How to Find a Maclaurin Series

To learn how to find Maclaurin series manually on a homework or exam, follow these five steps. The series is a Taylor series centered at x = 0: f(x) = Σ_{n=0}^{∞} [f^{(n)}(0) / n!] x^n. This formula is your blueprint.

1. Write the general Maclaurin formula. You need f(x) = Σ_{n=0}^{∞} [f^{(n)}(0) / n!] x^n. This is a Taylor series centered at zero.
2. Compute derivatives. Find the first several derivatives of f(x). Start with f(x) itself (the 0th derivative) and compute until you have as many terms as you need. For a series up to x⁴, you need derivatives up to the 4th.
3. Evaluate at x = 0. Plug x = 0 into each derivative. Write down these numbers: f(0), f'(0), f''(0), f'''(0), and so on.
4. Plug into the formula. For each n, compute the term: [f^{(n)}(0) / n!] * x^n. Remember 0! = 1.
5. Write the series. List terms from n = 0 to the desired number. Simplify coefficients when possible (like 1/2! = 1/2). Then look for a pattern to write the general term in sigma notation.

The failure case: skipping the factorial denominator. For n = 3, the term must be divided by 3! = 6, not 3. Get this wrong and the entire series is invalid.

Compute the Maclaurin series of e^x up to the x⁴ term.

Derivatives: f'(x) = e^x, f''(x) = e^x, f'''(x) = e^x, f⁴(x) = e^x. All derivatives are e^x.

Evaluate at 0: e^0 = 1 for all derivatives.

Terms: n = 0: 1/0! * x⁰ = 1; n = 1: 1/1! * x¹ = x; n = 2: 1/2! * x² = x²/2; n = 3: 1/3! * x³ = x³/6; n = 4: 1/4! * x⁴ = x⁴/24.

Series: e^x ≈ 1 + x + x²/2 + x³/6 + x⁴/24.

General term in sigma notation: The coefficient for xⁿ is 1/n!. So the infinite series is Σ_{n=0}^{∞} x^n / n!. This is one of the most common Maclaurin series, used as a building block for substitution.

Compute the Maclaurin series of sin x up to the x⁷ term. Derivatives cycle every 4 steps: sin → cos →, sin →, cos → sin.

Derivatives: f(x) = sin(x), f'(x) = cos(x), f''(x) =, sin(x), f'''(x) =, cos(x), f⁴(x) = sin(x).

Evaluate at 0: sin(0) = 0, cos(0) = 1,, sin(0) = 0,, cos(0) =, 1, sin(0) = 0, and the pattern repeats.

Terms: n = 0: 0; n = 1: 1/1! * x¹ = x; n = 2: 0; n = 3: (, 1)/3! * x³ =, x³/6; n = 4: 0; n = 5: 1/5! * x⁵ = x⁵/120; n = 6: 0; n = 7:, 1/7! * x⁷ =, x⁷/5040.

Series: sin x ≈ x, x³/6 + x⁵/120, x⁷/5040.

General term: Only odd powers appear, with alternating signs. For odd n = 2k+1, the term is (, 1)ᵏ x²ᵏ⁺¹ / (2k+1)!.

Compute the Maclaurin series of ln(1+x) up to the x⁴ term. This function has a more complex derivative pattern than e^x or sin x.

Derivatives: f(x) = ln(1+x); f'(x) = 1/(1+x); f''(x) = -1/(1+x)²; f'''(x) = 2/(1+x)³; f⁴(x) = -6/(1+x)⁴.

Evaluate at 0: f(0) = ln(1) = 0; f'(0) = 1; f''(0) = -1; f'''(0) = 2; f⁴(0) = -6.

Terms: n = 0: 0; n = 1: 1/1! * x¹ = x; n = 2: (-1)/2! * x² = -x²/2; n = 3: 2/3! * x³ = x³/3; n = 4: (-6)/4! * x⁴ = -x⁴/4.

Series: ln(1+x) ≈ x - x²/2 + x³/3 - x⁴/4.

General term: The pattern is Σ_{n=1}^{∞} (-1)^{n-1} x^n / n.

After computing terms, find the pattern to write the general term in sigma notation. This step is essential for determining the radius of convergence using the ratio test.

For e^x: coefficient is 1/n!, no sign changes.

For sin x: only odd powers, alternating signs: (, 1)ⁿ x²ⁿ⁺¹ / (2n+1)!.

For ln(1+x): all powers, alternating signs: (, 1)^{n-1} xⁿ / n.

The failure case: off-by-one errors in sigma notation. For ln(1+x), the sum starts at n = 1, not n = 0. Check your index before using the ratio test.

Use this table to quickly find derivative values at x = 0 for the three worked examples.

n	e^x	sin x	ln(1+x)
0	1	0	0
1	1	1	1
2	1	0	-1
3	1	-1	2
4	1	0	-6

The pattern for sin x cycles: 0, 1, 0, -1, 0, 1, etc. For ln(1+x), the pattern is (, 1)^{n-1} (n-1)! for n ≥ 1.

• Forgetting factorials. The denominator n! is not optional. The term for n = 3 must be divided by 3! = 6, not 3. This is the most frequent error on exams.
• Sign errors. When derivatives involve negative signs (like sin and cos), track them carefully. Write each derivative evaluation explicitly.
• Incorrect derivative order. The nth term uses the nth derivative. Taking the wrong number of derivatives gives the wrong series.
• Confusing Maclaurin with Taylor series. A Maclaurin series is a Taylor series centered at 0. If you center elsewhere, you are computing a general Taylor series, not a Maclaurin series.
• Assuming the series equals the function outside the interval of convergence. The Maclaurin series for ln(1+x) converges only for (-1, 1]. At x = 2, the series diverges even though the function is defined.

The most practical thing to do next: when you finish computing a Maclaurin series, check your work by testing a simple x value (like x = 0.5) on both the polynomial and the original function. If they are close, your series is likely correct.