Taylor Series Expansion for f(t) and G(x) with Convergence Analysis"

[hereiscassie]

Let f be the function given by f(t) = 4/ (1 + t^2) and G be the function given by G(x) = {Integral from 0 to x} f(t)dt .

(a) Find the first four nonzero terms and the general term for the power series expansion of f(t) about t = 0.

(b) Find the first four nonzero terms and the general term for the power series expansion of G(x) about x = 0.

(c) Find the interval of convergence of the power series in part (b). (Your solution must include an analysis that justifies your answer.)


for part a I got:
4 - 4t^2 + 4t^4 - 4t^6 +...+ [(-1)^n](4)t^2n +...

and for part b I got:
4x - (4x^3)/3 + (4x^5)/5 - (4x^7)/7 +...+ [(-1)^n](4)(t^2n)/(2n + 1) +...

how do u do part c? I don't know what to do

 

[Dick]

Apply a ratio test to your series to get the interval of convergence. Your general term for b should have an x in it and I don't think the power is quite right.

 

[vela]

Use the ratio test or nth-root test on your series and determine the interval over which the series will converge. Finally, figure out somehow whether the series converges on the endpoints of that interval.

 

[Anonymous217]

For your general term in part (b), it's x, not t. And yes, apply the ratio test, and then test the endpoints using AST.

 

[hereiscassie]

I fixed part b and i got:
4x - (4x^3)/3 + (4x^5)/5 - (4x^7)/7 +...+ [(-1)^n](4)(x^(2n +1))/(2n + 1) +...

and I am still slightly confused in part c...
I did the ratio test and i can't get past

\stackrel{lim}{n \rightarrow \infty} \left|\frac{(-1)x^2(2n +1)}{2n +3}|

srry, its supposed all in absolute value, i still don't know how to work latex

 

[vela]

Remember that the x acts like a constant here since you're taking the limit as n, not x, goes to infinity. Also, you have an absolute value so you can just erase the (-1)^n factor.

 

[hereiscassie]

oh right!
so after that i get
x^2 < 1 and then x will equal -1 and 1 right?

and then the IOC would be -1 < x < 1

but then how would i test these boundaries using AST?? do I use the sum i had with these?

 

[elibj123]

substitute x with 1 or -1 in the sum, and see if it is convergent (as a sum of a sequence)

for example for x=1 your sum is:

S=\sum^{\infty}_{0}\frac{4(-1)^{n}}{2n+1}

 

[hereiscassie]

Thank you so much! you helped me a lot!:)