Maclaurin Series Calculator
Free Maclaurin series Calculator for sequences. Enter values to get step-by-step solutions with formulas and graphs. Get results you can export or share.
Formula
f(x) = sum of f^(n)(0) * x^n / n!
The Maclaurin series expands f(x) as an infinite sum of terms involving the nth derivative of f evaluated at 0, multiplied by x^n/n!. It is a Taylor series centered at a = 0. The series converges to f(x) within the radius of convergence R.
Worked Examples
Example 1: Approximating e^1 with 8 Terms
Problem: Calculate e using the first 8 terms of the Maclaurin series for e^x at x = 1.
Solution: e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! + x^6/6! + x^7/7!\nAt x = 1: 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040\n= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 + 0.000198\n= 2.71825\nExact e = 2.71828...\nError = 0.00003 (5 digits of accuracy with 8 terms)
Result: 8-term approximation = 2.71825 | Exact = 2.71828 | Error = 3e-5
Example 2: Computing sin(0.5) via Maclaurin Series
Problem: Approximate sin(0.5) using the Maclaurin series with 5 non-zero terms.
Solution: sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9!\nsin(0.5) = 0.5 - 0.125/6 + 0.03125/120 - 0.0078125/5040 + 0.001953125/362880\n= 0.5 - 0.020833 + 0.000260 - 0.00000155 + 0.0000000054\n= 0.479426\nExact sin(0.5) = 0.479426...\nThe series converges extremely fast for |x| < 1
Result: sin(0.5) = 0.479426 | Full precision reached with just 4 terms
Frequently Asked Questions
What is a Maclaurin series and how does it differ from a Taylor series?
A Maclaurin series is a special case of the Taylor series centered at x = 0. The general Taylor series of f(x) centered at x = a is the sum of f^(n)(a) * (x-a)^n / n! for n = 0, 1, 2, and so on. When a = 0, this becomes the Maclaurin series: sum of f^(n)(0) * x^n / n!. Named after Scottish mathematician Colin Maclaurin, this series represents a function as an infinite polynomial around the origin. The Maclaurin series is particularly useful because many common functions have clean, memorable series when expanded at zero. Both Taylor and Maclaurin series converge to the function within the radius of convergence, providing polynomial approximations of arbitrary precision.
What are the most important Maclaurin series to memorize?
The essential Maclaurin series include: e^x = 1 + x + x^2/2! + x^3/3! + ... (converges for all x). sin(x) = x - x^3/3! + x^5/5! - ... (all x). cos(x) = 1 - x^2/2! + x^4/4! - ... (all x). ln(1+x) = x - x^2/2 + x^3/3 - ... (for -1 < x <= 1). 1/(1-x) = 1 + x + x^2 + x^3 + ... (for |x| < 1). arctan(x) = x - x^3/3 + x^5/5 - ... (for |x| <= 1). (1+x)^k = 1 + kx + k(k-1)x^2/2! + ... (the binomial series, for |x| < 1). These series are building blocks for deriving more complex expansions through substitution, differentiation, and multiplication.
How do you determine the radius of convergence of a Maclaurin series?
The radius of convergence R determines the interval (-R, R) where the series converges. The ratio test is the most common method: R = lim |a_n / a_(n+1)| as n approaches infinity, where a_n is the coefficient of x^n. Alternatively, the root test gives 1/R = lim |a_n|^(1/n). For e^x, R = lim (n+1)!/n! = lim (n+1) = infinity, so it converges everywhere. For ln(1+x), R = lim n/(n+1) = 1, so it converges for |x| < 1. The series may or may not converge at the endpoints x = R and x = -R, which must be checked separately. Functions with singularities in the complex plane have R equal to the distance from the center to the nearest singularity.
How do you derive a new Maclaurin series from known ones?
Several techniques generate new series from known ones. Substitution: replacing x with g(x) in a known series gives the series for f(g(x)). For example, e^(-x^2) = 1 - x^2 + x^4/2! - x^6/3! + ... by substituting -x^2 into the e^x series. Differentiation: differentiating term by term gives the series for f'(x). Since 1/(1-x) = 1 + x + x^2 + ..., differentiating gives 1/(1-x)^2 = 1 + 2x + 3x^2 + ... Integration: integrating 1/(1+x) = 1 - x + x^2 - ... gives ln(1+x) = x - x^2/2 + x^3/3 - ... Multiplication: multiplying two series gives the series for the product function. These techniques avoid computing derivatives directly.
How accurate is a partial sum of a Maclaurin series?
The accuracy depends on three factors: the number of terms used, the value of x, and the function being approximated. Near x = 0, even a few terms provide excellent accuracy. For e^1, 10 terms give 7 digits of accuracy. For sin(0.1), just 3 terms give 12 digits of accuracy. However, accuracy degrades as |x| increases. For e^10, you need about 40 terms for 10-digit accuracy. The error of an n-term partial sum is bounded by the Lagrange error bound: |error| <= M * |x|^(n+1) / (n+1)! where M bounds the (n+1)th derivative. For alternating series, the error is bounded by the absolute value of the first omitted term, which is often a tighter bound.
What is the relationship between Maclaurin series and Euler's formula?
Euler's formula e^(ix) = cos(x) + i*sin(x) connects the exponential, sine, and cosine Maclaurin series in a profound way. Substituting ix into the e^x series: e^(ix) = 1 + ix - x^2/2! - ix^3/3! + x^4/4! + ix^5/5! - ... Separating real and imaginary parts: the real terms give 1 - x^2/2! + x^4/4! - ... = cos(x), and the imaginary terms give x - x^3/3! + x^5/5! - ... = sin(x). The special case x = pi gives Euler's identity e^(i*pi) + 1 = 0, often called the most beautiful equation in mathematics. This relationship unifies exponential and trigonometric functions through complex numbers and their Maclaurin series.