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Applications of Series

Calculus 2 · Chapter 8: Power Series & Taylor SeriesSection 8.6 · lesson 44 of 5410 min · free

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In this lesson

Three problems that series solve better than anything else.

  • Evaluate indeterminate limits by expanding and cancelling
  • Integrate functions with no elementary antiderivative
  • Compute a value to a stated precision
  • Recognise which cue signals which strategy

Transcript

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MAT-265 - Calculus 2 · Section 8.6

Applications of Series

You've built the machinery. Power series, Taylor series, the five standard expansions, the operations that transform them, and the error bounds that certify them. This section puts all of it to work. Three kinds of problem where series aren't merely an alternative — they're the best available tool. And sometimes the only one. Evaluating indeterminate limits. Integrating functions with no elementary antiderivative. And computing values to a guaranteed precision.

Application 1

Indeterminate Limits, Without L'Hopital

First application. Indeterminate limits. Evaluate the limit as x approaches zero, of sine x minus x, over x cubed. Substituting x equals zero gives zero minus zero, over zero. Which is zero over zero. Indeterminate. So direct substitution fails. Now, you could use L'Hopital's Rule here. But watch what the series does instead. Recall the Maclaurin series for sine, valid for every real x. Sine x equals x, minus x cubed over three factorial, plus x to the fifth over five factorial, and so on. Subtract x from both sides. And notice — the leading terms cancel exactly. What's left is minus x cubed over six, plus x to the fifth over one twenty, and so on. Now divide by x cubed, dividing each term separately. Each division reduces the exponent by three. x cubed over x cubed is one. x to the fifth over x cubed is x squared. And so on. So the expression is now an ordinary power series. With no division difficulty at all. Minus one sixth, plus x squared over one twenty, minus x to the fourth over five thousand forty. Let x go to zero. Every term containing a positive power of x approaches zero. And only the constant survives. The limit is minus one sixth. About minus zero point one six six seven. Now compare the effort. L'Hopital's Rule requires three separate applications here. Each demanding a fresh derivative of both numerator and denominator. With the indeterminate form persisting until the third round. The series computation reached the answer in one pass. And as a bonus, it revealed the behavior near the limit. The function sits slightly above minus one sixth for x not zero — because the next term, x squared over one twenty, is positive. L'Hopital gives you the number. The series gives you the number and the shape.

Figure 8.6.1

The Series Fills the Hole

The picture shows exactly what that means. The function sine x minus x, over x cubed, is undefined at x equals zero. You can see the hollow circle. That's a removable hole. The function approaches a perfectly definite value from both sides — it just isn't defined at the single point itself. And the dashed line marks minus one sixth. The curve dips down toward that line from both directions, touching it in the limit. The series computation supplies exactly the value needed to fill that hole. And notice that the curve sits above the line on both sides — which is precisely what the positive x squared term predicted.

Application 2

Integrals With No Elementary Antiderivative

Second application. Integrals with no elementary antiderivative. Consider the integral from zero to one, of e to the minus x squared. Substitution fails. Integration by parts fails. Partial fractions and trig substitution don't apply. And not because they're being applied badly. Because no elementary antiderivative exists. It has been proven that no combination of polynomials, exponentials, logarithms or trig functions differentiates to e to the minus x squared. A series representation sidesteps the obstacle entirely. Step one. Take the series for e to the t, and substitute t equals minus x squared. Because that series converges for all real numbers, the substitution is valid for every x, with no restriction. You get one, minus x squared, plus x to the fourth over two factorial, minus x to the sixth over three factorial. The signs alternate because minus x squared to the n is minus one to the n, times x to the two n. And the exponents are even because each power of x squared doubles. Step two. Integrate term by term, from zero to one. The power rule gives x to the two n, integrating to x to the two n plus one, over two n plus one. Evaluating from zero to one leaves simply one over two n plus one — since every term vanishes at the lower limit. So the integral equals the sum of minus one to the n, over n factorial times two n plus one. Look at the picture. That shaded region has a perfectly well-defined area. About zero point seven four six eight two four. Even though no formula in elementary functions produces it.

The Numbers

Partial Sums Bracket the Answer

Let's compute the terms and watch what happens. For n equals zero: one over one times one. Plus one point zero. n equals one: minus one over one times three. Minus zero point three three three three. n equals two: plus one over two times five. Plus zero point one. n equals three: minus one over six times seven. Minus zero point zero two three eight. And so on, with the terms shrinking rapidly because of the factorial. Now look at the partial sums column. One point zero. Then zero point six six six seven. Then zero point seven six six seven. Then zero point seven four two nine. Then zero point seven four seven five. Zero point seven four six seven. Zero point seven four six eight. See the pattern? The partial sums bracket the true value from alternating sides. Each one overshoots, then the next undershoots — by a smaller amount each time. That's the hallmark of an alternating series. And it's what makes this estimate trustworthy. The true value, zero point seven four six eight two four, is trapped between every consecutive pair of partial sums. So the answer is certified without ever knowing it in advance.

Application 3

A Value to a Stated Precision

Third application. Computing a value to a stated precision. An engineer rarely wants an exact symbolic answer. They want a decimal, correct to a stated tolerance, with a guarantee attached. And series supply both the terms and the guarantee. For an alternating series with decreasing terms, the Alternating Series Estimation Theorem from section seven point three says the error after N terms is smaller than the first omitted term. So the procedure is simple. Sum terms until that next term drops below your target. For the integral we just computed, reaching six-decimal accuracy means summing until one over n factorial times two n plus one, falls below ten to the minus six. That first happens at n equals nine. So nine terms suffice. And the actual error stays beneath the bound at every single stage — as it must. Crucially, the process needs no knowledge of the true value at any point. The bound alone certifies the answer. Let me summarize the three applications, with the cue that identifies each. Indeterminate limit. Signal: the form zero over zero at the target point. Strategy: expand, cancel, divide by the lowest power. The alternative is L'Hopital's Rule. Non-elementary integral. Signal: no antiderivative found by substitution or parts. Strategy: expand the integrand and integrate term by term. The alternative is a numerical rule like Simpson's. Function value to fixed precision. Signal: a decimal answer with a stated tolerance. Strategy: sum terms until the bound is met. The alternative is a built-in calculator routine. In each case, series either replace the competing technique — or explain why the calculator's routine works at all.

Key Takeaways

What to Carry Forward

Five takeaways. For an indeterminate limit, expand both parts, cancel the leading terms, and divide by the lowest surviving power. One pass replaces repeated applications of L'Hopital's Rule. The series also reveals behavior near the limit. The sign of the next term tells you which side the function approaches from — information L'Hopital never gives you. For an integral with no elementary antiderivative, expand the integrand and integrate term by term. The result is an exact series for a definite integral that no closed form can express. Alternating partial sums bracket the true value. So the answer is certified without knowing it in advance. And to hit a tolerance, sum until the first omitted term drops below it. The guarantee rests on the bound. Never on the unknown true value. That closes Chapter Eight. Next up is Chapter Nine — Parametric Equations and Polar Coordinates.

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FAQ

Questions students ask us

About Applications of Series

How do you use Taylor series to evaluate a limit?
Expand the expression as a Maclaurin series, cancel the leading terms, divide by the lowest surviving power of x, and then let x \to 0. For \lim_{x\to 0} \frac{\sin x - x}{x^3}, write \sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \cdots. Dividing by x^3 gives -\frac{1}{6} + \frac{x^2}{120} - \cdots. Every term with a positive power of x vanishes, so the limit is -\frac{1}{6}.
How do you integrate e^(-x^2) using a power series?
Replace e^{-x^2} with its power series and integrate term by term. Substituting t=-x^2 into e^t=\sum t^n/n! gives e^{-x^2}=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}, which is valid for all real x. Integrating from 0 to 1 yields \int_0^1 e^{-x^2}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)}\approx 0.746824. This works even though e^{-x^2} has no elementary antiderivative.
How many terms of a series do you need to reach a given accuracy?
For an alternating series with decreasing terms, keep adding terms until the first omitted term is smaller than your tolerance. The Alternating Series Estimation Theorem guarantees the error is less than that term. For \sum \frac{(-1)^n}{n!(2n+1)}, six-decimal accuracy requires a term below 10^{-6}. That first happens at n=9, so the nine terms from n=0 through n=8 suffice. You never need to know the true value.
When should you use series instead of L'Hopital's Rule or other methods?
Use series when you see one of three cues: a 0/0 limit, an integral that resists substitution and integration by parts, or a request for a decimal to a stated tolerance. For limits, one expansion often replaces several rounds of L'Hopital's Rule. For example, \frac{\sin x - x}{x^3} needs three rounds. The series also shows which side the function approaches from, through the sign of the next term.

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