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Math Help - prove that this function differentiable endles times..

  1. #1
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    prove that this function differentiable endles times..

    prove that this function differentiable endles times on x=0 ??

    i tried to prove for the first derivative
    but i dont get a final limit here
    <br />
f(x)=e^\frac{-1}{x^2} \\<br />
    <br />
f'(0)=\lim_{x->0^+}\frac{e^\frac{-1}{x^2}-0}{x-0} \\<br />
    <br />
f'(0)=\lim_{x->0^-}\frac{e^\frac{-1}{x^2}-0}{x-0}<br />
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    Quote Originally Posted by transgalactic View Post
    prove that this function differentiable endles times on x=0 ??

    i tried to prove for the first derivative
    but i dont get a final limit here
    <br />
f(x)=e^\frac{-1}{x^2} \\<br />
    <br />
f'(0)=\lim_{x->0^+}\frac{e^\frac{-1}{x^2}-0}{x-0} \\<br />
    note that this is the same as \lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}

    now apply L'Hopital's

    <br />
f'(0)=\lim_{x->0^-}\frac{e^\frac{-1}{x^2}-0}{x-0}<br />
    coming from the left, the function is zero. you have the limit of (0 - 0)/x = 0
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    <br />
\lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}<br />
    after the lhopital law i still dont have a finite limit
    <br />
\lim_{x \to 0^+} \frac {-\frac{1}{x^2}}{-2\frac{1}{x^3}e^{1/x^2}}<br />
    ??
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    Quote Originally Posted by transgalactic View Post
    <br />
\lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}<br />
    after the lhopital law i still dont have a finite limit
    <br />
\lim_{x \to 0^+} \frac {-\frac{1}{x^2}}{-2\frac{1}{x^3}e^{1/x^2}}<br />
    ??
    well, we could simplify to get \frac x{2e^{1/x^2}} and use the fact exponentials grow faster than polynomials to conclude the limit goes to zero.

    if you do not like that, there is always the option of expressing e^{-1/x^2} as a power series from the begining, or in the new expression we got from L'Hopital's. the same conclusion follows
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    ok we prove the this function is differentiable for the first derivative

    ho to proove that its differentiable on every other derivative too

    how to find the n'th derivative??

    should i use macloren series for that?
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    Quote Originally Posted by transgalactic View Post
    ok we prove the this function is differentiable for the first derivative

    how to prove that its differentiable on every other derivative too

    how to find the n'th derivative??

    should i use macloren series for that?
    yes, that is fine. once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable, and we can find it's derivative by differentiating the power series term by term. do you have to actually come up with a formula for the nth derivative? if so, it is still beneficial to start with the power series
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    Exclamation

    I think there is something wrong in the question !
    Every function is differentiable endles times !!

    f(x)=3x^2
    f`(x)=6x
    f``(x)=6
    f```(x)=0
    ..=0
    ..=0 ------> Because 0 is a costant !! so its DIFF is 0
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    Quote Originally Posted by TWiX View Post
    I think there is something wrong in the question !
    Every function is differentiable endles times !!

    f(x)=3x^2
    f`(x)=6x
    f``(x)=6
    f```(x)=0
    ..=0
    ..=0 ------> Because 0 is a costant !! so its DIFF is 0
    not necessarily if we consider a certain point as we are doing here. recall, for instance, |x| is NOT differentiable at zero... not even once. and in general, a function is not differentiable where it is not continuous
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    Yes am talking about the conts. function

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  10. #10
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    the makloren formula is
    <br />
f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n)<br />

    how to take the n'th derivative from here??

    where to put the formula in and get the n'th derivative??

    this function
    is for approximating so in order for me to get to the n'th members approximation
    first i need to do manually one by one n times derivative
    so its not helping me
    ??
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    Quote Originally Posted by TWiX View Post
    Yes am talking about the conts. function

    yes, i realize that. but continuity is not enough. the example i gave, |x|, is continuous everywhere

    Quote Originally Posted by transgalactic View Post
    the makloren formula is
    <br />
f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n)<br />

    how to take the n'th derivative from here??

    where to put the formula in and get the n'th derivative??

    this function
    is for approximating so in order for me to get to the n'th members approximation
    first i need to do manually one by one n times derivative
    so its not helping me
    ??
    no approximations, use the actual power series. once you have it, differentiate it a few times to see a pattern
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  12. #12
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    i cant see any pattern here
    ??
    <br />
      f(x)=e^\frac{-1}{x^2} \\<br />
    <br />
      f'(x)=\frac x{2e^{1/x^2}}\\<br />
          <br />
f''(x)=\frac{2e^{1/x^2}+4xe^{1/x^2}\frac{1}{x^3}}{4e^{2/x^2}}=\frac{1+2\frac{1}{x^2}}{2e^{1/x^2}}<br />
    Last edited by transgalactic; February 8th 2009 at 12:29 PM.
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    Quote Originally Posted by transgalactic View Post
    i cant see any pattern here
    ??
    <br />
      f(x)=e^\frac{-1}{x^2} \\<br />
    <br />
      f'(x)=\frac x{2e^{1/x^2}}\\<br />
          <br />
f''(x)=\frac{2e^{1/x^2}+4xe^{1/x^2}\frac{1}{x^3}}{4e^{2/x^2}}=\frac{1+2\frac{1}{x^2}}{2e^{1/x^2}}<br />
    as i said, it's probably easier to use the power series.

    if you want to continue doing it the way you are doing it, going only to the second derivative is hardly ever enough to find a pattern
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  14. #14
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    so you say substitute e^x with its approximation taylor series??
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    Quote Originally Posted by transgalactic View Post
    so you say substitute e^x with its approximation taylor series??
    not approximation, the actual Taylor series that converges to e^{-1/x^2}
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