\left(\sqrt[3]{\frac{t+1}{t+2}}-\sqrt[6]{\frac{t+1}{t+2}}+1\right)}{\sqrt[6]{\frac{t+1}{t+2}}-1}\,dt$$
Now
$$\sqrt[6]{\frac{t+1}{t+2}}=y \implies t=\frac{1-2 y^6}{y^6-1}$$ makes
$$I=-12 \int\frac{ y^5 (y^2-y+1)}{(y-1) \left(y^6-1\right)^2}\,dy=-12 \int \frac{y^5}{y^5-y^3-y^2+1}\,dy$$
$$y^5-y^3-y^2+1=(y-1)^2 (y+1) \left(y^2+y+1\right)$$
Partial fraction decomposition of the integrand leads to
$$-\frac{12y^5}{y^5-y^3-y^2+1}=\frac{4 (y+1)}{y^2+y+1}-\frac{7}{y-1}-\frac{2}{(y-1)^2}+\frac{3}{y+1}-12$$ and the final result (hoping no mistake)
$$I=\log \left(\frac{(y+1)^3 \left(y^2+y+1\right)^2}{(y-1)^7}\right)-12
y+\frac{2}{y-1}+\frac{4 }{\sqrt{3}}\tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What does it mean to take the ratio of two equations?
The line joining the origin and the point of intersection of the curves $ax^2+2hxy+by^2+2gx=0$ and $a_1x^2+2h_1xy+b_1y^2+2g_1x=0$ will be mutally perpendicular if $g(a_1+b_1)=g_1(a+b)$
This is solved in my reference as
$$
ax^2+2hxy+by^2=-2gx\\
a_1x^2+2h_1xy+b_1y^2=-2g_1x\\
\color{red}{\frac{ax^2+2hxy+by^2}{a_1x^2+2h_1xy+b_1y^2}=\frac{g}{g_1}}\\
x^2(ag_1-a_1g)+2xy(hg_1-h_1g)+y^2(bg_1-b_1g)=0\\
\text{lines are perpendicular }\implies ag_1-a_1g+bg_1-b_1g=0\\
(a+b)g_1=(a_1+b_1)g
$$
Mathematical steps are fine but I really do not understand the logic behind it, particularly the first two steps where we take the ratio of the two equations of the given curves ?
Intuition
Say, we have two lines $x+y=1$ and $x-y=-2$, we can solve it by substituting $y=x+2$ in $x+y=1\implies x+y=x+x+2=2x+2=2(x+1)=1\implies x=\dfrac{1}{2}-1=\dfrac{-1}{2}\implies y=\dfrac{3}{2}$.
But, If I do
$$
y=1-x\quad;\quad y=x+2\\
1=\frac{-x+1}{x+2}\implies x+2=-x+1\implies2x=-1\implies x=-1/2\\
y=3/2
$$
So what does it mean to take the ratio of two equations ?
| The condition to be deduced involves the ratio $g/g_1.$ The equations each have the same multiple of $g$ and $g_1$ on their RHSs. Thus, taking their ratio begins to get one in the direction wanted.
| {
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Prove that summation is less than 1 Prove that $m! \times \sum_{n=m+1}^\infty \frac{1}{n!} < 1$
I started proving this by simplifying it to:
$= m! \times \left(\frac{1}{(m+1)!} + \frac{1}{(m+2)!} + ...\right)$
$= \left(\frac{1}{(m+1)} + \frac{1}{(m+2)(m+1)} + ...\right)$
I'm not too sure how to show that the last step sum is less than 1.
| If $m\ge 1$
$m!(\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+\frac{1}{(m+3)!}+...)=\frac{1}{m+1}+\frac{1}{(m+2)(m+1)}+\frac{1}{(m+3)(m+2)(m+1)}+...\le \frac{1}{2}+\frac{1}{2^2}+...=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse of a skew-symmetric matrix For $a, x, y, z \in \mathbb R$, let $$M= \left(
\begin{array}{cccc}
\cos(a) & \sin(a) \, x & \sin(a)\, y & \sin(a) \, z \\
-\sin(a) \, x & \cos(a) & \sin(a) \,z & -\sin(a)\, y \\
-\sin(a) \, y & -\sin(a) \, z & \cos(a) & \sin(a) \, x \\
-\sin(a) \, z & \sin(a) \, y & -\sin(a)\, x & \cos(a)
\end{array}
\right).$$
Without doing much calculation, why the matrix $M-I_4$ is invertible and why its inverse is given $$(M-I_4)^{-1}=\,\frac{-1}{2} I_{4} - \frac{\cot(\sqrt{x^2+y^2+z^2} )}{2\sqrt{x^2+y^2+z^2}} A,$$
where $A$ is the skew-symmetric matrix given by
$$A=\left(
\begin{array}{cccc}
0 & x & y & z \\
-x & 0 & z & -y \\
-y & -z & 0 & x \\
-z & y & -x & 0
\end{array}
\right).$$
Thank you in advance
| $A$ has the form $A = \begin{pmatrix}J&L\\-L&J\end{pmatrix}$, where $JL+LJ = 0$, $J^2 = -x^2I$, and $L^2 = (y^2+z^2)I$. Therefore,
$$
A^2 = \begin{pmatrix}J&L\\-L&J\end{pmatrix}\begin{pmatrix}J&L\\-L&J\end{pmatrix} = \begin{pmatrix}J^2-L^2&0\\0&J^2-L^2\end{pmatrix} = -r^2I_4,
$$
where $r := \sqrt{x^2+y^2+z^2}$. Hence, for $z\in\mathbb C$ we have $(A-z)(A+z) = A^2-z^2I = -(r^2+z^2)I$ and thus
$$
(A-z)^{-1} = -\frac{A+z}{r^2+z^2}.
$$
Now, as I wrote in the comments, we have $M-I = (\cos(a)-1)I + \sin(a)A$. So, for $a = 2k\pi$ the matrix $M-I$ is singular, whereas for $a=(2k+1)\pi$ we have $M-I = -2I$. If $a$ is not one of these values,
$$
(M-I)^{-1} = \left[\sin(a)\left(A - \frac{1-\cos(a)}{\sin(a)}\right)\right]^{-1} = \frac 1{\sin(a)}\cdot\frac{A+\alpha}{r^2+\alpha^2},
$$
where $\alpha = \frac{1-\cos(a)}{\sin(a)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$ The question is given below:
My questions are:
1- How can I find the general form of the multiplicative inverse of each element?
2-How can I find the multiplicative identity?
3-Is the only difference between the field and the subfield definition is that in the case of a subfield every nonzero element has an additive identity but in the field every element not only nonzero ones?
Could anyone help me in understanding these questions, please?
EDIT:
I have found this solution on the internet:
My question: is that a fully acceptable answer to the question? I guess yes.
| Let $F$ be a sub-field of the real numbers $\Bbb R$. If for some integer $k \gt 1$
$\quad \sqrt k \notin F$
then using elementary algebra it can be shown that the set
$\tag 1 \{ s_1 + s_2 \sqrt {k} \;\; | \, s_1,s_2 \in F\}$
is a field and that
$$\tag 2 ( s_1 + s_2 \sqrt {k})^{-1} = \frac{s_1 - s_2 \sqrt k}{s_1^2 - ks_2^2}$$
Since
$\tag 3 a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} = (a + b \sqrt {2}) + (c + d \sqrt{2}) \sqrt 3$
the OP is indeed dealing with a subfield of $\Bbb R$.
To arrive at a general formula for the inverse of $a + b \sqrt {2} + c \sqrt{3} + d\sqrt{6}$ in the form $a' + b' \sqrt {2} + c' \sqrt{3} + d'\sqrt{6}$ is unwieldy but can be done by applying (when necessary) $\text{(2)}$ twice.
Since the OP's question has the linear algebra tag, note that the general form for the multiplicative inverse is given by applying the inverse of
$$
\begin{bmatrix}a & 2b & 3c & 6d\\ b & a & 3d & 3c \\ c & 2d & a & 2b\\d & c & b & a\end{bmatrix}
$$
to the vector
$$
\begin{bmatrix}1 \\ 0 \\ 0 \\0 \end{bmatrix}
$$
to calculate
$$
\begin{bmatrix}a' \\ b' \\ c' \\d' \end{bmatrix}
$$
ANSWER:
$\frac{1}{a^4 - 2 a^2 (2 b^2 + 3 c^2 + 6 d^2) + 48 a b c d + 4 b^4 - 12 b^2 (c^2 + 2 d^2) + 9 (c^2 - 2 d^2)^2} \times$
\begin{bmatrix}a^3 - 2 b^2 a - 3 c^2 a - 6 d^2 a + 12 b c d \\2 b^3 - a^2 b - 3 c^2 b - 6 d^2 b + 6 a c d \\ 3 c^3 - a^2 c - 2 b^2 c - 6 d^2 c + 4 a b d \\6 d^3 - a^2 d - 2 b^2 d - 3 c^2 d + 2 a b c \end{bmatrix}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Stem and leaf diagrams I have the following data:
$2.6$ $ $ $3.3$ $ $ $2.4$ $ $ $1.1$ $ $ $0.8$ $ $ $3.5$ $ $ $3.9$ $ $ $1.6$ $ $ $2.8$ $ $ $2.6$ $ $ $3.4$ $ $ $4.1$ $ $ $2.0$ $ $ $1.7$ $ $ $2.9$ $ $ $1.9$ $ $ $2.9$ $ $ $2.5$ $ $ $4.5$ $ $ $5.0$
Built stem and leaf plot:
$Stem$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $Leaf$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $f$
$0$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $8$ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ $$ $ $ $$ $$ $ $ $
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$1$
$1$ $ $ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $1$ $ $ $ $ $ $ $6$ $ $ $ $ $ $ $7$ $ $ $ $ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $4$
$2$ $ $ $ $ $ $ $ $ $ $ $ $ $0$ $ $ $4$ $ $ $5$ $ $ $6$ $ $ $6$ $ $ $8 \ $ $9$ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $8$
$3$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $3$ $ $ $ $ $4$ $ $ $ $ $5$ $ $ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $4$
$4$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $1$ $ $ $ $ $ $ $ $ $ $ $5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $2$
$5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $0$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $1$
My lecturer said that this is true:
$P$ $(2\le$$X\le3)$ = ${8\over 20}$
$P$ $(1\le$$X\le6)$ = $1$ - ${1\over 20}$ = ${19\over 20}$
So, my question is:
Why is he not adding the last number into the range?
To me, since it is $"less $ $ than $ $ or $ $ equal $ $ to"$ this has to be:
$P$ $(2\le$$X\le3)$ = ${12\over 20}$
$P$ $(1\le$$X\le6)$ = $1$ - ${0\over 20}$ = ${20\over 20}$
Does it have something to do with continuous or discrete data? Cause in another example he added the last number into the range.
| The stem-and-leaf plot is equivalent to the following frequency and relative frequency table:
$$\begin{array}{c|l|c|c|с}
Stem&Leaf&f&&X&f&P(X)\\
\hline
0&8&1&&[0,1) \text{ or $0\le X<1$}&\color{red}1&\color{red}1/\color{blue}{20} \\
1&1679&4&&[1,2) \text{ or $1\le X<2$}&\color{red}4&\color{red}4/\color{blue}{20}\\
2&04566899&8&&[2,3) \text{ or $2\le X<3$}&\color{red}8&\color{red}8/\color{blue}{20}\\
3&3459&4&&[3,4) \text{ or $3\le X<4$}&\color{red}4&\color{red}4/\color{blue}{20}\\
4&15&2&&[4,5) \text{ or $4\le X<5$}&\color{red}2&\color{red}2/\color{blue}{20}\\
5&0&1&&[5,6) \text{ or $5\le X<6$}&\color{red}1&\color{red}1/\color{blue}{20}\\
\hline
&&20&& &\color{blue}{20}\\
\end{array}$$
Note that:
$$P(2\le X\le 3)=P(2\le X<3)+P(X=3)=\frac8{20}+\frac0{20}=\frac8{20}.$$
Similarly:
$$\begin{align}P(1\le X\le 2)&=P(1\le X<2)+P(X=2)=\frac4{20}+\frac1{20}=\frac5{20},\\
P(1\le X\le 6)&=\color{blue}{P(0\le X\le 6)}-P(0\le X<1)=\\
&=\color{blue}{P(0\le X<6)+P(X=6)}-P(0\le X<1)=\\
&=\frac{20}{20}+\frac0{20}-\frac1{20}=\frac{19}{20}.\end{align}$$
Exercise: 1) $P(3\le X\le 5)$; 2) $P(2.5\le X<3.5)$.
Answers (see the hidden area):
1) $\frac7{20}$. 2) $\frac8{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determining convergence of $\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$ I've tried to solve this problem about convergence:
$\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$
it's supposed to be solved using ratio, root tests or by testing the limit of the sumand. Anyways, I've tried both 3 and I had no success:
I get to a point where I'm stucked on:
$$lím_{n\rightarrow \infty} \left[\frac{n+1}{n} \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3\right]$$
Any suggestions? what would you usually do with therms like $(n+2i)^3$? I tried assuming non imaginary n values (because of the sum) and converting to polar form: $\sqrt{n^2+4}e^{i tan^{-1}(2/n)}$.
Also I tried expanding:
$$\left|\left(\frac{n+2i}{n+1+2i}\right)^3\right| = \frac{|n+2i||n+2i||n+2i|}{|n+1+2i|^3} = \left(\sqrt\frac{n^2 +4}{n^2 +2n+5}\right)^3$$
Gelp
| I just learned something new, I'd like to post it to ratify if it's true:
$$\sum \frac{3n i^n}{(n+2i)^3} = \sum 3\frac{n i^n (n-2i)^3}{(n+2i)^3(n-2i)^3} = 3\sum \frac{ni^n (n-2i)^3}{(n^2+4)^3} $$
Expanding terms:
= $$3\sum\frac{ni^n(n^3-4n^2i-4n-2n^2i-8n+8i}{(n^2+4)^3} = 3\sum\frac{n^4 i^n-6n^3 i^{n+1} -12n^2 i^n+8n i^{n+1}}{(n^2+4)^3}$$
as every term must converge, applying the ratio test on:
$$\sum\frac{i^n n^k}{(n^2+4)^3} \rightarrow \lim_{n\rightarrow \infty} \left|\frac{(n+1)^k}{((n+1)^2+4)^3}\right| \left| \frac{(n^2+4)^3}{n^k}\right| = \lim_{n\rightarrow \infty} \left|\frac{(n+1)^k}{n^k}\right| \left| \frac{(n^2+4)^3}{((n+1)^2+4)^3}\right|$$
First limit is 1, the second one goes to 0 (edit: to 1) because the n powers in the denominator are the same.
Is this too much going around?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$m$ and $n$ are odd numbers such that $|m^2 - n^2 + 1| \mid (n^2 - 1)$. Prove that $|m^2 - n^2 + 1|$ is a square number.
$m$ and $n$ are odd numbers such that $$\large |m^2 - n^2 + 1| \mid (n^2 - 1)$$. Prove that $|m^2 - n^2 + 1|$ is a square number.
I have provided the solution below, but I am uncertain about the structure of my solution, it's kind of clunky in my opinion.
| Let $p = \dfrac{n^2 - 1}{m^2 - n^2 + 1} + 1 = \dfrac{m^2}{m^2 - n^2 + 1} \implies (m^2 - n^2 + 1)(p - 1) = (n^2 - 1)$
We need to prove that $p$ is a perfect square.
Let $S$ be a set of integers pairs $(x, y)$ in that order satisfying $$(4xy + 1)p = (x + y)^2 \iff \frac{m^2}{m^2 - n^2 + 1} = \frac{(x + y)^2}{4xy + 1}$$
Let $x = \dfrac{m + n}{2}$ and $y = \dfrac{m - n}{2}$, we have that $$\frac{(x + y)^2}{4xy + 1} = \frac{\left(\dfrac{m + n}{2} + \dfrac{m - n}{2}\right)^2}{4 \cdot \dfrac{m + n}{2} \cdot \dfrac{m - n}{2} + 1} = \frac{m^2}{m^2 - n^2 + 1}$$
Therefore, $\left(\dfrac{m + n}{2}, \dfrac{m - n}{2}\right) \in S \implies S \ne \varnothing$. Let $$a = \min\{|x| \colon \exists(x, y) \in S\}$$
Pointing out $a = 0$ would lead to $p = y^2$ and $p$ is a perfect square.
If $(x, y) \in S$ then $(-x, -y) \in S$.
consequently, there exists $(a, y) \in S$ such that $$(4ay + 1)p = (a + y)^2 \implies y^2 - (4ap - 2a)y + (a^2 - p) = 0$$
Roots $b_1$ and $b_2$ of the equation are integers such that $|b_1| \ge a$ and $|b_2| \ge a$.
(because $(a, b) \in S$, it is also known that $(b, a) \in S$)
On the other hand, we have that $b_1 + b_2 = 4ap - 2a$ and $b_1b_2 = a^2 - p$
We consider two cases.
Case 1: $p < 0$
We have that $b_1 + b_2 \le 0$ and $b_1b_2 > 0 \implies b_1, b_2 < 0$
Because $p < 0$, $a = 0$.
Case 2: $p > 0$ and $a > 0$
We have that $b_1 + b_2 > 0$ and $(a + b_1)(a + b_2) > 0 \implies b_1, b_2 > 0$
Through the selection of $a$, $a^2 \le b_1b_2$, which is contradictory with the equation $b_1b_2 = a^2 - p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$.
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inverse.
I have tried picking two arbitrary elements, $m,n$ in our set. Let $m= x+y\alpha+z\alpha^2$ and let $n=a+b\alpha+c\alpha^2$. Then setting our product $m\cdot n$ equal to 1 gets us:
$$(ax+2zb+2yc) + (bx+ay+2zc)\alpha + (cx + yb +za)\alpha^2 = 1.$$
I was wondering if this could lead me to my proof?
| Recall the algebraic identity
$$x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2+z^2 - xy -yz - zx)\\
\implies
\frac{1}{x+y+z} = \frac{x^2+y^2+z^2 - xy-yz-zx}{x^3+y^3+z^3 - 3xyz}
$$
Substitute $(x,y,z)$ by $(x,y\alpha,z\alpha^2)$, one get
$$\frac{1}{x + y\alpha + z\alpha^2}
= \frac{(x^2-2yz) + (2z^2-xy)\alpha + (y^2-zx)\alpha^2}{x^3+2y^3 + 4z^3 -6xyz}$$
Aside from justifying why $x^3+2y^3+4z^3-6xyz \ne 0$, this gives you
a formula for the multiplicative inverse of $x + y\alpha + z\alpha^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trouble proving $3^2 + 3^3 + ... 3^n = 9 \cdot \frac{3^{n-1} - 1}2$ by induction So i'm supposed to prove by mathematical induction that this formula: $3^2 + 3^3 + ... 3^n = 9 \cdot \dfrac{3^{n-1} - 1}2$ holds true for all numbers greater than 2.
I started with the base case and just plugged in 2, it worked.
Then I assumed k was true, and then considered k+1.
What I ended up with is:
$9(3^{k-1} -1/ 2) + 3^{k+1} = 9((3^k - 1)/2)$
I tried making the left handside equal to the right handside but I couldn't. Did I do something wrong along the way?
| Why did you write it as
$9(3^{k-1} -1/ 2) + 3^{k+1} = 9((3^k - 1)/2)$
Rather than $9((3^{k-1} -1)/ 2) + 3^{k+1} = 9((3^k - 1)/2)$?
Anyway
$9\frac {3^{k-1} - 1}2 + 3^{k+1}=$
$9\frac {3^{k-1} - 1}2 + \frac {2*3^{k+1}}2=$
$9\frac {3^{k-1} - 1}2 + \frac {2*9*3^{k-1}}2 = $
$9\frac {3^{k-1} -1 + 2*3^{k-1}}2=$
$9\frac {3*3^{k-1} -1}2 =$
$9\frac {3^k -1}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hyperplane conversion between vector form and cartesian form Vector form to Cartesian form
In $\mathbb{R^2}$ a $\mathbb{R^1}$ object can be expressed as
$$\text{vector form: }s\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}$$
$$\text{cartesian form: }\begin{vmatrix}a&b\\x-c&y-d\end{vmatrix}=0$$
In $\mathbb{R^3}$ a $\mathbb{R^2}$ object can be expressed as
$$\text{vector form: }s\begin{pmatrix}a\\b\\c\end{pmatrix}+t\begin{pmatrix}d\\e\\f\end{pmatrix}+\begin{pmatrix}g\\h\\i\end{pmatrix}$$
$$\text{cartesian form: } \begin{vmatrix}a&b&c\\d&e&f\\x-g&y-h&z-i\end{vmatrix}=0$$
$\vdots$
In $\mathbb{R^n}$ a $\mathbb{R^{n-1}}$ object can be expressed as
$$\text{vector form: }s_1\begin{pmatrix}a_1\\a_2\\ \vdots\\a_n\end{pmatrix}+s_2\begin{pmatrix}b_1\\b_2\\ \vdots\\b_n\end{pmatrix}+\dots+s_{n-1}\begin{pmatrix}c_1\\c_2\\ \vdots\\c_n\end{pmatrix}+\begin{pmatrix}d_1\\d_2\\ \vdots\\d_n\end{pmatrix}$$
$$\text{cartesian form: } \begin{vmatrix}a_1&a_2&\dots&a_n\\b_1&b_2&\dots&b_n\\\vdots&&&\vdots\\c_1&c_2&\dots&c_{n}\\x_1-d_1&x_2-d_2&\dots&x_n-d_n\end{vmatrix}=0$$
I always use this fomular to do the conversion, but is it true for all $n\in[2,\infty)\cap\mathbb{N}$.
What I tried is using induction, for Base case, It's not hard to show it hold when n=2, but i don't know how to write the inductive step, or is this easier to do with direct prove?
Any help or hint or suggestion would be appreciated.
Cartesian form to Vector form
If given cartesian form in $\mathbb{R^2}$:
$$ax+by+c=0$$
when $c\neq0$, $X,Y$ are intersections of the line and aixs:
$$X=\begin{pmatrix}-\frac{c}{a}\\0\end{pmatrix} \text{ and } Y=\begin{pmatrix}0\\-\frac{c}{b}\end{pmatrix}$$
Therefore the line can be expressed as:
$$\text{vector form: }s(X-Y)+X$$
Since $X=Y$ when $c=0$, the method above no longer works.
But $N_1=\begin{pmatrix}-\frac{1}{a}\\\frac{1}{b}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a}\\-\frac{1}{b}\end{pmatrix}$ are normal to the normal of the line.
Choose any 1 as direction vector have:
$$\text{vector form: }sN_1$$
If given cartesian form in $\mathbb{R^3}$:
$$ax+by+cz+d=0$$
when $d\neq0$, $X,Y,Z$ are intersections of the plane and aixs:
$$X=\begin{pmatrix}-\frac{d}{a}\\0\\0\end{pmatrix}, Y=\begin{pmatrix}0\\-\frac{d}{b}\\0\end{pmatrix},Z=\begin{pmatrix}0\\0\\-\frac{d}{c}\end{pmatrix}$$
Therefore the plane can be expressed as:
$$\text{vector form: }s(X-Y)+t(X-Z)+X$$
Since $X=Y=Z$ when $d=0$, the method above no longer works.
But $N_1=\begin{pmatrix}-\frac{2}{a}\\\frac{1}{b}\\\frac{1}{c}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a}\\-\frac{2}{b}\\\frac{1}{c}\end{pmatrix}, N_3=\begin{pmatrix}\frac{1}{a}\\\frac{1}{b}\\-\frac{2}{c}\end{pmatrix}$ are normal to the normal of the plane.
Choose any 2 as direction vectors have:
$$\text{vector form: }sN_1+tN_2$$
$\vdots$
More generally, If given cartesian form in $\mathbb{R^n}$:
$$a_1x_1+\dots+a_nx_n+b=0$$
Have intersections on $x_1$ that $X_1=\begin{pmatrix}-\frac{b}{a_1}\\0\\\vdots\\0\end{pmatrix}$
And $N_1=\begin{pmatrix}-\frac{n-1}{a_1}\\\frac{1}{a_2}\\\vdots\\\frac{1}{a_n}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a_1}\\-\frac{n-1}{a_2}\\\frac{1}{a_3}\\\vdots\\\frac{1}{a_n}\end{pmatrix},\dots, N_n=\begin{pmatrix}\frac{1}{a_1}\\\frac{1}{a_2}\\\vdots\\\frac{1}{a_{n-1}}\\-\frac{n-1}{a_n}\end{pmatrix}$ are normal to the normal.
Choose any $n-1$ as direction vectors have: $$\text{vector form: }s_1N_1+\dots+s_{n-1}N_{n-1}+X_1$$
| Note that the vector form and the cartesian form are equivalent only if the $n-1$ vectors $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T$ are linearly independent. Otherwise, the vector form would produce something of degree less than $n-1,$ while the cartesian form would produce the complete $\mathbb{R}^n.$ So we assume the linear independence of $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T$ in the following.
If the cartesian form is given, we know that the $n$ vectors $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T,\;(x_1-d_1\;\ldots\;x_n-d_n)^T$ are linearly dependent, which means that there are $n$ real numbers $t_1,\ldots, t_n$, at least one of them $\neq 0,$ such that
$$
t_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+t_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+t_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
+t_{n} \begin{pmatrix} x_1-d_1 \\ \vdots \\ x_n-d_n\end{pmatrix}
=0
$$
Assume $t_n=0.$ Then one of the numbers $t_1,\ldots, t_{n-1}$ must be $\neq 0,$ because we assumed the $t_i\neq 0$ for at least one $i\in \{1,\ldots,n\}.$ This would mean that we had the following:
$$
t_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+t_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+t_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
=0
$$
But this is a contradiction to the linear independence of $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T.$ Therefore, $t_n\neq 0.$
We can now set $t_i = -t_n s_i$ for $i\in\{1,\ldots, n-1\}$, divide the equation by $-t_n$ and we get
$$
\begin{pmatrix} x_1 \\ \vdots \\ x_n\end{pmatrix}
=
s_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+s_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+s_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
+\begin{pmatrix} d_1 \\ \vdots \\ d_n\end{pmatrix}
$$
which shows that the given element of the hyperplane can also be expressed in terms of the vector form.
If the vector form is given, we have
$$
s_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+s_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+s_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
=
\begin{pmatrix} x_1-d_1 \\ \vdots \\ x_n-d_n\end{pmatrix}
$$
which shows that the rows of the determinant are linearly dependent, which in turn shows that the value of the determinant is $0.$
| {
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The $n^\text{th}$ term in the sequence $\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\frac{1}{4},\frac{2}{3},\dots$ $\begin{array} {|r|r|}\hline R\text{ \ }C & 1 & 2 & 3 & \dots \\ \hline 1 & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} & \\ \hline 2 & \frac{2}{1} & \frac{2}{2} & \frac{2}{3} & \\ \hline 3 & \frac{3}{1} & \frac{3}{2} & \frac{3}{3} & \\ \hline \vdots & & & & \ddots \\ \hline \end{array}$
Considering the sequence of fractions that is made by moving diagonally, with respect to the entries above, in the following manner:
$$\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\dots$$
where the entries are obtained by dividing the value of the row $R$ by the value of the column $C$.
So the next few terms are $\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1},\frac{1}{5},\dots$
Now I have two question:
What is the $n^\text{th}$ term in the sequence (including all
fractions)?
What is the $n^\text{th}$ term in the sequence (excluding repeated
values)?
For example, $\frac{4}{2}$ has to be excluded since $\frac{2}{1}$ is already considered. Similarly, $\frac{2}{2},\frac{3}{3},\frac{4}{4},\dots$ are to be excluded since $\frac{1}{1}$ is already there.
What I observed is; for any natural number $m$, the $(2m^2+2m+1)^\text{th}$
term is to be excluded since that term is equal to $1$ which is already considered.
Also, it is easy to find an expression in $m$ to exclude other fractions
I have no idea to solve these two questions.
Any help would be appreciated. THANKS!
| § 1 Frame of reference
It is useful to adopt a two index identification of the fractions.
Decompose the fractions into groups according to a scheme of numerators and denominators as shown for the first 4 groups below
$$(1)(1 2)(1 2 3)(1 2 3 4)...$$
$$(1)(2 1)(3 2 1)(4 3 2 1)...$$
The position $p_s(k)$ of the start of the k-th group in a linear sequence of fractions is given by
$$p_s=1+\frac{1}{2} k(k-1), k=1, 2, …\tag{1}$$
which is the sequence $1, 2,4,7,11,16,...$.
We can write the fractions in group $k$ as
$$f(k,m) = \frac{m}{k-m+1}, m=1..k\tag{2}$$
The position of fraction $f(k,m)$ in the sequence of all fractions is given by
$$p(k,m) = p_s+m-1=m+\frac{1}{2} k(k-1) \tag{3}$$
This can be inverted by
$$k(p)=\lfloor \left( \frac{1}{2}+\sqrt{\frac{1}{4}+2p-2} \right) \rfloor \tag{4a}$$
$$m(p) =p-\frac{k(p)}{2}(k(p)-1)\tag{4b}$$
Here $\lfloor . \rfloor$ is the floor function.
§ 2 Drop repeated fractions
If we wish to avoid double counting we have to check if the fraction is reducible. In other words we have to drop fractions with $GCD(m, k-m+1)\gt 1$.
I don't know how to express this condition in an explicit formula. Maybe others can.
Here is a related result:
Let $r(n)$ be the number of different rationals in our set of groups of fractions up to the element with the fraction $\frac{n}{1}$
Then $r(n)$ can be found by direct calculation to start like this
$$r(n) = {1, 3, 5, 9, 11, 17, 21, 27, 31, 41, 45, 57, 63, 71, 79, 95, 101}$$
This sequence is easily found in OEIS: http://oeis.org/A015614. And the formula is
$$r(n) = -1 + \sum_{i=1..n} \phi(i)\tag{5}$$
where $\phi(i)$ is Euler's totient function (https://en.wikipedia.org/wiki/Euler%27s_totient_function).
The asymptotic behaviour is
$$r(n\to \infty) \simeq \frac{3}{\pi^2}n^2+O(n \log(n))$$
Since the number of primarily different fractions is $1+2+3+...+n = \frac{1}{2}n(n+1)$ the percentage of different fractions is given asymptotically by $\frac{6}{\pi^2} \simeq 0.607927$.
| {
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Find the probability of getting the second ace as the $n^{th}$ card is picked from $52$ cards.
Cards are drawn one by one at random from a well shuffled full pack of $52$ cards. Find the probability of the second ace being the $n^{th}$ card.
For this what I did was, for the $n-1^{th}$ positions, we need to make a selection of $n-2$ card from $48$ cards and then choose $1$ ace out of $4$. This can be done in $^{48}C_{n-2} × 4$ . Now for $n^{th}$ position, I can have an ace in $3$ ways. So total no of favourable case is $^{48}C_{n-2} × 4× 3$. For total cases I select n card from $52$ in $^{52}C_{n}$. So $$P=\frac{^{48}C_{n-2} × 4× 3}{^{52}C_{n}}$$
But this is not quite the answer. The answer instead is $$P=\frac{^{48}C_{n-2} ×\color{red}{n} × 4× 3}{^{52}C_{n}}$$, which is quite similar but differs only by an $n$ . Can someone please explain what have I missed.
| We must multiply the probability that exactly one ace is selected among the first $n - 1$ cards by the probability that the $n$th card selected is an ace.
We can select $n - 1$ of the $52$ cards in the deck in $\binom{52}{n - 1}$ ways. If these cards include exactly one ace, we must select one of the four aces and $n - 2$ of the remaining $48$ cards. Hence, the probability that the first $n - 1$ cards selected contain exactly one ace is
$$\frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dbinom{52}{n - 1}}$$
Since $n - 1$ cards have been selected, there are $52 - (n - 1) = 53 - n$ cards left in the deck. Three of these are aces. Hence, the probability that the $n$th card is an ace given that exactly one ace has been selected among the first $n - 1$ cards is
$$\frac{3}{53 - n}$$
Therefore, the probability that the second ace is the $n$th card selected is
\begin{align*}
\Pr(\text{second ace is $n$th card selected}) & = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dbinom{52}{n - 1}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)![52 - (n - 1)]!}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)!(53 - n)!}} \cdot \frac{3}{53 - n}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}}{\dfrac{52!}{(n - 1)!(52 - n)!}} \cdot 3\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{\dfrac{52!n}{n(n - 1)!(52 - n)!}}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{\dfrac{52!n}{n!(52 - n)!}}\\
& = \frac{\dbinom{4}{1}\dbinom{48}{n - 2}\dbinom{3}{1}}{n\dbinom{52}{n}}
\end{align*}
so the stated answer is incorrect. That factor of $n$ should be in the denominator.
Your error was not accounting for the fact that the second ace must be in the $n$th position. That requires two separate events. There must be one ace among the first $n - 1$ cards and the $n$th card selected must be an ace. While you accounted for this in your numerator, you did not do so in your denominator.
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Divisiblity proof with coprimes of 6 Can someone please tell me if this proof is okay? Define "okay" as acceptable and rigorous. I wanted to avoid being to wordy so I did not prove a few trivial aspects. I feel the proof is quite verbose but I think as concise as I could get it without skipping too many details.
Suppose $(6,a) = (6,b) = 1$. We wish to prove $24 \mid (a^2-b^2)$. We will use the division algorithm for this proof. If when $a$ is divided by $6$ there is a remainder of $2$, then a is some multiple of $6$ plus $2$, then $2 \mid a$. The same applies for a remainder of $4$. If when a is divided by $6$ there is a remainder of $3$, then $a$ is a multiple of $6$ plus $3$, so a would be divisible by $3$. $a$ is not divisible by $2$ or $3$, therefore the only possible remainders when $a$ is divided by $6$ are $1$ and $5$. The same applies to $b$. so $a = 6p+1$ or $a = 6k+5$ and $b = 6m + 1$ or $b = 6n + 5$ for $p,k,m,n \in \mathbb{Z}$. Note that $a^2 - b^2 = (a+b)(a-b)$. We will prove for all 4 cases that $24 \mid (a+b)(a-b)$. Case 1: $(a+b)(a-b) = (6p + 1 + 6m + 1)(6p-6m) = 12(3(p+m) + 1)(p-m)$. Suppose $p$ and $m$ are both even or odd, then $p-m$ is even. Suppose one of $p, m$ is even and the other is odd. Then $p+m$ is odd, so $p+m = 2x + 1$ for some $x \in \mathbb{Z}$. So $3(p+m)+1 = 3(2x+1)+1 = 2(3x+2)$, so $3(p+m)+1$ is even. So $24 \mid (a^2-b^2)$. Case 2: $(a+b)(a-b) = (6p+1 +6n+5)(6p+1 - 6n - 5) = 12(p + n + 1)(3(p - n) -2)$. Suppose both $p$, $n$ are odd or even, then $p-n$ is even so $p-n = 2x$ for some $x \in \mathbb{Z}$. So $3(p-n)-2 = 2(3x - 1)$ which is even. Suppose one of $p,m$ are even and the other odd, then $p+n$ is odd so $p+n+1$ is even. So $24 \mid a^2-b^2$. Case 3: $(a+b)(a-b) = (6k+5 + 6m + 1)(6k+5 - 6m - 1) = 12(k + m + 1)(3k-3m+2)$. The proof of case 3 is identical to case 2, so $24 \mid a^2-b^2$. Case 4: $(a+b)(a-b) = (6k+5 + 6m +5)(6k-6m) = 12(3(k+m) + 5)(k-m)$. Suppose both k, m are even or odd. Then $k-m$ is even. Suppose one of $k,m$ is even and the other is odd. Then $k+m$ is odd. So $k+m = 2x+1$ for some $x \in \mathbb{Z}$. Then $3(k+m) + 5 = 2(3x+4)$ which is even. So $24 \mid a^2-b^2$. In all cases of a,b, we have proven that $24 \mid a^2 - b^2$. $\square$
| May I suggest a shorter proof? (With $a \equiv_n \pm 1$ is meant: $a \equiv 1 \mod n$ or $a \equiv -1 \mod n$ etc.)
$$3\not | a,b \Rightarrow a\equiv_3 \pm 1\mbox{ and } b\equiv_3 \pm 1 \Rightarrow a^2\equiv_3 b^2 \equiv_3 1 \Rightarrow a^2-b^2 \equiv_3 0$$
Hence, $\boxed{3|a^2-b^2}$
$$2 \not | a,b \Rightarrow a\equiv_4 \pm 1\mbox{ and } b\equiv_4 \pm 1 \Rightarrow a^2\equiv_4 b^2 \equiv_4 1 \Rightarrow a^2-b^2 \equiv_4 0$$
$$a^2-b^2 \equiv_4 (a-b)(a+b) \equiv 0 \Rightarrow a-b \equiv_4 0 \mbox{ or } a+b \equiv_4 0 $$
$$2 \not | a,b \Rightarrow a+b\equiv_2 0 \mbox{ and } a-b \equiv_2 0$$
From the last two lines follows: $4 | (a+b) \mbox{ and } 2|(a-b)$ or $2 | (a+b) \mbox{ and } 4|(a-b) \Rightarrow \boxed{8|a^2-b^2}$
| {
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Is this a family of similar matrices $\left(\begin{smallmatrix} 0&x\\ 0&0 \end{smallmatrix}\right)$? Is matrix $A = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ similar to matrix $B=\begin{pmatrix} 0&2\\ 0&0 \end{pmatrix}$? If so, how do I prove this?
I came here from following the answer to this question:
Do similar matrices have equal singular values?
| If they are similar then
$\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}= \begin{pmatrix} 0&2\\ 0&0 \end{pmatrix} \begin{pmatrix} a&b\\ c&d \end{pmatrix} $
for some invertible matrix $P=\begin{pmatrix} a&b\\ c&d \end{pmatrix}$.
Try to calculate such matrix $P$ from the above condition.
| {
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Continuity of $y \sin(\frac{1}{x})$ Problem: for which values of $\alpha \in \mathbb{R}$ is the function
\begin{equation*}
f: \mathbb{R}^2\to\mathbb{R}^2 \qquad \begin{pmatrix} x \\ y\end{pmatrix} \mapsto \begin{cases} y \sin(\frac{1}{x}) \quad &\text{if $x\neq 0$}\\0 \quad &\text{if $x= 0$}\end{cases}
\end{equation*}
continuous at $a = \begin{pmatrix} 0 \\ \alpha\end{pmatrix}$? Prove your answer.
Attempted solution: I suppose the answer is: $\left\{ \begin{pmatrix} 0 \\ 0\end{pmatrix} \right\}$. I tried to reason as follows, but I am not sure if this is technically correct.
$f$ is continuous at $a$
$\iff \lim_{x\to 0} f\begin{pmatrix} x \\ \alpha\end{pmatrix} = \begin{pmatrix} 0 \\ 0\end{pmatrix}$
$\iff \lim_{x\to 0} \alpha \sin\left(\frac{1}{x} \right)= \begin{pmatrix} 0 \\ 0\end{pmatrix}$
$\iff \alpha = 0$
since $\sin$ is periodic, and thus diverges.
Is this correct?
| Your solution looks fine to me.
Notice that $|y\sin{\frac{1}{x}}|=\sqrt{y^2}|\sin{\frac{1}{x}}| \leq \sqrt{x^2+y^2}$
So indeed the function is continuous at $(0,0)$ and only at this point because $\lim_{x \to 0}\sin{\frac{1}{x}}$ does not exist.
| {
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A system of equations Olympiad question Find all non-zero real numbers $x,y,z$ which satisfy the system of equations:
\begin{align}
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)&=xyz,\\(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4)&=x^3y^3z^3
\end{align}
It's an indian olympiad question. Can you guys help me out in solving it please ?
edit:
I have tried to write the first equation as:
$(\frac{x^3-y^3}{x-y})(\frac{y^3-z^3}{y-z})(\frac{z^3-x^3}{z-x})$ = xyz
And second equation as:
$(\frac{x^6-y^6}{x^2-y^2})(\frac{y^6-z^6}{y^2-z^2})(\frac{z^6-x^6}{z^2-x^2}) = x^3y^3z^3$
Then divided the two to get:
$(\frac{x^3+y^3}{x+y})(\frac{y^3+z^3}{y+z})(\frac{z^3+x^3}{z+x}) = x^2y^2z^2$
After that I have no sign what to do??
| You obtained $$\left(\frac{x^3+y^3}{x+y}\right)\!\!\left(\frac{y^3+z^3}{y+z}\right)\!\!\left(\frac{z^3+x^3}{z+x}\right) = x^2y^2z^2$$
which is equivalent to
$$\prod_\mathrm{cyc}(x^2-xy+y^2)=x^2y^2z^2.$$
Dividing throughout by $xy\cdot yz\cdot zx$, we get
$$\prod_{\mathrm{cyc}}\left(\frac{x}{y}-1+\frac{y}{x}\right)=1.$$
Now, we use the hint given by achille hui in the comments. For every nonzero $u\in\mathbb R$, we have $|u+u^{-1}-1|\geq 1$ with equality exactly when $u=1$. To see this, note that if $u>0$, then $u+u^{-1}\geq2$ by AM-GM inequality, whereas if $u<0$ then by the same logic, $u+u^{-1}\leq -2$. Now, it follows from the multiplicity of $|\cdot|$ that
$$\prod_{\mathrm{cyc}}\left|\frac{x}{y}-1+\frac{y}{x}\right|=1.$$
If any one of the terms in the cyclic product is $>1$, then that forces one of the other terms to be $<1$, which we know to be impossible. This implies that all the terms are exactly one, so that $x/y=1$ and $x=y$. Similarly $y=z$, so $x=y=z$. The rest is easy.
| {
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Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:
*
*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$
*
*$y \equiv 0\mod 2\Rightarrow y+1=3 \Leftrightarrow z=15$, contradiction since $15$ is not prime
*$y \equiv 1\mod 2\Rightarrow y+1\geq 2^2 \Leftrightarrow y+1=2*3, z+1=2^3$ or $y+1=2^2*3, z+1=2^2 \Leftrightarrow y=5, z=7$ or $y=11, z=3$
And since the equation is symmetric the solutions are the permutations of the latter ones. Similarly, through casework, we find the same solutions if $x \equiv 1\mod 2$ (if I haven't made a mistake). My question is if there exists a more simple way to solve this problem besides lots of casework (and if there exists another triplet that satisfies the given equation).
| Denote $x' := x + 1$ and analogously. As you point out, solutions are closed under permutation, so we may suppose w.l.o.g. that $x' \leq y' \leq z'$.
Since the factors $x', y', z'$ are all at least $3$, we must have $z + 1 \leq \frac{144}{3^2} = 16$. But the factors of $144 = 2^4 \cdot 3^2$ no larger than $16$ are
$1, 2, 3, 4, 6, 8, 9, 12, 16$, and the numbers among these $1$ larger than a prime are $\{3, 4, 6, 8, 12\}$. We can reduce the number of cases to check by observing that $5^3 < 144 < 6^3$, which implies that $x' \in \{3, 4\}$ and $z' \in \{6, 8, 12\}$.
*
*If $z' = 6$, then $x' y' = \frac{144}{6} = 24$ and $\sqrt{24} \leq y' \leq 6$, so $y' = 6$: $(x', y', z') = (4, 6, 6)$.
*If $z' = 8$, then $x' y' = \frac{144}{8} = 18$ and $\sqrt{18} \leq y' \leq 8$, but the only factor of $18$ in that range is $6$: $(x', y', z') = (3, 6, 8)$.
*If $z' = 12$, then $x' y' = \frac{144}{12} = 12$, so $x' \leq \sqrt{12}$ and thus $x' = 3$: $(x', y', z') = (3, 4, 12)$.
The solutions are: $$(x, y, z) = (2, 3, 11), (2, 5, 7), (3, 5, 5) .$$
It seems the error in the original post was in the unwritten details of the case $x \equiv 1 \pmod 2$.
| {
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Find argmax $\frac{1}{2b}(\frac{x}{Na})^{\frac{N-1}{2}}e^{-\frac{1}{2b}(x+Na)}\frac{1}{\sqrt{2\pi(N-1)}}(\frac{\frac{e}{b}\sqrt{Nax}}{2(N-1)})^{N-1}$ I want to find an expression for:
$$ (1) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\Big( \frac{\sqrt{Nax}}{b} \Big) \Big \} $$
Where $a,b,x \in \mathbb{R_{++}}$, $N \in \mathbb{N}$, and $I_{N-1}(\cdot)$ is the modified Bessel function of the first kind of order $N-1$.
Since this function is concave in $x$, all I need to do is take the partial derivative w.r.t $x$, set it equal to zero, and solve for $x$. Unfortunately, this seems like a pretty intractable problem. Try computing the $x$ partial in Wolfram to see what I mean.
So, I am going to approximate the answer instead. I know from NIST that for large orders:
$$I_{N-1}(z) \sim \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{e z}{2 (N-1)} \Big)^{N-1}$$
So then I have:
$$ (2) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} \frac{1}{\sqrt{2 \pi (N-1)}} \Big( \frac{\frac{e}{b} \sqrt{Nax} }{2 (N-1)} \Big)^{N-1} \Big \} $$
This problem is easy. After taking the logarithm, expanding terms, taking the derivative, and setting it to zero, I get:
$$ x^{\ast} = 2b (N-1) $$
This is exact for (2), but nowhere near the truth for (1), especially as $N$ gets large and $a,b \approx 1$. This can be checked numerically. I think this is because the NIST expansion doesn't account for the argument growing as well.
Questions:
*
*How can I solve (1) analytically? Can you solve it exactly via the derivative?
*Is there an expansion, similar to the one used in (2), for when both the order and argument of the Bessel function are growing?
*Any other ideas on how I can approach this problem? Maybe approximate the log of the Bessel function directly?
Thanks!
| We simplify slightly the expression by writing
\begin{align}
&\frac{1}{2b} \left(\frac{x}{Na}\right)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)=\frac{1}{2b}\left( Na \right)^{\frac{1-N}{2}}e^{-\frac{Na}{2b}}f(x)\\
&f(x)=x^{\frac{N-1}{2}}e^{-\frac{x}{2b}} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)
\end{align}
Then, we want to obtain the solutions of $f'(x)=0$. The derivative reads
\begin{equation}
f'(x)=\frac{N-1}{2}\frac{f(x)}{x}-\frac{1}{2b}f(x)+\frac{\sqrt{Na}}{2b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}f(x)
\end{equation}
The equation to solve is thus
\begin{equation}
\frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}}\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=0
\end{equation}
When both the order and the argument are large, a uniform asymptotic expansion can be used for $I_{N-1}$ and $I_{N-1}$:
\begin{align}
I_{N-1}\left((N-1) z\right)&\sim\frac{e^{(N-1)\eta}}{(2\pi(N-1))^{\frac{1}{2}}(1+z^{2%
})^{\frac{1}{4}}}\sum_{k=0}^{\infty}\frac{U_{k}(p)}{(N-1)^{k}}\\
I_{N-1}'\left((N-1) z\right)&\sim\frac{(1+z^{2})^{\frac{1}{4}}e^{(N-1)\eta}}{(2\pi%
(N-1))^{\frac{1}{2}}z}\sum_{k=0}^{\infty}\frac{V_{k}(p)}{(N-1)^{k}}
\end{align}
where
\begin{equation}
\eta=(1+z^{2})^{\frac{1}{2}}+\ln\frac{z}{1+(1+z^{2})^{\frac{1}{2}}}
\end{equation}
Here, we choose
\begin{equation}
z=\frac{\sqrt{Nax}}{b(N-1)}
\end{equation}
Keeping the term $k=0$ only, we find (with $U_0(p)=V_0(p)=1$)
\begin{equation}
\frac{I'_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}{I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)}=\frac{\sqrt{1+z^2}}{z}
\end{equation}
The equation to be solved becomes
\begin{equation}
\frac{N-1}{x}-\frac{1}{b}+\frac{\sqrt{Na}}{b\sqrt{x}}
\sqrt{1+\frac{b^2(N-1)^2}{Nax}}=0
\end{equation}
or
\begin{equation}
N-1-\frac{x}{b}+
\sqrt{\frac{Nax}{b^2}+(N-1)^2}=0
\end{equation}
which has a simple non-zero solution
\begin{equation}
x=Na+2(N-1)b
\end{equation}
which seems to be numerically correct. For $a=1,b=1,N=20$, the approximation gives $x=58$ while a numerical evaluation of the solution is $x=57.49$. For $a=1,b=1,N=50$ we find $148$ to be compared to $147.496$.
| {
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$ab+ac+bc \equiv 1 \bmod abc$ or "easy chinese remainder theorem problems" When teaching students about the Chinese remainder theorem, it is traditional to ask them questions like: "An integer $n$ is equivalent to $r_1 \bmod m_1$, to $r_2 \bmod m_2$ and to $r_3 \bmod m_3$. Compute $n \bmod m_1 m_2 m_3$." For example,
There are certain things whose number is unknown. If we count them by threes, we have two left over; by fives, we have three left over; and by sevens, two are left over. How many things are there? -- Sunzi Suanjing, 3rd century CE.
If it happens that $m_1 m_2 \equiv 1 \bmod m_3$, $m_1 m_3 \bmod m_2$ and $m_2 m_2 \bmod m_1$, then the question is particular easy to answer: One has $n = r_1 m_2 m_3 + r_2 m_1 m_3 + r_3 m_1 m_2$. I noticed this when preparing for my class today and planning to ask such a question with $(m_1, m_2, m_3) = (2,3,5)$, which has this property. Note that we can rewrite this property as $m_1 m_2 + m_1 m_3 + m_2 m_3 \equiv 1 \bmod m_1 m_2 m_3$.
So, for the fun of it, here is my question:
Can we describe all $k$-tuples of integers $(m_1, m_2, \ldots, m_k)$ such that $$m_1 m_2 \cdots m_{j-1} m_{j+1} \cdots m_k \equiv 1 \bmod m_j$$ for $1 \leq j \leq k$?
| $(1,1,1)$, $(1,1,m)$ for $m \geq 2$ and $(2,3,5)$ and are the only solutions with three positive integers. I'll limit myself to positive solutions in the following.
Note that $(m_1, m_2, \ldots, m_k, 1)$ obeys the conditions if and only if $(m_1, \ldots, m_k)$ do, so we may reduce to studying solutions with $m_j \geq 2$. (Or $|m_j| \geq 2$ if we allow negative solutions.) Thus, we want to show the only solutions with $k \leq 3$ are $\emptyset$, $(m)$ and $(2,3,5)$.
We can rewrite
$$ \sum m_1 m_2 \cdots m_{j-1} m_{j+1} \cdots m_k = N m_1 \cdots m_k + 1$$
as
$$\sum \frac{1}{m_j} = N + \frac{1}{m_1 \cdots m_k}.$$
Note that the $m_j$ must be relatively prime and $\geq 2$, so we can bound $N$ by $1/p_1+1/p_2 + \cdots + 1/p_k$, where $p_j$ is the $j$-th prime.
When $k=0$ we get $\emptyset$. When $k=1$, any $m$ works. When $k=2$ we must have $N < 1/2+1/3 =5/6$ so $N=0$, but $1/m_1 + 1/m_2$ is never of the form $1/n$. When $k=3$, we must have $N < 1/2+1/3+1/5 = 1+1/30$, so $N=0$ or $1$, and $N=0$ doesn't work as above. Moreover, the only case of $1/a+1/b+1/c > 1$ with $(a,b,c)$ pairwise relatively prime is $(2,3,5)$.
A few thoughts on larger $k$. There is an infinite sequence of examples as follows: Define $x_1 = 2$ and $x_{j+1} = x_1 x_2 \cdots x_j +1$, so the $x$'s start off $2$, $3$, $7$, $43$, $1807$, ... Then $(x_1, x_2, \ldots, x_{k-1}, x_k-2)$ is always a solution.
For any finite $k$, there are only finitely many solutions, and we can find them with a backtracking search. For example, if $k=4$, then $1/a+1/b+1/c+1/d \leq 1/2+1/3+1/5+1/7 = 1+37/210$, so $N=0$ or $1$, and there are no solutions with $N=0$. Assume that $a<b<c<d$. If $a \geq 3$, then $1/a+1/b+1/c+1/d \leq 1/3+1/5+1/7+1/8 <1$, so that doesn't work and $a$ must be $2$. If $b \geq 5$ then $1/a+1/b+1/c+1/d \leq 1/2+1/5 + 1/7 + 1/8<1$, so we must have $b=3$ (as $\mathrm{GCD}(2,4) \neq 1$). When $b=3$, if $c>12$ then $1/2+1/3+1/c+1/d < 1$, so instead $c \leq 12$ and we have $(2,3,7,\ast)$ or $(2,3,11,\ast)$. Each of these completes to one value of $d$: $(2,3,7,41)$ and $(2,3,11,13)$.
Saying anything about all $k$ at once, or about negative $m_j$, seems hard.
| {
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Generating functions (with symbol method) of special partitions of natural numbers I want to show that the generating function of the number of partitions where every summand appears at most twice of every natural number n equals the number of paritions of n into summands which are not divisible by 3.
I know that for the partitions $\mathcal{\tilde{P}}$ where every summand appears at most once it holds that:
$\tilde{\mathcal{P}} = (\epsilon + 1)\times (\epsilon + 2) \times (\epsilon + 3) \times \dotsc$ and therefore $P(z) = (1+z)(1+z^2)(1+z^3) + \dotsc$.
So now I'm wondering if the partitions $\mathcal{P}$ where every summand appears at motst twice is described as
1) $\mathcal{P} = (\{\epsilon\} + \{1\} + \{1,1\})\times (\{\epsilon\} + \{2\} + \{2,2\}) \times (\{\epsilon\} + \{3\} + \{3,3\}) \times \dotsc$ and therefore corresponds with
1a)
$P(z) = (1+z+2z)(1+z^2+2z^2)+(1+z^3+2z^3) \dotsc$ or with
1b)
$P(z) = (1+ z + z)(1+ z^2 + z^4)(1+z^3+z^6)\dotsc$
or if the description is
2) $\mathcal{P} = (\epsilon + 1)\times(\epsilon + 1) \times (\epsilon + 2) \times (\epsilon + 2) \times \dotsc$ and therefore corresponds with
$P(z) = (1+z)^2(1+z^2)^2(1+z^3)^2\dotsc$?
The second partition, the one into parts that are not divisible by 3, I think the description is:
$\mathcal{P}^{\ast} = SEQ(1) \times SEQ(2) \times SEQ(4) \times \dotsc$
and therefore $P^{\ast}(z) = \frac{1}{1-z}\frac{1}{1-z^2}\frac{1}{1-z^4}\frac{1}{1-z^5}\dotsc$.
So is any of the options described in 1) correct, and if so, which one or why not, and is 2) correct?
| The GF of the partitions where every summand appears at most twice is
$$f(x)=(1+x+x^2)(1+x^2+x^4)\cdot\ldots = \prod_{k\geq 1}(1+x^k+x^{2k})$$
while the GF of the partitions into summands which are not divisible by 3 is
$$ g(x) = \frac{1}{1-x}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x^4}\cdot\frac{1}{1-x^5}\cdot\ldots = \prod_{\substack{k\geq 1\\3\nmid k}}\frac{1}{1-x^k}. $$
It is not difficult to check that $f(x)=g(x)$. Indeed
$$ f(x)=\prod_{k\geq 1}\frac{1-x^{3k}}{1-x^k} = \prod_{k\geq 1}\frac{1}{1-x^k}\prod_{k\geq 1}(1-x^{3k}) = \prod_{n\geq 1}\frac{1}{1-x^n}\prod_{\substack{n\geq 1\\ 3\mid n}}(1-x^n) = g(x).$$
| {
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Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$ When $a=1,2,3,...,2010,2011$ the roots of the equation $x^2 -2x-a^2-a=0$ are $(m_1,n_1 ), (m_2,n_2 ), (m_3,n_ 3),..., (m_{2010},n_{2010} ), (m_{2011},n_{2011 }) $ respectively. Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2010}} +\frac{1}{n_{2010}} +\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$
**My attempt **$$$$
$ \frac{1}{m_a} +\frac{1}{n_a}$can be manipulated to give
$$\frac{m_a+n_a}{m_an_a}$$then w.l.o.g $m_a\ge n_a$
And since$$m_a=2\frac{2+2\sqrt{a^2+a+1}}{2}$$
And $$n_a= 2\frac{2-2\sqrt{a^2+a+1} }{2}$$
We get
$$\frac{m_a+n_a}{m_an_a}=\frac{-2}{a(a+1)}$$
But I dont know what to do next,
Suggestions and solutions would be appreciated
Taken from the 2011 IWYMIC
| You don’t need to compute the roots $n_a, m_a$. Just note that: $\dfrac{1}{n_a}+\dfrac{1}{m_a}=\dfrac{n_a+m_a}{n_am_a}=\dfrac{S_a}{P_a}$ where $S_a$ and $P_a$ denote the sum and the product of those two roots. And:
$\dfrac{S_a}{P_a}=\dfrac{-b}{c}=-\dfrac{2}{a(a+1)}=2\left(\dfrac{1}{a+1}-\dfrac1a\right)$. Finally:
$$\sum_{a=1}^{2011}{\dfrac{S_a}{P_a}}=2\sum_{a=1}^{2011}{\left(\dfrac{1}{a+1}-\dfrac1a\right)}=2\left(\dfrac{1}{2012}-\dfrac11\right)$$
| {
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Help with $\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} \sinh^{-1}(1)^2$ I've been trying to do the above integral using elementary methods. So far, I've reduced the integral down to evaluating $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{\sqrt{1-x^2}} \; \mathrm{d}x$ or $\displaystyle \int_0^1 \frac{\sin^{-1}(x)}{1+x^2} \; \mathrm{d}x$ through a u-sub and IBP, but neither of these integrals seem to yield an elementary method.
Any help would be great.
| Substitute $t=\sin x$ and then IBPs,
$$\hspace{-1cm}
\int_0^{\frac{\pi}{2}} \frac{x \cos x}{\sin^2x+1}dx
=\int_0^1\frac{\sin^{-1}t}{1+t^2}dx=\frac{\pi^2}8-I,\>\>\>\>
I=\int_0^1\frac{\tan^{-1}t}{\sqrt{1-t^2}}dt \tag1$$
Express $\tan^{-1}t=\int_0^1\frac t{1+y^2t^2}dy$ and use
$$\int_0^1\frac {t\>dt}{\sqrt{1-t^2}(1+y^2t^2)}=\frac{\sinh^{-1}y}{y\sqrt{1+y^2}}\tag 2$$
to integrate $I$,
$$\begin{align}
\hspace{-8mm}
I & \hspace{-3mm}
=\int_0^1\hspace{-4mm} \int_0^1\hspace{-3mm} \frac{tdydt }{\sqrt{1-t^2}(1+y^2t^2)}
=\int_0^1\frac{\sinh^{-1}y}{y\sqrt{1+y^2}}dy
\overset{u=\frac1y}=\int_1^\infty\frac{\text{csch}^{-1}u}{\sqrt{1+u^2}}du \\
& \hspace{-4mm}= \int_1^\infty d(\sinh^{-1}u)\>\text{csch}^{-1}u
=\sinh^{-1}u\>\text{csch}^{-1}u|_1^\infty+ \int_1^\infty\frac{\text{sinh}^{-1}u}{u\sqrt{1+u^2}}du \\
& \hspace{-4mm}= -[\sinh^{-1}(1)]^2 + \int_0^\infty\frac{\text{sinh}^{-1}u}{u\sqrt{1+u^2}}du-I \tag3 \\
\end{align}$$
Evaluate the remaining integral with (2)
$$\hspace{-1cm}
\int_0^\infty\frac{\text{sinh}^{-1}udu }{u\sqrt{1+u^2}}
=\int_0^\infty\hspace{-4mm} \int_0^1\frac {t dtdu}{\sqrt{1-t^2}(1+u^2t^2)}
=\frac\pi2 \int_0^1\frac {dt}{\sqrt{1-t^2}}= \frac{\pi^2}4 \\ \tag4
$$
Plug (4) into (3) to obtain $I =\frac{\pi^2}8-\frac12[\sinh^{-1}(1)]^2$. Then, plug it into (1) to obtain
$$\int_0^{\frac{\pi}{2}} \frac{x \cos x}{\sin^2x+1}dx = \frac{\pi^2}8-I=\frac12[\sinh^{-1}(1)]^2$$
| {
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Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts ,
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But it got me no where.
Any hints.
| Hint: $$\frac{x^2+1}{x^4+1}=\frac1{2\left(x^2+\sqrt2x+1\right)}+\frac1{2 \left(x^2-\sqrt2x+1\right)}.$$
| {
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest possible value of $x+y$?
The book says "the greatest value of $4$ is divided by $6$, which produces a remainder of $4$. The greatest value of $y$ is when $2$ is divided by, which produces a remainder of $2$. Therefore, the greatest value of $x+y$ is $6$."
I think what's throwing me off is the phrase "multiple of $4$" bc it makes me think that any multiple of $4$ can be divisible by $6$ (i.e. $24/6 = 4$). The books answer doesn't use multiples, just the $4$ and $2$, respectively. I don't understand how this works. Can someone please clarify?
|
The greatest value of $\color{red}{x \text{ is when}}$ 4 is divided by 6, which produces a remainder of 4.
Note that $4$ divided by $6$ produces the remainder $4$. In general, division by $6$ can produce remainders $0,1,2,3,4,5$. A multiple of $4$ divided by $6$ can not produce the remainder $1,3,5$, but only $0,2,4$ (why?), hence $4$ is the greatest possible remainder.
The greatest value of $y$ is when 2 is divided by $\color{red}3$, which produces a remainder of 2.
Note that $2$ divided by $3$ produces the remainder $2$. In general, division by $3$ can produce remainders $0,1,2$. A multiple of $2$ divided by $3$ can produce $0,1,2$, hence $2$ is the greatest possible remainder.
| {
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Inequality with a rational function Let $0< a_1\leq a_2\leq a_3\leq a_4\leq a_5$. Define the function $$f(x,y,z)=\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}.$$
Show that for all $x,y,z\in \mathbb R$ we have $f(x,y,z)\leq \cfrac{1}{a_1+a_2+a_3}.$
Information about the problem:
*
*The function forms an asymptote at $\cfrac{1}{a_1+a_2+a_3}$.
*We have $f(1,0,0)=\cfrac{1}{a_1+a_2+a_3}$.
*As a function of $x$, it seems there are only three critical points. Same for $y$ and $z$.
*Perhaps it helps to show that
$$\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}\leq \cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_3(y+z)^3+a_3(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_3(y+z)^2+a_3(z)^2 )^3}.$$
| Not an answer.
First, I would like to give the equivalent problem:
Problem 1: Let $0 < a_1 \le a_2 \le a_3 \le a_4 \le a_5$. Let $x, y, z \in \mathbb{R}$ such that
$$a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2 = a_1 + a_2 + a_3.\tag{1}$$ Prove that
$$a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3 \le a_1 + a_2 + a_3.$$
Remark: The constraint (1) is an ellipsoid.
Second, if $x, y, z$ is non-negative, the inequality is true as Lemma 1 stated. One may use the Buffalo Way to prove Lemma 1 (easily with computer). So, if one can prove that the optimal $x, y, z$ in the original problem are non-negative, the desired result follows.
Lemma 1: Let $x, y, z\ge 0$ not all zero. Then
$$\frac{(a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3)^2}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^3}
\le \frac{1}{a_1+a_2+a_3}.$$
| {
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Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\
&=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\
&=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx.
\end{align}
I can't go further from here, please help.
| As has been hinted, polar coordinates are the way to go. But if you really need to continue, the first thing I would note is that your integrand is an even function of $x$ over a symmetric integral. Therefore, you can simplify a tad to
$$\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx. $$
Now, whenever I see something like $\sqrt{a^2-x^2},$ I think trig substitution. In this case, you draw a triangle with $a$ as the hypotenuse, and one of the other sides as $x$. The third side will be $\sqrt{a^2-x^2}.$ I'll just put $\theta$ as the angle opposite $x,$ so that we have the following:
\begin{align*}
\frac{x}{a}&=\sin(\theta)\\
x&=a\sin(\theta)\\
dx&=a\cos(\theta)\,d\theta\\
\frac{\sqrt{a^2-x^2}}{a}&=\cos(\theta).
\end{align*}
For $x$ going from $0$ to $a,$ we can write that $0\le\theta\le\pi/2.$ Our integral becomes
\begin{align*}
\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx
&=\dfrac{4}{3}\int_{0}^{\pi/2}a\cos(\theta)\cdot\left(2a^2\sin^2(\theta)+a^2\right)a\cos(\theta)\,d\theta\\
&=\frac{4a^4}{3}\int_0^{\pi/2}\cos^2(\theta)\left(2\sin^2(\theta)+1\right)d\theta.
\end{align*}
This integral breaks up into two pieces, each of which succumbs to the usual methods for integrating products of $\sin(\theta)$ and $\cos(\theta).$ Can you take it from here? The result should be
$$\frac{4a^4}{3}\cdot\frac{3\pi}{8}=\frac{\pi a^4}{2}.$$
Check with polar version:
\begin{align*}
\iint_R\big(x^2+y^2\big)\,dA
&=\int_0^{2\pi}\int_0^a r^2\cdot r\,dr\,d\theta\\
&=2\pi\,\frac{r^4}{4}\bigg|_0^a\\
&=\frac{\pi a^4}{2},
\end{align*}
as required.
| {
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Question aboutPartial Fractions for example:
$$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$$
first method is:
$${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$$
but it is hard and take much time to find A,B,C
second method(Substituting the roots, or "zeros") is:
\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}
it is better and easier method for Partial Fractions. My question is why this method works? can you prove this method?
| Your "second method" is the next step of the first method. At each of the zeroes of $$x(x+2)(3x-2)$$ only one of the ABC terms is non-zero so you can solve for each ABC easily.
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"answer_id": 1
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Determinant of Large Matrix $A=\left[\begin{array}{ccccc}{-2} & {-1} & {} & {\cdots} & {-1} \\ {-1} & {-2} & {-1} & {\cdots} & {-1} \\ {} & {} & {\ddots} & {} & {} \\ {-1} & {\cdots} & {-1} & {-2} & {-1} \\ {-1} & {\cdots} & {} & {-1} & {-2}\end{array}\right] \in \mathbb{R}^{53 \times 53}$
So we want to find determinant of this big matrix. I tried for some cases I got the pattern like for even dimension determinant is $n+1$ and and for odd dimension it is $-n-1$ so answer should be $-54$ ;I guess. But what is formal method to do this calculation ; idea I have in mind is to find eigenvalue and then product will give me determinant.
| We can work with general dimension $n$. The wanted determinant is $(-1)^n$ times the determinant of
$$
B=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & { 2}\end{array}\right].
$$
Do the combination $C_n\leftarrow \sum_{i=1}^nC_i$. Then $\det(B)=\det(C)$, where
$$
C=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 2+n-1} \\ { 1} & { 2} & { 1} & {\cdots} & { 2+n-1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & {2+n- 1} \\ { 1} & {\cdots} & {} & { 1} & { 2+n-1}\end{array}\right],
$$
hence
$$\det(A)=(-1)^n(n-1)\det \left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & {1}\end{array}\right].
$$
Finally, do the substitutions $C_i\leftarrow C_i-C_n$, $1\leqslant i\leqslant n$ to get that the last determinant is one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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