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Power sums questions.

  1. Jun 2, 2006 #1
    1) develop the function f(x)=(e^x)sin(x) into a power sum over the point 0.
    2) find the convergence radius R of [tex]\sum_{\substack{0<=n<\infty}}\frac{(n!)^2}{(2n)!}x^n[/tex] and say if it converges or diverges at x=-R, x=R.

    about the second question i got that R=4, through hadamard test, but i didnt succeed in asserting if at x=R it diverges or converges, at x=-R i think it converges because it's an alternating sign sum, and according to leibnitz theorem it does.

    about the first question here what i got:
    i needed to find an equation for the derivative of [tex]f^{(n)}(x)[/tex], here what i got:
    [tex]f^{(n)}(x)=(g(x)h(x))^{(n)}=\binom{n}{n}g^{(n)}(x)h(x)+\binom{n}{n-1}g^{(n-1)}(x)h'(x)+...+\binom{n}{n-1}g'(x)h^{(n-1)}(x)+\binom{n}{n}g(x)h^{(n)}(x)[/tex] which i employed at the function which i got, is this equation correct?

    thanks in advance.
    Last edited: Jun 2, 2006
  2. jcsd
  3. Jun 2, 2006 #2
    for the first question, it is probably easiest to find the power series for e^x and for sin(x) about 0 and then multiply them together.
  4. Jun 2, 2006 #3
    i thought about it, but i wasn't sure, it would be accaptable.
    but on a second thought it does make a perfect sense.

    what about my second question?
    about my first question, how do i represent the product of the sums of e^x and sin(x) as one sum?
  5. Jun 2, 2006 #4


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    You could use the fact that:

    [tex]\sin x = \frac{1}{2i} (e^{ix}-e^{-ix})[/tex]

    so we can write:

    [tex]e^x \sin x = \frac{1}{2i}(e^{(1+i)x}-e^{(1-i)x})[/tex]

    Then, for example,

    [tex] e^{(1+i)x} = 1+ (1+i)x+ \frac{1}{2}(1+i)^2 x^2+...[/tex]

    To compute powers of [itex]1 \pm i[/itex], it is probably easiest to rewrite it as [itex]r e^{i\theta}[/itex] for an appropriate choice of [itex]r[/itex] and [itex]\theta[/itex].
    Last edited: Jun 2, 2006
  6. Jun 3, 2006 #5


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    [tex]e^{x}\sin{x} = \left( \sum_{n=0}^{\infty} \frac{x^{n}}{n!}\right) \left( \sum_{m=0}^{\infty} \frac{x^{2m+1}}{(2m+1)!}\right) = \sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{x^{n-k}}{(n-k)!} \frac{x^{2k+1}}{(2k+1)!} = \sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{x^{n+k+1}}{(n-k)!(2k+1)!}[/tex]
  7. Jun 3, 2006 #6
    can someone help on the other question, does it converge or diverge at x=4, and how to prove it?

  8. Jun 3, 2006 #7


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    Well the product of to absolutely convergent series is absolutely convergent, and the series used converge for all [tex]-\infty < x<\infty[/tex]: so, yes, it does converge at x=4.
  9. Jun 3, 2006 #8
    but R doesnt equal [tex]\infty[/tex], i know that for every |x|<R the sum converges but here i need to find what happens when x=R.
  10. Jun 3, 2006 #9


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    Find the ratio of successive terms at x=4. Do they get bigger or smaller?
  11. Jun 3, 2006 #10
    you mean, to use d'almbert test, ok, thanks.
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