1. The Question Inverted: From Functions to Series
The Big Idea: In the previous topic we started with a power series and asked "what function does this represent?" Now we reverse the question: given a function $f$, can we build a power series that equals it?
1.1 Deriving the Coefficient Formula
Suppose a function $f(x)$ does have a power series representation centered at $x = a$:
$$ f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots = \sum_{n=0}^{\infty} c_n(x-a)^n $$We want to discover what the coefficients $c_n$ must be. The strategy is beautifully simple: differentiate repeatedly and evaluate at $x = a$.
Step 0 — Set $x = a$:
$$ f(a) = c_0 + 0 + 0 + \cdots \implies c_0 = f(a) $$Step 1 — Differentiate once, then set $x = a$:
$$ f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \cdots $$ $$ f'(a) = c_1 \implies c_1 = f'(a) $$Step 2 — Differentiate twice:
$$ f''(x) = 2c_2 + 3 \cdot 2 \, c_3(x-a) + 4 \cdot 3 \, c_4(x-a)^2 + \cdots $$ $$ f''(a) = 2c_2 \implies c_2 = \frac{f''(a)}{2!} $$Step $n$ — Differentiate $n$ times:
$$ f^{(n)}(a) = n! \, c_n \implies \boxed{c_n = \frac{f^{(n)}(a)}{n!}} $$
Key Insight (Uniqueness Payoff): We proved earlier that a power series representation, if it exists, is unique. The coefficients are therefore forced by the function itself — they are derived, not decreed.
1.2 The Taylor Series Definition
Definition — Taylor Series
If $f$ has derivatives of all orders at $x = a$, then the Taylor series of $f$ centered at $a$ is:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots $$