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Math Help - Can you solve this !!

  1. #1
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    Can you solve this !!

    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.
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  2. #2
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    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.
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  3. #3
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    Beautiful.

    That's what you need to do.

    You have a trained eye, or u must have seen this before.
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  4. #4
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    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
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    I've been doing math for a while
    Cool


    Now about the second one .

    Experience won't help.
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  6. #6
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    Amazing no one can solve the second one !
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  7. #7
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    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.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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  9. #9
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    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.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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  11. #11
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    Quote Originally Posted by ╔(σ_σ)╝ View Post
    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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