find the taylor series for f(x)=lnx about x=2 and the radius of convergence for this series

Question

find the taylor series for f(x)=lnx about x=2 and the radius of convergence for this series

Answer 1

Alyssa:

This one's tricky!

f(x) = ln(x)

f '(x) = 1/x

f ''(x) = -1/x²

f '''(x) = ((-1)*(-2))/x³

f ^iv(x) = ((-1)*(-2)*(-3))/x⁴

f^v(x) = ((-1)*(-2)*(-3)*(-4))/x⁵

Remember n! = n(n-1)*(n-2)*....1

So f ⁿ(x) = ((-1)ⁿ⁺¹ *(n-1)!) /xⁿ for n ≥ 1

So f ⁿ(2) = ((-1)ⁿ⁺¹* (n-1)! )/ 2ⁿ for n≥1

Taylor expansion of ln(x) about a=2 gives

ln(2) + ∑((f ⁿ(2) * |x-2|ⁿ) / n!) we don't include 1st term, ln(2), in the sum.

Now f ⁿ(2)/n! is [ ((-1)ⁿ⁺¹* (n-1)! )/ 2ⁿ ] / n! Using the fact that (n-1)! / n! = 1/n

fⁿ(2)/n! is equal to [(-1)ⁿ⁺¹/(n*2ⁿ)]

So our taylor series is ln(2) + ∑(|x-2|ⁿ / n*2ⁿ). [n goes from n to ∞]

To find radius of convergence, find the radius of convergence of ∑(|x-2|ⁿ / n*2ⁿ). Forget the ln(2)

I let you figure out the radius of convergence. This problem is quite involved, so please ask me help if you need additional help!

Answer 2

Taylor series about x = 2 looking like this: f(2) + f'(2)/1!·(x - 2) + f''(2)/2!(x - 2)² + ...+ f⁽ⁿ⁾(2)/n!(x - 2)ⁿ + ...

f(2) = ln2; f'(x) = 1/x = x^-1, f'(2) = 1/2, f''(x) = - x^-2; f''(2) = - 1/4, f'''(x) = 2x^-3, f'''(2) = 2·2^-3 = 1/4;...

f⁽ⁿ⁾(x) = (-1)^n-1(n-1)!x^{- n}_; f⁽ⁿ⁾(2) = (- 1)^n-1(n-1)!·2^-n. So, Taylor series:

ln2 + 1/2(x - 2) - 1/8(x - 2)² + 1/24(x - 2)³ + ... + (-1)^{n - 1}·2^-n/n·(x - 2)ⁿ + ....

To get radius of convergence consider limit ratio test:

lim n→∞abs(u_{n + 1}/u_n) = lim n→∞ (n/(n + 1)·1/2)abs(x - 2) = 1/2abs(x - 2) < 1; from here - 2 < x - 2 < 2 or 0 < x < 4 This is interval of convergence and radius of convergence = 2