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Taylor Series Problem

  1. Feb 28, 2010 #1
    Let f be the function given by f(t) = 4/ (1 + t^2) and G be the function given by G(x) = {Integral from 0 to x} f(t)dt .

    (a) Find the first four nonzero terms and the general term for the power series expansion of f(t) about t = 0.

    (b) Find the first four nonzero terms and the general term for the power series expansion of G(x) about x = 0.

    (c) Find the interval of convergence of the power series in part (b). (Your solution must include an analysis that justifies your answer.)


    for part a I got:
    4 - 4t^2 + 4t^4 - 4t^6 +...+ [(-1)^n](4)t^2n +...

    and for part b I got:
    4x - (4x^3)/3 + (4x^5)/5 - (4x^7)/7 +...+ [(-1)^n](4)(t^2n)/(2n + 1) +...

    how do u do part c??? I dunno what to do
     
  2. jcsd
  3. Feb 28, 2010 #2

    Dick

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    Apply a ratio test to your series to get the interval of convergence. Your general term for b should have an x in it and I don't think the power is quite right.
     
  4. Feb 28, 2010 #3

    vela

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    Use the ratio test or nth-root test on your series and determine the interval over which the series will converge. Finally, figure out somehow whether the series converges on the endpoints of that interval.
     
  5. Feb 28, 2010 #4
    For your general term in part (b), it's x, not t. And yes, apply the ratio test, and then test the endpoints using AST.
     
  6. Mar 1, 2010 #5
    I fixed part b and i got:
    4x - (4x^3)/3 + (4x^5)/5 - (4x^7)/7 +...+ [(-1)^n](4)(x^(2n +1))/(2n + 1) +...

    and I am still slightly confused in part c....
    I did the ratio test and i can't get past

    [tex]\stackrel{lim}{n \rightarrow \infty}[/tex] [tex]\left|[/tex][tex]\frac{(-1)x^2(2n +1)}{2n +3}[/tex]|

    srry, its supposed all in absolute value, i still dunno how to work latex
     
  7. Mar 1, 2010 #6

    vela

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    Remember that the x acts like a constant here since you're taking the limit as n, not x, goes to infinity. Also, you have an absolute value so you can just erase the (-1)^n factor.
     
  8. Mar 1, 2010 #7
    oh right!!!
    so after that i get
    x^2 < 1 and then x will equal -1 and 1 right?

    and then the IOC would be -1 < x < 1

    but then how would i test these boundaries using AST?? do I use the sum i had with these?
     
    Last edited: Mar 1, 2010
  9. Mar 1, 2010 #8
    substitute x with 1 or -1 in the sum, and see if it is convergent (as a sum of a sequence)

    for example for x=1 your sum is:

    [tex]S=\sum^{\infty}_{0}\frac{4(-1)^{n}}{2n+1}[/tex]
     
  10. Mar 1, 2010 #9
    Thank you so much!!! you helped me a lot!!:)
     
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