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Maclaurin series

Build the Maclaurin polynomial Pn(x)P_n(x) for exe^x, sin⁡x\sin x, cos⁡x\cos x, ln⁡(1+x)\ln(1+x) or arctan⁡x\arctan x one degree at a time, and see how far from x=0x = 0 it stays close to the curve.

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What you are looking at

The Maclaurin polynomial of degree nn is the polynomial that agrees with ff and its first nn derivatives at x=0x = 0: Pn(x)=f(0)+xf′(0)+x22!f′′(0)+⋯+xnn!f(n)(0)P_n(x) = f(0) + x f'(0) + \dfrac{x^2}{2!}f''(0) + \cdots + \dfrac{x^n}{n!}f^{(n)}(0). In the visualiser the curve is blue and PnP_n is yellow. Raising nn matches one more derivative, and the yellow curve stays close to the blue one over a wider stretch around the origin.

How close the two are at a chosen point is shown by the vertical gap at x0x_0, drawn as a yellow segment and given in the panel as ∣f(x0)−Pn(x0)∣|f(x_0) - P_n(x_0)|. For exe^x, sin⁡x\sin x and cos⁡x\cos x the series converges for every real xx, although any fixed PnP_n still peels away far from the origin. For ln⁡(1+x)\ln(1+x) and arctan⁡x\arctan x the series converges only on a bounded interval, shaded green; outside it, adding terms does not close the gap.

Try this

  1. 1With f(x)=exf(x) = e^x and x0=1x_0 = 1, step nn up from 11: Pn(1)P_n(1) runs 2, 2.5, 2.667, 2.708,…2,\ 2.5,\ 2.667,\ 2.708, \ldots and closes in on e≈2.718e \approx 2.718.
  2. 2Choose sin⁡x\sin x and press Play. The polynomial only changes on odd values of nn, because the Maclaurin series of sin⁡x\sin x has no even powers.
  3. 3Choose ln⁡(1+x)\ln(1+x), drag x0x_0 to 22 and raise nn. A banner reports that the point is outside the interval of convergence, −1<x≤1-1 < x \le 1, and the error grows as nn rises.
  4. 4Choose arctan⁡x\arctan x and set x0=1x_0 = 1, the edge of the interval. Even at n=15n = 15 the error is still about 0.030.03: the series converges there, but slowly.

Mistakes students make

  • Leaving out the factorials. The coefficient of xnx^n is f(n)(0)n!\dfrac{f^{(n)}(0)}{n!}, not f(n)(0)f^{(n)}(0).
  • Assuming more terms always help. Outside the interval of convergence, as for ln⁡(1+x)\ln(1+x) with x>1x > 1, the partial sums do not approach f(x)f(x) at all.
  • Confusing the degree with the number of terms. For sin⁡x\sin x, P3(x)=x−x36P_3(x) = x - \dfrac{x^3}{6} has only two non-zero terms, and P4P_4 is the same polynomial.

Where it is taught

  • IB Maths AA HL 5.19 · Maclaurin series
  • AP Calculus BC Unit 10 · Taylor polynomial approximations, interval of convergence and Maclaurin series (10.11, 10.13, 10.14)

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