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Taylor expansion help

  1. Jun 9, 2014 #1
    1. The problem statement, all variables and given/known data


    Suppose that [itex]f(x)=\sum_{n=0}^{\infty}c_{n}x^{n}[/itex]for all x.
    If [itex]\sum_{n=0}^{\infty}c_{n}x^{n} = 0[/itex], show that [itex]c_{n} = 0[/itex] for all n.

    2. Relevant equations
    3. The attempt at a solution
    I know, by using taylor expansion, [itex]c_{n}=\frac{f^{n}(0)}{n!}[/itex], and because [itex]\sum_{n=0}^{\infty}c_{n}x^{n}[/itex] is zero, [itex]c_{n}[/itex] have to be zero. But, I don't know how to write more logical proof to this.
     
  2. jcsd
  3. Jun 9, 2014 #2

    AlephZero

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    That is the right idea. Instead of "using the Taylor expansion", you can just differentiate n times.

    ##\frac{d^n}{dx^n} f(0) = 0##, and if you differentiate the series term-by-term that gives ##c_n = 0##.

    To make that idea into a rigorous proof, you have to say why it is valid to differentiate the series term by term. (Hint: convergence).
     
  4. Jun 9, 2014 #3
    @AlephZero
    What I want to know is why [itex]c_n = 0[/itex] have to be zero if [itex]\sum_{n=0}^{\infty}c_{n}x^{n} = 0
    [/itex], not 'IF' [itex] c_n=0 [/itex]. Would you make this more concrete, please?
     
  5. Jun 9, 2014 #4
    [itex]\frac{d}{dx} f(x)=\sum_{n=1}^{\infty} nc_nx^{n-1}[/itex]
    Then let x=0 to get:
    [itex]0=c_1+\sum_{n=2}^{\infty} nc_n 0^{n-1}[/itex]

    Can you generalize this?
     
  6. Jun 9, 2014 #5
    hmmm. How can I use that to prove [itex]\sum_{n=0}^{\infty}c_{n}x^{n}=0[/itex],then [itex]c_{n}=0[/itex] when x is all real numbers?
     
    Last edited: Jun 10, 2014
  7. Jun 9, 2014 #6

    D H

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    Hint: Use the transitivity of equality. If a=b and b=c then a=c.

    What does the above say about ##f(x)## given that ##f(x)=\sum_{n=0}^{\infty} c_n x^n## and that ##\sum_{n=0}^{\infty} c_n x^n = 0## for all x?
     
  8. Jun 10, 2014 #7

    Ray Vickson

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    The result is false as you have stated it: ##\sin(x) = x - x^3/3! + x^5/5! - \ldots## is equal to 0 for ##x = \pm \pi, \pm 2\pi, \ldots## but its coefficients are not all zero. Did you mean "=0 for all x..."?
     
  9. Jun 10, 2014 #8

    D H

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    He had to have meant "for all x". Otherwise it's patently false.
     
  10. Jun 10, 2014 #9
    Yes, for all x with the radius of convergence R>0.
     
  11. Jun 10, 2014 #10
    So the question would be, "why [itex]c_{n}[/itex] is 0 for all n when [itex]\sum_{n=0}^{\infty}c_{n}x^{n}=0[/itex] for all [itex]x[/itex] with the radius of convergence R>0?
     
  12. Jun 10, 2014 #11

    HallsofIvy

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    You have asked that question four times now. Have you given any thought to the responses you have been given? If [itex]f(x)= \sum c_nx^n[/itex], what is f(0)? what is f'(0)? f''(0)?
     
  13. Jun 10, 2014 #12
    Oh, oh. I finally got it. [itex]f'(x)[/itex] is zero because [itex]f(x) = 0[/itex] and [itex]f'(x)[/itex] is like differentiating zero, right?
     
  14. Jun 10, 2014 #13

    D H

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    Exactly. You never went from ##f(x)=\sum_n c_n x^n## and ##\sum_n c_n x^n=0## to ##f(x)=0## until just now. Once you have ##f(x)=0## for all x, it's a simple matter to show that ##f^{(n)}(x) = 0## for all x and for all n.

    You should of course prove that ##f^{(n)}(x) = 0## for all x and for all n.
     
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