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If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
I solved this problem as follow,
Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\frac12(\alpha-\frac1{\alpha})$ and $$(x+\sqrt{1+x^2})^3=\left[\frac12(\alpha-\frac1{\alpha}) +\sqrt{1+\frac14(\alpha^2+\frac1{\alpha^2}-2)}\right]^3=\left[(\frac{\alpha}2-\frac1{2\alpha}) +\sqrt{\frac{\alpha^2}4+\frac1{4\alpha^2}+\frac12)}\right]^3=\left(\frac{\alpha}2-\frac1{2\alpha}+\left|\frac{\alpha}2+\frac1{2\alpha}\right|\right)^3=\alpha^3=2009$$
I'm wondering, is it possible to solve this problem with other approaches?
|
Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\dfrac12\left(\alpha-\dfrac1{\alpha}\right)$
We also have:
$$
\begin{align}
x = \frac{1}{2}\left(x+\sqrt{1+x^2} \;+\; x-\sqrt{1+x^2}\right) = \frac{1}{2}\left(x+\sqrt{1+x^2} \;-\; \frac{1}{x+\sqrt{1+x^2}}\right)
\end{align}
$$
Comparing with $\,x = \dfrac12\left(\alpha-\dfrac1{\alpha}\right)\,$ it follows that $\,x+\sqrt{1+x^2} = \alpha\,$, because the function $\,f(t) = \dfrac{1}{2}\left(t - \dfrac{1}{t}\right)\,$ is injective on $\,\mathbb R^+\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Analytical Solution to a nonlinear system of equations Is it possible to analytically solve for x,y,z in the following system, and if so how would one go about it?
$a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x = 0$
$a_{1} (z - 2y + x) - y + 2y^{3} - a_{2}y = 0$
$a_{1} (x - 2z + y) - z + 2z^{3} - a_{2}z = 0$
I've tried numerous methods such as cyclic permutations of the three equations and substitution but none seem to work, hence any advice would be greatly appreciated.
| Notice the symmetry in the equations. Adding them all together, we find that
$$-(x+y+z)+2(x^3+y^3+z^3)-a_2(x+y+z) = 0$$
$$2(x^3+y^3+z^3) = (1+a_2)(x+y+z).$$
Now, we utilize the convenient factored form:
$$x^3+y^3+z^3 - (x+y+z) = (x+y+z)(x+y+z+1)(x+y+z-1),$$
to write the above equation as
$$(x+y+z)(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}(x+y+z).$$
Assuming $x+y+z \ne 0$,
$$(x+y+z+1)(x+y+z-1) = \frac{a_2-1}{2}.$$
Thus,
$$z = \pm \sqrt{\frac{a_2+1}{2}}-x-y.$$
Notice that if we substitute this into the original first equation, we find
$$a_1(\sqrt{\frac{a_2+1}{2}}-3x)+2x^3-a_2x = 0.$$
This is a cubic in $x$, which you can solve.
Proceed in a similar fashion in the second original equation to find a cubic in terms only of $y$. Finally, solve $z = \pm \sqrt{\frac{a_2+1}{2}}-x-y$ to find $z$.
Note, you will also have to check for solution when $x+y+z = 0$, but I think this will at least give you a good start.
By using $s = x+y+z$ and looking at first part of dxiv's answer, this case is very simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Interpreting the ratio test for $\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $ We want to find the limit of this.
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $$
I would have done it as follows:
$$\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = \lim_{n\to\infty}\frac{\frac{n+1}{(n+1)^2+n}}{\frac{n}{n^2+n}} =\lim_{n\to\infty} \frac{n+1}{(n+1)^2+n} \cdot \frac{n^2+n}{n} = \lim_{n\to\infty}\frac{n^3+2n^2+n}{n^3+3n^2+n} \\
=\lim_{n\to\infty} \frac{n^3 \cdot \bigl(1+\frac{2}{n} + \frac{1}{n^2} \bigr)}{n^3\cdot\big(1+\frac{3}{n} + \frac{1}{n^2} \bigr)} = \frac{1}{1} = 1$$
According to the ratio test, the series converges if $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| <1$ and it diverges if it's $> 1$.
But since we get $1$ here, the series converges towards that value, no? But according to the ratio test, we can't make a statement about the series of the limit of $\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = 1$
What am I misunderstanding here?
| If you are familiar with harmonic numbers
$$a_n= \sum_{k=1}^n \frac{n}{n^2+k}=n \left(H_{n^2+n}-H_{n^2}\right)$$
Using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\right)$$ and continuing with Taylor series
$$a_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+\frac{1}{4
n^3}+O\left(\frac{1}{n^4}\right)$$
Use it for $n=10$ : the exact result is
$$a_{10}=\frac{11210403701434961}{11818204429243212}=0.948571$$ while the truncated series gives
$$\frac{11383}{12000}=0.948583$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Showing $\sum_{cyc} \frac{\cos(\frac{\alpha+\beta}{2})}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}=2$ when $\alpha+\beta+\gamma=\pi$ I saw this problem in a math magazine:
Let $\alpha,\beta$ and $\gamma$ be the angles of a triangle, so that
$$
\alpha+\beta+\gamma=\pi
$$
Show that
$$
\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\cos\left(\frac{\alpha+\gamma}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
I tried to rewrite $\alpha+\beta$ with $\pi-\gamma$ so that
$$
\cos\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)= \cos \frac{\pi}{2}\cos \frac{\gamma}{2}+ \sin\frac{\pi}{2}\sin \frac{\gamma}{2}= \sin\frac{\gamma}{2}
$$
Now the above equation equals
$$
\frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
But now I don't have ideas for continuation. Any hints to tackle this?
| One method is to use Computer Algebra Systems to factor rational expressions.
Let
$\, x = e^{i\alpha/2},
y = e^{i\beta/2}, z = e^{i\gamma/2}.\,$ Use
Euler's formula
to get
$$ \cos(\frac\alpha2) = \frac{x+\frac1x}2,
\quad \cos(\frac\beta2) = \frac{y+\frac1y}2,
\quad \cos(\frac\gamma2) = \frac{z+\frac1z}2,\\
\cos(\frac{\alpha+\beta}2) = \frac{xy+\frac1{xy}}2,
\cos(\frac{\alpha+\gamma}2) = \frac{xz+\frac1{xz}}2,
\cos(\frac{\beta+\gamma}2) = \frac{yz+\frac1{yz}}2.
$$
Then using a Computer Algebra System to do the algebra
factoring,
$$
\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\cos\left(\frac{\alpha+\gamma}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}} - 2 = \\
\frac{4(1+(xyz)^2)}{(1+x^2)(1+y^2)(1+z^2)}. $$
This is zero iff $(xyz)^2 = -1$ iff $\alpha+\beta+\gamma \equiv \pi \pmod{ 2\pi}.$
Alternatively, continuing from your approach (sign error fixed), you got
$$ \frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2.
$$
Multiply both sides by twice the product of the cosines and
use the known trig identity $\sin(2x) = 2\sin(x)\cos(x)$ to get
$$ \sin(\alpha) + \sin(\beta) + \sin(\gamma) = 4\cos(\frac\alpha2)
\cos(\frac\beta2)\cos(\frac\gamma2) $$ which is a known trig identity
for triangles as in the Wikipedia article
List of trigonometric identites.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A question from Integration Bee 2022: Evaluate $\int^2_{1/2}\ln(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})})dx$ I am trying to evaluate the integral $$\int^2_{\frac{1}2}\ln\left(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}\right)dx$$ by separating it as $$\int^2_{\frac{1}2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^2_{\frac{1}2}\ln\ln\left(\left(x-\frac{1}{2}\right)^2+4\right)dx$$
I think the substitution $x=1/u$ should be applied for the first item but I don't know how to go on. Please help. Thank you. (The final answer is $-\frac{3}{2}\ln2$)
|
$$I=\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^2_{1/2}\ln\ln\left(\left(x-\frac{1}{2}\right)^2+4\right)dx$$
Let $t=x-\frac{1}2$
$$I=\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^{3/2}_{0}\ln\ln\left(t^2+4\right)dt=I_1-I_2$$
Let $t=x-\frac{1}x,~~dt=dx-d\left(\frac{1}x\right),~~t^2+4=(x+\frac{1}x)^2$
$$\begin{align}
I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2\left(dx-d\left(\frac{1}x\right)\right)\\
\\
I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2 dx-\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2d\left(\frac{1}x\right)~~~~~\text{let}~u=\frac{1}x\\
\\
I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2 dx-\int^{1/2}_{1}\ln\ln\left(u+\frac{1}u\right)^2du\\
\\
I_2&=\int^{2}_{1/2}\ln\ln\left(x+\frac{1}x\right)^2 dx
\end{align}$$
$$I=I_1-I_2=\int^{2}_{1/2} \ln\left(\frac{1}2\right) dx=-\frac{3}{2}\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$
| Even if there is an antiderivative (have a look here), you could write
$$\log(\sin(x))=\log(x)-\sum_{n=1}^\infty(-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n}$$ where appear Bernoulli numbers.
Integrate termwise
$$\int_0^{\frac \pi 2}x\,\log(x)=-\frac{1}{16} \pi ^2 (1+2 \log (2)-2 \log (\pi ))$$
$$\int_0^{\frac \pi 2} (-1)^{n+1}\, \frac{ 2^{2 n-1}\, B_{2 n} }{n \,(2 n)!}x^{2 n+1}=(-1)^{n+1}\, \frac{\pi ^{2 n+2} B_{2 n}}{16 n (n+1) (2 n)!}$$ The infinite summation gives
$$-\frac{1}{16} \left(7 \zeta (3)+\pi ^2-2 \pi ^2 \log (\pi )\right)$$ and then the result (with a sign difference)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x+y+z=4 \\
x^2+y^2+z^2=14 \\
x^3+y^3+z^3=34
\end{split}
\end{equation}$$
My work:
I found out that $$xy+yz+xz=1$$ and $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=22$$
After this I'm stuck. Any help is greatly appreciated.
EDIT This not a duplicate. I'm looking for a detailed solution and not a solution just by inspection. Also I thought of a new idea. Maybe e should consider a cubic polynomial whose roots are $x,y,z$
| We observe that,
$$xy=\frac{(4-z)^2-(14-z^2)}{2}=f(z)$$
and
$$\begin{align}(4-z)(14-z^2)=xy(x+y)+x^3+y^3\\
=f(z)\times (4-z)+34-z^3\end{align}$$
After finding $z$, for the final step we need to solve
$$\begin{cases}x+y=4-z\\ xy =f(z)\end{cases}$$
By the Vieta's formulas, we have
$$t^2-(4-z)t+f(z)=0$$
where, $$t_1=x,\, t_2=y \,\,\, \text{or}\,\,\,t_1=y,\, t_2=x $$
Thus for every $z$, we obtain the following solutions:
$$(z,x,y) \,\,\,\text{and}\,\,\, (z,y,x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$ I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$
Using partial fration f(z) can written as:
$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$
since we have from Maclurin's series,
$$\sum_0^\infty \frac{1}{1+z} = (-z)^n for |z| <1$$
so ,
$$f(z) = \frac{1}{2} [\sum (-1)^n z^n - \frac{1}{1 + (z + 2)}]$$
How to proceed from here?
| For the first term, it's already done: $$\dfrac 12\dfrac 1{z+1}$$ on $\mid z+1\mid\gt0$.
For the other one, $$\dfrac 1{z+3}=\dfrac 1{(z+1)+2}=\dfrac 1{2--(z+1)}=\dfrac 12\dfrac 1{1--\dfrac {z+1}2}=\dfrac 12\sum_{n\ge0}(-\dfrac {z+1}2)^n$$ on $\mid z+1\mid\lt2$.
So to finish we have $$\dfrac 12\dfrac 1{z+1}-\dfrac 14\sum_{n\ge0}(-\dfrac{z+1}2)^n$$, on the overlap $0\lt\mid z+1\mid\lt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding a solution to the generalised Pell's equation $x^2-31y^2=5$ I'm trying to find a solution to the generalised Pell's equation $x^2-31y^2=5$. So far I have found the fundamental solution $x=1520,y=273$ to the equation $x^2-31y^2=1$, obtained from calculating the convergents $p_n$ and $q_n$ in the continued fraction expansion of $\sqrt{31}$. It's also possible to find e.g. solutions to $x^2-31y^2=4$ just by dividing both sides of this equation by 4 and using the fundamental solution.
Since $5$ is prime, though, there doesn't seem to be an obvious way to generate a solution using e.g. Brahmagupta's identity by multiplying solutions with a factor of $5$ on the right hand side. How can you find a solution to this?
| The same continued fraction that told you $1520^2 - 31 \cdot 273^2 = 1$ also tells you $6^2 - 31 \cdot 1^2 = 5$ and $657^2 - 31 \cdot 118^2 = 5$
I also have a program somewhere that typesets the whole business below in Latex, for now here is the bald version:
./Pell_Lubin 31
31
0. 5 : { sqrt(31) - 5} / 1 ;; 5 / 1 ::: 5^2 - 31 * 1^2 = -6
1. 1 : { sqrt(31) - 1} / 6 ;; 6 / 1 ::: 6^2 - 31 * 1^2 = 5
2. 1 : { sqrt(31) - 4} / 5 ;; 11 / 2 ::: 11^2 - 31 * 2^2 = -3
3. 3 : { sqrt(31) - 5} / 3 ;; 39 / 7 ::: 39^2 - 31 * 7^2 = 2
4. 5 : { sqrt(31) - 5} / 2 ;; 206 / 37 ::: 206^2 - 31 * 37^2 = -3
5. 3 : { sqrt(31) - 4} / 3 ;; 657 / 118 ::: 657^2 - 31 * 118^2 = 5
6. 1 : { sqrt(31) - 1} / 5 ;; 863 / 155 ::: 863^2 - 31 * 155^2 = -6
7. 1 : { sqrt(31) - 5} / 6 ;; 1520 / 273 ::: 1520^2 - 31 * 273^2 = 1
8. 10 : { sqrt( 31 ) - 5 } / 1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove inequality with complicated square roots I've been trying to prove the following inequality but the square roots really give me a hard time.
$$ x \cdot (\sqrt{4-x^2} +2x -\sqrt2)\leq\sqrt{-(x^2-4x+2)(x^2+4x+2)} $$ for $ 1 \leq x\leq\sqrt2$. Considering both sides as a function and using WolframAlpha,I found the intersection points $ x= \frac{\sqrt2}{2} $ and $ x=\sqrt2 $ but I couldn't prove that there is no other intersection point within this interval. Since the derivatives are also quite complicated I'm stuck.
| Let $P := -(x^2 - 4x + 2)(x^2 + 4x + 2) = (4x)^2 - (x^2 + 2)^2$.
It suffices to prove that
$$x(\sqrt{4 - x^2} - \sqrt 2)
\le \sqrt{P} - 2x^2$$
or
$$x\cdot \frac{2 - x^2}{\sqrt{4 - x^2} + \sqrt 2 }
\le \frac{P - (2x^2)^2}{\sqrt{P} + 2x^2}$$
or
$$x\cdot \frac{2 - x^2}{\sqrt{4 - x^2} + \sqrt 2 }
\le \frac{(2 - x^2)(5x^2 - 2)}{\sqrt{P} + 2x^2}.$$
Since $\sqrt{4 - x^2} + \sqrt 2
\ge \sqrt{4 - 2} + \sqrt 2 > 2$ and
$$P
\le (4x)^2 - 4 \cdot x^2 \cdot 2
= 8x^2 \le 9x^2,$$
it suffices to prove that
$$x\cdot \frac{2 - x^2}{2}
\le \frac{(2 - x^2)(5x^2 - 2)}{3x + 2x^2}$$
or
$$x/2
\le \frac{5x^2 - 2}{3x + 2x^2}$$
or
$$-2x^3+7x^2-4 \ge 0$$
or
$$(x-1)(-2x^2+5x+5) + 1 \ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Digit sum formula $n - 9 \sum_{i\ge1} \left \lfloor \frac{n}{10^i} \right \rfloor$ Let $S(n)$ be the sum of digits of n
Prove that $S(n) = n - 9 \sum_{i\ge 1} \left \lfloor \frac{n}{10^i} \right \rfloor$ for all natural numbers n
I started with induction which works easily if $10$ doesn't divide $n+1$, because then $S(n+1)$ is just $S(n) + 1$.
However, if $10|(n+1)$ then $S(n+1)$ could be anything. Moreover, I have no idea how to use that fact.
If I write e.g. $n = 10k + 9$ then $\sum_{i\ge1} \left \lfloor \frac{10k+10}{10^i} \right \rfloor$ = $\sum_{i\ge1} \left \lfloor \frac{k+1}{10^{i-1}} \right \rfloor$ and I don't know how to proceed
| Using strong induction, assume for all $n = 1,2, \ldots, k$, the sum of digits formula holds:
$$S(n) = n - 9\sum_{i\ge 1} \left\lfloor\frac n{10^i}\right\rfloor$$
The induction step would be to prove that $S(k+1)$ follows the same formula.
For the cases that $10\mid (k+1)$, let $k+1 = 10q$, (and so $1\le q \le k$)
$$\begin{align*}
S(k+1) &= S\left(\frac{k+1}{10}\right)\\
&= S(q)\\
&= q - 9\sum_{i\ge 1} \left\lfloor\frac q{10^i}\right\rfloor\\
&= \frac{k+1}{10} - 9 \sum_{i\ge 1} \left\lfloor\frac {(k+1)/10}{10^i}\right\rfloor\\
&= 10\cdot\frac{k+1}{10}-9\cdot\frac{k+1}{10} - 9 \sum_{j\ge 2} \left\lfloor\frac {k+1}{10^j}\right\rfloor && (j=i+1)\\
&= (k+1) - 9 \sum_{j\ge 1} \left\lfloor\frac {k+1}{10^j}\right\rfloor\\
&= RHS
\end{align*}$$
For the other cases that $10\not\mid(k+1)$, OP already mentioned in the question that $S(k+1) = S(k)+1$.
It is possible to merge the two cases with a slight change, by also proving the base case $n=0$ and including it into the assumption.
For the $n=k+1$ case, let $k+1 = 10q + r$, where $q = \left\lfloor \dfrac{k+1}{10}\right\rfloor$ and $r=(k+1)\bmod 10$.
$$\begin{align*}
S(k+1) &= S(q) + r\\
&= q - 9\sum_{i\ge 1} \left\lfloor\frac q{10^i}\right\rfloor + r\\
&= q - 9\sum_{i\ge 1} \left\lfloor\frac {10q/10}{10^i}\right\rfloor + r\\
&= 10q - 9\cdot \frac{10q}{10} - 9\sum_{j\ge 2} \left\lfloor\frac {10q}{10^j}\right\rfloor + r&&(j=i+1)\\
&= 10q - 9\sum_{j\ge 1} \left\lfloor\frac {10q}{10^j}\right\rfloor + r\\
&= 10q+r - 9\sum_{j\ge 1} \left\lfloor\frac {10q}{10^j} + \frac{r}{10^j}\right\rfloor\\
&= (k+1) - 9\sum_{j\ge 1} \left\lfloor\frac {k+1}{10^j}\right\rfloor\\
&= RHS
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Alternating sum of binomial coefficients $\sum_{k=0}^{49}(-1)^k \binom{99}{2k} = -2^{49}$ How can you prove that $$\sum_{k=0}^{49} \binom{99}{2k}(-1)^k = -2^{49}?$$
A more general formula seems to be $\sum_{k=0}^{n} \binom{2n + 1}{2k}(-1)^k$
For $n = 0:$ it equals $2^0$
For $n = 1: -2^1$
For $n = 2: -2^2$
For $n = 3: 2^3$
For $n = 4: 2^4$
For $n = 5: -2^5$
For $n = 6: -2^6$
For $n = 7: 2^7$
etc.
| How to write $(-1)^k = x^{2k}$, to match the $2k$ in the binomial coefficient? $x=i$ or $-i$.
Here only the even $2k$ terms appear; how to introduce the odd terms? By averaging the conjugates: $\frac12 \left[(1+i)^{99}+(1-i)^{99}\right] = \Re\left[(1+i)^{99}\right] = \ldots$
Combining these two ideas,
$$\begin{align*}
\sum_{k=0}^{49}\binom{99}{2k}(-1)^k &= \sum_{k=0}^{49}\binom{99}{2k}i^{2k}\\
&= \Re\left[\sum_{h=0}^{99}\binom{99}h i^h\right] &&(h=2k)\\
&= \Re\left[(1+i)^{99}\right]\\
&= \left(\sqrt2\right)^{96}\Re\left[(1+i)^3\right]\\
&= \left(\sqrt2\right)^{96}\left(1-3\right)\\
&= -2^{49}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving definite integral $\int_{0}^{1} \tan^{-1}(1-1/x)dx$ $$\int_{0}^{1} \tan^{-1}\left(1-\frac1x\right)dx$$
Here's what I have done so far. (the answer is given as $-\pi/4$)
Let
$$
I = \int_{0}^{1}\tan^{-1}\left(1-\frac1x\right)dx = \int_{0}^{1}\tan^{-1}\left(\frac{x-1}x\right)dx.
$$
Since, $\int_{0}^{1} f(x)dx = \int_{0}^{1} f(1-x)dx$ one has\begin{align}
I &= \int_{0}^{1}\tan^{-1}\left(1-\frac1{1-x}\right)dx\\
& = \int_{0}^{1}\tan^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\cot^{-1}\left(\frac x{x-1}\right)dx\\
& = \int_{0}^{1}\frac\pi2-\tan^{-1}\left(\frac{x-1}x\right)dx\\
& = \frac\pi2 - I,
\end{align}
Hence, $I = \dfrac\pi4$. The given answer is $-\dfrac\pi4$. Where have I gone wrong?
| $$\begin{align*}
\int_0^1 \arctan\left(1 - \frac1x\right) \, dx &= \int_{-\infty}^0 \frac{\arctan(x)}{(1-x)^2} \, dx & (1) \\[1ex]
&= -\int_{-\infty}^0 \frac{dx}{(1+x^2)(1-x)} & (2) \\[1ex]
&= -\int_0^\infty \frac{dx}{(1+x^2)(1+x)} & (3) \\[1ex]
&= -\frac12 \int_0^\infty \left(\frac1{1+x} - \frac{x}{1+x^2} + \frac1{1+x^2}\right) \, dx & (4) \\[1ex]
&= \frac12 \lim_{x\to-\infty} \left(\ln\left|\frac{1+x}{\sqrt{1+x^2}}\right| + \arctan(x)\right) & (5) \\[1ex]
&= \boxed{-\frac\pi4}
\end{align*}$$
$\begin{array}{cl}
(1) & \text{substitute }x\mapsto\frac1{1-x} \\
(2) & \text{integrate by parts} \\
(3) & \text{substitute }x\mapsto-x \\
(4) & \text{expand into partial fractions} \\
(5) & \text{apply the fundamental theorem of calculus}
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Show inequality for positive real numbers If $x,y$ are positive real numbers then we have that $$ \frac{1}{\sqrt{x+y}}<\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$$ right? But how can we show that?
I have tried the following but I don't think that this is the way we should go.
\begin{align*}\frac{1}{\sqrt{x + y}} < \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}=\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}} & \iff \left(\frac{1}{\sqrt{x + y}}\right)^2 < \left(\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}}\right)^2 \\ & \iff \frac{1}{x + y} <\frac{x+y+2\sqrt{xy}}{xy} \\ & \iff xy<(x+y)(x+y+2\sqrt{xy}) \\ & \iff xy<x^2+xy+2x\sqrt{xy}+xy+y^2+2y\sqrt{xy}\\ & \iff 0<x^2+2(x+y)\sqrt{xy}+xy+y^2\end{align*}
| The stronger inequality in the answer by Martin R can be proved by applying $a^2+b^2>2ab,$ for $0<a<b,$ twice. For $0<u<v$ we get $${1\over \sqrt{u^4+v^4}}<{1\over \sqrt{2}uv}={1\over 2\sqrt{2}}2{1\over u}{1\over v}<{1\over 2\sqrt{2}}\left ({1\over u^2}+{1\over v^2}\right )$$ Now plug in $x=u^4$ and $y=v^4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the inverse function of $f(x):=\sqrt x-\sqrt {4-x}+2$
Let $f(x):=\sqrt x-\sqrt {4-x}+2$.
Then:
$(i)$ Find the inverse function of $f(x)$
$(ii)$ What is the largest domain of $\sqrt {f(x)-f^{-1}(x)}$ ?
My work:
We know that
$$f:[0,4]\longrightarrow \mathbb [0,4]\\f^{-1}(x):[0,4]\longrightarrow [0,4]$$
Let $2-x=t$, then we have,
$$\begin{aligned}&f(2-t)-2=\sqrt {2-t}-\sqrt {2+t} \\
\implies &f^2(2-t)-4f(2-t)+4=4-2\sqrt {4-t^2}\\
\implies &\left(f^2(2-t)-4f(2-t)\right)^2=4(4-t^2)\\
\implies &t^2=\frac{16-\left(f^2(2-t)-4f(2-t)\right)^2}{4}\\
\implies &(2-x)^2=\frac{16-\left(f^2(x)-4f(x)\right)^2}{4}\\
\implies &f^{-1}(x)=2\pm \frac {\sqrt{16-(x^2-4x)^2}}{2}\end{aligned}$$
We see that the inverse function $f^{-1}(x)$ can be define
$$\begin{aligned}f^{-1}(x)&:=\begin{cases} 2-\frac {\sqrt{16-(x^2-4x)^2}}{2}\,,\text{if}\,\,\,0\leq x<2\\2+\frac {\sqrt{16-(x^2-4x)^2}}{2}\,,\text{if}\,\,\,\,2\leq x\leq 4
\end{cases}\end{aligned}$$
Finally, we need to solve
Case $-1:$
$$
\begin{aligned}&f(x)\ge f^{-1}(x),\,\sqrt x\ge \sqrt {4-x}\\
\implies &\sqrt x -\sqrt {4-x}\ge \frac {\sqrt{16-(x^2-4x)^2}}{2}\\
\implies &4-2\sqrt {4x-x^2}\ge \frac {16-(x^2-4x)^2}{4}\\
\implies &64(4x-x^2)\leq (4x-x^2)^4\\
\implies & 64u\leq u^4,\, u=4x-x^2 \\
\implies & x=2 \vee x=4\end{aligned}
$$
Case $-2: $
$$
\begin{aligned}&f(x)\ge f^{-1}(x),\,\sqrt x< \sqrt {4-x}\\
\implies &\sqrt x -\sqrt {4-x}\ge -\frac {\sqrt{16-(x^2-4x)^2}}{2}\\
\implies &4-2\sqrt {4x-x^2}\leq \frac {16-(x^2-4x)^2}{4}\\
\implies &64(4x-x^2)\ge (4x-x^2)^4\\
\implies & 64u\ge u^4,\\
\implies & 0\leq x<2\end{aligned}
$$
So, my answer is:
$$x\in[0,2]\cup \{4\}$$
My questions are:
$(i)$ Are my attempts/answer correct
$(ii)$ Is the inverse function I defined correct?
| Yes, you have done it well. Pictorially, $f(x)$ and $f^{-1}(x)$ are plotted below here in blue and red respectively. The green linr is $f(x)=x$.
Next $\sqrt{f(x)-f^{-1}(x)}$ is plotted below
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
evaluate beta function I know:
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1}{2}\cdot \mathrm B\left(11/2, 1/2\right)$$
and
$$\Gamma(1/2)=\sqrt{\pi}$$
But what is the following calculation based on?
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1\cdot3\cdot5\cdot7\cdot9\cdot\pi}{2\cdot4\cdot6\cdot8\cdot10\cdot2}=\frac{63\pi}{512}$$
https://www.goseeko.com/blog/what-are-beta-and-gamma-functions/
| Use the formula:
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
So we have:
$$B\left(\frac{11}2,\frac{1}2\right)=\frac{\Gamma\left(\frac{11}2\right)\Gamma\left(\frac{1}2\right)}{\Gamma\left(\frac{11}2+\frac{1}2\right)}=\frac{\Gamma\left(\frac{11}2\right)\Gamma\left(\frac{1}2\right)}{\Gamma(6)}$$
Use the formula:
$$\Gamma(z)=(z-1)\Gamma(z-1)$$
So we get:
$$\Gamma\left(\frac{11}2\right)=\frac{9}{2}\cdot \frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot\Gamma\left(\frac{1}2\right),~~~~\Gamma(6)=5!$$
Finally,
$$B\left(\frac{11}2,\frac{1}2\right)=\frac{\frac{9}{2}\cdot \frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot\Gamma\left(\frac{1}2\right)\cdot\Gamma\left(\frac{1}2\right)}{5!}=\frac{63\pi}{256}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
The coefficients of the power series $\frac{1}{1-z-z^2}$ centered at $0$ are the numbers in the Fibonacci sequence So I need to show the coefficients of the power series centered about $z=0$ of $\frac{1}{1-z-z^2}$ are $1,1,2,3,5,8,13,21,\dots$
Here's what I have thus far but I am unable to show all coefficients are equal to the coefficients of the series.
Let $F(z):= \sum_{n=0}^\infty a_nz^n$
Then
\begin{align}
\sum_{n=0}^\infty a_nz^n &= a_0 + a_1z + z^2 \sum_{n=0}^\infty (a_{n+2})z^n \\
&= 1+z+ z^2 \sum_{n=0}^\infty (a_{n+1}+a_n)z^n\\
&= 1+ z+ z \sum_{n=0}^\infty a_nz^n - 1 + z^2(\sum_{n=0}^\infty a_nz^n)\\
&= 1+ z(\sum_{n=0}^\infty a_nz^n)+z^2(\sum_{n=0}^\infty a_nz^n)
\end{align}
Then setting $F(z)= \sum_{n=0}^\infty a_nz^n$, bringing everything to the other side and dividing we obtain
$$F(z) = \frac{1}{1-z-z^2}$$
But I am stuck here. I saw another version of this asked but it uses the Cauchy Integral formula which we have not yet covered in the course. The hint was to write $\sum_{n=0}^\infty a_nz^n$ and use the recursive formula to rewrite the power series.
| One more way:
Let $a,b=\frac{1\pm\sqrt{5}}{2},a-b=\sqrt{5}$, we know that Fibonacci numbers are $f_n=\frac{a^n-b^n}{\sqrt{5}}.$
$$F(x)=\frac{1}{1-x-x^2}=\frac{1}{(1-ax)(1-bx)}=\frac{1}{x\sqrt{5}}\left[\frac{1}{1-ax}-\frac{1}{1-bx}\right].$$
Use IGP: $\frac{1}{1-z}=1+z+z^2+z^3+....., |z|<1$, then
$$F(x)=\sum_{k=0}^{\infty}\frac{(a^k-b^k)x^k}{\sqrt{5}}=\sum_{k=0}^{\infty}f_k ~x^k, \quad |x| <\frac{\sqrt{5}-1}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2*0)^2+(0^2-1)^2=(0^2+1)^2$
$(2*1)^2+(1^2-1)^2=(1^2+1)^2$
$(2*2)^2+(2^2-1)^2=(2^2+1)^2$
$(2*0)^2+(0-1)^2=(0+1)^2$
$(2*1)^2+(1-1)^2=(1+1)^2$
$(2*2)^2+(4-1)^2=(4+1)^2$
$0^2+1^2=1^2$
$2^2+0^2=2^2$
$4^2+3^2=5^2$
$0+1=1$
$4+0=4$
$16+9=25$
$1=1$
$4=4$
$25=25$
| We have the identity $a^{2}- b^{2}= \left ( a- b \right )\left ( a+ b \right )\!,$ so
$$\left ( m^{2}+ 1 \right )^{2}- \left ( m^{2}- 1 \right )^{2}\!=\!2m^{2}\cdot 2= \left ( 2m \right )^{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Choosing 4 people from 5 pairs My question is this:
$4$ students are chosen at random from $5$ pairs of $2$ students (i.e. $10$ students). What is the probability that no students are chosen from the same pair (each of the 4 chosen students is selected from a different group of two)?
I am very stuck on this question. I think it is $10C4$ - ways that both students of one group are chosen - ways that both students of $2$ groups are chosen, but I cannot figure out how to find the probability that both students of one group are chosen.
Any guidance would be helpful.
| Method 1: Ordered selections.
There are $10 \cdot 9 \cdot 8 \cdot 7$ ways to make an ordered selection of four of the ten students.
There are $10 \cdot 8 \cdot 6 \cdot 4$ ways to make an ordered selection of four students from different pairs.
Hence, the probability that no two students from the same pair are selected is
$$\frac{10 \cdot 8 \cdot 6 \cdot 4}{10 \cdot 9 \cdot 8 \cdot 7}$$
Method 2: Unordered selections.
There are $\binom{10}{4}$ ways to select four of the ten students.
There are $\binom{5}{4}$ ways to select four of the five pairs from which to select a student. For each such pair, there are two ways to select a student from that pair. Hence, the number of ways to select a student from four different pairs is
$$\binom{5}{4}2^4$$
Hence, the probability that no students from the same pair are selected is
$$\frac{\dbinom{5}{4}2^4}{\dbinom{10}{4}}$$
Method 3: Complementary counting.
As mentioned above, there are
$$\binom{10}{4}$$
ways to select four of the ten people.
We count selections in which both members of at least one pair of people are selected.
Both members of a pair are selected: There are five ways to select the pair from which both members are selected and $\binom{8}{2}$ ways to select two of the other eight people. Hence, there are
$$\binom{5}{1}\binom{8}{2}$$
such selections.
However, if we subtract this amount from the total, we will have subtracted too much since we will have subtracted each selection in which both members of two pairs are selected twice, once for each way we could have designated one of those two pairs as the pair from which both members were selected. We only want to subtract such cases once, so we must add the cases in which both members of two pairs are selected to our total.
Both members of two pairs are selected: There are
$$\binom{5}{2}$$
ways to select the two pairs from which both members are selected.
Hence, by the Inclusion-Exclusion Principle the number of favorable cases is
$$\binom{10}{4} - \binom{5}{1}\binom{8}{2} + \binom{5}{2}$$
Therefore, the probability that no two students from the same pair are selected is
$$\frac{\dbinom{10}{4} - \dbinom{5}{1}\dbinom{8}{2} + \dbinom{5}{2}}{\dbinom{10}{4}} = 1 - \frac{\dbinom{5}{1}\dbinom{8}{2} - \dbinom{5}{2}}{\dbinom{10}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral: $\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx$ (Context) While working on an integral for fun, I stumbled upon the perplexing conjecture:
$$\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx = 2\pi\operatorname{arccot}\left(\sqrt{3+2\sqrt{3}}\right).$$
(Attempt) I tried multiple methods. One method that stuck out to me was using the formula $$\arctan(\theta) = \frac{1}{2i}\ln{\left(\frac{1+i\theta}{1-i\theta}\right)}$$ so that my integral becomes
$$\frac{1}{2i}\int_{0}^{2\pi}\ln\left(1+i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right)dx-\frac{1}{2i}\int_{0}^{2\pi}\ln\left(1-i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right).$$
Both of these look similar to the integral
$$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos(x)\right)dx=\begin{cases}
0, &\text{for}\; |r|<1,\\
2\pi\ln \left(r^2\right), &\text{for}\; |r|>1,
\end{cases}\tag{2}$$
and its solution can be found here.
I tried to get my integrals to "look" like the above result but to no avail. Not wanting to give up, I searched on this site for any ideas, and it seems like a few people have stumbled upon the same kind of integral, such as here and here.
In the first link, the user @Startwearingpurple says,
"Now we have
\begin{align}
4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right)
\end{align}
with
$$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}."$$
I tried to replicate his method but even after doing messy algebra, I couldn't figure out how to manipulate the inside of my logarithm such that it looked like what he did. I also tried letting $\operatorname{arg}\left(1+i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right) \in \left(-\pi/2, \pi/2\right)$, if that helps.
(Another method I tried was noticing that the original integral's function is periodic, so I tried using residue theory by letting $z=e^{ix}$, but I wasn't able to calculate the residues.)
(Question) Can someone help me approach this integral (preferably finding a closed form)? Any methods are absolutely welcome. And if someone could figure out how to get my logarithms to look like $\ln{\left(1+r^2-2r\cos{(x)}\right)}$, that would be nice.
(Edit) After using @SangchulLee's integral,
$$ \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \pi \arg \left(1 + ia + \sqrt{b^2 + (1+ia)^2}\right), $$
found here, I was able to deduce that
$$\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx\ =\ 2\pi\operatorname{arccot}\left(\sqrt{3+2\sqrt{3}}\right).$$
I still have no idea how they proved it though.
| Here's a solution by Taylor series abuse. We have the useful Taylor series
\begin{equation}
\arctan\left(\frac{1+2y}{\sqrt{3}}\right)=\frac{\pi}{6}+\frac{\sqrt{3}}{2}\sum_{k=1}^\infty\frac{a_ky^k}{k}
\end{equation}
converging for $y\in [-1,1]$, where $(a_k)_k$ is a sequence of period $3$ with $(a_1,a_2,a_3)=(1,-1,0)$. We thus have that
\begin{equation}
I=\int_0^{2\pi}
\arctan\left(\frac{1+2\cos(x)}{\sqrt{3}}\right)dx=\frac{\pi^2}{3}+\frac{\sqrt{3}}{2}\sum_{k=1}^\infty\frac{a_k}{k}\int_0^{2\pi}\cos^k(x)dx
\end{equation}
Noting that
\begin{equation}
\int_0^{2\pi}\cos^k(x)dx=\frac{\pi}{2^{k-1}}{k\choose k/2}
\end{equation}
when $k$ is even, and the integral vanishes when $k$ is odd, we may simplify
\begin{equation}
I=\frac{\pi^2}{3}+\pi\sqrt{3}\sum_{k=1}^\infty\frac{a_{2k}}{2^{2k}(2k)}{2k\choose k}
\end{equation}
Note that $a_{2k}$ may also be expressed as
\begin{equation}
a_{2k}=-\frac{i\sqrt{3}}{3}\left[e^{i2k\pi/3}-e^{i4k\pi/3}\right]
\end{equation}
Using this fact, and noting that the following useful Taylor series
\begin{equation}
\sum_{k=1}^\infty\frac{y^k}{2^{2k}(2k)}{2k\choose k}=-\log\left(\frac{1+\sqrt{1-y}}{2}\right)
\end{equation}
converges on the $3$rd roots of unity, we thus may write
\begin{equation}
\begin{split}
I&=\frac{\pi^2}{3}-i\pi\sum_{k=1}^\infty\frac{1}{2^{2k}(2k)}{2k\choose k}\left[e^{i2k\pi/3}-e^{i4k\pi/3}\right]\\
&=\frac{\pi^2}{3}+i\pi\left[\log\left(\frac{1+\sqrt{1-e^{i2\pi/3}}}{2}\right)-\log\left(\frac{1+\sqrt{1-e^{i4\pi/3}}}{2}\right)\right]\\
&=\frac{\pi^2}{3}+2\pi\text{arctan}\left(\frac{\mathfrak{I}(1-e^{i2\pi/3})}{\mathfrak{R}(1-e^{i2\pi/3})}\right)\\
&=\frac{\pi^2}{3}+2\pi\arctan\left(-\frac{\sqrt{2\sqrt{3}-3}}{2+\sqrt{3+2\sqrt{3}}}\right)\\
&=2\pi\text{arccot}\left(\sqrt{3+2\sqrt{3}}\right)
\end{split}
\end{equation}
where the last equality can be obtained through the arctan addition formula, and by noting that $\frac{\pi^2}{3}=2\pi\arctan\left(\frac{1}{\sqrt{3}}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
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} |
Integrate $\int \frac{x}{(1-x^6)^{1/3}}\ dx$ My friend challenged me to integrate $\tan^{1/3}(t)$ without using the normal method of just letting $\tan(t) = x^3$ and proceeding...
I tried a different approach
$$\int\tan^{1/3}(t) dt = \int \frac{\sin^{1/3} t }{\cos^{1/3} t} dt$$
Putting $\cos t = x^3$ gives
$$\int \frac{x}{(1-x^6)^{1/3}} dx = \frac12 \int \frac{dy}{(1-y^3)^{1/3}}$$
where $y = x^2$.
Is there any way to solve it further?
| Letting $y^3=\frac 1{x^6}-1$, transforms the integral into
$$
I=-\frac{1}{2} \int \frac{y d y}{y^3+1}
$$
Resolving the integrand into partial fractions
$$\frac{y}{y^3+1}= -\frac{1}{3(y+1)}+\frac{y+1}{3\left(y^2-y+1\right)}$$
we have
$$
\begin{aligned}
\int \frac{y d y}{y^3+1} =& \int\left(-\frac{1}{3(y+1)}+\frac{y+1}{3\left(y^2-y+1\right)}\right) d y \\
=&-\frac{1}{3} \ln |y+1|+\frac{1}{6} \int \frac{(2 y-1)+3}{y^2-y+1} d y \\
=&-\frac{1}{3} \ln |y+1|+\frac{1}{6} \ln \left|y^2-y+1\right|+2 \int \frac{d y}{(2 y-1)^2+3} \\
=& \frac{1}{6}\left(\ln\frac{\left| y^2-y+1\right|}{(y+1)^2}+4 \sqrt{3} \tan ^{-1} \frac{2 y-1}{\sqrt{3}}\right)+c_1
\end{aligned}
$$
$$
I=\frac{1}{12}\left(\ln \frac{ (y+1)^2}{ \left|y^2-y+1\right|}-4 \sqrt{3} \tan ^{-1} \frac{2 y-1}{\sqrt{3}}\right)+C
$$
where $y^3=\frac 1{x^6}-1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992)
(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin B )$
should subtend a right angle at O is
$$a^2\cos A\cos B+b^2\sin A\sin B=0$$
(ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at
the point $(C\cos t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
subtend a right angle at $O$.
Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$
Part (ii) gives the result $x\cos t+y\sin t=C$
With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.
I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.
Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)
| Your idea of parametrising the ellipse works. We write a general point of the ellipse in the form $(a\cos\psi,b\sin\psi)$ for some angle $\psi$. Since the problem is invariant under scaling, we set $a^2+b^2=1$ without loss of generality, so $C=ab$. (This is just for convenience, the solution works just fine without this.) A generic point on the tangent line can be written as
$$ (C\cos t-\lambda \sin t,C\sin t+\lambda\cos t) $$
for some real parameter $\lambda$; since there are two points of intersections let them correspond to $\lambda_1,\lambda_2$. We have the equations
$$ \begin{cases}a\cos\psi = C\cos t -\lambda\sin t\\ b\sin\psi = C\sin t+\lambda\cos t.\end{cases} $$
Now, taking ratios, we find that
$$ \frac{b}{a}\tan\psi = \frac{C\sin t+\lambda\cos t}{C\cos t-\lambda\sin t} = \frac{\tan t+\lambda/C}{1-(\lambda/C)\tan t} = \tan(t+\phi) $$
where angle $\phi$ is defined so that $\tan\phi=\lambda/C$. By the given (i) of the question, we just need to show that the two values $\psi_1,\psi_2$ corresponding to the two points satisfy
$$ a^2\cos\psi_1\cos\psi_2+b^2\sin\psi_1\sin\psi_2=0 \iff 1+\frac{b^2}{a^2}\tan\psi_1\tan\psi_2=0. $$
Using the relation for $\psi$ we have found, we rewrite this as
$$ \left(\frac{b}{a}\tan\psi_1\right)\left(\frac{b}{a}\tan\psi_1\right)=-1 \iff \tan(t+\phi_1)\tan(t+\phi_2)=-1. $$
By standard properties of the tangent function, this is easily seen to be equivalent to the condition that $\phi_1=\phi_2-\pi/2$ (up to multiples of $\pi$). Taking $\tan$ of both sides, we see that it is equivalent to $x_1=-1/x_2$, where $x_i := \tan(\phi_i) = \lambda_i/C$ ($i=1,2$), i.e. $x_1x_2=-1$.
Now, we will endeavour to show this. By the equation of the ellipse, we have
$$ b^2(C^2\cos^2t-2C\lambda\cos t\sin t+\lambda^2\sin^2 t)+a^2(C^2\sin^2t+2C\lambda\cos t\sin t+ \lambda^2\cos^2 t)=a^2b^2. $$
Substituting $C=ab$ and $a^2+b^2=1$, simplifying, then collecting terms, with $x=\lambda/C$, we end up with the quadratic
$$ (b^2\sin^2 t+a^2\cos^2t)x^2 + 2(a^2-b^2)(\sin t\cos t)x + b^2\cos^2t+a^2\sin^2t-1=0. $$
Now, we finally find that the product of roots of this equation
$$ \begin{split}x_1x_2 &= \frac{b^2\cos^2t+a^2\sin^2t-1}{b^2\sin^2 t+a^2\cos^2t}\\ &= \frac{(1-a^2)(1-\sin^2t)+a^2\sin^2t-1}{(1-a^2)\sin^2t+a^2(1-\sin^2t)} \\ &= \frac{1-a^2-\sin^2t+2a^2\sin^2t-1}{\sin^2t+a^2-2a^2\sin^2t} = -1,\end{split} $$
as desired, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$
If $a+b+c=1$ and $a, b, c\geq0$,
what is the maximum value of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ ?
I found its answer by using CAS Calculator. The answer is $\frac{7}{18} (a=b=0, c=1)$.
But i wonder not only the answer but also the solution of it.
Please help me.
| We have, for all $x\in [0, 1]$,
$$\frac{1}{18}x + \frac{1}{9} - \frac{1}{x^2 - 4x + 9}
= \frac{x(x-1)^2}{18(x^2-4x+9)} \ge 0. \tag{1}$$
Using (1), we have
$$\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}
\le \frac{1}{18}(a + b + c) + \frac{1}{3} = \frac{7}{18}. $$
Also, when $a= b = 0, c = 1$,
we have $\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9} = \frac{7}{18}$.
Thus, the maximum of
$\frac{1}{a^2 - 4a + 9}
+ \frac{1}{b^2 - 4b + 9} + \frac{1}{c^2 - 4c + 9}$
is $7/18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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What is the probability of getting at least one six with three dice. Problem:
A person rolls $3$ six sided dice. What is the probability that the largest value will be $6$?
Answer:
Let $p$ be the probability that we seek. The probability that the first die comes up $6$ is $1/6$. The probability that the second die comes up $6$ is $1/6$. The same for the third die. As such, I might think the answer is:
$$ p = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2} $$
but that is wrong because I am counting the roll of all sixes $3$ times. Hence I claim that the
correct answer is less than $\dfrac{1}{2}$.
\begin{align*}
p &= \dfrac{ 3(5)^2 + 3(5)+1}{6^3} = \dfrac{75+15+1}{216} \\
p &= \dfrac{91}{216}
\end{align*}
To check the answer, I compute the probability that all the dice will be $5$ or less. Call this
probability $p_5$. Note that $p + p_5 = 1$.
\begin{align*}
p_5 &= \dfrac{ 5^3 }{6^3} \\
p &= 1 - p_5 = 1 - \dfrac{ 5^3 }{6^3} = \dfrac{ 6^3 - 5^3 }{6^3 } \\
p &= \dfrac{ 91}{216}
\end{align*}
Is my solution right? Is my first method correct?
| It is correct, and this can be generalized by the Binomial Theorem, since the numbers in the numerator of the first method are precisely the binomial coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Motivation behind the definition of multiplication in $ \mathbb{R}^2 $ I was reading Titu Andreescu’s book on complex numbers, Complex Numbers A to Z. In the book, the author defines multiplication in $ \mathbb{ R }^2 $ as,
$ (a,b)( c,d) = (ac - bd, ad + bc) $
I understand how this definition relates to multiplication in complex numbers, and I see how with the knowledge of complex numbers one would be motivated to define multiplication in $ \mathbb{R}^2 $. But what I am wondering is, how one would motivate this definition purely from the geometry of $ \mathbb{R}^2 $, from which we can then construct the field of complex numbers.
Can you recommend me any resources where I can find this approach ? Thank you very much in advance.
| You can define geometric rotation in $ \mathbb{R}^{2} $ like this:
$$
\begin{align*}
\operatorname{cis}(\theta) &= \cos(\theta) + \sin(\theta) \cdot \mathrm{i}\\
(a_{1} + b_{1} \cdot \mathrm{i}) (a_{2} + b_{2} \cdot \mathrm{i}) &= (\sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \operatorname{cis}(\arg(a_{1} + b_{1} \cdot \mathrm{i}))) \cdot (\sqrt{a_{2}^{2} + b_{2}^{2}} \cdot \operatorname{cis}(\arg(a_{1} + b_{1} \cdot \mathrm{i}))\\
(a_{1} + b_{1} \cdot \mathrm{i}) (a_{2} + b_{2} \cdot \mathrm{i}) &= \sqrt{(a_{1}^{2} + b_{1}^{2}) \cdot (a_{2}^{2} + b_{2}^{2})} \cdot \operatorname{cis}(\arg((a_{1} + a_{2}) + (b_{1} + b_{2}) \cdot \mathrm{i}))\\
&\Rightarrow (r_{1} \cdot \operatorname{cis}(\theta_{1})) \cdot (r_{2} \cdot \operatorname{cis}(\theta_{2})\\
&\Rightarrow (r_{1} \cdot r_{2}) \cdot \operatorname{cis}(\theta_{1} + \theta_{2})\\
\end{align*}
$$
You can also represent complex numbers as versors (similar to vectors):
$$
\begin{align}
x + y \cdot \mathrm{i} &\widehat{=} \left[\begin{matrix}
x\\ y \end{matrix}\right]\\
w + x \cdot \mathrm{i} + y \cdot \mathrm{j} + z \cdot \mathrm{k} &\widehat{=} \left[\begin{matrix}
w\\ x\\ y\\ z \end{matrix}\right]\\
\\
\text{quaternion: }\mathrm{i}^{2} = \mathrm{j}^{2} = \mathrm{k}^{2} = \mathrm{i} \cdot \mathrm{j} \cdot\mathrm{k} &= -1
\end{align}
$$
You also can show it with Brahmagupta–Fibonacci identity or Euler's four-square identity:
$$
\begin{align}
\left(a^2 + b^2\right)\left(c^2 + d^2\right) & {}= \left(ac-bd\right)^2 + \left(ad+bc\right)^2 & & (1) \\
& {}= \left(ac+bd\right)^2 + \left(ad-bc\right)^2. & & (2)
\end{align}
$$
$$
\begin{align}
\left(a_1^2+a_2^2+a_3^2+a_4^2\right)\left(b_1^2+b_2^2+b_3^2+b_4^2\right)
= {}
&\left(a_1 b_1 - a_2 b_2 - a_3 b_3 - a_4 b_4\right)^2 \\
&+ \left(a_1 b_2 + a_2 b_1 + a_3 b_4 - a_4 b_3\right)^2 \\
&+ \left(a_1 b_3 - a_2 b_4 + a_3 b_1 + a_4 b_2\right)^2 \\
&+ \left(a_1 b_4 + a_2 b_3 - a_3 b_2 + a_4 b_1\right)^2.
\end{align}
$$
While the Brahmagupta–Fibonacci identity has a direct relationship to the multiplication of complex numbers, i.e. in $ \mathbb{R}^{2} $, the Euler's four-square identity has a direct relationship to the quaternions (an extension of the complex numbers), i.e. in $ \mathbb{R}^{4} $ and $ \mathbb{R}^{2} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Find the first derivative of $y=\sqrt[3]{\frac{1-x^3}{1+x^3}}$ Find the first derivative of $$y=\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
The given answer is $$\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$ It is nice and neat, but I am really struggling to write the result exactly in this form. We have $$y'=\dfrac13\left(\dfrac{1-x^3}{1+x^3}\right)^{-\frac23}\left(\dfrac{1-x^3}{1+x^3}\right)'$$ The derivative of the "inner" function (the last term in $y'$) is $$\dfrac{-3x^2(1+x^3)-3x^2(1-x^3)}{\left(1+x^3\right)^2}=\dfrac{-6x^2}{(1+x^3)^2},$$ so for $y'$ $$y'=-\dfrac13\dfrac{6x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23$$ Can we actually leave the answer this way?
| You got $$y'=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23$$
You can leave till here if you want. If you want to match the given answer, multiply and divide by $\left(\dfrac{1+x^3}{1-x^3}\right)^\frac13$, thus,
$$y'=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)\left(\dfrac{1-x^3}{1+x^3}\right)^\frac13\\=\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
If you want to reach the given answer directly, you can take $\log$ on the given expression, thus,
$$\log y=\frac13\left(\log(1-x^3)+\log(1+x^3)\right)$$
Taking derivative,
$$\frac{y'}{y}=\frac13\left(\frac{-3x^2}{1-x^3}-\frac{3x^2}{1+x^3}\right)\\=x^2\left(\frac{-1-x^3-1+x^3}{1-x^6}\right)$$
Therefore, $$y'=\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve system of equations with modulo function? If I have a system of equations like
\begin{align}
X &= g^{(a+2)(b+2)} \bmod p, \\
Y &= g^{(a+3)(b+3)} \bmod p, \\
Z &= g^{(a+4)(b+4)} \bmod p;
\end{align}
where $p$ (a prime number), $g$ and $X$ are known and $g$ and $p$ are co-prime.
Can we find $Y$ and $Z$ in terms of $X$?
So far, I have reached here:
\begin{align}
X &= g^{(ab+2a+2b+4)} \bmod p \\
&= [g^{ab} \bmod p \cdot g^{2(a+b)} \bmod p \cdot g^4 \bmod p] \bmod p; \\
Y &= g^{(ab+3a+3b+9)} \bmod p \\
&= [g^{ab} \bmod p \cdot g^{3(a+b)} \bmod p \cdot g^9 \bmod p] \bmod p; \\
Z &= g^{(ab+4a+4b+16)} \bmod p \\
&= [g^{ab} \bmod p \cdot g^{4(a+b)} \bmod p \cdot g^{16} \bmod p] \bmod p.
\end{align}
Edit: as suggested if I simplify the exponents: $u=a+2$ and $v=b+2$, I get:
\begin{align}
X &= g^{uv} \bmod p, \quad\text{and} \\
Y &= g^{(u+1)(v+1)} \bmod p.
\end{align}
Still, I cannot express $Y$ in terms of $X$.
| The problem is equivalent (with the help of the comments above) to: given integers $p$ (prime) and $X,g$ (not dividible by $p$), find all $g^{u+v}$'s $\bmod p$ such that $X\equiv g^{uv}\bmod p$. Then, exhaustive search modulo the multiplicative order of $g\bmod p$ (of $w$, if any, such that $g^w\equiv X\bmod p$, and then of $u,v$ such that $w=uv$) seems the only way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $|x^y - y^x| > 2$
Prove that for any integers $x\neq y ,x,y>2, |x^y-y^x|>2$.
I know that the function $f(x) = \dfrac{\ln x}{x}$ is increasing for $x<e$ and decreasing for $x > e$. So $|x^y-y^x| = x^y - y^x \ge 1$ if $x < y$. I was thinking of proving something involving ratios (e.g. $f(x)/f(y)$ for $x < y$). I could possibly consider the case where x and y are consecutive and I could assume WLOG that $y>x$ so that it suffices to prove $x^y > y^x+2$ for $y>x>2.$ The inequality seems to be fairly weak; even for the smallest possible values of x and y, $y=4,x=3,$ we have $x^y = y^x+17.$ I think the sequence $\dfrac{x^{x+1}}{(x+1)^x} = x(1-\dfrac{1}{x+1})^x$ is an increasing function of x, and one could prove this using derivatives.
Source: A PuMAC 2008 problem.
| Let us prove that $|x^y - y^x| \ge 17$.
WLOG, assume that $3 \le x < y$.
Let $g(u) := x^u - u^x$ for $u \ge x + 1$. We have
$$g'(u) = x^u \ln x - u^{x-1} x
> x^u - u^{x-1}x
= x(x^{u-1} - u^{x-1}).
$$
Let $h(v) := \frac{\ln v}{v - 1}$.
We have $h'(v) = -\frac{v\ln v - v + 1}{v(v-1)^2}$.
Thus, $h(v)$
is strictly decreasing on $v \ge 3$. Thus, $\frac{\ln x}{x - 1}
> \frac{\ln u}{u - 1}$
which results in $x^{u - 1} > u^{x-1}$. Thus, $g'(u) > 0$
for all $u \ge x + 1$.
Thus, we have
$$x^y - y^x = g(y) \ge g(x+1) = x^{x+1} - (x + 1)^x.$$
Let $F(x) := x^{x+1} - (x + 1)^x$.
Using $(1 + 1/x)^x \le \mathrm{e} < x$, we have $(x + 1)^x < x^{x + 1}$. We have
\begin{align*}
F'(x) &= x^{x+1}\left(\ln x + \frac{x+1}{x}\right) - (x + 1)^x \left(\ln (x + 1) + \frac{x}{x + 1}\right)\\
&> x^{x+1}\left(\ln x + \frac{x+1}{x}\right) - x^{x+1} \left(\ln (x + 1) + \frac{x}{x + 1}\right)\\
&= x^{x+1}\left(\frac{1}{x} - \ln(1 + 1/x) + \frac{1}{x + 1}\right)\\
&> 0
\end{align*}
where we have used $\ln (1 + 1/x) \le 1/x$.
Thus, we have $F(x) \ge F(3) = 3^4 - 4^3 = 17$.
Thus, $x^y - y^x \ge 17$.
We are done.
| {
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Conditions for generalized projection matrix of size (2x2)? My results seem incorrect... I am trying to derive the general conditions that must be imposed upon the real-valued entries of a $2 \times 2$ orthogonal projection matrix. However, I am coming to a conclusion that seems wrong and hope someone can point me in the right direction.
The prompt reads as follows:
Find conditions on $a,b,c,d \in \mathbb{R}$ that guarantee the matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ defines a rank-1 orthogonal projection.
I begin by imposing the conditions that all projection matrices $P$ must fulfill, namely
(i) $P = P^T$
(ii) $P = P^2$
Applying condition (i),
$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ b & d\end{pmatrix}$
I get $b=c$ and so I continue working with $\begin{pmatrix} a & c \\ c & d\end{pmatrix}$ and move on to condition (ii):
$\begin{pmatrix} a & c \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ c & d\end{pmatrix}\begin{pmatrix} a & c \\ c & d\end{pmatrix}=\begin{pmatrix} a^2 +c^2 & ac + cd \\ ac +cd & d^2 + c^2\end{pmatrix}$
From that, I get three equations:
(1) $a = a^2 +c^2$
(2) $c=ac+cd=c(a+d)$
(3) $d=d^2 + c^2$
Further working out (2) by striking $c$ from each side, the system is then
(1) $a = a^2 +c^2$
(2) $1=a+d$
(3) $d=d^2+c^2$
I re-arrange (1) and (3) to get
(1) $c^2 = a - a^2$
(3) $c^2 = d - d^2$
and thus $a-a^2 = d-d^2$ from which I conclude that $a = d$. Returning to equation (2), I then get
(2) $1 = a+d = 2a$
and so $a = \frac{1}{2}$ and $d = \frac{1}{2}$.
Plugging either one of these into equation (1) or (3) then allows me to solve for $c$,
$\frac{1}{2} = \left ( \frac{1}{2} \right )^2+c^2$
or
$c^2 = \frac{1}{4}$
And thus $c=\pm \frac{1}{2}$.
So, according to these results, to be guaranteed an orthogonal projection matrix of rank-1, my matrix $P$ must be one of two possibilities (as indicated by $\pm$):
$P = \frac{1}{2} \begin{pmatrix} 1 & \pm 1 \\ \pm 1 & 1\end{pmatrix}$
This result is unsettling. First of all, I was expecting a broader range of possibilities for $a,b,c,d$ or at least for one or two variables. It seems odd that $a$ and $d$ are wholly constrained to one value and that $c$ only has two options.
Is this correct? It seems like there should be more here that is possible. Aren't matrices such as $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ also to be included here?
For example, when I apply $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ to a general vector $\begin{pmatrix} x \\ y \end{pmatrix}$, it's easy for me to see the orthogonal projection:
$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x \\ 0\end{pmatrix}$
But when I apply the resulting projection matrix from above to the same generalized vector, I get a result that does reveal orthogonal projection to me (using $c = + \frac{1}{2}$):
$\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \frac{1}{2} \begin{pmatrix} x+y \\ x+y\end{pmatrix}$
How can that be an orthogonal projection? Isn't that simply a change in length along the same direction? Am I not understanding the fundamental concept?
Thanks for any help you might be able to provide.
| There are two places where you lose some solutions:
*
*From $c = c(a+d)$ to $1 = a+d$, you theoretically lose some solutions with $c=0$: the diagonal matrices. This turns out to be unimportant because there are four diagonal solutions: the zero matrix and the identity matrix (which we throw out because they don't have rank 1) and the solutions where $\{a,d\} = \{0,1\}$ in some order (which happen to satisfy $1 = a+d$ anyway). But it's important to be careful about this sort of thing!
*More importantly, from $a - a^2 = d - d^2$ or $a(1-a) = d(1-d)$, we cannot conclude $a=d$; this is also possible by taking $a=1-d$, and in fact since we already know that $a+d=1$, we already know that $a=1-d$.
So let's return to the step where we have
\begin{align}
a &= a^2 + c^2 \\
1 &= a+d \\
d &= d^2 + c^2
\end{align}
Setting $d = 1-a$, we see that $1-a = (1-a)^2+c^2$ simplifies to $a = a^2+c^2$, so the third equation is redundant. We could at this point describe all solutions as
$$
P =
\begin{bmatrix}
a & \pm \sqrt{a-a^2} \\
\pm \sqrt{a-a^2} & 1-a
\end{bmatrix}.
$$
In order for $\sqrt{a-a^2}$ to be real, we want $0 \le a \le 1$, but any value of $a$ in this range works (and so does either choice of $\pm$, assuming we choose the same $\pm$ for both off-diagonal entries).
To resolve another point of confusion, the map $(x,y) \mapsto (\frac{x+y}{2}, \frac{x+y}{2})$ is an orthogonal projection: it's an orthogonal projection onto the diagonal line $x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
multiplication of two consecutive numbers a, Is it true, that if $n \gt 1$ odd number, then we can express ${n^2-1\over 4}$ as multiply of two consecutive numbers?
b, Is it true, that we can express ${n^4+2n^3+3n^2+2n\over 4}$ in case of n= all natural numbers, as multiply of two consecutive numbers?
What I did already: I know, that the multiply of two consecutive numbers can end to 0,or 2, or 6.
a. ${n^2-1\over 4} = {(3+2m)^2-1\over 4}= {9+2*3*2m+4m^2-1\over 4} = {8+12m+4m^2\over 4}=2+3m+m^2$
m= 0,1,2,3...
| a. $\frac{(2m+1)^2-1}{4}=m(m+1)$.
b. $\frac{n^4+2n^3+3n^2+2n}{4}=\left(\frac{n(n+1)}{2}\right)\left(\frac{n(n+1)}{2}+1\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of reciprocal of primes failed computation Set $$X:=\sum_p \dfrac{1}{p^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2} +\dfrac{1}{5^2}+\cdots$$
As $X$ is absolute convergent and less than $1$, we have (not sure for infinite rearrangement) naive calculation implies
$$\dfrac{\pi^2}{6}-1=\sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2}+\sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2}+\cdots,$$
where $\Omega(n)$ denotes the number of prime factors of $n$ counting multiple.
The right term above is
$$X+X^2+\cdots.$$
Thus, $X=1-6/\pi^2<0.4$, but with the answer in the following link this is false.
https://mathoverflow.net/questions/53443/sum-of-the-reciprocal-of-the-primes-squared
Is it possible to fill the gap of the calculation?
| It's not true that $\displaystyle \sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2} = \biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2$ (and so on). Indeed,
\begin{align*}
\biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2 = \biggl( \sum_{p\text{ prime}}\dfrac{1}{p^2} \biggr)^2 = \sum_{p,q\text{ prime}}\dfrac{1}{p^2q^2}
\end{align*}
counts every integer that's the product of two distinct primes twice, once as $pq$ and once as $qp$. The correct (but perhaps not useful) identity is
$$
\biggl( \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2} \biggr)^2 = 2\sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2} - \sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the real root of the almost symmetric polynomial $x^7+7x^5+14x^3+7x-1$
Find the real root of following almost symmetric polynomial by radicals $$p(x)=x^7+7x^5+14x^3+7x-1$$
Here are my attempts.
The coefficients of $p(x)$ are : $1,7,14,7,-1$.
I wanted to try possible factorizations. But Wolfram Alpha can not factorise this polynomial. This can be a reason of our case, so factorisation over $\Bbb R$ seems impossible.
The Rational root theorem also failed.
Again I tried
$$\begin{align}
x^7+7x^5+14x^3+7x-1
&=x^7+7x^5+7x^3+7x^3+7x-1 \\
&=x^7+7x^3(x^2+1)+7x(x^2+1)-1 \\
&=x^7+7x(x^2+1)^2-1
\end{align}$$
But, this manipulation also didn't work.
| Since
$$x^7+7x^5+14x^3+7x=2i\,\text T_7\left(\dfrac{ix}2\right)= 2i \cos\left(7\arccos\,\dfrac{ix}2\right)=1,\tag1$$
where $\;\text T_n(x)\;$ is the Chebyshev polynomial of the first kind (see also WA test),
then
$$x_k=-2i\cos\left(\dfrac17\arccos\,\left(-\dfrac i2\right)+\dfrac{2\pi k}7\right),\qquad(k=0,1,2,3,4,5,6)\tag2$$
(see also WA test),
wherein
$$\arccos\left(-\dfrac i2\right)=-i\ln\left(-\dfrac i2+i\sqrt{1-\left(-\dfrac i2\right)^2}\,\right)=-i\left(\ln e^{^{\Large\frac\pi2i}} + \ln\dfrac1\varphi\right),$$
$$\arccos\left(-\dfrac i2\right)=\dfrac\pi2+i\ln\varphi,\tag3$$
and $\;\varphi=\dfrac{\sqrt5+1}2\;$ is the golden ratio.
From $(2)-(3)$ should
$$x_k=-2i\cos\left(\dfrac{4\pi k+\pi+2i\ln\varphi}{14}\right), \qquad(k=0,1,2,3,4,5,6),\tag4$$
(see also WA test).
If $\,k=5,\,$ then $\dfrac{4k+1}{14}\pi=\dfrac32\pi,$ and we have result in radicals:
$$x_5= 2i\sin\,\left(\frac i7\ln\varphi\right)
=2i\cdot\sin\left(i \ln\sqrt[\large7]\varphi\right) =\sqrt[7]\varphi-\frac1{\sqrt[7]\varphi}\approx0.13759740974800.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{1-4\cos80^\circ}{\tan20^\circ}$ I am working on simplifying the value
$$\frac{1-4\cos80^\circ}{\tan20^\circ}$$
I am looking for it to be expressed in terms of tangent, because the value seems to be $\tan40^\circ$. However, I can't find any obvious way to do so. How would simplify it to express like such?
| Let's call $t=\tan(x)$ with $x=20°=\frac{\pi}9$
From tan(3x) formula we get $\ \tan(3x)=\sqrt{3}=\dfrac{3t-t^3}{1-3t^2}$
Squaring this to get rid of the root, we get that $t$ is solution of the integer polynomial
$3(1-3t^2)^2=(3t-t^3)^2\iff$
$$t^6-33t^4+27t^2-3=0$$
Now we can use the $\tan(\theta/2)$ conversions from the initial equation
*
*$1-4\cos(4x)=5-8\cos(2x)^2=5-8\left(\frac{1-t^2}{1+t^2}\right)^2$
*$\tan(40°)=\tan(2x)=\frac{2t}{1-t^2}$
And we get
$\begin{align}(1-4\cos(4x))-\tan(x)\tan(2x)&=5-8\left(\frac{1-t^2}{1+t^2}\right)^2-\frac{2t^2}{1-t^2}\\\\
&=\frac{t^6-33t^4+27t^2-3}{(1-t^2)(1+t^2)^2}\end{align}$
Which is precisely $0$ according to the work above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solving the Rossler equations and sketching the phase plane We are given the system
\begin{align}
\begin{cases}
\dot{x} &= -y\\
\dot{y} &= x + ay
\end{cases} \ , \ a >0
\end{align}
which, by the way, is called Rossler System. More info on this here. I am asked to "solve" them, ie. find the equations that fulfill it and then draw the phase plane.
For the first part we have a lot of methods to go about this. The usual one is by way of elimination. Deriving the first equation and substituting in the second we have the system
\begin{align}
\ddot{x}-a \dot{x}+x=0.
\end{align}
The characteristic equation is
\begin{align}
\lambda^2-a\lambda+1=0 &\implies \Delta=a^2-4.
\end{align}
Here's the first problem I am having: am I supposed to take every option into account?
Ie.
*
*If $\boxed{a=2}$ (negative root is not accepted since $a>0$), then we have
\begin{align}
\ddot{x} - 2\dot{x}+x=0 &\implies x(t)=c_1e^t+c_2te^t
\end{align}
and we must continue the system to find $y(t)$. Through a lot of calculations we can do
\begin{align}
y(t) = -c_1 t e^t - c_2 t^2 e^t.
\end{align}
In that specific case, what does the system "tell" us?
At any rate the phase portrait can be sketched by the eigenvalues of the matrix
\begin{align}
\begin{pmatrix}
0 & -1 \\ 1 & 2
\end{pmatrix}
\end{align}
which we have already (through the system) calculated to be $\lambda = 1$. Now we know that if we put $\lambda = 1$ in the characteristic polynomial, we get
\begin{align}
\begin{pmatrix}
-1 & -1\\
1 & 1
\end{pmatrix} \implies \begin{pmatrix} -1 \\ 1 \end{pmatrix} \ \text{null space.}
\end{align}
How can we draw this?
-Am I supposed to do the same now for $a \in (0,2)$ and $a \in (2, + \infty)$?
| I shall provide an intuitive method to sketch the phase portraits which I have learnt back at university years ago. In simpler terms, it shows the trajectory of your gradient field. It is relatively straightforward given a first-order $2 \times 2$ matrix linear ODE system. We shall leverage on the info you have provided above and some Linear Algebra to analyze the dynamics of the system.
We can rewrite our system of linear ODE in matrix form.
$$
\begin{bmatrix}
\dot x \\
\dot y \\
\end{bmatrix} =
\begin{bmatrix}
0 & -1 \\
1 & a \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
\end{bmatrix}
$$
$$
\implies A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{bmatrix} =
\begin{bmatrix}
0 & -1 \\
1 & a \\
\end{bmatrix}
$$
For a $2 \times 2$ system, the characteristic polynomial can be given by:
$$ p(\lambda) = \lambda^2-\text{tr}(A)\lambda + \text{det}(A) $$
The roots or the eigenvalues to the characteristic polynomial is given by setting $p(\lambda)=0$. Then:
$$ \lambda_{1,2}=\frac{\text{tr}(A)}{2} \pm \sqrt{\Big(\frac{\text{tr}(A)}{2}\Big)^2-\text{det}(A)} $$
If $A$ is diagonalized, ie. $A=Q\Lambda Q^{-1}$, where $Q$ consists of eigenvectors $q_i$ and eigenvalues in $\Lambda$, we can find our eigenvectors by computing:
$$q_i =
\begin{bmatrix}
-a_{12} \\
a_{11} - \lambda_i \\
\end{bmatrix}
$$
The eigenvectors will show the trajectory of the phase portrait.
From the discriminant $\Delta$ and $a > 0$, we will have to analyze the system for $a \in (0,2),$ $a \in (2,\infty),$ and $a =2$. I will skip the computations and present the concept directly. I leave it to you to DIY.
Case $1$: $a \in (0,2)$. This implies $\Delta<0$. The eigenvalues are complex. For simplicity, let us assume $a=1$.
$$\lambda_{1,2} = \frac{1}{2} \pm i \frac{\sqrt3}{2} $$
For this case, we shall just consider one of the eigenvectors, for the corresponding eigenvalue has a positive square-root. Its real and imaginary parts of the (complex) eigenvector will determine the trajectory of the field.
$$q_1 =
\begin{bmatrix}
-(-1) \\
0 - \big( \frac{1}{2} + i\frac{\sqrt3}{2} \big) \\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2} - i\frac{\sqrt3}{2} \\
\end{bmatrix}
$$
$$\implies q_{\text{re}} =
\begin{bmatrix}
1 \\
- \frac{1}{2}\\
\end{bmatrix}, -q_{\text{im}}
\begin{bmatrix}
0 \\
\frac{\sqrt 3}{2}\\
\end{bmatrix}
$$
As the eigenvalue is complex with $\text{Re}\{\lambda_{1,2}\} > 0$, it is an unstable focus. We shall sketch $q_{\text{re}}$ and $-q_{\text{im}}$ in the phase plane as follows.
Case $2$: $a \in (2,\infty)$. This implies $\Delta > 0$. The eigenvalues are real and not the same. For simplicity, let us assume $a=3$.
$$ \lambda_1 = \frac{3}{2} + \frac{\sqrt{5}}{2} = \frac{1}{2}(3 + \sqrt5) \implies
q_1 =
\begin{bmatrix}
-(-1) \\
0 - \frac{1}{2}(3 + \sqrt5)\\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2}(3 + \sqrt5)\\
\end{bmatrix}
$$
$$ \lambda_2 = \frac{3}{2} - \frac{\sqrt{5}}{2} = \frac{1}{2}(3 - \sqrt5) \implies
q_2 =
\begin{bmatrix}
-(-1) \\
0 - \frac{1}{2}(3 - \sqrt5)\\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2}(3 - \sqrt5)\\
\end{bmatrix}
$$
$$\lambda_i: \lambda_1 > \lambda_2 > 0 $$
The eigenvalue with the highest magnitude, i.e. $|\lambda_i|$ is known as the 'fast' eigenvalue. The trajectory will take the direction of the eigenvector corresponding to the 'fast' eigenvalue as it converges to (or diverges from) the origin. Since $\lambda_1 > \lambda_2 > 0$, it is an unstable node.
Case $3$: $a = 2$. This implies $\Delta = 0$. The eigenvalues are real and the same (repeated). This shows a degenerate system. This also means $A$ is not diagonalizable. The trajectory take straight lines through the origin. We can deduce the trajectory by multiplying $A$ with the standard basis vectors as follows:
$$
\lambda_1 = \lambda_2 = \frac{2}{2} = 1
$$
$$
q_1 =
\begin{bmatrix}
0 & -1 \\
1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix} =
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix}
$$
$$
q_2 =
\begin{bmatrix}
0 & -1 \\
1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix} =
\begin{bmatrix}
-1 \\
2 \\
\end{bmatrix}
$$
Since $\lambda_1 = \lambda_2 > 0$, it shows an unstable degenerate node.
Alternatively, you can check out this video on how to sketch phase portraits: https://www.youtube.com/watch?v=dpbRUQ-5YWc
Hope it helps. Cheers!
| {
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"url": "https://math.stackexchange.com/questions/4560826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to take the square root of a dual number ($\sqrt{a + b \cdot \varepsilon}$ with $a, ~b \in \mathbb{R}$ and $\varepsilon^{2} = 0$)? How to take the square root of a dual number:
$\sqrt{\Xi}$ with $\begin{align*}
a, ~b &\in \mathbb{R}\\
\varepsilon^{2} &= 0\\
\varepsilon &\ne 0\\
\\
\Xi &:= a + b \cdot \varepsilon\\
\end{align*}$
My first attempt worked (at least I think so), but the second one came to nothing, but it seems to me that taking the square root of binary numbers can also be derived in a different way, which is why I am here under the question in the question and the answers to show all possibilities to the question. (I wrote my first attempt as an answer under the question.)
Surely the question sounds strange, because why should one calculate that at all, but let's just ask ourselves how that would work?
My attempt $2$
My attempt $2$ was "try and hope that it works":
$$
\begin{align*}
\Xi^{2} &= \left( a + b \cdot \varepsilon \right)^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot + b \cdot \varepsilon + \left( b \cdot \varepsilon \right)^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot \varepsilon^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot 0\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + 0\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon \\
\Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
a + b \cdot \varepsilon &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
c := a^{2} &\wedge d := 2 \cdot a \cdot b \cdot \varepsilon\\
a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot a} \cdot \varepsilon^{-1}\\
a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\
\\
\Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
a + b \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon} \quad\mid\quad a = \sqrt{c} \wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1} \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot 1 &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c + d \cdot \varepsilon} &= \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}}\\
\sqrt{a + b \cdot \varepsilon} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \quad\mid\quad \left( ~~ \right)^{2}\\
a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \right)^{2}\\
a + b \cdot \varepsilon &= a + 2 \cdot \sqrt{a} \cdot \frac{b}{2 \cdot \sqrt{a}} + \frac{b^{2}}{4 \cdot a}\\
a + b \cdot \varepsilon &= a + b + \frac{b^{2}}{4 \cdot a}\\
b \cdot \varepsilon &= b + \frac{b^{2}}{4 \cdot a}\\
\end{align*}
$$
But that makes no sense...
| You are making things too complicated. Let $w=a+b\epsilon$ be a dual number (with $a,b$ real). We want to find the dual numbers $z$ such that $z^2=w$.
We write $z=x+y\epsilon$, with $x$ and $y$ real numbers. A short computation shows that $z^2 = x^2 + 2xy\epsilon$. Since $(1,\epsilon)$ is a base of dual numbers (as a vector space over the real numbers), we get the system
$$
\left\{
\begin{array}{lcl}
x^2 & = & a \\
2xy & = & b\\
\end{array}
\right.
$$
*
*The first thing to remark is that, if $a<0$, then there is no solution.
*If $a=0$ but $b\neq 0$, then the first equation implies that $x=0$, so the second equation becomes $0=b$, so no solution.
*If $a=b=0$, then $x=0$ and $y$ can be any real number. So $0=0+0\epsilon$ has infinitely many square roots in the dual numbers.
*Finally, if $a>0$, then $x=\pm \sqrt{a}$ and $y=\frac{b}{2x}=\pm \frac{b}{2\sqrt{a}}$, so $\omega=a+b\epsilon$ has two opposite square roots: $\pm(\sqrt{a}+\frac{b}{2\sqrt{a}}\epsilon)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{{{\sec }^2}A - {{\tan }^2}A}}{{\sec A + \tan A}}} \right) \Rightarrow \sqrt 3 \sin C = \sec A + \left( {\frac{1}{{\sec A + \tan A}}} \right)$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{1}{{\cos A}} + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{{1 + \sin A + 1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
$\sqrt 3 \sin C = \frac{{2 + \sin A - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
Not able to proceed further
| We had $\sqrt{3}\sin C\cos A+\sin A=2$.
It is well-known that the maximum of $a\cos\theta+b\sin\theta$ is $\sqrt{a^2+b^2}$ (and also its minumum is $-\sqrt{a^2+b^2}$).
So, the maximum of $\sqrt{3}\sin C\cos A+\sin A$ is $\sqrt{3\sin^2 C+1}$. It is clear that $\sqrt{3\sin^2 C+1}$ can be at most $2$ and it occurs at $C=90^{\circ}$ in the interval $(0^{\circ},180^{\circ})$. So, $\sqrt{3}\sin C\cos A+\sin A=2$ only if $C=90^{\circ}$.
We also can find $A$: $\sqrt{3}\cos A+\sin A=2\implies \sqrt{3}\cos A=2-\sin A\implies 3\cos^2A=4-4\sin A+\sin^2A\implies$$ 4\sin^2A-4\sin A+1=0\implies (2\sin^2 A-1)=0\implies \sin A=\frac{1}{2}\implies A=30^{\circ}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
} |
What's the Probability when throwing $2$ dice, the sum equals $12$ before $7$? When throwing $2$ dice and taking their sum what's the probability the first $12$ occurs before the first $7$?
My approach:
Let $A$ denote the sum equalling $12$ and $B$ equalling $7$
\begin{align}
P(A|A\text{ or }B) &= \frac{P(A \text{ and } (A \text{ or } B))}{P(A \text{ or } B)}\\
&= \frac{P(A)}{P(A\text{ or }B)}\\
&= \frac{1/36}{1/36 + 1/6}\\
&= 0.14
\end{align}
Have I gone wrong somewhere?
| Let me demonstrate a different approach leading to the same answer.
Let the sum of two throws be denoted by $X$.
We're doing multiple experiments in a sequence: $X_1, X_2, X_3, \cdots$
Each $X_i$ can take integer values from $2$ to $12$ given by the following distribution
$$\begin{gather*}
P(x) = \begin{cases}
\dfrac{1}{36} & \text{if $x \in \{2,12\}$}\\
\dfrac{2}{36}=\dfrac{1}{18} & \text{if $x \in \{3,11\}$}\\
\dfrac{3}{36}=\dfrac{1}{12} & \text{if $x \in \{4,10\}$}\\
\dfrac{4}{36}=\dfrac{1}{9} & \text{if $x \in \{5,9\}$}\\
\dfrac{5}{36} & \text{if $x \in \{6,8\}$}\\
\dfrac{6}{36} =\dfrac{1}{6} & \text{if $x = 7$}\\
\end{cases}
\end{gather*}$$
$\color{white}{.}$
$$ \begin{align}
&\;\;\;\;\ P (\text{sum equals 12 before 7}) \\
&= P(X_1=12) + P(X_1\neq 12, 7 ; X_2=12) + P(X_1, X_2\neq 12, 7 ; X_3=12) + \ldots \\
&= (1/36) + (1-(1/36+1/6))(1/36) + (1-(1/36+1/6))^2(1/36) + \ldots \\
&= \dfrac{1/36}{1-(1-(1/36+1/6))} \\
&= \dfrac{1/36}{1/36+1/6} \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of the expression $\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$ Let $x, y, z$ be positive real numbers such that $x + y + z = xyz$.
Find the minimum value of the expression $$\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$$
My attempt:
By using Cauchy-Schwarz inequality and AM-GM inequality,
$$\sqrt{\frac{1}{3}t^4+1}\cdot\sqrt{3+1}\geq t^2+1$$
$$(x^2+y^2+z^2)(x+y+z)\geq3(xyz)^{\frac{2}{3}}\cdot 3(xyz)^{\frac{1}{3}}=9xyz$$
$$\therefore \sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}\geq \frac{1}{2}(x^2+y^2+z^2)+\frac{3}{2}$$
$$\geq\frac{1}{2}\left (\frac{9xyz}{x+y+z}+\frac{3}{2} \right)=6$$
with equality if and only if $x=y=z=\sqrt{3}$
What are the alternative methods to solve this question?
| You already have a good method, here is an alternative with only AM-GM. First, we have
$xyz = x+y+z \geqslant 3\sqrt[3]{xyz}\implies xyz \geqslant 3\sqrt3$. Further, again using AM-GM,
$$\sqrt{\frac{t^4}9+\frac{t^4}9+\frac{t^4}9+1}\geqslant \sqrt{\frac4{3\sqrt3}t^3}=\frac2{3^{3/4}}t^{3/2}$$
$\begin{align}
\implies \sqrt{\frac13x^4+1}+\sqrt{\frac13y^4+1}+\sqrt{\frac13z^4+1} &\geqslant \frac2{3^{3/4}}(x^{3/2}+y^{3/2}+z^{3/2})
\\ &\geqslant \frac2{3^{3/4}}\cdot 3\,(xyz)^{1/2}
\\ &\geqslant \frac6{3^{3/4}}\,(3\sqrt3)^{1/2}=6
\end{align}
$
with equality possible iff $x=y=z=\sqrt3$, so we have the minimum.
P.S. Yet another way would be to show the estimate $\sqrt{t^4/3+1} \geqslant \sqrt3 t -1$ holds, and then $\sum \sqrt{x^4/3+1} \geqslant \sqrt3(\sum x)-3 = \sqrt3(xyz)-3 \geqslant \sqrt3\cdot 3\sqrt3-3 = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $k$ is the greatest integer such that $k < \sqrt{n+2}$ and $k|n$, prove that $n$ is a perfect square
If $n>2$ an odd number and $k \in \mathbb{Z}$ such that $k$ is the greatest integer for which $k<\sqrt{n+2}$ and, furthermore, $k|n$ holds, prove that $n$ is a perfect square.
My idea of thinking was that, because $k$ is the greatest integer for which this inequality holds, we must have that $k < \sqrt{n+2} \leq k+1\Rightarrow k^2 < n+2 < (k+1)^2 \Rightarrow k^2 -2 < n < (k+1)^2 -2$. If I try to write $n=ak$ for some $a \in \mathbb{Z}$ (all of them are positive because of the assertion that $k$ is the greatest integer such that the inequality holds, and it holds for positive integers.) I get some intervals for $a$ but it doesn't help me get any further. Any suggestions?
| We first note the following:
Claim 1: The condition that $n$ is odd implies that $n \not = (a)(a+1)$ for all integer $a$, because as always, one of $a,a+1$ is even.
Write $n = ab$, with $1<a \le b < n$. Then if $a$ and $b$ are the same, then $n=a^2=b^2$, and we are done. So now it remains to consider the case $a$ and $b$ are not the same.
Then if $a$ and $b$ are not the same however, this and Claim 1 gives $b \ge a+2$. Thus
$$b^2 \ge b(a+2)$$ $$= ba +2b = n + 2b > n+2,$$ or in particular, $$b^2 > n+2,$$ or equivalently, $$b > \sqrt{n+2}.$$ This contradicts the hypothesis that $b \le \sqrt{n+2}$ however, which implies that $a$ and $b$ must be the same after all, which implies that $n=a^2=b^2$ must be a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\prod_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$ Show that
$$\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi}\sin{\left(\frac{2\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{k+2}\right)}=\frac{\pi}{2}$$
Context:
Inspired by this question, I considered the product of the areas of every regular odd-gon inscribed in a circle of area $1$. This is $\prod\limits_{k=1}^\infty \frac{2k+1}{2\pi} \sin{\frac{2\pi}{2k+1}}=0.18055...$ which I guess does not have a closed form.
On a whim, I divided this number by the Kepler-Bouwkamp constant, $\prod\limits_{k=1}^\infty \cos{\frac{\pi}{k+2}}=0.11494...$, and numerical calculation suggests that the result is $\pi/2$.
My attempt:
I used the double angle formula for sine to get $\prod\limits_{k=1}^\infty \frac{2k+1}{\pi}\sin{\left(\frac{\pi}{2k+1}\right)}\sec{\left(\frac{\pi}{2k+2}\right)}$ but this seems just as intractable as the original form of the product.
| Interesting product. Let us define the Kepler-Bouwkamp constant as $c$. Then
$$c=\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)$$
Let $\text{Si}(x)=\frac{\sin(x)}{x}$. Then, $\cos(x)=\frac{\text{Si}(2x)}{\text{Si}(x)}$. Plugging this in, we get
$$c=\prod_{k=1}^\infty\frac{\text{Si}\left(\frac{2\pi}{k+2}\right)}{\text{Si}\left(\frac{\pi}{k+2}\right)}$$
Now, observe that every term in the denominator cancels with a term in the numerator because $\frac{\pi}{k+2}=\frac{2\pi}{2(k+1)}$. All that remains is the second term in the numerator and the odd terms of the numerator, viz
$$c=\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2+2}\right)\text{Si}\left(\frac{2\pi}{2k+1}\right)=\text{Si}\left(\frac{\pi}{2}\right)\prod_{k=1}^\infty \text{Si}\left(\frac{2\pi}{2k+1}\right)$$
Simplifying, and plugging in the actual product expression for $c$, we have
$$\prod_{k=1}^\infty \cos\left(\frac{\pi}{k+2}\right)=\frac{2}{\pi}\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)$$
Rearranging gives us that $$\frac{\pi}{2}=\prod_{k=1}^\infty\frac{2k+1}{2\pi}\sin\left(\frac{2\pi}{2k+1}\right)\sec\left(\frac{\pi}{k+2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Solve the differential equation: $\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
Solve the differential equation: $\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$
My Attempt:
$$\frac{2xdx-2ydy}{2x^2\frac{xdy-ydx}{x^2}}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\\ \frac{d(x^2-y^2)}{2x^2d(\frac yx)}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\\ \frac{\frac{d(x^2-y^2)}{x^2-y^2}}{2\frac{x^2}{x^2-y^2}d(\frac yx)}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\\ \frac1{x^2-y^2}\sqrt{\frac{x^2-y^2}{1+x^2-y^2}}d(x^2-y^2)=2\frac{1}{1-(\frac yx)^2}d(\frac yx)$$
Put $x^2-y^2=p, \frac yx=q$
$$\frac1p\frac{\sqrt p}{\sqrt{1+p}}dp=\frac{2dq}{1-q^2}\\ \frac{dp}{\sqrt{p^2+p}}=\frac{2dq}{1-q^2}\\ \frac{dp}{\sqrt{(p+\frac12)^2-\frac14}}=\frac{2dq}{1-q^2}\\ \ln|p+\frac12+\sqrt{p^2+p}|=\ln\frac{1+q}{1-q}+\ln c\\ \implies x^2-y^2+\frac12+\sqrt{x^2-y^2}\sqrt{1+x^2-y^2}=c\frac{x+y}{x-y}$$
The answer given is $\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}=c\frac{x+y}{\sqrt{x^2-y^2}}$
Can we reach the answer with the approach that I followed?
The other approach is mentioned here.
| In the other approach, this step is wrong.
$$\ln\left|\color{red}1+\sqrt{1+r^2}\right| = \ln \left|\sec \theta +\tan \theta\right|+\ln \mathcal{C}$$
Should be
$$\ln\left|\color{red}r+\sqrt{1+r^2}\right| = \ln \left|\sec \theta +\tan \theta\right|+\ln \mathcal{C}$$
Then
$$\ln\left|\sqrt{x^2-y^2}+\sqrt{1+x^2-y^2}\right| = \ln C\left|\frac{x+y}{\sqrt{x^2-y^2}}\right|$$
Note that
$$(r+\sqrt{1+r^2})^2=r^2+r^2+1+2r\sqrt{r^2+1}=2(r^2+\frac{1}{2}+\sqrt{r^4+r^2})=2(p+\frac{1}{2}+\sqrt{p^2+p})$$
$$(\sec \theta +\tan \theta)^2=\frac{(1+\sin \theta)^2}{\cos ^2\theta}=\frac{(1+\sin \theta)^2}{1- \sin ^2\theta}=\frac{1+\sin \theta}{1-\sin \theta}=\frac{1+q}{1-q}$$
I can match your terms to the terms in the other approach and your given anwser. Square the given answer, you can get back your answer form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$? I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}.$
Answer is $\frac1{24}$.
The first equation has two solutions however the latter one has only one solution. I lossed up one solution but still, I'm interested in solving it.
I wish if someone could help me in solving any of the above equations.
Here's my work:
$$\begin{align}x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} &= 2\\x + \frac{1}{2}\left(2 \sqrt{x}\sqrt{x+1} + 2\sqrt{x+1}\sqrt{x+2} + 2\sqrt{x}\sqrt{x+2}\right) &=2\tag{1}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (x + x + 1 + x + 2)\Big]& = 2\tag{2}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 2\\2x + \Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 4\\(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 & = x + 7\\\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} & = \sqrt{x+7}\end{align}$$
Moving from $(1)$ to $(2)$, I used $2(ab + bc + ca) = (a+b+c)^2 - (a^2+b^2+c^2)$.
In general, is it possible to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+k}$ by hand?
| There is a fairly simple approach to this:
$\sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+7}$
$\sqrt{x}+\sqrt{x+1}=\sqrt{x+7}-\sqrt{x+2}$
Now, squaring both sides, we get:
$2x+1+2\sqrt{x(x+1)}=2x+9-2\sqrt{(x+2)(x+7)}$
$\sqrt{x(x+1)}=4-\sqrt{(x+2)(x+7)}$
Squaring once again:
$x(x+1)=16-8\sqrt{(x+2)(x+7)}+(x+2)(x+7)$
$x^2+x=16-8\sqrt{(x+2)(x+7)}+x^2+9x+14$
$8\sqrt{(x+2)(x+7)}=30+8x$
$4\sqrt{(x+2)(x+7)}=15+4x$
Squaring once again:
$16(x+2)(x+7)=225+120x+16x^2$
$16(x^2+9x+14)=225+120x+16x^2$
From here, the solution is very straight forward with a simple quadratic equation that you can solve easily. Make sure to verify the roots you get by plugging them in and checking if they satisfy the original equation. Some of them could be extraneous due to squaring of both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solving $\int \sin^3 2 \theta \sqrt{\cos 2 \theta}\:d\theta$ using Integration by Parts. We can solve the following problem by using substitution and trigonometric identities. Such is the solution: $$
\begin{aligned}
& \int \sin ^3 2 \theta \sqrt{\cos 2 \theta} d \theta \\
=& \int \sin ^3 2 \theta(\cos 2 \theta)^{\frac{1}{2}} \\
=& \int \sin ^2 2 \theta \cdot \sin 2 \theta(\cos 2 \theta)^{1 / 2} \\
=& \int\left(1-\cos ^2 2 \theta\right)(\cos 2 \theta)^{1 / 2} \sin 2 \theta d \theta \\
\text { let } & u=\cos 2 \theta \\
& d u=-2 \sin 2 \theta d \theta \\
& d \theta=-\frac{d u}{2 \sin 2 \theta} \\
=& \int\left(1-u^2\right)(u)^{1 / 2} \cdot \sin 2 \theta \cdot \frac{d u}{2 \sin 2 \theta} \\
=&-\frac{1}{2} \int\left(1-u^2\right)(u)^{1 / 2} d u \\
=&-\frac{1}{2} \int\left(u^{1 / 2}-u^{3 / 2}\right) d u \\
&=-\frac{1}{2}\left[\frac{2}{3} u^{3 / 2}-\frac{2}{5} u^{5 / 2}\right] \\
&=-\frac{1}{3} u^{3 / 2}+\frac{1}{5} u^{5 / 2}+C \\
&=-\frac{1}{3} \cos ^{3 / 2} 2 \theta+\frac{1}{5} \cos ^{5 / 2} 2 \theta+C
\end{aligned}$$
We are trying to find out how to solve the problem using Integration by Parts (The one where it says $\int udv = uv - \int vdu$). We have arrived at the solution below but we do not know how to continue from here until we get the desired result which is $-\frac{1}{3} \cos ^{3 / 2} 2 \theta+\frac{1}{5} \cos ^{5 / 2} 2 \theta+C$. Any help would be appreciated.
$$
\begin{aligned}
&\int \sin ^3 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\int \sin ^2 2 \theta \sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\text { Let } u=\sin ^2 2 \theta, d v=\sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&d u=\sin 4 \theta d \theta, \quad v=\int \sin 2 \theta \sqrt{\cos 2 \theta} d \theta \\
&\text { Let } x=\cos 2 \theta \\
&d x=-2 \sin 2 \theta d \theta \\
&\therefore d \theta=\frac{d x}{-2 \sin 2 \theta} \\
&\therefore v=-\frac{1}{2} \int \sqrt{x} d x \\
&=-\frac{1}{2}\left[\frac{2}{3} u^{3 / 2}\right]=-\frac{1}{3} \cos ^{3 / 2} 2 \theta \\
&\therefore \sin ^2 2 \theta\left(-\frac{1}{3} \cos ^{3 / 2} 2 \theta\right)-\int\left(-\frac{1}{3} \cos ^{3 / 2} 2 \theta\right) \cdot \sin 4 \theta d \theta \\
&
\end{aligned}
$$
| There are a couple of mistakes in your solutions:
*
*In your first method you write $(1-u^2) u^{\frac12} =u ^{\frac12} - u ^{\color{red}{\frac32}}$, but the second exponent should be $2 + \frac{1}{2} = \frac{4}{2}+ \frac{1}{2} = \color{red}{\frac{5}{2}}$. With this correction, the antiderivative you get is
$$
-\frac{1}{3}\cos^{\frac32}(2\theta) + \frac{1}{7}\cos^{\frac72}(2\theta)+C
$$
*On your second solution you write $u = \sin^2(2\theta)$ implies $\mathrm{d}u = \sin(4\theta)$, but the derivative of $\sin^2(2\theta)$ is $\color{green}{2}\sin(4\theta)$ (you probably forgot to apply the chain rule on the inner-most $2\theta$).
Since $\sin(2x) = 2\sin(x) \cos(x)$ we get
$$
\int \cos^{\frac32}(2\theta) \sin(4\theta) \, \mathrm{d}\theta = 2\int \cos^{\frac52}(2\theta) \sin(2\theta)\, \mathrm{d} \theta \overset{\color{blue}{u = \cos(2\theta)}}{=} - \int u^{\frac52}\, \mathrm{d}u = - \frac27 \cos^{\frac72}(2\theta) \tag{1}
$$
Then, from your last (corrected) equation, we get
\begin{align*}
\int\sin^3(2\theta)\sqrt{\cos(2\theta)}\, \mathrm{d} \theta & \overset{\text{IBP}}{=} -\frac{\color{purple}{\sin^2(2\theta)}\cos^{\frac32}(2\theta)}{3} + \frac{\color{green}{2}}{3} \int \cos^{\frac32}(2\theta) \sin(4\theta) \, \mathrm{d}\theta \\
& \overset{(1)}{=} -\frac{\color{purple}{\left(1-\cos^2(2\theta)\right)}\cos^{\frac32}(2\theta)}{3} - \frac{4}{21} \cos^{\frac72}(2\theta) +C\\
& = -\frac{1}{3}\cos^{\frac32}(2\theta) + \frac{1}{7}\cos^{\frac72}(2\theta)+C
\end{align*}
as desired.
| {
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"url": "https://math.stackexchange.com/questions/4582259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove an asymptotic estimate for $n \choose k$, where $k = \mathcal{o}(n^{2/3})$?
Let $k = k(n)$ such that $k \rightarrow \infty$ as $n \rightarrow \infty$, but $k = \mathcal{o}(n^{2/3})$. Use Stirling's formula $$n! = (1+\mathcal{o}(1))\sqrt{2\pi n} \biggl(\frac{n}{e}\biggr)^n$$
and the Taylor expansion of $\ln(1+x)$ to show that
$${n\choose k} = (1+\mathcal{o}(1))\frac{1}{\sqrt{2\pi n}} \biggl(\frac{n}{k}\biggr)^k \begin{cases} \exp(k) \quad \text{ for } k = \mathcal{o}(n^{1/2}) \\ \exp \biggl(k-\frac{k^2}{2n}\biggr) \quad \text{ for } k = \Omega(n^{1/2})\end{cases} .$$
I understand that by definition we have
$${n\choose k} = \frac{n!}{k! (n-k)!},$$
but I guess we can not just simply use Stirling's formula on $n!$, $k!$ and $(n-k)!$ at the same time, so I do not see how to proceed. Could you please give me a hint?
| It follows from the main result of this paper that
$$
\log ((n + a)!) = \left( {n + a + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{{6a^2 + 6a + 1}}{{12n}} + \mathcal{O}\!\left( {\frac{{\max (\left| a \right|^3 ,1)}}{{n^2 }}} \right)
$$
provided $n + a+1 \ge 0$ and $\left| a+1 \right| < \frac{3}{5}n$, where the implied constant does not depend on $n$ or $a$. Then
$$
\log (n!) = \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + o(1),
$$
$$
\log (k!) = \left( {k + \frac{1}{2}} \right)\log k - k + \frac{1}{2}\log (2\pi ) + o(1),
$$
$$
\log ((n - k)!) = \left( {n - k + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{k^2}{{2n}} + o(1),
$$ as $n,k\to +\infty$ with $k=o(n^{2/3})$. Accordingly,
$$
\log \binom{n}{k} = - \frac{1}{2}\log (2\pi k) + k\log \left( {\frac{n}{k}} \right) + k - \frac{k^2}{{2n}} + o(1)
$$
or
$$
\binom{n}{k} = \frac{1}{{\sqrt {2\pi k} }}\left( {\frac{n}{k}} \right)^k \exp\! \left( {k - \frac{k^2}{{2n}}} \right)(1 + o(1)),
$$
as $n,k\to +\infty$ with $k=o(n^{2/3})$. Note that the factor involves $\sqrt{2\pi k}$ and not $\sqrt{2\pi n}$.
| {
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"url": "https://math.stackexchange.com/questions/4582704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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absolute fraction of polynomials I'm trying to solve inequality:
$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg|\gt 1$$
The first step I do is to conclude how the function behaves (to see when it's negative):
$$\frac{x^2+2x-36}{x^2-4} = 0$$
My approach is to solve the nominator and denominator separately (as separate second degrees polynomials). The denominator is easy (opens upward parabola):
$$x^2-4=0 \\(x-2)(x+2)=0\iff x=-2 \vee x=2
$$
The issue I have is with the nominator:
$$x^2+2x-36 = 0$$
Since the standard approach with quadratic formulas doesn't look sensible to me.
Could you share some hints? How I should approach that?
| $$
\begin{aligned}
& \left|\frac{x^2+2 x-36}{x^2-4}\right|>1, \text { where } x \neq \pm 2 \\
\Rightarrow \quad& \left|x^2+2 x-36\right|>\left|x^2-4\right| \\
\Rightarrow \quad&\left(x^2+2 x-36\right)^2>\left(x^2-4\right)^2 \\
\Rightarrow \quad& \left(x^2+2 x-36\right)^2-\left(x^2-4\right)^2>0 \\
\Rightarrow \quad& \left(x^2+2 x-36+x^2-4\right)\left(x^2+2 x-36-x^2+4\right)>0 \\
\Rightarrow \quad& 4\left(x^2+x-20\right)(x-16)>0 \\
\Rightarrow \quad& 4(x+5)(x-4)(x-16)>0 \\
\Rightarrow \quad& -5<x<4 \text { or } x>16
\end{aligned}
$$
However $x\ne \pm 2$, therefore the solutions are
$$\boxed{-5<x<-2, \quad-2<x<2, \quad 2<x<4 \quad\textrm{ and }x>16.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral $\int\ln(1−x)dx$ by parts I have an exercise that explicitly asks for the indefinite integral of the following equation using integral by parts.
$$
\int\ln(1-x)\,dx
$$
I tried to define: $ u = \ln(1-x) $ so,
$$
\frac{\,du}{\,dx} = \frac{-1}{1-x}\\
\,du = \frac{-\,dx}{1-x}
$$
and, $ \,dv = \,dx $ resulting in $ v = x $ and finally applying: $ \int{u\,dv} = uv - \int{v\,du} $
$$
\begin{align}
\int\ln(1-x)\,dx&= x \ln(1-x) -\int x \, \frac{-\,dx}{1-x}\\
&=x\ln(1-x) - \int \frac{-x}{1-x} \,dx\\
&=x\ln(1-x) - \int \frac{1-x-1}{1-x} \,dx\\
&=x\ln(1-x) - \int \left(\frac{1-x}{1-x} - \frac{1}{1-x} \right)\,dx\\
&=x\ln(1-x) - \int \left(1 - \frac{1}{1-x} \right)\,dx\\
&=x\ln(1-x) - \left(\int{1\,dx} - \int{\frac{1}{1-x}}\,dx\right)\\
&=x\ln(1-x) - x -\left(- \int{\frac{1}{1-x}}\,dx\right)\\
&=x\ln(1-x) - x + \int{\frac{1}{1-x}}\,dx\\
&=x\ln(1-x) - x + \ln(1-x) + c\\
&=(x + 1)\ln(1-x) - x + c
\end{align}
$$
that does not match the answer in the book:
$(x - 1)\ln(1-x) - x + c$
I have found this other solution that I believe is not valid considering that the exercise asks to utilize integral by parts.
| Your steps all ALL correct except the last second step as @J.G. said.
On the other hand, using integration by parts wisely, we get
$$
\begin{aligned}
\int \ln (1-x) d x & =-\int \ln (1-x) d(1-x) \\
& =(1-x) \ln (1-x)+\int(1-x) \frac{1}{1-x} d x \\
& =(1-x) \ln (1-x)+x+C
\end{aligned}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$ without calculus How would you prove this without calculus?
$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$
In this post it is proved with calculus, and I was wondering if there was an easier way to prove it. I tried proving it with induction, by holding $n$ constant, however I didn't seem to get anywhere.
| Induction works.
Let
$$a(n,m)=x^{n+1}\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^k$$
$$b(n,m)=(1-x)^{m+1}\sum_{k=0}^{n}\binom{m+k}{k}x^k$$
$$f(n,m)=a(n,m)+b(n,m)$$
So we have to prove $f(n,m)=1$ for all $n,m\ge0$.
Let's prove this by induction on $m$.
First, $f(n,0)=1$ for all $n\ge0$, as $a(n,0)=x^{n+1}$ and $b(n,0)=(1-x)\sum_{k=0}^nx^k=1-x^{n+1}$.
Below, $n\ge0$ is fixed.
Let's assume $f(n,m)=1$ for some $m\ge0$. Then:
$$a(n,m+1)=a(n,m)+x^{n+1}\binom{n+m+1}{m+1}(1-x)^{m+1}$$
$$b(n,m+1)=(1-x)^{m+2}\sum_{k=0}^n\binom{m+k+1}{k}x^k=(1-x)^{m+2}\left\{\sum_{k=0}^n\binom{m+k}{k}x^k+\sum_{k=1}^n\binom{m+k}{k-1}x^k\right\}\\=(1-x)(1-x)^{m+1}\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)^{m+2}\sum_{k=1}^n\binom{m+k}{k-1}x^k\\=(1-x)b(n,m)+(1-x)^{m+2}\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}$$
Therefore
$$f(n,m+1)=f(n,m)+(1-x)^{m+1}\left\{\binom{n+m+1}{m+1}x^{n+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\right\}$$
We want to prove that the expression inside the curly brackets is zero:
$$\binom{n+m+1}{m+1}x^{n+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k+(1-x)\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\\=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k-x\sum_{k=0}^{n-1}\binom{m+k+1}{k}x^{k+1}\\
=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-x\sum_{k=0}^n\binom{m+k}{k}x^k-x\sum_{k=1}^{n}\binom{m+k}{k-1}x^{k}\\
=\sum_{k=0}^n\binom{m+k+1}{k}x^{k+1}-\sum_{k=0}^n\binom{m+k}{k}x^{k+1}-\sum_{k=1}^{n}\binom{m+k}{k-1}x^{k+1}\\
=x-x+\sum_{k=1}^n\left\{\binom{m+k+1}{k}-\binom{m+k}{k}-\binom{m+k}{k-1}\right\}x^{k+1}\\=0$$
Where the $x-x$ in the last line comes from the index $k=0$ in the first two sums of the preceding line.
Therefore the induction step is proved: if $f(n,m)=1$, then $f(n,m+1)=1$. And by induction, the equality is true for all $m\ge0$. Since this is proved for any $n\ge0$, the equality $f(n,m)=1$ is true for all $n,m\ge0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
To prove an identity related to Gamma function
Question: How to prove the following identity for all positive integers $k$ and $n$:
\begin{align}\tag{1}
(k+1)(2k+1) \cdots (nk+1) = \sum_{i=1}^n & \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \left[ (k+1)(2k+1) \cdots ((i-1)k+1) \right] \\
&\ \times \left[ (k+1)(2k+1) \cdots ((n-i)k+1) \right].
\end{align}
I could check by enumeration in $n$:
\begin{align}
n&=1, & k+1 &= \binom{1}{1} (k+2-1), \\
n&=2, & (k+1)(2k+1) &= \binom{2}{1} \left( \frac{1}{2}(k+2)-1 \right) (k+1) + \binom{2}{2} \left( (k+2)-1 \right) (k+1) \\
& & & = k(k+1) + (k+1)^2, \\
n&=3, & (k+1)(2k+1)(3k+1) &= \binom{3}{1} \left( \frac{1}{3}(k+2)-1 \right) (k+1)(2k+1) \\
& & &\quad + \binom{3}{2} \left( \frac{2}{3}(k+2)-1 \right) (k+1)^2 \\
& & &\quad + \binom{3}{3} \left( (k+2)-1 \right) (k+1)(2k+1) \\
& & & = (k-1)(k+1)(2k+1) + (2k+1)(k+1)^2 + (k+1)^2(2k+1), \\
& \cdots & &
\end{align}
But I failed to prove it by induction in $n$:
Set
\begin{equation}
I(n;k) := \sum_{i=1}^n \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \left[ (k+1) \cdots ((i-1)k+1) \right] \left[ (k+1) \cdots ((n-i)k+1) \right].
\end{equation}
Then
\begin{equation}
\begin{split}
I(n+1;k) - I(n;k) =&\ (k+1) \left[ (k+1) \cdots (nk+1) \right] \\
&\ + \sum_{i=1}^n \binom{n}{i} \left[ (k+1) \cdots ((i-1)k+1) \right] \left[ (k+1) \cdots ((n-i)k+1) \right] \\
&\ \qquad\quad \times \underbrace{\left[ \frac{n+1}{n+1-i} \left( \frac{i}{n+1}(k+2)-1 \right) ((n+1-i)k+1) - \left( \frac{i}{n}(k+2)-1 \right) \right]}_{=:J(n;k,i)}
\end{split}
\end{equation}
If the followin equation holds,
\begin{equation}\tag{2}
J(n;k,i) = \left( \frac{i}{n}(k+2)-1 \right)(nk-1),
\end{equation}
then using the inductive hypothesis $I(n;k) = (k+1) \cdots (nk+1)$, we have
\begin{equation}
\begin{split}
I(n+1;k) - I(n;k) =&\ (k+1) \left[ (k+1) \cdots (nk+1) \right] + (nk-1) I(n;k) = (n+1)k I(n;k)
\end{split}
\end{equation}
which yields $I(n+1;k) = ((n+1)k+1) I(n;k) = (k+1) \cdots (nk+1) ((n+1)k+1)$ as desired.
But, unfortunately, equation $(2)$ does not hold...
I was also thinking that the question may be related to Gamma function, because
$$
(k+1)(2k+1) \cdots (nk+1) = k^n \frac{\Gamma(n+1+\frac{1}{k})}{\Gamma(1+\frac{1}{k})}
$$
so that identity $(1)$ turns to
\begin{equation}\tag{1'}
k \Gamma\left(n+1+\frac{1}{k}\right) \Gamma\left(1+\frac{1}{k}\right) = \sum_{i=1}^n \binom{n}{i} \left( \frac{i}{n}(k+2)-1 \right) \Gamma\left(i+\frac{1}{k}\right) \Gamma\left(n-i+1+\frac{1}{k}\right).
\end{equation}
But still, I have no clue to prove $(1')$.
Could anyone help on it? Any hint or comment will be appreciated. TIA...
| Let $$A_m=(k+1)(2k+1)...(mk+1)$$
Let $$B_{n+1}=\sum_{i=1}^{n+1}\left[(k+2){n\choose i-1}-{n+1\choose i}\right]A_{i-1}A_{n+1-i}$$
Break both the binomial coefficients according to Pascal's rule. The terms with $(k+2)$ factors can then be paired up again, giving terms of the form
$${n-1\choose i-1}(A_{i-1}A_{n+1-i}+A_iA_{n-i})\\
={n-1\choose i-1}A_{i-1}A_{n-i}((n+1)k+2)$$
The other terms can be combined similarly, but there is an $A_n$ left over from the $i=1$ term. From comparison with $B_n$, we get $$B_{n+1}=((n+1)k+2)B_n-A_n$$
So, by induction $B_n=A_n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine the number of odd binomial coefficients in the expansion of $(x+y)^{1000}$.
Determine the number of odd binomial coefficients in the expansion of $(x+y)^{1000}$. Hint: The number of odd coefficients in any finite binomial expansion is a power of $2$.
Is there a way to prove this without using something like Lucas's theorem or any other non-trivial result? It's a problem from a problem solving book and they haven't introduced any theorems in it. They have just given that it should be a power of $2$, but nothing on how to go about finding it. Computing a few smaller terms shows that $(x+y)^0$ has $1$ odd coefficient, $(x+y)^1$ has $2$, $(x+y)^2$ has $2$, $(x+y)^3$ has $4$, $(x+y)^4$ has 2 and $(x+y)^5$ has $4$, but no pattern seems to emerge.
| Hints:
Employing $\text{modulo 2}$ polynomial calculations we are pleased to find that
$$ 512 + 256 + 128 + 64 + 32 + 8 = 1000$$
and we don't have to expand
$$ (x^{512}+ y^{512}) (x^{256}+ y^{256}) (x^{128}+ y^{128}) (x^{64}+ y^{64}) (x^{32}+ y^{32}) (x^{8}+ y^{8})$$
since $1 \times 1 = 1$ and there is no need to collect like terms after combining/adding the exponents .
Now just set both $x$ and $y$ to $1$.
ANS: $64$
Note: If you start calculating
$\; (x+y)^1$
$\; (x+y)^2 \equiv x^2 + y^2$
$\; (x+y)^3 \equiv (x^2 + y^2)(x+y)$
$\; (x+y)^4 \equiv (x^4 + y^4)$
$\; (x+y)^5 \equiv (x^4 + y^4)(x+y)$
$\; (x+y)^6 \equiv (x^4 + y^4)(x^2+y^2)$
$\; (x+y)^7 \equiv (x^4 + y^4)(x^2+y^2)(x+y)$
$\; (x+y)^8 \equiv (x^8 + y^8)$
$\; (x+y)^9 \equiv (x^8 + y^8)(x+y)$
$\; \dots$
You will discover the pattern given in Oscar Lanzi's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to LHS but did not succeed in solving by this means.
Please provide a solution.
| Because $$\sum_{cyc}\left(\frac{a^2+bc}{b+c}-a\right)=\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{\prod\limits_{cyc}(a+b)}=\frac{\sum\limits_{cyc}(a^2-b^2)^2}{2\prod\limits_{cyc}(a+b)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Computing $\int_{\Bbb{B}^3\times\Bbb{B}^3}\frac{\rm{d}^3x\,\rm{d}^3y}{\|x-y\|}=\frac{32}{15}\pi^2$, where $\Bbb{B}^3$ is the 3-dimensional unit ball Let's call $\mathbb{B}^3$ the three-dimensional unit ball, then how to compute the six fold
$$
\int_{\mathbb{B}^3\times \mathbb{B}^3}\frac{\text{d}^3x\,\text{d}^3y}{\|x-y\|}=\frac{32}{15}\pi^2
$$
I saw this question on a forum, but no one seemed willing to answer it so I put it here for help. Any useful help would be greatly appreciated.
| We can focus on one integral first - in this case consider only
$$\int_{B(0,1)}\frac{d^3x}{|x-y|} = \int_{B(y,1)}\frac{d^3x}{|x|}$$
by translating the origin to the other variable. We can always rotate to a coordinate system for $x$ where $y$ always lies on the $x_3$ axis, giving us the equation of the ball
$$x^2-2|y|x_3 = 1-y^2$$
Converting to spherical coordinates, we obtain
$$r^2-2|y|r\cos\theta = 1-y^2 \implies \begin{cases}\theta = \cos^{-1}\left(\frac{r^2+y^2-1}{2|y|r}\right)\\ r = |y|\cos\theta+\sqrt{1-y^2\sin^2\theta}\end{cases}$$
Of course this $\theta$ upper bound is only valid when $r>1-|y|$, so we need to split the integral into two
$$\int_{B(y,1)}\frac{r^2\sin\theta \: d(r,\theta,\varphi)}{r}$$
$$ = \int_0^{2\pi}\int_0^{1-|y|}\int_0^{\pi}r \sin\theta \:d\theta\:dr\:d\varphi + \int_0^{2\pi}\int_{1-|y|}^{1+|y|}\int_0^{\cos^{-1}\left(\frac{r^2+y^2-1}{2|y|r}\right)}r \sin\theta \:d\theta\:dr\:d\varphi$$
$$= 2\pi(1-|y|)^2 + 2\pi\int_{1-|y|}^{1+|y|}\frac{1-y^2}{2|y|}+r-\frac{r^2}{2|y|}\:dr$$
$$= 2\pi - 4\pi|y| + 2\pi y^2 +4\pi |y| - \frac{8\pi}{3} y^2 = 2\pi -\frac{2\pi}{3}y^2$$
We can plug that polynomial into the integral over a unit ball and get our final answer
$$2\pi\int_{B(0,1)}d^3y - \frac{2\pi}{3} \int_{\Bbb{S}^2}d\Omega \int_0^1 r^4\:dr $$
$$= 2\pi\cdot\left(\frac{4\pi}{3}\right)-\frac{2\pi}{3}\cdot(4\pi)\cdot\left(\frac{1}{5}\right) = \boxed{\frac{32\pi^2}{15}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Asymptotic expansion of a coefficients given functional equation of the generating function Given a functional equation ($0<q<1$)
$$ g(x)= \frac{1}{1-x} + \frac{q^2}{(1-(1-q) x)^3} g\left(\frac{q x}{1-(1-q) x}\right) $$
of the analytic function
$$ g(x) = \sum_{n=0}^\infty g_n x^n, $$
how can we extract the behaviour of $g(x)$ near $x=1$, and more importantly, to get the asymptotics for $g_n$ as $n$ is large?
I was able, using Mathematica, to express $g(x)$ in terms of QPolyGamma function, but I was unable to extract the asymptotics.
EDIT:
Motivation: Consider a game where you have $n$ players. Each player throw a dice. The players with number 6 $(q = 1/6)$ continue into the next round, until there is only one player. If no player scores 6, then the round is repeated for those players left. $g_{n-2}$ corresponds to expected number of rounds when there are $n$ players in the beginning. A harder problem is the following: Instead of repeating rounds, we will say winners going to the next round are those with highest scored number (1 to 5). This only changes $\alpha$ as the asymptotic is the same. However, I was not able yet to determine its precise value.
| After couple of hours, I think I have it:
$$g(x) \approx - \frac{1}{\ln q}\left(\frac{\ln\frac{1}{1-x}}{1-x}\right) + \frac{\alpha}{1-x}$$
from which the asymptotics should be
$$g_n \approx -\frac{H_{n+2}}{\ln q}+\alpha,$$
where
$$\alpha = \lim_{p\to 0}\left(-\frac{\ln p}{\ln
q}+\sum _{k=0}^{\infty } \frac{q^{2 k}}{\left(p+q^k\right)^2}\right).\tag{0}$$
I. PROOF
The correct asymptotics ansatz is
$$g(x) = \gamma \frac{\ln\frac{1}{1-x}}{1-x} + \frac{\alpha }{1-x}+\beta \ln \left(\frac{1}{1-x}\right)\tag{1}$$
to see that, we substitute this into
$$g(x) = \frac{1}{1-x} + \frac{q^2}{(1-(1-q)x)^3}g\left(\frac{x}{1-(1-q)x}\right)\tag{2}$$
which becomes, expanding in the neighbourhood of the singular point $x=1$:
$$ \frac{\gamma \ln \left(\frac{1}{1-x}\right)}{1-x}+\frac{\alpha }{1-x}+\beta \ln \left(\frac{1}{1-x}\right)=\frac{\gamma \ln
\left(\frac{1}{1-x}\right)}{1-x}+\frac{1+\alpha +\gamma \ln (q)}{1-x}+\frac{(\beta -2 (1-q) \gamma ) \ln
\left(\frac{1}{1-x}\right)}{q}+O(1).$$
Comparing:
$$\gamma = -\frac{1}{\ln q},\qquad \alpha \text{ arbitrary}, \qquad \beta = -\frac{2}{\ln q}.$$
Via Taylor expansion then from $(1)$
$$g_n \approx -\frac{H_{n}}{\ln q} + \alpha - \frac{2}{n \ln q} + O(1/n^2).$$
However, since $$H_n = \ln n+\gamma + \frac{1}{2 n} + O(1/n^2)$$ and thus
$$H_{n+2} = \ln n+\gamma + \frac{5}{2 n} + O(1/n^2),$$
we can write
$$g_n \approx -\frac{H_{n+2}}{\ln q} + \alpha + O(1/n^2).$$
II. ALPHA FORMULA
To see what $\alpha$ is equal to, based on $(2)$, we write ansatz expansion of $g(x)$ in the following form
$$g(x) = \sum_{j=0}^\infty \frac{x^j}{(1-x)^{j+3}} \psi_j. \tag{3}$$
Inserting this anzatz into $(2)$, we get the following series expansion.
$$\sum_{j=0}^\infty \frac{x^j}{(1-x)^{j+3}} \psi_j = \frac{1}{1-x}+ \sum_{j=0}^\infty \frac{x^j}{(1-x)^{j+3}} q^{j+2} \psi_j. \tag{4}$$
To extract the coefficients $\psi_j$, we write $x=z/(1+z)$, so $(4)$ becomes
$$\sum_{j=0}^\infty z^j \psi_j = \frac{1}{(1+z)^2}+ \sum_{j=0}^\infty q^{j+2} z^j \psi_j. \tag{5}$$
Hence, comparing $j-$th coefficient in $z$ small in $(5)$, we get the exact value for $\psi_j$
$$\psi_j = \frac{(-1)^j (j+1)}{1-q^{j+2}}.$$
Inserting this into $(3)$, and expanding in $q$,
$$g(x) = \sum_{j=0}^\infty \frac{x^j}{(1-x)^{j+3}} \frac{(-1)^j (j+1)}{1-q^{j+2}} = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{x^j}{(1-x)^{j+3}} (-1)^j (j+1) q^{k(j+2)} = \sum_{k=0}^\infty \frac{q^{2 k}}{(1-x) \left(x q^k-x+1\right)^2}.\tag{6}$$
Again, by writing $x = z/(1+z)$, we get from $(6)$,
$$g\left(\frac{z}{1+z}\right) = \sum_{k=0}^\infty \frac{(z+1)^3 q^{2 k}}{\left(z q^k+1\right)^2}.$$
Note that since $0<q<1$, the series converges for any $z\geq 0$. On the other hand, inserting $x=z/(1+z)$ into the ansatz asymptotics $(1)$, for $z$ large (and thus near $x = 1$),
$$ g\left(\frac{z}{1+z}\right) = \alpha (z+1)+\log (z+1) (\beta +\gamma +\gamma z) = z \gamma \log (z)+\alpha z+(\beta +\gamma ) \log (z)+O(1).$$
By comparison at $z \to \infty$, we get the following relation for $\alpha$
$$\alpha = \lim_{z\to \infty} \left(\frac{\ln z}{\ln q} + \left(\frac{1+z}{z}\right)^3 \sum_{k=0}^\infty \frac{q^{2k}}{\left(q^ k + \frac{1}{z}\right)^2} \right),$$
which is, after writing $z=1/p$, the same limit as in $(0)$. Restating,
$$\alpha = \lim_{p\to 0}\left(-\frac{\ln p}{\ln
q}+\sum _{k=0}^{\infty } \frac{q^{2 k}}{\left(p+q^k\right)^2}\right).$$
III. EXACT ALPHA
Moreover, we are able to find the value $\alpha$ exactly. Via Euler-MacLaurin formula
$$\sum_{k=a}^b f(k) = \int_a^b f(x) \, \mathrm{d}x + \frac{f(a)+f(b)}{2} + \sum_{m=1}^\infty \frac{B_{2m}}{(2m)!}(f^{(2m-1)}(b)-f^{(2m-1)}(a)),$$
for
$$f(k) = \frac{q^{2k}}{\left(p+q^k\right)^2},$$
we have
$$\int_0^\infty f(x) \,\mathrm{d}x = \frac{\frac{1}{p+1}+\ln \left(\frac{p}{p+1}\right)}{\ln q}, \qquad f(0)=\frac{1}{(1+p)^2}, \qquad f(\infty) = 0,$$
so
$$ \sum _{k=0}^{\infty } \frac{q^{2 k}}{\left(p+q^k\right)^2} = \frac{\frac{1}{p+1}+\ln \left(\frac{p}{p+1}\right)}{\ln q} + \frac{1}{2(1+p)^2} + O(p)$$
from which we immediately get $$\alpha = \frac12 + \frac{1}{\ln q},$$
thus
$$g_n \approx -\frac{H_{n+2}}{\ln q} + \frac12 + \frac{1}{\ln q}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$ So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$
The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas WolframAlpha used L'Hopital's Rule in solving it.
My first instinct to solving the question when I read it was like this-
$$ \lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \frac{1}{x^2} - \frac{x}{x^3} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{x}{x^3} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{1}{x^2} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} $$
$$ \implies 1 \times 0 $$
$$ \implies 0 $$
I do not understand, where I am going wrong with this procedure. I would really appreciate it if someone can answer.
Thank you.
For reference here is the solution given in Cengage.
| Assuming the limit exists, let
$$l=\lim_{x \to 0}\frac{\sin x-x}{x^3} $$
$$ l=\lim_{x \to 0}\frac{\sin x-x}{x^3}\overset{x=2t}{=}\lim_{t \to 0}\frac{\sin 2t-2t}{8t^3}=\lim_{x \to 0}\frac{2\sin x \cos x-2x}{8x^3} $$
Hence
$$4l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3} $$
Now consider
$$3l=4l-l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3}-\lim_{x \to 0}\frac{\sin x-x}{x^3}=\lim_{x \to 0}\frac{\sin x \cos x-\sin x}{x^3}\\ 3l=\lim_{x \to 0}\frac{\ \sin x}{x} \lim_{x \to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2} $$
Finally
$$ \boxed{l=\lim_{x \to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded. Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded.
What I've tried: Since
$\frac{1}{2^4}+\frac{1}{3^4}\leq \frac{1}{2^4}+\frac{1}{2^4}=\frac{1}{2^3}$
$\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}\leq \frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}=\frac{1}{4^3}$
and so on. So the series in question is less than $1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}...$. If I can show the latter is bounded I'm done. But I can't so maybe there is and easier solution?
| Without resorting to known p-series, do the following: You've already done part of the work in comparing subsets of terms via a Cauchy-condensation-type move. The next step is to realize that $4 = 2^2, 8 = 2^3,$ and so on. You then have
$$ \sum_{j \geq 1} \frac{1}{(2^3)^{j-1}} = \sum_{j \geq 1} \left( \frac{1}{8} \right)^{j-1} = \frac{8}{7}$$
(you should check this by using the geometric series formula). Thus by comparison we have
$$ \sum_{j \geq 1} \frac{1}{j^4} \leq \frac{8}{7}.$$
Notice that this estimate is off by $-0.060533$ from the actual value of the sum of $\frac{\pi^4}{90}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\lim_\limits{x \to \infty} x\cdot \ln\left(1 + \frac{1}{x}\right) = 1$ using Taylor series We need to prove that $\lim_\limits{x \to \infty} \ln((1 + \frac{1}{x})^x) = \lim_\limits{x \to \infty} x \cdot \ln(1 + \frac{1}{x}) = 1$ and would like to use Taylor series.
To do this, when we expand $\ln(1 + \frac{1}{x})$, we should get a series like $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$, multiplying by $x$ would make it $1 + \frac{a_1}{x} + \frac{a_2}{x^2} + ...$ which is $1$ if $x \to \infty$.
The Taylor expansion of $f(x) = \ln(1 + \frac{1}{x})$ at some $a$ is $\sum_{n = 0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + ...$.
$f'(a) = -\frac{1}{a^2 + a}$, $f'(a) \cdot (x-a) = - \frac{x-a}{a^2 + a}$
$f''(a) = \frac{2a + 1}{(a^2 + a)^2}$, $f''(a) \cdot (x-a) = \frac{(2a + 1)(x-a)}{(a^2 + a)^2}$
So the Taylor series begins like this:
$\ln(1 + \frac{1}{x}) = \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ...$
Since the first term doesn't involve $x$ and we want it to be $\frac{1}{x}$, we would like it to be $0$ and the next term to be $\frac{1}{x}$. To make the first term $0$, $a$ needs to be approaching infinity.
$\lim_\limits{a \to \infty} \left( \ln(1 + \frac{1}{a}) - \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right) = \lim_\limits{a \to \infty} \left(- \frac{x-a}{a^2 + a} + \frac{(2a + 1)(x-a)}{(a^2 + a)^2} - ... \right)$.
Now, this isn't in the form $\frac{1}{x} + \frac{a_1}{x^2} + \frac{a_2}{x^3} + ...$ that we wanted.
Do we need to pick some other $a$?
I found https://math.stackexchange.com/a/1071689/1095885 which says
When x is very large, using Taylor
$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$
The series $\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{x^3} + ...$, which would be perfect, seems to be a Laurent series though.
Edit: When substituting $u = \frac{1}{x}$ and taking the Taylor series of $\ln(1 + u)$, why don't we have to use the chain rule?
| An elementary way to do
the discrete version of this
is to show that
if
$a_n=(1+1/n)^n$
and
$b_n=(1+1/n)^{n+1}$
then
$a_n$ is increasing and
$b_n$ is decreasing
and,
as $n \to \infty$,
$a_n$ and $b_n$
have a common limit,
which
I suggest
we call $e$.
An elementary proof of this,
which only uses the
arithmetic mean-geometric mean inequality,
can be found in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563.
| {
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How to simplify $\sin(n\frac{\pi}{2})$ as we do with $\cos(n\pi)$ When calculating Fourier series we almost always get some expression that look like:
\begin{equation} ...\left[\cos(nx) \right]_0^\pi.
\end{equation}
Then we use $\cos(n\pi) = (-1)^n$.
However, lately I have come across expressions like this:
\begin{equation} ...\left[\sin(nx) \right]_0^{\pi/2}.
\end{equation}
Which is equal to:
\begin{array}{|c|c|c|c|}
\hline
n& 1 & 2 & 3 & 4 \\ \hline
\sin(nx)&\sin(\frac{\pi}{2}) & \sin(\pi) & \sin(\frac{3\pi}{2})& \sin(2\pi)\\ \hline
val &1 &0 &-1 &0\\ \hline
\end{array}
How do we simplify this so that it looks like the $\cos(n\pi) = (-1)^n$.
| Note that $e^{\pm in\pi / 2} = \cos n\pi/2 \pm i\sin n\pi/2$, so
$$\sin n\pi/2 = \tfrac{1}{2i}(e^{in\pi / 2} - e^{-in\pi / 2})$$
$$= \tfrac{1}{2i}(i^n - i^{-n})$$
$$=\frac{i^{2n}-1}{2i^{n+1}}$$
$$= \frac{(-1)^n - 1}{2i^{n+1}}$$
$$= \boxed{\frac{1 - (-1)^n}{2i^{n-1}}}\tag{mult top/bottom by $i^{-2} = -1$}$$
This expression vanishes for even $n$, and for odd $n = 2k+1$ it becomes
$$\frac{1 - (-1)^{2k+1}}{2i^{2k+1-1}} = \frac{2}{2(-1)^k} = (-1)^k$$
as desired.
| {
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In how many ways to NOT take a red ball in an urn? There are 25 balls in five different colors in an urn. The balls are numbered 1 to 5 in each color. I.e. they are all unique.
In how many ways can I take 10 balls without returning any of them and I DON'T get any red ball.
Maybe I make this harder than it is. I've trying to solve it by inclusion-exclusoin and generating functions but I don't really get any reasonable answer.
By generating functions i tried to solve that I take at least one red ball and subtract that from $25 \choose 10$ that is the total ways of choosing 10 balls.
My first try was something like this where every x represent the different colors and $x_1$ is the red.
$x_1 + x_2 + x_3 + x_4 + x_5 = 10$
$1 \leq x_1 \leq 5, \\ 0 \leq x_i \leq 5 , \\ 2 \leq i \leq 5$
From this we get
$5 + 9 - 1 \choose 9$ = 715.
I didnt like that answer so I tried something else.
$x(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)^4=10$
And from that, by $\frac{1-x^{n+1}}{1-x}$ we get
$x \cdot (\frac{1-x^5}{1-x})(\frac{1-x^6}{1-x})^{^4} = 10 $
calculating the coefficient of 10 we get 505 and that doesn't either feel right.
By inclusion - exclusion:
${25 \choose 10} - {24 \choose 9} + {23 \choose 8} - + - \dots $ But here I dont take respect to the red balls in any way.
My thought is to perhaps divide the problem into different cases, like
I take one red ball
I take two red balls
$\vdots$
And make something from that but now I am pretty confused generally.
| If you do not select any red balls, then you must select $10$ of the remaining $25 - 5 = 20$ balls, which can be done in
$$\binom{20}{10}$$
ways.
As a sanity check, observe that the number of ways of selecting exactly $k$ of the five red balls and $10 - k$ of the remaining $20$ balls is
$$\binom{5}{k}\binom{20}{10 - k}$$
Hence, the total number of selections is
$$\binom{5}{0}\binom{20}{10} + \binom{5}{1}\binom{20}{9} + \binom{5}{2}\binom{20}{8} + \binom{5}{3}\binom{20}{7} + \binom{5}{4}\binom{20}{6} + \binom{5}{5}\binom{20}{5} = \binom{25}{5}$$
which can be checked by direct calculation or by applying Vandermonde's identity.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$
Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.
Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.
| Just another way, we can use a lesser used but quite useful generalisation of the Radon inequality (see here for e.g.). Reproducing for ease of reference, for positive reals $a_i, b_i, r,s$ s.t. $r\geqslant s+1$ it says:
$$\sum_{i=1}^n \frac{a_i^r}{b_i^s} \geqslant \frac{\left(\sum_i a_i \right)^r}{n^{r-s-1}\;\left( \sum_i b_i\right)^s}$$
It can now be easily noted that for our problem$$\sum \frac{a^3}{b+c+d}\geqslant \frac{(a+b+c+d)^3}{4\cdot3\cdot(a+b+c+d)}=\frac1{12}(a+b+c+d)^2$$
Now by AM-GM, $(a+b+c+d)^2\geqslant 4(a+c)(b+d)=4$ finishes this.
| {
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"timestamp": "2023-03-29T00:00:00",
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For the polynomial $P(x) = x^{2004}$, finding $Q(0)Q(1)$ Let's assume that $P(x)\in \mathbb{R}[x]$ is a polynomial such that $P(x) = x^{2004}$, and that $Q(x)$ is the quoitent of the division of $P$ by $x^2-1$. How could we find $Q(0)Q(1)$?
$$x^2\equiv 1\pmod{x^2-1}$$
$$P(x) = x^{2004}\equiv 1\pmod{x^2-1}$$
Then, $P(x) = (x^2-1)Q(x)+1$ and
$$\begin{align}\frac{P(x)-1}{x^2-1} = \frac{x^{2004}-1}{x^2-1} = \frac{\biggr(x^{1002}-1\biggr)\biggr(x^{1002}+1\biggr)}{x^{2}-1} &= \frac{\biggr(x^{501}-1\biggr)\biggr(x^{501}+1\biggr)\biggr(x^{1002}-1\biggr)}{x^{2}-1} \\ &= \cdots\end{align}$$
Which will get progressively worse.
| As you already showed, $Q(x)=\frac{x^{2004}-1}{x^2-1}.$ Therefore:
*
*$Q(0)=1$ and
*$Q(1)=\lim_{y\to1}\frac{y^{1002}-1}{y-1}=f'(1)$ where $f(y)=y^{1002},$ hence
$Q(1)=1002.$ "Alternatively", $Q(\sqrt{1+h})=\frac{(1+h)^{1002}-1}h=\sum_{k=1}^{1002}\binom{1002}kh^{k-1}$ hence $Q(1)=\binom{1002}1=1002.$
| {
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Inequality of norms $\| \frac{a}{d}-b\|$ < $\| \frac{a}{d}-c\| \Leftrightarrow \| a-bd\|$ < $\| a-cd\|$ How would one show or disproof the following statement
$\forall a,b,c,d \in \mathbb{C}\setminus \{0\} :$
$\left\| \frac{a}{d}-b\right\|_p$ < $\left\| \frac{a}{d}-c \right\|_p \Leftrightarrow \left\| a-bd\right\|_p$ < $\left\| a-cd \right\|_p$
| p-norm is defined as
$$
||{\mathbf x}||_p \equiv \left( {\sum_{n \mathop = 0}^d {|x_n|}^p} \right)^{1/p}
$$
For $x_n \in \mathbb{C}$, $|x_n|$ is usually defined by complex module
$$
|x_n| = |A + Bi| \equiv \sqrt{A^2 + B^2}
$$
Because $a,b,c,d \in \mathbb{C}$ in the question, I assume all the p-norm values in the inequation are the p-norms of the one-dimension vector which are scalers. First, we have to prove $||ad||_p = ||a||_p ||d||_p$.
For $A, B, C, D \in \mathbb{R}$
$$\begin{align*}
||(A + Bi)(C + Di)||_p
&= ||(AC - BD) + (AD + BC)i||_p \\
&= \left( |(AC - BD) + (AD + BC)i|^p \right)^{1/p} \\
&= |(AC - BD) + (AD + BC)i|
&& (|(AC - BD) + (AD + BC)i| \in \mathbb{R}) \\
&= \sqrt{(AC - BD)^2 + (AD + BC)^2} \\
&= \sqrt{A^2 C^2 + B^2 D^2 - 2ABCD + A^2 D^2 + B^2 C^2 + 2ABCD} \\
&= \sqrt{A^2 C^2 + B^2 D^2 + A^2 D^2 + B^2 C^2} \\
&= \sqrt{(A^2 + B^2)(C^2 + D^2)} \\
&= \sqrt{(A^2 + B^2)} \sqrt{(C^2 + D^2)} \\
&= |A + Bi| |C + Di| \\
&= ||A + Bi||_p ||C + Di||_p
\end{align*}$$
Using the identity above, we can prove the inequality.
$$\begin{align*}
\left\| a-bd\right\|_p \lt \left\| a-cd\right\|_p
&\Rightarrow \left\| \frac{a}{d}-b\right\|_p \left\|d\right\|_p \lt \left\| \frac{a}{d}-c\right\|_p \left\|d\right\|_p \\
&\Rightarrow \left\| \frac{a}{d}-b\right\|_p \lt \left\| \frac{a}{d}-c\right\|_p
&& \left( \left\|d\right\|_p \gt 0 \right)
\end{align*}$$
You can use the same method to prove the inversed part.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving y(t) from a differential equation Given the initial problem
$$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0, y(1) =3 $$
I got the solution to be
$$\frac{t^2}{4y^4}-\frac{3}{2y^2} = -\frac{53}{324} $$
But how do I write it in terms of $y(t)$
| $$\frac{3y^2-t^2}{y^5} \frac{dy}{dt} + \frac{t}{2y^4}=0$$
$$\frac{dt}{dy}=-2\frac{3y^2-t^2}{ty} $$
$$\frac{2tdt}{2ydy}=-2\frac{3y^2-t^2}{y^2} $$
$T=t^2\quad;\quad Y=y^2$
$$\frac{dT}{dY}=2\frac{T}{Y} -6$$
First order linear ODE easy to solve :
$$T=6Y+c\:Y^2$$
Condition : $y(1)=3\quad\implies\quad Y(1^2)=3^2$
$1^2=6*(3^2)+c\:3^4\quad\implies\quad c=-\frac{53}{81}$
$$T=6Y-\frac{53}{81}Y^2$$
Solve it for $Y$ . Then replace $Y$ by $y^2$ and replace $T$ by $t^2$.
| {
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Closed Form of the Chebyshev Polynomials of the First Kind [Proof Request] $$T_n(x) = \frac{n}{2} \sum_{r=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-1)^r}{n-r} \binom{n-r}{r} (2x)^{n-2r}$$
Searching on the web yielded no results, and the result is given without proof on OEIS and Wolfram MathWorld
| It is defined by $T_n(x)=\frac{u^{n}+u^{-n}}{2}$ where $x=\frac{u+u^{-1}}{2}.$ This implies that $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$. So, it is essentially enough to check this recursive relation at $n+1-2r$-the power term:
$$\frac{n+1}{2}\frac{(-1)^r}{n+1-r}{n+1-r\choose r}2^{n+1-2r}=2\frac{n}{2}\frac{(-1)^{r}}{n-r}{n-r\choose r}2^{n-2r}-\frac{n-1}{2}\frac{(-1)^{r-1}}{n-r}{n-r\choose r-1}2^{n+1-2r}$$
which is equivalent to
$$\frac{n+1}{n+1-r}{n+1-r\choose r}=\frac{n}{n-r}{n-r\choose r}+\frac{n-1}{n-r}{n-r\choose r-1}$$
and it is true.
| {
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Prove that $(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\ge \frac{3}{4}$ I would appreciate if somebody could help me with the following problem:
Q: Prove that where $a,b,c \in [0,1]$, $a+b+c=1$
$$(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\ge
\frac{3}{4}$$
My work: The first method tried was to use the arithmetic-geometric mean because of the positive number condition. However, the expression of the problem becomes more complicated, but the solution is not visible.
| The function $f(x) = \frac{1}{x(x+1)}$ is convex on $x > 0$ (note: $f''(x) = \frac{6x^2 + 6x + 2}{x^3(x+1)^3} > 0$).
Thus, by Jensen's inequality, we have
\begin{align*}
\frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}
&\ge f(ab + bc + ca)\\
&= \frac{1}{(ab+bc+ca)(ab+bc+ca+1)}.
\end{align*}
Thus, we have
\begin{align*}
&(ab+bc+ca)\left( \frac{a}{b(b+1)}+\frac{b}{c(c+1)}+\frac{c}{a(a+1)}\right)\\
\ge{}& (ab+bc+ca)\cdot \frac{1}{(ab+bc+ca)(ab+bc+ca+1)}\\
={}& \frac{1}{ab+bc+ca+1}\\
\ge{}& \frac{1}{1/3 + 1}\\
={}& \frac34
\end{align*}
where we have used $ab + bc + ca \le (a + b + c)^2/3 = 1/3$.
We are done.
| {
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} |
Show that ${\sum\limits_{k=1}^n\frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^n\frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$ Let $X_1,X_2,\dots$ be independent random variables with $\lim\limits_{n\to\infty}\mathbb E[X_n]=c\in\mathbb R$ and $\sum\limits_{n=1}^\infty \frac1{\sigma_n^2}=\infty$.
I need to show that as $n\to\infty$
$${\sum\limits_{k=1}^n\frac{X_k}{\sigma_k^2}}\big/{\sum\limits_{k=1}^n\frac{1}{\sigma_k^2}}\overset{a.s.}\longrightarrow c.$$
I know that $\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \frac{\mathbb V[X_k]}{\sigma_k^4} = \frac{1}{\sigma_k^2}$ from which follows with independence and the condition above that $\sum\limits_{k=1}^\infty\mathbb V\left[\frac{X_k}{\sigma_k^2}\right] = \infty$. How can I continue?
| By Chebyshev Inequality,
$$P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \frac{Var\left( \frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} \right) }{\epsilon^2}\leq \frac{\left( \frac{\sum_{k=1}^n \frac{1}{\sigma_k^2}}{\left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)^2} \right) }{\epsilon^2} \leq \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)}$$
$$\lim_{n \rightarrow \infty} P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \lim_{n \rightarrow \infty} \frac{1}{\epsilon^2 \left(\sum_{k=1}^n\frac{1}{\sigma_k^2}\right)} = 0$$
If $$\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}}$$ exists then,
For all, $ n \geq N_{\delta,\epsilon,m}$ with $\lim_{m \rightarrow \infty} N_{\delta,\epsilon,m} = \infty$ such that
$$ P \left(\left|\frac{\sum_{k=1}^n \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^n\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right) \leq \frac{\delta}{2^{m}} \implies$$
$$ P \left( \cup_{m=1}^{\infty} \left \{ \left|\frac{\sum_{k=1}^{N_{\delta,\epsilon,m}} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{N_{\delta,\epsilon,m}}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) \leq \delta \implies$$
$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) \leq \delta \implies$$
Since $\delta$ is arbitrary,
$$ P \left( \left \{ \left|\frac{\sum_{k=1}^{\infty} \frac{X_k}{\sigma_k^2}}{\sum_{k=1}^{\infty}\frac{1}{\sigma_k^2}} -c\right|^2 > \epsilon^2 \right \} \right) = 0$$
| {
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Proof by contraposition. Let $x$ be an integer. If $8$ does not divide $x^2-1$, then $x$ is even. Prove by contraposition: Let $x$ be an integer. If $8$ does not divide $x^2-1$, then $x$ is even.
Assume $x$ is odd.
Prove $8|x^2-1$
So, $x=2n+1$ for some integer $n$.
Then $x^2-1=4(n^2+n)$ for some $n^2+n$ integer by closure of Z.
By algebra, multiply through by $2$ to get $2(x^2-1)=8(n^2+n)$.
Then, $x^2-1=8(\frac{n^2+n}{2})$
I get stuck here, because I cannot say that the fraction is an integer because $\mathbb{Z}$ isn't closed for division. Help!
| How many ways are there to choose two people from $n+1$ people? It is $${n+1 \choose 2} = \frac{n(n+1)}{2} = \frac{n^2 +n}{2} $$
This is the quantity you are interested in. But the number of ways to choose two people from $n+1$ people is an integer. Can you use this to conclude that your expression is an integer?
| {
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Solving a triangle by angle, circumradius, and area The problem itself is as follows.
Consider a triangle $ABC$. The cosine of one of its interior angles is $m$, its circumradius is equal to $R$, and its area is equal to $S$. Solve this triangle: find its sides and angles.
By applying the sine theorem, I've got that the length of a side, opposite to the angle with the known cosine, which is $2R\sqrt{1-m^2}$. But I had no luck in advancing further, as it either results in too many unknowns. Most I could do is getting the product of sines of two unknown angles, from the formula $S=2R^2\sin\angle A\sin\angle B\sin\angle C$.
I'm looking for proofs that such triangle exists and is clearly defined, as well as the way to solve it if it is.
| Below is my own answer on the problem I came up with. There may exist more elegant ways, but this is the most straightforward way of which I know.
Consider a triangle $ABC$ with circumradius $R$, area $S$, and $\cos\angle C=m$.
*
*Find the sine of $\angle C$ by using the famous trigonometric identity $\sin^2\alpha+\cos^2\alpha=1$. As all interior angles of the triangle have positive sines, we take the positive value, thus $\sin\angle C=\sqrt{1-\cos^2\angle C}=\sqrt{1-m^2}$.
*Find the length of side $AB$, opposite to $\angle C$, by using the law of sines. Thus, $AB=2R\sin\angle C=2R\sqrt{1-m^2}$.
*Find the product of two other sides. For that, find the altitude of triangle $ABC$ to side $AB$ by using the formula $h_{AB}=\dfrac{2S}{AB}$. Now, if we equate two formulas for the area of a triangle, we get $\dfrac{AB\cdot BC\cdot AC}{4R}=\dfrac{1}{2}AB\cdot h_{AB}$, which we can transform further:
$$\begin{align*}
\dfrac{AB\cdot BC\cdot AC}{4R} = \dfrac{1}{2}AB\cdot h_{AB}\ \ &\Leftrightarrow\ \ \dfrac{BC\cdot AC}{2R} = h_{AB} \\
&\Leftrightarrow\ \ \dfrac{BC\cdot AC}{2R} = \dfrac{2S}{AB} \\
&\Leftrightarrow\ \ BC\cdot AC = \dfrac{4RS}{AB}
\end{align*}$$
Thus, $BC = \dfrac{4RS}{AB\cdot AC}$.
*By using the law of cosines, we get $AB^2=BC^2+AC^2-2\cdot BC\cdot AC\cos\angle C$. Here, we can substitute $BC\cdot AC = \dfrac{4RS}{AB}$, and $BC = \dfrac{4RS}{AB\cdot AC}$. We get:
$$\begin{align*}
AB^2=\left(\dfrac{4RS}{AB\cdot AC}\right)^2+AC^2-2\cdot \dfrac{4mRS}{AB}
\end{align*}$$
Here, we know everything except $AC$, and we can find both unknown sides by simplifying the last formula into the biquadratic equation, positive roots of which are sides $BC$ and $AC$. Finally, to find the other angles, one can use the law of cosines if the given angle is obtuse ($m < 0$), as there is only one obtuse angle in the triangle, or the law of cosines in all other cases.
| {
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Evaluate $\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$ (from MIT Integration Bee 2023) This is from the final round of MIT Integration Bee 2023.
$$\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$$
The given answer is $7\pi$.
I found a way to do it using contour integrals (see answer below), but this is not a calculation I can finish within 4 minutes (the time limit in the competition). I am still looking for other elegant methods, possibly without using contour integrals.
| Let $f(x)$ denote the egregious fraction of sines inside the square. Writing $w = e^{2ix}$, we have
\begin{align*}
f(x)
&:= \frac{(\sin 2x)(\sin 3x)(\sin 5x)(\sin 30x)}{(\sin x)(\sin 6x)(\sin 10x)(\sin 15x)} \\
&= w^{-4} \underbrace{ \frac{(w^2 - 1)(w^3 - 1)(w^5 - 1)(w^{30} - 1)}{(w - 1)(w^6 - 1)(w^{10} - 1)(w^{15} - 1)} }_{=:g(w)}.
\end{align*}
Writing the fraction part in the last line as $g(w)$, algebraic manipulation using the finite geometric series formula yields
\begin{align*}
g(w)
= \frac{(w + 1)(w^{15} + 1)}{(w^3 + 1)(w^5 + 1)}
= \frac{w^{10} - w^5 + 1}{w^2 - w + 1}
= w^8 + w^7 - w^5 - w^4 - w^3 + w + 1,
\end{align*}
where we utilized the long division in the last step.1) Now, we write
$$ g(w) = \sum_{k\geq 0} a_k w^k $$
for simplicity. Then by noting that $f(x)$ is real-valued,
\begin{align*}
\int_{0}^{\pi} f(x)^2 \, \mathrm{d}x
&= \int_{0}^{\pi} f(x)\overline{f(x)} \, \mathrm{d}x \\
&= \int_{0}^{\pi} \left( e^{-8ix} g(e^{2ix}) \right) \overline{\left( e^{-8ix} g(e^{2ix}) \right)} \, \mathrm{d}x \\
&= \sum_{j,k} a_j \overline{a_k} \int_{0}^{\pi} e^{2ix(j-k)} \, \mathrm{d}x \\
&= \pi \sum_{k \geq 0} |a_k|^2 \\
&= 7\pi.
\end{align*}
1) Alternatively, if OP is familiar with some abstract algebra, we can proceed as below:
\begin{align*}
g(w)
= \prod_{d \mid 30} (w^d - 1)^{\mu(30/d)}
= \Phi_{30}(w)
= w^8 + w^7 - w^5 - w^4 - w^3 + w + 1,
\end{align*}
where $\mu$ is the Möbius function and $\Phi_n$ is the cyclotomic polynomial.
| {
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Reference for, and/or proof of, $\prod_{n=1}^\infty(\frac{4n+1}{4n-1})^{4n}(\frac{2n^2-2n+1}{2n^2+2n+1})^n=\sqrt2\cosh(\pi/2)e^{-2G/\pi}$ Context:
I have derived some infinite products that I think are not well known. This is the easiest of them:
$$\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=\sqrt{2}\cosh(\pi/2)\,e^{-\frac{2G}{\pi}},\tag{1}$$
where $G$ is Catalan's constant. After some searching I can't find any reference to it.
Take a look at WolframAlpha.
Question 1:
Do you know any reference to this or something like that?
Question 2:
Could you find a solution to $(1)$?
| My goal being to obtain the partial product and then use asymptotics
$$P_1=\prod_{n=1}^p (4n+1)^{4n}=\frac{e^{\frac{1}{24}-\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma
\left(\frac{1}{4}\right)}{\sqrt{A}\,\, \Gamma \left(p+\frac{5}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right) \right)$$
$$P_2=\prod_{n=1}^p (4n-1)^{4n}=\frac{e^{\frac{1}{24}+\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma
\left(p+\frac{3}{4}\right)}{\sqrt{A}\,\, \Gamma \left(\frac{3}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{3}{4}\right) \right)$$
$$\color{blue}{\frac{P_1}{P_2}=\sqrt{\pi }\, e^{-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,\times} $$ $$\color{blue}{\exp\left(4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta
^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)\right)}$$
For large values of $p$
$$4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta
^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)=$$ $$2 p \log (p)+(\log (p)+1)+\frac{3}{16 p}+O\left(\frac{1}{p^2}\right)$$
$$\color{blue}{\frac{P_1}{P_2}\sim\sqrt{\pi }\, e^{1-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,p^{2p+1}}$$
This was the easy part.
Now, writing
$$\frac{2 n^2-2 n+1}{2 n^2+2 n+1}=\frac {(n-a_1)(n-a_2) } {(n+a_1)(n+a_2) }$$ where
$$a_1=\frac {1-i}2 \qquad \text{and}\qquad a_2=\frac {1+i}2$$
$$Q_a=\prod_{n=1}^p (n+a)^n=e^{A_a}$$
$$A_a=a \left(\zeta
^{(1,0)}(0,a+1)-\zeta ^{(1,0)}(0,a+p+1)\right)+$$
$$\left(\zeta ^{(1,0)}(-1,a+p+1)-\zeta ^{(1,0)}(-1,a+1)\right)$$ and using the tedious expansion of $A_a$ for large $p$ the expansion of the second product is
$$\color{blue}{\prod_{n=1}^p \left(\frac{2 n^2-2 n+1}{2 n^2+2 n+1}\right)^n = e^{-(2 p+1)} \sqrt{2 (1+\cosh (\pi ))}+O\left(\frac{1}{p^2}\right)}$$
All the above make that, at the limit, the infinite product is
$$\color{red}{\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=e^{-\frac{2 C}{\pi }} \sqrt{1+\cosh (\pi )}}$$ which is your formula.
In terms of asymptotics, it would be
$$\sqrt{2}\, e^{-\frac{2 C}{\pi }} \cosh \left(\frac{\pi }{2}\right) \exp\left( -\frac{3}{8 p}+\frac{18}{97 p^2}-\frac{13}{188 p^3}+O\left(\frac{1}{p^4}\right)\right)$$
For $p=9$, the relative error is less than $0.01$% and for $p=18$ it reduces to $0.001$% .
| {
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"answer_count": 2,
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What is the probability that Carlos's purchase of 5 CDs includes at least one rap, country, and heavy metal CD out of a total of 12? Full Problem
Carlos has chosen $12$ different CDs he would like to buy: $4$ are rap music, $5$ are country music, and $3$ are heavy metal music. (Carlos has very eclectic tastes in music!) Unfortunately, he has only enough money to afford to buy $5$ of them (they all cost the same price). So he selects $5$ of them at random. What is the probability that his purchase includes at least one CD from each of the three categories?
My Response
First, there are a total of $\dbinom{12}5$ total ways for Carlos to choose, without order, $5$ CDs from $12$ CDs. Then, there are a total of $\dbinom41 \dbinom51 \dbinom31 \dbinom92$ ways for Carlos to choose at least one CD from each category. This simplifies to $\dfrac{30}{11}$, which is obviously not correct. What went wrong in my process?
| From another view point:
Because we calculate a probability, each item should be seen as distinct, even if they are in same category. So, let's write their generating functions forms thinking this fact !
*
*Generating function of selecting at least one rap music : $$\binom{4}{1}x^1+\binom{4}{2}x^2+\binom{4}{3}x^3+\binom{4}{4}x^4=(1+x)^4 -1$$
*Generating function of selecting at least one country music : $$\binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5=(1+x)^5 -1$$
*Generating function of selecting at least one heavy metal music : $$\binom{3}{1}x^1+\binom{3}{2}x^2+\binom{3}{3}x^3=(1+x)^3 -1$$
Now , find the coefficient of $x^5$ in the expansion of $$[(1+x)^4-1][(1+x)^5-1][(1+x)^3-1]$$
Calculation:
$$\frac{[x^5]([(1+x)^4-1][(1+x)^5-1][(1+x)^3-1])}{\binom{12}{5}}=\frac{590}{792}=0,7449...$$
| {
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Maximize $(1-a)(1-c)+(1-b)(1-d)$ over $a^2+b^2=c^2+d^2=1$. Let $a,b,c,d$ be real numbers such that $a^2+b^2=c^2+d^2=1$. Find the maximum value of $(1-a)(1-c)+(1-b)(1-d)$.
I tried substituting $a=\sin x, b =\cos x, c = \sin y, d=\cos y$, then expanded $(1-a)(1-c)+(1-b)(1-d)$. However this just leads to an ugly expression, and I'm not sure how to proceed
| By the Cauchy-Schwarz inequality, we have:
$$1 \times 1=(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \implies 1 \geq ac+bd;$$
and,
$$(a^2+b^2)(1^2+1^2)\geq (a+b)^2 \implies \sqrt 2 \geq -a-b, $$
similarly, $\sqrt 2 \geq -c-d$.
Therefore,
$$(1-a)(1-c)+(1-b)(1-d) \\ =2+(-a-b)+(-c-d)+(ac+bd) \\ \leq 2+2\sqrt 2+1=3+2\sqrt 2.$$
The equality case happens at $a=b=c=d=-\frac{\sqrt 2}{2}.$
| {
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Solve the system of equations: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$ Given system of equations:
$$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$$
$$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$$
$$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$$
I'll use matrices to solve these. We can write the system as:
$$AX=B$$
$$A^{-1}AX=A^{-1}B$$
$$X=A^{-1}B$$ where $X$ is our variable matrix. For simplicity, let $\frac{1}{x}=\alpha, \frac{1}{y}=\beta,\frac{1}{z}=\gamma $
Thus we have to solve:
$$2\alpha+3\beta+10\gamma=4$$
$$4\alpha-6\beta+5\gamma=1$$
$$6\alpha+9\beta-20\gamma=2$$
$A$ is simply the matrix formed using the coefficients of the variables:
$$A=\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}$$
$X$ is the variable matrix
$$X=\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}$$
$B$ is the matrix formed by right hand side values of the respective equations:
$$B=\begin{bmatrix}4\\1\\2\end{bmatrix}$$
Thus, our system of equations becomes:
$$\begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}\begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\begin{bmatrix}4\\1\\2\end{bmatrix}$$
Now, we just need to find $A^{-1}.$ Finding cofactors for A
Cofactor $ A=\begin{bmatrix}75&110&72\\150&-100&0\\75&30&-24\end{bmatrix}$
$A^{-1}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}$
$\Rightarrow \begin{bmatrix}\alpha\\\beta\\\gamma\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix}$
$\alpha=\frac{1}{2},\beta=\frac{1}{3},\gamma=\frac{1}{5}$
$\Rightarrow x=2,y=3,z=5$
My problem- This method seems unnecessarily lengthy. What better ways are there to solve such systems of equations?
| Following the comments, subtracting thrice the first equation from the third gives $-50\gamma=-10$ or $\gamma=\frac15$. Subtracting twice the first equation from the second and substituting (and negating) gives
$12\beta=4$ or $\beta=\frac13$. Substituting the known values into the first equation finally gives $\alpha=\frac12$, so $(x,y,z)=(2,3,5)$.
| {
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prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$ Prove $\forall x \in (0, \frac{\pi}{2})$: $x^2 \cos^2 (x) < 2$
It is clear, that:
*
*$x^2 \cos^2 (x) \to 0, x \to \frac{\pi}{2}$, since cosine goes to 0.
*$x^2 \cos^2 (x) \to 0, x \to 0$, since $x$ goes to 0.
Of course, we can split $(0, \frac{\pi}{2})$ into $A_1 = (0, \frac{\pi}{4}]$, $ A_2 = [\frac{\pi}{4}, \frac{\pi}{3}]$ and $A_3 = [\frac{\pi}{3}, \frac{\pi}{2})$
(sorry for intersection, just don't want to write limits. It doesn't affect anything)
Then we take max and min values from $A_i$ and evaluate max value at $x^2$ and min value at $\cos^2 x$:
$A_1$: $\max(A_1)^2 = \frac{\pi^2}{16} \in (0, 1); \cos^2 \min(A_1) = 1 \implies \max(A_1)^2 \cos^2 \min(A_1) < 2$
$A_2$: $\max(A_2)^2 = \frac{\pi^2}{9} \in (1, 2) ; \cos^2 \min(A_2) = \frac{1}{2} \implies
\max(A_2)^2 \cos^2 \min(A_2) < 2$
$A_3$: $\max(A_3)^2 = \frac{\pi^2}{4} \in (2, 3) ; \cos^2 \min(A_3) = \frac{1}{4} \implies
\max(A_3)^2 \cos^2 \min(A_3) < 2$
Actually, here we can see that upper bound is even lower (not bigger that 1).
Unfortunately, I am not satisfied with this writing (somehow I am sure there are much better proofs).
But what are they? Thank you.
| This is based on the inequality $x< \tan x$ for $x\in\left(0, \frac \pi2\right)$.
(Geometrically, this can be obtained by comparing the areas of 1) a sector with unit radius and angle $x$, and 2) a right-angled triangle with adjacent side $1$ and opposite side $\tan x$.)
$$\begin{align*}
x &< \tan x\\
x^2 \cos^2 x &< \tan^2x\cos ^2x\\
&= \sin^2 x < 1 < 2
\end{align*}$$
| {
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Finding two angles between vectors (analytic-geometry, vectors)
The vector $\vec a$ makes an angle of $π/3$ with the $x$−axis, an angle of $π/4$ with the $y$−axis, and some sharp angle with the $z$−axis. The vector $\vec b$ forms an angle of $π/6$ with the $x$−axis. What are the angles that vector $\vec b$ makes with $y$ and $z$−axis if it is known that vectors $\vec a$ and $\vec b$ are perpendicular?
Here is what I know:
$\cos^2(\alpha) + \cos^2(\beta)+\cos^2(\gamma)=1 \\ \rightarrow\cos^2(π/3)+ \cos^2(π/4) +\cos^2(\gamma)=1 \\ \implies\cos^2(\gamma)=\pm1/2$
Vectors $\vec a$ and $\vec b$ are perpendicular that means angle between $\vec a$ and $\vec b$ is $π/2$ and $\vec a\cdot\vec b=0$. I don't know what to do next and how to find angles between vector $b$ and $y$ and $z$ axis. I would be thankful for any help.
| $\def\uv#1{\hat{\mathbf#1}}$
WLOG, consider the normalised vectors of $\vec a$ and $\vec b$, respectively $\uv a$ and $\uv b$.
Let $\uv i$, $\uv j$ and $\uv k$ be the standard unit vectors along the $x$-, $y$- and $z$-axes respectively. Then decompose $\uv a$ into its scalar components:
$$\begin{align*}
\uv a &= a_x \uv i + a_y \uv j + a_z \uv z\\
&= \cos\frac\pi3\cdot \uv i + \cos \frac\pi4\cdot \uv j + a_z \uv k\\
&= \frac12 \uv i + \frac1{\sqrt 2} \uv j + a_z \uv k
\end{align*}$$
Like what OP already did in the question, $\uv a$ has length $1$:
$$\begin{align*}
\left(\frac12\right)^2 + \left(\frac1{\sqrt2}\right)^2 + a_z^2 &= 1\\
a_z^2 &= 1-\frac14 - \frac12\\
&= \frac14\\
a_z &= \frac12
\end{align*}$$
Assuming the "sharp" angle that $\vec a$ makes with the $z$-axis means an "acute" angle, $a_z$ is taken to be positive.
Similarly, decompose $\uv b$ into its scalar components:
$$\begin{align*}
\uv b &= \cos\frac\pi6 \cdot \uv i + b_y \uv j + b_z \uv k\\
&= \frac{\sqrt3}2 \uv i + b_y \uv j + b_z \uv k
\end{align*}$$
$\vec a$ and $\vec b$ are perpendicular, so $\uv a \cdot \uv b = 0$:
$$\begin{align*}
\frac12 \cdot \frac{\sqrt3}2 + \frac1{\sqrt2} b_y + \frac12 b_z&= 0\\
b_z &= -\frac{\sqrt3}2 - \sqrt2 b_y
\end{align*}$$
$\uv b$ has length $1$:
$$\begin{align*}
\left(\frac{\sqrt3}2\right)^2 + b_y ^2 + b_z^2 &= 1\\
\left(\frac{\sqrt3}2\right)^2 + b_y ^2 + \left(-\frac{\sqrt3}2 - \sqrt2 b_y\right)^2 &= 1\\
\frac{3}4 + b_y ^2 + \frac{3}4 + \sqrt 6 b_y +2 b_y^2 &= 1\\
3b_y^2 + \sqrt6 b_y +\frac12 &= 0\\
b_y &= \frac{-\sqrt6 \pm\sqrt{6-4\cdot 3/2}}{2\cdot 3}\\
&= -\frac{1}{\sqrt 6}\\
b_z &= -\frac{\sqrt3}{2} + \sqrt2 \cdot \frac{1}{\sqrt6}\\
&= -\frac{1}{2\sqrt3}\\
\uv b &= \frac{\sqrt3}2 \uv i - \frac1{\sqrt6} \uv j - \frac{1}{2\sqrt3}\uv k
\end{align*}$$
So $\vec b$ makes the angle $\arccos b_y = \arccos\left(- \frac1{\sqrt6}\right)$ with the $y$-axis, and makes the angle $\arccos b_z = \arccos\left(\frac{1}{2\sqrt3}\right)$ with the $z$-axis.
| {
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"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Maximizing volume of a cylinder with $2$ cones given surface area $A$ A cylinder with a cone on either side of it. The cones have the same radius as the cylinder.
Cylinder area without the sides:
$$A_{cyl}=2\pi r h_1$$
Cone area without the base:
$$A_{cone}=\pi r \sqrt{h_2^2+r^2}$$
Total area of body:
$$A_{tot}=2\pi r h_1 + 2\pi r \sqrt{h_2^2+r^2}$$
Volume:
$$V_{tot}=\pi r^2 h_1 +\frac{2\pi r^2 h_2}{3}=\pi r^2 (h_1 + \frac{2}{3}h_2)$$
I have managed to calculate the maximum volume given surface area for the cone and the cylinder separately but together there are three variables. How would I go about calculating $r,h_1,h_2$ to maximize the volume?
| Since $\frac{A_{tot}}{2\pi}\geq \max\{rh_1+rh_2,r^2\}$, it follows that $r$ is bounded and the volume has to be finite:
$$V= \pi r\left(rh_1+\frac{2}{3}rh_2\right)\leq \pi r \frac{A_{tot}}{2\pi}\leq \frac{A_{tot}^{3/2}}{2\sqrt{2\pi}}.$$
Moreover the above inequality implies that $V\to 0$ as $r\to 0$. So we may assume that $r\geq r_0$ where $r_0>0$ is sufficiently small. Then
also the positive variables $h_1$ and $h_2$ have to be bounded.
Now we apply the method of Lagrange multipliers to the function
$$V(r,h_1,h_2)=\pi r^2 h_1 +\frac{2\pi r^2 h_2}{3}=\pi r^2 \left(h_1 + \frac{2}{3}h_2\right)$$
subject to the constraint
$$A(r,h_1,h_2)=2\pi r \left(h_1 + \sqrt{h_2^2+r^2}\right)=A_{tot}.$$
Assuming $r\geq r_0$, $h_1>0$, $h_2>0$ we find
just one constrained stationary point (which has to be the maximum):
$$\begin{cases}
\frac{\partial V}{\partial r}=\lambda\frac{\partial A}{\partial r}\\
\frac{\partial V}{\partial h_1} =\lambda\frac{\partial A}{\partial h_1}\\
\frac{\partial V}{\partial h_2}=\lambda\frac{\partial A}{\partial h_2}
\end{cases}
\implies
\begin{cases}
r \left(h_1 + \frac{2}{3}h_2\right)=\lambda \Big(h_1 + \sqrt{h_2^2+r^2}+\frac{r^2}{\sqrt{h_2^2+r^2}}\Big)\\
r^2 =\lambda 2r\\
\frac{r^2}{3} =\lambda \frac{r h_2}{\sqrt{h_2^2+r^2}}
\end{cases}
\implies
\begin{cases}
\lambda=\frac{r}{2}\\h_1=h_2=\frac{2r}{\sqrt{5}}
\end{cases}$$
where $r$ is given by
$$A_{tot}=2\pi r \left(\frac{2r}{\sqrt{5}} + \sqrt{\frac{4r^2}{5}+r^2}\right)\implies r=\sqrt{\frac{A_{tot}}{2\pi\sqrt{5}}}.$$
Hence, given the surface area $A_{tot}$, the maximum volume is
$$V=\pi r^2 \frac{2r}{\sqrt{5}}\left(1 + \frac{2}{3}\right)=
\frac{2\pi\sqrt{5} }{3}r^3=
\frac{A_{tot}^{3/2}}{3\sqrt{2\pi}\,5^{1/4}}.$$
Note that the maximal shape has an inscribed sphere of radius $r$:
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
"answer_id": 0
} |
Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $ Find the sum to n terms of the series $\frac{1} {1\cdot2\cdot3\cdot4} + \frac{1} {2\cdot3\cdot4\cdot5} + \frac{1} {3\cdot4\cdot5\cdot6}\ldots $
Please suggest an approach for this task.
| HINT $\rm\displaystyle\ \frac{1}{(k+1)(k+2)(k+3)(k+4)} = \frac{1}{6(k+1)} - \frac{1}{2(k+2)}+\frac{1}{2(k+3)}-\frac{1}{6(k+4)}$
$\rm\ f(k+1)-f(k)\: = $ above $\rm\displaystyle\ \Rightarrow\ f(k) \:=\: c_0 + \frac{c_1}{k+1}\ \:+\:\ \frac{c_2}{k+2}\ \:+\:\ \frac{c_3}{k+3}$
Calculating yields $\rm\ c_0,c_1,c_2,c_3 \ =\ 1/18,\ -1/6,\ 1/3,\ -1/6$.
For remarks on the group theory behind rational indefinite summation see my post here
| {
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For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$
| $\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &\in (4n+1, 4n+2).\end{align*}$
(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n < GM(n,n+1) < AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 < AM(n,n+1)-GM(n,n+1) < 1/2$, and the third line follows.)
But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.
| {
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"answer_id": 3
} |
Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$ is to multiply by $\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$.
I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \,\mathrm d \theta$. Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$?
| Here is another variation on a theme. It relies on the following two double angle formulae for sine and cosine, namely
$$\sin 2\theta = 2 \sin \theta \cos \theta \qquad \text{and} \qquad \cos 2 \theta = \cos^2 \theta - \sin^2 \theta,$$
two obvious substitutions, and a simple partial fraction decomposition. It is similar to making the famous $t$-substitution of $t = \tan \frac{x}{2}$ without having to rely on knowing this.
If the substitution $x = 2u$ is made, we have
\begin{align}
\int \sec x \, dx &= 2 \int \sec 2u \, du\\
&= 2 \int \frac{du}{\cos 2u}\\
&= 2 \int \frac{du}{\cos^2 u - \sin^2 u}\\
&= 2 \int \frac{du}{\cos^2 u(1 - \tan^2 u)}\\
&= 2 \int \frac{\sec^2 u}{1 - \tan^2 u} \, du.
\end{align}
Now let $t = \tan u, dt = \sec^2 u \, du$. Thus
\begin{align}
\int \sec x \, dx &= 2 \int \frac{dt}{1 - t^2}\\
&= 2 \int \frac{dt}{(1 - t)(1 + t)}\\
&= \int \left [\frac{1}{1 - t} + \frac{1}{1 + t} \right ] dt\\
&= \ln \left |\frac{1 + t}{1 - t} \right | + C\\
&= \ln \left |\frac{1 + \tan u}{1 - \tan u} \right | + C \qquad \text{since} \,\, t = \tan u\\
&= \ln \left |\frac{\cos u + \sin u}{\cos u - \sin u} \right | + C\\
&= \ln \left |\frac{\cos u + \sin u}{\cos u - \sin u} \cdot \frac{\cos u + \sin u}{\cos u + \sin u} \right | + C\\
&= \ln \left |\frac{\cos^2 u + \sin^2 u + 2\sin u \cos u}{\cos^2 u - \sin^2 u} \right | + C\\
&= \ln \left |\frac{1 + \sin 2u}{\cos 2u} \right | + C\\
&= \ln \left |\frac{1 + \sin x}{\cos x} \right | + C \qquad \text{since} \,\, x = 2u\\
&= \ln |\sec x + \tan x| + C,
\end{align}
as expected.
| {
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"answer_id": 7
} |
Generalized Pell's Equation: Why is 4 special? I know how to generate solutions to $x^2-Dy^2=\pm 1$ from the fundamental solution to $x^2-Dy^2=\pm 4$. But how do we know that all the solutions are generated from this fundamental solution?
| I had written out an answer previously, but to make it work I need to assume a little more. Namely, let me assume that $D$ is not congruent to $1 \pmod 4$. (I'll think a little more about what's going on in general, and someone else is of course welcome to jump in and help me out.)
One of the running assumptions of the Pell equation is that $D$ is squarefree. In particular, it is not $0 \pmod 4$. So if you reduce the equation $x^2 - Dy^2 = \pm 4$
modulo $4$, you get $x^2 - Dy^2 \equiv 0 \pmod 4$. Since the squares mod $4$ are $0$ and $1$, running through the four possibilities for $x^2$ and $y^2$ mod $4$ one sees that the only solutions are when $x^2 \equiv y^2 \equiv 0$, i.e., when $x$ and $y$ are even. [N.B.: here I am using that $D \not \equiv 1 \pmod 4$; otherwise $1^2 - D \cdot 1^2 \equiv 0 \pmod 4$.]
That is, every integral solution to $x^2 - Dy^2 = \pm 4$ occurs with $x$ and $y$ even, so we may write $x = 2X$, $y = 2Y$ and substitute, getting
$(2X)^2 - D (2Y)^2 = 4(X^2 - DY^2) = \pm 4$.
Dividing by $4$, we get
$X^2 - DY^2 = \pm 1$.
The process can also be reversed: if $X^2 - DY^2 = \pm 1$, then $(2X)^2 - D(2Y)^2 = \pm 4$. Thus we have found a bijection between the sets of solutions to $x^2 - Dy^2 = \pm 4$ and $X^2 - DY^2 = \pm 1$.
So this is the most direct answer to your question. One can still wonder whether this occurs for integers other than $4$, but for other $k > 1$ it can happen that there are solutions to $x^2 - Dy^2 = \pm k$ in which $x$ and $y$ are not both divisible by $k$. Try it and see. So $4$ is special here because the theory of quadratic equations modulo $4$ is especially simple.
Added: See http://sites.google.com/site/tpiezas/008 for some further information on going from $x^2 - dy^2 = \pm 4$ to $x^2 - dy^2 = \pm 1$.
| {
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Euler angles and gimbal lock Can someone show mathematically how gimbal lock happens when doing matrix rotation with Euler angles for yaw, pitch, roll? I'm having a hard time understanding what is going on even after reading several articles on Google.
Is the only work-around to use quaternions?
| Many answers have been given that are mathematically very sound and correct. Here is another way, using only high school mathematics:
Suppose, we use the ZXY Euler rotations: first about the Z-axis (yaw), then about the X-axis (roll) and finally about the y-axis(pitch).
$R_z(\psi) = \left(\begin{array}{ccc}
\cos \left(\psi \right) & -\sin \left(\psi \right) & 0\\
\sin \left(\psi \right) & \cos \left(\psi \right) & 0\\
0 & 0 & 1
\end{array}\right)$
$R_x(\phi) = \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & \cos \left(\phi \right) & -\sin \left(\phi \right)\\
0 & \sin \left(\phi \right) & \cos \left(\phi \right)
\end{array}\right)$
$R_y(\theta) = \left(\begin{array}{ccc}
\cos \left(\theta \right) & 0 & \sin \left(\theta \right)\\
0 & 1 & 0\\
-\sin \left(\theta \right) & 0 & \cos \left(\theta \right)
\end{array}\right)$
The rotation matrix in ZXY configuration would be:
$R = R_z(\psi)R_x(\phi)R_y(\theta)$
$R = \left(\begin{array}{ccc}
\cos \left(\psi \right)\,\cos \left(\theta \right)-\sin \left(\phi \right)\,\sin \left(\psi \right)\,\sin \left(\theta \right) & -\cos \left(\phi \right)\,\sin \left(\psi \right) & \cos \left(\psi \right)\,\sin \left(\theta \right)+\cos \left(\theta \right)\,\sin \left(\phi \right)\,\sin \left(\psi \right)\\
\cos \left(\theta \right)\,\sin \left(\psi \right)+\cos \left(\psi \right)\,\sin \left(\phi \right)\,\sin \left(\theta \right) & \cos \left(\phi \right)\,\cos \left(\psi \right) & \sin \left(\psi \right)\,\sin \left(\theta \right)-\cos \left(\psi \right)\,\cos \left(\theta \right)\,\sin \left(\phi \right)\\
-\cos \left(\phi \right)\,\sin \left(\theta \right) & \sin \left(\phi \right) & \cos \left(\phi \right)\,\cos \left(\theta \right)
\end{array}\right)$
Now, let's replace $\phi =\frac{\pi }{2}$
You will get R as:
$R = \left(\begin{array}{ccc}
\cos \left(\psi \right)\,\cos \left(\theta \right)-\sin \left(\psi \right)\,\sin \left(\theta \right) & 0 & \cos \left(\psi \right)\,\sin \left(\theta \right)+\cos \left(\theta \right)\,\sin \left(\psi \right)\\
\cos \left(\psi \right)\,\sin \left(\theta \right)+\cos \left(\theta \right)\,\sin \left(\psi \right) & 0 & \sin \left(\psi \right)\,\sin \left(\theta \right)-\cos \left(\psi \right)\,\cos \left(\theta \right)\\
0 & 1 & 0
\end{array}\right)$
Using trigonometric identities, we get:
$R = \left(\begin{array}{ccc}
\cos \left(\psi +\theta \right) & 0 & \sin \left(\psi +\theta \right)\\
\sin \left(\psi +\theta \right) & 0 & -\cos \left(\psi +\theta \right)\\
0 & 1 & 0
\end{array}\right)$
There is infinite $\psi$ and $\theta$ that create the same rotation matrix.
For $\phi =-\frac{\pi }{2}$:
$R = \left(\begin{array}{ccc}
\cos \left(\psi \right)\,\cos \left(\theta \right)+\sin \left(\psi \right)\,\sin \left(\theta \right) & 0 & \cos \left(\psi \right)\,\sin \left(\theta \right)-\cos \left(\theta \right)\,\sin \left(\psi \right)\\
\cos \left(\theta \right)\,\sin \left(\psi \right)-\cos \left(\psi \right)\,\sin \left(\theta \right) & 0 & \cos \left(\psi \right)\,\cos \left(\theta \right)+\sin \left(\psi \right)\,\sin \left(\theta \right)\\
0 & -1 & 0
\end{array}\right)$
$R = \left(\begin{array}{ccc}
\cos \left(\psi -\theta \right) & 0 & -\sin \left(\psi -\theta \right)\\
\sin \left(\psi -\theta \right) & 0 & \cos \left(\psi -\theta \right)\\
0 & -1 & 0
\end{array}\right)$
You can use Symbolic math tools to create these simplifications. I have used MATLAB:
syms psi
R_z = [ cos(psi), -sin(psi), 0;
sin(psi), cos(psi), 0;
0 , 0 , 1
];
syms phi
R_x = [ 1, 0, 0;
0, cos(phi), -sin(phi);
0, sin(phi), cos(phi)
];
syms theta
R_y = [ cos(theta), 0, sin(theta);
0 , 1, 0;
-sin(theta), 0, cos(theta)
];
R = R_z*R_x*R_y;
R_new_1_expr = subs(R, phi, pi/2)
R_new_1_simplified = simplify(R_new_1_expr);
display(R_new_1_simplified)
$\left(\begin{array}{ccc}
\cos \left(\psi +\theta \right) & 0 & \sin \left(\psi +\theta \right)\\
\sin \left(\psi +\theta \right) & 0 & -\cos \left(\psi +\theta \right)\\
0 & 1 & 0
\end{array}\right)$
| {
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} |
Why is $\frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$ valid in a triangle? This is extracted from my module :
In a $\displaystyle\bigtriangleup ABC$,
$$\displaystyle \frac{\sin (A) + \sin(B)}{\sin(A) -\sin(B)} = \frac{k \cdot(\sin (A) + \sin(B))}{k \cdot(\sin(A) -\sin(B))} = \frac{a+b}{a-b}$$
where $A,B,C$ are the angles and $a,b,c$ are the opposite sides,I can't figure out how this step holds valid?
| The sine rule for a triangle goes as
$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$, where $R$ is the radius of the circumcircle.
From this, we get $\sin(A) = \frac{a}{2R}$ and $\sin(B) = \frac{b}{2R}$
The problem has $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)}$. Plug in the expressions for $\sin(A)$ and $\sin(B)$ from the above and we get,
$\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{a}{2R}-\frac{b}{2R}} = \frac{\frac{a+b}{2R}}{\frac{a-b}{2R}} = \frac{a+b}{a-b}$
Hence, $\frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{a+b}{a-b}$
| {
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"answer_count": 4,
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How to prove this inequality without use of computers? With help from Maple, I got
$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2-(c-a)^2-(c-b)^2$$
equal to
$$\frac{(c(x^3+y^3+z^3)+(a-c)(x^2y+y^2z+z^2x)+(b-c)(x^2z+y^2x+z^2y)-3(a+b-c)xyz)^2}{(x-y)^2(y-z)^2(x-z)^2}$$ which of course is $\ge 0$.
But with no help from a computer algebra, how would one prove:$$\left(\frac{ax+by+cz}{x-y}\right)^2+\left(\frac{ay+bz+cx}{y-z}\right)^2+\left(\frac{az+bx+cy}{z-x}\right)^2\ge (c-a)^2+(c-b)^2 ?$$
| We may express it in matrix form. Let $u = [a, \ b, \ c]^\mathsf{T}$,
$p_1 = [x, \ y, \ z]^\mathsf{T}$, $p_2 = [y, \ z, \ x]^\mathsf{T}$,
$p_3 = [z, \ x, \ y]^\mathsf{T}$,
$q_1 = [-1, \ 0, \ 1]^\mathsf{T}$ and $q_2 = [0, \ -1, \ 1]^\mathsf{T}$.
Let
$$S = \frac{1}{(x-y)^2} p_1p_1^\mathsf{T}
+ \frac{1}{(y-z)^2} p_2p_2^\mathsf{T} + \frac{1}{(z-x)^2} p_3p_3^\mathsf{T}
- q_1q_1^\mathsf{T} - q_2q_2^\mathsf{T}.$$
We have $\mathrm{LHS} - \mathrm{RHS} = u^\mathsf{T}S u$.
It suffices to prove that $S$ is positive semidefinite.
Note that all $2\times 2$ minors of $S$ are zero
(e.g., $S_{1,1} S_{2,2}-S_{1,2}S_{2,1} = 0$, etc.). Thus, $\mathrm{rank}(S)\le 1$.
Also, $S_{1,1} = x^2/(x-y)^2+y^2/(y-z)^2+z^2/(z-x)^2-1 > 0$
(the proof is not difficult).
Thus, $S$ is positive semidefinite. We are done.
| {
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"source": "stackexchange",
"question_score": "8",
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AM-GM application How to show that
$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$
$a_1,a_2,a_3$ are all of same algebraic sign.
| Hint: $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 9 \cdot\frac{1}{3}(a+b+c)\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
Assuming $a,b,c$ are positive, you can use the AM-GM inequality twice.
| {
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"source": "stackexchange",
"question_score": "2",
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How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$ $$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
| Moron's and J.M.'s solutions are nice.
Hopefully this solution is simpler.
Without loss of generality we may assume that $1\gt x\gt 0$.
Put $x:=\sqrt{y}$, $1\gt y\gt 0$. Then we obtain
$$
\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx=\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy.
$$
Introduce the new variable
$$
t:=\frac{1+y}{1-y},\qquad 1\lt t \lt \infty.
$$
Then we have
$$
y=\frac{-1+t}{1+t},
$$
$$
\mathrm dy=\frac{2}{(1+t)^2}\,\mathrm dt.
$$
Substituting back we obtain
$$
\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy=\int{\frac{t}{2\sqrt{1+\left(\dfrac{-1+t}{1+t} \right)^2}\sqrt{\dfrac{-1+t}{1+t}}}\frac{2}{(1+t)^2}\,\mathrm dt}
$$
$$
=\frac{1}{\sqrt{2}}\int{\frac{t}{\sqrt{t^4-1}}}\mathrm dt
$$
$$
=\frac{1}{2\sqrt{2}}\ln(t^2+\sqrt{t^4-1})+C.
$$
Putting back everything we obtain
$$
\frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "44",
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If $2^{n+1} - 1$ is composite, is $2^{n+2} - 1$ prime? In 1556, Tartaglia claimed that the sums
1 + 2 + 4
1 + 2 + 4 + 8
1 + 2 + 4 + 8 + 16
are alternative prime and composite. Show that his conjecture is false.
With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?
| If $2^{n}-1$ is prime, then $n$ must be odd, otherwise, we could factor
$2^{n}-1$ as $(2^{n/2}+1)(2^{n/2}-1).$
So at least every other number in the series $2^n-1,n=1,2,3...$
must be composite, by the conjugacy rule.
EDIT: Reread the question, but you can do the same for $a^3-b^3 = (a-b)(a^2+ab+b^2)$ so whenever $n$ is divisible by 3, you can do a factorization as above. In general, $a^n-b^n$ is divisible by $a-b,$
so if $n$ is not prime, $n=pq$, and we have
$$2^{n}-1 = (2^q)^p - 1^p = (2^q-1)Q$$
where $Q$ is an integer.
| {
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$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots \biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$ How does one find all integer solutions to this equation: $$ \biggl\lfloor{\frac{x}{1!}\biggr\rfloor} + \biggl\lfloor{\frac{x}{2!}\biggr\rfloor} + \cdots +\biggl\lfloor{\frac{x}{10!}\biggr\rfloor}=1001$$
Can't think about methods for solving this. Though i did something i am not sure of it.
| We are given $\lfloor x/1!\rfloor + \lfloor x/2! \rfloor + \dots + \lfloor x/10!\rfloor = 1001$.
Now,
$$\begin{align}
\lfloor x/1!\rfloor \leq 1001 &\Longrightarrow x \leq 1001
\\
&\Longrightarrow x/7! \leq 1001/7!
\\
&\Longrightarrow x/7! \leq 1001/5040
\\
&\Longrightarrow \lfloor x/7! \rfloor = 0
\\
&\Longrightarrow \lfloor x/k! \rfloor = 0, \quad\text{ for } k \in\{ 7,8,9,10 \}
\end{align}$$
It suffices to solve $\lfloor x/1! \rfloor + \dots + \lfloor x/6! \rfloor = 1001$. Since $y - 1 < \lfloor y \rfloor \leq y$, we have
$$\begin{align}
x(1/1! + \dots 1/6!) - 6 < \lfloor x/1!\rfloor + \dots + \lfloor x/6! \rfloor \leq x(1/1! + \dots + 1/6!)
\end{align}$$
and thus $582.635 \leq x < 585.546$. Among the integers $583, 584$, and $585$, the integer $584$ satisfies the equation.
| {
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Prove a number is composite How can I prove that $$n^4 + 4$$ is composite for all $n > 5$?
This problem looked very simple, but I took 6 hours and ended up with nothing :(. I broke it into cases base on quotient remainder theorem, but it did not give any useful information.
Plus, I try to factor it out:
$$n^4 - 16 + 20 = ( n^2 - 4 )( n^2 + 4 ) - 5\cdot4$$
If a composite is added to a number that is a multiple of $5$, is there anything special? A hint would suffice.
Thanks,
Chan
| This is a special case of a class of cyclotomic factorizations due to Aurifeuille, Le Lasseur and Lucas, the so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^3}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
| {
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Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
&\quad = A(x+1)^2(x^2+2)+B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2
\end{align*}
Substitute in $x=0$ to get $4=A(1)(2)$, so $A = 2$
$$6x^4+11x^3+10x^2+14x = B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2$$
Substitute in $x=-1$ to get
$$6-11+10-14 = C(-1)(1+2)$$
so $-9=-3C$, thus $C=3$.
Leaving me what I have below:
Which brings me to where I am currently stuck.
$$6x^4 +8x^3 +10x^2+8x = B(x)(x+1)(x^2+2) + (Dx + E) (x) (x+1)^2$$
Is the next best move to use substitution to solve for $B$?
| First, I trust you used the correct partial fraction decomposition:
$$\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{Dx+E}{x^2+2}.$$
This leads to
\begin{align*}
&8x^4 + 15x^3 + 16x^2 + 22x + 4\\
&\qquad = A(x+1)^2(x^2+2) + Bx(x+1)(x^2+2) + Cx(x^2+2) + (Dx+E)x(x+1)^2.
\end{align*}
A useful "trick" is to evaluate at the zeros of the linear factors to get some information; I suspect you evaluated at $x=0$ to get $2A = 4$, from which you got $A=2$.
You can then evaluate at $x=-1$ to get $-3C = -9$, which is how you got $C=3$. Looking good.
Then you used that to simplify.
$$2(x^2+2x+1)(x^2+2) +3x(x^2+2) = 2x^4 + 7x^3 + 6x^2 + 14x + 4,$$
which subtracted from $8x^4 + 15x^3 + 16x^2 + 22x + 4$ gave you
$$6x^4 + 8x^3 + 10x^2 + 8x = Bx(x+1)(x^2+2) + (Dx+E)x(x+1)^2.$$
Hmmm... Which is not quite what you have. Did you use the correct decomposition, or did you forget about being careful with that $(x+1)^2$?
Anyway: here's where you go stuck because you are used to being able to solve the partial fractions problems using only the evaluation trick. But when you have irreducible quadratic factors or powers of linear factors (or worse, both), the trick doesn't get you all the way there.
Here, we can factor out $x$ from both sides to get
$$6x^3 + 8x^2 + 10x + 8 = B(x+1)(x^2+2) + (Dx+E)(x+1)^2.$$
(We factored out $x$ from both sides and cancelled; that's how we dropped from fourth power to cube).
Edit.
We can further factor out $x+1$ from both sides:
$$(x+1)(6x^2 + 2x + 8) = (x+1)(B(x^2+2) + (Dx+E)(x+1))$$
to get
$$6x^2 + 2x + 8 = B(x^2+2) + (Dx+E)(x+1).$$
Contrary to your claim before, now that we had all the right terms, we cannot simply conclude that $D=6$, because there are two quadratic terms: $Bx^2$ and $Dx^2$.
You can, however, evaluate at $x=-1$ to get $12 = 3B$, or $B=4$; from this you go to
$$6x^2 + 2x + 8 = 4x^2 + 8 + (Dx+E)(x+1)$$
or
$$2x^2 + 2x = (Dx+E)(x+1).$$
Noting that the constant term on the right is $E$, and $0$ on the left, you get $E=0$. This gives
$$2x(x+1) = Dx(x+1)$$
which, cancelling $x(x+1)$ yields $D=2$.
Alternatively, from $2x^2+2x = (Dx+E)(x+1)$, we can factor the left hand side completely to get
$$2x(x+1) = (Dx+E)(x+1)$$
from which we immediately get $Dx+E = 2x$, so $D=2$ and $E=0$.
So, in summary, $A=2$, $B=4$, $C=3$, $D=2$, $E=0$.
| {
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"source": "stackexchange",
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} |
Proofs of the Cauchy-Schwarz Inequality? How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
| Yet another proof can be done by AM-GM inequality---
In all that follows,all summations run from $i=1$ to $n$.
Let $X^2=\sum{x_i}^2 , Y^2=\sum{y_i}^2\tag{1}$
By Arithmetic-mean-Geometric-mean(AM-GM) inequality(i.e. AM$\ge$GM ),we have
$\frac{a+b}{2}\ge\sqrt{ab}\tag{2}$
Putting $a=x_i^2/X^2$ and $b=y_i^2/Y^2$ in $(2)$,we get
$$\frac{(x_i^2/X^2+y_i^2/Y^2)}{2}\ge \frac{x_iy_i}{XY}$$
Taking summation on both sides, we obtain
$$\sum\frac{(x_i^2/X^2+y_i^2/Y^2)}{2}\ge \sum\frac{x_iy_i}{XY}\tag{3}$$
Since X and Y are already summed over all $i$s,they are independent of the individual $x_i$s and $y_i$s and can be taken outside the summations.Therefore,the quantity on LHS can be written as,
$$\frac{\sum x_i^2}{2X^2}+\frac{\sum y_i^2}{2Y^2}=\frac{X^2}{2X^2}+\frac{Y^2}{2Y^2}=1/2+1/2=1$$ where the first equality is a consequence of $(1)$.
$\therefore$ LHS of (3) is 1. Hence,$(3)$becomes,
$$1 \ge \sum\frac{x_iy_i}{XY}$$
Arranging the inequality properly and putting in X and Y from $(1)$ and finally squaring both sides,we get
$$\left( \sum_i x_i y_i \right)^2 \le \left( \sum_i x_i^2 \right) \left( \sum_i y_i^2 \right)
$$
which is the Cauchy-Schwarz inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?
Thanks,
| HINT $\rm\ \ x\ p(x)\ =\ 2\ x^4 + 7\ x^2 - 4\ x\ \equiv\ 2\ (x-1)^2\ \ (mod\ 5)\ $ for $\rm\ x\not\equiv 0\ $ since then $\rm\ x^4\equiv 1\ $
| {
"language": "en",
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"source": "stackexchange",
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Coefficients of characteristic polynomial of a matrix For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.
If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$
that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?
| Use the fact that $\begin{vmatrix} a & b+e \\
c & d+f \end{vmatrix} =
\begin{vmatrix} a & b \\
c & d \end{vmatrix} +
\begin{vmatrix} a & e \\
c & f \end{vmatrix}
$
We can use this fact to separate out powers of $\lambda$. Following is an example for $2 \times 2$ matrix.
$$
\begin{vmatrix} a-\lambda & b \\
c & d-\lambda \end{vmatrix} =
\begin{vmatrix} a & b \\
c & d-\lambda \end{vmatrix} +
\begin{vmatrix} -\lambda & b \\
0 & d-\lambda \end{vmatrix} =
\begin{vmatrix} a & b \\
c & d \end{vmatrix} + %%
\begin{vmatrix} a & 0 \\
c & -\lambda \end{vmatrix} + %%
\begin{vmatrix} -\lambda & b \\
0 & d \end{vmatrix} +
\begin{vmatrix} -\lambda & 0 \\
0 & -\lambda \end{vmatrix}
$$
This decompose $det$ expression into sum of various powers of $\lambda$.
Now try it with a $3 \times 3$ matrix and then generalize it.
| {
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How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
| Factor out the $\frac{2}{3}$. Then write $$\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{1}{3^n} + \sum_{n=2}^{\infty} \frac{1}{3^n} + \sum_{n=3}^{\infty} \frac{1}{3^n} + \cdots$$
It is easy to show that $$\sum_{n=m}^{\infty} \frac{1}{3^n} = \frac{3}{2} \left(\frac{1}{3} \right)^m$$
and so
$$\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{2} \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n $$
which you can sum. Don't forget to put the $\frac{2}{3}$ back in.
| {
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$(7a+1)x^3+(7b+2)y^3+(7c+4)z^3+(7d+1)xyz=0$ does not have integer solutions Let $a,b,c,d$ be integers. How can I prove that the equation
$$(7a+1)x^3+(7b+2)y^3+(7c+4)z^3+(7d+1)xyz=0$$
Does not have an integer solution $(x,y,z)$ such that $\gcd(x,y,z)=1$?
| modulo $7$, the equation is $x^3 + 2y^3 + 4z^3 + xyz = 0$.
Cubes modulo $7$ are $0,1,-1$.
If one of $x,y,z$ is $0$, for example if $z = 0$, then you get $x^3 + 2y^3 = 0$, which is only possible if $x=y=0$.
So either $x=y=z=0$ or they are all nonzero.
If they are non zero, then $x^6 = y^6 = z^6 = 1$.
If you take $xyz = -(x^3 + 2y^3 + 4z^3)$ and cube it, you get $(xyz)^3 = -5(x^3 + y^3 + z^3)+(xyz)^3$, which implies that $x^3+y^3+z^3 = 0$. But since cubes are $1$ or $-1$, this is again impossible.
Thus the only solution modulo $7$ is $x=y=z=0$, so for any solution, $x,y,z$ have to be multiples of $7$. In particular there can't be any primitive solution, so there is no solution at all.
| {
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Summation by parts of $\sum_{k=0}^{n}k^{2}2^{k}$ I want to evaluate this sum
$$\sum_{k=0}^{n}k^{2}2^{k}$$ by summation by parts (two times) and I need to know, if my approach was right.
I know the formula for summation by parts is $$\sum u\Delta v=uv-\sum\left(Ev\right)\Delta u$$ where $E\left(v\left(x\right)\right)=v\left(x+1\right)$ and $(v(x))=v(x+1)-vx)$.
Further if $f$ is a Antiderivative of $g$ I know the formula
$$\sum_{k=a}^{b}g\left(k\right)=f\left(b+1\right)-f\left(a\right)$$
First I choose $u=x^{2}$. This means $\Delta u=\Delta x^{2}=2x+1$. Choose $v=2^{x}$. That means $\Delta v=2^{x}$. Then I get
$$\begin{eqnarray*}
\sum_{k=0}^{n}k^{2}2^{k} &=& \sum_{0}^{n+1}x^{2}2^{x} \\
&=& \left.x^{2}*2^{x}\right|_{0}^{n+1}-\sum_{0}^{n+1}2^{x+1}(2x+1) \\
&=& (n+1)^2 2^{n+1}-\sum_{0}^{n+1}2x2^{x+1}+2^{x+1} \\
&=& (n+1)^2 2^{n+1}-4\sum_{0}^{n+1}x2^{x} - 2\sum_{0}^{n+1}2^{x}
\end{eqnarray*}$$
Now choose $u(x)=x$. This means $\Delta u=\Delta x=1$. Chose $v=2^{x}$. This means $\Delta v=2^{x}$ and $E(v(x))=2^{x+1}$. I get then
$$\begin{eqnarray*}
\sum_{x=0}^{n+1}x2^{x+1}
&=& \left[x2^{x}-2^{x+1}\right]_{0}^{n+1} \\
&=& ((n+1)-1)2^{(n+1)+1}+2 \\
&=& n2^{n+2}+2
\end{eqnarray*}$$
Putting those two calculations together resolves in
$$\begin{eqnarray*}
\sum_{k=0}^{n}k^{2}2^{k}
&=&(n+1)^2 2^{n+1}-2\sum_{0}^{n+1}x2^{x+1} - \sum_{0}^{n+1}2^{x+1} \\
&=& (n+1)^2 2^{n+1}-2(n2^{n+2}+2) - (n+2)2^{x} \\
&=& 2^{n+1}(n^2+2n+1) -4(n2^{n+2}+2) - 2(n+2)2^{x} \\
&=& 2^{n+1}(n^2+2n+1-8n-(n+2))-8 \\
&=& 2^{n+1}(n^2-7n-1)-8 \\
\end{eqnarray*}$$
But Mathematica says
$$2^{n+1} ((n-2) n+3)-6 = 2^{n+1} (n^2-2n+3)-6$$
Where is my fault?
Update
I placed the right calculations as an answer below.
| You wrongly calculated the difference of $x^2$. Plug it directly into the definition.
(And you can always check your calculation by looking at small values of $n$. In particular, you should have 0 for $n=0$. To locate the error, "binary search" is advisable: Put $n=0$ somewhere in the middle to see if the error was in the first half or the second half, etc.)
| {
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How to calculate $\lim_{x \to 0}\left(\frac1{x} + \frac{\ln(1-x)}{x^2}\right)$? How to calculate the following limit?
$$\lim_{x \to 0}\left(\frac1{x} + \frac{\ln(1-x)}{x^2}\right)$$
| You have
\begin{eqnarray*}
\lim_{x\to 0} \left(\frac{1}{x}+\frac{\ln(1-x)}{x^2}\right) &=& \lim_{x\to 0} \frac{x+\ln(1-x)}{x^2},
\end{eqnarray*}
note that
$$ \lim_{x\to 0} x+\ln(1-x)=0,\: \lim_{x\to 0} x^2= 0, $$
then by the L'Hospital's rule
$$\begin{align*}
\lim_{x\to 0} \left(\frac{1}{x}+\frac{\ln(1-x)}{x^2}\right)&= \lim_{x\to 0} \frac{\frac{d}{dx}(x+\ln(1-x))}{\frac{d}{dx}x^2}\\
&= \lim_{x\to 0} \frac{1-\frac{1}{1-x}}{2x}\\
&= \lim_{x\to 0}\frac{\frac{1-x - 1}{1-x}}{2x}\\
&= \lim_{x\to 0} \frac{\frac{x}{x-1}}{2x}\\
&= \lim_{x\to 0}\frac{1}{2(x-1)}\\
&= -\frac{1}{2}.
\end{align*}$$
| {
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
| Let $\wp(n) = (n-1)^3 + n^3 + (n+1)^3$. Observe that $\wp(1) = 9$, and so is divisible by $9$. To complete this inductive proof, we want to prove that $\wp(n+1) - \wp(n)$ is divisible by $9$.
$\wp(n+1) - \wp(n) = n^3 + (n+1)^3 + (n+2)^3 - (n-1)^3 - n^3 - (n+1)^3 $
$= (n+2)^3 - (n-1)^3 = 9n^2 + 9n + 9 = 9(n^2 + n +1)$
and so is also divisible by $9$. We have thus shown that $\wp(1) = 0 \mod 9$, and that $\wp(n+1) - \wp(n) = 0 \mod 9$, and so $\wp(2) = \wp(1) + \wp(2) - \wp(1) = 0 \mod 9$, and inductively proceed from there...
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"timestamp": "2023-03-29T00:00:00",
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Subtracting rational functions I'm trying to find out how to solve this:
$$\frac{x-2}{x^2+2x} - \frac{x+2}{x^2-2x} - \frac{4x}{x^2-4}$$
The answer is $\displaystyle \frac{-4}{x-2}$
What is this called? And is there any video of it on http://www.khanacademy.org ?
Thanks!
| Added: The method is to transform the sum of the given rational fractions (the numerator and denominator consists of polynomials) into a
single equivalent fraction.
The properties used are
*
*$$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$$
*$$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$
I would calculate as follows, starting with the factorization of denominators
$$\begin{eqnarray*}
&&\frac{x-2}{x^{2}+2x}-\frac{x+2}{x^{2}-2x}-\frac{4x}{x^{2}-4} \\
&=&\frac{x-2}{x(x+2)}-\frac{x+2}{x\left( x-2\right) }-\frac{4x}{\left(
x-2\right) \left( x+2\right) } \\
&=&\frac{\left( x-2\right) ^{2}}{x(x+2)\left( x-2\right) }-\frac{\left(
x+2\right) ^{2}}{x\left( x-2\right) \left( x+2\right) }-\frac{4x^{2}}{%
x\left( x-2\right) \left( x+2\right) } \\
&=&\frac{\left( x-2\right) ^{2}-\left( x+2\right) ^{2}-4x^{2}}{x\left(
x-2\right) \left( x+2\right) } \\
&=&\frac{x^{2}-4x+4-x^{2}-4x-4-4x^{2}}{x\left( x-2\right) \left( x+2\right) }
\\
&=&\frac{-4x^{2}-8x}{x\left( x-2\right) \left( x+2\right) } \\
&=&-\frac{4x\left( x+2\right) }{x\left( x-2\right) \left( x+2\right) } \\
&=&-\frac{4}{x-2}.
\end{eqnarray*}$$
so that I reduce the rational fractions to a common denominator first to allow me to add (subtract) them, and simplify the numerator thereafter. Finally I divide by the common factors to both the numerator and denominator. This is valid iff $x\ne -2$ and $x\ne 0$, because you cannot divide by zero. Also for $x=2$ the fraction is not defined.
See "Simplifying Rational Expressions 1, 2 and 3", "Adding and Subtracting Rational Expressions 1, 2 and 3" on http://www.khanacademy.org.
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Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
| Another way (by Euler, I think), starting from the geometric sum:
$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$
Differentiate both sides:
$$1 + 2 x + 3 x^2 + \cdots + n x^{n-1} = \frac{n x^{n+1}-(n+1) x^{n} +1}{(x-1)^2} \tag{2}$$
Multiply by $x$ and differentiate once more. We get on the LHS
$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$
which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to multiply by $x$ the RHS of $(2)$, compute the (eg, with L'Hopital rule) and evaluate it at $x \to 1$.
It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.
(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let
$$\begin{align}
g(x)&=\frac{x^{n+1}-1}{x-1}\\
&=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\
&={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4}
\end{align}$$
Differentiating, multiplying by $x$ and differentiating again we get $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$
... which evaluated at $x=1$ gives the desired answer:
$${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6} $$
We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the series $\sum\limits_{ n=\lfloor\frac{m}{2} \rfloor}^m \frac1{n}$ The floor function is confusing me a bit.
| Hint. m even $\implies n = \frac{m}{2} = \frac{2k}{2} = k$, m odd $\implies n = \frac{m-1}{2}$ since $\frac{m}{2} = \frac{2k+1}{2} = k + \frac{1}{2}$ for some k.
Edit:
Let $H(m) = \sum_{n=1}^{m} \frac{1}{n}$. For even m = 2k, then $\sum_{ n=\lfloor\frac{m}{2} \rfloor}^{m} \frac{1}{n} = H(2k) - H(k)$.
Now H(2k) - H(k) is equivalent to the alternating sum = $1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2k}$, and so approaches ln(2). This can be proved using induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quick way to expand $\cos^{-1}(\cos^2 x)$ up to $O(x^2)$ For a part of a question, I need to expand $\cos^{-1}(\cos^2 x)$ up to $O(x^2)$ about $x=0$. It took me quite a while to get an incorrect answer. What are some quick and efficient offline (i.e, no alpha) ways to get a good approximation?
EDIT: To elaborate a bit more, the first order term can be confidently stated to be $\sqrt{2}x$ as this expression is the length of a hypotenuse for a right spherical triangle, which must reduce to the flat triangle hypotenuse. No "reasoning" like this could work for second order term so I had to take the cumbersome way.
| We have $\cos^2 x = \left( 1 - \frac{x^2}{2} + O(|x|^4) \right)^2 = 1 - x^2 + O(|x|^4)$. Since $\cos x = 1 - \frac{x^2}{2} + O(|x|^4)$, it follows that $\cos^{-1} \left( 1 - \frac{x^2}{2} \right) = x + O(|x|^2)$, hence
$$\cos^{-1} \left( \cos^2 x \right) = \sqrt{2} x + O(|x|^2).$$
Edit: Jyrti seems to be right. Writing $\cos^{-1} \left( \cos^2 x \right) = \sqrt{2} (x + h)$, we get
$$\cos^2 x = 1 - x^2 + O(|x|^4) = 1 - (x + h)^2 + O(|x|^4)$$
so we can get $h = O(|x|^3)$. Just for fun, writing $h = ax^3 + O(|x|^4)$ we get
$$1 - x^2 + \frac{x^4}{3} + O(|x|^6) = 1 - x^2 - 2ax^4 + \frac{x^4}{6} + O(|x|^6)$$
hence $a = - \frac{1}{12}$ and
$$\cos^{-1} \left( \cos^2 x \right) = \sqrt{2} x - \sqrt{2} \frac{x^3}{12} + O(|x|^4).$$
Note that these power series computations are much faster than repeated computation of derivatives using the chain rule. The manipulations I'm doing make sense in the ring of formal power series modulo $x^4$ or $x^5$ or whatever precision I care about, and in that generality there's no need to think in terms of complicated sums of products of functions (which implicitly carry information about the derivatives to all orders, which is irrelevant).
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$? Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$
| Expanding your problem:
$$ \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{3^k }}} = \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots $$
$$ = 1 + \left (\frac{1}{3} + \frac{1}{3} \right) + \left(\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} \right) + \left(\frac{1}{3^3} + \frac{1}{3^3}+ \frac{1}{3^3}+ \frac{1}{3^3}\right) + \dots$$
This can be grouped into:
$$ = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$
$$ \left(\frac{1}{3^3} + \dots\right) + \dots $$
Using the fact that $ S = \sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{3}{2}$:
$$ = \frac{3}{2} + $$
$$ \frac{3}{2} - (1) + $$
$$ \frac{3}{2} - \left(1 + \frac{1}{3} \right) + $$
$$ \frac{3}{2} - \left( 1 + \frac{1}{3} + \frac{1}{3^2} \right ) + \dots $$
The partial sum $S_k$ is computed as: $S_k = \sum_{n=0}^k \frac{1}{3^n} = \frac{3}{2} - \frac{1}{2}\left(\frac{1}{3}\right)^k$
Hence,
$$ = \frac{3}{2} + \left(\frac{3}{2} - S_0 \right) + \left(\frac{3}{2} - S_1 \right) + \left(\frac{3}{2} - S_2 \right) \dots$$
$$ = \frac{3}{2} + \frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots \right) $$
$$ = \frac{3}{2} + \frac{1}{2}S = \frac{3}{2} + \frac{1}{2} \frac{3}{2}$$
$$ = \mathbf{\frac{9}{4}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.