Taylor Series

116. Taylor Series

Taylor Series (Expansion)

Taylor polynomial

Approximate a complicated function near a point $𝑎$ by a polynomial made from the function’s derivatives at that point

𝑓 (𝑥) = \sum_{𝑛 = 0}^{\infty} \frac{𝑓^{(𝑛)} (𝑎)}{𝑛!} {(𝑥 - 𝑎)}^{𝑛}

𝑓 (𝑥) = \underset{* * *}{\underset{⏟}{\underset{* *}{\underset{⏟}{\underset{*}{\underset{⏟}{𝑓 (𝑥_{0})}} + 𝑓^{'} (𝑥_{0}) (𝑥 - 𝑥_{0})}} + \frac{1}{2!} 𝑓^{″} (𝑎) (𝑥 - 𝑎)}} + \dots + \frac{1}{𝑛!} 𝑓^{(𝑛)} (𝑥_{0}) (𝑥 - 𝑎),

Where:

$*$ : Ensures value of the polynomial matches with the value of $𝑓$ when $𝑥 = 𝑎$
$* *$ : Ensures slope of the polynomial matches with the slope of $𝑓$ when $𝑥 = 𝑎$
$* * *$ : Ensures rate of change of the polynomial matches with the rate of change of $𝑓$ when $𝑥 = 𝑎$

Where:

Example

Setup

𝑓 (𝑥) = 𝑥^{2}

Expand it around the point $𝑎 = 1$

Step 1: Compute Derivatives

Step 2: Evaluate at $𝑥 = 1$

Step 3: Write the Taylor expansion

\begin{aligned} 𝑓 (𝑥) & = 𝑓 (1) + 𝑓^{'} (1) (𝑥 - 1) + \frac{𝑓^{″} (1)}{2!} {(𝑥 - 1)}^{2} \\ = 1 + 2 (𝑥 - 1) + \frac{2}{2} {(𝑥 - 1)}^{2} \\ = 1 + 2 (𝑥 - 1) + {(𝑥 - 1)}^{2} \end{aligned}

Step 3: Expand

1 + 2 (𝑥 - 1) + {(𝑥 - 1)}^{2} = 1 + 2 𝑥 - 2 + 𝑥^{2} + 1 = 𝑥^{2}

It gives back the original function exactly

Example