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Representation of Functions as Power Series

Calculus 2 · Chapter 8: Power Series & Taylor SeriesSection 8.2 · lesson 40 of 549 min · free

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In this lesson

Three operations on one seed series.

  • Force a function into the template 1/(1-u)
  • Substitute, multiply, and re-derive the interval
  • Differentiate and integrate a series term by term
  • Build a reference library from a single seed

Transcript

Everything said in the video, section by section (1,287 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 8.2

Functions as Power Series

Section eight point one asked where a given power series converges. This section runs that question backwards. Given a function, find a power series that equals it. And here's the remarkable thing. Nearly every answer descends from one series you already know. The geometric series. One over one minus x equals the sum of x to the n, for absolute x less than one. You've seen that formula since Chapter Six. But there it was read left to right — a series that adds up to a number. Read it right to left, and it becomes something else entirely. A function, rewritten as an infinite polynomial.

The Seed

One Series to Build Everything From

So that's our seed. One over one minus x equals one plus x plus x squared plus x cubed, and so on. Valid on minus one to one. Now here's the productive part. Wherever the letter x appears in that formula, any expression u of x may be substituted. Provided the convergence condition is restated as: absolute u of x less than one. And multiplying the entire representation by a constant, or by a power of x, is likewise permitted. Because those operations distribute across every term of the sum. Two moves. Substitution, and multiplication. Those alone generate a large family of representations.

Substitution

Force the Function Into the Template

Let's find a power series for one over one plus x squared. Step one. Force the expression into the template form, one over one minus u. The template has a minus sign. But our function has a plus. So rewrite the denominator. One plus x squared equals one minus, the quantity minus x squared. That's the key move. Whenever the sign is wrong, absorb it into u. Step two. Substitute u equals minus x squared into the seed. You get the sum of minus x squared, to the n. Step three. Simplify the general term. Apply the exponent law: a b to the n equals a to the n, b to the n. So minus x squared to the n is minus one to the n, times x to the two n. Step four. Write the result. One over one plus x squared equals the sum of minus one to the n, x to the two n. Expanded: one minus x squared plus x to the fourth minus x to the sixth, and so on. Step five. Determine the interval. The requirement is absolute minus x squared less than one. Since that's just x squared, the condition is x squared less than one. Taking square roots, absolute x less than one. So the interval is minus one to one. This particular representation is worth retaining — because integrating it later produces the arctangent series.

Calculus on Series

Differentiate and Integrate Term by Term

Now two more operations, and they're more powerful still. Inside its interval of convergence, a power series may be differentiated or integrated one term at a time. And the radius R is unchanged by either operation. Only the endpoints may shift. Let's differentiate the seed. The derivative of one over one minus x is one over one minus x, squared. So differentiate the series side term by term. The derivative of one plus x plus x squared plus x cubed is zero plus one plus two x plus three x squared, and so on. In summation form — since the constant term vanished, the index begins at n equals one. One over one minus x squared equals the sum of n x to the n minus one. Let's verify numerically at x equals one half. The closed form gives one over one half squared. Which is one over one quarter. Four. The series gives one, plus two times one half, plus three times one quarter, plus four times one eighth, plus five times one sixteenth. That's one plus one plus zero point seven five plus zero point five plus zero point three one two five. Three point five six two five. Climbing toward four as more terms are added. Confirmed. Now integration. Find a power series for log of one minus x. By the chain rule, the derivative of log one minus x is minus one over one minus x. So log one minus x is minus the integral of the seed. Integrate term by term. You get C plus the sum of x to the n plus one, over n plus one. Attach the negative sign from step one. Now solve for C by substituting x equals zero. The left side is log one, which is zero. Every term of the sum contains a positive power of x, so the entire sum is zero. Hence C equals zero. Now check the endpoints — and here's where something interesting happens. The radius remains one. At x equals minus one, the series becomes the alternating harmonic series. Which converges. At x equals one, it becomes the negative of the harmonic series. Which diverges. So the interval is minus one, inclusive, to one, exclusive. One endpoint wider than the original geometric series. Integration widened the interval. Which is exactly why endpoints must always be retested after differentiating or integrating.

Figure 8.2

Three Operations, One Seed

Let's see the whole structure in one picture. At the top is the seed. One over one minus x equals the sum of x to the n. Three operations branch off it. Substitute — replace x by u of x. That gives one over one plus x squared. Differentiate term by term. That gives one over one minus x squared. Integrate term by term. That gives log of one minus x. And notice what each operation does mechanically. Substitution changes what is raised to the nth power. Differentiation multiplies each coefficient by its exponent, and lowers the exponent by one. Integration divides by the new exponent, and raises it by one. Three simple mechanical rules. Applied to one seed.

Reference

A Catalog Built From One Series

Applying those three operations in various combinations produces a compact reference library. One over one minus x. The seed itself. One over one plus x. Substitute u equals minus x. One over one plus x squared. Substitute u equals minus x squared. One over one minus x, squared. Differentiate the seed. Log of one minus x. Integrate the seed, then negate. And arctangent x. Integrate the third row — the one over one plus x squared representation. That gives the sum of minus one to the n, x to the two n plus one, over two n plus one. Now look at the intervals in the last two rows. Log of one minus x is valid on minus one inclusive, to one exclusive. Arctangent x is valid on minus one to one, both endpoints inclusive. Integration widened the interval in both cases. Which is precisely why endpoints must always be tested separately, rather than assumed. One piece of advice. Memorizing the seed and the three operations is far more efficient than memorizing these entries. And it lets you build entries this table doesn't even contain.

Key Takeaways

What to Carry Forward

Five takeaways. Everything descends from one over one minus x equals the sum of x to the n, on absolute x less than one. The geometric series read backwards. Substitution. Replace x by any expression u of x, and restate the condition as absolute u of x less than one. Force the function into the template first — minus sign included. Multiplication by a constant, or by a power of x, distributes across every term and leaves the interval unchanged. Term-by-term differentiation and integration are valid inside the interval, and both preserve the radius R. But they can change the endpoints. Integration in particular can widen the interval. So always retest the endpoints afterwards. Next up is section eight point three — Taylor and Maclaurin Series. Which build a series for a function directly from its derivatives, rather than from the geometric seed.

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FAQ

Questions students ask us

About Representation of Functions as Power Series

What does it mean to represent a function as a power series?
It means writing a function as an infinite polynomial that equals the function on some interval. Most basic examples come from the geometric series read backwards: \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots for |x|<1. From this one seed, you can build many other representations by substitution, by multiplying by constants or powers of x, and by differentiating or integrating term by term.
How do you find a power series for 1/(1+x^2)?
Force it into the template \frac{1}{1-u}, then substitute into the geometric series. Absorb the plus sign into u by writing 1+x^2 = 1-(-x^2), so u=-x^2. This gives \sum_{n=0}^{\infty}(-x^2)^n = \sum_{n=0}^{\infty}(-1)^n x^{2n} = 1 - x^2 + x^4 - \cdots. The condition |u|<1 becomes x^2<1, which means |x|<1. The interval of convergence is therefore (-1,1).
How do you get the power series for ln(1-x)?
Integrate the geometric series term by term, then negate. Since \int \frac{1}{1-x}\,dx = -\ln(1-x) + C, integrating \sum x^n gives -\ln(1-x) = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}, and plugging in x=0 shows C=0. So \ln(1-x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}. The radius stays 1. The series converges at x=-1 but diverges at x=1, so the interval is [-1,1).
What is a common mistake when differentiating or integrating a power series term by term?
The most common mistake is assuming the interval of convergence stays exactly the same. Term-by-term differentiation and integration preserve the radius R, but the endpoints can change. For example, the series for \frac{1}{1+x^2} converges on (-1,1). Its integral, \arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}, converges on [-1,1]. Always retest each endpoint separately after either operation.

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