Calculus I, II, and III: A Problem-Based Approach with Early Transcendentals:

Section Power Series

From the first semester of calculus, we know that the best linear (\(1^{st}\) degree polynomial) approximation to the function, \(f(x) = e^x\) at the point \(P=(2,e^2)\) is the tangent line to the function at this point. Let's call that approximation \(L\) and note that \(L(2) = f(2)\) and \(L'(2) = f'(2)\text{.}\) That is, \(L\) and \(f\) intersect at \(P\) and \(L\) and \(f\) share the same derivative at \(P\text{.}\) What is the best quadratic (\(2^{nd}\) degree polynomial) approximation? If the best first degree approximation to the curve agrees at the point and in the first derivative, then the best second degree approximation should agree with the function at the point, in the first derivative, and in the second derivative.

From our work on geometric series, we know that

for \(|t| \lt 1\text{.}\) Therefore the function \(\dsp g(t) = \frac{1}{1-t} \mbox{ where } |t| \lt 1\) can be written as a series,

Replacing \(t\) by \(-x^2\text{,}\) we have

Based on these two observations, we see that we can write at least two rational functions as infinite series (or infinite polynomials). And we can write every polynomial as an infinite series, since

where \(a_i = 0\) for \(i > N\text{.}\) Our goal is a systematic way to write any differentiable function (like \(sin\) or \(cos\) or \(\ln\)) as an infinite series.

Notation. To keep from confusing the powers of \(f\) with the derivatives of \(f\text{,}\) we use \(f^{(n)}\) to represent \(f\) if \(n=0\) and the \(n^{th}\) derivative of \(f\) if \(n \ge 1.\) Therefore when \(n \geq 1\text{,}\) \(f^{(n)}(c)\) means the \(n^{th}\) derivative of \(f\) evaluated at the number \(c\text{.}\)

We now can write certain functions as infinite series, but we know that not every infinite series converges. Therefore, our expressions only make sense for the values of \(x\) for which the series converges. For this reason, every time we write down an infinite series to represent a function, we need to determine the values of \(x\) for which the series converges. This is called interval of convergence (or domain) of the series.

The next theorem says that the interval of convergence of a power series expanded at the point \(c\) is either (i) only the one point, \(c\) (bad, since this means the series is useless), (ii) an interval of radius \(R\) centered at \(c,\) (good, and the endpoints might be contained in the domain), or (iii) all real numbers (very good).

When we write the power series for a function (like \(e^x\)) and it converges for all values of \(x\) then we simply have a different way to write the function that turns out to be both computationally useful, because of convergence, and theoretically useful. Speaking loosely, \(e^x\) is just an infinite polynomial — how cool is that?!

The following theorem says that if you have a power series, then its derivative is exactly what you want it to be! Just take the derivative of the series term by term and the new infinite series you get is the derivative of the first and is defined on the same domain.

How does the calculator on your phone approximate numbers like \(\sqrt{5}\) or \(\sin(\pi/7)\text{?}\) It uses Taylor series. For example, if we want to approximate \(\sqrt{5}\) accurate to 3 decimal places, we could compute the Taylor series for \(f(x) = \sqrt{x}\) expanded about \(c=4\) because 4 is the integer closest to 5 for which we actually know the value of \(\sqrt{4}\text{.}\) Then we can use this polynomial to approximate \(\sqrt{5}.\) But there's a problem. While the Taylor series approximates the function, it's pretty hard to store an infinite polynomial in a phone. So, how can we determine what degree Taylor polynomial we need to compute in order to assure a given accuracy? We look at the remainder of the series. For any positive integer, \(N,\) \(f\) can be rewritten as the sum of the Taylor polynomial of degree \(N\) plus the remainder of the series by writing,

or

where

and

Since \(f(x) = T_N(x) + R_N(x),\) for all \(x\) in the interval of convergence, the error for any \(x\) in the interval of convergence is just \(| f(x) - T_N(x)| = |R_N(x)|.\) If we could just approximate the size of the remainder term, we would know how accurate the \(N^{th}\) degree Taylor polynomial was. The next theorem says that the remainder, even though it is an infinite sum, can be written completely in terms of the \((N+1)^{st}\) derivative, so if we can find a bound for the \((N+1)^{st}\) derivative then we can get a bound on the error of the Taylor series.