How do you find the taylor series series for #f(x)=lnx# at a=2?

Question

Charles Arocha · Accepted Answer

First, we can start with the general definition of the Taylor series expansion, which is: #sum_(n=0)^N f^((n))(a)/(n!)(x-a)^n# where #f^((n))(a)# is the #n#th derivative of #f(x)# evaluated at #x -> a#, #n# varies, #a# does not, and #n!# is #1xx2xx3xxcdotsxx(n-1)xxn#. Note that #x->a# does not apply to #(x-a)^n# (yes, I've seen that happen). Thus you have to take the derivative up to some #n#th order term. Let's say #n = 4#. Then you have: #f^((0))(x) = f(x) = lnx#
#f'(x) = 1/x#
#f''(x) = -1/x^2#
#f'''(x) = 2/(x^3)#
#f''''(x) = -6/(x^4)# Now you have the final result to simplify: #sum_(n=0)^(4) = (f(2))/(0!)(x-2)^0 + (f'(2))/(1!)(x-2)^1 + (f''(2))/(2!)(x-2)^2 + (f'''(2))/(3!)(x-2)^3 + (f''''(2))/(4!)(x-2)^4 + ...# #= (ln2) + (1/2)(x-2) + (-1/4)/(2)(x-2)^2 + (2/8)/(6)(x-2)^3 + (-6/16)/(24)(x-2)^4 + ...# #= color(blue)(ln2 + 1/2(x-2) - 1/8(x-2)^2 + 1/24(x-2)^3 - 1/64(x-2)^4 + ...)# Just remember the following:
- #x->a# only for #f(a)#, not #(x-a)^n#
- #n# varies, but #a# does not
- It's probably wise to take the #n#th derivatives first so you aren't doing them as you are writing out the next step from the point you write out the general series formula

Charles Anctil · Answer

#ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3-1/64(x-2)^4+\cdots#

Use the following expression for the Taylor series of an infinitely differentiable function at #x=a#:

#f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+(f''''(a))/(4!)(x-a)^4\cdots#

Since #f(x)=ln(x)#, we get #f'(x)=1/x=x^{-1}#, #f''(x)=-x^{-2}#, #f'''(x)=2x^{-3}#, #f''''(x)=-6x^{-4}#, etc...

Since #a=2#, we calculate #f(2)=ln(2)#, #f'(2)=1/2#, #f''(2)=-1/4#, #f'''(2)=2/8=1/4#, #f''''(2)=-6/16=-3/8#, etc...

Therefore, we can write the answer as

#ln(2)+1/2(x-2)-1/8(x-2)^2+1/24(x-2)^3-1/64(x-2)^4+\cdots#

This series happens to equal #ln(x)# for #0 < x < 4# (the "radius of convergence" is 2 and it equals the function for these values as well).

Evelyn Agard · Answer

undefined To find the Taylor series for $ f(x) = \ln(x) $ at $ a = 2 $, we start by finding the derivatives of $ f(x) = \ln(x) $ and evaluating them at $ x = 2 $ to obtain the coefficients of the Taylor series.

1. First derivative of $ f(x) = \ln(x) $: $ f'(x) = \frac{1}{x} $
2. Second derivative of $ f(x) = \ln(x) $: $ f''(x) = -\frac{1}{x^2} $
3. Third derivative of $ f(x) = \ln(x) $: $ f'''(x) = \frac{2}{x^3} $
4. Fourth derivative of $ f(x) = \ln(x) $: $ f''''(x) = -\frac{6}{x^4} $

Now, we evaluate these derivatives at $ x = 2 $:

1. $ f'(2) = \frac{1}{2} $
2. $ f''(2) = -\frac{1}{4} $
3. $ f'''(2) = \frac{1}{4} $
4. $ f''''(2) = -\frac{3}{16} $

Since $ f(a) = \ln(2) $ when $ a = 2 $, the first term of the Taylor series is $ \ln(2) $.

Therefore, the Taylor series for $ f(x) = \ln(x) $ at $ a = 2 $ is:

$$ \ln(x) = \ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + \frac{1}{12}(x - 2)^3 - \frac{1}{32}(x - 2)^4 + \cdots $$