Introduction

Taylor Series are a remarkable piece of mathematical machinery that every aspiring theoretical physicist should know about! The reason they are so powerful is because their purpose is to transform an awkward function into a nice, polynomial one.

They are also useful in providing the rigor required to say that $\sin(x) \approx \tan(x) \approx x$, a common hand-wavy trick that makes physicists get on mathematicians' nerves.

Definition & Intuition

Given an $n$-times differentiable function $f(x)$, the polynomial

is called a Taylor Polynomial, and its limit as you add infinitely many terms,

is called the Taylor Series of the function $f(x)$ centered around $x = a$.

If $S(x)$ converges and equals $f(x)$ everywhere, it is said that the Taylor Expansion of $f(x)$ is $S(x)$.

To get an intuition of how this works, consider $f(x) = \sin(x)$. Close to the origin, the graph of $\sin(x)$ looks very similar to the linear function $y = x$. The reason for their similarity is that their first derivatives match near the origin: differentiating $\sin(x)$ gives $\cos(x)$, and at $x = 0$, $\cos(0) = 1$ — the same as the slope of $y = x$.

Can we improve on this? Since $\sin(x)$ is an odd function ($\sin(-x) = -\sin(x)$), our polynomial approximation must also be odd, ruling out all even powers of $x$. Our next best guess is to add a cubic term: $y = x - cx^3$. Differentiating:

For $y = \sin(x)$ at $x = 0$:

For the third derivatives to match, we need $c = \dfrac{1}{6}$, giving us $y = x - \dfrac{x^3}{6}$.

And that's the core idea of Taylor Series! Each time we match a higher derivative, our polynomial approximates the original function better and better. To model the function around some point $x = a$ instead of $x = 0$, compute the derivatives at $a$ and replace each $x^k$ with $(x - a)^k$.

Common Taylor Series

Binomial Expansion

You might have learned this in school and be thinking "this isn't a Taylor Series!" — and you're right, but since these two ideas share many similar principles, it's fitting to discuss them together.

Given a binomial raised to a power, its expanded form is:

where $\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$ is the binomial coefficient.

To see why this is true, consider $x = 0$. On the RHS, all terms vanish except the $k = 0$ term, leaving $\dbinom{n}{0}a^n = a^n$, which matches the LHS. Taking derivatives of both sides and evaluating at $x = 0$ likewise confirms that each successive term is correct — the binomial expansion can be understood as a Taylor polynomial.

Trigonometric Functions

Here are the classic Taylor Expansions of trigonometric functions:

Exponential & Logarithmic Functions

Note that the logarithmic series starts at $k = 1$ to avoid division by zero.

Power Series

A power series $P(x)$ has the form:

where each coefficient $a_k$ is a constant. A familiar example is the power series of $\dfrac{1}{1-x}$, where all coefficients equal 1:

Power series appear throughout higher mathematics and physics, from generating functions in combinatorics to solving the Schrödinger equation for hydrogen-like atoms. One typically uses them to construct solutions to equations by plugging in the series and deducing what the coefficients must be.

Example — Simple Harmonic Motion: Consider the SHM equation $y'' = -\omega^2 y$. Let $y = \sum_{k=0}^{\infty} a_k t^k$ be a solution. Substituting and reindexing:

Matching coefficients of $x^k$ gives the recurrence:

Since this connects coefficients two indices apart, we get two independent families — one for even $k$ and one for odd $k$:

Substituting back into the power series and recognizing the resulting sums as the Taylor series for cosine and sine:

Applications

Approximations

For the power series of $\dfrac{1}{1-x}$, successive terms differ by a factor of $x$. When $x$ is very small (e.g. $x = 0.01$), each additional term contributes less and less. In physics, measurements are only useful up to a certain precision, so we can approximate:

Other famous small-angle or small-$x$ approximations include:

All of which only hold when $|x| \ll 1$.

Problems & Practice

Problem 1: Show that both of the following definitions of $e^x$ are equivalent:

$e^x = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n$

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

Problem 2: Find the Taylor expansion of $\arctan x$.

Problem 3: Consider the Taylor expansion of $\ln\!\left(\dfrac{1}{1-x}\right)$.

(a) Does the Taylor expansion match the function well everywhere? What happens as you add more and more terms near the boundaries?

(b) Under what circumstances will the series converge? (Hint: think about ratios of successive terms.)

(c) Find the range of values $x$ for which the infinite series converges. This is called the radius of convergence. (Careful about the boundary inequalities!)

Problem 4: Prove Euler's Formula:

$e^{ix} = \cos(x) + i\sin(x)$

where $i$ is the imaginary unit satisfying $i^2 = -1$.

Solutions

Solution 1: Apply the binomial theorem to $\left(1 + \dfrac{x}{n}\right)^n$ and simplify as $n \to \infty$ to recover the series definition.

Solution 2: Differentiating repeatedly and matching derivatives at $x = 0$:

$\arctan x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}$

Solution 3:

(a) The expansion diverges at the sides; it only converges on some central interval.

(b) The series converges when successive terms get smaller, i.e. $\left|\dfrac{T_{n+1}}{T_n}\right| < 1$:

$\left|\frac{\frac{1}{n+1}x^{n+1}}{\frac{1}{n}x^n}\right| = \left|\frac{n}{n+1}x\right| < 1$

(c) Taking $n \to \infty$ in the inequality above gives $|x| < 1$, i.e. $-1 < x < 1$. Checking the boundary $x = -1$: the series becomes $\sum \dfrac{(-1)^{k+1}}{k}$, which converges (to $\ln 2$). So the full radius of convergence is:

$\boxed{-1 \leq x < 1}$

Solution 4: Use the Taylor expansion of $e^x$ and substitute $x \to ix$:

$e^{ix} = \sum_{k=0}^{\infty} \frac{(ix)^k}{k!}$

Separating real and imaginary parts recovers the Taylor series for $\cos x$ and $\sin x$ respectively.