Expert: Socrates - 11/28/2005

Question
I have to write a Taylor series for the function:

f(x)=sin3x   where c= pi/12

i also have to  find the interval of convergence and the radius of convergence?

Answer
Correction!

I just realized that I omitted the n! term from the denominator of Taylor series. This changes the series, the radius of convergence and the interval of convergence . Here is a corrected answer to your question.


We find

f(pi/12)= 1/2^1/2

f'(pi/12)= 3/2^1/2

f''(pi/12)= -9/2^1/2

f'''(pi/12)= -27/2^1/2

f''''(pi/12)= 81/2^1/2

f'''''(pi/12)= 243/2^1/2

this pattern shows that the nth derivative of f at pi/4 is ((-1)^[n/2])(3^n)/2^1/2

where [n/2] is the greatest integer of n/2.

The Taylor series for f at pi/12 is thus the sum for n=0 to infinity of the term
((-1)^[n/2])((3^n)/n!2^1/2)(x-pi/12)^n  

To find the radius of convergence, we first find lim of
((3^n)/n!2^1/2)^1/n as n goes to infinity. This simplifies to lim 3/(n!2^1/2)^1/n = 0. The radius of convergence, r,  is the reciprocal of this limit, so the radius of convergence is + infinity. The interval of convergence is given by (pi/12 - r , pi/12 + r), and since r is infinite, the interval of convergence is (-infinity,+infinity), in other words, the Taylor series converges for every real number.