Represent a function as an infinite polynomial centered at a point \(a\).
The Taylor series of \(f\) centered at \(a\) is the unique power series that matches all derivatives of \(f\) at \(x=a\).
Tip: Compute successive derivatives, evaluate at \(a\), and build the coefficients \(\frac{f^{(n)}(a)}{n!}\).
Most of the important functions in math and science — \(e^x\), \(\sin x\), \(\ln x\), \(\sqrt{1+x}\), etc. — are not polynomials. Polynomials are easy to add, multiply, differentiate, and integrate. Taylor series let us turn complicated functions into (infinite) polynomials that we can actually work with.
Approximation & Calculators
Your calculator does not magically “know” the value of \(\sin(1.3)\). It evaluates a carefully chosen Taylor polynomial. The same idea is used in almost every computer program that needs accurate function values.
Solving Hard Problems
Many differential equations that appear in physics and engineering have no simple closed-form answer. Writing the solution as a Taylor series is one of the standard ways to find it.
Understanding Behavior
The first few terms tell us the local shape of a graph (slope, concavity, etc.). The radius of convergence tells us where the approximation is trustworthy, and the remainder term tells us how accurate it is.
Bridge to Advanced Math
Taylor series are the doorway to Fourier series, complex analysis, numerical methods, and many topics you will see in college-level math, physics, and engineering.
In short: Taylor series let us replace hard functions with polynomials that we can compute, analyze, and manipulate. That single idea is one of the most practical tools in all of calculus.
Find the Taylor series for \(f(x) = e^x\) centered at \(a = 2\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= e^x \\ f'(x) &= e^x \\ f”(x) &= e^x \\ f”'(x) &= e^x \\ &\vdots \\ f^{(n)}(x) &= e^x \end{align*} \]
Step 2: Evaluate at \(a = 2\)
\[ f^{(n)}(2) = e^2 \quad \text{for every } n \]
Step 3: Build the first few terms
\[ \begin{align*} n=0: &\quad \dfrac{e^2}{0!}(x-2)^0 = e^2 \\ n=1: &\quad \dfrac{e^2}{1!}(x-2) = e^2(x-2) \\ n=2: &\quad \dfrac{e^2}{2!}(x-2)^2 = \dfrac{e^2}{2}(x-2)^2 \\ n=3: &\quad \dfrac{e^2}{6}(x-2)^3 \end{align*} \]
Beginning of the series: \[ e^x = e^2 + e^2(x-2) + \dfrac{e^2}{2}(x-2)^2 + \dfrac{e^2}{6}(x-2)^3 + \cdots \]
Step 4: Write the general series
\[ e^x = \sum_{n=0}^{\infty} \dfrac{e^2}{n!}(x-2)^n = e^2\sum_{n=0}^{\infty} \dfrac{(x-2)^n}{n!} \]
\( e^x = e^2\sum_{n=0}^{\infty} \dfrac{(x-2)^n}{n!} \)
Basic Taylor expansions about a non-zero center
Find the first four non-zero terms of the Taylor series for \(f(x) = \ln x\) centered at \(a = 1\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \ln x \\ f'(x) &= \dfrac{1}{x} \\ f”(x) &= -\dfrac{1}{x^2} \\ f”'(x) &= \dfrac{2}{x^3} \\ f^{(4)}(x) &= -\dfrac{6}{x^4} \end{align*} \]
Step 2: Evaluate at \(a = 1\)
\[ \begin{align*} f(1) &= 0 \\ f'(1) &= 1 \\ f”(1) &= -1 \\ f”'(1) &= 2 \\ f^{(4)}(1) &= -6 \end{align*} \]
Step 3: Build the first four non-zero terms
\[ \begin{align*} n=1: &\quad \dfrac{1}{1!}(x-1) = (x-1) \\ n=2: &\quad \dfrac{-1}{2!}(x-1)^2 = -\dfrac{(x-1)^2}{2} \\ n=3: &\quad \dfrac{2}{3!}(x-1)^3 = \dfrac{(x-1)^3}{3} \\ n=4: &\quad \dfrac{-6}{4!}(x-1)^4 = -\dfrac{(x-1)^4}{4} \end{align*} \]
Step 4: Write the series
\[ \ln x = (x-1) – \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} – \dfrac{(x-1)^4}{4} + \cdots \]
\( (x-1) – \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} – \dfrac{(x-1)^4}{4} + \cdots \)
Find the Taylor series for \(f(x) = \cos x\) centered at \(a = \pi\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \cos x \\ f'(x) &= -\sin x \\ f”(x) &= -\cos x \\ f”'(x) &= \sin x \\ f^{(4)}(x) &= \cos x \\ f^{(5)}(x) &= -\sin x \\ f^{(6)}(x) &= -\cos x \\ &\vdots \end{align*} \]
Step 2: Evaluate at \(a = \pi\)
\[ \begin{align*} f(\pi) &= -1 \\ f'(\pi) &= 0 \\ f”(\pi) &= 1 \\ f”'(\pi) &= 0 \\ f^{(4)}(\pi) &= -1 \\ f^{(5)}(\pi) &= 0 \\ f^{(6)}(\pi) &= 1 \end{align*} \]
All odd-powered derivatives are zero at \(x=\pi\).
Step 3: Build the first four non-zero terms
\[ \begin{align*} n=0: &\quad -1 \\ n=2: &\quad \dfrac{1}{2}(x-\pi)^2 \\ n=4: &\quad -\dfrac{1}{24}(x-\pi)^4 \\ n=6: &\quad \dfrac{1}{720}(x-\pi)^6 \end{align*} \]
\[ \cos x = -1 + \dfrac{1}{2}(x-\pi)^2 – \dfrac{1}{24}(x-\pi)^4 + \dfrac{1}{720}(x-\pi)^6 – \cdots \]
Step 4: Write the general series
\[ \cos x = -\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}(x-\pi)^{2n} \]
\( \cos x = -\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}(x-\pi)^{2n} \)
Find the Taylor series for \(f(x) = \sin x\) centered at \(a = \dfrac{\pi}{2}\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= \sin x \\ f'(x) &= \cos x \\ f”(x) &= -\sin x \\ f”'(x) &= -\cos x \\ f^{(4)}(x) &= \sin x \\ &\vdots \end{align*} \]
Step 2: Evaluate at \(a = \pi/2\)
\[ \begin{align*} f(\pi/2) &= 1 \\ f'(\pi/2) &= 0 \\ f”(\pi/2) &= -1 \\ f”'(\pi/2) &= 0 \\ f^{(4)}(\pi/2) &= 1 \end{align*} \]
Step 3: Build the first four non-zero terms
\[ \begin{align*} n=0: &\quad 1 \\ n=2: &\quad -\dfrac{1}{2}\Bigl(x-\dfrac{\pi}{2}\Bigr)^2 \\ n=4: &\quad \dfrac{1}{24}\Bigl(x-\dfrac{\pi}{2}\Bigr)^4 \\ n=6: &\quad -\dfrac{1}{720}\Bigl(x-\dfrac{\pi}{2}\Bigr)^6 \end{align*} \]
Step 4: Write the general series
\[ \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}\Bigl(x-\dfrac{\pi}{2}\Bigr)^{2n} \]
\( \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!}\bigl(x-\tfrac{\pi}{2}\bigr)^{2n} \)
Find the Taylor series for \(f(x) = e^x\) centered at \(a = 1\).
Step 1: List the derivatives
Every derivative of \(e^x\) is \(e^x\).
Step 2: Evaluate at \(a = 1\)
$$f^{(n)}(1) = e \quad \text{for every } n$$
Step 3: Build the first few terms
\[ e^x = e + e(x-1) + \dfrac{e}{2}(x-1)^2 + \dfrac{e}{6}(x-1)^3 + \cdots \]
Step 4: Write the general series
\[ e^x = e\sum_{n=0}^{\infty} \dfrac{(x-1)^n}{n!} \]
\( e^x = e\sum_{n=0}^{\infty} \dfrac{(x-1)^n}{n!} \)
More derivatives or algebraic manipulation required
Find the Taylor polynomial of degree 3 for \(f(x) = \sqrt{x}\) centered at \(a = 4\).
Step 1: List the derivatives
\[ \begin{align*} f(x) &= x^{1/2} \\ f'(x) &= \dfrac{1}{2}x^{-1/2} \\ f”(x) &= -\dfrac{1}{4}x^{-3/2} \\ f”'(x) &= \dfrac{3}{8}x^{-5/2} \end{align*} \]
Step 2: Evaluate at \(a = 4\)
\[ \begin{align*} f(4) &= 2 \\ f'(4) &= \dfrac{1}{4} \\ f”(4) &= -\dfrac{1}{32} \\ f”'(4) &= \dfrac{3}{256} \end{align*} \]
Step 3: Build the degree-3 polynomial
\[ \begin{align*} n=0: &\quad 2 \\ n=1: &\quad \dfrac{1}{4}(x-4) \\ n=2: &\quad \dfrac{-1/32}{2}(x-4)^2 = -\dfrac{1}{64}(x-4)^2 \\ n=3: &\quad \dfrac{3/256}{6}(x-4)^3 = \dfrac{1}{512}(x-4)^3 \end{align*} \]
Step 4: Write the polynomial
\[ P_3(x) = 2 + \dfrac{1}{4}(x-4) – \dfrac{1}{64}(x-4)^2 + \dfrac{1}{512}(x-4)^3 \]
\( P_3(x) = 2 + \dfrac{1}{4}(x-4) – \dfrac{1}{64}(x-4)^2 + \dfrac{1}{512}(x-4)^3 \)
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