- #1
rbpl
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Consider the power series (n=1 to infinity) [tex]\Sigma[/tex] (x^n)/(n*3^n).
(a) Find the radius of convergence for this series.
(b) For which values of x does the series converge? (include the discussion
of the end points).
(c) If f(x) denotes the sum of the series, find f'(x) as explicitly as possible.
I know how to do parts a and b and for the radius I got 3 and the interval looks like this:
-3≤x<3
But I don't understand the last part (c)
I know that since (n=0 to infinity)f(x)= [tex]\Sigma[/tex] c_n(x-a)^n then (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1)
I was looking thru various materials and first I was thinking about finding the representation for the power series but I was not sure how this would help me here since I don't have a function to begin with just the series. Then I was thinking about differentiating by terms but I didn't know how to incorporate the radius of convergence. Finally I did something that I pretty sure is wrong:
(n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1) = [tex]\Sigma[/tex] n*(1/(n*3^n))*(x^(n-1))=(1/(3^n))*(x^(n-1))=(x^n)/(x*3^n) for -3≤x<3
But like I said it doesn't seem correct. Any suggestions?
(a) Find the radius of convergence for this series.
(b) For which values of x does the series converge? (include the discussion
of the end points).
(c) If f(x) denotes the sum of the series, find f'(x) as explicitly as possible.
I know how to do parts a and b and for the radius I got 3 and the interval looks like this:
-3≤x<3
But I don't understand the last part (c)
I know that since (n=0 to infinity)f(x)= [tex]\Sigma[/tex] c_n(x-a)^n then (n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1)
I was looking thru various materials and first I was thinking about finding the representation for the power series but I was not sure how this would help me here since I don't have a function to begin with just the series. Then I was thinking about differentiating by terms but I didn't know how to incorporate the radius of convergence. Finally I did something that I pretty sure is wrong:
(n=1 to infinity)f'(x)=[tex]\Sigma[/tex] n*c_n(x-a)^(n-1) = [tex]\Sigma[/tex] n*(1/(n*3^n))*(x^(n-1))=(1/(3^n))*(x^(n-1))=(x^n)/(x*3^n) for -3≤x<3
But like I said it doesn't seem correct. Any suggestions?