Difficulty: Intermediate | Prerequisites: Calculus I (derivatives), basic power series and convergence tests (Ratio Test, Alternating Series Test).

Taylor series let you rewrite any sufficiently smooth function as an infinite polynomial centred at a chosen point. This is one of the central ideas in Calculus II: it connects differentiation, integration, and infinite series into a single framework. If you can take derivatives and you understand what a power series is, you are ready for this material. If the words "radius of convergence" or "ratio test" are unfamiliar, review those first.

TL;DR: A Taylor series expresses a function as an infinite sum of terms built from its derivatives at a single point. You can derive these by computing derivatives and spotting a pattern, or by substituting into one of five standard series you should memorise. The ratio test tells you where the series converges.

Key Terms

Taylor series

The representation of a function f(x) as an infinite sum of terms calculated from the values of its derivatives at a single point a. In simple terms, you are building a polynomial that matches the function's behaviour near a, term by term.

Maclaurin series

A Taylor series centred at a = 0. It is just the special case, not a separate concept. When someone says "find the Taylor series centred at 0," they mean the Maclaurin series.

Centre (or centre of expansion), a

The point around which the series is built. Every (x − a) factor in the series measures distance from this point.

Interval of convergence (I.O.C.)

The set of x-values for which the Taylor series converges to a finite sum. Think of it as the range of inputs where the infinite polynomial actually works as a stand-in for the original function.

Radius of convergence, R

Half the width of the interval of convergence. If the series converges for |x − a| < R, then R is the radius. The series always diverges for |x − a| > R; the endpoints need separate checking.

Ratio test

A convergence test: compute lim (n→∞) |a_{n+1}/a_n|. If the limit is less than 1 the series converges absolutely; if greater than 1 it diverges. This is the standard tool for finding the radius of convergence of a Taylor series.

Double factorial (n!!)

The product of every other positive integer down to 1 or 2. For odd n: n!! = n × (n−2) × (n−4) × … × 1. For even n: n!! = n × (n−2) × … × 2. Example: 7!! = 7 × 5 × 3 × 1 = 105; 8!! = 8 × 6 × 4 × 2 = 384. It appears in the Taylor series for functions like 1/√x.

Alternating Series Test (A.S.T.)

Used to check convergence at endpoints of the interval. If the terms alternate in sign, decrease in absolute value, and tend to zero, the series converges. This is how you determine whether an endpoint is included in (or excluded from) the interval of convergence.

Core Content: The Taylor Series Formula

For a function f(x) that is infinitely differentiable near x = a, the Taylor series centred at a is:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x - a)^n

Written out, the first several terms look like this:

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Each term captures one more layer of the function's behaviour at the point a. The zeroth term matches the function's value, the first term matches its slope, the second term matches its concavity, and so on.

The recipe for deriving a Taylor series from scratch:

• Compute f(a), f'(a), f''(a), f'''(a), and enough higher derivatives to spot a pattern in f^(n)(a)

• Write out the general term f^(n)(a)/n! multiplied by (x − a)^n

• Simplify the general term into a clean closed-form expression for the nth term

• Use the ratio test to find the interval of convergence

Worked Example: ln(1 + x) Centred at a = 0

Step 1: Compute derivatives and evaluate at a = 0

Step 2: Spot the pattern

The general formula for the nth derivative at 0 is:

f^{(n)}(0) = (-1)^{n+1}(n-1)! \quad \text{for } n \geq 1

This works for n = 1 as well, since 0! = 1 by definition. The n = 0 term is simply f(0) = ln(1) = 0, so it drops out.

Step 3: Write the series

Plugging into the Taylor formula and noting the n = 0 term is zero:

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}\,x^n = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\,x^n

The simplification (n−1)!/n! = 1/n is the key step.

Written out:

\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots

Alternative derivation (via integration)

You can also reach this series by integrating the geometric series. Start from 1/(1 + x) = 1/(1 − (−x)) = Σ(−x)^n, then integrate term by term:

\ln(1+x) = \int \frac{1}{1+x}\,dx = \int \sum_{n=0}^{\infty}(-1)^n x^n\,dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} + C

Re-indexing (set m = n + 1) and noting C = 0 gives the same result.

Step 4: Interval of convergence

Apply the ratio test:

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{n}{n+1}|x| = |x|

This is less than 1 when |x| < 1, so the radius of convergence is R = 1 and the open interval is (−1, 1).

Check the endpoints separately:

• At x = −1: the series becomes Σ (−1)^(n+1)(−1)^n / n = −Σ 1/n, which diverges (harmonic series)

• At x = 1: the series becomes Σ (−1)^(n+1)/n, which converges by the Alternating Series Test (the terms b_n = 1/n decrease and tend to zero)

The interval of convergence is (−1, 1].

Worked Example: 1/√x Centred at a = 1

This example introduces the double factorial and a non-zero centre.

Step 1: Rewrite and compute derivatives

Write f(x) = x^(−1/2). The derivatives follow the pattern:

Step 2: Spot the pattern

The general formula uses the double factorial:

f^{(n)}(1) = \frac{(-1)^n\,(2n-1)!!}{2^n} \quad \text{for } n \geq 1

Recall that (2n−1)!! = 1 × 3 × 5 × … × (2n−1). For example, when n = 3: (2·3−1)!! = 5!! = 5 × 3 × 1 = 15.

Step 3: Write the series

The n = 0 term contributes f(1)·(x−1)^0 = 1. For n ≥ 1:

\frac{1}{\sqrt{x}} = 1 + \sum_{n=1}^{\infty} \frac{(-1)^n\,(2n-1)!!}{2^n\,n!}\,(x-1)^n

Written out:

\frac{1}{\sqrt{x}} = 1 - \frac{1}{2}(x-1) + \frac{3}{8}(x-1)^2 - \frac{5}{16}(x-1)^3 + \cdots

Step 4: Interval of convergence

Apply the ratio test:

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(2n+1)}{2(n+1)}\,|x-1| = |x-1|

The limit (2n+1)/(2n+2) tends to 1, so the overall limit is |x − 1|. This is less than 1 when |x − 1| < 1, giving 0 < x < 2.

The interval of convergence is (0, 2). (Endpoints can be checked separately, but the notes indicate this is left as an exercise.)

Finding the Interval of Convergence

The interval of convergence (I.O.C.) tells you where a Taylor series is valid. The standard approach has three steps.

Step 1: Apply the ratio test

For a series Σ a_n, compute:

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

Set L < 1 and solve for x. This gives you the open interval and the radius of convergence R.

Step 2: Check the endpoints

The ratio test is inconclusive when L = 1 (the endpoints). Plug each endpoint into the original series and test convergence with a different test:

• If the endpoint produces a p-series (Σ 1/n^p), it converges when p > 1 and diverges when p ≤ 1

• If it produces an alternating series, use the Alternating Series Test: check that the terms b_n decrease and b_n → 0

• The harmonic series Σ 1/n diverges; the alternating harmonic series Σ (−1)^(n+1)/n converges

Step 3: Write the interval

Use parentheses for excluded endpoints and brackets for included endpoints.

Summary of the two worked examples:

Formulas to Know

Taylor series (general form)

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

Ratio test for interval of convergence

\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 \implies \text{convergence}

Derivative pattern for ln(1 + x)

f^{(n)}(0) = (-1)^{n+1}(n-1)! \quad (n \geq 1)

Derivative pattern for 1/√x at a = 1

f^{(n)}(1) = \frac{(-1)^n(2n-1)!!}{2^n} \quad (n \geq 1)

Key simplification

(n − 1)!/n! = 1/n. This is why the ln(1 + x) series simplifies so cleanly.

Common Misconceptions

• Students often forget that the n = 0 term in the Taylor formula is just f(a), not zero. For ln(1 + x) it happens to be zero, but that is a special case, not the rule.

• The ratio test does not settle convergence at the endpoints. It returns L = 1 there, which is inconclusive. You must test the endpoints individually with another method.

• Confusing the centre a with a variable. Once you choose a, it is a fixed constant. The only variable in the series is x.

• Writing (−1)^n when you mean (−1)^(n+1) (or vice versa). The sign of the first term of the series tells you which exponent is correct. If the first nonzero term is positive, make sure your formula gives a positive sign at the starting index.

Real-World Applications

Taylor series are how calculators and computers evaluate functions like sin, cos, exp, and log internally: they truncate the infinite series to a finite number of terms and use the remainder to bound the error. In physics and engineering, replacing a complicated function with the first few terms of its Taylor expansion (a "linearisation" or "small-angle approximation") is one of the most common problem-solving moves. The small-angle approximation sin(θ) ≈ θ is just the first term of the Taylor series for sin.

Why It Matters / Exam Flags

⚠️ "Derive the Taylor series from scratch" is a classic exam question. Be ready to compute several derivatives, spot the pattern, and write the general term.

⚠️ Finding the interval of convergence almost always appears alongside a Taylor series question. Do not skip the endpoint checks.

⚠️ The simplification (n−1)!/n! = 1/n appears frequently. Recognise it instantly.

⚠️ Know when to use the Alternating Series Test at endpoints versus when to recognise a p-series or harmonic series.

Quick Self-Test

1. True or False: A Maclaurin series is a Taylor series centred at a = 1.

1. Fill in the blank: In the Taylor series formula, the nth term is f^(n)(a) divided by ______, multiplied by (x − a)^n.

1. True or False: The ratio test can determine convergence at the endpoints of the interval.

1. Fill in the blank: 5!! = ______.

1. True or False: The Taylor series for ln(1 + x) centred at 0 converges at x = 1.

(Answers: 1. False, a = 0. 2. n! 3. False, it is inconclusive there. 4. 15. 5. True, by the Alternating Series Test.)

Practice Q&A

Q: Write the Taylor series formula for a general function f(x) centred at x = a.

A: f(x) = Σ (n=0 to ∞) f^(n)(a)/n! × (x − a)^n.

Q: Find the first four nonzero terms of the Taylor series for ln(1 + x) centred at a = 0.

A: x − x²/2 + x³/3 − x⁴/4.

Q: What is the interval of convergence for the Taylor series of ln(1 + x) centred at 0?

A: (−1, 1]. The series diverges at x = −1 (harmonic series) and converges at x = 1 (alternating harmonic series).

Q: Explain why the n = 0 term vanishes in the Taylor series for ln(1 + x) centred at 0.

A: Because f(0) = ln(1 + 0) = ln(1) = 0.

Q: Define the double factorial 7!! and compute its value.

A: 7!! = 7 × 5 × 3 × 1 = 105. The double factorial multiplies every other integer down to 1 (for odd numbers) or 2 (for even numbers).

Q: Find the radius of convergence for the Taylor series of 1/√x centred at a = 1.

A: R = 1. The ratio test gives |x − 1| < 1, so the series converges for 0 < x < 2.

Connections to Other Topics

This material builds directly on power series and convergence tests from earlier in the course. The ratio test, alternating series test, and p-series test all reappear here as tools for finding intervals of convergence.

The alternative derivation of ln(1 + x) via integration connects Taylor series back to the geometric series 1/(1 − x) = Σ x^n. That link between known series and new ones is the foundation of the substitution technique covered in the companion notes.

Taylor series also connect forward to approximation and error bounds (Taylor's remainder theorem), which you will need for numerical methods in later courses.

Related Terms / Search Tags

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