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Math Help - Which is bigger: π^e or e^π? Prove it.

  1. #1
    Newbie BraveHeart's Avatar
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    Which is bigger: π^e or e^π? Prove it.

    Which is bigger: \pi^e or e^{\pi}?
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  2. #2
    Super Member Bacterius's Avatar
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    We know that \pi \neq e.

    Let us conjecture that {\pi}^e < e^{\pi}.

    This means that \ln{({\pi}^e)} < \ln{(e^{\pi})}, that is, e \times \ln{(\pi)} < \pi. Therefore, if our conjecture is true, we must have e < \frac{\pi}{\ln{(\pi)}}.

    Let us consider the function f(x) = \frac{x}{\ln{(x)}}. Find its minimum on (1, +\infty). You will find that the minimum of f is e, and this minimum is reached when x = e.

    But we know that \pi \neq e, thus the minimum e is never reached with f(\pi).

    \therefore e < f(\pi). You can follow the steps backwards to finally prove that {\pi}^e < e^{\pi}

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    \Rightarrow That was the nearly-complete proof. Now, you only have to show that the minimum of f(x) is e when x = e, and you are done
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bacterius View Post
    We know that \pi \neq e.

    Let us conjecture that {\pi}^e < e^{\pi}.

    This means that \ln{({\pi}^e)} < \ln{(e^{\pi})}, that is, e \times \ln{(\pi)} < \pi. Therefore, if our conjecture is true, we must have e < \frac{\pi}{\ln{(\pi)}}.

    Let us consider the function f(x) = \frac{x}{\ln{(x)}}. Find its minimum on (1, +\infty). You will find that the minimum of f is e, and this minimum is reached when x = e.

    But we know that \pi \neq e, thus the minimum e is never reached with f(\pi).

    \therefore e < f(\pi). You can follow the steps backwards to finally prove that {\pi}^e < e^{\pi}

    ________________________


    \Rightarrow That was the nearly-complete proof. Now, you only have to show that the minimum of f(x) is e when x = e, and you are done

    Here is another way I just thought of. Bacterius just showed us that this is equivalent to showing that e<\frac{\pi}{\ln(\pi)}\implies 0<\frac{\pi}{\ln(\pi)}-e ...BUT! this is the same as

    0<\frac{\pi}{\ln(\pi)}-\frac{e}{\ln(e)}=\int_e^{\pi}\left(\frac{x}{\ln(x)  }\right)'\text{ }dx and then we have to do is show that \left(\frac{x}{\ln(x)}\right)'=\frac{\ln(x)-1}{\ln^2(x)}\geqslant 0 for e\leqslant x\leqslant \pi but this is clear since for this interval we have that

    \ln(x)\geqslant1\implies \ln(x)-1\geqslant 0 and so \frac{\ln(x)-1}{\ln^2(x)}\geqslant 0. TA-DA! No calculations needed!
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  4. #4
    Super Member Bacterius's Avatar
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    Lol, I wasn't sure my proof was going to pass since this is the algebra forum and my proof is already half-way into calculus, but Drex actually brought in the integral !
    Nice proof though, everything is already trivially proved and one just has to put the bits together
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Bacterius View Post
    Lol, I wasn't sure my proof was going to pass since this is the algebra forum and my proof is already half-way into calculus, but Drex actually brought in the integral !
    Nice proof though, everything is already trivially proved ...
    Haha, I moved the thread to Calculus. I'd like to see someone use pre-university algebra to solve this one!
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  6. #6
    Super Member Bacterius's Avatar
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    I'd like to see someone use pre-university algebra to solve this one!
    Is it actually possible ?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Another way to do it using derivatives is let f(x)=e^x x^{-e}. We have that f'(e)=0 and it's easy to show it's a minimum. So then f(\pi)\geqslant f(e)=1.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by BraveHeart View Post
    Which is bigger: \pi^e or e^{\pi}?
    See here.
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  9. #9
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    Another one:
    \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi

    Note that this can be generalized: for all m > n \ge e, it is true that n^m > m^n
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  10. #10
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    Quote Originally Posted by Unbeatable0 View Post
    \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e
    Can you please elaborate this step? I tried to see how the LHS is smaller than the RHS. It's not somehow apparent to me.
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  11. #11
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    Quote Originally Posted by Jhevon View Post
    I'd like to see someone use pre-university algebra to solve this one!
    This would work for me.
    Attached Thumbnails Attached Thumbnails Which is bigger: &#960;^e or e^&#960;? Prove it.-untitled.gif  
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  12. #12
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    Quote Originally Posted by Captcha View Post
    Can you please elaborate this step? I tried to see how the LHS is smaller than the RHS. It's not somehow apparent to me.
    We know that \lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^x = e^a. It can also be verified that if x>y>0 and a>0 then \left(1+\frac{a}{x}\right)^x > \left(1+\frac{a}{y}\right)^y. From these two results we can conclude that for all a,x\in \mathbb{R}^+, \left(1+\frac{a}{x}\right)^x < e^a (and in our example, for x=e and a=\pi-e)
    Last edited by Unbeatable0; January 16th 2010 at 02:57 AM.
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    This would work for me.
    graphing actually occured to me, but i don't like this since actually graphing that graph without the use of calculus would be a pain. if you asked a pre-university student to do this, they'd probably just use a graphing utility or plot points with a calculator, in which case they wouldn't need the graph to find the answer to the problem, they'd just use the calculator. i think the idea here is to manually do the problem.
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  14. #14
    Newbie I4talent's Avatar
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    Lots of thanking in this thread. I guess \pi and e are beloved.
    Quote Originally Posted by Unbeatable0 View Post
    \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi
    Simple & Beautiful!
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