# Math Challenge - October 2020

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Mentor
Summary:: Functional Analysis, Algebras, Measure Theory, Differential Geometry, Calculus, Optimization, Algorithm, Integration. Lie Algebras.

1. (solved by @julian ) Let ##(a_n)\subseteq\mathbb{R}## be a sequence of real numbers such that ##a_n \leq n^{-3}## for all ##n\in \mathbb{N}.## Given the family ##\mathcal{A}## of functions ##f_n\, : \,[0,1]\longrightarrow \mathbb{R}## defined by ##f_n(x)=\sum_{k=n}^\infty a_k\sin(kx)## for ##n\in \mathbb{N},## show that every sequence ##(g_n)\subseteq\mathcal{A}## contains a subsequence ##(g_{n_k})## which converges uniformly on ##[0,1].##

2. (a) Show that if a ##*##-algebra ##A## admits a complete ##C^*##-norm, then it is the only ##C^*##-norm on ##A##.
(b) Let ##A## be a ##*##-algebra. Show that there is a ##*##-isomorphism ##M_n(\Bbb{C}) \otimes A \cong M_n(A)##.
(c) Deduce that the ##C^*##-algebra ##M_n(\Bbb{C})## is nuclear for all ##n \geq 1##. (MQ)

3. Give an example of a Riemann integrable function that is not Borel measurable. (MQ)

4. Let ##C## be the Cantor set. Show that ##\frac{1}{2}C + \frac{1}{2}C = [0,1]##. Deduce that the sum of sets of measure ##0## must not have measure ##0##. (MQ)

5. Let ##\pi\, : \,\mathbb{R}^n\longrightarrow \mathbb{T}^n## be the canonical projection and ##f:=\pi|_{[0,1]^n}## its restriction on the closed unit cube. Show with the help of ##f\, : \,[0,1]^n\longrightarrow\mathbb{T}^n##, that a quotient map in general doesn't have to be open.

6. Let ##D=\{\,z\in \mathbb{C}\, : \,|z|<1\,\}## be the complex open unit disk and let ##0<a<1## be a real number. Suppose ##f\, : \,D\longrightarrow \mathbb{C}## is a holomorphic function such that ##f(a)=1## and ##f(-a)=-1.##
(a) Prove that ##\sup_{z\in D}\{|f(z)|\}\geq \dfrac{1}{a}.##
(b) Prove that if ##f## has no root, then ##\sup_{z\in D}\{|f(z)|\}\geq \exp\left(\dfrac{1-a^2}{4a}\,\pi\right).##

7. (solved by @julian )
(a) Let ##0<p\leq a,b,c,d,e \leq q## and show that
$$(a+b+c+d+e)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}\right) \leq 25+6\left(\sqrt{\dfrac{p}{q}}-\sqrt{\dfrac{q}{p}}\right)^2.$$
(b) This is a special case of a general inequality. Which is the general case and how is it proven?

8. Let ##n>1## be an integer. There are ##n## lamps ##L_0,\ldots,L_{n-1}## arranged in a circle. Each lamp is either ON ##(1)## or OFF ##(0)##. A sequence of steps ##S_0,\ldots,S_i,\ldots## is carried out. Step ##S_j## affects the state of ##L_j## only (leaving the states of all other lamps unaltered) as follows:
If ##L_{j-1}## is ON, ##S_j## changes the state of ##L_j## from ON to OFF or from OFF to ON.
If ##L_{j-1}## is OFF, ##S_j## leaves the state of ##L_j## unchanged.
The lamps are labeled modulo ##n##, that is ##L_{-1}=L_{n-1}, L_0=L_n,## etc. Initially all lamps are ON.
Show that
(a) (solved by @Jarvis323 ) there is a positive integer ##M(n)## such that after ##M(n)## steps all the lamps are ON again;
(b) if ##n=2^k,## then all lamps are ON after ##(n^2-1)## steps;
(c) if ##n=2^k+1,## then all lamps are ON after ##(n^2-n+1)## steps.

9. (solved by @Fred Wright ) The pseudosphere is the rotational surface of the tractrix, e.g. parameterized by
$$f\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}^3\; , \;f(x,y)=\begin{bmatrix} \cos (y)/ \cosh (x)\\ \sin (y)/ \cosh (x)\\x- \tanh (x)\end{bmatrix}.$$
Show that the pseudosphere has a constant negative Gauß curvature.

10. Let ##\mathfrak{g}## be a Lie algebra with trivial center ##\mathfrak{Z(g)}=\{0\}## over a field of characteristic not equal two and
\begin{align*}
\mathfrak{A(g)}&=\{\varphi:\mathfrak{g}\stackrel{linear}{\longrightarrow}\mathfrak{g}\,|\,[\varphi(X),Y]=[\varphi(Y),X]\text{ for all }X,Y\in \mathfrak{g}\}\\
&=\operatorname{lin}\{\alpha,\beta\neq 0\,|\,[\alpha,\beta]=\alpha\beta-\beta\alpha=\beta\}
\end{align*}
Show that image ##\operatorname{im}\beta## and kernel ##\operatorname{ker}\beta## of ##\beta## are ideals in ##\mathfrak{g}.##
Hint: ##\mathfrak{A(g)}## is a ##\mathfrak{g}##-module by ##X.\varphi =[\operatorname{ad}X,\varphi].##

High Schoolers only

11.
(solved by @etotheipi ) Let ##I## and ##J## be bounded open intervals in ##\Bbb{R}## with ##I \cap J \neq \emptyset## and the length of ##J## is greater than the length of ##I##. Show that ##I \subseteq 3*J##, where ##3*J## is the interval with length ##3## times the length of ##J## and with the same centre as ##J##. (MQ)

12. Given a positive integer ##n##. Assume that ##4^n## and ##5^n## start with the same digit in the decimal system.
Show that this digit has to be ##2## or ##4.##

13. A parcel service charges a price proportional to the sum height plus length plus width per box.
Could it be, that there is a case where it is cheaper to put a more expensive package into a cheaper box?

14. Let ##a## be a positive integer and ##(a_n)_{n\in \mathbb{N}_0}## the sequence defined by
$$a_0:=1\; , \;a_n:=a+\prod_{k=0}^{n-1}a_k \quad (n\geq 1)$$
(a) There are infinitely many primes which divide at least one number of the sequence.
(b) There is a prime which does not divide any of the numbers in the sequence

15. Let ##a,b,c## be positive real numbers such that ##a+b+c+2=abc.##
Show that ##(a+1)(b+1)(c+1)\geq 27.## Under which condition does equality hold?

Last edited:
JD_PM, StoneTemplePython, Jarvis323 and 3 others

Office_Shredder
Staff Emeritus
Gold Member
Not a high schooler, but very confused by 11

isn't I=(-10,1) and J=(0,2) a counterexample? It feels like you needed to include something about the radius of I vs J

member 587159, PeroK and Infrared
wrobel
Probl. 1
The described sequence of functions is uniformly continuous sinse all the derivatives are estimated as follows ##|f'_k(x)|\le \sum_n\frac{1}{n^2}##
The Arzelà–Ascoli theorem finishes the proof

Mentor
Probl. 1
The described sequence of functions is uniformly continuous sinse all the derivatives are estimated as follows ##|f'_k(x)|\le \sum_n\frac{1}{n^2}##
The Arzelà–Ascoli theorem finishes the proof
Too thin.

member 587159
Not a high schooler, but very confused by 11

isn't I=(-10,1) and J=(0,2) a counterexample? It feels like you needed to include something about the radius of I vs J

Thanks. There was an assumption missing: that the length of ##I## is smaller than the length of ##J##.

wrobel
Probl 6 b

It looks like a consequence from the Schwarz–Pick theorem https://en.wikipedia.org/wiki/Schwarz_lemma

Indeed, take a branch of a function ##w(z)=\log f(z)## such that ##w(-a)=i\pi,\quad w(a)=0##.

Applying the Schwarz–Pick theorem to the function $$F(z)=\frac{w(z)}{\sup_{|z|<1}|w(z)|},\quad z_1=a,\quad z_2=-a$$

we get

$$\sup_{|z|<1}|w(z)|\ge \frac{(1+a^2)\pi}{2a}.$$

It looks like the required estimate but it is not the same. I am confused little bit.

Mentor
I am confused little bit.
$$f(z)=-i\,\exp\left(\dfrac{iz-a^2}{iz+1} \cdot \dfrac{\pi}{2a}\right).$$

etotheipi
Not sure if I still qualify for high-school, but I'll take a shot at Math's question anyway
11. Let ##I## and ##J## be bounded open intervals in ##\Bbb{R}## with ##I \cap J \neq \emptyset## and the length of ##J## is greater than the length of ##I##.
Show that ##I \subseteq 3*J##, where ##3*J## is the interval with length ##3## times the length of ##J## and with the same centre as ##J##. (MQ)

Let ##J = (a-d, a+d)## and ##3*J = (a-3d, a+3d)##. Also, let ##I = (b-e, b+e)##. Since ##I \cap J \neq \emptyset##, we either have ##b-e < a+d## or ##a-d < b+e##, or both. Then, since it's given that ##d > e##,\begin{align*}\text{Case 1:} \quad b-e < a+d &\implies b+e < a + d + 2e < a + 3d \\ \text{Case 2:} \quad b+e > a-d &\implies b-e > a - d - 2e > a - 3d\end{align*}In the first case ##\text{sup}(I) < \text{sup}(3*J)##, and also since ##I \cap J \neq \emptyset \implies \text{inf}(I) > a - d - 2e > a - 3d = \text{inf}(3*J)##, we have ##\text{inf}(I) > \text{inf}(3*J)##. Then ##I \subseteq 3*J##.

In the second case ##\text{inf}(I) > \text{inf}(3*J)##, and also since ##I \cap J \neq \emptyset \implies \text{sup}(I) < a + d + 2e < a + 3d = \text{sup}(3*J)##, we have ##\text{sup}(I) < \text{sup}(3*J)##. Then ##I \subseteq 3*J##.

member 587159
wrobel
yes, I see the mistake:)

Mentor
yes, I see the mistake:)
Schwarz was the correct idea, but the function which he is applied to is a bit tricky.

member 587159
Not sure if I still qualify for high-school, but I'll take a shot at Math's question anyway
Let ##J = (a-d, a+d)## and ##3*J = (a-3d, a+3d)##. Also, let ##I = (b-e, b+e)##. Since ##I \cap J \neq \emptyset##, we either have ##b-e < a+d## or ##a-d < b+e##, or both. Then, since it's given that ##d > e##,\begin{align*}\text{Case 1:} \quad b-e < a+d &\implies b+e < a + d + 2e < a + 3d \\ \text{Case 2:} \quad b+e > a-d &\implies b-e > a - d - 2e > a - 3d\end{align*}In the first case ##\text{sup}(I) < \text{sup}(3*J)##, and also since ##I \cap J \neq \emptyset \implies \text{inf}(I) > a - d - 2e > a - 3d = \text{inf}(3*J)##, we have ##\text{inf}(I) > \text{inf}(3*J)##. Then ##I \subseteq 3*J##.

In the second case ##\text{inf}(I) > \text{inf}(3*J)##, and also since ##I \cap J \neq \emptyset \implies \text{sup}(I) < a + d + 2e < a + 3d = \text{sup}(3*J)##, we have ##\text{sup}(I) < \text{sup}(3*J)##. Then ##I \subseteq 3*J##.

Looks ok to me! For reference, you can use this exercise to prove the Vitali covering theorem in ##\mathbb{R}##. More general versions of this exercise can also be proven.

Btw, you still count as a high school student until you have had your first course at the university.

etotheipi
benorin
Homework Helper
For 4) what is the meaning of a constant times a set? Point-wise multiplication and perhaps a shift?

member 587159
For 4) what is the meaning of a constant times a set? Point-wise multiplication and perhaps a shift?
$$\frac{1}{2}C + \frac{1}{2}C := \left\{\frac{1}{2}c + \frac{1}{2}c': c,c' \in C\right\}$$

mathwonk
Homework Helper
2020 Award
question about 6b: so does the assumption "no root" mean that f has no zero in D?

Mentor
question about 6b: so does the assumption "no root" mean that f has no zero in D?
Yes, non zero on ##D##. This is necessary to have a logarithm.

mathwonk
Homework Helper
2020 Award
Hint for 6a:

It seems to me that Schwarz Lemma (plus the fact the automorphism group of the disc acts transitively) implies any holomorphic map from the disc to itself with more than one fix point is the identity. Hence if we multiply the function f by a, and assume the sup (of |a.f|) is < 1, we get a contradiction.

mathwonk
Homework Helper
2020 Award
about post #6: do you really get to specify two values of your log (your map w)? I.e. isn't a lift of a map, on a connected space, through a covering map (like exp), determined by the choice of just one value?

Mentor
Hint for 6a:

It seems to me that Schwarz Lemma (plus the fact the automorphism group of the disc acts transitively) implies any holomorphic map from the disc to itself with more than one fix point is the identity. Hence if we multiply the function f by a, and assume the sup (of |a.f|) is < 1, we get a contradiction.
My proof is easier. It essentially uses only the triangle inequality.

mathwonk
Homework Helper
2020 Award
Nice. Does the easy proof also identify the function as f(x) = (1/a).z, when the sup equals 1/a?

Mentor
Nice. Does the easy proof also identify the function as f(x) = (1/a).z, when the sup equals 1/a?
The maximum principle is used as well, same as in Schwarz.

wrobel
about post #6: do you really get to specify two values of your log (your map w)? I.e. isn't a lift of a map, on a connected space, through a covering map (like exp), determined by the choice of just one value?
this is exactly the point of mistake I have already acknowledged

Problem 9.
We have the parameterized function for the surface,
$$r(x,y)=\begin{pmatrix}\\ \frac{\cos(y)}{\cosh(x)}\\ \frac{\sin(y)}{\cosh(x)}\\ x-\tanh(x) \end{pmatrix}$$
The Gaussian curvature can be expressed as,
$$K=\frac{eg - f^2}{EG-F^2}$$
where e, f, and g are the coefficients of the first fundamental form and E, F, and G are the coefficients of the second fundamental form, such that,
$$E=r_{xx}\cdot\hat n$$
$$G=r_{yy}\cdot\hat n$$
$$F=r_{xy}\cdot\hat n=f_{yx}\cdot\hat n\\$$
$$e=r_{x}\cdot\hat n_x$$
$$g=r_{y}\cdot\hat n_y$$
$$f=r_{y}\cdot\hat n_x=r_{x}\cdot\hat n_y$$
where the subscripts indicate partial differentiation and the normal to the surface is,
$$\hat n = \frac{r_x\times r_y}{|r_x\times r_y|}$$
We have,
$$r_x=\begin{pmatrix}\\ \frac{-\cos(y)\sinh(x)}{\cosh^2(x)}\\ \frac{-\sin(y)\sinh(x)}{\cosh^2(x)}\\ \tanh^2(x) \end{pmatrix}$$
$$r_y=\begin{pmatrix}\\ \frac{-\sin(y)}{\cosh(x)}\\ \frac{\cos(y)}{\cosh(x)}\\ 0 \end{pmatrix}$$
$$\hat n=\begin{pmatrix}\\ -\cos(y)\tanh(x)\\ -\sin(y)\tanh(x)\\ \frac{1}{\cosh(x)} \end{pmatrix}$$
$$\hat n_x=\begin{pmatrix}\\ \frac{-\cos(y)}{\cosh^2(x)}\\ \frac{-\sin(y)}{\cosh^2(x)}\\ \frac{\sinh(x)}{\cosh^2(x)} \end{pmatrix}$$
$$\hat n_y=\begin{pmatrix}\\ -\sin(y)\tanh(x)\\ \cos(y)\tanh(x)\\ 0 \end{pmatrix}$$
$$r_{xx}=\begin{pmatrix}\\ \frac{\cos(y)(\sinh^2(x)-1)}{\cosh^3(x)}\\ \frac{\sin(y)(\sinh^2(x)-1)}{\cosh^3(x)}\\ \frac{2\sinh(x)}{\cosh^3(x)} \end{pmatrix}$$
$$r_{yy}=\begin{pmatrix}\\ \frac{-\cos(y)}{\cosh(x)}\\ \frac{-\sin(y)}{\cosh(x)}\\ 0 \end{pmatrix}$$
$$r_{yx}=r_{xy}=\begin{pmatrix}\\ \frac{\sin(y)\sinh(x)}{\cosh^2(x)}\\ \frac{-\cos(y)\sinh(x)}{\cosh^2(x)}\\ 0 \end{pmatrix}$$
$$E=r_{xx}\cdot\hat n=-\frac{\sinh(x)(\sinh^2(x)+1)}{\cosh^4(x)}$$
$$e=r_{x}\cdot\hat n_x=\frac{\sinh(x)(\sinh^2(x)+1)}{\cosh^4(x)}=-E$$
$$G=r_{yy}\cdot\hat n=\frac{\sinh(x)}{\cosh^2(x)}$$
$$g=r_{y}\cdot\hat n_y=\frac{\sinh(x)}{\cosh^2(x)}=G$$
$$F=r_{xy}\cdot\hat n=f_{yx}\cdot\hat n=0$$
$$f=r_{y}\cdot\hat n_x=r_{x}\cdot\hat n_y=0$$
and thus the Gaussian curvature,
$$K=-1$$

benorin and fresh_42
I'm confused by problem 1.
What's the difference between $f_0$ and $f_1$?

Mentor
I'm confused by problem 1.
What's the difference between $f_0$ and $f_1$?
You are right. This is a typo. I corrected it.

There must be a simpler way to prove this, but I think this works.

Problem 8a.
Let a configuration ##C_t## (of the given process) be defined by the ON/OFF states of each lamp before applying the ##t^{th}## step, combined with the current lamp index ##I_t = t \mod n##.

Because there is a finite number of configurations, some configuration must be repeated eventually.

Let ##C_a## be the first repeating configuration and assume for contradiction that ##a \neq 0##. Let ##C_b## be the configuration preceding the second occurrence of a configuration equal to ##C_a##. Then ##C_{a} = C_{b+1}## but ##C_{a-1} \neq C_b## because if ##C_{a-1} = C_{b}##, then ##C_{a-1}## is a repeating configuration that appears before ##C_a##.

Since ##C_{b}## and ##C_{a-1}## both precede a configuration equal to ##C_a##, ##I_{(b \mod n)} = I_{((a-1) \mod n)}##. Thus ##C_b## and ##C_{a-1}## must have different lamp states. Since a single step can only change the state of at most ##1## lamp, and both configurations are ##1## step away from being equal, if follows that they must differ only in ##L_{((a-1) \mod n)} \equiv L_{(b \mod n)}##. Thus ##L_{((a-2) \mod n)} \equiv L_{((b-1) \mod n)}## must be in the same state in both ##C_{b-1}## and ##C_{a-2}##.

Case 1: That state is OFF. Then neither lamp states (neither in ##C_{b}## nor ##C_{a-1}##) would change in their next step. But they both differ, and are both assumed to directly precede a configuration equal to ##C_{a}##, which is a contradiction.

Case 2: That state is ON. Then, the next step would change both lamp states ( both in ##C_{b}## and ##C_{a-1}##), namely by flipping ON/OFF ##L_{ (b \mod n} )##, but since they both differed in ##L_{ b (\mod n} )## prior to this step, afterwards they would still differ in ##L_{ (b \mod n} )##.

All cases are covered, and each contradict the assumption that ##a\neq 0##

It follows that the first repeating configuration must be the initial configuration.

##Q.E.D.##

EDIT: Tried to make the notation and definitions more consistent.

Last edited:
fresh_42