Difficulty: Intermediate | Prerequisites: Taylor series definition (see companion notes), basic derivatives, familiarity with the five standard Maclaurin series.

Once you know the five standard Taylor series (e^x, sin x, cos x, 1/(1−x), ln(1+x)), you can build the Taylor series for a huge range of other functions without computing a single derivative. The trick is substitution: replace the variable in a known series with an expression, then simplify. This is faster, cleaner, and far less error-prone than deriving from scratch. This set of notes covers the five series you need to memorise and three worked examples of the substitution technique.

TL;DR: Memorise e^x, sin x, cos x, 1/(1−x), and ln(1+x) as series. To find the Taylor series for a new function, substitute into the closest known series rather than computing derivatives. You can also multiply a known series by x or a constant, or factor out a constant like e^2 to shift the centre.

Key Terms

Substitution (in the context of Taylor series)

Replacing the variable in a known Taylor series with a different expression. For instance, substituting (−x²) for x in the e^x series gives the series for e^(−x²). The interval of convergence transforms accordingly.

Maclaurin series

A Taylor series centred at a = 0. All five standard series in this set of notes are Maclaurin series.

Re-indexing

Changing the starting value of the summation index without changing the series itself. For example, after multiplying a series by x, the powers shift up by one, and you can re-index to keep the expression tidy.

Shifting the centre

Writing a function in a form that lets you use a known Maclaurin series even when the centre is not zero. For e^x centred at a = 2, you write e^x = e^2 · e^(x−2), then expand e^(x−2) as a Maclaurin series in the variable (x − 2).

The Five Essential Taylor Series

These are the building blocks. Memorise them cold.

1. Exponential function

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots \qquad x \in (-\infty, \infty)

Converges everywhere. This is the most useful series in all of mathematics.

2. Sine

\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\,x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \qquad x \in (-\infty, \infty)

Only odd powers appear. The signs alternate.

3. Cosine

\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \qquad x \in (-\infty, \infty)

Only even powers appear. The signs alternate. Notice that cos is the derivative of sin, and differentiating the sin series term by term gives exactly this.

4. Geometric series

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + x^4 + \cdots \qquad x \in (-1, 1)

Converges only for |x| < 1. This is the series you integrate to get ln(1 + x).

5. Natural logarithm

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\,x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \qquad x \in (-1, 1]

Note the asymmetric interval: it includes x = 1 but not x = −1.

The Substitution Method

The idea is simple: if you know a Taylor series for g(x), and you want the series for g(h(x)) where h(x) is some expression, just replace every x in the known series with h(x).

The general recipe:

• Identify which of the five standard series is closest to the function you need

• Determine what expression plays the role of x in that standard series

• Substitute that expression in place of x throughout the series

• Simplify powers and constants

• If the original function also has a factor outside the series (like a leading x or a constant multiplier), multiply the entire series by that factor

Why this works: A Taylor series is a power series identity. If e^x = Σ x^n/n! is valid for all x, then it remains valid when you replace x with any expression, provided the result falls within the interval of convergence.

How the interval of convergence transforms: If the original series converges for |x| < R, and you substitute u(x) for x, then the new series converges wherever |u(x)| < R. For example, substituting (−x²) into e^x: since e^x converges for all x, and (−x²) is always finite, the new series also converges for all x.

Worked Example: e^(−x²) Centred at a = 0

Why not derive from scratch? Computing derivatives of e^(−x²) gets ugly fast. f'(x) = −2x e^(−x²), f''(x) = (−2 + 4x²)e^(−x²), and the pattern only gets worse. The substitution method avoids all of this.

Step 1: Start from the known series

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

This converges for all x. In particular, it converges when x is replaced by any expression.

Step 2: Substitute (−x²) for x

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

Step 3: Simplify

(−x²)^n = (−1)^n x^(2n). So:

e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\,x^{2n} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots

The series converges for all x, since the original e^x series does.

Worked Example: x cos(4x) Centred at a = 0

This example combines two moves: substitution and multiplication by x.

Step 1: Start from the cosine series

\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

This converges for all x.

Step 2: Substitute (4x) for x

\cos(4x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,(4x)^{2n} = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \cdots

Since (4x)^(2n) = 4^(2n) x^(2n) = 16^n x^(2n), this simplifies to:

\cos(4x) = \sum_{n=0}^{\infty} \frac{(-1)^n\,16^n}{(2n)!}\,x^{2n} = 1 - \frac{16x^2}{2!} + \frac{256x^4}{4!} - \frac{4096x^6}{6!} + \cdots

Which simplifies further to:

\cos(4x) = 1 - 8x^2 + \frac{32x^4}{3} - \frac{256x^6}{45} + \cdots

Step 3: Multiply the entire series by x

x\cos(4x) = x \cdot \sum_{n=0}^{\infty} \frac{(-1)^n\,16^n}{(2n)!}\,x^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n\,16^n}{(2n)!}\,x^{2n+1}

Written out:

x\cos(4x) = x - 8x^3 + \frac{32x^5}{3} - \frac{256x^7}{45} + \cdots

The series converges for all x, since the original cosine series does.

Optional re-indexing: You can also write this as Σ (n=1 to ∞) (−1)^(n−1) x^n / (2n−2)! by setting m = n − 1, though the form above is perfectly acceptable on an exam.

Worked Example: e^x Centred at a = 2

This example shows how to handle a non-zero centre using a known Maclaurin series.

The goal: We want e^x = Σ c_n (x − 2)^n, but we need to find the coefficients c_n.

Step 1: Factor the function

Since x = (x − 2) + 2, we can write:

e^x = e^{(x-2)+2} = e^2 \cdot e^{(x-2)}

This splits e^x into a constant (e^2) times a function of (x − 2).

Step 2: Use the known Maclaurin series for e^u, with u = (x − 2)

e^{(x-2)} = \sum_{n=0}^{\infty} \frac{(x-2)^n}{n!}

Step 3: Multiply by the constant e^2

e^x = e^2 \sum_{n=0}^{\infty} \frac{(x-2)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x-2)^n

The series converges for all x, since the exponential series converges everywhere.

Verification via the direct approach: If you used the Taylor formula directly, you would compute f^(n)(2) for f(x) = e^x. Since every derivative of e^x is e^x, you get f^(n)(2) = e^2 for all n. Plugging into the Taylor formula: Σ e^2/n! × (x − 2)^n. Same answer, confirming the substitution method works.

Common Misconceptions

• Students sometimes substitute into the wrong standard series. cos(4x) starts from the cosine series, not the sine series. Check which function you are actually dealing with before substituting.

• Forgetting to raise the entire substituted expression to the nth power. When you put (4x) into cos, the term is (4x)^(2n), not 4·x^(2n). The 4 gets raised to the power as well.

• Dropping the constant factor when shifting the centre. For e^x centred at a = 2, the e^2 out the front is not optional. Every coefficient in the series carries that factor.

• Confusing the interval of convergence after substitution. If the original series converges for |x| < 1 and you substitute 4x, the new series converges for |4x| < 1, which means |x| < 1/4. The radius shrinks by a factor of 4.

Why It Matters / Exam Flags

⚠️ "Find the Taylor series for [function] centred at a = 0" is one of the most common Calc II exam questions. The expected approach is substitution, not computing derivatives.

⚠️ You will almost certainly need to reproduce at least two of the five standard series from memory. e^x and cos(x) are the most frequently tested.

⚠️ Multiplying a series by x (or x^2, etc.) after substitution is a common follow-up step. Do not forget it.

⚠️ The e^x at a = 2 example is a classic "trick" question. Recognise the factoring pattern e^x = e^a · e^(x−a) for any centre a.

Quick Self-Test

1. True or False: The Taylor series for sin(x) contains only even powers of x.

1. Fill in the blank: The geometric series 1/(1 − x) = Σ x^n converges for x in ______.

1. True or False: To find the Maclaurin series for e^(3x), substitute 3x into the series for e^x.

1. Fill in the blank: e^x centred at a = 5 has coefficients c_n = ______ / n!.

1. True or False: Multiplying a Taylor series by x changes every x^n term to x^(n+1).

(Answers: 1. False, only odd powers. 2. (−1, 1). 3. True. 4. e^5. 5. True.)

Practice Q&A

Q: Write the Maclaurin series for e^x, sin(x), and cos(x) from memory, including their intervals of convergence.

A: e^x = Σ (n=0 to ∞) x^n/n!, converges for all x. sin(x) = Σ (n=0 to ∞) (−1)^n x^(2n+1)/(2n+1)!, converges for all x. cos(x) = Σ (n=0 to ∞) (−1)^n x^(2n)/(2n)!, converges for all x.

Q: Find the Taylor series for e^(−x²) centred at 0. State the first four nonzero terms.

A: Substitute (−x²) for x in the e^x series: 1 − x² + x⁴/2! − x⁶/3! + … The general term is (−1)^n x^(2n)/n!.

Q: Find the first three nonzero terms of the Maclaurin series for x cos(4x).

A: Start with cos(4x) = 1 − (4x)²/2! + (4x)⁴/4! − … = 1 − 8x² + 32x⁴/3 − … Then multiply by x: x − 8x³ + 32x⁵/3 − …

Q: Find the Taylor series for e^x centred at a = 2.

A: Write e^x = e^2 · e^(x−2). Then e^(x−2) = Σ (x−2)^n/n!. So e^x = Σ e^2/n! × (x−2)^n, converging for all x.

Q: If the Maclaurin series for 1/(1−x) converges for |x| < 1, for what values of x does the series for 1/(1−3x) converge?

A: Substitute 3x for x: the series converges when |3x| < 1, so |x| < 1/3. The interval of convergence is (−1/3, 1/3).

Connections to Other Topics

The five standard series connect directly to the geometric series you learned earlier in Calc II: 1/(1−x) = Σ x^n is the starting point, and integrating it gives ln(1+x). Differentiating it gives 1/(1−x)², and so on.

The substitution technique reappears when you study Fourier series and Laplace transforms in later courses. The core idea (rewrite a complicated function in terms of a simpler one whose expansion you already know) is the same.

The companion notes cover deriving Taylor series from scratch and finding the interval of convergence via the ratio test.

Related Terms / Search Tags

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