Operations with Power Series: Learn It 4

Uniqueness of Power Series

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique?

In other words, if you find a power series for function [latex]f[/latex] at point [latex]a[/latex] using one method, could a different technique give you a completely different power series for the same function at the same point?

The answer is no—power series representations are unique. This shouldn’t be too surprising if you think of power series as “infinite polynomials.”

Intuitively, if two power series are equal:

[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots ={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\cdots[/latex]

for all values [latex]x[/latex] in some open interval [latex]I[/latex] around zero, then the coefficients must match: [latex]c_n = d_n[/latex] for all [latex]n \geq 0[/latex].

We now state this result formally in Uniqueness of Power Series.

theorem: uniqueness of power series

If two power series [latex]\sum_{n=0}^{\infty} c_n (x-a)^n[/latex] and [latex]\sum_{n=0}^{\infty} d_n (x-a)^n[/latex] are both convergent and satisfy:

[latex]\sum_{n=0}^{\infty} c_n (x-a)^n = \sum_{n=0}^{\infty} d_n (x-a)^n[/latex]

for all [latex]x[/latex] in an open interval containing [latex]a[/latex], then the coefficients are identical:

[latex]c_n = d_n[/latex] for all [latex]n \geq 0[/latex]

Proof

Let

[latex]\begin{array}{cc}\hfill f\left(x\right)& ={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+{c}_{3}{\left(x-a\right)}^{3}+\cdots \hfill \\ & ={d}_{0}+{d}_{1}\left(x-a\right)+{d}_{2}{\left(x-a\right)}^{2}+{d}_{3}{\left(x-a\right)}^{3}+\cdots .\hfill \end{array}[/latex]

 

Then [latex]f\left(a\right)={c}_{0}={d}_{0}[/latex]. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,

[latex]\begin{array}{cc}\hfill {f}^{\prime }\left(x\right)& ={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\cdots \hfill \\ & ={d}_{1}+2{d}_{2}\left(x-a\right)+3{d}_{3}{\left(x-a\right)}^{2}+\cdots ,\hfill \end{array}[/latex]

 

and thus, [latex]{f}^{\prime }\left(a\right)={c}_{1}={d}_{1}[/latex]. Similarly,

[latex]\begin{array}{cc}\hfill f^{\prime\prime}\left(x\right)& =2{c}_{2}+3\cdot 2{c}_{3}\left(x-a\right)+\cdots \hfill \\ & =2{d}_{2}+3\cdot 2{d}_{3}\left(x-a\right)+\cdots \hfill \end{array}[/latex]

 

implies that [latex]f^{\prime\prime} \left(a\right)=2{c}_{2}=2{d}_{2}[/latex], and therefore, [latex]{c}_{2}={d}_{2}[/latex]. More generally, for any integer [latex]n\ge 0,{f}^{\left(n\right)}\left(a\right)=n\text{!}{c}_{n}=n\text{!}{d}_{n}[/latex], and consequently, [latex]{c}_{n}={d}_{n}[/latex] for all [latex]n\ge 0[/latex].

[latex]_\blacksquare[/latex]

In this section, we’ve explored how to find power series representations using algebraic operations, differentiation, and integration. However, we’re still limited in the types of functions we can represent this way.

Next, we’ll dramatically expand our toolkit by introducing Taylor series—a method that allows us to find power series representations for many more functions.