Taylor Series

A Taylor series is simply a power series built to imitate one particular function near one particular point. (It's named after Brook Taylor, the English mathematician who published the general method in 1715.) It's how calculators find $\sin 0.3$ or $e^{1.7}$: they add up a few polynomial terms. The formula $\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$ looks heavy, but it's just $f^{(n)}(a)$ (the function's value, slope, bend and so on at the center), $n!$ (a correction factor that undoes the numbers differentiating $(x-a)^n$ produces), and $(x - a)^n$ (the powers we met last lesson). It's less scary than it looks, so bear with it.

Where the formula comes from. You already know the first step of this idea: the tangent line. It's the straight line that has the same value and the same slope as $f$ at $a$, and near $a$ it's a decent stand-in for $f$. So why stop at matching the slope? Match the curvature too, and the next derivative, and so on.

Here's how that pins down the coefficients. Suppose $f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots$. Setting $x = a$ kills every term but the first, so $c_0 = f(a)$. Differentiate and set $x = a$ again: $c_1 = f'(a)$. Differentiate once more and the $(x-a)^2$ term leaves behind $2c_2$, so $2c_2 = f''(a)$. After $n$ derivatives, $(x - a)^n$ has turned into $n!$, so $n!\,c_n = f^{(n)}(a)$. That's where the factorial comes from.

Taylor polynomials. Stopping at degree $n$ gives $T_n(x)$. $T_1$ is the tangent line, $T_2$ also matches the curvature, and so on. Each extra term copies one more feature of the function, so it hugs the function over a wider stretch.

Series to know (all centered at 0):

Does the series equal the function? $f(x) = T_n(x) + R_n(x)$, where $R_n$ is whatever error is left over. So the series equals $f$ exactly when $R_n(x) \to 0$, and Taylor's inequality is how you check. For $e^x$, $\sin x$ and $\cos x$ the remainder goes to 0 for every $x$, because $(n+1)!$ eventually crushes any power $|x|^{n+1}$.

A common mistake. It's tempting to think that matching every derivative at one point guarantees the series equals the function everywhere. It doesn't. The series for $\frac{1}{1-x}$ matches it perfectly at $0$ but is useless at $x = 3$, outside the radius of convergence. Even inside it, you need the remainder to go to 0. For the functions in the table it does, on the intervals listed.

Build new series from old ones. For $x^2 e^{-x}$ or $\sin(3x)$, substitute into a known series and multiply, exactly like last lesson. That's far faster than computing derivatives, and since a function has only one power series at a given center, you're guaranteed to get the same answer.

Where this leads: Euler's formula. Look at the first three rows of the table together. Substitute $x = i\theta$ into the series for $e^x$: the powers of $i$ cycle through $i, -1, -i, 1$, so the terms sort themselves into the cosine series plus $i$ times the sine series, and $e^{i\theta} = \cos\theta + i\sin\theta$. That one line is why sines and cosines appear when a differential equation has complex roots: a spring or a circuit oscillates because of how these three series fit together.

Physics runs on Taylor polynomials: $\sin\theta \approx \theta$ gives the simple pendulum, and expanding energy around an equilibrium gives Hooke's law. Newton's method and gradient descent are built on first- and second-order Taylor approximations.