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Can anyone remind me how to rewrite a summation?

  1. Apr 7, 2017 #1
    1. The problem statement, all variables and given/known data
    "Given: ##sin(t)=Σ\frac{(-1)^nt^{2n+1}}{(2n+1)!}##
    Prove: ##L[sin(t)]=\frac{1}{s^2+1}##."

    2. Relevant equations
    ##∑ar^k=\frac{a}{1-r}##

    3. The attempt at a solution
    ##L[sin(t)]=L[Σ\frac{(-1)^nt^{2n+1}}{(2n+1)!}]=L[t]-\frac{1}{3!}L[t^3]+\frac{1}{5!}L[t^5]+...+\frac{1}{k!}L[t^k]-\frac{1}{(k+2)!}L[t^{k+2}]##
    ##L[sin(t)]=\frac{1}{s^2}-\frac{1}{s^4}+\frac{1}{s^6}+...+\frac{1}{s^{k+1}}-\frac{1}{s^{k+3}}=\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})##
    ##\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})=\frac{-1}{1-s^2}=\frac{1}{s^2-1}##

    I'm stuck on primarily the last part. How would I get from the summation above to the Laplace transform of ##sin(t)##?
     
    Last edited: Apr 7, 2017
  2. jcsd
  3. Apr 7, 2017 #2
    Hold on a second... ##\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})=-\sum_{i=1}^\infty 1(-\frac{1}{s^2})^i=-\frac{1}{s^{-2}+1}##

    Something's still not right, here...
     
    Last edited: Apr 7, 2017
  4. Apr 11, 2017 #3

    DrClaude

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    Staff: Mentor

    You forgot to set the limits here.
     
  5. Apr 13, 2017 #4
    ##\sum_{k=0}^\infty ar^k=\frac{a}{1-r}##

    In any case, I had to turn in the homework today, so I don't really need to solve this problem anymore.
     
  6. Apr 13, 2017 #5

    DrClaude

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    Staff: Mentor

    For others stumbling on the thread: this sum starts from 0, while the one from the Laplace transform starts at 1. Taking this into account solves the problem.

    @Eclair_de_XII: It is actually best not to make the first reply to your thread as it removes it from the unanswered thread list, and many helpers will miss it.
     
  7. Apr 13, 2017 #6
    I see. I will keep that in mind for the future. Thank you.
     
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