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Thread: 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: $\displaystyle \pi^e$ or $\displaystyle e^{\pi}$?
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  2. #2
    Super Member Bacterius's Avatar
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    We know that $\displaystyle \pi \neq e$.

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

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

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

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

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

    ________________________


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

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

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

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

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

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

    ________________________


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

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

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

    $\displaystyle \ln(x)\geqslant1\implies \ln(x)-1\geqslant 0$ and so $\displaystyle \frac{\ln(x)-1}{\ln^2(x)}\geqslant 0$. 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 $\displaystyle f(x)=e^x x^{-e}$. We have that $\displaystyle f'(e)=0$ and it's easy to show it's a minimum. So then $\displaystyle f(\pi)\geqslant f(e)=1$.
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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: $\displaystyle \pi^e$ or $\displaystyle e^{\pi}$?
    See here.
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  9. #9
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    Another one:
    $\displaystyle \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$

    Note that this can be generalized: for all $\displaystyle m > n \ge e$, it is true that $\displaystyle n^m > m^n$
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  10. #10
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    Quote Originally Posted by Unbeatable0 View Post
    $\displaystyle \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 $\displaystyle \lim_{x\rightarrow \infty} \left(1+\frac{a}{x}\right)^x = e^a$. It can also be verified that if $\displaystyle x>y>0$ and $\displaystyle a>0$ then $\displaystyle \left(1+\frac{a}{x}\right)^x > \left(1+\frac{a}{y}\right)^y$. From these two results we can conclude that for all $\displaystyle a,x\in \mathbb{R}^+$, $\displaystyle \left(1+\frac{a}{x}\right)^x < e^a$ (and in our example, for $\displaystyle x=e$ and $\displaystyle a=\pi-e$)
    Last edited by Unbeatable0; Jan 16th 2010 at 01:57 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 $\displaystyle \pi $ and $\displaystyle e$ are beloved.
    Quote Originally Posted by Unbeatable0 View Post
    $\displaystyle \pi^e = \left(\frac{\pi}{e}\right)^e e^e = \left(\frac{e+\pi-e}{e}\right)^e e^e = \left(1+\frac{\pi-e}{e}\right)^e e^e < e^{\pi-e} e^e = e^\pi$
    Simple & Beautiful!
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