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Math Help - Taylor Series/Limits/Convergence Test

  1. #1
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    Taylor Series/Limits/Convergence Test

    Hey everyone,

    I am having some trouble working through these problems, and I was hoping that someone can help me out, or at least put me on the right track. I managed to get through all of them but I'm pretty sure I messed them up.

    #1
    Define the function f(x) in two pieces by
    f(x) = 0 if x ≤ 0, and f(x) = e^(-1/x^2) if x > 0.
    Show that the function f(x) is continuous at x = 0. Furthermore, show that f`(0) exists. Your life will be made easier if rather than considering
    lim d/dx e^(-1/x^2)
    x->0+
    you explain why we may instead consider
    lim d/dt e^(-t^2)
    t-> ∞
    Then show that f(ⁿ)(0) exists for all n.
    Write the Taylor Series for f(x) centered at 0. Show that this series converges everywhere, but for x > 0 does not converge to f(x).

    #2
    Suppose that you have the series

    an,
    n = 1
    and assume that
    lim |an|^(1/n) = L.
    n->∞
    Show that if L < 1, then the series converges (absolutely), and if L > 1, the series diverges. Finally, give a specific example of a series where L = 1 and the series converges, and another example where L = 1 and the series diverges.

    #3
    Assume that the Taylor series for e^z converges for not just for all real numbers z, but for all complex numbers z as well. By setting z = ix, where i^2 = −1 and x is real, show that:
    e^ix = cos(x) + i sin(x).

    Use this formula to derive Euler’s formula:
    e^iπ + 1 = 0.


    Thanks in advance.
    I apologize that this doesn't look very nice, as well. Not really sure how to do the symbols properly in a text form.
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  2. #2
    Super Member Deadstar's Avatar
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    Any chance you could maybe tell us where you got stuck so we don't go over stuff you've got sorted. Just small notes below...

    1. t=\frac{1}{x} as x->0, t->\infty. Hence e^{-t^2} -> 0 so conclude convergence using that etc.
    For f^n(0) you could differentiate a few times to get a general formula on how its going to turn out. Should always have e^{\frac{-1}{x^2}} in each term so hence each term -> 0.

    2. There's a proof of that here Root test - Wikipedia, the free encyclopedia

    What are you stuck on?
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  3. #3
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    Well, do I start the last one using the Taylor Series for e^z, and substitute z=ix into that series? Like:

    [(ix)^n]/n!
    n=1

    I guess turning that into cos(x) + i sin(x) is kinda confusing to me right now. It makes me think that I need to use the Fourier Series. After spending so much time on it, I am feeling a little brain dead.

    As for the part about Euler's Formula, is that just as simple as substituting x=π?
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  4. #4
    Super Member Deadstar's Avatar
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    expand the taylor series of e^{iz} in the same way you would e^x to get...

     e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots

     = 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \cdots

    = (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots )

    Which are the expansions of cos(z) and sin(z)

    And for the second part yes just sub in \pi
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  5. #5
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    Wow, I guess I was trying to make that last one wayyy too complex. As soon I saw the "expand it out" it clicked. Thanks for your help!
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