2.6667
1
1
0.5
0.1667
0.1667
2.6667
1
1
0.5
0.1667
0.1667
The Power Series Calculator evaluates a truncated power series at any given point. Enter up to four coefficients, a center point $$c$$, and an evaluation point $$x$$, and the calculator computes $$S(x) = a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3$$ along with each individual term.
A power series is an infinite series of the form $$\sum_{n=0}^{\infty} a_n (x - c)^n$$, where $$c$$ is the center and the $$a_n$$ are coefficients. Power series are the backbone of mathematical analysis, providing a way to represent functions as infinite polynomials. Every analytic function — including exponentials, trigonometric functions, logarithms, and Bessel functions — has a power series representation within its radius of convergence.
The most familiar power series are Taylor series and Maclaurin series (Taylor series centered at $$c = 0$$). For example, the exponential function has the Maclaurin series $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$, with coefficients $$a_n = 1/n!$$. The default values in this calculator ($$a_0 = 1, a_1 = 1, a_2 = 0.5, a_3 \approx 0.1667$$) approximate the first four terms of $$e^x$$.
Power series are essential in numerical methods. When a function cannot be evaluated in closed form, its Taylor polynomial (a truncated power series) provides an approximation whose accuracy improves with more terms. In physics, perturbation theory expresses solutions as power series in a small parameter. In engineering, transfer functions and signal analysis rely on power series and their close cousin, the Laurent series. In statistics, the moment-generating function is a power series whose coefficients encode all moments of a distribution.
This calculator displays each term individually, allowing you to see how the contributions decrease (or grow) as the power increases. The magnitude of the last term provides a rough estimate of the approximation error when the series is convergent. For best results, use this calculator with $$|x - c|$$ small relative to the radius of convergence of your series.
Experiment with the default coefficients to approximate $$e^1 \approx 2.7183$$ using just four terms, or enter your own coefficients for any power series you need to evaluate.
Step 1: Compute the displacement. The key variable is $$\Delta x = x - c$$, the signed distance from the evaluation point to the series center.
Step 2: Evaluate each term.
$$T_0 = a_0$$
$$T_1 = a_1 (x - c)$$
$$T_2 = a_2 (x - c)^2$$
$$T_3 = a_3 (x - c)^3$$
Step 3: Sum all terms.
$$S(x) = T_0 + T_1 + T_2 + T_3 = a_0 + a_1(x-c) + a_2(x-c)^2 + a_3(x-c)^3$$
This is a third-degree polynomial approximation to the function represented by the full power series.
Step 4: Last term magnitude.
$$|T_3| = |a_3| \cdot |x - c|^3$$
In a convergent series where terms decrease monotonically, the magnitude of the last included term provides an upper bound on the truncation error (by the alternating series estimation theorem for alternating series, or as a heuristic for general series).
S(x) is the value of the truncated power series at the chosen point. Compare this to the exact function value (if known) to assess the approximation quality. For the exponential series defaults at $$x = 1$$: $$S(1) = 1 + 1 + 0.5 + 0.1667 = 2.6667$$, compared to $$e^1 = 2.7183$$.
Individual terms show the contribution of each power. In a well-behaved convergent series, the terms should decrease in magnitude. If the terms are growing, the series may be diverging at that point, or you may be outside the radius of convergence.
Last Term Magnitude gives a rough sense of the truncation error. If $$|T_3|$$ is very small (say, $$< 10^{-6}$$), the four-term approximation is likely very accurate. If $$|T_3|$$ is comparable to the sum, more terms would be needed.
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The Maclaurin series for eˣ uses aₙ = 1/n!. With four terms: S(1) = 1 + 1 + 0.5 + 0.1667 = 2.6667. The exact value e¹ ≈ 2.7183, so the error is about 0.0516 (1.9%). Adding the next term (1/4! = 0.0417) would bring S closer to 2.7083.
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cos(x) = 1 − x²/2 + x⁴/24 − … Setting a₀=1, a₁=0, a₂=−0.5, a₃=0 gives the quadratic approximation. At x=0.5: S = 1 + 0 + (−0.5)(0.25) + 0 = 0.875. Exact cos(0.5) ≈ 0.8776, error ≈ 0.003.
A power series is an infinite sum of the form $$\sum_{n=0}^{\infty} a_n (x - c)^n$$, where $$a_n$$ are the coefficients, $$c$$ is the center, and $$x$$ is the variable. Each term is a coefficient times a power of $$(x - c)$$. Power series generalize polynomials to allow infinitely many terms.
This calculator provides a quick evaluation tool for the first four terms of any power series. For many applications, especially near the center $$c$$, four terms give a good approximation. The individual term display helps you judge whether more terms would significantly change the result.
Every power series has a radius of convergence $$R$$ such that the series converges for $$|x - c| < R$$ and diverges for $$|x - c| > R$$. The behavior at $$|x - c| = R$$ must be checked separately. This calculator does not compute $$R$$, but if the terms grow rather than shrink, you are likely outside the radius of convergence.
For a Taylor series of $$f(x)$$ centered at $$c$$, the coefficients are $$a_n = f^{(n)}(c) / n!$$. For example, for $$\sin(x)$$ at $$c = 0$$: $$a_0 = 0$$, $$a_1 = 1$$, $$a_2 = 0$$, $$a_3 = -1/6 \approx -0.1667$$.
The center $$c$$ shifts the expansion point. A Taylor series centered at $$c$$ approximates the function best near $$x = c$$. At $$x = c$$, the series value is simply $$a_0$$. The farther $$x$$ is from $$c$$, the more terms are needed for accuracy.
Yes. A Maclaurin series is simply a Taylor series centered at $$c = 0$$. Set the center to 0, enter the Maclaurin coefficients, and evaluate at any $$x$$. The default values approximate the Maclaurin series for $$e^x$$.
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