How Do You Solve a Calculus Free Response Question on Taylor Series?

[SWFanatic]

Homework Statement

The function f has a Taylor series about x=2 that converges tp f(x) for all x in the interval of convergence. The nth derivative of f at x=2 is given by f^(n)(2)=((n+1)!)/3^n for n>=1, and f(2) =1.
(a). write the first four terms and the general term of the Taylor series for f about x=2.
(b). find the radius of convergence for the Taylor series for f about x=2.

Homework Equations

The Attempt at a Solution

(a). F(x) = 1 + (2/3)(x-2) + (2/3)(x-2)^2 + (8/9)(x-2)^3 +...+ ((n+1)!(x-2)^n)/(3^n)
This seems correct, however I am not sure, because when I atempt part b it doesn't really work.
(b) Standard Ratio Test for the general term in part a = abs((n+2)(x-2))/3 <1
Does this not mean its divergent then? or am i all mixed up? Thanks for any help.

 

[HallsofIvy]

Homework Statement

The function f has a Taylor series about x=2 that converges tp f(x) for all x in the interval of convergence. The nth derivative of f at x=2 is given by f^(n)(2)=((n+1)!)/3^n for n>=1, and f(2) =1.
(a). write the first four terms and the general term of the Taylor series for f about x=2.
(b). find the radius of convergence for the Taylor series for f about x=2.

Homework Equations

The Attempt at a Solution

(a). F(x) = 1 + (2/3)(x-2) + (2/3)(x-2)^2 + (8/9)(x-2)^3 +...+ ((n+1)!(x-2)^n)/(3^n)
This seems correct, however I am not sure, because when I atempt part b it doesn't really work.

This is completely wrong but since you don't say how you got those coefficients, I don't what you did wrong. Did you forget the n! in the denominator of the formula for the coefficients?

(b) Standard Ratio Test for the general term in part a = abs((n+2)(x-2))/3 <1
Does this not mean its divergent then? or am i all mixed up? Thanks for any help.

 

[SWFanatic]

yes, i did forget to use n! in the denominator.
New Equation(for part a): f(x) = 1 + (2(x-2))/3*1!) + (6(x-2)^2)/(9*2!) + (24(x-2)^3)/(27*3!)+...+ ((x-2)^n)/((3^n)*(n!))
I got this by using the given nth derivative formula f^n(2)= (n+1)!/3^n for the f1 f2 f3 derivative parts of the formula for series (f(2) + f1(x-2) + (f2(2)(x-2)^2)/2! + (f3(2)(x-2)^3)/3! +...
I still do not think its correct however because for part b:
I use the ration test lim(x>0)|((x+2)^(n+1)/(3^(n+1) *(n+1)!)) * ((3^n(n!))/(x-2)^n)| = lim(x>0) |(x-2)/3(n+1)|
From here i don't know what to do because if i made it less than 1, wouldn't it mean that the series always converges? Is my general term off? Thanks again

 

[HallsofIvy]

You have the correct derivatives in
F(x) = 1 + (2/3)(x-2) + (2/3)(x-2)^2 + (8/9)(x-2)^3 +...+ ((n+1)!(x-2)^n)/(3^n)k
but did not divide by n! Since (n+1)!/n!= n+ 1, the correct series is
F(x) = 1 + (2/3)(x-2) + (1/3)(x-2)^2 + (4/27)(x-2)^3 +...+ ((n+1)(x-2)^n)/(3^n)
Somehow, you have put the "n+1" in the denonimator, not the numerator.

Now, use the ratio test: take the limit of [(n+1)/n]|x-2|/3 as n goes to infinity.