0.4794255386
0.4794255386
0
2.547402e-9
%
0.4794255386
0.4794255386
0.4794255386
0
2.547402e-9
%
0.4794255386
The Maclaurin Series Calculator computes polynomial approximations of functions expanded about the origin ($$a = 0$$). A Maclaurin series is the special case of a Taylor series centered at zero:
$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$
Named after the Scottish mathematician Colin Maclaurin (1698–1746), though the concept was known earlier to James Gregory and Brook Taylor, this series provides the most natural and widely used polynomial representations of elementary functions.
The following expansions are among the most important in all of mathematics. Each is derived by computing successive derivatives at $$x = 0$$:
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}$$
$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}$$
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \quad \text{for } -1 < x \le 1$$
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \quad \text{for } |x| < 1$$
$$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \quad \text{for } |x| \le 1$$
The exponential, sine, and cosine Maclaurin series converge for all real numbers—their radius of convergence is infinite. This means that with enough terms, the polynomial can approximate these functions to any desired accuracy anywhere on the real line. The logarithmic and geometric series, however, converge only within a finite radius, so the approximation is meaningful only within that interval.
Maclaurin series are fundamental in physics for deriving small-angle approximations ($$\sin\theta \approx \theta$$), in electrical engineering for analyzing circuits with nonlinear components, in computer science for implementing math library functions, and in statistics for moment-generating functions. Euler's identity $$e^{i\pi} + 1 = 0$$ itself follows from combining the Maclaurin series for $$e^x$$, $$\sin(x)$$, and $$\cos(x)$$.
Select a function and enter the value of $$x$$ where you want to evaluate it. Choose the number of terms (1–6) for the polynomial. The calculator computes the partial sum and compares it to the exact function value, showing both the absolute and relative errors.
The approximation is the value of the truncated Maclaurin polynomial at your chosen $$x$$. A smaller absolute error means the polynomial is closely tracking the true function. Relative error expresses this as a percentage of the exact value. For best accuracy, use values of $$x$$ near zero and include more terms.
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Using 1 − x²/2 + x⁴/24 − x⁶/720 at x = 0.5 gives an error of only about 2.4×10⁻⁷, demonstrating rapid convergence near 0.
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The ln(1+x) series converges slower than trig series. At x = 0.5 with 5 terms the error is about 0.45%, still a reasonable approximation.
A Maclaurin series is simply a Taylor series with the center point $$a = 0$$. Every Maclaurin series is a Taylor series, but not every Taylor series is a Maclaurin series. The Maclaurin form is preferred when expanding about the origin simplifies the coefficients.
The Maclaurin series for $$\ln(1+x)$$ has radius of convergence $$R = 1$$, so it converges only for $$-1 < x \le 1$$. Outside this interval the partial sums diverge. At $$x = 1$$ the series converges conditionally to $$\ln(2)$$, and at $$x = -1$$ it diverges.
The small-angle approximation $$\sin(\theta) \approx \theta$$ is simply the 1-term Maclaurin series for sine. Similarly, $$\cos(\theta) \approx 1 - \theta^2/2$$ is the 2-term Maclaurin series. These approximations are widely used in physics and engineering when angles are small.
Yes, any function that is infinitely differentiable at $$x = 0$$ has a Maclaurin series. However, this calculator provides pre-built expansions for the six most common functions. For other functions, you would compute derivatives $$f^{(k)}(0)$$ manually and build the polynomial.
Because $$\arctan(x)$$ is an odd function: $$\arctan(-x) = -\arctan(x)$$. Odd functions have Maclaurin series with only odd-powered terms, just as even functions (like $$\cos(x)$$) have only even-powered terms.
Most implementations use argument reduction to map $$x$$ into a small interval near 0, then evaluate a truncated Maclaurin-like polynomial (often a minimax polynomial) with 7–10 terms. This gives full double-precision accuracy in just a few multiplications and additions.
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