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Taylor Series

  1. Apr 10, 2004 #1


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    Staff Emeritus

    This problem has been bugging me and I can't seem to figure it out:

    y'' = e^y where y(0)= 0 and y'(0)= -1

    I'm supposed to get the first 6 nonzero terms

    I know the form is:

    y(x) = y(0) + y'(0)x/1! + y''(0)x^2/2! + y'''(0)x^3/3! +......

    and I know the first two terms are

    y(x) = 0 - x +.....

    But whats troubling is the e^y. How would I go about getting the y''(0) term. I tried a subsitution of u = y', but the integrals gets very messy. I'm thinking there is either a typo in the book or there is a simpler way to get the answer.

    All I need is an example of how to get the next term. I can figure out the rest.
  2. jcsd
  3. Apr 10, 2004 #2

    matt grime

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    y''(0) = e^{y(0)} = e^0=1

    if you don't mind I'll also say that y''' = (y'')' = d/dx(e^y) = y'e^y
  4. Apr 10, 2004 #3


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    Staff Emeritus

    Ah, ok, thanks alot. I understand now.
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