Challenge Math Challenge - November 2020

[fresh_42]

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 only11. (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 $\left(0,\dfrac{\pi}{2}\right)$

 

[Keith_McClary]

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

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

 

[fresh_42]

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]

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 $\sigma^2(t)\approx 2Dt.$

Spoiler: #6

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

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

 

[Office_Shredder]

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.

Spoiler

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.

 

[fresh_42]

Spoiler: #6

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 $|x|>\varepsilon$?

 

[fresh_42]

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.

Spoiler

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]

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.

 

[Office_Shredder]

What about closure? This is the trick I wanted to see.

Whoop!

Spoiler

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.

 

[fresh_42]

Whoop!

Spoiler

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]

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?

 

[fresh_42]

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]

Alright let me try

Spoiler

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

 

[fresh_42]

Alright let me try

Spoiler

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]

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

Spoiler

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)

 

[fresh_42]

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]

Now that @etotheipi is at university, there's no one left who can do the high school problems!

 

[fresh_42]

@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.

 

[berkeman]

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

 

[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

 

[fresh_42]

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.

 

[berkeman]

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...

 

[songoku]

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]

@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.

 

[fresh_42]

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.

 

[Mayhem]

There should be some intermediate questions for early undergraduates who haven't taken any advanced math yet.

 

[member 587159]

@etotheipi
Moreover, Taylor and MVT are not really school level. Induction is.

I saw both Taylor and MVT in high school, but only one induction proof.

 

[fresh_42]

There should be some intermediate questions for early undergraduates who haven't taken any advanced math yet.

We tried, but it didn't work. What is intermediate? Taylor and MVT in high schools (see post #24), or are even induction and limits too difficult? Especially school systems vary significantly all over the world.

And on the other end: Problems #4 and #5 are not very sophisticated. I would have called them "intermediate", and even the algebraic questions are not so difficult. We even had questions which could have been solved by looking up some definitions on Wikipedia, and even this was too much to do. People get scared by technical terms, but our questions are already heavily limited to only easy concepts. I tried to find some more interesting questions in my books ... no chance! I would need to hold a lecture beforehand, regardless whether a question would had been easy or difficult, simply to set up the language.

We have a thread with solutions to many questions:
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/
where I think can be found problems at any level.

 

[Fred Wright]

And on the other end: Problems #4 and #5 are not very sophisticated. I would have called them "intermediate", and even the algebraic questions are not so difficult. We even had questions which could have been solved by looking up some definitions on Wikipedia, and even this was too much to do.

I agree. I found problem #5 could be solved easily in a few minutes by referring to https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections. I like this problem because it showed me a beautiful connection between linear algebra and geometry. Thank you for making the effort to post the monthly challenges, I learn a lot from them

 

[etotheipi]

What is a Euclidean normal form, and what do we mean by $(x,y)^{\tau}$?

 

[member 587159]

What is a Euclidean normal form, and what do we mean by $(x,y)^{\tau}$?

I believe it means the vector $(x,y)$ transposed, so written as column vector $\begin{pmatrix}x \\ y\end{pmatrix}$.

 

[fresh_42]

What is a Euclidean normal form, and what do we mean by $(x,y)^{\tau}$?

Yes, it is only a habit in this case to write vectors or coordinates as column. The Euclidean normal form is, when the axis of the conic sections are the coordinate axis, and its center the origin of the Cartesian coordinate system: rotations, translations; basically such that all interesting quantities can directly be read.

 

[etotheipi]

So if I understand correctly,

Spoiler: P5

$$A_Q = \begin{pmatrix} 34 & 12 & 10 \\ 12 & 41 & 55 \\ 10 & 55 & 50\end{pmatrix}$$and $f(x,y) = \mathbf{x}^{\tau}A_Q \mathbf{x}$, $\text{det}(A_Q) = -31250$, and $\text{det}(A_{33}) = 1250$, and $\frac{\text{det}(A_Q)}{\text{det}(A_{33})} = -25$, so I guess when you say you want it in normal form, since $A_{33}$ has eigenvalues $\lambda = 25, 50$, it will be$$50u^2 + 25v^2 = 25 \iff 2u^2 + v^2 = 1 \iff \left(\frac{u}{1/\sqrt{2}}\right)^2 + \left(\frac{v}{1}\right)^2 = 1$$so vertices at $(u,v) = \left( \pm 1/\sqrt{2}, 0\right), \left( 0, \pm 1\right)$ and foci at $(0, \pm \frac{1}{2})$? Is that what you had in mind, @Fred Wright?

 

[fresh_42]

So if I understand correctly,

Spoiler: P5

$$A_Q = \begin{pmatrix} 34 & 12 & 10 \\ 12 & 41 & 55 \\ 10 & 55 & 50\end{pmatrix}$$and $f(x,y) = \mathbf{x}^{\tau}A_Q \mathbf{x}$, $\text{det}(A_Q) = -31250$, and $\text{det}(A_{33}) = 1250$, and $\frac{\text{det}(A_Q)}{\text{det}(A_{33})} = -25$, so I guess when you say you want it in normal form, since $A_{33}$ has eigenvalues $\lambda = 25, 50$, it will be$$50u^2 + 25v^2 = 25 \iff 2u^2 + v^2 = 1 \iff \left(\frac{u}{1/\sqrt{2}}\right)^2 + \left(\frac{v}{1}\right)^2 = 1$$so vertices at $(u,v) = \left( \pm 1/\sqrt{2}, 0\right), \left( 0, \pm 1\right)$ and foci at $(0, \pm \frac{1}{2})$? Is that what you had in mind, @Fred Wright?

Well, you could have shown how you did it. It's a bit difficult to learn anything from your solution. And the foci are $(0,\pm 1/\sqrt{2})$ if I wasn't mistaken.

 

[Office_Shredder]

As far as making questions more accessible, you could have just asked something like construct a sequence of functions that are differentiable whose limit is not differentiable, and not scared off people that haven't heard of a Banach space.

 

[fresh_42]

As far as making questions more accessible, you could have just asked something like construct a sequence of functions that are differentiable whose limit is not differentiable, and not scared off people that haven't heard of a Banach space.

That isn't what the problem asks for.

 

[etotheipi]

Well, you could have shown how you did it. It's a bit difficult to learn anything from your solution. And the foci are $(0,\pm 1/\sqrt{2})$ if I wasn't mistaken.

Yes, sorry, I forgot to square root. Anyway, I'll give a more complete method here for anyone that wants it. First we need to find the centre of $$f(x,y) = 34x^2 + 24xy +41y^2 + 20x + 110y + 50=0$$Set $f_y = 0$ and $f_x = 0$ to get$$\begin{align*}68x + 24y + 20 &= 0 \\
82y + 24x + 110&=0
\end{align*}$$this is solved by $\mathbf{x}_c = (x_c,y_c)^{\tau} = (\frac{1}{5}, \frac{-7}{5})^{\tau}$, i.e. this is the centre of the ellipse. Now we perform a coordinate transformation $\mathbf{x}' = \mathbf{x} -\mathbf{x}_c$ so that the centre is at our new origin $\mathcal{O}'$, i.e.$$34(x' + \frac{1}{5}) + 24(x' + \frac{1}{5})(y' - \frac{7}{5}) + 41(y' - \frac{7}{5})^2 + 20(x' + \frac{1}{5}) + 110(y' - \frac{7}{5}) + 50 = 0$$ $$34x'^2 + 24x'y' + 41y'^2 = 25$$Now we perform a further coordinate transformation, this time a rotation (preserving the origin, $\mathcal{O}'' = \mathcal{O}'$), so that the new axis intersect the ellipse at the vertices. This will be $\mathbf{x}'' = R\mathbf{x}'$, where $R = \begin{pmatrix} c & s \\ -s & c\end{pmatrix}$, and $c=\cos{\theta}$, $s = \sin{\theta}$ for some angle $\theta$ to be determined. Running through the algebra with $x' = cx'' - sy''$ and $y' = sx'' + cy''$ yields$$34(cx'' -sy'')^2 + 24(cx'' - sy'')(sx'' + cy'') + 41(sx'' + cy'') = 25$$ $$x''^2 [ 34 c^2 + 41s^2 + 24cs] + y''^2 [41c^2 + 34s^2 - 24cs] + x''y''[24c^2 - 24s^2 + 14cs]$$We require that the coefficient of $x''y''$ is brought to be zero, i.e. that $s = 4/5$ and $c=3/5$. Now it's just a case of substituting those back into the equation,$$50x''^2 + 25y''^2 = 25 \iff 2x''^2 + y''^2 = 1$$and for brevity I'll re-label $x'' \equiv u$ and $y'' \equiv v$. And that completes the transformation!

 

[Mayhem]

Yes, sorry, I forgot to square root. Anyway, I'll give a more complete method here for anyone that wants it. First we need to find the centre of $$f(x,y) = 34x^2 + 24xy +41y^2 + 20x + 110y + 50=0$$Set $f_y = 0$ and $f_x = 0$ to get$$\begin{align*}68x + 24y + 20 &= 0 \\
82y + 24x + 110&=0
\end{align*}$$this is solved by $\mathbf{x}_c = (x_c,y_c)^{\tau} = (\frac{1}{5}, \frac{-7}{5})^{\tau}$, i.e. this is the centre of the ellipse. Now we perform a coordinate transformation $\mathbf{x}' = \mathbf{x} -\mathbf{x}_c$ so that the centre is at our new origin $\mathcal{O}'$, i.e.$$34(x' + \frac{1}{5}) + 24(x' + \frac{1}{5})(y' - \frac{7}{5}) + 41(y' - \frac{7}{5})^2 + 20(x' + \frac{1}{5}) + 110(y' - \frac{7}{5}) + 50 = 0$$ $$34x'^2 + 24x'y' + 41y'^2 = 25$$Now we perform a further coordinate transformation, this time a rotation (preserving the origin, $\mathcal{O}'' = \mathcal{O}'$), so that the new axis intersect the ellipse at the vertices. This will be $\mathbf{x}'' = R\mathbf{x}'$, where $R = \begin{pmatrix} c & s \\ -s & c\end{pmatrix}$, and $c=\cos{\theta}$, $s = \sin{\theta}$ for some angle $\theta$ to be determined. Running through the algebra with $x' = cx'' - sy''$ and $y' = sx'' + cy''$ yields$$34(cx'' -sy'')^2 + 24(cx'' - sy'')(sx'' + cy'') + 41(sx'' + cy'') = 25$$ $$x''^2 [ 34 c^2 + 41s^2 + 24cs] + y''^2 [41c^2 + 34s^2 - 24cs] + x''y''[24c^2 - 24s^2 + 14cs]$$We require that the coefficient of $x''y''$ is brought to be zero, i.e. that $s = 4/5$ and $c=3/5$. Now it's just a case of substituting those back into the equation,$$50x''^2 + 25y''^2 = 25 \iff 2x''^2 + y''^2 = 1$$and for brevity I'll re-label $x'' \equiv u$ and $y'' \equiv v$. And that completes the transformation!

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

 

[etotheipi]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

Coordinate transformations I learned from Douglas Gregory, Classical Mechanics! It's not so bad, you just need to be careful about when you're leaving the vector alone but changing its representation i.e. w.r.t. a new coordinate system (passive), or if you're actually transforming the vector (active), and not getting the signs mixed up!

 

[fresh_42]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

It is not too difficult:

Given the red one, asked for the blue one. You have to rotate the curve, stretch it, and shift the center into the origin. You can either handle this with linear algebra (hint in post #29, or wait for the long version in my solution manual), or simply perform coordinate transformations (post #44) until it fits. We want to have an ellipse at the end: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.

Imagine the ellipse is a crystal instead, and the red one is what you see under the microscope, whereas you need the blue version to identify it by the pictures in the book.

 

[LCSphysicist]

How did you learn this stuff when you're a first year uni student? I'm at the top of my first year math course for chemistry, but I haven't even seen half of this stuff before.

In physics i risk to say there is change of coordinates in all place, mainly in mechanic, since it is important to all the laws be the same in all coordinates, i think the college starts to talk about it very early when the course is physics. For example, in my first semester of physics i saw it, probably the same with ethoteipi

 

[fresh_42]

In physics i risk to say there is change of coordinates in all place, mainly in mechanic, since it is important to all the laws be the same in all coordinates, i think the college starts to talk about it very early when the course is physics. For example, in my first semester i saw it.

I would even go as far as to say: physics is all about coordinates. Without coordinates, how do you measure? And it is the reason why general relativity is hard to understand: all of a sudden, coordinates become dependent on the observer! Mathematically you just want to know whether it is an ellipse, a parabola or an hyperbola.

 

[etotheipi]

Problem 5, I can do the first two parts today, too tired to attempt the harder final parts,$$\begin{align*}

\text{det}\begin{bmatrix}5 - \lambda &0&1&6\\3&3-\lambda&5&2\\0&0&3-\lambda&0\\6&0&3&-\lambda\end{bmatrix} &= \text{det}\begin{bmatrix}0&0&3-\lambda&0\\6&0&3&-\lambda\\5 - \lambda &0&1&6\\3&3-\lambda&5&2\end{bmatrix} \\ \\

&= (3-\lambda) \text{det} \begin{bmatrix} 6 & 0 & -\lambda \\ 3-\lambda & 0 & 6 \\ 3 & 3-\lambda & 2 \end{bmatrix} \\ \\

&= (3-\lambda)\left( 36(\lambda -3) -\lambda(\lambda-3)^2 \right) \\

&= (\lambda - 3)^2 (\lambda + 4)(\lambda -9) = 0
\end{align*}$$so $\lambda_1 = 3$, $\lambda_2 = -4$, $\lambda_3 = 9$ are the eigenvalues of $A$. Now for the eigenvectors, $\vec{u}, \vec{v}$ and $\vec{w}$,$$\begin{align*}\begin{bmatrix} -4 & 0 & 1 & 6 \\ 3 & -6 & 5 &2 \\ 0 & 0 & -6 & 0 \\ 6 & 0 & 3 & 9\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff u_1 = \frac{3}{2} u_4, u_2 = \frac{13}{12}u_4, u_3 = 0 \\ \\\begin{bmatrix} 9 & 0 & 1 & 6 \\ 3 & 7 & 5 &2 \\ 0 & 0 & 7 & 0 \\ 6 & 0 & 3 & 4\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff v_1 = -\frac{2}{3}v_4, v_2 =0, v_3 = 0 \\ \\ \begin{bmatrix} 2 & 0 & 1 & 6 \\ 3 & 0 & 5 &2 \\ 0 & 0 & 0 & 0 \\ 6 & 0 & 3 & -3\end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff w_1 = w_3 = w_4 = 0\end{align*}$$

So $\vec{u} = \begin{bmatrix} 18 \\ 13 \\ 0 \\ 12 \end{bmatrix}$, $\vec{v} = \begin{bmatrix} -2 \\ 0 \\ 0 \\ 3 \end{bmatrix}$, $\vec{w} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

 

[fresh_42]

Problem 5, I can do the first two parts today, too tired to attempt the harder final parts,$$\begin{align*}

\text{det}\begin{bmatrix}5 - \lambda &0&1&6\\3&3-\lambda&5&2\\0&0&3-\lambda&0\\6&0&3&-\lambda\end{bmatrix} &= \text{det}\begin{bmatrix}0&0&3-\lambda&0\\6&0&3&-\lambda\\5 - \lambda &0&1&6\\3&3-\lambda&5&2\end{bmatrix} \\ \\

&= (3-\lambda) \text{det} \begin{bmatrix} 6 & 0 & -\lambda \\ 3-\lambda & 0 & 6 \\ 3 & 3-\lambda & 2 \end{bmatrix} \\ \\

&= (3-\lambda)\left( 36(\lambda -3) -\lambda(\lambda-3)^2 \right) \\

&= (\lambda - 3)^2 (\lambda + 4)(\lambda -9) = 0
\end{align*}$$so $\lambda_1 = 3$, $\lambda_2 = -4$, $\lambda_3 = 9$ are the eigenvalues of $A$. Now for the eigenvectors, $\vec{u}, \vec{v}$ and $\vec{w}$,$$\begin{align*}\begin{bmatrix} -4 & 0 & 1 & 6 \\ 3 & -6 & 5 &2 \\ 0 & 0 & -6 & 0 \\ 6 & 0 & 3 & 9\end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff v_1 = -\frac{2}{3} v_4, v_2 = \frac{13}{12}u_4, v_3 = 0 \\ \\\begin{bmatrix} 9 & 0 & 1 & 6 \\ 3 & 7 & 5 &2 \\ 0 & 0 & 7 & 0 \\ 6 & 0 & 3 & 4\end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff u_1 = \frac{3}{2}u_4, u_2 =0, u_3 = 0 \\ \\ \begin{bmatrix} 2 & 0 & 1 & 6 \\ 0 & 7 & 5 &2 \\ 0 & 0 & 0 & 0 \\ 6 & 0 & 3 & -3\end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \iff w_1 = w_3 = w_4 = 0\end{align*}$$So $\vec{u} = \begin{bmatrix} 18 \\ 13 \\ 0 \\ 12 \end{bmatrix}$, $\vec{v} = \begin{bmatrix} -2 \\ 0 \\ 0 \\ 3 \end{bmatrix}$, $\vec{w} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

Sleep over it. It is problem 4. And what is the difference between 3 and -4? And what are 9,12,13,18?

 

[etotheipi]

Sleep over it. It is problem 4. And what is the difference between 3 and -4? And what are 9,12,13,18?

Think I got the numbers mixed up, they should be ok now!

 

[fresh_42]

Think I got the numbers mixed up, they should be ok now!

Let's agree on a single set of representatives, not on many at the same time.

 

[mathwonk]

historical hints: 3-weierstrass, inverse mapping theorem, 7-sylow, 9-nakayama.

 

[songoku]

Spoiler: Question 11

Let $f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x$

1. Finding the stationary points of $f(x) \rightarrow f'(x)=0$
$4x^3 + 18x^2 + 22x + 6 = 0$
$(2x+3)(4x^2+12x+4)=0$
$x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}$

Checking the sign diagram of first derivative:
a. For $x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)$

b. For $\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)$

c. For $-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)$

d. For $x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)$

$x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 - \sqrt{5}}{2} , -1)$ is local minimum

$x=-\frac{3}{2} \rightarrow y=\frac{9}{16}$ so $(-\frac{3}{2} , \frac{9}{16})$ is local maximum

$x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1$ so $(\frac{-3 + \sqrt{5}}{2} , -1)$ is local minimum2. Checking the end behaviour of $f(x)$
$\lim_{x \to \pm \infty} {f(x)=\infty}$

So the local minimum found above is also global minimum

3.
$f(x)=a$ will have no solution if $a$ is below global minimum $\rightarrow a < -1$

No $a$ will result in $f(x)$ has one solution

$f(x)=a$ will have two solutions if $a$ is global minimum or $a$ is above global maximum $\rightarrow a=-1$ or $a>\frac{9}{16}$

$f(x)=a$ will have three solutions if $a$ is global maximum $\rightarrow a=\frac{9}{16}$

$f(x)=a$ will have four solutions if $a$ is between local minimum and local maximum $\rightarrow -1<a<\frac{9}{16}$

I am not sure how to write good conclusion for final answer

 

[songoku]

Spoiler: Question13

Let $a=65536$ and $b=32768$ so $z=\frac{(a-3)^3 +(a-2)^3 + (a-1)^3 + a^3 + (a+1)^3 (a+2)^3 + (a+3)^3}{(b-3)(b-2)+(b-1).b+b.(b+1)+(b+2)b+3)}$

After expanding everything: $z=\frac{7a^3+84a}{4b^2 +12}$

Since $a=2b$ :
$z=\frac{56b^3 +168b}{4b^2 +12}$

$=\frac{56b(b^2+3)}{4(b^2+3)}$

$=14b$

 

[fishturtle1]

Spoiler: question 7

Note $351 = 3^3\cdot 13$. Let $n_3$ denote the number of Sylow $3$ subgroups of $G$ and $n_{13}$ denote the number of Sylow $13$ subgroups of $G$. For a given prime $p$ that divides $\vert G \vert$, the Sylow $p$ subgroups are conjugates. So, it's enough to show $n_3 = 1$ or $n_{13} = 1$. By Sylow's theorem, $n_{13} \vert 27$ and $n_{13} = 1 \text{mod 13}$. So, $n_{13} = 1$ or $27$. If $n_{13} = 1$, we are done. Suppose $n_{13} = 27$. This gives $27\cdot 12 = 324$ distinct elements of order $13$. This leaves $351 - 324 = 27$ elements unaccounted for. Each Sylow $3$ subgroup has order $27$, and we know $n_3 \vert 13$. (And no element of a Sylow $3$ subgroup contains an element of order $13$). Combining the last two lines, we have $n_3 = 1$. []
[\spoiler]