How do you find the taylor series expansion of #f(x) =x/(1+x)# around x=0?

Question

Luke Alvarez · Accepted Answer

You know that #1/(1-x) = 1 +x + x^2 + x^3 + ...# so, #1/(1+x) = 1 - x + x^2 - x^3 + ....# therefore, #x/(1+x) = x - x^2 + x^3 - x^4 + ... = sum_{k=0}^\infty (-1)^(k)x^(k+1)#

Cameron Alexander · Answer

undefined To find the Taylor series expansion of $ f(x) = \frac{x}{1+x} $ around $ x = 0 $, you can follow these steps:

1. Find the derivatives of $ f(x) $ at $ x = 0 $ up to the desired order.
2. Evaluate each derivative at $ x = 0 $.
3. Use the Taylor series formula to construct the expansion.

The Taylor series formula is:

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

where $ f(a) $ is the value of the function at the expansion point $ a $, and $ f'(a), f''(a), f'''(a), $ etc., are the derivatives of $ f(x) $ evaluated at $ a $.

Now, let's find the derivatives of $ f(x) = \frac{x}{1+x} $ and evaluate them at $ x = 0 $:

$$ f(x) = \frac{x}{1+x} $$
$$ f'(x) = \frac{1}{(1+x)^2} $$
$$ f''(x) = \frac{-2}{(1+x)^3} $$
$$ f'''(x) = \frac{6}{(1+x)^4} $$
$$ f^{(4)}(x) = \frac{-24}{(1+x)^5} $$

Evaluate each derivative at $ x = 0 $:

$$ f(0) = 0 $$
$$ f'(0) = 1 $$
$$ f''(0) = -2 $$
$$ f'''(0) = 6 $$
$$ f^{(4)}(0) = -24 $$

Plug these values into the Taylor series formula:

$$ f(x) = 0 + 1 \cdot x - 2 \cdot \frac{x^2}{2!} + 6 \cdot \frac{x^3}{3!} - 24 \cdot \frac{x^4}{4!} + \dotsb $$

Simplify the expression:

$$ f(x) = x - x^2 + x^3 - x^4 + \dotsb $$

This is the Taylor series expansion of $ f(x) = \frac{x}{1+x} $ around $ x = 0 $.