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

  • Thread starter blursotong
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  • #1
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[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.
 
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Answers and Replies

  • #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?
 
  • #3
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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 ??
 
  • #4
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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 ??
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]
 
  • #5
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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??
 
  • #6
Dick
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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??
That's it. If you meant to write x^(n+1) in the numerator instead of (x+1)^n.
 
  • #7
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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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