Where the coefficients come from, why the fit spreads outwards from zero, and how to build new series from old ones.
A polynomial is nothing like sin x. It has no periodicity and it runs off to infinity. And yet: add terms one at a time and watch it grip the curve, further out each time.
The match starts at the origin and spreads outwards. That is what "a series about zero" means.
f(x) = f(0) + f′(0)x + f″(0)2!x² + f‴(0)3!x³ + …
Each coefficient is a derivative at zero, divided by a factorial. Nothing more. The polynomial is built to agree with f at the origin in value, then in gradient, then in curvature, and so on down the derivatives.
| f(x) | series |
|---|---|
| ex | 1 + x + x²⁄2! + x³⁄3! + … |
| sin x | x − x³⁄3! + x⁵⁄5! − … |
| cos x | 1 − x²⁄2! + x⁴⁄4! − … |
| ln(1 + x) | x − x²⁄2 + x³⁄3 − … |
| arctan x | x − x³⁄3 + x⁵⁄5 − … |
| (1 + x)p | 1 + px + p(p−1)2!x² + … |
Notice the pattern. sin x is odd, so only odd powers survive. cos x is even, so only even powers do. ln(1 + x) and arctan x have plain denominators, not factorials, which is an easy thing to mix up under pressure.
Deriving a series from scratch is slow. Nearly every question wants you to start from one of the six above and transform it, and saying which one you started from is part of the answer.
These are series about zero. They are accurate near the origin and get worse further out, which the animation shows plainly. Using a three-term sin series at x = 5 is not a small error; it is a wrong answer.
1. What is the coefficient of x³ in the Maclaurin series for sin x?
2. Using 1 + x + x²⁄2, estimate e1.
3. The quickest way to get the series for ex² is to:
Stating which standard series you are starting from, and what you are doing to it. "Replacing x by x² in the series for ex" is a method mark.
Keeping terms up to the required power and no further, and saying so. A question asking for terms up to x⁴ does not want x⁵.
Factorials in the right places. The e, sin and cos series have them; the ln and arctan series do not.
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