Can you solve this !!

**Banned for attempted hacking** · February 8th 2009, 02:33 PM

Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.

**Danny** · February 8th 2009, 03:34 PM

Originally Posted by ╔(σ_σ)╝

Prove that
( x^100) / e^(x^70) + e^(x^70) /(x^100) > 2

for every positive x.

Prove that
cos 20 =/= a/b where a and b are integers .

Nice questions i did them i just wanted to see other solutions.

This should be easy for a lot of you i looked around the forum there are a lot of beautiful mathematicians around.

For the first one, let $u = \frac{x^{100}}{e^{x^{70}}}$ , then

$\frac{x^{100}}{e^{x^{70}}} + \frac{e^{x^{70}} }{x^{100}} = u + \frac{1}{u}$

so you want to show $u + \frac{1}{u} > 2$ which is to show that

$u^2 - 2u + 1 > 0$ or $(u-1)^2 > 0$ which is true provide that $u \ne 1$ . Now u is bounded above (using some calculus) so $u \le u_{max} < 1$ establishing the inequality.

**Banned for attempted hacking** · February 8th 2009, 03:43 PM

Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.

**Danny** · February 8th 2009, 03:47 PM

Originally Posted by ╔(σ_σ)╝

Beautiful.

That's what you need to do.

You have a trained eye, or u must have seen this before.

I've been doing math for a while

**Banned for attempted hacking** · February 8th 2009, 03:56 PM

Originally Posted by danny arrigo

I've been doing math for a while

Cool

Now about the second one

.

Experience won't help.

**Banned for attempted hacking** · February 8th 2009, 05:06 PM

Amazing no one can solve the second one !

**ThePerfectHacker** · February 8th 2009, 05:33 PM

Originally Posted by ╔(σ_σ)╝

Prove that
cos 20 =/= a/b where a and b are integers .

This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$ .
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.

**Banned for attempted hacking** · February 8th 2009, 05:37 PM

Originally Posted by ThePerfectHacker

This is an idea, I did not even try doing this.
By the triple angle identity we have a cubic polynomial equaling $\cos 60 = \frac{\sqrt{3}}{2}$ .
Now if we square both sides of the polynomial to make everything rational we just need to argue the polynomial has no rational solutions.

$\cos 60 =/= \frac{\sqrt{3}}{2}$

$\cos 60 = \frac{\sqrt{1}}{2}$

Finish it.

**ThePerfectHacker** · February 8th 2009, 05:47 PM

Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$ .
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.

**Banned for attempted hacking** · February 8th 2009, 06:15 PM

Originally Posted by ThePerfectHacker

Let $x=\cos 20$ then $4x^3 - 3x = \tfrac{\sqrt{3}}{2}$ .
If $x$ was rational then $4x^3 - 3x$ would be rational but it is not.
Thus, $x$ must be irrational.

Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2

**ThePerfectHacker** · February 8th 2009, 06:19 PM

Originally Posted by ╔(σ_σ)╝

Why do you keep saying this ?
$\cos 60 = \frac{\sqrt{3}}{2}$ -- This is wrong

cos 60 = 1/2

Sorry! That was a trivial error.
But the idea still works because: $4x^3 - 3x = \tfrac{1}{2} \implies 8x^3 - 6x -1 = 0$ .
Apply rational roots theorem and conclude there are no rational solutions.

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