Challenge Math Challenge - May 2021

[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. ?

 

[fresh_42]

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. ?

What is the difference?

 

[fishturtle1]

What is the difference?

I think I get it now, I had to look up the definition of GL(V). So, $\rho(g^{-1})$ is an automorphism of $V$ and $\tau(g)$ is an automorphism of $W$ and $\varphi$ is a k linear map from $V$ to $W$.

 

[fresh_42]

I think I get it now, I had to look up the definition of GL(V). So, $\rho(g^{-1})$ is an automorphism of $V$ and $\tau(g)$ is an automorphism of $W$ and $\varphi$ is a k linear map from $V$ to $W$.

Yes. As functions, it is the composition, which in coordinates is matrix multiplication.

 

[fishturtle1]

One other question, what is $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$? I'm pretty sure $\text{Hom}_{\mathbb{K}}(V, W)$ is the set of all $\mathbb{K}$-linear maps from $V$ to $W$. And we can make $V$ into a $\mathbb{K}G$ module by defining $g \cdot v = \rho(g)v$ I think?? So is $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))$ the set of $\mathbb{K}G$ homomorphisms from $V$ to $W$?

 

[fresh_42]

One other question, what is $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$? I'm pretty sure $\text{Hom}_{\mathbb{K}}(V, W)$ is the set of all $\mathbb{K}$-linear maps from $V$ to $W$. And we can make $V$ into a $\mathbb{K}G$ module by defining $g \cdot v = \rho(g)v$ I think?? So is $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))$ the set of $\mathbb{K}G$ homomorphisms from $V$ to $W$?

I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ means the homomorphisms of representations as defined. We have $\rho: G \longrightarrow \operatorname{GL}(V)$ and $\tau : G \longrightarrow \operatorname{GL}(W)$. I do not see any functions from $G$ to $\mathbb{K} .$ The condition of the characteristic simply allows us to divide by $|G|$ in the symmetry operator.

 

[fishturtle1]

I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ means the homomorphisms of representations as defined. We have $\rho: G \longrightarrow \operatorname{GL}(V)$ and $\tau : G \longrightarrow \operatorname{GL}(W)$. I do not see any functions from $G$ to $\mathbb{K} .$ The condition of the characteristic simply allows us to divide by $|G|$ in the symmetry operator.

that clears things up, thank you!

 

[fishturtle1]

Spoiler: problem 4

\textbf{Claim 1.} $\text{Sym}(\varphi)$ is a linear map from $V \to W$.

Proof:
First, we have $\text{char} \mathbb{K} \not\vert \vert G \vert$. So, $\vert G \vert \neq 0$ and $\frac{1}{\vert G \vert}$ is defined. For each $g \in G$, we have $(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W$. Since $W$ is a vector space, it is closed under addition and scalar multiplication. So, $(\text{Sym}\varphi)(v) \in W$. Next, we check that $\text{Sym}(\varphi)$ is $\mathbb{K}$-linear. Let $u, v \in V$ and $\lambda \in \mathbb{K}$. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows $\text{Sym}\varphi$ is $\mathbb{K}$-linear.
[]

\textbf{Claim 2.} The map $\text{Sym}$ is a $\mathbb{K}$-linear mapping.

Proof:
Let $\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)$ and $\lambda \in \mathbb{K}$. For any $g \in G$, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude $\text{Sym}$ is $\mathbb{K}$-linear.
[]

\textbf{Claim 3.} $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace$

Proof:
$(\subseteq)$: Let $\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. By definition, we have $\theta \circ \rho(g) = \tau(g) \circ \theta$ for all $g \in G$. In particular, $\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta$.
\\

$(\supseteq)$: Suppose for all $g \in G$, $\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta$. Multiplying both sides by $\rho(g)$, we have $\tau(g) \circ \theta = \theta \circ \rho(g)$. This shows $\supseteq$.
[]\textbf{Claim 4.} $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ is a subspace of $\text{Hom}_{\mathbb{K}}(V, W)$.

Proof:
Consider the map that sends every element in $V$ to $0$. This map is contained in $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. So, $\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)$. Let $\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}$ and $g \in G$.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, $\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, $\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. We can conclude $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ is a subspace of $\text{Hom}_{\mathbb{K}}(V, W)$.

[]

\textbf{Claim 5.} $\text{Sym}$ is a projection onto $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$.

Proof:
Let $\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. Then $\text{Sym}(\varphi) = \varphi$. In particular, $\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)$. It follows that $\text{Sym}^2 = \text{Sym}$ on $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. This proves Claim 5.
[]

 

[fresh_42]

problem 4
\textbf{Claim 1.} $\text{Sym}(\varphi)$ is a linear map from $V \to W$.

(1) $\operatorname{Im(Sym)} \subseteq \operatorname{Hom}(V,W).$ You haven't pointed out that well-definition is one of the 5 claims, as has been already mentioned by @Infrared. But since you mentioned it implicitly in your proof, I take the following for well-definition, too. (2)

Proof:
First, we have $\text{char} \mathbb{K} \not\vert \vert G \vert$. So, $\vert G \vert \neq 0$ and $\frac{1}{\vert G \vert}$ is defined. For each $g \in G$, we have $(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W$. Since $W$ is a vector space, it is closed under addition and scalar multiplication. So, $(\text{Sym}\varphi)(v) \in W$. Next, we check that $\text{Sym}(\varphi)$ is $\mathbb{K}$-linear. Let $u, v \in V$ and $\lambda \in \mathbb{K}$. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows $\text{Sym}\varphi$ is $\mathbb{K}$-linear.
[]

\textbf{Claim 2.} The map $\text{Sym}$ is a $\mathbb{K}$-linear mapping.

Linearity. (3) This is basically clear from the definition.

Proof:
Let $\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)$ and $\lambda \in \mathbb{K}$. For any $g \in G$, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}

Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude $\text{Sym}$ is $\mathbb{K}$-linear.
[]

\textbf{Claim 3.} $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace$

This is no claim. It is the definition of a homomorphism of representations. You have to prove that it holds for the symmetry operator as we defined it, i.e. that all $\operatorname{Sym}(\varphi )$ "commute" with the representations. It is actually one of two points where an argument is necessary. (The projection is the other one.)

Proof:
$(\subseteq)$: Let $\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. By definition, we have $\theta \circ \rho(g) = \tau(g) \circ \theta$ for all $g \in G$. In particular, $\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta$.
\\

$(\supseteq)$: Suppose for all $g \in G$, $\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta$. Multiplying both sides by $\rho(g)$, we have $\tau(g) \circ \theta = \theta \circ \rho(g)$. This shows $\supseteq$.
[]\textbf{Claim 4.} $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ is a subspace of $\text{Hom}_{\mathbb{K}}(V, W)$.

I actually did not count this as a claim, since the linear spaces are already included by definition. We have some homomorphisms with an additional condition
$$
\{\vartheta :V\longrightarrow W\,|\,\forall_{g\in G}\, : \,\tau(g)\circ\vartheta\circ \rho(g^{-1})=\vartheta \}
$$
and all homomorphisms $\operatorname{Hom}(V,W)$ on the other hand. That it is a subspace follows from the linearity in the condition, which is already contained in your claim (2) if you drop the sums.

Proof:
Consider the map that sends every element in $V$ to $0$. This map is contained in $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. So, $\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)$. Let $\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}$ and $g \in G$.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, $\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, $\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. We can conclude $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$ is a subspace of $\text{Hom}_{\mathbb{K}}(V, W)$.

[]

\textbf{Claim 5.} $\text{Sym}$ is a projection onto $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$.

Proof:
Let $\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. Then $\text{Sym}(\varphi) = \varphi$. In particular, $\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)$. It follows that $\text{Sym}^2 = \text{Sym}$ on $\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))$. This proves Claim 5.
[]

This is wrong. $\operatorname{Sym}(\varphi ) \stackrel{i.g.}{\neq } \varphi .$ You have to calculate that $\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )$.

You missed both crucial points which actually require some "proof":

• $\tau(h)\circ\operatorname{Sym}(\varphi)=\operatorname{Sym}(\varphi)\circ \rho(h)$
• $\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )$

 

[fresh_42]

Correction to the last point: I overlooked that you have chosen $\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))$. So $\operatorname{Sym}(\varphi )=\varphi$ indeed, but why?

 

[fishturtle1]

Correction to the last point: I overlooked that you have chosen $\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))$. So $\operatorname{Sym}(\varphi )=\varphi$ indeed, but why?

I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.

 

[fresh_42]

I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.

Now for the last one: why does $\operatorname{Sym}(\varphi )$ commute with representation matrices?

 

[fishturtle1]

For $\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))$, we have $\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)$

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}

 

[fresh_42]

For $\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))$, we have $\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)$

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}

Yes, but $\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}(V,W)$ and we want to show that actually $\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}((\rho,V),(\tau,W))$. That is, we have an arbitrary homomorphism $\varphi \, : \,V\longrightarrow W$, and only its image satisfies the additional condition, which we want to show.

$\operatorname{Sym}$ is a projection, i.e. maps something from bigger to smaller.

Hint: We haven't used that $\rho$ and $\tau$ are representations, yet.

 

[fishturtle1]

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that $\rho$ and $\tau$ are homomorphisms.

 

[fresh_42]

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that $\rho$ and $\tau$ are homomorphisms.

... and the fact that left multiplication in a group is a bijection $L_{h^{-1}}\, : \,g \longmapsto h^{-1}g$.

Just saying, because the difficulty of the problem was mainly to identify all those seemingly clear facts. It is sometimes more difficult to see what has to be shown than it is to show it.

 

[kshitij]

Problem 11
$$\begin{align}

f(a,b)&=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\

f(a,b)&=\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)^2-2-\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\
\end{align}$$
replace $\frac a b+\frac b a$ with $t$
we get,
$$\begin{align}
f(t)&=(t^2-2)^2-2-t^2+2+t\nonumber\\

f(t)&=t^4-5t^2+t+4\nonumber
\end{align}$$
and as we know that, $a,b\in \mathbb{R}$ and $a,b>0$
So, using $A.M\geq G.M$ we get,
$$\begin{align}
\frac{\dfrac{a}{b}+\dfrac{b}{a}} {2}\geq&\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}\nonumber\\
\dfrac{a}{b}+\dfrac{b}{a}\geq&2\nonumber\\
t\geq2\nonumber
\end{align}$$
Also, $f(t)$ is always increasing for $t\geq2$ because $f'(t)\gt0$ for $t\geq2$

Thus, the minimum value of $f(t)=2$, when $t=2$ or $a=b\space \forall\space a,b\gt0 \in \mathbb{R}$

 

[fresh_42]

Problem 11
$$f(a,b)=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}$$
replace $\frac a b+\frac b a$ with $t$
we get,
$$f(t)=t^4-5t^2+t+4$$
and as we know that, $a,b\in \mathbb{R}$ and $a,b>0$
So, using $A.M\geq G.M$ we get, $$t\geq2$$
Also, $f(t)$ is always increasing for $t\geq2$ because $f'(t)\gt0$ for $t\geq2$

Thus, the minimum value of $f(t)=2$, when $t=2$ or $a=b\space \forall\space a,b\gt0 \in \mathbb{R}$

Can you show your calculations?

 

[kshitij]

Can you show your calculations?

For what?
How did I get $f(t)$?

 

[kshitij]

For what?
How did I get $f(t)$?

$$\begin{align}
f(a,b)&=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\
f(a,b)&=\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)^2-2-\left(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-2\right)+\dfrac{a}{b}+\dfrac{b}{a}\nonumber\\
f(t)&=(t^2-2)^2-2-t^2+2+t\nonumber\\
f(t)&=t^4-5t^2+t+4\nonumber
\end{align}$$

 

[fresh_42]

For what?
How did I get $f(t)$?

Yes. How did you get the polynomial, and how did you get $t\geq 2$ from AM > GM.

 

[kshitij]

Yes. How did you get the polynomial, and how did you get $t\geq 2$ from AM > GM.

I edited that in my response, sorry I didn't include them initially, it takes a lot of effort for me to type this as I'm still new to this.

 

[kshitij]

Yes. How did you get the polynomial, and how did you get $t\geq 2$ from AM > GM.

Also you didn't check my response for Problem 12, have I got that right?

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 [edit] $t=5,-5$ are [edit] 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] $x=-9,-8,-7,-1,0,1$ [edit]

So ordered pairs $(x,y)$ are [edit] $(-9,-12);(-9,12);(-8,0);(-7,0);(-1,0);(0,0);(1,12);(1,-12)$ [edit]

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

Yet another edit: I forgot the $t=-5$ cases here.

Edit (iii): I wrote 144 instead of 12 for some reasons.

 

[fresh_42]

I edited that in my response, sorry I didn't include them initially, it takes a lot of effort for me to type this as I'm still new to this.

In case you will have to write LaTeX more often, here or IRL, it is convenient to download a little script program (I use AutoHotKey), write the script, and run it. It makes LaTeX typing really easy.

Just leave out the keys you need for other purposes like Ctrl+C/V/X/A or Alt+D. But the program allows you to suspend it with a single click in case you need the regular shortcuts.

E.g. if I want to write
\begin{align*}

\end{align*}

I only hit Alt+I which in the script is:
!i::
Send, \begin{{}align*{}}{Enter}{Enter}\end{{}align*{}}{Up}
Return

It may take a while to define shortcuts that you're comfortable with, but if you wrote
\left. \dfrac{\partial }{\partial }\right|_{} for the tenth time, you will appreciate a new shortcut.

 

[kshitij]

In case you will have to write LaTeX more often, here or IRL, it is convenient to download a little script program (I use AutoHotKey), write the script, and run it. It makes LaTeX typing really easy.

Just leave out the keys you need for other purposes like Ctrl+C/V/X/A or Alt+D. But the program allows you to suspend it with a single click in case you need the regular shortcuts.

E.g. if I want to write
\begin{align*}

\end{align*}

I only hit Alt+I which in the script is:
!i::
Send, \begin{{}align*{}}{Enter}{Enter}\end{{}align*{}}{Up}
Return

It may take a while to define shortcuts that you're comfortable with, but if you wrote
\left. \dfrac{\partial }{\partial }\right|_{} for the tenth time, you will appreciate a new shortcut.

Thank you so much for this, I was wondering how you all type all these so fast. I thought maybe it was just a practice thing.

 

[fresh_42]

Also you didn't check my response for Problem 12, have I got that right?

Yet another edit: I forgot the $t=-5$ cases here.

Edit (iii): I wrote 144 instead of 12 for some reasons.

Sorry, that slipped through somehow.

Besides the square instead of the number, you also didn't find all solutions. The ones you have are correct though (with 12 for 144).

 

[fresh_42]

Thank you so much for this, I was wondering how you all type all these so fast. I thought maybe it was just a practice thing.

I also have the problems and their solutions in a TeX file, so I can just copy and paste it. The time-consuming typing is so invisible. I do this in case someone solves an old problem from previous challenges and I don't remember the solution and in order to provide a solution manual if people want to practice and compare their solutions with mine.
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/

 

[kshitij]

you also didn't find all solutions

I'll think about them then.

 

[kshitij]

I'll think about them then.

One more case that I found is with $t=0$, i.e., $x=-4$ & $y=\pm12$

 

[fresh_42]

One more case that I found is with $t=0$, i.e., $x=-4$ & $y=\pm12$

These are two. So you have 8/10 now.

 

[kshitij]

These are two. So you have 8/10 now.

I'll keep thinking then

 

[kshitij]

These are two. So you have 8/10 now.

I think I already have 10,
$$(x,y)=(-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)$$

Edit: I'll add an explanation for t=0 as well.

 

[kshitij]

Problem #12 (second attempt)
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
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 $t=0,\pm3,\pm4,\pm5$

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

if we use $t^2-16=t^2-9$, then we won't get any solutions,

From $t^2-16=l^2\space \text{&} \space t^2-9=m^2$
we get,
$t^2=16+l^2=m^2+9$
clearly $t=5,-5$ are the only possibility as 3,4,5 are pythagorean triplets.

And from $t^2-16=-p^2\space \text{&} \space t^2-9=-q^2$
we get,
$t^2+p^2=16$ & $t^2+q^2=9$
as mentioned earlier, $3^2+4^2=5^2$ is the only possible triplet with 3 & 4, for the above expression to be true, we have either $(t=0,p=\pm4)$ or $(t=\pm4,p=0)$ and similarly, $(t=0,q=\pm3)$ or $(t=\pm3,q=0)$

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

So putting the obtained values back in $t=x+4$ we get $x=-9,-8,-7,-4,-1,0,1$

So ordered pairs $(x,y)$ are $(-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)$

 

[kshitij]

Problem #12 (alternate method)
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute $x+4 \rightarrow t$
then the equation becomes,
$$\begin{align}
y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\
y^2&=(t^2-16)\cdot(t^2-9)\nonumber\\
y^2&=t^4-25t^2+144\nonumber\\
y^2&=\left(t^2-\frac {25} 2\right)^2+144-\frac{625} 4\nonumber\\
4y^2&=(2t^2-25)^2-49\nonumber
\end{align}$$
from this we get,$$\begin{align}(7)^2+(2y)^2&=(2t^2-25)^2\nonumber\end{align}$$
But we know that there is only one possible pythagorean triplet with 7, i.e.,
$$(7)^2+(24)^2=(25)^2$$
So, now there are only two possible cases,

either,
$y=0$ and $2t^2-25=\pm7\Rightarrow t=\pm3,\pm4$

or,
$y=\pm12$ and $2t^2-25=\pm25\Rightarrow t=0,\pm5$

Putting the respective values of $t$ in $x=t-4$ we get,
$$(x,y)\equiv (-9,-12);(-9,12);(-8,0);(-7,0);(-4,-12);(-4,12);(-1,0);(0,0);(1,12);(1,-12)$$

 

[julian]

Do stupid things when V. sleep deprived, but have a go at #3 anyway:

\begin{align*}
I = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\end{align*}

From the standard half angle formula substitution:

\begin{align*}
\int_0^{\pi} f ( \cos x ) dx = \int_0^\infty \frac{2}{1 + t^2} f ( \frac{1 - t^2}{1 + t^2} ) dt
\end{align*}

(didn't notice this was given as a hint until after written this up!). So we can write

\begin{align*}
I & = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\nonumber \\
& = \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{1 - \frac{1 - t^2}{1 + t^2} \frac{1 - u^2}{1 + u^2} \frac{1 - v^2}{1 + v^2}} \frac{2}{1 + t^2} \frac{2}{1 + u^2} \frac{2}{1 + v^2}
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 + u^2 + v^2 + t^2 u^2 v^2} dt du dv
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 u^2 + u^2 v^2 + v^2 t^2 + 1} dt du dv
\end{align*}

where in the last step we made the substitution $t \rightarrow 1/t$, $u \rightarrow 1/u$, $v \rightarrow 1/v$.

Write $p = tu$, $q = uv$, $r = vt$.

\begin{align*}
J (p,q,r) & =
\left|
\frac{( \partial t , \partial u , \partial v) }{( \partial p , \partial q , \partial r) }
\right|
\nonumber \\
& = 1/
\left|
\frac{ ( \partial p , \partial q , \partial r) }{( \partial t , \partial u , \partial v)}
\right|
\nonumber \\
& = 1/
\begin{vmatrix}
u & t & 0 \\
0 & v & u \\
v & 0 & t
\end{vmatrix}
\nonumber \\
&= 1 / 2 tuv
\nonumber \\
&= \frac{1}{2 \sqrt{pqr}}
\end{align*}

So $J (p,q,r) = \frac{1}{2 \sqrt{pqr}}$ with $0 \leq p < \infty$, $0 \leq q < \infty$, $0 \leq r < \infty$:

\begin{align*}
I & = 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{p^2 + q^2 + r^2 + 1} \frac{dp dq dr}{\sqrt{pqr}}
\nonumber \\
&= \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{x + y + z + 1} \frac{dx dy dz}{(xyz)^{3/4}} \qquad (\text{used } x = p^2 \text { etc})
\nonumber \\
& = \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{(xyz)^{3/4}} \left( \int_0^\infty e^{- \alpha (x + y + z + 1)} d \alpha \right) dx dy dz
\nonumber \\
& = \frac{1}{4} \int_0^\infty e^{- \alpha} \left( \int_0^\infty \frac{1}{x^{3/4}} e^{- \alpha x} dx \right)^3 d \alpha
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty \frac{e^{- \alpha}}{\alpha^{3/4}} d \alpha \right) \left( \int_0^\infty \frac{1}{x^{3/4}} e^{-x} dx \right)^3
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty e^{-x} x^{\frac{1}{4} - 1} dx \right)^4
\nonumber \\
& = \frac{1}{4} \left[ \Gamma \left( \frac{1}{4} \right) \right]^4
\end{align*}

 

[fresh_42]

Do stupid things when V. sleep deprived, but have a go at #3 anyway:

\begin{align*}
I = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\end{align*}

From the standard half angle formula substitution:

\begin{align*}
\int_0^{\pi} f ( \cos x ) dx = \int_0^\infty \frac{2}{1 + t^2} f ( \frac{1 - t^2}{1 + t^2} ) dt
\end{align*}

(didn't notice this was given as a hint until after written this up!). So we can write

\begin{align*}
I & = \int_0^{\pi} \int_0^{\pi} \int_0^{\pi} \frac{1}{1 - \cos x \cos y \cos z} dx dy dz
\nonumber \\
& = \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{1 - \frac{1 - t^2}{1 + t^2} \frac{1 - u^2}{1 + u^2} \frac{1 - v^2}{1 + v^2}} \frac{2}{1 + t^2} \frac{2}{1 + u^2} \frac{2}{1 + v^2}
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 + u^2 + v^2 + t^2 u^2 v^2} dt du dv
\nonumber \\
& = 4 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{t^2 u^2 + u^2 v^2 + v^2 t^2 + 1} dt du dv
\end{align*}

where in the last step we made the substitution $t \rightarrow 1/t$, $u \rightarrow 1/u$, $v \rightarrow 1/v$.

Write $p = tu$, $q = uv$, $r = vt$.

\begin{align*}
J (p,q,r) & =
\left|
\frac{( \partial t , \partial u , \partial v) }{( \partial p , \partial q , \partial r) }
\right|
\nonumber \\
& = 1/
\left|
\frac{ ( \partial p , \partial q , \partial r) }{( \partial t , \partial u , \partial v)}
\right|
\nonumber \\
& = 1/
\begin{vmatrix}
u & t & 0 \\
0 & v & u \\
v & 0 & t
\end{vmatrix}
\nonumber \\
&= 1 / 2 tuv
\nonumber \\
&= \frac{1}{2 \sqrt{pqr}}
\end{align*}

So $J (p,q,r) = \frac{1}{2 \sqrt{pqr}}$ with $0 \leq p < \infty$, $0 \leq q < \infty$, $0 \leq r < \infty$:

\begin{align*}
I & = 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{p^2 + q^2 + r^2 + 1} \frac{dp dq dr}{\sqrt{pqr}}
\nonumber \\
&= \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{x + y + z + 1} \frac{dx dy dz}{(xyz)^{3/4}} \qquad (\text{used } x = p^2 \text { etc})
\nonumber \\
& = \frac{1}{4} \int_0^\infty \int_0^\infty \int_0^\infty \frac{1}{(xyz)^{3/4}} \left( \int_0^\infty e^{- \alpha (x + y + z + 1)} d \alpha \right) dx dy dz
\nonumber \\
& = \frac{1}{4} \int_0^\infty e^{- \alpha} \left( \int_0^\infty \frac{1}{x^{3/4}} e^{- \alpha x} dx \right)^3 d \alpha
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty \frac{e^{- \alpha}}{\alpha^{3/4}} d \alpha \right) \left( \int_0^\infty \frac{1}{x^{3/4}} e^{-x} dx \right)^3
\nonumber \\
& = \frac{1}{4} \left( \int_0^\infty e^{-x} x^{\frac{1}{4} - 1} dx \right)^4
\nonumber \\
& = \frac{1}{4} \left[ \Gamma \left( \frac{1}{4} \right) \right]^4
\end{align*}

The integral is known as Watson integral. Its value is
$$
\int_0^\pi \int_0^\pi \int_0^\pi \dfrac{1}{1-\cos x\,\cos y\,\cos z}\,dx\,dy\,dz=\dfrac{1}{4}\,\Gamma\left(\dfrac{1}{4}\right)^4=2\pi {\overline\omega}^2=2G^2\pi^3\approx 43.198
$$
with the Gauß constant $G=\displaystyle{\dfrac{2}{\pi}}\int_0^1\dfrac{ds}{\sqrt{1-s^4}}\,.$

 

[kshitij]

I have been trying problem #14 for a long long time, I hope that I finally got it right,

$$f(x)=a_nx^n+\ldots+a_1x+a_0$$

(case I)

Let $a_n>0$, then $f(x)>0 \space \forall \space x \in\mathbb{R}$
And $f(x) \to +\infty$ when $x \to \pm \infty$ {as $f(x)$ has no real roots, thus n is even}

Also, the coefficient of $x^n$ in $F(x)$ is also $a_n$
$\therefore F(x) \to +\infty$ when $x \to \pm \infty$ {as n is even}
$\therefore$ maximum value of F(x) is $+\infty$

Now,
$$\begin{align}
F(x)&=f(x)+h\cdot f'(x)+h^2\cdot f''(x)+\ldots+h^n\cdot f^{(n)}(x)\nonumber\\
F(x)&=f(x)+h\left(f'(x)+h\cdot f''(x)+\ldots+h^{(n-1)}\cdot f^{(n)}(x)\right)\nonumber\\
F(x)&=f(x)+h\cdot F'(x)
\end{align}$$

Let the minimum value of $F(x)$ is at $x=a$, thus $F'(a)=0$, so using equation (1),
$$F(a)=f(a)>0\space \left(\text{as}\space f(x)>0 \space \forall \space x \in\mathbb{R}\right)$$
$\therefore$ minimum value of $F(x)$ is $F(a)$ which is greater than zero.
$\therefore F(x)>0 \space \forall \space x \in\mathbb{R}$

(case II)

Similarly, if $a_n<0$, then $f(x)<0 \space \forall \space x \in\mathbb{R}$
And $f(x) \to -\infty$ when $x \to \pm \infty$

$\therefore F(x) \to -\infty$ when $x \to \pm \infty$
$\therefore$ minimum value of F(x) is $-\infty$

Let the maximum value of $F(x)$ is at $x=b$, thus $F'(b)=0$, and using equation (1),
$$F(b)=f(b)<0\space \left(\text{as}\space f(x)<0 \space \forall \space x \in\mathbb{R}\right)$$
$\therefore$ maximum value of $F(x)$ is $F(b)$ which is less than zero.
$\therefore F(x)<0 \space \forall \space x \in\mathbb{R}$

Thus, we can see that $F(x)$ is never equal to zero for any real value of $x$
$\therefore$ it doesn't have any real zeros.

 

[kshitij]

Problem 13
$$\underbrace{\left|x-\dfrac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\right|}_{=:f(x)}\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$$
let,
$$\begin{align}
f(x)&=x-\frac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\nonumber\\
f'(x)&=1-\frac{\cos(x)(14+\cos(x))(9+6\cos(x))-\sin^2(x)(9+6\cos(x))+6\sin^2(x)(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{(9+6\cos(x))^2-(14\cos(x)+\cos^2(x))(9+6\cos(x))+(1-\cos^2(x))(9+6\cos(x))-6(1-\cos^2(x))(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{81+36\cos^2(x)+108\cos(x)-126\cos(x)-84\cos^2(x)-9\cos^2(x)-6\cos^3(x)+9+6\cos(x)-9\cos^2(x)-6\cos^3(x)-84-6\cos(x)+84\cos^2(x)+6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{6-18\cos(x)+18\cos^2(x)-6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{6(1-\cos(x))^3}{(9+6\cos(x))^2}\nonumber
\end{align}$$
We can see that $f'(x)>0 \space \forall \space x \in \mathbb{R}$
$\therefore f(x)$ is always increasing

Thus, the minimum value of $f(x)\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$ is 0 at $x=0$
And, maximum value is,
$$\begin{align}
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{\frac{1}{\sqrt2}\left(14+\frac{1}{\sqrt2}\right)}{(9+3\sqrt2)}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)}{(6(3+\sqrt2))}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)(3-\sqrt2)}{42}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{42\sqrt2-28+3-\sqrt2}{42}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{41\sqrt2}{42}+\frac{25}{42}\nonumber
\end{align}$$
On putting the values of $\pi$ and $\sqrt2$, we get
$$f(x)_{max}=9.7261908 \times 10^{-5}$$
$\therefore f(x)\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$

 

[fresh_42]

I have been trying problem #14 for a long long time, I hope that I finally got it right,

$$f(x)=a_nx^n+\ldots+a_1x+a_0$$

(case I)

Let $a_n>0$, then $f(x)>0 \space \forall \space x \in\mathbb{R}$
And $f(x) \to +\infty$ when $x \to \pm \infty$ {as $f(x)$ has no real roots, thus n is even}

Also, the coefficient of $x^n$ in $F(x)$ is also $a_n$
$\therefore F(x) \to +\infty$ when $x \to \pm \infty$ {as n is even}
$\therefore$ maximum value of F(x) is $+\infty$

Now,
$$\begin{align}
F(x)&=f(x)+h\cdot f'(x)+h^2\cdot f''(x)+\ldots+h^n\cdot f^{(n)}(x)\nonumber\\
F(x)&=f(x)+h\left(f'(x)+h\cdot f''(x)+\ldots+h^{(n-1)}\cdot f^{(n)}(x)\right)\nonumber\\
F(x)&=f(x)+h\cdot F'(x)
\end{align}$$

Let the minimum value of $F(x)$ is at $x=a$, thus $F'(a)=0$, so using equation (1),
$$F(a)=f(a)>0\space \left(\text{as}\space f(x)>0 \space \forall \space x \in\mathbb{R}\right)$$
$\therefore$ minimum value of $F(x)$ is $F(a)$ which is greater than zero.
$\therefore F(x)>0 \space \forall \space x \in\mathbb{R}$

(case II)

Similarly, if $a_n<0$, then $f(x)<0 \space \forall \space x \in\mathbb{R}$
And $f(x) \to -\infty$ when $x \to \pm \infty$

$\therefore F(x) \to -\infty$ when $x \to \pm \infty$
$\therefore$ minimum value of F(x) is $-\infty$

Let the maximum value of $F(x)$ is at $x=b$, thus $F'(b)=0$, and using equation (1),
$$F(b)=f(b)<0\space \left(\text{as}\space f(x)<0 \space \forall \space x \in\mathbb{R}\right)$$
$\therefore$ maximum value of $F(x)$ is $F(b)$ which is less than zero.
$\therefore F(x)<0 \space \forall \space x \in\mathbb{R}$

Thus, we can see that $F(x)$ is never equal to zero for any real value of $x$
$\therefore$ it doesn't have any real zeros.

You can abbreviate the second case by simply mention that we can use $-f(x)$ instead.

 

[kshitij]

You can abbreviate the second case by simply mention that we can use $-f(x)$ instead.

As I said that I was trying this problem for a long time, I didn't think much after I got an idea about how to prove it, I was just too excited to post it here

 

[fresh_42]

Problem 13
$$\underbrace{\left|x-\dfrac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\right|}_{=:f(x)}\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$$
let,
$$\begin{align}
f(x)&=x-\frac{\sin(x)(14+\cos(x))}{9+6\cos(x)}\nonumber\\
f'(x)&=1-\frac{\cos(x)(14+\cos(x))(9+6\cos(x))-\sin^2(x)(9+6\cos(x))+6\sin^2(x)(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{(9+6\cos(x))^2-(14\cos(x)+\cos^2(x))(9+6\cos(x))+(1-\cos^2(x))(9+6\cos(x))-6(1-\cos^2(x))(14+\cos(x))}{(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{81+36\cos^2(x)+108\cos(x)-126\cos(x)-84\cos^2(x)-9\cos^2(x)-6\cos^3(x)+9+6\cos(x)-9\cos^2(x)-6\cos^3(x)-84-6\cos(x)+84\cos^2(x)+6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{6-18\cos(x)+18\cos^2(x)-6\cos^3(x)} {(9+6\cos(x))^2}\nonumber\\
f'(x)&=\frac{6(1-\cos(x))^3}{(9+6\cos(x))^2}\nonumber
\end{align}$$
We can see that $f'(x)>0 \space \forall \space x \in \mathbb{R}$
$\therefore f(x)$ is always increasing

Thus, the minimum value of $f(x)\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$ is 0 at $x=0$
And, maximum value is,
$$\begin{align}
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{\frac{1}{\sqrt2}\left(14+\frac{1}{\sqrt2}\right)}{(9+3\sqrt2)}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)}{(6(3+\sqrt2))}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{(14\sqrt2+1)(3-\sqrt2)}{42}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{42\sqrt2-28+3-\sqrt2}{42}\nonumber\\
f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{41\sqrt2}{42}+\frac{25}{42}\nonumber
\end{align}$$
On putting the values of $\pi$ and $\sqrt2$, we get
$$f(x)_{max}=9.7261908 \times 10^{-5}$$
$\therefore f(x)\leq 10^{-4}\text{ for } x\in \left[0,\dfrac{\pi}{4}\right]$

You should have used the approximations that I gave in the problem statement, not a calculator so that the final conclusion would be

Now $\pi/4= 0.7853975+\dfrac{\delta}{4} < 0.7854$ and $\dfrac{41\sqrt{2}-25}{42}=\dfrac{41\cdot 1.41421-25+41 \varepsilon }{42}>\dfrac{32.98261}{42}>0.7853\,,$ i.e. $0\leq f(x)<0.7854-0.7853=10^{-4}.$

but, yes, this is correct.

The reason is: If you use a calculator, then you make implicitly the assumption, that it is more precise than the values you have been given. Well, this is probably correct, as long as you didn't use a slide rule. Nevertheless, it is an assumption about a device you have no control of and you should be aware of it, e.g. if you write a protocol of an experiment.