Finding Maclaurin Series by Substitution

Building a maclaurin series by substitution from a known series is faster than computing derivatives from scratch. For example, to get the maclaurin series of e^-x^2, replace every x in the e^x series (Σ xⁿ/n!) with −x². The result is Σ (−1)ⁿ x²ⁿ/n!, converging for all real x because the e^x series converges everywhere. Four shortcut techniques work: substitution, multiplication by powers of x, term-by-term differentiation, and term-by-term integration. Each method relies on the nine series in Stewart §11.10, Table 1, and works only inside the interval of convergence of the original series.

Substitution works when the target function is a known series with its variable replaced by an algebraic expression. The process: identify a known series from Stewart Table 1, replace x with the expression, and simplify.

Example 1: Maclaurin Series for e^{−x²}

Start with e^x = Σ_{n=0}^{∞} xⁿ/n!. Replace x by −x²: e^{−x²} = Σ_{n=0}^{∞} (−x²)ⁿ/n! = Σ_{n=0}^{∞} (−1)ⁿ x²ⁿ/n!. The interval of convergence remains (−∞, ∞) because the e^x series converges for all real x. No derivative computation needed.

Example 2: Maclaurin Series for 1/(1+4x³)

Use the geometric series 1/(1−u) = Σ_{n=0}^{∞} uⁿ for |u| < 1. Write 1/(1+4x³) = 1/(1−(−4x³)). Then substitute u = −4x³: 1/(1+4x³) = Σ_{n=0}^{∞} (−4x³)ⁿ = Σ_{n=0}^{∞} (−1)ⁿ 4ⁿ x³ⁿ, valid when |−4x³| < 1, i.e., |x| < 4^{−1/3}.

Substitution fails if the replacement makes the expression outside the original radius. For 1/(1−x) series at x = 2, substituting x=2 yields divergence.

If a function is xᵏ times a known series, multiply each term of the known series by xᵏ. This shifts indices but does not change the radius of convergence.

Example 3: Maclaurin Series for x² sin x

sin x = Σ_{n=0}^{∞} (−1)ⁿ x²ⁿ⁺¹/(2n+1)!. Multiply term‑by‑term by x²: x² sin x = Σ_{n=0}^{∞} (−1)ⁿ x²ⁿ⁺³/(2n+1)!. The radius remains ∞ because sin x converges for all real x. The series starts at n=0 with term x³/1! = x³.

This technique is useful for power series manipulation when preparing a series for integration or substitution.

Differentiating a known series term‑by‑term gives the series for the derivative of the function, valid inside the same interval of convergence.

Example 4: Maclaurin Series for 1/(1−x)²

Start with 1/(1−x) = Σ_{n=0}^{∞} xⁿ for |x| < 1. Differentiate both sides: d/dx [1/(1−x)] = 1/(1−x)² = Σ_{n=1}^{∞} n x^{n−1}. Re‑index to start at n=0: let k = n−1 → 1/(1−x)² = Σ_{k=0}^{∞} (k+1) xᵏ. Radius remains 1. The series for 1/(1−x)² is not in Stewart Table 1, but term‑by‑term differentiation produces it easily.

This method applies to any function whose Maclaurin series is known. The ratio test on the differentiated series gives the same radius, but endpoint convergence may change.

Integration term‑by‑term works when the derivative of the target function matches a known series.

Example 5: Maclaurin Series for arctan x

Note that d/dx(arctan x) = 1/(1+x²). Use the geometric series for 1/(1+u) = Σ_{n=0}^{∞} (−1)ⁿ uⁿ for |u| < 1, with u = x²: 1/(1+x²) = Σ_{n=0}^{∞} (−1)ⁿ x²ⁿ. Integrate term‑by‑term from 0 to x: arctan x = ∫₀ˣ 1/(1+t²) dt = Σ_{n=0}^{∞} (−1)ⁿ ∫₀ˣ t²ⁿ dt = Σ_{n=0}^{∞} (−1)ⁿ x²ⁿ⁺¹/(2n+1). The series converges for |x| ≤ 1 (Stewart Table 1). At x = 1, the alternating series gives arctan 1 = π/4 = 1 − 1/3 + 1/5 − 1/7 + … .

This integration technique is a standard power series manipulation for functions whose derivatives are simpler.

Maclaurin polynomials replace functions near zero, turning limits and definite integrals into polynomial problems.

Evaluating Limits

Replace each function with its first few Maclaurin terms. For lim_{x→0} (sin x − x)/x³, use sin x = x − x³/6 + O(x⁵).The limit equals −1/6 exactly.

Evaluating Integrals

For ∫₀¹ e^{−x²} dx, substitute the series: ∫₀¹ Σ (−1)ⁿ x²ⁿ/n! dx = Σ (−1)ⁿ/(n!(2n+1)). Truncate after n=3: 1 − 1/3 + 1/(2·5) − 1/(6·7) = 1 − 0.3333 + 0.1 − 0.0238 ≈ 0.7429. The alternating series error bound guarantees the error is less than the magnitude of the next term (n=4: 1/(24·9) ≈ 0.0046).

The failure case: using too few terms near the edge of the interval of convergence. For x close to the radius, include more terms or check endpoint convergence first.