Math Challenge - May 2021

[fresh_42]

I get

$$I=36\sum_{k=0}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

as a series solution for the integral using integral transformations and series expansions and Beta functions to finish it off.

This is wrong. I don't know the series expansion of the solution, since it involves $\pi$ and there are so many series for $\pi$. But I checked with $\lambda =0$ and have definitely another number.

Are we allowed to swap summation and integration if we use the series expansion? It's better to swap integration with something else.

 

[benorin]

This is wrong. I don't know the series expansion of the solution, since it involves $\pi$ and there are so many series for $\pi$. But I checked with $\lambda =0$ and have definitely another number.

Are we allowed to swap summation and integration if we use the series expansion? It's better to swap integration with something else.

I think I see my error, the Beta function integral is not valid for $k=0$ term. So it would have been$$I=\underbrace{4\int_0^1\int_0^1 (1-u)^{-\tfrac{1}{3}}(1-v)^{-\tfrac{2}{3}}\, du\, dv}_{=18} + 36\sum_{k=1}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

if this is also incorrect I guess my calculus prowess is what it was back in college :p

 

[fresh_42]

I think I see my error, the Beta function integral is not valid for $k=0$ term. So it would have been$$I=4\int_0^1\int_0^1 (1-u)^{-\tfrac{1}{3}}(1-v)^{-\tfrac{2}{3}}\, du\, dv + 36\sum_{k=1}^\infty (-1)^k\tfrac{\lambda ^{3k}}{(3k-1)(3k-2)}$$

if this is also incorrect I guess my calculus prowess is what it was back in college :p

I have no idea. I am looking for a cute little solution. (cp. post #27)

The solution is a little beauty!

 

[benorin]

Refresh page and look again, I edited
@fresh_42

 

[fresh_42]

Refresh page and look again, I edited
@fresh_42

No. The approach via series makes it harder to see the trick. Another hint which is a good one for any integral is: eliminate what disturbs the most.

 

[benorin]

To start with separate the integrals like this

$$I=\int_0^\infty\int_0^\infty e^{-\tfrac{\lambda ^3}{xy}}x^{-\tfrac{2}{3}}y^{-\tfrac{1}{3}}\, dx\, dy \cdot \underbrace{\left( \int_0^\infty e^{-x}\, dx\right) ^2}_{=4 \Gamma ^2 (2)=4}$$

both converge as they should. Then to work on the left hand integrals by transformations. Will edit with the rest later.

 

[fresh_42]

This happens, dear students, if you do not note the dependencies! It has to be "for all $\varepsilon$ there is an $N(\varepsilon )$" and not "for all $\varepsilon$ there is an $N$". Sloppiness has its price!

 

[benorin]

I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows

 

[fresh_42]

I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows

But if you find the idea, then it is really beautiful. @Fred Wright has been close, even with the series, but this isn't the nice solution. Forget series.

 

[Infrared]

I think Problem 2 can be done without any fiddling with commutators:

Consider the function $\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.$ By the universal property of free groups, this uniquely extends to a homomorphism $F_n\to\mathbb{Z}^n.$ Note that $x\in F_n$ lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator $w_i$ is zero).

Since $\mathbb{Z}^n$ is abelian, this homomorphism descends to the quotient $F_n/[F_n,F_n]\to\mathbb{Z}_n.$ We check that this map is injective by constructing a left inverse: since both groups are abelian and $\mathbb{Z}^n$ is free abelian, the function $\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]$ taking $e_i$ to the class of $w_i$ (uniquely) extends to a homomorphism $\mathbb{Z}^n\to F_n/[F_n,F_n]$ which is clearly a left inverse. (In fact, the map $F_n/[F_n,F_n]\to\mathbb{Z}^n$ is also clearly surjective so it is an isomorphism).

Now, by injectivity, $\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].$ On the other hand, we already know that the kernel of $F^n\to\mathbb{Z}^n$ is the set of elements in $F_n$ whose exponents satisfy the given condition.

(Here, $e_i$ is the vector whose $i$-th component is $1$ and all others are zero.)

 

[Infrared]

Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that $\text{Sym}$ is well-defined: Why can you divide by $|G|$? Why is $\text{Sym}(\varphi)$ actually an element of $\text{Hom}_{\mathbb{K}}(V,W)$?

2. Show that $\text{Sym}$ is $\mathbb{K}$-linear.

3. Why is $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$ a subspace of $\text{Hom}_{\mathbb{K}}(V,W)?$

4. Why is $\text{Sym}(\varphi)$ an element of the above space?

5. Why is $\text{Sym}$ a projection onto $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$? That is, show that $\text{Sym}^2=\text{Sym}$ and that the image of $\text{Sym}$ is this space.Also, problem $5$ confuses me. To me, it looks like $f^n(x)$ is a polynomial of odd degree $>1$ for all $n$ (not just even). So, $f^n(x)-x$ is also an odd degree polynomial and thus has a root, which is a fixed point of $f^n$. Am I missing something? Maybe the base field is $\mathbb{Q}$ instead of $\mathbb{R}$?

 

[fresh_42]

I think Problem 2 can be done without any fiddling with commutators:

Consider the function $\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.$ By the universal property of free groups, this uniquely extends to a homomorphism $F_n\to\mathbb{Z}^n.$ Note that $x\in F_n$ lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator $w_i$ is zero).

Since $\mathbb{Z}^n$ is abelian, this homomorphism descends to the quotient $F_n/[F_n,F_n]\to\mathbb{Z}_n.$ We check that this map is injective by constructing a left inverse: since both groups are abelian and $\mathbb{Z}^n$ is free abelian, the function $\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]$ taking $e_i$ to the class of $w_i$ (uniquely) extends to a homomorphism $\mathbb{Z}^n\to F_n/[F_n,F_n]$ which is clearly a left inverse. (In fact, the map $F_n/[F_n,F_n]\to\mathbb{Z}^n$ is also clearly surjective so it is an isomorphism).

Now, by injectivity, $\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].$ On the other hand, we already know that the kernel of $F^n\to\mathbb{Z}^n$ is the set of elements in $F_n$ whose exponents satisfy the given condition.

(Here, $e_i$ is the vector whose $i$-th component is $1$ and all others are zero.)

Does this count as an attempt at an answer?

The idea is correct, although you can capture all this within 4 lines. Consider for an arbitrary group $G\longrightarrow G/[G,G],$ set $G=F_n$, done.

Since you had all points in your answer I count this as solved. But I want to mention that the direct path via induction along word length is only marginally longer and still shorter than the long answer above.

 

[fresh_42]

Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that $\text{Sym}$ is well-defined: Why can you divide by $|G|$? Why is $\text{Sym}(\varphi)$ actually an element of $\text{Hom}_{\mathbb{K}}(V,W)$?

2. Show that $\text{Sym}$ is $\mathbb{K}$-linear.

3. Why is $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$ a subspace of $\text{Hom}_{\mathbb{K}}(V,W)?$

4. Why is $\text{Sym}(\varphi)$ an element of the above space?

5. Why is $\text{Sym}$ a projection onto $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$? That is, show that $\text{Sym}^2=\text{Sym}$ and that the image of $\text{Sym}$ is this space.

Not quite sure, whether your point 4 equals my 'homomorphism of representations' property, but I think so.

Also, problem $5$ confuses me. To me, it looks like $f^n(x)$ is a polynomial of odd degree $>1$ for all $n$ (not just even). So, $f^n(x)-x$ is also an odd degree polynomial and thus has a root, which is a fixed point of $f^n$. Am I missing something? Maybe the base field is $\mathbb{Q}$ instead of $\mathbb{R}$?

Can happen with self-constructed problems. I was so busy to find a good polynomial, that I missed the obvious. I will correct it and make it difficult.

 

[Mayhem]

As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?

 

[fresh_42]

As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?

The integration is usually inside out according to the order of the $dx \,(1)\,dy\,(2)\,dz\,(3)$ terms.

Hint: In this case, however, nested integrals are not necessary.

 

[Office_Shredder]

I feel really dumb reading #4, because I don't even see five separate claims to prove?

 

[kshitij]

Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

edit: In the above answer, $a^2+b^2\neq 0$

 

[fresh_42]

Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

Why does $x=y=0$ imply $z=0$?

 

[kshitij]

Why does x=y=0 imply z=0?

I used that z=0 imply x=y=0

as we have $2cz=(a^2+b^2)z^2$ from here we get one value of z as zero and substituting that we get x=y=0

 

[fresh_42]

I used that z=0 imply x=y=0

as we have $2cz=(a^2+b^2)z^2$ from here we get one value of z as zero and substituting that we get x=y=0

Sure, but what if $z\neq 0$? You missed a possibility.

 

[kshitij]

Sure, but what if $z\neq 0$?

then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
as we had,
$$\begin{align}

2x=z(a+b)\nonumber\\

2y=z(a-b)\nonumber

\end{align}$$
and putting $2cz=(a^2+b^2)z^2$ ⇒ $z=\frac {2c} {a^2+b^2}$in these two equations, we get the above values of (x,y,z)

 

[kshitij]

Problem #12
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute $x+4 \rightarrow t$
then the equation becomes,
$$\begin{align}
y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\
y^2&=(t^2-16)\cdot(t^2-9)\nonumber
\end{align}$$
so, for the R.H.S to be a perfect square,
the only possibilities are $t=3,4,5$

as for product of two numbers (say $a,b$) to be a perfect square, the only possibilities are,
if $a=b,a=0,b=0\space or\space a=l^2,b=m^2$ (where l,m are any real numbers)

if we use $t^2-16=t^2-9$, then we won't get any solutions, so we'll have to use
$t^2-16=l^2\space \text{&} \space t^2-9=m^2$
from this we get,
$t^2=16+l^2=m^2+9$
clearly $t=5$ is the only possibility as 3,4,5 are pythagorean triplets.

Also we can have either $t^2-9=0$ or $t^2-16=0$
from this we get [edit] $t=3,4,-3,-4$ [edit]

So putting the obtained values back in $t=x+4$ we get [edit] $x=-8,-7,-1,0,1$ [edit]

So ordered pairs $(x,y)$ are [edit] $(-8,0);(-7,0);(-1,0);(0,0);(1,144);(1,-144)$ [edit]

*Edited the answer to include all values of (x,y)

 

[fresh_42]

then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
as we had,
$$\begin{align}

2x=z(a+b)\nonumber\\

2y=z(a-b)\nonumber

\end{align}$$
and putting $2cz=(a^2+b^2)z^2$ ⇒ $z=\frac {2c} {a^2+b^2}$in these two equations, we get the above values of (x,y,z)

This works only for $a^2+b^2\neq 0$. Your first post was already correct, except for one special case.

 

[kshitij]

This works only for ab≠0.

Why? What is the problem if ab=0, we should still have
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

Edit: I see that if both a & b are 0 then this is wrong. So yes both a & b shouldn't be zero , but one of them can be right?

 

[kshitij]

Your first post was already correct, except one special case.

Is a=b=0 the special case you were talking about here?

 

[fresh_42]

Is a=b=0 the special case you were talking about here?

Yes. $a=b=c=0$ and $x=y=0$ is a possibility, where $z$ doesn't have to be zero.

 

[kshitij]

Yes. $a=b=c=0$ and $x=y=0$ is a possibility, where $z$ doesn't have to be zero.

But if c=0 then z is also 0

 

[fresh_42]

But if c=0 then z is also 0

No. If $a=b=c=0$ and $x=y=0$ then $z=1$ is a solution, as is any arbitrary value for $z$.

 

[kshitij]

No. If $a=b=c=0$ and $x=y=0$ then $z=1$ is a solution, as is any arbitrary value for $z$.

Yes I missed that, I was looking at this expression
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
Didn't even notice the question, my bad.

 

[fishturtle1]

Sorry for the dumb question but do the $\circ$'s in 4) mean function composition or matrix multiplication? As I understand it, $\tau(g)$ and $\rho(g^{-1})$ are matrices in $GL(W)$ and $GL(V)$ resp. ?