Math Challenge - May 2021

[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.

 

[kshitij]

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

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.

I'll keep that in mind next time

 

[fresh_42]

I'll keep that in mind next time

It is not important in a case like this. I still have the dream that people can learn from those problems now and then, and the problem was all about precision. Hence the lesson to be learned is, that the precision of a result is always as good as the precision of the measurement, or calculating devices is. My remark was meant to sharpen your senses, not to criticize anything.

 

[graphking]

7. Let $\alpha$ be an algebraic number of degree $n\geq 1.$ Then there is a real number $c>0$ such that for all $\mathbb{Q}\ni\dfrac{p}{q}\neq \alpha$
$$
\left|\alpha -\dfrac{p}{q}\right|\geq \dfrac{c}{q^n}
$$

Hi guys, I didn't read all your post but it seems that no one say anything about Q7. I remember that is a beautiful theorem called Liouville Theorem to describe algebraic number and transcendental number a little bit. Here's some hint:
let f(x) be the polynomial for $\alpha$ . consider the nearest p/q to $\alpha$ , when p is variable and q is settled. Then try f($\alpha$)-f(p/q), factorize it and get $\alpha - p/q$.

 

[julian]

I may have done it, must go to bed now:

The $\alpha$ is the root of an $n-$th order polynomial with integer coefficients:

\begin{align*}
f (x) = \sum_{i=0}^n a_i x^i
\end{align*}

Put

\begin{align*}
\tilde{\alpha} = \frac{p}{q} .
\end{align*}

If $f (x)$ has rational roots then denote them as $\{ r_1 , \dots , r_k \}$.

Case (a) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| \leq 1$.

We have the identity

\begin{align*}
f (x) & = \sum_{i=1}^n a_i x^i - \sum_{i=1}^n a_i a^i
\nonumber \\
& = \sum_{i=1}^n a_i (x - a) (x^{i-1} + x^{i-2} a + \cdots + a^{i-1})
\nonumber \\
& = (x - a) \sum_{i=1}^n a_i \sum_{j=0}^{i-1} x^{i-1-j} a^j
\end{align*}

Application of the triangle inequality $|\tilde{\alpha}| = |\tilde{\alpha} - \alpha + \alpha| \leq |\alpha - \tilde{\alpha}| + |\alpha|$ together with $|\alpha - \tilde{\alpha}| \leq 1$ implies $|\tilde{\alpha}| \leq 1 + |\alpha|$. We use this to obtain the inequality:

\begin{align*}
|f (\alpha) - f (\tilde{\alpha})| & \leq |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1-j} \tilde{\alpha}^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1}| | \alpha^{-j} (1 + |\alpha|)^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \sum_{j=0}^{i-1} \left( 1 + \frac{1}{|\alpha|} \right)^j
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \frac{\left( 1 + \frac{1}{|\alpha|} \right)^i - 1}{\left( 1 + \frac{1}{|\alpha|} \right) - 1}
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right)
\nonumber \\
& = |\alpha - \tilde{\alpha}| C_\alpha
\end{align*}

where $C_\alpha = \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right) > 0$. We have

\begin{align*}
|f ( \tilde{\alpha} )| & = \left|\frac{a_0 q^n + a_1 q^{n-1} p + \cdots + a_n p^n}{q^n} \right| \geq \frac{1}{q^n}
\end{align*}

as the numerator is a nonzero integer, and so

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \frac{1}{C_\alpha} |f ( \tilde{\alpha} )| \geq \frac{1}{C_\alpha} \frac{1}{q^n}
\end{align*}

Case (b) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| > 1$.

\begin{align*}
|\alpha - \tilde{\alpha}| & > 1 \geq \frac{1}{q^n}
\end{align*}

Case (c) $\alpha \not\in \{ r_1 , \dots , r_k \}$, $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$.

Choose $C_r = \min_i | \alpha - r_i| > 0$. Then

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq C_r \geq \frac{C_r}{q^n}
\end{align*}

(If there are no rational roots, ignore case (c)).

Case (d) Say $\alpha \in \{ r_1 , \dots , r_k \}$ and $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ (in which case $k > 1$). Set $\overline{C}_r = \min_{i \not= j}|r_i - r_j| > 0$,

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \overline{C}_r \geq \overline{C}_r \frac{1}{q^n}
\end{align*}

(If there are no rational roots, ignore case (d)).

Finally, we can write

\begin{align*}
c = \min \left\{
\begin{matrix}
\frac{1}{C_\alpha} & : |\alpha - \frac{p}{q}| \leq 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
1 & : |\alpha - \frac{p}{q}| > 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
C_r & : \alpha \not\in \{ r_1 , \dots , r_k \} , \quad \frac{p}{q} \in \{ r_1 , \dots , r_k \} \\
\overline{C}_r & : \alpha , \frac{p}{q} \in \{ r_1 , \dots , r_k \}
\end{matrix}
\right.
\end{align*}

then

\begin{align*}
\left| \alpha - \frac{p}{q} \right| & \geq \frac{c}{q^n}
\end{align*}

where $c > 0$.

 

[fresh_42]

I may have done it, must go to bed now:

The $\alpha$ is the root of an $n-$th order polynomial with integer coefficients:

\begin{align*}
f (x) = \sum_{i=0}^n a_i x^i
\end{align*}

Put

\begin{align*}
\tilde{\alpha} = \frac{p}{q} .
\end{align*}

If $f (x)$ has rational roots then denote them as $\{ r_1 , \dots , r_k \}$.

Case (a) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| \leq 1$.

We have the identity

\begin{align*}
f (x) & = \sum_{i=1}^n a_i x^i - \sum_{i=1}^n a_i a^i
\nonumber \\
& = \sum_{i=1}^n a_i (x - a) (x^{i-1} + x^{i-2} a + \cdots + a^{i-1})
\nonumber \\
& = (x - a) \sum_{i=1}^n a_i \sum_{j=0}^{i-1} x^{i-1-j} a^j
\end{align*}

The $a$ on the right-hand side should be an $\alpha .$

Application of the triangle inequality $|\tilde{\alpha}| = |\tilde{\alpha} - \alpha + \alpha| \leq |\alpha - \tilde{\alpha}| + |\alpha|$ together with $|\alpha - \tilde{\alpha}| \leq 1$ implies $|\tilde{\alpha}| \leq 1 + |\alpha|$. We use this to obtain the inequality:

\begin{align*}
|f (\alpha) - f (\tilde{\alpha})| & \leq |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1-j} \tilde{\alpha}^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \sum_{j=0}^{i-1} |\alpha^{i-1}| | \alpha^{-j} (1 + |\alpha|)^j |
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \sum_{j=0}^{i-1} \left( 1 + \frac{1}{|\alpha|} \right)^j
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i \alpha^{i-1}| \frac{\left( 1 + \frac{1}{|\alpha|} \right)^i - 1}{\left( 1 + \frac{1}{|\alpha|} \right) - 1}
\nonumber \\
& = |\alpha - \tilde{\alpha}| \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right)
\nonumber \\
& = |\alpha - \tilde{\alpha}| C_\alpha
\end{align*}

The first equality sign has to be less or equal.

where $C_\alpha = \sum_{i=1}^n |a_i| \left( (|\alpha| + 1)^i - |\alpha|^i \right) > 0$. We have

\begin{align*}
|f ( \tilde{\alpha} )| & = \left|\frac{a_0 q^n + a_1 q^{n-1} p + \cdots + a_n p^n}{q^n} \right| \geq \frac{1}{q^n}
\end{align*}

as the numerator is a nonzero integer, and so

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \frac{1}{C_\alpha} |f ( \tilde{\alpha} )| \geq \frac{1}{C_\alpha} \frac{1}{q^n}
\end{align*}

Case (b) $\tilde{\alpha} \not\in \{ r_1 , \dots , r_k \}$ and $|\alpha - \tilde{\alpha}| > 1$.

\begin{align*}
|\alpha - \tilde{\alpha}| & > 1 \geq \frac{1}{q^n}
\end{align*}

Sloppy, since $q^n < 0$ isn't ruled out and $c$ cannot swallow the sign.

Case (c) $\alpha \not\in \{ r_1 , \dots , r_k \}$, $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$.

I assume this is meant to be the other way around since we already covered all cases of $\alpha \not\in \{ r_1 , \dots , r_k \}.$

The good news is that it is irrelevant because we may assume $f(x)$ to be irreducible over $\mathbb{Q}$.

Choose $C_r = \min_i | \alpha - r_i| > 0$. Then

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq C_r \geq \frac{C_r}{q^n}
\end{align*}

(If there are no rational roots, ignore case (c)).

Case (d) Say $\alpha \in \{ r_1 , \dots , r_k \}$ and $\tilde{\alpha} \in \{ r_1 , \dots , r_k \}$ (in which case $k > 1$). Set $\overline{C}_r = \min_{i \not= j}|r_i - r_j| > 0$,

\begin{align*}
|\alpha - \tilde{\alpha}| & \geq \overline{C}_r \geq \overline{C}_r \frac{1}{q^n}
\end{align*}

(If there are no rational roots, ignore case (d)).

Finally, we can write

\begin{align*}
c = \min \left\{
\begin{matrix}
\frac{1}{C_\alpha} & : |\alpha - \frac{p}{q}| \leq 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
1 & : |\alpha - \frac{p}{q}| > 1, \quad \frac{p}{q} \not\in \{ r_1 , \dots , r_k \} \\
C_r & : \alpha \not\in \{ r_1 , \dots , r_k \} , \quad \frac{p}{q} \in \{ r_1 , \dots , r_k \} \\
\overline{C}_r & : \alpha , \frac{p}{q} \in \{ r_1 , \dots , r_k \}
\end{matrix}
\right.
\end{align*}

then

\begin{align*}
\left| \alpha - \frac{p}{q} \right| & \geq \frac{c}{q^n}
\end{align*}

where $c > 0$.

It would have been shorter to write $f(x)=(x-\alpha )g(x)\in \mathbb{C}[x]$ and then discuss the neighborhood of $\alpha$ by continuity of $g(x)$.