If #f(x) = ln(1+2x)#, where a = 2 and n = 3 how do you approximate f by a Taylor polynomial with degree n at the number a?

1 Answer
Feb 21, 2017

#= ln 5 + 2/5 (x-2) - 2/25 (x-2)^2 + 8/375 (x-2)^3 #

Explanation:

The value of #f(a)# is straightforward ... it's #ln (1 + 2a)#.

So maybe you want to approximate the behaviour of the function around #x = a#?

If that is the case, you can apply the Taylor Series formulation to third degree:

#f(x-a) = f(a) + ((x-a) f'(a))/(1!) + ((x-a)^2 f''(a))/(2!) + ((x-a)^3 f'''(a))/(3!)#

And so we need these:

#f = ln (1 + 2x)#

#f' = 2/(1 + 2x)#

#f'' = - 4/(1 + 2x)^2#

#f''' = 16/(1 + 2x)^3#

We can then say that:

#f(x-2) = ln 5 + ((x-2) (2/5))/(1!) + ((x-2)^2 (- 4/25))/(2!) + ((x-2)^3 (16/125))/(3!)#

#= ln 5 + 2/5 (x-2) - 2/25 (x-2)^2 + 8/375 (x-2)^3 #