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Math Help - prove this is differentiable everywhere

  1. #1
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    prove this is differentiable everywhere

    f(x) = p(1/x) * e^(-1/x^2) , x/=0
    0 , x=0


    p(t) is a nonzero polynomial in t


    prove this is differentiable everywhere, i cant prove this limit::

    limit( (p(1/h)*e^(-1/h^2))/h, h goes to 0)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    f(x) = p(1/x) * e^(-1/x^2) , x/=0
    0 , x=0


    p(t) is a nonzero polynomial in t


    prove this is differentiable everywhere, i cant prove this limit::

    limit( (p(1/h)*e^(-1/h^2))/h, h goes to 0)
    Do you mean \lim_{h\to{0}}p\bigg(\frac{1}{h}\bigg)\cdot\frac{e  ^{\frac{-1}{h^2}}}{h}?

    Alternatively you could consider the fact that since e^x is analytic everywhere it is thus differentiable everywhere and work at it from the power series point of view
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  3. #3
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    yes

    yes. what i did is : show f(x) is differentiable when x/=0,
    now i m trying to show that limit =0, therefore, the whole function is differentiable.

    please show how to do that limit..
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    yes. what i did is : show f(x) is differentiable when x/=0,
    now i m trying to show that limit =0, therefore, the whole function is differentiable.

    please show how to do that limit..
    Did you try using Maclaurin series on the limit...Besides that there is nothing I can readily think of...I will report back if I do
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    ok

    ok....i m waiting for u
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    \lim_{h\to{0}}p \left ( \frac{1}{h} \right ) \cdot \frac{e^{\frac{-1}{h^2}}}{h}

    You know that p is a non-zero polynomial, so this limit is of the form
    \sum_n a_n \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}

    So we need to find
    \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}
    for some arbitrary positive integer n.

    Let
    x = \frac{1}{h}

    Then
    h^n = \frac{1}{x^n}

    So
    \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n} = \lim_{x \to \infty} x^n e^{-x^2}

    = \lim_{x \to \infty} \frac{x^n}{e^{x^2}}

    which is in a slightly more "doable" form.

    -Dan
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Do you mean \lim_{h\to{0}}p\bigg(\frac{1}{h}\bigg)\cdot\frac{e  ^{\frac{-1}{h^2}}}{h}?

    Alternatively you could consider the fact that since e^x is analytic everywhere it is thus differentiable everywhere and work at it from the power series point of view
    Rewriting this as \bigg(\frac{1}{h^{n+1}}\bigg)\cdot{e^{\frac{-1}{h^2}}}

    we get \lim_{h\to{0}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...\pm\frac{(\frac{1}{h^2})^{n-1}}{(n-1)!}\mp\frac{(\frac{1}{h^2})^n}{n!}}{h^{n+1}}

    can you go from there
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  8. #8
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Rewriting this as \bigg({\color{red}\frac{1}{h^{n+1}}}\bigg)\cdot{e^  {\frac{-1}{h^2}}}
    huh ?

    we get \lim_{n\to{\infty}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...}{h^{n+1}}

    can you go from there
    Ok, and where did the limit of h go ??
    Plus, you, can you go from there ?


    *After l'H˘pital's rule, power series, here comes the newly trend : MacLaurin series !*

    (am kidding)
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    huh ?


    Ok, and where did the limit of h go ??
    Plus, you, can you go from there ?


    *After l'H˘pital's rule, power series, here comes the newly trend : MacLaurin series !*

    (am kidding)
    I misread...it should be let p\bigg(\frac{1}{h}\bigg)=\sum_{n=0}^{c}a_n\bigg(\f  rac{1}{h}\bigg)^{n}=a_0+a_1\bigg(\frac{1}{h}\bigg)  +a_2\bigg(\frac{1}{h}\bigg)^2+...a_n\bigg(\frac{1}  {h}\bigg)^n
    Last edited by Mathstud28; May 15th 2008 at 03:19 PM.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    huh ?


    Ok, and where did the limit of h go ??
    Plus, you, can you go from there ?


    *After l'H˘pital's rule, power series, here comes the newly trend : MacLaurin series !*

    (am kidding)
    No I have always liked power series more it was just I was not adept enough at LaTeX yet to use them...now I am...and yes you could go from there...technically the answer would be ∞ due to an infinite amount of ∞ + an infinite amount of 0's...but as you pointed out I read the question wrong so that limit does not apply
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