Active Calculus 2nd Ed

In Preview Activity 8.4.1, we consider an infinite geometric series and explore the function to which the series is related and that function’s Taylor polynomials.

From our work in Activity 8.3.4 and Preview Activity 8.4.1, we have seen two examples where the Taylor polynomials of a given function lead to a corresponding infinite Taylor series that is actually equal to the original function itself. This suggests that one reason Taylor series are important is that they give rise to polynomial-like representations of non-polynomial functions.

Next we formally define the Taylor series of a function and introduce a tool for determining where its Taylor series converges.

In the special case where \(a=0\) in Definition 8.4.1, the Taylor series is also sometimes called the Maclaurin series for \(f\text{.}\)

It turns out that while most Taylor series are not geometric series, we can compare any Taylor series to a geometric series in order to determine an interval of \(x\)-values for which the Taylor series converges. To see how to approach this issue, we consider the following example with a particular function with a known Taylor series and determine the \(x\)-values for which its Taylor series should converge.

Example 8.4.2.

Let \(f(x) = \ln(1+x)\text{.}\) By investigating how close the terms of the Taylor series of \(f\) of resemble a geometric series, determine the \(x\)-values for which the Taylor series appears to converge.

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Solution.

In Example 8.2.5, we found that the general degree \(n\) Taylor polynomial centered at \(a = 0\) for \(f(x) = \ln(1+x)\) is

\begin{equation*} T_n(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1}\frac{1}{n}x^n\text{.} \end{equation*}

This pattern thus allows us to see that the Taylor series for \(f(x)\text{,}\) centered at \(a = 0\text{,}\) is

\begin{align*} T_f(x) \amp= \sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k}x^k\\ \amp= x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \cdots + (-1)^{n-1}\frac{1}{n}x^n + \cdots\text{.} \end{align*}

Unlike the Taylor series for the function \(\frac{1}{1-x}\text{,}\) the Taylor series for \(f(x) = \ln(1+x)\) turns out to not be geometric since there is not a common ratio between the terms, so we introduce some new ideas to think about the \(x\)-values for which we expect the series to converge.

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For a series to be geometric, we must be able to construct the next term in the series by multiplying the prior term by a constant ratio. In \(T_f(x)\text{,}\) the general \(n^{\text{th}}\) term is

\begin{equation*} a_n(x) = (-1)^{n-1}\frac{1}{n}x^n \end{equation*}

and the \((n+1)^{\text{st}}\) term is

\begin{equation*} a_{n+1}(x) = (-1)^{n}\frac{1}{n+1}x^{n+1} \text{.} \end{equation*}

If we consider the ratio of these consecutive terms and call that quantity \(r_n(x)\text{,}\) we find that

\begin{equation*} r_n(x) = \frac{a_{n+1}(x)}{a_n(x)} = \frac{(-1)^{n}\frac{1}{n+1}x^{n+1}}{(-1)^{n-1}\frac{1}{n}x^n}\text{.} \end{equation*}

We can rewrite this fraction algebraically and simplify it considerably. Grouping the similar terms between the numerator and denominator, we see

\begin{equation*} r_n(x) = \frac{(-1)^{n}}{(-1)^{n-1}} \cdot \frac{n}{n+1} \cdot \frac{x^{n+1}}{x^n}\text{,} \end{equation*}

which allows us to simplify and conclude that

\begin{equation*} r_n(x) = (-1) \cdot \frac{n}{n+1} \cdot x\text{.} \end{equation*}

This tells us the series \(T_f(x)\) is not geometric since the ratio of consecutive terms depends on \(n\text{.}\) For example, if we take \(n = 2\text{,}\) we have

\begin{equation*} r_2(x) = -\frac{2}{3}x\text{,} \end{equation*}

which is the ratio between the third and second terms of \(T_f(x)\text{,}\) while

\begin{equation*} r_3(x) = -\frac{3}{4}x \end{equation*}

is the ratio between the fourth and third terms. Since the ratio \(r_n(x)\) depends on \(n\text{,}\) the series lacks a common ratio between terms.

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However, as \(n\) gets large, the series \(T_f(x)\) is almost geometric. This is because the quantity \(\frac{n}{n+1}\) approaches \(1\) as \(n\) increases without bound, and therefore for large \(n\text{,}\)

\begin{equation*} r_n(x) \approx -x\text{.} \end{equation*}

This suggests that the series \(T_f(x)\) converges for \(|r_n(x)| \approx |-x| = |x| \lt 1\text{,}\) since the series is very close to being geometric with a common ratio \(-x\text{.}\)

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Graphical evidence supports the conclusion that for \(|x| \lt 1\text{,}\) the series \(T_f(x)\) converges in the same way that a geometric one would, and also that the interval of convergence is \(-1 \lt x \lt 1\text{.}\) In Figure 8.4.3, we plot \(f(x) = \ln(1+x)\text{,}\) \(T_{10}(x) = x - \frac{1}{2}x^2 + \cdots - \frac{1}{10}x^{10}\text{,}\) and \(T_{20}(x) = x - \frac{1}{2}x^2 + \cdots - \frac{1}{20}x^{20}\text{.}\)

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In Figure 8.4.3 we see that on the interval \(-0.8 \lt x \lt 0.8\text{,}\) \(f(x)\text{,}\) \(T_{10}(x)\text{,}\) and \(T_{20}(x)\) are nearly indistinguishable. Furthermore, we see that if we widen the interval to \(-0.9 \lt x \lt 0.9\text{,}\) \(T_{20}(x)\) remains a very good approximation of \(f(x)\text{.}\) For example, we can do some computations and find that

\begin{equation*} f(-0.8) - T_{10}(-0.8) \approx -0.030563, f(-0.8) - T_{20}(-0.8) \approx -0.001091\text{,} \end{equation*}

and

\begin{equation*} f(-0.9) - T_{10}(-0.9) \approx -0.183837, f(-0.9) - T_{20}(-0.9) \approx -0.032564 \text{.} \end{equation*}

Increasing the degree of the approximation further allows us to do a bit better, but since \(f(x) = \ln(1+x)\) is undefined with a vertical asymptote at \(x = -1\text{,}\) no polynomial function can agree with \(f(x)\) at \(x = -1\text{.}\) Thus, it appears to be the case that \(T_f(x)\) converges for \(x\)-values that satisfy \(-1 \lt x \lt 1\text{,}\) and on that interval the Taylor series converges to \(f(x) = \ln(1+x)\text{.}\)

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We next introduce a more formal way to test “how geometric” a Taylor series is in an effort to understand where the Taylor series is guaranteed to converge.

In essence, the Ratio Test is a way to find \(x\) values for which \(T(x)\) behaves like a geometric series with ratio \(r(x)\) such that \(|r(x)| \lt 1\text{;}\) for those \(x\)-values, the infinite series converges to a finite value. There is a subtle issue that arises for \(x\)-values for which \(|r(x)| = 1\text{:}\) it’s possible for the infinite series to either converge or diverge at such \(x\)-values, but we will normally not concern ourselves with this issue.

In the next activity, we conduct a similar investigation for \(f(x) = e^x\text{.}\)

While there is considerable technical detail involved in finding Taylor series and deciding where they converge, it’s important not to miss the big picture. For certain “nice” functions that are infinitely differentiable, such as \(f(x) = \ln(1+x)\) and \(f(x) = \sin(x)\text{,}\) not only can we find the Taylor series for each and determine an open interval of \(x\)-values at which the Taylor series converges, but for those values the Taylor series converges to the function itself. This means that for certain \(x\)-values (and sometimes all \(x\)-values), we can represent a non-polynomial function as being equal to an infinite polynomial.

A function such as \(f(x) = \sin(x)\) that is infinitely differentiable and has a power series that converges on some interval to the function itself is said to be analytic. Analytic functions are amazing and among the nicest functions in all of mathematics: they are completely determined by what happens at a single point.

Almost all functions are not analytic. For example, if a function measures your location as a function of time on a highway, that function cannot be analytic, since what happens at 12:01 is independent of what happens at 12:00 due to other circumstances such as traffic or construction. At the same time, the mathematical universe is better to us than we deserve in that many important, common functions are analytic, including several we’ve mentioned (\(\frac{1}{1-x}\text{,}\) \(\ln(1+x)\text{,}\) \(\sin(x)\text{,}\) \(\cos(x)\text{,}\) and \(e^x\)) and most of those represented by buttons on a calculator.

The five Taylor series that we’ve found in our work to date are extremely important and will be used extensively in Section 8.5, so we summarize them in the following table.