Taylor Series Derivation

TL;DR

Taylor series let you approximate any smooth function with an infinite sum of polynomial terms. You derive these terms by matching the function's derivatives at a specific point. Each term uses a higher derivative and a factorial to ensure accuracy around that point.

1. The Mental Model

Imagine you want to predict a function's behavior near a point using just a polynomial. A Taylor series builds this polynomial by making sure it matches the function's value, slope, concavity, and so on, at that single point.

2. The Core Material

You know that polynomials are easy to work with (differentiate, integrate, evaluate). The goal of a Taylor series is to represent a more complex function, say $f(x)$, as an infinite polynomial:

$f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \dots$

The key is to find the coefficients $c_0, c_1, c_2, \dots$ so that the polynomial matches $f(x)$ and all its derivatives at a specific "center" point, $x=a$.

Finding the Coefficients ($c_n$)

Photo by Brett Jordan on Pexels

Let's evaluate $f(x)$ and its derivatives at $x=a$:

1. At $x=a$:
   $f(a) = c_0 + c_1(a-a) + c_2(a-a)^2 + \dots$
   $f(a) = c_0$
   So, $c_0 = f(a)$.

2. First Derivative:
   $f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \dots$
   Evaluate at $x=a$:
   $f'(a) = c_1 + 2c_2(a-a) + 3c_3(a-a)^2 + \dots$
   $f'(a) = c_1$
   So, $c_1 = f'(a)$.

3. Second Derivative:
   $f''(x) = 2c_2 + 3 \cdot 2 c_3(x-a) + 4 \cdot 3 c_4(x-a)^2 + \dots$
   Evaluate at $x=a$:
   $f''(a) = 2c_2 + 3 \cdot 2 c_3(a-a) + \dots$
   $f''(a) = 2c_2$
   So, $c_2 = \frac{f''(a)}{2}$.

4. Third Derivative:
   $f'''(x) = 3 \cdot 2 c_3 + 4 \cdot 3 \cdot 2 c_4(x-a) + \dots$
   Evaluate at $x=a$:
   $f'''(a) = 3 \cdot 2 c_3$
   So, $c_3 = \frac{f'''(a)}{3 \cdot 2} = \frac{f'''(a)}{3!}$.

Do you see a pattern emerging? For the $n$-th derivative, $f^{(n)}(a) = n! c_n$.

This gives you the general formula for the coefficients: $c_n = \frac{f^{(n)}(a)}{n!}$.

The Taylor Series Formula

Photo by Andrea Piacquadio on Pexels

Substituting these coefficients back into the polynomial series, you get the Taylor Series for $f(x)$ centered at $x=a$:

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

This can be written in summation notation:

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

Maclaurin Series

Photo by Andrea Piacquadio on Pexels

A Maclaurin series is just a special case of a Taylor series where the center point $a$ is 0.

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots$

Here's how the derivation process flows:

graph TD
    A["Start with desired polynomial form: Σ c_n (x-a)^n"] --> B["Assume f(x) = the polynomial"]
    B --> C["Evaluate f(x) at x=a"]
    C --> D["Solve for c_0 = f(a)"]
    D --> E["Differentiate f(x) once"]
    E --> F["Evaluate f'(x) at x=a"]
    F --> G["Solve for c_1 = f'(a)"]
    G --> H["Differentiate f(x) again"]
    H --> I["Evaluate f''(x) at x=a"]
    I --> J["Solve for c_2 = f''(a)/2!"]
    J --> K["Repeat for n-th derivative"]
    K --> L["Discover pattern: c_n = f^(n)(a)/n!"]
    L --> M["Substitute c_n back into polynomial"]
    M --> N["Result: Taylor Series Formula"]

3. Worked Example

Let's derive the Taylor series for $f(x) = e^x$ centered at $a=0$ (a Maclaurin series).

1. Find the function and its derivatives:
   $f(x) = e^x$
   $f'(x) = e^x$
   $f''(x) = e^x$
   $f'''(x) = e^x$
   ...
   $f^{(n)}(x) = e^x$

2. Evaluate at $a=0$:
   $f(0) = e^0 = 1$
   $f'(0) = e^0 = 1$
   $f''(0) = e^0 = 1$
   $f'''(0) = e^0 = 1$
   ...
   $f^{(n)}(0) = e^0 = 1$

3. Apply the Maclaurin Series formula:
   $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$
   $e^x = \frac{1}{0!}x^0 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots$
   $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots + \frac{x^n}{n!} + \dots$

This is the well-known Maclaurin series for $e^x$.

4. Key Takeaways

• Taylor series approximate a function using an infinite polynomial based on its derivatives at a single point.
• The coefficients $c_n$ are found by ensuring the polynomial's derivatives match the function's derivatives at the center point.
• The general formula for a Taylor series centered at $a$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$.
• A Maclaurin series is simply a Taylor series centered at $a=0$.
• Each term in the series gets more accurate by matching a higher-order derivative.

• The factorial in the denominator (n!) comes from the repeated differentiation of the polynomial terms.

• Common mistakes to avoid:

  • Forgetting the factorial in the denominator.
  • Incorrectly evaluating derivatives at the center point $a$.
  • Mixing up $x$ and $a$ in the $(x-a)^n$ term.
  • Forgetting the $(x-a)$ term in general, especially when $a \ne 0$.

5. Now Try It

Derive the Taylor series for $f(x) = \sin(x)$ centered at $a=0$ (Maclaurin series). Find at least the first four non-zero terms.

What success looks like: You'll have a series that starts with $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ and you can clearly show how you got each coefficient.