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Does anyone know how to find radius of convergence for sin x and e^x

  1. Mar 21, 2010 #1
    [sloved]Does anyone know how to find radius of convergence for sin x and e^x

    We know that to find radius of convergence we use ratio test (ie lim {a_n+1} /{a_n})
    Can this method be used for sin x and e^x? ( whose radius of convergence is -infinity and infinity)
    if radius of convergence is -infinity<x< infinity right? it means when we use ratio test the result is zero?

    but we see for e^x,
    i get |e| when using ratio test, which implies that it diverge?

    Im confused. Can anyone can help?
    thanks.
     
    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 21, 2010 #2

    Dick

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    You are very confused. You have to use the ratio test on the coefficients of the Taylor series for e^x and sin(x). How do you get |e| for e^x?
     
  4. Mar 22, 2010 #3
    cause i use lim (e^(n+1)/ e^(n)) and i got e?
    do you mean i use the coefficient of taylor series for e^x = 1 + x + (x^2)/2 + .....
    then e^(x+1) = 1 + (x+1) + (x+1)^2/2 +...
    then divide ??
     
  5. Mar 22, 2010 #4

    Mark44

    Staff: Mentor

    No, use the general term in whatever series you're investigating. For ex, the Maclaurin series is
    [tex]\sum_{n = 0}^{\infty} \frac{x^n}{n!}[/tex]
     
  6. Mar 22, 2010 #5
    wow..thats a great hint.
    so its lim of [ {(x+1)^n /(n+1)!} * {(n!)/(x)^n} ] which gives lim {x/ (n+1} and when n tends to infinity the limit becomes zero ...so the radius of convergence is - infinity to + infinity??
     
  7. Mar 22, 2010 #6

    Dick

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    Science Advisor
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    That's it. If you meant to write x^(n+1) in the numerator instead of (x+1)^n.
     
  8. Mar 22, 2010 #7
    opps..haha..yup..
    thanks alot for your help!!
    anw, can you help me in another multiple integration question too?
    no one replied yet=(
     
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