Find the Taylor polynomial Tn(x) for the given function at the number a.

Question

Find the Taylor polynomial T_n(x) for the given function at the number a.

f(x)=tan^-1(2x)

a=1/2

n=3

T₃(x) = [?]

Answer 1

Sinon, thanks for asking an amazing question!

You may already be familiar that the Taylor series for a function f(x) centered at x = a is equal to:

T_n(x) = f (a) + f ' (a) (x - a) + f " (a) / 2! *(x - a) ^ 2 + ..... + (nth derivative of f(x) at x = a) / n! * (x - a) ^ n

First, let's make a third degree Taylor series for f(x) = arctan(x) at a = 0.5:

T3(x) = arctan(0.5) + f ' (0.5) (x - 0.5) + f " (0.5) / 2! *(x - 0.5) ^ 2 + f "' (0.5) / 3! *(x - 0.5) ^ 3

Find the first, second, and third derivatives of arctan(x) at a = 0.5:

f ' (0.5) = 0.8

f " (0.5) / 2!= -0.32

f "' (0.5) / 3! = -0.04267

T3(x) = arctan(0.5) + 0.8 * (x - 0.5) - 0.32 * (x - 0.5) ^ 2 - 0.04267 * (x - 0.5) ^ 3

However, since the problem wants the x to be replaced to 2x, simply substitute every x in the series above to 2x (no need to change the derivatives and numbers, only change the x's to 2x):

T3(x) = arctan(0.5) + 0.8 * (2x - 0.5) - 0.32 * (2x - 0.5) ^ 2 - 0.04267 * (2x - 0.5) ^ 3 <<<< (ANSWER)

Let me know if you have additional questions for me! Hope this helped!