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Power series & Taylor series

  1. Apr 28, 2007 #1
    My exam is coming up, I have 2 questions on infinite series. Any help is appreciated!:smile:

    Quesetion 1) [​IMG]
    For part a, I got:
    g(x)= Sigma (n=0, infinity) [(-1)^n * x^(2n)]

    For part b, I got:
    ∫ tan^-1 (t^2) dt = Sigma (k=0, infinity) [(-1)^k * x^(4k+3)] / [(2k+1) (4k+3)]
    But the part b, they also ask for the radius of convergence, how can I find it? Should I apply the ratio test to this series expansion (colored in red) to find the radius of convergence? Is there any faster way?

    Question 2) Suppose f(x)= x cos(x^2), find f^(4101) (0).
    [f^(n) (0) is the "n"th derivative evaluated at 0]

    Should I use the fact that "On its interval of convergence, a power series is the Taylor series of its sum" to do this question?
    So is it true that the power series of x cos(x^2) is EQUAL to the Taylor series of x cos(x^2) for ALL real numbers x?
    Is there any difference between power series and Taylor series?

    Thanks for your help!:smile:
    Last edited: Apr 28, 2007
  2. jcsd
  3. Apr 29, 2007 #2

    Gib Z

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    Homework Helper

    Your solutions to Question 1 a and b are correct. To see the radius of convergence, notice that the radius of convergence of sine's taylor series is the whole real number line. Is this summation smaller term by term???

    2) Yes use that fact, it is correct :) For trigonometric functions, the taylor series and power series are the same. Good work :)
  4. Apr 30, 2007 #3
    1) But the geometric series has radius of convergence 1, will the radius of convergence of the integral arctan (t^2) also be the same (1) ? Why?

    2) Why are the Taylor series and power series the same thing, but having different names? I don't get it...

  5. May 1, 2007 #4

    Gib Z

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    1) Ahh sorry about that your correct. Since the series from which the integral is calculated only conveges for |x|< 1, the integral does so as well.

    2) They aren't the same thing, they just happen to give the same formula for the trig functions, but not for all functions :)


    The article gives the Power series of exp(x) and sin (x) and states they happen to be their taylor series as well. However, there exist power series which are not the Taylor series of any function, for instance

    [tex]\sum_{n=0}^{\infty} n! x^n = 1 + x + 2! x^2 + 3! x^3 + \cdots.[/tex]
  6. May 1, 2007 #5


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    Any Taylor's series is a power series, of course, but the other way is not true. It is possible to find a power series that is NOT the Taylor's series of any function. For example,
    [tex]\Sum_{n=0}^\infty (n!)x^n[/tex]
    cannot be the Taylor's series of any function because it does not converge for any x except 0. Power series of a function are unique: IF a function has a Taylor's series that converges to that function at every point within its radius of convergence, then every power series that converges to that function must be identical to the Taylor's series.
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