Table of Common Maclaurin Series
The Maclaurin series to memorise: e^x, sin x, cos x, ln(1+x), 1/(1−x), arctan x, (1+x)^k, sinh and cosh, each with sigma form and convergence interval.
Table of Common Maclaurin Series
For any function with derivatives of all orders at 0, the Maclaurin series is Σ [f^(n)(0)/n!] x^n. But you rarely compute that from scratch. The common Maclaurin series listed below, from Stewart §11.10 Table 1, are the building blocks for nearly every series problem in Calculus II and AP Calculus BC. Each entry gives the function, its series in expanded and sigma form, and the interval of convergence. Copy this table onto a sheet you can keep open while you work. Newcomers most often get wrong that the series equals the function only within its interval of convergence, not for all x, and that the factorial denominator grows faster than the numerator, ensuring convergence for many functions.
| Function | Series (First 4 Terms) | Sigma Form (n from 0 to ∞) | Interval of Convergence |
|---|---|---|---|
| 1/(1-x) | 1 + x + x² + x³ + … | Σ xⁿ | (-1, 1) |
| eˣ | 1 + x + x²/2! + x³/3! + … | Σ xⁿ/n! | (-∞, ∞) |
| sin x | x – x³/3! + x⁵/5! – x⁷/7! + … | Σ (-1)ⁿ x²ⁿ⁺¹/(2n+1)! | (-∞, ∞) |
| cos x | 1 – x²/2! + x⁴/4! – x⁶/6! + … | Σ (-1)ⁿ x²ⁿ/(2n)! | (-∞, ∞) |
| ln(1+x) | x – x²/2 + x³/3 – x⁴/4 + … | Σ (-1)ⁿ⁻¹ xⁿ/n (n from 1 to ∞) | (-1, 1] |
| arctan x | x – x³/3 + x⁵/5 – x⁷/7 + … | Σ (-1)ⁿ x²ⁿ⁺¹/(2n+1) | [-1, 1] |
| (1+x)ᵏ | 1 + kx + k(k-1)x²/2! + k(k-1)(k-2)x³/3! + … | Σ (k choose n) xⁿ | (-1, 1) |
| sinh x | x + x³/3! + x⁵/5! + x⁷/7! + … | Σ x²ⁿ⁺¹/(2n+1)! | (-∞, ∞) |
| cosh x | 1 + x²/2! + x⁴/4! + x⁶/6! + … | Σ x²ⁿ/(2n)! | (-∞, ∞) |
Maclaurin Series Table: Patterns That Help You Memorise
Memorise three anchors, then derive the rest. First, 1/(1-x) = Σ xⁿ, the geometric series. Its interval (-1, 1) is open because the series diverges at both endpoints. Second, eˣ = Σ xⁿ/n! converges for all x because n! grows faster than any power of x. Third, cos x and sin x are the even- and odd-power halves of eⁱˣ: cos x uses only even powers with alternating signs; sin x uses odd powers with alternating signs. Both converge for all x.
What changes with the alternating sign pattern? For ln(1+x) and arctan x, the sign alternates and the denominator is a simple integer, not a factorial. That limits the radius of convergence to 1. The interval for ln(1+x) is half-open: the series converges conditionally at x=1 (by the alternating series test) but diverges at x=-1. For arctan x, convergence is absolute at both endpoints, giving the closed interval [-1, 1].
Which series have no factorial in the denominator? Only the geometric series and the binomial series (1+x)ᵏ. The binomial coefficient (k choose n) = k(k-1)...(k-n+1)/n! disguises a factorial in its denominator, but the pattern of terms is a product of linear factors, not a factorial alone. For non-integer k, the binomial series is infinite; for integer k, it terminates as the ordinary binomial theorem.
List of Maclaurin Series: Deriving New Series From the Table
Once you know the list of Maclaurin series in the table, you generate new series without recomputing derivatives. Three operations work: substitution, term-by-term differentiation, and term-by-term integration. All are valid inside the interval of convergence of the original series.
Substitution
To get the series for e^(x²), replace every x in the series for eˣ with x². You get Σ (x²)ⁿ/n! = Σ x²ⁿ/n!, converging for all x because the original radius is ∞. For sin(2x), substitute 2x into the sin series: Σ (-1)ⁿ (2x)²ⁿ⁺¹/(2n+1)! = Σ (-1)ⁿ 2²ⁿ⁺¹ x²ⁿ⁺¹/(2n+1)!. Radius stays ∞.
Failure case: substituting into 1/(1-x) when |x|≥1 produces a divergent series. Always check that the substituted expression lies inside the original interval of convergence. For example, 1/(1 - x²) = Σ (x²)ⁿ converges only for |x|<1, not for |x|≥1.
Integration
To derive the series for arctan x from 1/(1+x²), integrate the geometric series for 1/(1+u) term by term. Set u = x²: 1/(1+x²) = 1 - x² + x⁴ - x⁶ + … for |x|<1. Integrate term by term from 0 to x: arctan x = x - x³/3 + x⁵/5 - x⁷/7 + …, interval [-1,1] after checking endpoints. This is faster than computing derivatives of arctan x directly.
Differentiation works the same way. Differentiate the series for 1/(1-x) to get the series for 1/(1-x)². You get Σ (n+1)xⁿ, still converging for |x|<1.
Taylor Series Table: Writing a Printable Version
The Taylor series table here is designed to fit on one printed sheet. Copy it exactly for closed-book exams: list the function, the first three or four terms in expanded form, the sigma notation with n starting index, and the interval. Double-check each entry against Stewart §11.10 Table 1. The most common student error is writing the sign pattern wrong for sin x (starts positive for n=0: term is x) or for ln(1+x) (sigma form uses n from 1 to ∞, not 0).
For self-learners: print the table and keep it next to your problem set. Every time you use a series, verify the interval. If the problem asks you to approximate a function at x=0.9, the series for ln(1+x) works (0.9 is inside (-1,1]), but the series for 1/(1-x) also works (0.9 is in (-1,1)). If x=1.2, neither series converges, and you must use a different representation or a Taylor series centered at a nonzero point.
The single thing that most often goes wrong: assuming all series converge for |x|<1. Check the factorial denominators. eˣ, sin x, cos x, sinh x, cosh x converge for all x. The rest converge only for |x|<1, with endpoint behaviour that you must test separately using the alternating series test or p-series comparison.
Printable Version and Next Steps for the Maclaurin Series Table
The most practical thing you can do: print the table above on one page, then derive at least two new series from it by substitution and by integration before your next exam. Check that your derived series converge on the expected interval. When you get stuck on a problem, ask yourself which series from the table the function resembles, then apply substitution or term-by-term calculus. If the ratio test gives L=1, test the endpoints immediately; that step is where most points are lost.
Common Questions
How do I find the interval of convergence when the ratio test gives L=1?
The ratio test gives the radius of convergence, not the interval. When L=1, the test is inconclusive. You must test the endpoints x = ±R separately using the p-series test, alternating series test, or direct comparison. For example, the series for ln(1+x) has radius 1. At x=1, the alternating harmonic series converges conditionally. At x=-1, the harmonic series diverges. So the interval is (-1, 1].
Why does the Lagrange remainder sometimes give a bound 100 times bigger than the actual error?
The Lagrange remainder |R_n(x)| ≤ M |x|^(n+1)/(n+1)! uses M as the maximum of |f^(n+1)(z)| on the entire interval from 0 to x. In practice, the (n+1)th derivative may be much smaller at most points, and the maximum M is a worst-case bound. For sin x on a small interval, M is 1, so the bound is not too loose. For functions like eˣ on an interval containing x=2, M = e² ≈ 7.4, and the true error can be several times smaller.
Can I use the Maclaurin series for ln(1+x) at x=1?
Yes, the series converges conditionally at x=1 to ln 2. The alternating series test applies because terms decrease in magnitude and approach zero. The error after n terms is bounded by the first omitted term. At x=1, the series is 1 - 1/2 + 1/3 - 1/4 + … and converges to ln 2 ≈ 0.693. At x=-1, the series diverges to negative infinity.
What is the error if I truncate after 3 terms of sin x for x=0.5?
The Maclaurin polynomial of degree 3 for sin x is x - x³/6. At x=0.5, that gives 0.5 - 0.5³/6 = 0.5 - 0.0208333 = 0.4791667. The true sin(0.5) = 0.4794255. The error is about 0.000259, which is less than the next term x⁵/120 = 0.0002604. The alternating series error bound guarantees |error| ≤ 0.00026.
How do I derive the series for arctan x from the series for 1/(1+x²)?
Start with 1/(1+u) = Σ (-1)ⁿ uⁿ for |u|<1.Check endpoints: at x=±1, the alternating series converges by the alternating series test, giving interval [-1,1].
What is the pattern for the Maclaurin series of (1+x)^p when p is not an integer?
The binomial series (1+x)^p = Σ (p choose n) xⁿ, where (p choose n) = p(p-1)...(p-n+1)/n!. For non-integer p, the series is infinite and the coefficients never become zero. The terms do not follow a simple factorial pattern in the numerator, but the ratio test still gives radius 1. For p = 1/2, the series is 1 + (1/2)x - (1/8)x² + (1/16)x³ - (5/128)x⁴ + …
How many terms do I need for 4-digit accuracy in e^x for x=2?
Use Taylor's inequality: |R_n(2)| ≤ e² |2|^(n+1)/(n+1)! ≈ 7.39 × 2^(n+1)/(n+1)!. For 4-digit accuracy, you need error < 0.00005. Test n=6: 7.39 × 128/5040 ≈ 0.188; n=8: 7.39 × 512/362880 ≈ 0.0104; n=10: 7.39 × 2048/39916800 ≈ 0.00038; n=11: 7.39 × 4096/479001600 ≈ 0.000063; n=12: 7.39 × 8192/6227020800 ≈ 0.0000097, so 12 terms suffice.