# Math Challenge - November 2020

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• Featured
Mentor
Summary: Diffusion Equation, Sequence Space, Banach Space, Linear Algebra, Quadratic Forms, Population Distribution, Sylow Subgroups, Lotka-Volterra, Ring Theory, Field Extension.

1. Let ##u(x,t)## satisfy the one dimensional diffusion equation ##u_t=Du_{xx}## in a space-time rectangle ##R=\{0\leq x\leq l,0\leq t\leq T\}##, then the maximum value of ##u(x,t)## is assumed either on the initial line ##(t=0)## or on the boundary lines ##(x=0 \,or\,x=l )##. ##D > 0.##

2. (solved by @Office_Shredder, second proof possible ) Show that ## M= \{ (a_n)\in \ell^2(\mathbb{C})\,|\,\forall \,n: |a_n| \leq n^{-1} \} \subseteq \ell^2(\mathbb{C})## is bounded and compact.

3. (solved by @Office_Shredder ) Show by two different methods that the normed space ##\mathcal{C}:=(C^1([0,1]),\|.\|_\infty)## is not a Banach space.

4. (solved by @PeroK ) Let
$$A:=\begin{bmatrix}5&0&1&6\\3&3&5&2\\0&0&3&0\\6&0&3&0\end{bmatrix} \in \mathbb{M}_4(\mathbb{Z}_7)$$
(a) Determine the characteristic polynomial ##\chi_A(x)## of ##A.##
(b) Determine bases of the eigenspaces.
(c) Determine a matrix ##S\in\operatorname{GL}_4(\mathbb{Z}_7)## such that ##S^{-1}AS## is a diagonal matrix. Which one?
(d) Calculate ##A^{31}.##

5. (solved by @etotheipi ) Let ##f(x,y)=34x^2+24xy+41y^2+20x+110y+50.## Determine the Euclidean normal form of the conic section $$Q_f=\{(x,y)^\tau\in\mathbb{R}^2\,|\,f(x,y)=0\}.$$ What are its foci and vertices in the normal form?

6. (solved by @benorin ) Let ##u(x,t)## be a solution of the one dimensional diffusion equation ##u_t=Du_{xx}.## Assume that
$$C:=\int_{-\infty}^{\infty} u(x,t)\,dx$$
is independent of ##t,## which corresponds to a constant population, and ##u## is small at infinity, which means that
$$\lim_{x \to \pm\infty} xu(x,t) = 0 = \lim_{x \to \pm \infty} x^2 \dfrac{\partial}{\partial x}u(x,t)$$
If
$$\sigma^2(t)=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2u(x,t)\,dx$$
then
$$\sigma^2(t)=2Dt\,+\,\sigma^2(0)$$
In the special case of an initial population (i.e. for ##t = 0##) which is concentrated near ##x = 0## (like a ##\delta ##-function) then we get ##\sigma^2(t)\approx 2Dt.##

7. (solved by @fishturtle1 ) Let ##G## be a group of order ##351.## Show that ##G## has a non trivial normal subgroup.

8. Show that the diffusional Lotka-Volterra system ##(a>0)##
\begin{align}
u_t&\, = \,u(1-v)+D\Delta u\\
v_t&\, = \,av(u-1)+D\Delta v
\end{align}
with equal diffusion coefficient ##D>0## and homogeneous Neumann boundary conditions
$$\dfrac{\partial u}{\partial n}(x,t)=0=\dfrac{\partial v}{\partial n}(x,t)$$
for ##x\in \partial\Omega\, , \,\Omega\subseteq \mathbb{R}^n## of finite volume and ##n## outward normal, ##\Delta ## the Laplace operator, tends to a spatially uniform state for ##t \to \infty,## i.e.
$$\lim_{t \to \infty} \nabla u = \lim_{t \to \infty} \nabla v = 0$$
Hint: Consider the energy of the system ##s=a(u-\log u)+(v - \log v).##

9. (a) (solved by @disregardthat ) Let ##R## be a Notherian local commutative ring with ##1## and maximal ideal ##M.## If ##A \trianglelefteq R## is an ideal in ##R## such that ##A/MA\cong_R \{0\},## then ##A=(0).##
9. (b) (solved by @disregardthat ) Let ##R## be an integral domain, and ##\dim R_P=0## for all ##P\in \operatorname{Spec}(R)##, then ##R## is a field. The dimension is the Krull dimension.

10. (solved by @Office_Shredder ) Let ##\alpha\in \mathbb{C}## a root of the polynomial ##f(x)=x^3-3x-1\in \mathbb{Q}[x]##. Show that ##f(x)## is irreducible, and that there is an automorphism ##\sigma\in \operatorname{Aut}(\mathbb{Q}(\alpha)/\mathbb{Q})## with ##\sigma(\alpha)=2-\alpha^2.## If ##\alpha ## is chosen closest to zero, what is ##+\sqrt{12-3\alpha^2}## in the splitting field of ##f(x)##? This means in terms of a polynomial in ##\alpha,## not numerical.

High Schoolers only

11.
(solved by @songoku ) Determine all ##a\in \mathbb{R}## such that
$$x(x+1)(x+2)(x+3)=a$$
has no real solution, a unique real solution, exactly two, three, or four real solutions, more than four real solutions.

12. (solved by @songoku ) An international conference has ##30## scientists who speak English, Russian or Spanish. The number of people who speak exactly two languages is more than twice as big, but less than thrice as much as the number of people who speak only one language, which are as many as speak all three languages. Those who speak only English are more than those who speak only Russian, but less than those who speak only Spanish. The number of those who speak only English is less than thrice the number of people who speak only Russian. How many people do speak only English, Russian, Spanish, and how many all three languages? (The conference language is French.)

13. (solved by @songoku ) Calculate (manually!)
$$z=\dfrac{65533^3+65534^3+65535^3+65536^3+65537^3+65538^3+65539^3}{32765\cdot 32766+32767\cdot 32768+32768\cdot 32769+32770\cdot 32771}$$

14. (solved by @songoku ) Show that (##n\in \mathbb{N}_0##)
$$f_n(x)=1+x+\dfrac{x^2}{2!}+\ldots+\dfrac{x^n}{n!}$$
has at most one real zero.

15. (solved by @songoku ) Find all ##\lambda\in \mathbb{R}## such that
$$\sin^4x-\cos^4x=\lambda(\tan^4x-\cot^4x)$$
has no, exactly one, exactly two, more than two real solutions in ##\left(0,\dfrac{\pi}{2}\right)##

Last edited:
LCSphysicist and Delta2

Keith_McClary
Gold Member
In 5 I think I know the meaning of Euclidean normal form and its foci and vertices.

Do I need to understand the notation ##(x,y)^\tau## ?

Mentor
In 5 I think I know the meaning of Euclidean normal form and its foci and vertices.

Do I need to understand the notation ##(x,y)^\tau## ?
Old habit to write vectors as columns, so rows must be transposed. Just a reflex.

benorin
Homework Helper
6. Let ##u(x,t)## be a solution of the one dimensional diffusion equation ##u_t=Du_{xx}.## Assume that
$$C:=\int_{-\infty}^{\infty} u(x,t)\,dx$$
is independent of ##t,## which corresponds to a constant population, and ##u## is small at infinity, which means that
$$\lim_{x \to \pm\infty} xu(x,t) = 0 = \lim_{x \to \pm \infty} x^2 \dfrac{\partial}{\partial x}u(x,t)$$
If
$$\sigma^2(t)=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2u(x,t)\,dx$$
then
$$\sigma^2(t)=2Dt\,+\,\sigma^2(0)$$
eIn the special case of an initial population (i.e. for ##t = 0##) which is concentrated near ##x = 0## (like a ##\delta ##-function) then we get ##\sigma^2(t)\approx 2Dt.##
We have
$$\begin{gathered}\sigma^2(t)=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2u(x,t)\,dx=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2\int_{0}^{t}u_{t}(x,\tau )\,d\tau \,dx+\sigma^2 (0) \\ = \dfrac{D}{C}\int_{-\infty}^{+\infty} x^2\int_{0}^{t}u_{xx}(x,\tau )\,d\tau \,dx+\sigma^2 (0) \\ = \dfrac{D}{C} \int_{0}^{t} \int_{-\infty}^{+\infty}x^2 u_{xx}(x,\tau ) \, dx \,d\tau+\sigma^2 (0) \\ \end{gathered}$$

Let ##u_1^{\prime}=x^2\Rightarrow du_1^{\prime}=2x\, dx## and ##dv_1^{\prime}=u_{xx}(x,\tau )\, dx\Rightarrow v_1^{\prime}=u_{x}(x, \tau )##, then

$$\begin{gathered}\sigma^2(t) = \dfrac{D}{C} \int_{0}^{t}\underbrace{\left[ x^2 u_{x}(x, \tau )\right|_{x=-\infty}^{+\infty}}_{=0}\, d\tau- \dfrac{2D}{C} \int_{0}^{t} \int_{-\infty}^{+\infty}x u_{x}(x,\tau ) \, dx \,d\tau+\sigma^2 (0) \\ \end{gathered}$$

Let ##u_2^{\prime}=x\Rightarrow du_2^{\prime}=dx## and ##dv_2^{\prime}=u_{x}(x,\tau )\, dx\Rightarrow v_2^{\prime}=u(x, \tau )##, then

$$\begin{gathered}\sigma^2(t) = \dfrac{D}{C} \int_{0}^{t}\underbrace{\left[ x u(x, \tau )\right|_{x=-\infty}^{+\infty}}_{=0}\, d\tau + \dfrac{2D}{C} \int_{0}^{t} \underbrace{\int_{-\infty}^{+\infty} u(x,\tau ) \, dx}_{=C} \,d\tau+\sigma^2 (0) \\ = 2D \int_{0}^{t}\,d\tau + \sigma^2 (0) \boxed{=2Dt + \sigma^2 (0)} \\ \end{gathered}$$

Last edited:
Delta2
Office_Shredder
Staff Emeritus
Gold Member
Very cool solution for 6 (and very cool problem). I think you typod and have an ##x^2## in the last integration by parts that is supposed to be an ##x##.

Number 2. I'm a little embarrassed by the length of the proof and wonder if I wasn't supposed to use this result.
by this neat theorem
https://calculus7.org/2017/04/27/compact-sets-in-banach-spaces/

We have to show it's bounded and flat.

Let ##a=(a_i) \in M##. Then
$$||a|| = \sum_i |a_i|^2 \leq \sum_i \frac{1}{n^2}$$.

That last sum converges, so ##M## is bounded.

Now we need to show it is flat. Since that sum above converges, for any ##\epsilon >0##, there exists ##r## such that
$$\sum_{i=r+1}^{\infty} \frac{1}{n^2} < \epsilon$$.

Let ##V_r \subset l_2## be the space of all sequences for which only the first ##r## elements are non-zero. Then ##V_r## is flat, and any element ##a## of ##M## is at most ##\epsilon## away from ##V_r## by taking the sequence which equals ##a## for the first ##r## elements and then 0 afterwards. Therefore it's flat and hence compact.

Mentor
We have
$$\begin{gathered}\sigma^2(t)=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2u(x,t)\,dx=\dfrac{1}{C}\int_{-\infty}^{+\infty} x^2\int_{0}^{t}u_{t}(x,\tau )\,d\tau \,dx+\sigma^2 (0) \\ = \dfrac{D}{C}\int_{-\infty}^{+\infty} x^2\int_{0}^{t}u_{xx}(x,\tau )\,d\tau \,dx+\sigma^2 (0) \\ = \dfrac{D}{C} \int_{0}^{t} \int_{-\infty}^{+\infty}x^2 u_{xx}(x,\tau ) \, dx \,d\tau+\sigma^2 (0) \\ \end{gathered}$$
Let ##u_1^{\prime}=x^2\Rightarrow du_1^{\prime}=2x\, dx## and ##dv_1^{\prime}=u_{xx}(x,\tau )\, dx\Rightarrow v_1^{\prime}=u_{x}(x, \tau )##, then
$$\begin{gathered}\sigma^2(t) = \dfrac{D}{C} \int_{0}^{t}\underbrace{\left[ x^2 u_{x}(x, \tau )\right|_{x=-\infty}^{+\infty}}_{=0}\, d\tau- \dfrac{2D}{C} \int_{0}^{t} \int_{-\infty}^{+\infty}x u_{x}(x,\tau ) \, dx \,d\tau+\sigma^2 (0) \\ \end{gathered}$$
Let ##u_2^{\prime}=x\Rightarrow du_2^{\prime}=dx## and ##dv_2^{\prime}=u_{x}(x,\tau )\, dx\Rightarrow v_2^{\prime}=u(x, \tau )##, then
$$\begin{gathered}\sigma^2(t) = \dfrac{D}{C} \int_{0}^{t}\underbrace{\left[ x^2 u(x, \tau )\right|_{x=-\infty}^{+\infty}}_{=0}\, d\tau + \dfrac{2D}{C} \int_{0}^{t} \underbrace{\int_{-\infty}^{+\infty} u(x,\tau ) \, dx}_{=C} \,d\tau+\sigma^2 (0) \\ = 2D \int_{0}^{t}\,d\tau + \sigma^2 (0) \boxed{=2Dt + \sigma^2 (0)} \\ \end{gathered}$$
It's a bit easier if you consider ##\dfrac{d\sigma^2}{dt}## and pull the differentiation into the integrals, but ok. What about the last question if ##u(x,0)=0## for ##|x|>\varepsilon##?

Mentor
Very cool solution for 6 (and very cool problem). I think you typod and have an ##x^2## in the last integration by parts that is supposed to be an ##x##.

Number 2. I'm a little embarrassed by the length of the proof and wonder if I wasn't supposed to use this result.
by this neat theorem
https://calculus7.org/2017/04/27/compact-sets-in-banach-spaces/

We have to show it's bounded and flat.

Let ##a=(a_i) \in M##. Then
$$\|a\| = \sum_i |a_i|^2 \leq \sum_i \frac{1}{n^2}$$.

That last sum converges, so ##M## is bounded.

Now we need to show it is flat. Since that sum above converges, for any ##\epsilon >0##, there exists ##r## such that
$$\sum_{i=r+1}^{\infty} \frac{1}{n^2} < \epsilon$$.

Let ## V_r \subset l_2 ## be the space of all sequences for which only the first ## r ## elements are non-zero. Then ## V_r ## is flat, and any element ## a ## of ## M ## is at most ## \epsilon ## away from ## V_r ## by taking the sequence which equals ## a ## for the first ## r ## elements and then 0 afterwards. Therefore it's flat and hence compact.
What about closure? This is the trick I wanted to see.

benorin
Homework Helper
It's a bit easier if you consider ##\dfrac{d\sigma^2}{dt}## and pull the differentiation into the integrals, but ok. What about the last question if ##u(x,0)=0## for ##|x|>\varepsilon##?
Taking the ##\delta -##function we have ##u(x,0)=\tfrac{1}{2\varepsilon}## for ##|x|\leq \varepsilon## whence

$$\begin{gathered} \sigma ^2(0)=\tfrac{1}{C}\int_{-\infty}^{+\infty} x^2 u(x,0) \, dx = \int_{-\varepsilon }^{+\varepsilon}x^2 \tfrac{1}{2\varepsilon} \, dx \\ = \tfrac{1}{6\varepsilon} \left[ x^3 \right|_{-\varepsilon}^{+\varepsilon} = \tfrac{\varepsilon ^2}{3} \rightarrow 0^{+} \\ \end{gathered}$$
since ##C=1## by definition of the ##\delta -##function.

Delta2 and fresh_42
Office_Shredder
Staff Emeritus
Gold Member
What about closure? This is the trick I wanted to see.

Whoop!

To show ##M## is closed, we can show the complement is open. Let ##b=(b_i) \notin M##. Then there exists some ##k## such that ##|b_k|>1/k##.

Pick##\epsilon>0## such that ##\epsilon < |b_k| - \frac{1}{k}##. Let ##c\in B_{\epsilon}##. Then ##|c_k|<\epsilon##, and hence by the triangle inequality ##|b_k+c_k|> |b_k|-\epsilon > \frac{1}{k}##.

Hence ##b+c \notin M##, and therefore the compliment of ##M## is open.

Mentor
Whoop!

To show ##M## is closed, we can show the complement is open. Let ##b=(b_i) \notin M##. Then there exists some ##k## such that ##|b_k|>1/k##.

Pick##\epsilon>0## such that ##\epsilon < |b_k| - \frac{1}{k}##. Let ##c\in B_{\epsilon}##. Then ##|c_k|<\epsilon##, and hence by the triangle inequality ##|b_k+c_k|> |b_k|-\epsilon > \frac{1}{k}##.

Hence ##b+c \notin M##, and therefore the compliment of ##M## is open.
For all who want to prove it without the theorem @Office_Shredder quoted:
Show that ##M## is sequence compact.

Office_Shredder
Staff Emeritus
Gold Member
Just wanted to check - for #1, "space-time" rectangle doesn't have some special definition from physics, it just means a rectangle in the domain of u, right?

Mentor
Just wanted to check - for #1, "space-time" rectangle doesn't have some special definition from physics, it just means a rectangle in the domain of u, right?
Right. It's a spatial coordinate ##x## and a time coordinate ##t##. Classical of course. I guess it would remain true in more than one spatial coordinate and norms, but we have only one. The statement even has a name.

Office_Shredder
Office_Shredder
Staff Emeritus
Gold Member
Alright let me try

it's called the diffusion equation. If the statement wasn't true it would mean mass is gathering, not diffusing.

I think it needs a little work

Mentor
Alright let me try

it's called the diffusion equation. If the statement wasn't true it would mean mass is gathering, not diffusing.

I think it needs a little work
... or a little variation calculus.

Office_Shredder
Staff Emeritus
Gold Member
Number 3., I have one method. How different does the second way need to be?

Define ##f_n(x)## to be ##\sqrt{(x-1/2)^2 +1/n}##. Then each ##f_n## is differentiable, and in ##C^{0}## with the sup norm converges to ##|x-1/2|##. In particular it's a cauchy sequence in ##C^1##. But the limit is not in ##C^1##, so it's not complete as required.

I assume picking another cauchy sequence that doesn't converge in ##C^1## doesn't count as a second proof? Every idea I have boils down to constructing another such sequence (e.g. show the graph of the derivative operator is not a closed subset of the product topology boils down to constructing a sequence like the above)

mathwonk
Mentor
Number 3., I have one method. How different does the second way need to be?

Define ##f_n(x)## to be ##\sqrt{(x-1/2)^2 +1/n}##. Then each ##f_n## is differentiable, and in ##C^{0}## with the sup norm converges to ##|x-1/2|##. In particular it's a cauchy sequence in ##C^1##. But the limit is not in ##C^1##, so it's not complete as required.
This is correct, but you could be a little more detailed for lurkers and users who are still in the learning process.
I assume picking another cauchy sequence that doesn't converge in ##C^1## doesn't count as a second proof?
Well, yes, no.
Every idea I have boils down to constructing another such sequence (e.g. show the graph of the derivative operator is not a closed subset of the product topology boils down to constructing a sequence like the above)
I thought about an operator instead. In a way the complete abstraction of the sequence method above.

PeroK
Homework Helper
Gold Member
2020 Award
Now that @etotheipi is at university, there's no one left who can do the high school problems!

member 587159 and etotheipi
Mentor
@etotheipi

Since you are no longer at school, I have to remove your post. Well, I think it is still a bit early in the month. Let me hide it and see if someone else tries. If not, I will undelete it later on and give you the credits.

Moreover, Taylor and MVT are not really school level. Induction is, but you have to rephrase the statement first, i.e. prove a slightly stronger version.

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berkeman
Mentor
Since you are no longer at school, I have to remove your post. Well, I think it is still a bit early in the month. Let me hide it and see if someone else tries. If not, I will undelete it later on and give you the credits.
I was wondering about that. He is in his first term at uni, so technically still a super-senior from high school. But given his extraordinary intellectual gifts, maybe it's an unfair competition against folks still in high school. Still very nice work by @etotheipi

fresh_42
etotheipi
Yeah okay sorry, I won't post any more solutions. The HS problems are good practice but I'm still too dumb to do any of the big boy questions. Maybe that's just more motivation to study more, so that I can start doing the challenges again

Mentor
Yeah okay sorry, I won't post any more solutions. The HS problems are good practice but I'm still too dumb to do any of the big boy questions. Maybe that's just more motivation to study more, so that I can start doing the challenges again
You can do 4, 5 and maybe 7.

etotheipi
berkeman
Mentor
The HS problems are good practice but I'm still too dumb to do any of the big boy questions
You have just received an infraction for posting false information. Your infraction point total is an imaginary number, however, so I'm not sure if the system will ban you or not...

etotheipi
Now that @etotheipi is at university, there's no one left who can do the high school problems!
Not a problem. I will take his place. All I need to do is just posting all the questions in homework section and let the helpers do the rest

pbuk
Gold Member
@etotheipi
Since you are no longer at school, I have to remove your post. Well, I think it is still a bit early in the month. Let me hide it and see if someone else tries. If not, I will undelete it later on and give you the credits.
Yeah, foul!

Moreover, Taylor and MVT are not really school level.
In the UK these are covered in the Further Maths A-level, usually at age 16-17.

Mentor
In the UK these are covered in the Further Maths A-level, usually at age 16-17.
Well, this is not everywhere the case. I only wanted to encourage other high schoolers that this heavy machinery isn't necessary.

pbuk