Maclaurin Series Calculator

Enter a function f(x) and choose the number of terms to get its Maclaurin series expansion — a power series centered at x = 0. The calculator returns the polynomial approximation, lists each term coefficient, and shows a chart comparing the original function to its approximation over a range of x values.

Select the function to expand as a Maclaurin series around x = 0.

How many non-zero terms to include in the expansion.

Point at which to compare the series approximation to the exact value.

Results

Series Approximation at x

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Exact Value at x

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Absolute Error

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Terms Used

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f(x) vs Maclaurin Approximation

Results Table

Frequently Asked Questions

What is a Maclaurin series?

A Maclaurin series is a special case of the Taylor series where the expansion point is x = 0. It expresses a function as an infinite sum of terms based on the function's derivatives at zero: f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + … This allows complex functions to be approximated by simple polynomials.

How is the Maclaurin series different from the Taylor series?

The Taylor series expands a function around any point a, while the Maclaurin series is specifically centered at a = 0. Every Maclaurin series is a Taylor series, but not every Taylor series is a Maclaurin series. When a = 0 is substituted into the Taylor formula, you get the Maclaurin form.

How do I use the Maclaurin Series Calculator?

Select your function from the dropdown (e.g. sin(x), e^x, cos(x)), choose the number of terms n, and enter an x value to evaluate the approximation at. The calculator instantly returns the polynomial expansion, the approximation value, the exact value, and the error between them.

What does the 'number of terms' control?

The number of terms n determines how many terms of the infinite series are summed. More terms generally give a more accurate approximation, especially for x values farther from 0. For example, sin(x) with 5 terms gives a very accurate result near x = 0 but may diverge for large x.

Which functions can be expanded using a Maclaurin series?

Any function that is infinitely differentiable at x = 0 can be expanded. Common examples include sin(x), cos(x), e^x, ln(1+x), 1/(1−x), arctan(x), sinh(x), and cosh(x). Functions with singularities at x = 0, like ln(x) or 1/x, cannot be expanded there.

Why does the approximation get worse for large values of x?

The Maclaurin series converges best near x = 0 — the point of expansion. As x moves further from zero, more terms are needed for an accurate approximation, and some series (like ln(1+x)) only converge within a limited radius. The region where the series converges is called its radius of convergence.

What is the radius of convergence?

The radius of convergence R is the range of x values for which the Maclaurin series converges to the actual function value. For e^x, sin(x), and cos(x) it is infinite (all real x). For ln(1+x) and 1/(1−x) it is R = 1, meaning the series only reliably works for |x| < 1.

How is the absolute error calculated?

The absolute error is simply the difference |f(x) − P_n(x)|, where f(x) is the exact function value and P_n(x) is the Maclaurin polynomial sum using n terms. A smaller error means the polynomial is a better approximation at that x value with that many terms.

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