11.9: Representation of Functions as a Power Series

Representing Functions as Power Series

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

Consider again the geometric series

\[1+x+x^2+x^3+\ldots =\sum_{n=0}^∞x^n \nonumber \]

Recall that the geometric series

\[a+ar+ar^2+ar^3+\ldots \nonumber \]

converges if and only if \(|r|<1.\) In that case, it converges to \(\dfrac{a}{1−r}\). Therefore, if \(|x|<1\), the series in Example \(\PageIndex{1}\) converges to \(\dfrac{1}{1−x}\) and we write

\[1+x+x^2+x^3+\ldots =\dfrac{1}{1−x} \text{ for }|x|<1 \nonumber \]

As a result, we are able to represent the function \(f(x)=\dfrac{1}{1−x}\) by the power series

\[1+x+x^2+x^3+\ldots \text{ when }|x|<1 \nonumber \]

We now show graphically how this series provides a representation for the function \(f(x)=\dfrac{1}{1−x}\) by comparing the graph of \(f\) with the graphs of several of the partial sums of this infinite series.

Illustration \(\PageIndex{1}\): Graphing a Function and Partial Sums of its Power Series

Below, we sketched a graph of \(f(x)=\dfrac{1}{1−x}\) and the graphs of the corresponding partial sums \( \displaystyle S_N(x)=\sum_{n=0}^Nx^n\) for \(N=2,4,6\) on the interval \((−1,1)\). 

From the graph in Figure \(\PageIndex{2}\), you see that as \(N\) increases, \(S_N\) becomes a better approximation for \(f(x)=\dfrac{1}{1−x}\) for \(x\) in the interval \((−1,1)\).

Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

Term‑by‑Term Differentiation and Integration of Power Series

The real power of representing a function by a power series lies in how easily we can differentiate or integrate the series. For a polynomial, we differentiate or integrate term by term. The same is true for power series inside their interval of convergence.

This means we can differentiate or integrate a known power series representation to obtain new power series for related functions without recomputing coefficients from scratch.  For example, we know

\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad \text{for } |x|<1 \nonumber \]

Differentiate both sides term-by-term:

\[ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} \nonumber \]

\[ \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right) = \sum_{n=0}^{\infty} \frac{d}{dx} \left(  x^{n} \right)= \sum_{n=0}^{\infty} n x^{n-1} \nonumber \]

Thus, we know that these quantities are equivalent:

\[ \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} nx^{n-1} \quad \text{for } |x|<1 \nonumber  \]

The series on the right doesn't look precisely like a power series. It is worth noting that the first term (when \(n=0\)) is \(0x^{-1}=0\), so we can start the sum at \(n=1\) without losing any terms. Now we can do substitution on the index, letting \(m=n-1\). Just like in an integral, this changes the variables from \(n-1\) into \(m\), and the limits of the sum from \(n=1\) to \(m=0\) and from \(n=\infty\) to \(m=\infty\).

\[ \sum_{n=0}^{\infty} n x^{n-1} = \sum_{n=1}^{\infty} n x^{n-1} = \sum_{m=0}^{\infty} (m+1) x^{m} \nonumber \]

Therefore, we have found a new power series relationship:

\[ \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1) x^n \quad \text{for } |x|<1 \nonumber  \]

Similarly, we can integrate both sides of our power series representation for \(\dfrac{1}{1-x}\) term-by-term:

\[ \int_0^x \frac{1}{1-t}\,dt = -\ln\left|1-t\right|\bigg|_0^x =-\ln|1-x| \nonumber \]

\[ \int_0^x \left( \sum_{n=0}^{\infty} t^n \right)\,dt = \sum_{n=0}^{\infty} \int t^n\,dt  =\sum_{n=0}^\infty \dfrac{1}{n+1}t^{n+1}\bigg|_0^x = \sum_{n=0}^\infty \dfrac{x^{n+1}}{n+1} \nonumber \]

After another re-indexing substitution and multiplying both sides by \(-1\), we see that:

\[ \ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}\quad \text{for } |x|<1 \nonumber  \] 

In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.

Key Concepts

• The geometric series \(\displaystyle \sum_{n=0}^∞x^n=\dfrac{1}{1−x}\) for \(|x|<1\) allows us to represent certain functions using geometric series.
• Inside its interval of convergence, a power series can be differentiated and integrated term‑by‑term, and the resulting series has the same radius of convergence.
• Differentiation and integration of known power series (like the geometric series) provide easy ways to obtain new series representations for related functions.

Key Equations

• Geometric Power Series center at \(x=0\)

\[ \sum_{n=0}^∞x^n=\dfrac{1}{1-x} \text{ when } -1<x<1 \nonumber \]

• Term‑by‑term differentiation (center at \(x=a\))

\[ \frac{d}{dx} \left( \sum_{n=0}^{\infty} c_n (x-a)^n \right) = \sum_{n=1}^{\infty} n c_n (x-a)^{n-1} \nonumber \]

• Term‑by‑term integration (center at \(x=a\))

\[ \int_a^x \left( \sum_{n=0}^{\infty} c_n (t-a)^n \right)\, dt = \sum_{n=0}^{\infty} c_n \frac{(x-a)^{n+1}}{n+1} \nonumber \]