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Taylor and Maclaurin Series

Calculus 2 · Chapter 8: Power Series & Taylor SeriesSection 8.3 · lesson 41 of 549 min · free

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In this lesson

Building a series from a function's derivatives.

  • State the Taylor series and its a=0 special case
  • Derive the five standard Maclaurin series
  • Distinguish radius from interval
  • Use the reference table instead of re-deriving

Transcript

Everything said in the video, section by section (1,337 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 8.3

Taylor and Maclaurin Series

Section eight point two built power series by manipulating one geometric seed. Substitute, differentiate, integrate, multiply. And that produced a useful library. But it only works when a function can be forced into the template one over one minus u. And e to the x cannot. Neither can sine, or cosine. So we need a construction that works for any function — one that doesn't depend on recognizing a geometric pattern. Taylor's construction does exactly that. It builds the series directly from the function's derivatives at a point.

The Definition

Coefficients Come From Derivatives

Here's the definition. Let f be a function possessing derivatives of every order at a point a. The Taylor series of f centered at a is the sum, from n equals zero to infinity, of the nth derivative of f at a, divided by n factorial, times x minus a, to the n. Written out, that reads: f of a, plus f prime of a times x minus a, plus f double prime of a over two factorial, times x minus a squared, and so on. With the conventions that the zeroth derivative is f itself, and zero factorial is one. When the center is a equals zero, the series is called the Maclaurin series. So a Maclaurin series is not a separate object. It's the special case a equals zero of a Taylor series. Now, one caution belongs here, and it matters. This definition constructs a series from a function. But it does not by itself guarantee that the series sums back to the function. That equality holds on the interval where the remainder term of Taylor's Theorem tends to zero — which is the subject of the next section. For every function we treat here, that equality does hold throughout the stated interval.

Derivation

The Same Four Steps Every Time

Five Maclaurin expansions arise so persistently in physics, engineering, statistics and numerical work that they should be committed to memory. And each is derived by the same four-step procedure. Compute derivatives at zero. Detect the pattern. Assemble the series. Find the interval with the Ratio Test. Start with e to the x. The easiest case. The derivative of e to the x is e to the x. So every derivative is the function itself. Evaluating at the center zero, and using e to the zero equals one, gives one for every n. So every coefficient is one over n factorial. e to the x equals the sum of x to the n over n factorial. One plus x, plus x squared over two factorial, plus x cubed over three factorial. For the radius, apply the Ratio Test. The absolute ratio is absolute x over n plus one. For any fixed x that approaches zero. Since the limit is zero, less than one, for every real x — the series converges absolutely everywhere. The radius is infinite. Now sine. Here the derivatives cycle with period four. Sine, cosine, minus sine, minus cosine, and back to sine. Evaluated at zero, that gives zero, one, zero, minus one — repeating. So every even-indexed derivative vanishes, killing all the even powers. And the odd ones alternate in sign. The result is sine x equals the sum of minus one to the n, x to the two n plus one, over two n plus one factorial. Cosine works the same way, but the cycle starts one step later — so the odd powers vanish and the even ones survive. Both have infinite radius. And notice the symmetry. Sine is an odd function, and its series contains only odd powers. Cosine is even, and its series contains only even powers. The symmetry of the function is visible in the formula. One more thing worth understanding rather than memorizing. Why the factorials? Because differentiating x to the n exactly n times produces n factorial. Dividing by n factorial is precisely what makes the nth derivative of the series match the nth derivative of f at the center. The factorials aren't decoration. They're what makes the construction work.

Figure 8.3

Partial Sums Closing In

The picture shows what those partial sums actually do. The dark curve is sine x. The dashed curves are Taylor polynomials of increasing degree. T one is just x. And near the origin that's a genuinely good approximation — which is why the small-angle approximation sine theta equals theta works in physics. But it diverges from the curve almost immediately. T three adds the minus x cubed over six term. Now the polynomial tracks sine noticeably further. T five extends the agreement further still. And T seven follows the curve well past x equals pi. So each additional term extends the range over which the polynomial tracks the function. That's the general pattern. Convergence is comfortable near the center, and degrades as you move away from it. Which is exactly why the choice of center matters.

Reference

The Five Standard Maclaurin Series

The five results are collected in the reference table. e to the x. One plus x plus x squared over two factorial, and so on. Radius infinite, interval all reals. Sine x. x minus x cubed over three factorial, plus x to the fifth over five factorial. Radius infinite. Cosine x. One minus x squared over two factorial, plus x to the fourth over four factorial. Radius infinite. One over one minus x. The geometric seed. Radius one, interval minus one to one, both endpoints excluded. And log of one plus x. x minus x squared over two, plus x cubed over three. Radius one, interval minus one to one — with the right endpoint included. Note that only log of one plus x includes an endpoint. The three series built from e, sine and cosine cover the entire real line. The two tied to the geometric expansion are confined to a radius of one about the center. Memorizing this table pays immediate dividends. Because most series you meet later are obtained from these five — by substitution, differentiation, integration or multiplication. Not by computing derivatives from scratch.

A Distinction

Radius Versus Interval

One distinction is worth stating precisely, because the two terms get confused constantly. The radius of convergence R is a single nonnegative number, or infinity. It measures how far from the center the series remains valid. The interval of convergence is the actual set of x values. With endpoints included or excluded as the arithmetic dictates. And here's the key fact. The Ratio Test determines R. But it is always inconclusive at exactly the distance R. Which is why endpoints require separate testing, every time. So two series can share a radius and still have different intervals. One over one minus x, and log of one plus x, both have radius one. Yet one excludes both endpoints, and the other includes one. Finally, a word on why the center matters at all. The series agrees with f most accurately near a. And it degrades as the distance from a grows. So a Taylor series centered somewhere else — say at a equals one rather than a equals zero — is a genuinely different series for the same function. Better suited to inputs near that new center.

Key Takeaways

What to Carry Forward

Five takeaways. The Taylor series of f at a has coefficients: the nth derivative of f at a, divided by n factorial. And a Maclaurin series is simply the case a equals zero. The factorial denominators exist so that the series' nth derivative matches f's nth derivative at the center. They're structural, not decorative. Five standard series should be memorized. e to the x, sine x, cosine x, one over one minus x, and log of one plus x. e to the x, sine and cosine converge on the entire real line. The geometric-family series are confined to radius one — and only log of one plus x captures an endpoint. And radius is a number, while interval is a set. The Ratio Test gives you R, but never settles the endpoints. Those must be tested individually. Next up is section eight point four — Taylor Polynomials and the Remainder. Which measures exactly how far a truncated series sits from the function it approximates.

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FAQ

Questions students ask us

About Taylor and Maclaurin Series

What is the difference between a Taylor series and a Maclaurin series?
A Maclaurin series is simply a Taylor series centered at a=0. The Taylor series of f at a is \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n, using the conventions f^{(0)}=f and 0!=1. Setting a=0 gives \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n. A series centered elsewhere, such as a=1, is a different series for the same function. It is most accurate for inputs near that center.
How do you find the Maclaurin series of e^x?
Every derivative of e^x is e^x, so f^{(n)}(0)=1 for all n and every coefficient is 1/n!. That gives e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots. This follows the standard four steps: compute derivatives at zero, detect the pattern, assemble the series, and apply the Ratio Test. The ratio \frac{|x|}{n+1} tends to 0 for every x, so the radius is infinite.
What are the five Maclaurin series you should memorize?
The five standard series are for e^x, \sin x, \cos x, \frac{1}{1-x}, and \ln(1+x). They are \sin x=\sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \cos x=\sum \frac{(-1)^n x^{2n}}{(2n)!}, \frac{1}{1-x}=\sum x^n on (-1,1), and \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots on (-1,1]. The series for e^x=\sum \frac{x^n}{n!}, \sin x, and \cos x converge for all real x. Most other series come from these by substitution, differentiation, integration, or multiplication.
Is the radius of convergence the same as the interval of convergence?
No, the radius R is a single nonnegative number (or \infty), while the interval is the actual set of x values where the series converges. The Ratio Test finds R but is always inconclusive at the endpoints, so each endpoint must be tested separately. For example, \frac{1}{1-x} and \ln(1+x) both have R=1. However, the first converges on (-1,1) and the second converges on (-1,1].

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