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?

Question

Samantha Adkison · Accepted Answer

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

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 #

Ezekiel Alexander · Answer

undefined To approximate $ f(x) = \ln(1 + 2x) $ by a Taylor polynomial with degree $ n = 3 $ at the number $ a = 2 $, we need to compute the first three derivatives of $ f(x) $ and evaluate them at $ x = 2 $.

1. $ f(x) = \ln(1 + 2x) $
2. First derivative: $ f'(x) = \frac{1}{1 + 2x} $
3. Second derivative: $ f''(x) = -\frac{2}{(1 + 2x)^2} $
4. Third derivative: $ f'''(x) = \frac{8}{(1 + 2x)^3} $

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

1. $ f(2) = \ln(1 + 2 \times 2) = \ln(5) $
2. $ f'(2) = \frac{1}{1 + 2 \times 2} = \frac{1}{5} $
3. $ f''(2) = -\frac{2}{(1 + 2 \times 2)^2} = -\frac{2}{25} $
4. $ f'''(2) = \frac{8}{(1 + 2 \times 2)^3} = \frac{8}{125} $

The Taylor polynomial of degree $ n = 3 $ at $ a = 2 $ is given by:

$$ P_3(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 $$

Substituting the values we found:

$$ P_3(x) = \ln(5) + \frac{1}{5}(x - 2) - \frac{1}{25}(x - 2)^2 + \frac{1}{125}(x - 2)^3 $$

This polynomial approximates $ f(x) = \ln(1 + 2x) $ near $ x = 2 $ with an error that decreases as $ n $ increases.