Building New Series from the Ones You Know

You’ll be able to

Recall the four required Maclaurin series and their intervals of convergence
Produce a new series by substitution, multiplication, differentiation or integration of a known one
Use a series to evaluate a limit or an integral that resists other methods

Four series, memorized, generate almost everything

The CED expects four Maclaurin series on recall: e^x = Σ xⁿ/n!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, and 1/(1 − x) = Σ xⁿ for |x| < 1. The first three converge for all x; the last only on (−1, 1). Nearly every series question that is not a convergence test is one of these four, transformed. Deriving them from scratch each time is a use of exam minutes you will want back.

Four legitimate transformations

Substitute an expression for x: e^(−x²) comes from e^x by replacing x with −x², giving Σ(−1)ⁿx^(2n)/n!. Multiply by a power or a constant: x·sin x shifts every exponent up by one. Differentiate term by term. Integrate term by term, remembering the constant. Each operation applies to the whole series, and the radius of convergence is preserved (endpoint behavior can change under integration) — for a substitution, the interval transforms with the substitution, so 1/(1 − 2x) converges for |2x| < 1, i.e. |x| < 1/2.

Series as a computational tool

Two exam uses justify the effort. A limit that resists L'Hôpital often falls immediately to series: lim(x→0) (sin x − x)/x³ becomes lim (−x³/6 + …)/x³ = −1/6 by inspection. And an integral with no elementary antiderivative, such as ∫e^(−x²) dx, can be integrated term by term to any desired accuracy. In both cases the series is not the answer — it is the route to a number.

The four required Maclaurin series

e^x = Σ xⁿ/n! (all x) · sin x = Σ (−1)ⁿx^(2n+1)/(2n+1)! (all x) · cos x = Σ (−1)ⁿx^(2n)/(2n)! (all x) · 1/(1−x) = Σ xⁿ (|x| < 1)

sin has odd powers and cos has even ones — checking the parity of the exponents catches a swapped pair instantly.

Worked example

Find the first three nonzero terms of the Maclaurin series for f(x) = x·cos(2x), and use a series to evaluate lim(x→0) [cos(2x) − 1]/x².

1.Start from cos u = 1 − u²/2! + u⁴/4! − … and substitute u = 2x: cos(2x) = 1 − 4x²/2 + 16x⁴/24 − … = 1 − 2x² + (2/3)x⁴ − …
2.Multiply through by x: x·cos(2x) = x − 2x³ + (2/3)x⁵ − …, which gives the first three nonzero terms.
3.For the limit, use the same expansion: cos(2x) − 1 = −2x² + (2/3)x⁴ − …
4.Divide by x²: [cos(2x) − 1]/x² = −2 + (2/3)x² − …
5.As x → 0 every remaining term vanishes, so the limit is −2.

Answer: x·cos(2x) = x − 2x³ + (2/3)x⁵ − …, and the limit is −2. The series turns a 0/0 limit into reading off a constant term.

Tip

When substituting into a known series, substitute into the general term as well as the written-out terms. A question asking for "the general term" after a substitution is common, and reconstructing it from three expanded terms is much harder than transforming Σ(−1)ⁿu^(2n)/(2n)! directly.

Checkpoint

What is the Maclaurin series for e^(−x²)?

Checkpoint

The series for 1/(1 − x) converges for |x| < 1. For what x does the series for 1/(1 − 3x) converge?

On the exam

Series free-response questions are usually four parts that all rest on one expansion: write the series, differentiate or integrate it, bound the error, then state the interval. Getting the first part right carries the rest, so spend the time there.

Answer the 2 checkpoints as you read.