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Find Taylor Series from a function and its interval of convergence

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  1. Apr 8, 2016 #1
    Let f(x) = (1+x)-4
    Find the Taylor Series of f centered at x=1 and its interval of convergence.

    [tex] \sum_{n=0}^\infty f^n(c)\frac{(x-c)^n}{n!}[/tex] is general Taylor series form

    My attempt

    I found the first 4 derivatives of f(x) and their values at fn(1). Yet from here I do not know how to find the taylor series, from there I should be able to finish it myself. Any ideas on how to find the Taylor series here?
     
    Last edited: Apr 8, 2016
  2. jcsd
  3. Apr 8, 2016 #2

    PeroK

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    How did you find the derivatives?
     
  4. Apr 8, 2016 #3
    f0(x)= [tex] \frac{(1)}{(1+x)^4}[/tex]
    f1(x)= [tex] \frac{(-4)}{(1+x)^5}[/tex]
    f2(x)= [tex] \frac{(20)}{(1+x)^6}[/tex]
    f3(x)= [tex] \frac{(-120)}{(1+x)^7}[/tex]
    f4(x)= [tex] \frac{(840)}{(1+x)^8}[/tex]

    I think that I got the Taylor Series of [tex] f^n(x)=\frac{(-1)^(n+1)*(n-1)!}{2^(n+4)} [/tex]
     
    Last edited: Apr 8, 2016
  5. Apr 8, 2016 #4

    PeroK

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    Can you fix that latex? In any case, it doesn't look right. It looks like a mixture of parts of the solution.
     
  6. Apr 8, 2016 #5
    Fixed
     
  7. Apr 8, 2016 #6

    PeroK

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    You can see that your general formula doesn't match the first four derivatives.
     
  8. Apr 8, 2016 #7
    True, I can see that now. I think the issue is the (n-1)! in the numerator. I can't find something that works because I want it to be n! but to stop at 4. like 7*6*5*4, but not include *3*2*1. To do this would I just divide that by 6?
     
  9. Apr 8, 2016 #8

    PeroK

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    For the time being, just calculate ##f^{(n)}(x)##. You're not aiming at anything.

    And, yes, ##7*6*5*4 = 7!/6##
     
  10. Apr 8, 2016 #9
    I think that I have [tex] f^n(x)=\frac{(-1)^(n)*\frac{(n-1)!}{6}}{(1+x)^(n+4)} [/tex]
     
  11. Apr 8, 2016 #10

    PeroK

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    You need to check that ##(n-1)!##.
     
  12. Apr 8, 2016 #11
    Does the above look better?
     
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