Taylor Series

f(x) = ∑ f^(n)(a)n!(x-a)^n — polynomials become any smooth function.

The idea

A Taylor series expresses a function as a sum of powers of $(x - a)$, built entirely from the function's derivatives at the point $a$, so that its partial sums serve as polynomial approximations to the function. Polynomials are the functions we can compute with directly — evaluating one requires only addition and multiplication — while $e^{x}$, $\sin x$ and $\ln(1 + x)$ are not polynomials.

The sum is chosen to match $f$ at $a$ to every order: the same value, the same first derivative, the same second derivative, and so on. These conditions force the coefficients. Differentiating $(x - a)^{n}$ exactly $n$ times leaves the constant $n!$, so for the $n$-th derivative at $a$ to equal $f^{(n)}(a)$, the coefficient of $(x - a)^{n}$ must be $f^{(n)}(a)/n!$. The result is $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^{n},$ the Taylor series of $f$ at $a$. Centered at $a = 0$ it is also called the Maclaurin series.

The factorial denominators drive the later terms down, so truncating after a few terms leaves a genuine approximation. And the construction uses information at the single point $a$ only, so the approximation is sharpest near $a$ and degrades as $x$ moves away.

Ways to work on it

• Walkthrough. Build a Taylor series at zero from derivatives and use it to approximate a value.
• Practice. Approximate a function's value from the first three terms of its Taylor series at zero.
• Hardest. Build a function's linear approximation at zero from scratch.

Not sure where to start? Take the ten-question placement test.