Taylor Series Expansion

• Any smooth function can be locally approximated by its Taylor series expansion

1D Function f(x)

• The Taylor series expansion is: $f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 +... + \frac{f^{(n)}(x)}{n!}h^n + R_n$
• $R_n$ represents the remainder.

Proof of Taylor Series Expansion

• $f(x+h) = f(x) + \int_x^{x+h} f'(\xi) d\xi$

Integration by Parts

• $\int_x^{x+h} f'(\xi) d\xi = f'(\xi) \cdot \xi |_x^{x+h} - \int_x^{x+h} f''(\xi) \cdot \xi d\xi$
• $= f'(x+h)(x+h) - f'(x)x - \int_x^{x+h} f''(\xi) \cdot \xi d\xi$
• $= f'(x)h + \int_x^{x+h} f''(\xi) (x+h-\xi) d\xi$

Next Step in Integration by Parts

• $\int_x^{x+h} f''(\xi) (x+h-\xi) d\xi = f''(\xi)(x+h-\xi)(-\frac{1}{2}(x+h-\xi)^2)|_x^{x+h} - \int_x^{x+h} f'''(\xi) \frac{1}{2}(x+h-\xi)^2 d\xi$
• $= \frac{f''(x)}{2!}h^2 + \int_x^{x+h} f'''(\xi) \frac{1}{2}(x+h-\xi)^2 d\xi$

Continuing the Pattern

• $\int_x^{x+h} f'''(\xi) \frac{1}{2}(x+h-\xi)^2 d\xi = \frac{f'''(x)}{3!}h^3 + \int_x^{x+h} f^{iv}(\xi) \frac{1}{3!}(x+h-\xi)^3 d\xi$

Resulting Taylor Series

• $f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 +... + \frac{f^{(n)}(x)}{n!}h^n + R_n$
• $R_n = \int_x^{x+h} f^{(n+1)}(\xi) \frac{1}{n!}(x+h-\xi)^n d\xi$

Approximating f'(x)

• The goal is to approximate $f'(x)$ using $f(x)$, $f(x+h)$, and $f(x+2h)$

Expansions of f(x+h) and f(x+2h)

• $f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 +...$
• $f(x+2h) = f(x) + f'(x)2h + \frac{f''(x)}{2!}(2h)^2 + \frac{f'''(x)}{3!}(2h)^3 +...$

General Form

• $f'(x) = a f(x) + b f(x+h) + c f(x+2h)$

Substituting Series

• $f'(x) = a f(x) + b (f(x) + f'(x)h + \frac{f''(x)}{2!}h^2 + \frac{f'''(x)}{3!}h^3 +...) + c (f(x) + f'(x)2h + \frac{f''(x)}{2!}(2h)^2 + \frac{f'''(x)}{3!}(2h)^3 +...)$

Collecting Terms

• $f(x): a + b + c = 0$
• $f'(x): bh + 2ch = 1$
• $f''(x): b\frac{h^2}{2} + c\frac{4h^2}{2} = 0$

Solving for a, b, and c

• From the f''(x) equation: $b + 4c = 0$, which means $b = -4c$
• Substituting into the f'(x) equation: $-4ch + 2ch = 1$, which simplifies to $-2ch = 1$. Thus, $c = -\frac{1}{2h}$
• Solving for b: $b = \frac{4}{2h} = \frac{2}{h}$
• Solving for a: $a = -b - c = -\frac{2}{h} + \frac{1}{2h} = -\frac{3}{2h}$

Final Approximation Formula

• $f'(x) = -\frac{3}{2h}f(x) + \frac{2}{h}f(x+h) - \frac{1}{2h}f(x+2h)$
• $f'(x) = \frac{-3f(x) + 4f(x+h) - f(x+2h)}{2h}$