Math Challenge - July 2020

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1. (solved by @nuuskur ) Let ##V## be an infinite dimensional topological vector space. Show that the weak topology on ##V## is not induced by a norm. (MQ)

2. The matrix groups ##U(n)## and ##SL_n(\mathbb{C})## are submanifolds of ##\mathbb{C}^{n^2}=\mathbb{R}^{2n^2}##. Do they intersect transversely? (IR)

3. (solved by @zinq ) Let ##f:\mathbb{R}\to \mathbb{R},\quad f>0## be a continuous and ##1-##periodic function. Show that
$$\int_0^1\frac{f(x+a)}{f(x)}dx\ge 1$$ for any ##a\in\mathbb{R}##. (WR)

4. (solved by @nuuskur ) Prove that ##C[0,1]## is not dual to a Banach space. (WR)

5. (solved by @julian , @mathwonk ) Let ##(M,g)## be a Riemannian manifold. Let ##f:M\to\mathbb{R}## be a smooth function such that ##|\nabla f|=1## everywhere on ##M##. Show that all integral curves of ##\nabla f## are geodesics. (IR)

6. (solved by @Incand ) Let ##f: (a,b) \to \mathbb{R}## be a continuous function that is midpoint convex, i.e. ##f(\frac{1}{2}x + \frac{1}{2}y) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)## for all ##x,y \in (a,b)## . Show that ##f## is convex, i.e. ##f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)## for all ##x,y \in (a,b)## and ##0 \leq t \leq 1##. (MQ)

7. (solved by @nuuskur ) Consider ##D:=\{f: \mathbb{R} \to \mathbb{R}\mid \mathrm{\ f \ is \ not \ continuous}\}##. What is the cardinality of ##D##? (MQ)

8. (solved by @zinq ) Let ##X## be a compact manifold such that ##\pi_1(X)## (the fundamental group of ##X##) is finite and nontrivial. Show that ##\pi_k(X)## is also non-trivial for some ##k\geq 2.## (IR)

9. (solved by @mathwonk ) Let ##(X,d)## be a compact metric space and ##f:X\to X## be a mapping onto. Assume that ##d(f(x),f(y))\le d(x,y),\quad \forall x,y\in X.## Show that ##d(f(x),f(y))= d(x,y),\quad \forall x,y\in X.## (WR)

10. (solved by @etotheipi ) Calculate the electrostatic potential ##U(a)## of a surface ##S=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=z^2,\,0\leq z\leq 1\,\}## charged with a field of homogeneous density ##\rho## at the point ##a=(0,0,1)##. (FR)



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11.
(solved by @Isaac0427 , @ItsukaKitto ) Prove that the product of a finite number of sums of two integers squares is again a sum of two integers squared.
$$
(a_1^2+b_1^2)\cdot (a_2^2+b_2^2)\cdot \ldots \cdot (a_n^2+b_n^2)=a^2+b^2
$$

12. (solved by @etotheipi ) Given a positive integer in decimal representation without zeros. We build a new integer by concatenation of the number of even digits, the number of odd digits, and the number of all digits (the sum of the former two). Then we proceed with that number.
Determine whether this algorithm always comes to a halt. What is or should be the criterion to stop?

13. (solved by @Lament ) List all real functions ##f\, : \,\mathbb{R}\longrightarrow \mathbb{R}## with the following properties:
\begin{align*}
f(xy)&=f(x)f(y)-f(x)-f(y)+2\\
f(x+y)&=f(x)+f(y)+2xy-1\\
f(1)&=2
\end{align*}

14. (solved by @ItsukaKitto ) Find all real solutions ##(x,y)## such that
$$
\sin^4x = y^4+x^2y^2-4y^2+4\, , \,\cos^4x=x^4+x^2y^2-4x^2+1
$$

15. (solved by @etotheipi , @Adesh) Prove
$$
\dfrac{(2n)!}{(n!)^2}>\dfrac{4^n}{n+1}
$$
for all natural numbers ##n>1.##
 
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  • #2
The statement holds for ##n=2##, now assume for ##n=k## that$$\dfrac{(2k)!}{(k!)^2}>\dfrac{4^k}{k+1}$$Then for ##n=k+1##,$$\dfrac{(2(k+1))!}{((k+1)!)^2} = \frac{2(2k+1)}{k+1} \dfrac{(2k)!}{(k!)^2} > \frac{2(2k+1)}{k+1} \dfrac{4^k}{k+1} = \frac{(2\times 4^k)(2k+1)}{(k+1)^2}$$Now consider the expression $$\frac{(2\times 4^k)(2k+1)}{(k+1)^2} - \frac{4^{k+1}}{k+2} = 2 \times 4^k \left (\frac{2k+1}{(k+1)^2} - \frac{2}{k+2} \right) = 2 \times 4^k \left( \frac{k}{(k+1)^2 (k+2)} \right) > 0$$ since ##k>1##. That means that $$\dfrac{(2(k+1))!}{((k+1)!)^2} > \frac{4^{k+1}}{(k+1)+1}$$so if the statement is true for ##n=k## it also holds for ##n=k+1##. Hence the statement is true for all ##n \in \mathbb{N}##
 
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  • #3
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We study the inequality
\begin{equation}
f(tx+(1-t)y) \le tf(x)+(1-t)f(y).
\end{equation}
Take points ##x,y \in (a,b)##. We know the inequality is satisfied for ##t=1/2##.

By using the midpoint inequality for the points ##x,y,(x+y)/2## we also get that the inequality is true for ##t=1/4,3/4##. Proceeding iteratively this way we get that the inequality is true for any ##t## of the form
\begin{equation}
t = \frac{m}{2^n},
\end{equation}
for ##m,n\in \mathbb{Z}^+## with ##m\le 2^n##. (Cf. binary search.)

Next we show that the fractions of the above form are dense in ##[0,1]##. For any ##r\in [0,1]## and ##n\in \mathbb{Z}^+## we have with an appropriate choice of ##m## that
\begin{equation}
|r-\frac{m}{2^n}| \le \frac{1}{2^n},
\end{equation}
which we can make arbitrary small by increasing ##n##.

Since ##\mathbb{R}## is Hausdorff, ##f## is determined by its valued on a dense subset. This forces the inequality for all ##t\in [0,1]##.
 
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  • #4
We study the inequality
\begin{equation}
f(tx+(1-t)y) \le tf(x)+(1-t)f(y).
\end{equation}
Take points ##x,y \in (a,b)##. We know the inequality is satisfied for ##t=1/2##.

By using the midpoint inequality for the points ##x,y,(x+y)/2## we also get that the inequality is true for ##t=1/4,3/4##. Proceeding iteratively this way we get that the inequality is true for any ##t## of the form
\begin{equation}
t = \frac{m}{2^n},
\end{equation}
for ##m,n\in \mathbb{Z}^+## with ##m\le 2^n##. (Cf. binary search.)

Next we show that the fractions of the above form are dense in ##[0,1]##. For any ##r\in [0,1]## and ##n\in \mathbb{Z}^+## we have with an appropriate choice of ##m## that
\begin{equation}
|r-\frac{m}{2^n}| \le \frac{1}{2^n},
\end{equation}
which we can make arbitrary small by increasing ##n##.

Since ##\mathbb{R}## is Hausdorff, ##f## is determined by its valued on a dense subset. This forces the inequality for all ##t\in [0,1]##.

Yes! Well done! The key is indeed to realize that the so called dyadic rationals
$$D:= \left\{\frac{m}{2^n}\mid 0 \leq n; 0 \leq m \leq 2^n\right\}$$ are dense in ##[0,1]##.
 
  • #5
Summary:: Functional Analysis, Topology, Differential Geometry, Analysis, Physics
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

High Schoolers only
I just graduated high school...do I count?
 
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  • #6
Delta2
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I find the formulation of 10. a bit strange. @fresh_42 Does it mean to find the potential U(a) at point a, created by the given surface that is charged with a uniform surface charge density ##\rho##?
Also for 2. what's the definition of 1-periodic.
 
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  • #7
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Also for 2. what's the definition of 1-periodic.
I confirmed it with some people and they said it means a function with period as ##1##, that is ##f(x+1)=f(x)##.
 
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  • #8
I just graduated high school...do I count?

Yes. Definitely.
 
  • #9
I'm pretty sure this is massively incorrect, but in any case I'll write it up.

A ring-shaped surface element on the cone at a height ##z## has charge ##\rho dA = 2\sqrt{2} \pi \rho z dz##, and the distance ##d## from this element to the point ##a## is ##d = \sqrt{2z^2 -2z + 1}##. That means $$dU = \frac{2\sqrt{2} \pi \rho z}{4\pi \epsilon_0 \sqrt{2z^2 -2z + 1}}dz$$ $$U = \frac{\rho}{2 \epsilon_0} \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$Then we let ##u = z-\frac{1}{2}##, so that$$\begin{align*}

U = \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du &= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}} \\

&= \frac{\rho}{2\epsilon_0} \text{arsinh}(1)

\end{align*}
$$
 
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  • #10
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I'm pretty sure this is massively incorrect, but in any case I'll write it up.

A ring-shaped surface element on the cone at a height ##z## has charge ##\rho dA = 2\sqrt{2} \pi \rho z dz##, and the distance ##d## from this element to the point ##a## is ##d = \sqrt{2z^2 -2z + 1}##. That means $$dU = \frac{2\sqrt{2} \pi \rho z}{4\pi \epsilon_0 \sqrt{2z^2 -2z + 1}}dz$$ $$U = \frac{\rho}{2 \epsilon_0} \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$Then we let ##u = z-\frac{1}{2}##, so that$$\begin{align*}

U = \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du &= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}} \\

&= \frac{\rho}{2\epsilon_0} \text{arsinh}(1)

\end{align*}
$$
I consider the density includes permittivity, but this isn't important. What is your result?
 
  • #11
I consider the density includes permittivity, but this isn't important. What is your result?

If the density includes the permittivity, then I get about ##U = \frac{1}{2}\text{arsinh}(1)\rho \approx 0.441 \rho##...
 
  • #12
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If the density includes the permittivity, then I get about ##U = \frac{1}{2}\text{arsinh}(1)\rho \approx 0.441 \rho##...
I meant the exact real number. Your potential integral looks ok, modulo constant factors, but what is ##x## in ##U(a)=x\cdot \rho\;##?

##0.4...## is wrong, but as you didn't show anything I cannot tell where or whether you were mistaken anywhere. It looks as if you failed on the lower bound of the integral.
 
  • #13
I meant the exact real number. Your potential integral looks ok, modulo constant factors, but what is ##x## in ##U(a)=x\cdot \rho\;##?

##0.4...## is wrong, but as you didn't show anything I cannot tell where or whether you were mistaken anywhere.

I guess in that case I would have ##x = \frac{\ln{(1+\sqrt{2}})}{2}##... it's very likely I've messed up so I'll have another look. Numerically that's the same as what I get if I calculate the first integral involving ##z##
 
  • #14
fresh_42
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I guess in that case I would have ##x = \frac{\ln{(1+\sqrt{2}})}{2}##... it's very likely I've messed up so I'll have another look.
Close ... but wrong denominator and missing ##\pi##.

I hope I didn't make the mistake. But where did you get the zero from in the lower bound?
 
  • #15
and missing ##\pi##.

That is strange... for me the ##\pi##'s cancel :nb)

I hope I didn't make the mistake. But where did you get the zero from in the lower bound?

For that I just used the fact that the second term in the integrand is an even function.
 
  • #16
fresh_42
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That is strange... for me the ##\pi##'s cancel :nb)



For that I just used the fact that the second term in the integrand is an even function.
The ##\pi## comes in by the parametrization of the surface. O.k. it may cancel depending on the definition of ##\rho##. I used the formula
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$
##\operatorname{arsinh}## isn't even and I haven't your second term at all. But as you didn't show us what you have done, we discuss it here instead.
 
  • #17
##\operatorname{arsinh}## isn't even.
I meant$$\begin{align*}
U &= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}}

\end{align*}
$$because ##\frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}}## is even and ##\frac{u}{\sqrt{u^2 + \frac{1}{4}}}## is odd.
 
  • #18
fresh_42
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I have ##U(a)=2\,\pi\rho\,\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt \,,## too, but then we differ. Maybe I used a different split than you. I have no idea what you did between the above integral and the ##u-##substitution.
 
  • #19
fresh_42
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I meant$$\begin{align*}
U &= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{u}{\sqrt{u^2 + \frac{1}{4}}} + \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2 \epsilon_0} \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{u^2 + \frac{1}{4}}} \, du \\

&= \frac{\rho}{2\epsilon_0} \left[\text{arsinh}({2u}) \right]_{0}^{\frac{1}{2}}

\end{align*}
$$because ##\frac{\frac{1}{2}}{\sqrt{u^2 + \frac{1}{4}}}## is even and ##\frac{u}{\sqrt{u^2 + \frac{1}{4}}}## is odd.
As you basically solved it and we are only discussing the integration, would you agree if I gave you the credits and post my calculation instead, so that people can read the solution in one step instead of spread all over the place?
 
  • #20
As you basically solved it and we are only discussing the integration, would you agree if I gave you the credits and post my calculation instead, so that people can read the solution in one step instead of spread all over the place?

I don't mind 😁, but I think we agree on the integration of ##\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt##, it's just that your leading factor was ##2\pi \rho## and mine was ##\frac{\rho}{2}## (or with the ##\epsilon_0## added in, ##\frac{\rho}{2\epsilon_0}##).
 
  • #21
fresh_42
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The general formula (Coulomb) for the potential is
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$
With the parameterization ##\Phi(t,\varphi)=(t\cos\varphi,t\sin\varphi,t)## we get ##\Phi_t=(\cos \varphi,\sin \varphi ,1)## and ##\Phi_\varphi=(-t\sin \varphi,t\cos\varphi,0)##. The fundamental quantities are $$E=\Phi_t\cdot \Phi_t\, , \,F=\Phi_t\cdot \Phi_\varphi\, , \,G=\Phi_\varphi \cdot \Phi_\varphi$$
These are in our case ##E=2,\,F=0,\,G=t^2## and the scalar surface element is ##dO=\sqrt{EG-F^2}\,dt\,d\varphi= \sqrt{2}t\,dt\,d\varphi## since ##t\geq 0.## Thus
\begin{align*}
U(a)&=\int_0^1 \int_0^{2\pi} \dfrac{\rho}{\sqrt{(t\cos \varphi - 0)^2+(t\sin\varphi-0)^2+(t-1)^2}} \sqrt{2}t\,dt\,d\varphi \\
&=\int_0^1 dt\int_0^{2\pi}d\varphi\, \,\dfrac{\sqrt{2}\rho\, t}{\sqrt{t^2+(t-1)^2}}=2\,\pi\rho\,\int_0^1 \dfrac{t}{\sqrt{t^2-t+\frac{1}{2}}}\,dt\\
&=2\,\pi\rho\,\int_0^1 \dfrac{1}{2}\left(\dfrac{2t-1}{\sqrt{t^2-t+\frac{1}{2}}} + \dfrac{1}{2\sqrt{t^2-t+\frac{1}{2}}}\right)\\
&=\pi\rho \left[\sqrt{t^2-t+\frac{1}{2}}\right]_0^1 +\pi\rho \int_0^1\dfrac{dt}{\sqrt{t^2-t+\frac{1}{2}}}\\
&\stackrel{\tau=t-1/2}{=}\pi\rho \int_{-1/2}^{1/2}\dfrac{d\tau}{\sqrt{\tau^2+\frac{1}{4}}}=\pi\rho \int_{-1/2}^{1/2}\dfrac{d(2\tau)}{\sqrt{(2\tau)^2+1}}\\
&=\pi\rho \left[\operatorname{arsinh}(2\tau)\right]_{-1/2}^{1/2}=\pi\rho \left[\log\left(2\tau + \sqrt{(2\tau)^2+1}\right)\right]_{-1/2}^{1/2}\\
&=\pi\rho \log\left(\dfrac{1+\sqrt{2}}{-1+\sqrt{2}}\right)=(\log(3+2\sqrt{2}))\pi\rho
\end{align*}
 
  • #22
The general formula (Coulomb) for the potential is
$$
U(a)=\int\int_S \dfrac{\rho}{|x-a|}\,dO
$$

I think it is because I am using the form of Coulomb's law equivalent to $$U(a) = \int \int_S \frac{\rho}{4 \pi \epsilon_0 |x - a|} dO$$That would explain the factor of ##4## between our answers, in addition to the un-cancelled ##\pi## in yours.
 
  • #23
fresh_42
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I think it is because I am using the form of Coulomb's law equivalent to $$U(a) = \int \int_S \frac{\rho}{4 \pi \epsilon_0 |x - a|} dO$$That would explain the factor of ##4## between our answers, in addition to the un-cancelled ##\pi## in yours.
Sure, but I think you've made an integration error anyway. You skipped what was my lines 3 and 4 in the above calculation and that's where the problems with your solution started. As a result we had different lower limits in the final term.
 
  • #24
Sure, but I think you've made an integration error anyway. You skipped what was my lines 3 and 4 in the above calculation and that's where the problems with your solution started. As a result we had different lower limits in the final term.

When I do $$I = \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$ on my calculator (which you have in line 2), I get ##I = 0.881##, which is the same as ##\text{arsinh}(1)## in my answer.

And the ratio of our answers is $$\frac{\ln{(3+2\sqrt{2})}\pi \rho}{\frac{\rho}{2\epsilon_0} \text{arsinh}(1)} = 4 \pi \epsilon_0$$So I think everything just comes down to the form of Coulomb's law.
 
  • #25
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When I do $$I = \int_0^1 \frac{z}{\sqrt{z^2 -z+\frac{1}{2}}} dz$$ on my calculator (which you have in line 2), I get ##I = 0.881##, which is the same as ##\text{arsinh}(1)## in my answer.

And the ratio of our answers is $$\frac{\ln{(3+2\sqrt{2})}\pi \rho}{\frac{\rho}{2\epsilon_0} \text{arsinh}(1)} = 4 \pi \epsilon_0$$So I think everything just comes down to the form of Coulomb's law.
I concentrated on the difference between my ##\operatorname{arsinh}(1)-\operatorname{arsinh}(-1)## and your ##\operatorname{arsinh}(1)-0##. But as ##\operatorname{arsinh}## is an odd function, this difference could well be hidden in a factor ##2## and we are debating about literally nothing. A typical confusion online versus us in front of a blackboard with a piece of chalk in hand each.
 
  • #26
Yes, I probably should have shown more steps given that I had quite a lot of ##2##'s flying around. Effectively I just changed ##\text{arsinh}(1) - \text{arsinh}(-1) \rightarrow 2\text{arsinh}(1)##.
and we are debating about literally nothing.
It's beginning to look to me too that this is the case 😁
 
  • #27
PeroK
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Re question 10. If we use SI units first. For a ring on the surface at neight ##z## we have:
$$dq = 2\pi R \rho dl = 2\pi z \rho (\sqrt 2 dz)$$
Where ##R = z## is the radius of a ring and ##dl = \sqrt 2 dz## is the length of the ring along the surface. Therefore, we have:
$$V = \int \frac{dq}{4\pi \epsilon_0 r} = \int_0^1 \frac{2\pi z \rho (\sqrt 2 dz)}{4\pi \epsilon_0 \sqrt{z^2 + (1-z)^2}} = \frac{\rho}{2\epsilon_0}\int_0^1 \frac{z dz}{\sqrt{(z - \frac 1 2)^2 + \frac 1 4}} $$
Using ##u = z - \frac 1 2##, we get:
$$V = \frac{\rho}{2\epsilon_0}\int_{- \frac 1 2}^{\frac 1 2} \frac{(u + \frac 1 2) du}{\sqrt{u^2 + \frac 1 4}} = \frac{\rho}{2\epsilon_0}\int_{0}^{\frac 1 2} \frac{du}{\sqrt{u^2 + \frac 1 4}} $$
The first term is an odd function, so the integral vanished and the second is even. This gives:
$$V = \frac{\rho}{2\epsilon_0}[\ln(\frac 1 2 + \sqrt{\frac 1 2}) - \ln(\frac 1 2)] = \frac{\rho}{2\epsilon_0}[\ln(1 + \sqrt 2)]$$
The answer in SI and cgs units should be:
$$V_{SI} = \frac{\rho}{2\epsilon_0}[\ln(1 + \sqrt 2)]$$
$$V_{cgs} = 2\pi \rho[\ln(1 + \sqrt 2)]$$
 
  • #28
This may or may not be the intended solution, but it's the first thing that came to mind when I saw the problem. Let's factor every term on the LHS:
$$(a_1^2+b_1^2)(a_2^2+b_2^2)(...)(a_n^2+b_n^2)=(a_1+b_1i)(a_1-b_1i)(a_2+b_2i)(a_2-b_2i)(...)(a_n+b_ni)(a_n-b_ni).$$
We can now regroup these terms in the following way:
$$(a_1+b_1i)(a_2+b_2i)(...)(a_n+b_ni)*(a_1-b_1i)(a_2-b_2i)(...)(a_n-b_ni).$$
The group of terms on the left is going to be some complex number with integer real and imaginary parts (assuming all ##a_n## and ##b_n## are integers). We will call this new complex number a+bi. But, if the group of terms on the left is a+bi, then the group of terms on the right will be a-bi, due to the property of complex conjugates
$$\left(z_1*z_2*...*z_n\right)^*=z_1^**z_2^**...*z_n^*.$$
Thus, our expression reduces to (a+bi)(a-bi) which is equal to ##a^2+b^2##.
 
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  • #29
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Uhh, been a while since I did cardinal arithmetic.. here goes..
Firstly, [itex]|\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|[/itex]. Indeed, since [itex]|\mathbb R| = |\mathbb R^2|[/itex] we have [itex]|\mathcal P(\mathbb R)| = |\mathcal P(\mathbb R^2)|[/itex]. Given [itex]f:\mathbb R\to\mathbb R[/itex], it is, as a relation, a subset
[tex]
\{(x,f(x)) \mid x\in\mathbb R\} \subseteq \mathbb R^2.
[/tex]
On the other hand, given [itex]F\subseteq \mathbb R[/itex], we can map it to its characteristic function [itex]\chi _F[/itex], so we have a bijection and thus [itex]|\mathcal P(\mathbb R)| = |2^{\mathbb R}|[/itex]. Certainly the [itex]\chi _F\in\mathbb R^{\mathbb R}[/itex], so we have
[tex]
|\mathcal P(\mathbb R)|=|2^{\mathbb R}|\leq |\mathbb R^{\mathbb R}| \leq \mathcal P(\mathbb R).
[/tex]
By Cantor-Bernstein-Schroeder, we have [itex]|\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|[/itex].

Let [itex]C[/itex] be the subset of continuous maps. On the one hand [itex]|\mathbb R| \leq |C| [/itex] since mapping to constant functions is injective. On the other hand, if two continuous maps coincide on [itex]\mathbb Q[/itex], then they coincide on [itex]\mathbb R[/itex], thus mapping [itex]f\mapsto f\vert_{\mathbb Q}[/itex] is injective. Thus, [itex]|C| \leq |\mathbb R^{\mathbb Q}| = |\mathbb R|[/itex]. Again, by CBS, [itex]|C| = |\mathbb R|[/itex].
By the axiom of choice, we have for infinite cardinals [itex]\kappa + \lambda = \max\{\kappa,\lambda\}[/itex], thus
[tex]
|\mathcal P(\mathbb R)| = |\mathbb R^{\mathbb R} \setminus C|+|C| = \max\{|\mathbb R^{\mathbb R} \setminus C|,|C|\}.
[/tex]
By Cantor's theorem [itex]|\mathbb R| < |\mathcal P(\mathbb R)|[/itex], so it must be [itex]|D|=|\mathbb R^{\mathbb R} \setminus C| = |\mathcal P(\mathbb R)|[/itex].
 
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  • #30
Uhh, been a while since I did cardinal arithmetic.. here goes..
Firstly, [itex]|\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|[/itex]. Indeed, since [itex]|\mathbb R| = |\mathbb R^2|[/itex] we have [itex]|\mathcal P(\mathbb R)| = |\mathcal P(\mathbb R^2)|[/itex]. Given [itex]f:\mathbb R\to\mathbb R[/itex], it is, as a relation, a subset
[tex]
\{(x,f(x)) \mid x\in\mathbb R\} \subseteq \mathbb R^2.
[/tex]
On the other hand, given [itex]F\subseteq \mathbb R[/itex], we can map it to its characteristic function [itex]\chi _F[/itex], so we have a bijection and thus [itex]|\mathcal P(\mathbb R)| = |2^{\mathbb R}|[/itex]. Certainly the [itex]\chi _F\in\mathbb R^{\mathbb R}[/itex], so we have
[tex]
|\mathcal P(\mathbb R)|=|2^{\mathbb R}|\leq |\mathbb R^{\mathbb R}| \leq \mathcal P(\mathbb R).
[/tex]
By Cantor-Bernstein-Schroeder, we have [itex]|\mathbb R^{\mathbb R}| = |\mathcal P(\mathbb R)|[/itex].

Let [itex]C[/itex] be the subset of continuous maps. On the one hand [itex]|\mathbb R| \leq |C| [/itex] since mapping to constant functions is injective. On the other hand, if two continuous maps coincide on [itex]\mathbb Q[/itex], then they coincide on [itex]\mathbb R[/itex], thus mapping [itex]f\mapsto f\vert_{\mathbb Q}[/itex] is injective. Thus, [itex]|C| \leq |\mathbb R^{\mathbb Q}| = |\mathbb R|[/itex]. Again, by CBS, [itex]|C| = |\mathbb R|[/itex].
By the axiom of choice, we have for infinite cardinals [itex]\kappa + \lambda = \max\{\kappa,\lambda\}[/itex], thus
[tex]
|\mathcal P(\mathbb R)| = |\mathbb R^{\mathbb R} \setminus C|+|C| = \max\{|\mathbb R^{\mathbb R} \setminus C|,|C|\}.
[/tex]
By Cantor's theorem [itex]|\mathbb R| < |\mathcal P(\mathbb R)|[/itex], so it must be [itex]|D|=|\mathbb R^{\mathbb R} \setminus C| = |\mathcal P(\mathbb R)|[/itex].

Yes, well done. To see that ##|\mathcal{P}(\mathbb{R})| = |\mathbb{R}^\mathbb{R}|##, one can also argue that
$$|\mathbb{R}^\mathbb{R}| = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_0. 2^{\aleph_0}} = 2^{2^{\aleph_0}}= |\mathcal{P}(\mathbb{R})|$$
where it was used that ## 2^{\aleph_0} = |\mathbb{R}|##.
 
  • #31
Adesh
735
188
@wrobel Сэр, я получаю равенство, а не неравенство, here is my attempt.

##f(x+a)## has period 1, ##f(x)## has also period 1, therefore ##g(x) = \frac{f(x+a)}{f(x)}## will have a period 1.

Trigonometric Fourier Series for ##g(x) = \frac{f(x+a)}{f(x)}## is
$$
\frac{f(x+a)}{f(x)} = a_0 \sum_{n=1}^{\infty} \left[ a_n \cos (2\pi n x) + b_n \sin (2\pi nx)\right]
$$
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = \int_{0}^{1} a_0 dx + 0 $$
$$\int_{0}^{1} \frac{f(x+a)}{f(x)} = a_0
$$
Fourier exponential series for ##f(x)## and ##f(x+a)##
$$
f(x+a) = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n (x+a) } = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n x} e^{i2\pi na}
$$
$$
f(x) = \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx}
$$
$$
g(x) = \frac{f(x+a)}{f(x)} = \frac{\sum_{n=-\infty}^{\infty} c_n e^{i2\pi nx} e^{i2\pi na} }
{ \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx} }
$$
$$
a_0 = \frac{c_0}{c'_0}
$$
##c_0## and ##c'_0## are the average values of ##f(x+a)## and ##f(x)##.
$$
c_0 = \int_{0}^{1} f(x+a) dx = \int_{a}^{1+a} f(u) du $$
##u = x+a##.
$$
c'_0 = \int_{0}^{1} f(x) dx
$$
Therefore, ##c_0 = c'_0##. Hence, ##a_0 = \frac{c_0}{c'_0} = 1##. So, finally we have
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = 1
$$

I don't know where the mistakes lies because of which I'm not getting an inequality.
 
  • #33
Adesh
735
188
here is a mistake:
$$a_0=\frac{c_0}{c_0'}$$
Okay, but what’s the mistake in that?
 
  • #34
@wrobel Сэр, я получаю равенство, а не неравенство, here is my attempt.

##f(x+a)## has period 1, ##f(x)## has also period 1, therefore ##g(x) = \frac{f(x+a)}{f(x)}## will have a period 1.

Trigonometric Fourier Series for ##g(x) = \frac{f(x+a)}{f(x)}## is
$$
\frac{f(x+a)}{f(x)} = a_0 \sum_{n=1}^{\infty} \left[ a_n \cos (2\pi n x) + b_n \sin (2\pi nx)\right]
$$
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = \int_{0}^{1} a_0 dx + 0 $$
$$\int_{0}^{1} \frac{f(x+a)}{f(x)} = a_0
$$
Fourier exponential series for ##f(x)## and ##f(x+a)##
$$
f(x+a) = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n (x+a) } = \sum_{n=-\infty}^{\infty} c_n e^{i 2\pi n x} e^{i2\pi na}
$$
$$
f(x) = \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx}
$$
$$
g(x) = \frac{f(x+a)}{f(x)} = \frac{\sum_{n=-\infty}^{\infty} c_n e^{i2\pi nx} e^{i2\pi na} }
{ \sum_{n=-\infty}^{\infty} c'_n e^{i2\pi nx} }
$$
$$
a_0 = \frac{c_0}{c'_0}
$$
##c_0## and ##c'_0## are the average values of ##f(x+a)## and ##f(x)##.
$$
c_0 = \int_{0}^{1} f(x+a) dx = \int_{a}^{1+a} f(u) du $$
##u = x+a##.
$$
c'_0 = \int_{0}^{1} f(x) dx
$$
Therefore, ##c_0 = c'_0##. Hence, ##a_0 = \frac{c_0}{c'_0} = 1##. So, finally we have
$$
\int_{0}^{1} \frac{f(x+a)}{f(x)} = 1
$$

I don't know where the mistakes lies because of which I'm not getting an inequality.

A more serious flaw is that not every continuous function has a (pointwise convergent) Fourier series so your argument does not work.
 
  • #35
wrobel
Science Advisor
Insights Author
997
859
Okay, but what’s the mistake in that?
take a function ##f(x)=2+\cos(2\pi x)## and try to prove this equality
 
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