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Manipulating Taylor Series

Calculus 2 · Chapter 8: Power Series & Taylor SeriesSection 8.5 · lesson 43 of 5410 min · free

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In this lesson

Six operations on a five-entry library.

  • Substitute and multiply by a power
  • Add and multiply two known series
  • Differentiate and integrate term by term
  • Integrate a function with no elementary antiderivative

Transcript

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MAT-265 - Calculus 2 · Section 8.5

Manipulating Taylor Series

Section eight point three derived five Maclaurin series by computing derivatives. That procedure works. But it's laborious. Consider getting the series for e to the minus three x squared, from the definition. You'd need six derivatives. Each demanding the chain rule. And from the second onward, the product rule as well. There's a much better way. Almost every series you'll ever need can be assembled from the five you already have — using operations far cheaper than differentiation. This section is that toolkit.

The Principle

Never Compute Derivatives Twice

Here's the governing principle, stated formally. If f of x equals the sum of c n x to the n, valid on absolute x less than R, then two operations are permitted. First. Substitution. You may replace x by any expression u of x — giving the sum of c n, times u of x, to the n. Valid wherever absolute u of x is less than R. Second. Multiplication. You may multiply by any constant k and any nonnegative power x to the m. That gives the sum of k c n, x to the n plus m. With the same interval of convergence. Those two moves alone handle a large fraction of the series you'll meet.

Substitution and Multiplication

The Two Cheapest Moves

Let's use them. First, find the Maclaurin series for e to the minus three x squared. Substitute u equals minus three x squared into the standard series for e to the x. You get the sum of minus three x squared, to the n, over n factorial. Expand the numerator. Apply a b c to the n equals a to the n, b to the n, c to the n. And x squared to the n equals x to the two n. So minus three x squared to the n is minus one to the n, times three to the n, times x to the two n. Assemble: the sum of minus one to the n, three to the n, x to the two n, over n factorial. Write out the first few terms. For n equals zero through three: one, minus three x squared, plus nine x to the fourth over two, minus nine x to the sixth over two. And the interval. The series for e to the u converges for every real u. And minus three x squared is real for every x. So the representation holds for all real x. Now compare that to the alternative. Six derivatives, each with chain and product rules. Versus one substitution and one exponent law. Second example. Find the Maclaurin series for x cubed, sine of two x. Start from the standard sine series and substitute u equals two x. Expand the power: two x to the two n plus one is two to the two n plus one, times x to the two n plus one. Now multiply by x cubed. Each exponent rises by three. So x to the two n plus one becomes x to the two n plus four. Verify with the first two terms. For n equals zero you get two x to the fourth. For n equals one, minus four x to the sixth over three. So the series begins two x to the fourth, minus four x to the sixth over three. Valid for all real x.

Combining Two Series

Add, or Multiply

Now combining two series. Addition is performed by collecting like powers. The only subtlety is bookkeeping. Two series may use different index patterns — one indexed by two n, another by two n plus one. So it's usually safest to write out several terms of each, and combine column by column. Let's do one. The hyperbolic cosine is defined as e to the x, plus e to the minus x, all over two. Find its Maclaurin series. Write both series explicitly. The standard series gives e to the x. And substituting u equals minus x gives e to the minus x — in which the odd powers acquire a negative sign. Now add them. Every odd power cancels. And every even power doubles. Divide by two, and the doubling undoes itself. What's left is the sum of x to the two n, over two n factorial. One, plus x squared over two factorial, plus x to the fourth over four factorial. And notice — that confirms from a completely different direction what section three point one told us. Cosh is an even function, and only even powers survive. Multiplication works differently. You multiply the two series as though they were long polynomials, collecting each power of x in turn. For e to the x times cosine x, the constant term is one. The x term is one. The x squared terms cancel exactly — you get minus one half from cosine, plus one half from e to the x. Zero. And the x cubed coefficient works out to minus one third. So the product begins one, plus x, plus zero x squared, minus x cubed over three. Note that multiplication gives you the first several terms, rather than a closed general formula. Which is usually enough — because the terms furthest out matter least near the center.

Calculus on Series

Differentiate and Integrate Term by Term

The last two operations are differentiation and integration, term by term. Let's confirm that differentiating the sine series produces the cosine series. Write the sine series in expanded form. x, minus x cubed over three factorial, plus x to the fifth over five factorial. Differentiate each term using the power rule. The derivative of x to the k over k factorial is k, times x to the k minus one, over k factorial. And since k factorial equals k times k minus one factorial, the factor k cancels — leaving x to the k minus one, over k minus one factorial. So the derivative of x is one. The derivative of x cubed over three factorial is x squared over two factorial. And so on. Reassemble with the original alternating signs, and you get one, minus x squared over two factorial, plus x to the fourth over four factorial. Which is exactly the cosine series. The known derivative relationship is reproduced precisely. Which is a useful consistency check on the library itself. The picture shows why these approximations behave the way they do. Near x equals zero, even a single term tracks sine closely. Each added term extends the range of good agreement outward, in both directions. And every series built from sine through these operations inherits that behavior.

The Payoff

Integrating a Nonelementary Integrand

Now the payoff that justifies the whole chapter. The function e to the minus x squared has no antiderivative expressible in elementary terms. Yet its definite integral is needed constantly in statistics — it's the normal distribution. Term-by-term integration solves the problem completely. Step one. Substitute u equals minus x squared into the standard series for e to the x. That gives the sum of minus one to the n, x to the two n, over n factorial. Step two. Integrate term by term, from zero to one. Each x to the two n integrates 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. So the integral equals the sum of minus one to the n, over n factorial times two n plus one. Step three. Compute the first five terms. One. Minus one third, which is minus zero point three three three. Plus one tenth. Minus one over forty-two, about minus zero point zero two four. Plus one over two sixteen, about zero point zero zero four six. Step four. Sum them. You get zero point seven four seven four eight seven. And now — how accurate is that? The series alternates with decreasing terms. So by the Alternating Series Estimation Theorem, from section seven point three, the error is smaller than the first omitted term. That term is one over one hundred twenty, times eleven. About zero point zero zero zero seven six. The true value is approximately zero point seven four six eight two four. Well within the bound. Notice what just happened. A Chapter Eight representation supplied the terms. And a Chapter Seven error bound certified the accuracy. Two chapters converging on one practical answer.

Key Takeaways

Matching the Cue to the Operation

Recognizing which operation to apply is largely a matter of comparing the target function to the five library entries — and asking what structural difference must be closed. If the variable is replaced by an expression: substitution. If a power of x multiplies the whole function: multiply by x to the m. If the function is a sum or difference of known pieces: add or subtract. If it's a product of two known pieces: multiply the series. If the target is the derivative of a known function: differentiate. And if it's an antiderivative — possibly nonelementary: integrate. Complex targets may require two operations in sequence. Substitution followed by multiplication, for instance. Four takeaways. Build from the five standard series rather than recomputing derivatives. Substitution and multiplication by a power are the cheapest moves. Addition collects like powers. Multiplication yields the first several terms rather than a closed formula — which is usually enough. Term-by-term differentiation and integration preserve the radius. And differentiating sine back into cosine is a good consistency check. And integration term by term evaluates integrals with no elementary antiderivative — with the alternating-series bound certifying the accuracy. Next up is section eight point six — Applications of Series. Putting all of this to work.

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FAQ

Questions students ask us

About Manipulating Taylor Series

How do you find a Maclaurin series by substitution instead of taking derivatives?
Replace x in a known series with an expression u(x), and the result is valid wherever |u(x)|<R. For example, substituting u=-3x^2 into the series for e^x gives \sum (-1)^n 3^n x^{2n}/n!, which begins 1-3x^2+\frac{9x^4}{2}-\frac{9x^6}{2}. Since e^u converges for every real u, this holds for all real x. Computing the same terms from the definition would take six derivatives.
What is the Maclaurin series for x^3 sin(2x)?
It is \sum (-1)^n 2^{2n+1} x^{2n+4}/(2n+1)!, valid for all real x. First substitute u=2x into the sine series, using (2x)^{2n+1}=2^{2n+1}x^{2n+1}. Then multiply by x^3, which raises every exponent by three and leaves the interval of convergence unchanged. The series begins 2x^4-\frac{4x^6}{3}. Checking the first couple of terms by hand catches exponent mistakes.
How do you multiply two power series, such as e^x times cos x?
Multiply them like long polynomials, collecting each power of x in turn. For e^x\cos x, the constant term is 1 and the x term is 1. The x^2 terms cancel, since -\frac12+\frac12=0. The x^3 coefficient is \frac16-\frac12=-\frac13. So the product begins 1+x+0x^2-\frac{x^3}{3}. This gives the first several terms rather than a general formula, which is usually enough near the center.
How do you differentiate a power series term by term?
Apply the power rule to each term separately; the radius of convergence stays the same. Because k!=k\,(k-1)!, the derivative of \frac{x^k}{k!} is \frac{x^{k-1}}{(k-1)!}. Differentiating the sine series x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots this way gives 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots. That is exactly the cosine series, which makes it a useful consistency check.

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