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Math Help - Which is greater: e^π or π^e?

  1. #1
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    Which is greater: e^π or π^e?

    My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

    Please contribute.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Jamin2112 View Post
    My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

    Please contribute.
    Could you write the instructions exactly as they were given please? I'm having trouble understanding what the problem is.

    And BTW, the taylor series for e^x converges for all x.
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  3. #3
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    This is very interesting, do you have to provide some sort of proof?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    \log(e^n)=n\log(e)

    \log(n^e)=e\log(n)

    What do you think?
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  5. #5
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    the answer is  e^x is greater


    let  f(x) = x - e\ln(x)

    we have  f(x) \to +\infty , when x \to 0

    and

     \lim_{x\to+\infty} x - e\ln(x) =   \lim_{x\to +\infty} x ( 1 - \frac{e\ln(x)}{x} ) = +\infty

    consider

     f'(x) = 1 - \frac{e}{x}

    we can see when  x = e ,  f'(x) = 0

    its second derivative gives  f''(e) = \frac{1}{e} > 0

    means  f(x) has only one local min.

    come back to  f(e) = e - e\ln(e) = e-e = 0

    from the above calculations .

    we can see  f(x) \geq 0

     \implies  x \geq e \ln(x)  \implies e^x \geq x^e

    equality holds when  x = e (obviously )

    ps . add a constraint  x > 0
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  6. #6
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    consider the function f(x)=\frac{\ln x}x, which is strictly decreasing for x>e.

    since \pi>e, we get \frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e, and that yields \ln\pi^e<\ln e^\pi, and finally since g(x)=e^x is a strictly increasing function, then e^{\ln \pi^e}<e^{\ln e^\pi}\implies \pi^e<e^\pi, and we're done.
    Last edited by Krizalid; December 7th 2009 at 03:09 AM.
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    consider the function f(x)=\frac{\ln x}x, which is strictly decreasing for x>e.

    since \pi>e, we get \frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e, and that yields \ln\pi^e<\ln e^\pi, and finally since g(x)=e^x is a strictly increasing function, then e^{\ln \pi^e}<e^{\ln e^\pi}\implies \pi^e<e^\pi, and we're done.
    Thanks, dude! I went through all your steps and now I understand it!
    Last edited by Krizalid; December 7th 2009 at 03:09 AM.
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