Challenge Math Challenge - June 2020

[fresh_42]

Questions

1. (solved by @nuuskur ) Let $H_1, H_2$ be Hilbert spaces and $T: H_1 \to H_2$ a linear map. Suppose that there is a linear map $S: H_2 \to H_1$ such that for all $x\in H_2$ and all $y \in H_1$ we have
$$\langle Sx,y \rangle = \langle x, Ty \rangle$$
Show that $T$ is continuous. (MQ)

2. (solved by @strangerep , @PeroK ) Let $A$ and $B$ be complex $n\times n$ matrices such that $AB-BA$ is a linear combination of $A$ and $B$. Show that $A$ and $B$ must have a common eigenvector. (IR)

3. (solved by @zinq ) Give an example of a vector space $X$ over $\mathbb{R}$ and a convex subset $Q\subset X,\quad Q\ne X$ such that $Q$ is not contained in any half-space of $X$. (WR)

4. (solved by @benorin , @Delta2 ) Let $S:=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=(2-z)^2,\,0\leq z\leq 2\,\}$ be the surface of a cone $C$ with a circular cross section and a peak at $(0,0,2)$. The orientation of $S$ be such, that the normal vectors point outwards. Calculate the flux through $S$ of the vector field (FR)
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}xy^2\\x^2y\\(x^2+y^2)(1-z)\end{pmatrix}.
$$

5. (solved by @nuuskur ) Let $X$ be a normed space and $f:X\to\mathbb{R}$ be a linear functional that is not bounded. Show that $\ker f$ is dense in $X$. (WR)

6.a.) Let $G$ be a group with a compact Hausdorff topology for which multiplication is continuous. Show that $G$ is a topological group, i.e. show that inversion is continuous. (MQ)
6.b.) Is this still true if we don't assume that $G$ is compact Hausdorff? If yes, give a proof. Otherwise, present a counterexample. (MQ)

7. (solved by @StoneTemplePython ) Let $A$ be an $n\times n$ matrix such that

1. the off-diagonal entries of $A$ are all positive
2. the sum of the entries in each row is negative.

Show that $A$ is invertible. (IR)

8. (solved by @Fred Wright , @benorin ) Calculate for $|\alpha|\geq 1$
$$
\int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx
$$
a.) without using series expansions
b.) by using serious expansions. (FR)

9. Let $X$ be a Banach space and let $Y$ be a normed space. There is a linear operator $A:X\to Y$.
As usual $X',Y'$ stand for the spaces of bounded linear functionals and the dual operator $A':Y'\to X'$ is defined. Perhaps $A'$ is not bounded but it is defined on the whole $Y'$.
Prove that $A$ is bounded. (WR)

10. (solved by @benorin ) Evaluate $\int_0^{\infty}\frac{\log(x)}{x^2-1}dx.$ (IR)

High Schoolers only

11. (solved by @Halc , @etotheipi ) You have two nails in a wall. Can you hang a rope around them in such a way that if either nail is removed, then the rope will fall to the floor?

12.a.) (solved by @etotheipi , @Adesh ) Show that $\sqrt{i^i} \in \mathbb{R}$ where $i$ is the imaginary unit $i=\sqrt{-1}.$
12.b.) (solved by @Adesh ) Which of the following equation signs is wrong and why?
$$
-1 \stackrel{(1)}{=}i\cdot i \stackrel{(2)}{=}\sqrt{-1}\cdot \sqrt{-1}\stackrel{(3)}{=}\sqrt{(-1)\cdot(-1)}\stackrel{(4)}{=}\sqrt{1}\stackrel{(5)}{=}1
$$
12.c.) Calculate all solutions of $z^3=1$ by three different methods.

13. (solved by @Adesh , @PhysicsBoi1908 ) Which is the smallest natural number $n\in \mathbb{N}_0$ such that there are no integers $a,b\in \mathbb{Z}$ with $3a^3+b^3=n?$

14. (solved by @etotheipi ) Is it possible to cover an equilateral triangle with two smaller equilateral triangles without a gap? It's not required that they are of equal area, nor that they won't overlap, only that they are smaller and together have a greater area than the original triangle.

15. (solved by @PhysicsBoi1908 ) Prove: Given $n$ different integers $\{\,a_1,\ldots,a_n\,\}$, then there exists a subset $\{\,a_{j_1},\ldots,a_{j_m}\,\}$ with $1\leq j_1 < \ldots < j_m \leq n$ such that $n$ divides $a_{j_1}+\ldots+a_{j_m}\,.$

 

[Adesh]

I just want to know if question number 8 requires complex analysis. Can we do it by elementary methods and other things?
(I will delete this post of mine if it is inappropriate to ask)

 

[Adesh]

Question 12 a.

Spoiler

Let $\sqrt{ i ^i} = z$. Now, we have
$$
i ^{ i/2} = z \\
\frac{i}{2} ln (i) = ln (z) \\
$$
Since, $e^{i\theta} = \cos \theta + i\sin \theta$, we have $e^{i \pi/2} = i$, therefore $ln(i) = i \pi/2$.

$$
\frac{i}{2} ~ i \frac{\pi}{2}= ln(z) \\
-\frac{\pi}{4} = ln(z) \\
z= e ^{-\pi /4}
$$
Hence, $z$ is a real number.

 

[fresh_42]

I just want to know if question number 8 requires complex analysis. Can we do it by elementary methods and other things?
(I will delete this post of mine if it is inappropriate to ask)

It can be done with real analysis and elementary methods. However, part one is rather tricky, whereas part two only requires to find the series expansion which fits best. Here is a hint for part a):
$$\displaystyle{\int_0^{\pi/2} \log(\sin(x))\,dx=-\frac{\pi}{2}\log(2)}$$

 

[fresh_42]

Question 12 a.

Spoiler

Let $\sqrt{ i ^i} = z$. Now, we have
$$
i ^{ i/2} = z \\
\frac{i}{2} ln (i) = ln (z) \\
$$
Since, $e^{i\theta} = \cos \theta + i\sin \theta$, we have $e^{i \pi/2} = i$, therefore $ln(i) = i \pi/2$.

$$
\frac{i}{2} ~ i \frac{\pi}{2}= ln(z) \\
-\frac{\pi}{4} = ln(z) \\
z= e ^{-\pi /4}
$$
Hence, $z$ is a real number.

If you use the complex logarithm, you have to manage the branches, i.e. prove why the formulas from the reals still hold. We don't even know what $\ln z$ is without additional explanations.

Can you give a solution without complex functions? (Hint: Euler's identity.)

 

[Adesh]

If you use the complex logarithm, you have to manage the branches, i.e. prove why the formulas from the reals still hold. We don't even know what lnzln⁡z\ln z is without additional explanations.

I’m sorry sir, but I couldn’t understand what you said. In the last step I just took the exponential on both sides, hence $ln (z)$ became $z$.

 

[fresh_42]

I’m sorry sir, but I couldn’t understand what you said. In the last step I just took the exponential on both sides, hence ln(z)ln(z) became zz.

Sure, but what is lnzln⁡z? It is not uniquely defined. The equation z=exz=ex has many solutions in the complex numbers.

One has to be especially careful if we apply real formulas to complex numbers, see part b.)

 

[Adesh]

Question 12 b.

Spoiler

Equality number 3 is wrong. There we have written
$$
\sqrt{-1} \cdot ~ \sqrt{-1} = \sqrt{ (-1) \cdot (-1) }$$
But this is true only when the radicand is positive. When they are negative we have to put a negative sign before the radical for combining them. Here is the proof:
$$
\sqrt{-a} \cdot \sqrt {-b}= \sqrt{a} i \cdot \sqrt{b} i \\
\sqrt{-a} \cdot \sqrt{-b}= -\sqrt{a}\sqrt{b} \\
\sqrt{-a} \cdot \sqrt{-b} = - \sqrt{ab}
$$
And putting a negative sign in after equality number 3 would result in a sound conclusion.

 

[Physics lover]

@fresh_42
For question 8th,I am getting iota (i) in my answer.Am I wrong or it is there?Can you please tell.

 

[etotheipi]

I’m sorry sir, but I couldn’t understand what you said. In the last step I just took the exponential on both sides, hence $ln (z)$ became $z$.

For number 12 I would just say $i^{\frac{i}{2}}=((e^{\frac{i\pi}{2}})^i)^{\frac{1}{2}} = (e^{-\frac{\pi}{2}})^\frac{1}{2} = e^{-\frac{\pi}{4}}$

 

[fresh_42]

Question 12 b.

Spoiler

Equality number 3 is wrong. There we have written
$$
\sqrt{-1} \cdot ~ \sqrt{-1} = \sqrt{ (-1) \cdot (-1) }$$
But this is true only when the radicand is positive. When they are negative we have to put a negative sign before the radical for combining them. Here is the proof:
$$
\sqrt{-a} \cdot \sqrt {-b}= \sqrt{a} i \cdot \sqrt{b} i \\
\sqrt{-a} \cdot \sqrt{-b}= -\sqrt{a}\sqrt{b} \\
\sqrt{-a} \cdot \sqrt{-b} = - \sqrt{ab}
$$
And putting a negative sign in after equality number 3 would result in a sound conclusion.

Yes. Here is a more formal way to show it:
\begin{align*}
\sqrt{-1}\cdot \sqrt{-1}& \stackrel{\text{Gauß}}{=} \left(\cos(\pi/2)+i \sin(\pi/2)\right) \cdot \left(\cos(\pi/2)+i \sin(\pi/2)\right) \\
&\stackrel{\text{Euler}}{=}e^{i\pi/2}\cdot e^{i\pi/2}\\
&=e^{i\pi}\\
&=\cos(\pi)+i \cdot\sin(\pi)\\
&=-1+i\cdot 0\\
&=-1
\end{align*}

 

[fresh_42]

@fresh_42
For question 8th,I am getting iota (i) in my answer.Am I wrong or it is there?Can you please tell.

The value of the integral is real.

 

[fresh_42]

For number 12 I would just say $i^{\frac{i}{2}}=((e^{\frac{i\pi}{2}})^i)^{\frac{1}{2}} = (e^{-\frac{\pi}{2}})^\frac{1}{2} = e^{-\frac{\pi}{4}}$

Yes. And with $i=0+1\cdot i = \cos(\pi/2)+i \sin(\pi/2)=e^{i\pi /2}$ it would have been perfect.

 

[etotheipi]

Yes. And with $i=0+1\cdot i = \cos(\pi/2)+i \sin(\pi/2)=e^{i\pi /2}$ it would have been perfect.

I left the rest of the proof in my username. "$i$ am $\text{e to the i pi (over two)}$ "

 

[Adesh]

Sure, but what is lnzln⁡z? It is not uniquely defined. The equation z=exz=ex has many solutions in the complex numbers.

One has to be especially careful if we apply real formulas to complex numbers, see part b.)

Okay, in my original solution I wrote $ln(i)$ as $e^{i\pi/2}$ so, sir, I claim that $ln(z)$ means the principle value natural logarithm of $z$.

 

[Adesh]

I left the rest of the proof in my username. "$i$ am $\text{e to the i pi (over two)}$ "

Oh My God! I used to pronounce you as “Ethopia” really. I never thought of that, I used to think that you had some relations with Ethopia.

 

[fresh_42]

Okay, in my original solution I wrote $ln(i)$ as $e^{i\pi/2}$ so, sir, I claim that $ln(z)$ means the principle value natural logarithm of $z$.

But this is the problem. You used a function but didn't explain its meaning. As you have seen in your solution of part b.), one has to be cautious with complex numbers. This caution was the main reason for the question.

 

[etotheipi]

Spoiler: 14

14. Is it possible to cover an equilateral triangle with two smaller equilateral triangles without a gap? It's not required that they are of equal area, nor that they won't overlap, only that they are smaller and together have a greater area than the original triangle.

My initial reaction was this; wherever you put it, a single smaller triangle can only cover either zero or one vertices of the larger triangle, since the separation of two vertices of the large triangle is larger than the side length of the smaller triangle. So the maximum number of vertices of the larger triangle you can cover with two smaller triangles is two; this means that at least one vertex of the large triangle remains uncovered wherever you put the other two.

 

[wrobel]

I am sorry, in problem 9 please replace "conjugated operator" with "dual operator"
Literal translation from Russian into English was a bad idea. I am sorry again

 

[fresh_42]

Spoiler: 14

My initial reaction was this; wherever you put it, a single smaller triangle can only cover either zero or one vertices of the larger triangle, since the separation of two vertices of the large triangle is larger than the side length of the smaller triangle. So the maximum number of vertices of the larger triangle you can cover with two smaller triangles is two; this means that at least one vertex of the large triangle remains uncovered wherever you put the other two.

Yes. The underlying principle of the reasoning is called "the pigeonhole principle". It is a major and important proof technique.

 

[StoneTemplePython]

7. Let $A$ be an $n\times n$ matrix such that

1. the diagonal entries of $A$ are all positive
2. the sum of the entries in each row is negative.

Show that $A$ is invertible. (IR)

$A:=\left[\begin{matrix}2 & 0 & -1 & -3\\0 & 2 & -3 & -1\\-1 & -3 & 2 & 0\\-3 & -1 & 0 & 2\end{matrix}\right]$
$A\mathbf 1 = -2\cdot \mathbf 1$

but $\det\big(A\big) = 0$
which is a contradiction... ?

 

[Infrared]

Thanks for the the catch @StoneTemplePython Condition 1 should be "The $\textbf{off-diagonal}$ entries of $A$ are all positive." I should have checked more carefully.

 

[benorin]

#8) a)

Let $I(\alpha ) := \int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx\quad (\alpha \geq 1)$, then

$$\begin{gathered} \tfrac{dI}{d\alpha } = \int_0^\pi \tfrac{\partial }{\partial \alpha} \log (1-2\alpha \cos(x)+\alpha ^2)\,dx = 2 \int_0^\pi \tfrac{\alpha -\cos (x) }{1-2\alpha \cos (x)+\alpha ^2} \,dx \\ = \tfrac{1}{\alpha} \int_0^\pi\left(1- \tfrac{1-\alpha ^2}{1-2\alpha \cos(x)+\alpha^2} \right) \,dx \\ \end{gathered}$$

I used a table of integrals here and got

$$\tfrac{dI}{d\alpha } =\tfrac{\pi}{\alpha }-\tfrac{2}{\alpha}\left[ \arctan \left(\tfrac{1+\alpha}{1-\alpha}\tan \left( \tfrac{x}{2}\right)\right)\right| _{x=0}^{\pi} =\tfrac{2\pi}{\alpha}$$

hence

$$I(\alpha )=2\pi \int\tfrac{d\alpha}{\alpha}= 2\pi\log | \alpha | + C$$

I'll leave the rest of the solution (determining the constant) to somebody else. Good luck!

Note: I've read this problem somewhere before.

 

[Adesh]

Summary:: Normed, Banach and Hilbert spaces, topology, geometry, linear algebra, integration, flux.
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

Prove: Given $n$ different integers {a1,…,an}{a1,…,an}\{\,a_1,\ldots,a_n\,\}, then there exists a subset {aj1,…,ajm}{aj1,…,ajm}\{\,a_{j_1},\ldots,a_{j_m}\,\} with 1≤j1<…<jm≤n1≤j1<…<jm≤n1\leq j_1 < \ldots < j_m \leq n such that nnn divides aj1+…+ajm

Although, it is clearly written “n different integers”, but then also I want to ask if all of them are different or not all of them are equal? I mean is there a possibility that two integers could be same?

 

[Adesh]

My attempt at Question 15:

Spoiler

Proof by Induction:-

Base Case: Let $n=1$, so we have a singleton set $\{a_1\}$ and the only subset of this is $\{a_1\}$ itself and it is of course, divisible by 1.

Let's say when we have $k$ different integers, $\{a_1, a_2, a_3, \cdots , a_k\}$ we take a subset $\{a_{j1}, a_{j2}, a_{j3}, \cdots , a_{jm}\}$ such that
$$
\frac{a_{j1} + a_{j2} + \cdots + a_{jm} }{k} = q
$$

Inductive Step: Let's add just one more integer to our previous set of integers, we add $q$. Now, we have $k+1$ integers. So, we have $\{a_1, a_2, \cdots , a_k, q\}$. Take the same subset that we took in the $k$th case along with $q$ in it, so we have $\{a_{j1}, a_{j2}, \cdots , a_{jm} , q\}$. Now,
$$
\frac{a_{j1} + a_{j2} + \cdots a_{jm} + q}{k+1} = \frac{kq + q}{k+1} \\
\frac{a_{j1} + a_{j2} + \cdots a_{jm} + q}{k+1} = q
$$

Hence, we can always choose a subset from the set of given $n$ integers such that $n$ divides the sum of members of subset.

A few words: I suspect the step when I said "let's add $q$ to our set and we will get $k+1$ integers" because I have to make sure that $q$ doesn't occur in the set $\{a_1, a_2, \cdots a_k\}$ (because we're asked in the question to prove for "$n$ different integers"). This is the only thing I doubt, however, question-posters may find some other mistakes, they are welcome to inform me about the flaws in the proof.

 

[benorin]

10. Evaluate $\int_0^{\infty}\frac{\log(x)}{x^2-1}dx.$ (IR)

Owing to the infinite discontinuities of the integrand at $x=0$ and $x=1$, I will split the integral up, $$I_1:=\int_0^{1}\frac{\log(x)}{x^2-1}dx\text{ and }I_2:=\int_1^{\infty}\frac{\log(x)}{x^2-1}dx$$

In $I_1$ substitute $x=e^{-t}\implies dx=-e^{-t}dt\wedge 0\leq t \leq \infty$ and the the integral becomes

$$\begin{gathered} I_1:=\int_0^{\infty}\frac{te^{-t}}{1-e^{-2t}}\, dt =\int_0^{\infty}t\sum_{k=0}^{\infty}e^{-(2k+1)t}\, dt \\ =\sum_{k=0}^{\infty}\int_0^{\infty}te^{-(2k+1)t}\, dt \\ \end{gathered} $$

let $u=(2k+1)t$ then we have
$$\begin{gathered} I_1=\sum_{k=0}^{\infty}\tfrac{1}{(2k+1)^2} \underbrace{\int_0^{\infty}te^{-u}\, dt}_{=\Gamma (2)=1} \\ =\sum_{k=0}^{\infty}\tfrac{1}{(2k+1)^2}=\lambda (2) \\ \end{gathered} $$

where $\lambda (z)$ is the Dirichlet Lambda Function.

The evaluation of the other integral, $I_2$, goes the same as $I_1$ except that the first change of variable is $x=e^t$, and the value is $I_2 =\lambda (2)$. Hence the value of the given integral is $2\lambda (2)$.

 

[wrobel]

I see that 12a has already been solved but I think that $i^i$ is a countable set and I have not seen that was stressed anywhere in the thread perhaps I see not carefully

 

[Adesh]

I see that 12a has already been solved but I think that $i^i$ is a countable set and I have not seen that was stressed anywhere in the thread perhaps I see not carefully

Sir I would very much like to know that. What it means to say “$i^i$ is a countable set” ? As for my knowledge till now, countable set is a set whose elements can be counted (it may be infinite but it is countably infinite). Please explain the point which needed to be stressed.

 

[wrobel]

$$i^i=\{e^{-\pi/2+2\pi n}\mid n\in\mathbb{Z}\}$$

 

[etotheipi]

My attempt at Question 15:

Spoiler

Proof by Induction:-

Base Case: Let $n=1$, so we have a singleton set $\{a_1\}$ and the only subset of this is $\{a_1\}$ itself and it is of course, divisible by 1.

Let's say when we have $k$ different integers, $\{a_1, a_2, a_3, \cdots , a_k\}$ we take a subset $\{a_{j1}, a_{j2}, a_{j3}, \cdots , a_{jm}\}$ such that
$$
\frac{a_{j1} + a_{j2} + \cdots + a_{jm} }{k} = q
$$

Inductive Step: Let's add just one more integer to our previous set of integers, we add $q$. Now, we have $k+1$ integers. So, we have $\{a_1, a_2, \cdots , a_k, q\}$. Take the same subset that we took in the $k$th case along with $q$ in it, so we have $\{a_{j1}, a_{j2}, \cdots , a_{jm} , q\}$. Now,
$$
\frac{a_{j1} + a_{j2} + \cdots a_{jm} + q}{k+1} = \frac{kq + q}{k+1} \\
\frac{a_{j1} + a_{j2} + \cdots a_{jm} + q}{k+1} = q
$$

Hence, we can always choose a subset from the set of given $n$ integers such that $n$ divides the sum of members of subset.

A few words: I suspect the step when I said "let's add $q$ to our set and we will get $k+1$ integers" because I have to make sure that $q$ doesn't occur in the set $\{a_1, a_2, \cdots a_k\}$ (because we're asked in the question to prove for "$n$ different integers"). This is the only thing I doubt, however, question-posters may find some other mistakes, they are welcome to inform me about the flaws in the proof.

I am not sure if this approach works, I think for the reason that you pointed out at the end that it needs to be proven for all possible sets. For instance, what if the new set of size $k+1$ has the new element $q+3$ instead of $q$?

To restate the problem in a simpler form I was thinking that we suppose the parent set has $n$ elements. Each element can be expressed in the form $an+b$, where $b \in \{0,1,2,\dots, n-1\}$. We then need to show that given any combination of $n$ numbers (repetition allowed) from $\{0,1,2,\dots, n-1\}$, we can always find $k$ of them such that their sum equals a multiple of $n$.

For example, if $n=3$ so that $b \in \{0,1,2\}$, for any combination of 3 numbers from that set (e.g. {0,0,0}, {1,2,2}, {1,2,3}, etc.) we must prove that we can always find a group in that combination that adds to a multiple of 3.

 

[fresh_42]

#8) a)

Let $I(\alpha ) := \int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx\quad (\alpha \geq 1)$, then

$$\begin{gathered} \tfrac{dI}{d\alpha } = \int_0^\pi \tfrac{\partial }{\partial \alpha} \log (1-2\alpha \cos(x)+\alpha ^2)\,dx = 2 \int_0^\pi \tfrac{\alpha -\cos (x) }{1-2\alpha \cos (x)+\alpha ^2} \,dx \\ = \tfrac{1}{\alpha} \int_0^\pi\left(1- \tfrac{1-\alpha ^2}{1-2\alpha \cos(x)+\alpha^2} \right) \,dx \\ \end{gathered}$$

I used a table of integrals here and got

$$\tfrac{dI}{d\alpha } =\tfrac{\pi}{\alpha }-\tfrac{2}{\alpha}\left[ \arctan \left(\tfrac{1+\alpha}{1-\alpha}\tan \left( \tfrac{x}{2}\right)\right)\right| _{x=0}^{\pi} =\tfrac{2\pi}{\alpha}$$

hence

$$I(\alpha )=2\pi \int\tfrac{d\alpha}{\alpha}= 2\pi\log | \alpha | + C$$

I'll leave the rest of the solution (determining the constant) to somebody else. Good luck!

Note: I've read this problem somewhere before.

The Feynman trick is the correct idea, but I'd like to see the complete integration, since this is a bit tricky to do. And what about $|\alpha|=1$?

 

[fresh_42]

Although, it is clearly written “n different integers”, but then also I want to ask if all of them are different or not all of them are equal? I mean is there a possibility that two integers could be same?

We consider $n$ different integers. Otherwise we would have $n-1$ many. It is the starting situation, not a resulting property.

 

[fresh_42]

A few words: I suspect the step when I said "let's add $q$ to our set and we will get $k+1$ integers" because I have to make sure that $q$ doesn't occur in the set $\{a_1, a_2, \cdots a_k\}$ (because we're asked in the question to prove for "$n$ different integers").

This is not the point. You cannot "add" an integer you like, since then you could always add $n$ and the statement becomes trivial. We have given any $n$ different integers.

Induction is the wrong idea, sum and division, however, was a good idea. Try different sums.

 

[fresh_42]

$$i^i=\{e^{-\pi/2+2\pi n}\mid n\in\mathbb{Z}\}$$

You can do this with every complex number: $z=re^{i\varphi}=re^{i(\varphi + 2n\pi)}.$ We usually assume $0\leq \varphi < 2\pi$ as representation.

 

[wrobel]

You can do this with every complex number: z=reiφ=rei(φ+2nπ).z=re^{i\varphi}=re^{i(\varphi + 2n\pi)}.

$e^{i(\varphi+2n\pi)}$ is one number while $i^i$ is the infinite set of numbers

 

[benorin]

Ignore this post, it contains an arithmetic error, for the correct solution, read post #40.

4. Let $S:=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=(2-z)^2,\,0\leq z\leq 2\,\}$ be the surface of a cone $C$ with a circular cross section and a peak at $(0,0,2)$. The orientation of $S$ be such, that the normal vectors point outwards. Calculate the flux through $S$ of the vector field (FR)
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}xy^2\\x^2y\\(x^2+y^2)(1-z)\end{pmatrix}.$$

I assume the cone has a base because it's oriented. Let $S_1$ denote the upper cone-shaped surface given by $z=g(x,y):=2-\sqrt{x^2+y^2},\quad ( 0\leq x^2+y^2\leq 2^2)$. Also denote the $x,y,$ and $z$ components of $\vec{F}$ by $P,Q,$ and $R$, respectively. Then

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\= \int_{0}^{2\pi}\int_0^2\left(\tfrac{r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\tfrac{32\pi}{5}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{8\pi}{5}}$.

 

[fresh_42]

I assume the cone has a base because it's oriented. Let $S_1$ denote the upper cone-shaped surface given by $z=g(x,y):=2-\sqrt{x^2+y^2},\quad ( 0\leq x^2+y^2\leq 2^2)$. Also denote the $x,y,$ and $z$ components of $\vec{F}$ by $P,Q,$ and $R$, respectively. Then

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\= \int_{0}^{2\pi}\int_0^2\left(\tfrac{r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\tfrac{32\pi}{5}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{8\pi}{5}}$.

This is the wrong answer. The mistake is in the first part of your calculation, which I do not really understand what you did. Cover the cone with a disc $D$ and calculate $\int_{D\cup S}F\vec{n}dS$.

 

[benorin]

I used a formula out of my old Stuart Calculus book, 4th ed., pg 1100: If $\vec{F}=\left< P,Q,R\right>$ and $S:=\left\{(x,y,z) | z=g(x,y)\forall (x,y)\in D\right\}$ then (with upward directed normal)
$$\iint_{S}\vec{F}\cdot d\vec{S}=\iint_{D}\left( -P \tfrac{\partial g}{\partial x}-Q \tfrac{\partial g}{\partial y}+R\right) \, dA$$

where the instance of $z$ in any of $P,Q,R$ is replaced by $z=g(x,y)$. This is how I learned to calcullate flux through surfaces. Is this incorrect? I checked my integration and arithmetic with WolframAlpha.

 

[benorin]

Oh I see that I "lost" a factor of 2 on a single term along the way. 1 sec.

 

[benorin]

Corrected arithmetic error from previous post, $\boxed{\boxed{\text{corrections are double boxed}}}$.

4. Let $S:=\{\,(x,y,z)\in \mathbb{R}^3\,|\,x^2+y^2=(2-z)^2,\,0\leq z\leq 2\,\}$ be the surface of a cone $C$ with a circular cross section and a peak at $(0,0,2)$. The orientation of $S$ be such, that the normal vectors point outwards. Calculate the flux through $S$ of the vector field (FR)
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}xy^2\\x^2y\\(x^2+y^2)(1-z)\end{pmatrix}.$$

I assume the cone has a base because it's oriented. Let $S_1$ denote the upper cone-shaped surface given by $z=g(x,y):=2-\sqrt{x^2+y^2},\quad ( 0\leq x^2+y^2\leq 2^2)$. Also denote the $x,y,$ and $z$ components of $\vec{F}$ by $P,Q,$ and $R$, respectively. Then

$$\begin{gathered} \iint_{S_1}\vec{F}\cdot d\vec{S} = \iint_{\text{Projection of } S_1\text{ in the }xy\text{-plane}}\left(-Pg_x-Qg_y+R\right)\, dA \\ = \iint_{x^2+y^2\leq 2^2}\left[\tfrac{2x^2y^2}{\sqrt{x^2+y^2}}+(x^2+y^2)\left(-1+\sqrt{x^2+y^2}\right)\right] \, dA\\ = \int_{0}^{2\pi}\int_0^2\left(\tfrac{\boxed{\boxed{-2}}r^4\sin ^2\theta\cos ^2\theta}{r}+r^2(r-1)\right) r\, dr d\theta\\ =\boxed{\boxed{-2}}\int_{0}^{2\pi}\sin ^2\theta\cos ^2\theta\, d\theta\int_0^2r^4\, dr+\int_{0}^{2\pi}d\theta \int_0^2 (r^4-r^3)\, dr=\boxed{\boxed{\tfrac{8\pi}{5}}}\\ \end{gathered} $$

Let $S_2$ denote the circular disk $z=h(x,y)=0,\quad (0\leq x^2+y^2\leq 2^2 )$ with downward pointing normal. Then

$$\begin{gathered} \iint_{S_2}\vec{F}\cdot d\vec{S}=-\iint_{x^2+y^2\leq 2^2}(x^2+y^2)\, dA\\ = -\int_0^{2\pi}\, d\theta \int_0^2 r^3\, dr=\boxed{-8\pi}\\ \end{gathered}$$

Hence, adding the fluxes, we get $\boxed{\boxed{ \iint_{S}\vec{F}\cdot d\vec{S}=-\tfrac{32\pi}{5}}}$.

 

[fresh_42]

Corrected arithmetic error from previous post, .
I assume the cone has a base because it's oriented. Let denote the upper cone-shaped surface given by . Also denote the and components of by and , respectively. ThenLet denote the circular disk with downward pointing normal. ThenHence, adding the fluxes, we get .

Divergence is zero, flux is not. And consider the correct sign! The orientation is given in the problem statement.

 

[benorin]

Ahh, yes. Seems it was a negative two that I fudged. I've edited the corrected post #40, please refresh the page @fresh_42 .

 

[benorin]

The 3b1b (3 blue 1 brown) guy said you never really stop making stupid little mistakes, or something close to that. I see some evidence of this claim. Lol.

 

[fresh_42]

The 3b1b (3 blue 1 brown) guy said you never really stop making stupid little mistakes, or something close to that. I see some evidence of this claim. Lol.

There is no fifth in the solution. It is quite simple if you use Gauß' theorem.

 

[Halc]

I need clarification on problem 11: How to get a rope to fall if either nail is removed.
From what I can tell, the trick is to not have the rope fall, which I suppose could be done by making a loop of it. Maybe the problem is suggesting a loop (the phrase 'hang the rope around' sort of suggests one).

Spoiler

Suppose a 3m rope draped symmetrically between a pair of nails separated by a meter. The rope hands straight down on either side and kind of a curve between. Remove either nail and the rope is over twice as long on one side of the remaining nail as the other, so it just pulls the short side up and falls.

If the rope is a loop, then just reduce a 6 meter loop to a 3 meter rope by doubling it up.

This is way too easy of an answer, and so I presume there's something unstated in the problem that I'm not getting.

 

[benorin]

I will circle back to the flux thing after this post.

The Feynman trick is the correct idea, but I'd like to see the complete integration, since this is a bit tricky to do. And what about $|\alpha|=1$?

Edit: The unabridged version.

#8) a) Let $I(\alpha ) := \int_0^\pi \log (1-2\alpha \cos(x)+\alpha^2)\,dx\quad ( | \alpha | > 1)$, then

$$\begin{gathered} \tfrac{dI}{d\alpha } = \int_0^\pi \tfrac{\partial }{\partial \alpha} \log (1-2\alpha \cos(x)+\alpha ^2)\,dx = 2 \int_0^\pi \tfrac{\alpha -\cos (x) }{1-2\alpha \cos (x)+\alpha ^2} \,dx \\ = \tfrac{1}{\alpha} \int_0^\pi\left(1- \tfrac{1-\alpha ^2}{1-2\alpha \cos(x)+\alpha^2} \right) \,dx \\ \end{gathered}$$

now do Karl Weierstass' trick and apply the change the change of variables $t=\tan \left(\tfrac{x}{2}\right) , 0\leq t\leq \infty\implies \cos (x)= \tfrac{1-t^2}{1+t^2}\text{ and } dx=\tfrac{2\, dt}{1+t^2}$ to get

$$\begin{gathered} \tfrac{dI}{d\alpha } = 2\int_0^\infty \tfrac{\alpha -\tfrac{1-t^2}{1+t^2}}{1-2\alpha \left(\tfrac{1-t^2}{1+t^2}\right)+\alpha ^2}\cdot \tfrac{2\, dt}{1+t^2} = 4\int_0^\infty \tfrac{(\alpha+1)t^2+(\alpha -1)}{\left[ (\alpha +1)^2t^2+(\alpha -1) ^2\right] (1+t^2)}\, dt\\ \end{gathered}$$

Aside: partial faction decomposition of the integrand. Let $A,B, C,$ and $D$ be rational functions of $\alpha$. Then set

$$\begin{gathered}\tfrac{(\alpha+1)t^2+(\alpha -1)}{\left[ (\alpha +1)^2 t^2+(\alpha -1) ^2\right] (1+t^2)}=\tfrac{At+B}{1+t^2}+\tfrac{Ct+D}{ (\alpha +1)^2 t^2+ (\alpha -1) ^2}\\ \implies (\alpha +1) t^2 + (\alpha -1)= (At+B) \left[ (\alpha +1)^2 t^2 + (\alpha -1) ^2 \right] + (Ct+D)(1+t^2)\\ \end{gathered}$$

solving this system of equations is quickest by plugging in enough values of $t$, namely,

$$\begin{gathered} \boxed{t=0} : \alpha -1 = B(\alpha -1)^2+D\implies D=(\alpha -1)\left[ 1-B(\alpha -1)\right] \quad (eqn\, 1) \\ \boxed{t=i} : -2 = (Ai+B)\left[ (\alpha -1)^2+(\alpha +1)^2\right]\implies A=0 \quad (eqn \, 2)\text{ and } B=\tfrac{1}{2\alpha}\quad (eqn \, 3)\\ \end{gathered} $$

Now $(eqn \, 2)$ implies $C=0$ and substituting $(eqn \, 3)$ into $(eqn \, 1)$ gives $D=\tfrac{1}{2}\left(\alpha - \tfrac{1}{\alpha}\right)$. End aside.

Hence we have

$$\begin{gathered} \tfrac{dI}{d\alpha } = 2\int_0^\infty\left( \tfrac{\tfrac{1}{\alpha}}{1+t^2}+\tfrac{\alpha -\tfrac{1}{\alpha}}{ (\alpha +1)^2 t^2+(\alpha -1) ^2}\right) \, dt \\ = \tfrac{2}{\alpha} \left[ \tan ^{-1} (t) \right| _{t=0}^{\infty}+2\tfrac{\alpha -\tfrac{1}{\alpha}}{(\alpha +1) ^2}\int_0^\infty \tfrac{dt}{ t^2+\left( \tfrac{\alpha -1}{\alpha +1}\right) ^2} \\= \tfrac{\pi}{\alpha} +2\tfrac{\alpha -\tfrac{1}{\alpha}}{(\alpha +1) ^2}\cdot \left( \tfrac{\alpha +1}{\alpha -1}\right)\left[ \tan ^{-1} \left( t\cdot \tfrac{\alpha +1}{\alpha -1}\right) \right| _{t=0}^{\infty} \\ = \boxed{\tfrac{2\pi}{\alpha}=\tfrac{dI}{d\alpha}}\\ \end{gathered}$$

hence

$$I(\alpha )=2\pi \int\tfrac{d\alpha}{\alpha}= 2\pi\log | \alpha | + C\quad (\text{for } | \alpha | > 1) $$

As for the case of $\alpha =1$, I know that $I(1)=0$ but infuriatingly enough I cannot prove it with any clever trick thus far. I think there's going to be something on par with the integral of a odd function over a symmetric interval vanishes--but I did trig, log, and basic integral identities for a half hour: no joy! I think I'm just tired. I hope it will come to me later...

Note: I've read this problem somewhere before.

Edit 2: Did I say after this post? Right. Well not right after...

 

[fresh_42]

I hope it will come to me later...

You're closing in. Your answer is almost right and the two cases $|\alpha|=1$ are missing.

 

[fresh_42]

I will circle back to the flux thing after this post.

Note my hint in post #37.

 

[Infrared]

@benorin Your answer is correct, but you can simplify: What is $2\lambda(2)$?

@Halc The condition "will fall down" means that the rope is not supported by a nail. You could take the rope to be a loop, if that's helpful.

 

[StoneTemplePython]

Thanks for the the catch @StoneTemplePython Condition 1 should be "The $\textbf{off-diagonal}$ entries of $A$ are all positive." I should have checked more carefully.

so the 'right answer' is to apply Gerschgorin Discs / strict diagonal dominance and the result falls out. Here's a different approach:

Spoiler

the problem implies all diagonal entries are negative
$A = -D +B$
where $B$ is strictly positive except the diagonal is zero'd out, and $D$ is a diagonal matrix with positive diagonal entries. Multiplication by an invertible matrix doesn't change rank and it is convenient to homogenize the diagonal, so it suffices to check the invertibility of

$D^{-1}A = -I + D^{-1}B = -I + C$
where $C$ is strictly positive except its diagonal is zero

the problem tells us
$A\mathbf 1 \leq -\alpha \mathbf 1$
for some $\alpha \gt 0$
where all inequalities are interpreted component-wise

re-scaling by positive numbers:
$\big(-I + C\big)\mathbf 1 = D^{-1}A\mathbf 1 \leq -\alpha D^{-1}\mathbf 1 \leq -\sigma \mathbf 1$
where $\sigma := \alpha \cdot \min_i (D^{-1}_{i,i})$ which implies $\sigma$ is positive. If for some reason $\sigma \gt 1$ then we can use an even looser upper bound and select $\sigma := \frac{1}{2}$. Thus we have $\sigma \in (0,1)$

re-arranging terms
$C\mathbf 1 \leq (1-\sigma)\mathbf 1$
and using the non-negativity in $C$
$C^2\mathbf 1 \leq (1-\sigma)C\mathbf 1 \leq (1-\sigma)^2\mathbf 1$
$\longrightarrow C^k\mathbf 1 \leq (1-\sigma) C^{k-1}\mathbf 1\leq ... \leq (1-\sigma)^k\mathbf 1$
- - - - -
to help with interpretation:
$\mathbf x := C^{k-1}\mathbf 1$
$\mathbf x = C^{k-1}\mathbf 1 \leq (1-\sigma)^{k-1} \mathbf 1$
so we know $0 \leq x_j \leq (1-\sigma)^{k-1}$ and by non-negativity of $C$, we can specialize to its jth column giving us
$x_j \mathbf c_j \leq (1-\sigma)^{k-1}\mathbf c_j$
summing over the bound:
$C^{k}\mathbf 1 = C \mathbf x = \sum_{j=1}^n x_j \mathbf c_j\leq \sum_{j=1}^n (1-\sigma)^{k-1} \mathbf c_j =(1-\sigma)^{k-1} \cdot \sum_{j=1}^n \mathbf c_j = (1-\sigma)^{k-1} C\mathbf 1 \leq (1-\sigma)^k \mathbf 1$
- - - - -

if $C$ had an eigenvalue with modulus $\geq 1$, we'd have

$1$
$\leq \big \Vert C^k\big \Vert_F$
$= \Big(\sum_{i=1}^n\sum_{j=1}^n \vert c^{(k)}_{i,j}\vert^2 \Big)^\frac{1}{2}$
$\leq \Big(\sum_{i=1}^n\sum_{j=1}^n \vert c^{(k)}_{i,j}\vert \Big) \text{ (triangle inequality)}$
$=\sum_{i=1}^n\sum_{j=1}^n c^{(k)}_{i,j}$
$= \mathbf 1^T C^k \mathbf 1$
$\leq (1-\sigma)^k \mathbf 1^T \mathbf 1$
$=(1-\sigma)^k \cdot n$

but the upper bound may be arbitrarily small for large enough k thus the spectral radius of $C$ is $\lt 1$, and in particular $C$ can't have an eigenvalue of 1 telling us
$\det\big(-I+C\big ) \neq 0\longrightarrow \det\big(A\big) \neq 0$