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.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Maclaurin Series Calculator
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Formula: f(x) = sum of f^(n)(0) * x^n / n!
Worked example — 8-term approximation = 2.71825 | Exact = 2.71828 | Error = 3e-5
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! At x = 1: 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040 = 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 + 0.000198 = 2.71825 Exact e = 2.71828... Error = 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! sin(0.5) = 0.5 - 0.125/6 + 0.03125/120 - 0.0078125/5040 + 0.001953125/362880 = 0.5 - 0.020833 + 0.000260 - 0.00000155 + 0.0000000054 = 0.479426 Exact sin(0.5) = 0.479426... The 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.
How are Maclaurin series used in numerical computing?
Scientific calculators and math libraries use Maclaurin (and Taylor) series internally to compute transcendental functions. To calculate sin(x), the input is first reduced to the range [-pi/4, pi/4] using periodicity and symmetry, then the Maclaurin series converges rapidly with just a few terms. For e^x, argument reduction writes e^x = 2^k * e^r where |r| is small, computes e^r via series, then multiplies by the power of 2. The IEEE 754 floating-point standard relies on polynomial approximations derived from truncated series. Compensated summation techniques (like Kahan summation) minimize rounding errors when adding many terms. Hardware implementations often use minimax polynomials optimized from the series.
What happens when a Maclaurin series is evaluated outside its radius of convergence?
When |x| exceeds the radius of convergence R, the series diverges: partial sums grow without bound or oscillate wildly. For 1/(1-x) = 1 + x + x^2 + ..., at x = 2 the partial sums 1, 3, 7, 15, ... grow exponentially even though 1/(1-2) = -1 is perfectly well-defined. At the boundary |x| = R, behavior varies: ln(1+x) converges at x = 1 (giving ln(2)) but diverges at x = -1. The function may be analytic beyond R, but the Maclaurin series cannot represent it there. To approximate the function outside R, use a Taylor series centered at a different point, Pade approximants, or analytic continuation techniques. Understanding convergence boundaries is crucial for avoiding incorrect numerical results.
What is the connection between Maclaurin series and polynomial approximation?
A Maclaurin polynomial of degree n is the partial sum of the first n+1 terms of the Maclaurin series, providing the best polynomial approximation near x = 0 in the sense that it matches the function and its first n derivatives at x = 0. However, it is not necessarily the best approximation over an interval. Chebyshev polynomials minimize the maximum error over an interval and often provide better global approximations with fewer terms. Pade approximants use ratios of polynomials and can approximate functions with poles that pure polynomials cannot handle. Minimax approximation finds the polynomial that minimizes the worst-case error. Despite these alternatives, Maclaurin polynomials remain the most theoretically elegant and widely taught polynomial approximations.
How do you multiply and divide Maclaurin series?
To multiply two series, use the Cauchy product: if f(x) = sum a_n * x^n and g(x) = sum b_n * x^n, then f(x)*g(x) = sum c_n * x^n where c_n = sum(a_k * b_(n-k), k=0..n). For example, e^x * (1/(1-x)) can be computed by multiplying their series term by term. To divide series, assume f(x)/g(x) = sum d_n * x^n, multiply both sides by g(x), and solve for d_n recursively by matching coefficients. For tan(x) = sin(x)/cos(x), this gives x + x^3/3 + 2x^5/15 + ... Series multiplication is used to find products of generating functions in combinatorics and to compute convolutions in signal processing. Keeping track of enough terms ensures desired accuracy.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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