Math Challenge - December 2019

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Summary:

Topology
Calculus
Group Theory

Main Question or Discussion Point

Questions

1.
Let ##(X,d)## be a metric space. The open ball with center ##z\in X## of radius ##r > 0## is defined as
$$
B_r(z) :=\{\,x\in X\,|\,d(x,z)<r\,\}
$$
a.) Give an example for
$$
\overline{B_r(z)} \neq K_r(z) :=\{\,x\in X\,|\,d(x,z)\leq r\,\}
$$
Does at least one of the inclusions ##\subseteq## or ##\supseteq## always hold?

b.) What are the answers in the previous case, if we additionally assume that ##(X,d)## has an inner metric?

An inner metric ##d_0## associated to ##d## is defined as the infimum of all lengths of rectified curves between two points:
Let ##\sigma \, : \,[0,1]\longrightarrow X## with ##\sigma(0)=x\, , \,\sigma(1)=y## a rectified curve with length
$$
L(\sigma)=\sup \left\{ \left. \sum_{k=1}^n d(\sigma(t_{k-1}),\sigma(t_k) )\,\right|\,0=t_0<t_1<\cdots < t_n=1\, , \,n\in \mathbb{N} \right\}
$$
Then ##d_0(x,y)=\inf L(\sigma)\,.##

2. Let ##f(z)=\dfrac{7z-51}{z^2-12z+27}## be a complex function.
a.) Determine the Laurent series of ##f(z)## and their radius of convergences around ##z=3## in the cases where ##0## is in the area of convergence, and ##10## is in the area of convergence.
b.) Determine ##\lim_{z \to 3}f(z)\, , \,\operatorname{Res}(f,3)## and the kind of singularity in ##z=3\,.##

3. Write the following groups as amalgamated products of cyclic groups:
a.) ##G=\langle x,y\,|\, x^3y^{-3},y^6 \rangle##
b.) ##H=\langle x,y\,|\, x^{30}, y^{70},x^3y^{-5} \rangle##

4. Prove that there are uncountably many groups, which are generated by two elements, and not finitely presented.
Hint: There are uncountably many non-isomorphic groups with two generators [Bernhard Neumann, 1937].

5. (solved by @archaic ) Let ##f : [1,\infty) \longrightarrow [0,\infty)## be a continuously differentiable function. Write ##S## for the solid of revolution of the graph ##y = f(x)## about the ##x-##axis. If the surface area of ##S## is finite, then so is the volume.

6. Calculate ##\sum_{k,j=1}^\infty \dfrac{1}{kj(k+j)^2}##

7. Calculate ##S:= \displaystyle{\sum_{n=0}^\infty}\,\displaystyle{\sum_{k=0}^n}\,\dfrac{3^k(2n-2k)!(2k)!}{2^k8^n[(n-k)!]^2[k!]^2(2n(1+2k)+(1-4k^2))}##

8. (solved by @Mastermind01 ) Solve ##y'x-y=\sqrt{x^2-y^2}##

9. Let ##(a_n)_{n\in \mathbb{N}}\, , \,(b_n)_{n\in \mathbb{N}} \subseteq \mathbb{R}_{\geq 0}## be two sequences of nonnegative numbers, where not all sequence elements vanish, and be ##p,q\in \mathbb{R}## with ##1<p,q<\infty\, , \,\frac{1}{p}+\frac{1}{q}=1\,.## Prove
$$
\sum_{n=1}^\infty \sum_{m=1}^\infty \dfrac{a_nb_m}{n+m} < \dfrac{\pi}{\sin (\pi/p)} \cdot \left(\sum_{n=1}^\infty a_n^p\right)^{\frac{1}{p}} \cdot \left(\sum_{m=1}^\infty b_m^q\right)^{\frac{1}{q}}
$$

10. Let ##f\, : \,\mathbb{R}_{\geq 0} \longrightarrow \mathbb{R}_{\geq 0}## be an integrable function and ##p>1\,.## Prove
$$
\int_0^\infty\left(\dfrac{1}{x}\int_0^x f(t)\,dt\right)^p\,dx \leq \left(\dfrac{p}{p-1}\right)^p \int_0^\infty (f(x))^p\,dx
$$
Hint: Substitute ##t=xu^{p/r}## and at the end ##r=p-1\,.##



1572565883019-png.png


High Schoolers only

11.
(solved by @Not anonymous ) Choose any odd prime, square it and subtract one. Show that the result is always divisible by twenty-four except for three.
What can be said, if we take the prime up to the power four, and subtract one?

12. In a square of side length ##4##, there is a circle of radius ##1## in each corner. In the center of the square is another circle that touches the other four. Analogously, in the three-dimensional case, in the center of a cube of edge length ##4##, there would be a sphere which would touch eight spheres of radius ##1## placed in the corners of the cube. In which dimension does the central hypersphere become so large that it touches all sides of the hypercube?
1575158642494.png


13. There is only one rule at Christmas at the world's richest family: The gifts have to be expensive, heavy and glamorous. So they all present statues of pure gold. It may be large figure, a tiger sculpture or an opulent candlestick. The eldest son who doesn't live at home anymore receives gifts of nine tons total, but none of which is heavier than a ton. He wants to bring home all of them, but only could rent trucks which can load three tons maximal. How many trucks are needed to at least be able to transport all gifts of gold at the same time?

14. Prove ##\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{3}{2}## for ##a,b,c >0##

15. Let ##\mathbf{x}=(x_1,\ldots ,x_n)\, , \,\mathbf{y}=(y_1,\ldots ,y_n)## be tuples of positive numbers. Prove
$$
\prod_{k=1}^{n} (x_k+y_k)^{1/n} \geq \prod_{k=1}^{n} x_k^{1/n} + \prod_{k=1}^{n} y_k^{1/n}
$$
 
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Answers and Replies

201
51
8. Solve ##y′x−y=\sqrt{x^2−y^2}##

Dividing by x throughout we get $$y' - \frac{y}{x} = \frac{\sqrt{x^2 - y^2}}{x} \implies y' - \frac{y}{x} = \sqrt{\frac{x^2 - y^2}{x^2}}$$.

Now using the substitution ##y = vx## and differentiating it throughout we get ##y' = xv'+ v##

Using the above relations and substituting them for ##y## and ##y'## respectively, we get ##xv' + v - v = \sqrt{1-v^2} \implies xv' =\sqrt{1-v^2}##
Now we seperate the variables i.e. ##x\frac{dv}{dx} = \sqrt{1-v^2} \implies \frac{dv}{\sqrt{1-v^2}} = \frac{dx}{x}## and integrate ##\int \frac{dv}{\sqrt{1-v^2}} = \int \frac{dx}{x}##. The integral on the R.H.S is simply ##\ln{|x|}## for the L.H.S we use the subsitution ##v = sin z## to get ##\int \frac{\cos{z} dz}{cos z} \implies z## Therefore the L.H.S integral is ##\arcsin{v}## (by substituting z for v). Using the integral values in the original equation we get ##v = \sin(\ln{|x|} + C)##. Replacing v, the final answer is ##y = x\sin(\ln{|x|} + C)##
 
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8. Solve ##y′x−y=\sqrt{x^2−y^2}##

Dividing by x throughout, ##y' - \frac{y}{x} = \sqrt{\frac{x^2 + y^2}{x^2}}##. Now using the subsitution ##y = vx## and ##y' = xv'+ v## we get ##xv' + v - v = \sqrt{1-v^2}## Seperating and integrating we get ##v = \sin(\ln{x} + C)##. Replacing v, the final answer is ##y = x\sin(\ln{x} + C)##
Can you write down the calculation for those less familiar with differential equations? And I assume you meant ##\ln|x|##! Nevertheless, it's only part of the answer.
 
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Can you write down the calculation for those less familiar with differential equations? And I assume you meant ##\ln|x|##! Nevertheless, it's only part of the answer.
I amended my post to explain things better and yes I missed the || on both lines
 
fresh_42
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I amended my post to explain things better and yes I missed the || on both lines
Still not the complete answer.
 
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What am I missing?
You cannot rule out that ##v(x_0)^2=1## for some point ##x_0##. This means we have ##\sqrt{1-v(x_0)^2}=0## or ##v(x_0)=\pm 1 ##, i.e. ##y=\pm x## are also solutions to the Jacobian Differential Equation ##y'x-y=\sqrt{x^2-y^2}.##
 
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201
51
You cannot rule out that ##v(x_0)^2=1## for some point ##x_0##. This means we have ##\sqrt{1-v(x_0)^2}=0## or ##v(x_0)=\pm 1 ##, i.e. ##y=\pm x## are also solutions to the Jacobian Differential Equation ##y'x-y=\sqrt{x^2-y^2}.##
Did not realise I was doing that. I do now. Thank you.
 
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64
6. Calculate ##\sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}##
$$\sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}= \frac {1}{2^2} + \frac {1}{3^2 6^2} + \frac{1}{4^2 8^2} + \frac{1}{5^2 10^2} + ...
\\ = \frac{1}{2^2} +\frac{1}{2^2 2^4} + \frac{1}{2^2 3^4} + \frac{1}{2^2 4^4} + \frac{1}{2^2 5^4} + ...
\\ = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^4}$$
I observe that a piecewise continuous function f can be expressed as a Fourier series,$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_ncos(\frac{n\pi x}{L}) + b_nsin(\frac{n\pi x}{L})
\\a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos(\frac{n\pi x}{L})dx
\\b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin(\frac{n\pi x}{L})dx$$ I also recall Parseval's identity,$$\frac{1}{\pi}\int_{-\pi}^{\pi}| f(x) |^2dx=\frac{a_{0}^2}{2}+\sum_{n=1}^{\infty} (a_{n}^2 + b_{n}^2)$$I assert without proof that for ##L=\pi## the Fourier series for ##x^2## is$$ x^2= \frac{\pi^2}{3} + 4(cos(x) -\frac{cos(2x)}{2^2} + \frac{cos(3x)}{3^2}-...
\\a_0=\frac{2\pi^2}{3}
\\a_n= \frac{4(-1)^{n}}{n^2}$$From Parseval's identity,$$\frac{1}{\pi}\int_{-\pi}^{\pi}x^4dx=\frac{(\frac{2\pi^2}{3})^2}{2} + \sum_{n=1}^{\infty}\frac{16}{n^4}
\\ \frac{2\pi^4}{5}=\frac{(\frac{2\pi^2}{3})^2}{2} + \sum_{n=1}^{\infty}\frac{16}{n^4}
\\ \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}
\\\sum_{k,j=1}^{\infty}\frac{1}{kj(k+j)^2}=\frac{\pi^4}{360}$$
 
fresh_42
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I cannot follow you. In your very first equation I get different terms:
$$
\sum_{k,j=^1}^\infty \dfrac{1}{kj(k+j)^2}=\underbrace{\dfrac{1}{1\cdot 2^2}}_{(1,1)}+\underbrace{\dfrac{2}{2\cdot 3^2}}_{(1,2)}+\underbrace{\dfrac{1}{4\cdot 4^2}}_{(2,2)}+\underbrace{\dfrac{2}{3\cdot 4^2}}_{(1,3)}+\underbrace{\dfrac{2}{6\cdot 5^2}}_{(2,3)}+\underbrace{\dfrac{1}{9\cdot 6^2}}_{(3,3)}+\ldots
$$
so you must have used a summation I cannot see. And the result is a different one.

The trick is indeed to rewrite the summands. My proof uses integrals.
 
11. Choose any odd prime, square it and subtract one. Show that the result is always divisible by twenty-four except for three.
What can be said, if we take the prime up to the power four, and subtract one?
Any odd prime ##p## other than 3 can we written as ##p = (2k + 3)## where ##k## is a positive integer and also meets the criterion ##k \neq 0\ mod\ 3##.

Let ##A## denote ##p^2 - 1##. ##A = p^2 - 1 = (p - 1)(p + 1) = (2k + 2)(2k + 4) = 4(k + 1)(k + 2)##. We now show that ##(k + 1)(k + 2)## must be a multiple of 6. ##(k + 1)(k + 2)## must be an even number since it is a multiple of 2 consecutive integers. If ##(k + 1)## is odd then ##(k + 2)## must be even and if ##(k + 1)## is even then ##(k + 1)## must be odd. Therefore ##(k + 1)(k + 2) \equiv 0\ mod\ 2##, i.e. it is a multiple of 2.

Secondly, we show that ##(k + 1)(k + 2)## must be a multiple of 3. ##k \neq 0\ mod\ 3## (as stated earlier) ##\Rightarrow k \equiv 1\ mod\ 3## or ##k \equiv 2\ mod\ 3 \Rightarrow (k + 1)(k + 2) \equiv (2\ mod\ 3)(3\ mod\ 3) = 0\ mod\ 3## or ##(3\ mod\ 3)(4\ mod\ 3) = 0\ mod\ 3##.

Combining the above 2 observations, it follows that ##(k + 1)(k + 2) = 0\ mod\ 6##, i.e. ##(k + 1)(k + 2)## is a multiple of 6. Therefore, ##4(k + 1)(k + 2)## must be a multiple of ##4 \times 6 = 24##. Hence proved.

For the second part of the question, we need to consider ##B = p^4 - 1 = (p^2 - 1)(p^2 + 1)##. It has already been proved that ## (p^2 - 1)## is a multiple of 24. Now, since ##p## is an odd number, ##(p^2 + 1)## must be ab even number, i.e. a multiple of 2. Therefore ##p^4 - 1## must be a multiple of ##24 \times 2 = 48##
 
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Any odd prime ##p## other than 3 can we written as ##p = (2k + 3)## where ##k## is a positive integer and also meets the criterion ##k \neq 0\ mod\ 3##.

Let ##A## denote ##p^2 - 1##. ##A = p^2 - 1 = (p - 1)(p + 1) = (2k + 2)(2k + 4) = 4(k + 1)(k + 2)##. We now show that ##(k + 1)(k + 2)## must be a multiple of 6. ##(k + 1)(k + 2)## must be an even number since it is a multiple of 2 consecutive integers. If ##(k + 1)## is odd then ##(k + 2)## must be even and if ##(k + 1)## is even then ##(k + 1)## must be odd. Therefore ##(k + 1)(k + 2) \equiv 0\ mod\ 2##, i.e. it is a multiple of 2.

Secondly, we show that ##(k + 1)(k + 2)## must be a multiple of 3. ##k \neq 0\ mod\ 3## (as stated earlier) ##\Rightarrow k \equiv 1\ mod\ 3## or ##k \equiv 2\ mod\ 3 \Rightarrow (k + 1)(k + 2) \equiv (2\ mod\ 3)(3\ mod\ 3) = 0\ mod\ 3## or ##(3\ mod\ 3)(4\ mod\ 3) = 0\ mod\ 3##.

Combining the above 2 observations, it follows that ##(k + 1)(k + 2) = 0\ mod\ 6##, i.e. ##(k + 1)(k + 2)## is a multiple of 6. Therefore, ##4(k + 1)(k + 2)## must be a multiple of ##4 \times 6 = 24##. Hence proved.

For the second part of the question, we need to consider ##B = p^4 - 1 = (p^2 - 1)(p^2 + 1)##. It has already been proved that ## (p^2 - 1)## is a multiple of 24. Now, since ##p## is an odd number, ##(p^2 + 1)## must be ab even number, i.e. a multiple of 2. Therefore ##p^4 - 1## must be a multiple of ##24 \times 2 = 48##
Yes, correct.

But ##48\,|\,p^4-1## can be made a bit stronger. Can someone show that in fact ##240\,|\,p^4-1## for ##p>5##?
 
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5. Let ##f : [1,\infty) \longrightarrow [0,\infty)## be a continuously differentiable function. Write ##S## for the solid of revolution of the graph ##y=f(x)## about the ##x##-axis. If the surface area of ##S## is finite, then so is the volume.
The solid of revolution about the ##x##-axis:
$$S=\pi\int_1^\infty f^2(x)dx$$
The surface area ##S##:
$$A=2\pi\int_1^\infty f(x)\sqrt{1+(f'(x))^2}dx$$
For ##A## to converge, ##\lim_{x\to\infty} f(x)## needs to be ##0## (if the limit is finite but non-zero, the limit of ##f'(x)## would be ##0##, but the limit of ##f(x)\sqrt{1+(f'(x))^2}## would have the limit of ##f(x)## and the integral will diverge). It follows that the limit of ##f'(x)=0## as ##x\to\infty## (since ##f(x)## is becoming constant).
We then have:
$$f(x)\sqrt{1+(f'(x))^2}dx\underset{\infty}{\sim}f(x)dx\Rightarrow \int_0^\infty f(x)dx \text{ converges as A converges.}$$
Since ##f(x)## is going to ##0##, then at some point ##f^2(x)\leq f(x)##, thus ##\int_1^\infty f^2(x)dx \leq \int_1^\infty f(x)dx## from which follows that if ##A## converges, so does ##S##.
 
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The solid of revolution about the ##x##-axis:
$$S=\pi\int_1^\infty f^2(x)dx$$
The surface area ##S##:
$$A=2\pi\int_1^\infty f(x)\sqrt{1+(f'(x))^2}dx$$
For ##A## to converge, ##\lim_{x\to\infty} f(x)## needs to be ##0## (if the limit is finite but non-zero, the limit of ##f'(x)## would be ##0##, but the limit of ##f(x)\sqrt{1+(f'(x))^2}## would have the limit of ##f(x)## and the integral will diverge). It follows that the limit of ##f'(x)=0## as ##x\to\infty## (since ##f(x)## is becoming constant).
We then have:
$$f(x)\sqrt{1+(f'(x))^2}dx\underset{\infty}{\sim}f(x)dx\Rightarrow \int_0^\infty f(x)dx \text{ converges as A converges.}$$
Since ##f(x)## is going to ##0##, then at some point ##f^2(x)\leq f(x)##, thus ##\int_1^\infty f^2(x)dx \leq \int_1^\infty f(x)dx## from which follows that if ##A## converges, so does ##S##.
Yes, correct. The object is known as Gabriel's Horn, where ##f(x)=1/x##.

A bit more formalized:
\begin{align*}
\lim_{t \to \infty} \sup_{x \geq t}f(x)^2-f(1) &= \limsup_{t \to \infty} \int_1^t \left( f(x)^2 \right)'\,dx\\
&\leq \int_1^\infty \left| \left( f(x)^2 \right)' \,\right|\,dx\\
&=2 \int_1^\infty f(x)|f'(x)|\,dx\\
&\leq 2 \int_1^\infty f(x)\sqrt{1+f'(x)^2}\,dx\\
&=\dfrac{A}{\pi} < \infty
\end{align*}
Hence there is a ##t_0\geq 1## such that ##\sup_{x\geq t_0}f(x) < \infty## and so is ##L:=\sup_{x\geq 1}f(x) < \infty ## because ##f(x)## is continuous with values in ##[0,\infty)##, i.e. bounded on ##[1,\infty)##. For the volume we have
\begin{align*}
V&=\int_1^\infty f(x)\cdot \pi f(x)\, dx \\ &\leq \int_1^\infty \dfrac{L}{2}\cdot 2\pi f(x)\,dx \\ &\leq \dfrac{L}{2}\int_1^\infty 2\pi f(x) \sqrt{1+f'(x)^2}\,dx \\ &= \dfrac{L}{2}\cdot A \\ &< \infty
\end{align*}
 
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The trick is indeed to rewrite the summands. My proof uses integrals.
Why do you need integrals? Is it wrong to say that, since they take the same value each time, we can replace all ##k##s and ##j##s by ##n##?
$$\sum_{k,\,j=1}^\infty\dfrac{1}{kj(k+j)^2}=\sum_{n=1}^\infty\dfrac{1}{nn(n+n)^2}$$
 
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Why do you need integrals? Is it wrong to say that, since they take the same value each time, we can replace all ##k##s and ##j##s by ##n##?
$$\sum_{k,\,j=1}^\infty\dfrac{1}{kj(k+j)^2}=\sum_{n=1}^\infty\dfrac{1}{nn(n+n)^2}$$
This equation you wrote is wrong (where has ##m## gone to?), the sum has to be done across both indices, so it's a Cauchy product. We sum up all combinations: ##(1,1),(1,2),(2,1),(2,2),(1,3),(3,1),(2,3),(3,2),(3,3), \ldots ## I don't know whether we need integrals, my solution uses them to get rid of the ##(k+j)## terms.
 
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This equation you wrote is wrong (where has ##m## gone to?), the sum has to be done across both indices, so it's a Cauchy product. We sum up all combinations: ##(1,1),(1,2),(2,1),(2,2),(1,3),(3,1),(2,3),(3,2),(3,3), \ldots ## I don't know whether we need integrals, my solution uses them to get rid of the ##(k+j)## terms.
Which ##m##?
l.PNG

But from what you say I guess you meant ##\sum_{k=1}^\infty\sum_{j=1}^\infty \dfrac{1}{kj(k+j)^2}##?
 
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(The ##m## in your formula on the RHS.)

Yes, ##\sum_{k,j=1}^\infty## is short for ##\sum_{k=1}^\infty \sum_{j=1}^\infty##.
 
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(The ##m## in your formula on the RHS.)

Yes, ##\sum_{k,j=1}^\infty## is short for ##\sum_{k=1}^\infty \sum_{j=1}^\infty##.
I was sceptic at first but maybe you are on a mobile phone? It's ##n*n## :D
 
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@Fred Wright considered the sum as I did, which simplifies to his expression.
 
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I was sceptic at first but maybe you are on a mobile phone? It's ##n*n## :D
Oh, I see. No, that's wrong. The summation involves all mixed terms, too. The shortest notation is ##\sum_{k,j}## but I added from ##1## to ##\infty ## (which refers to both).
 
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13. 3 trucks? since three of the gifts weight no more than 3 tons
 
12. In a square of side length 4, there is a circle of radius 1 in each corner. In the center of the square is another circle that touches the other four. Analogously, in the three-dimensional case, in the center of a cube of edge length 4, there would be a sphere which would touch eight spheres of radius 1 placed in the corners of the cube. In which dimension does the central hypersphere become so large that it touches all sides of the hypercube?
Is one-dimensional case to be considered valid for this question? In the 1-D case, instead of a square, we will have a line segment of length 4 on the X-axis (or the only axis in 1-D). And instead of the circles of the 2-D case, we will have 2 pairs of points, ##(X=x_1, X=x_2)## and ##(X=x_2, X=x_3)## such that ##x_2 - x_1 = 2## and ##x_3 - x_2 = 2##. And the 1-D point ##(X=x_2)## in the 1-D case is analogous to the central circle/hypersphere of higher dimensional cases. And this point happens to lie on the line segment ##(X=x_1)##-to-##(X=x_3)## on the X-axis and this line segment is the 1-D equivalent of an edge of square/hypercube in higher dimensions, thus meeting the criteria of central hypersphere touching all sides of the hypercube but admittedly it does not really meet the condition of being "large".
 

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