A Use Differentiation To Find A Power Series Representation For

a usedifferentiation to find a power series representation for

Introduction

In calculus, a power series is an infinite sum of the form [ \sum_{n=0}^{\infty} c_n (x-a)^n , ]

which converges to a function (f(x)) on an interval around the centre (a). One of the most powerful techniques for constructing such series is differentiation: if we already know a power‑series expansion for a simple function, we can differentiate that series term‑by‑term to obtain the series for its derivative. Conversely, integrating a known series yields the series for an antiderivative.

This article explains how differentiation can be employed to find a power‑series representation for a given function, outlines the theoretical justification, walks through a step‑by‑step procedure, provides concrete examples, highlights common pitfalls, and answers frequently asked questions. By the end, you should feel comfortable applying differentiation (and its counterpart, integration) to build series expansions for a wide variety of elementary functions.

Detailed Explanation

Why differentiation works

A power series that converges on an open interval (|x-a|<R) defines a function that is infinitely differentiable on that interval. Inside the radius of convergence, the series can be differentiated term‑by‑term, and the resulting series has the same radius (R). Formally, if

[ f(x)=\sum_{n=0}^{\infty} c_n (x-a)^n \quad\text{for }|x-a|<R, ]

then

[ f'(x)=\sum_{n=1}^{\infty} n c_n (x-a)^{n-1} =\sum_{n=0}^{\infty} (n+1)c_{n+1}(x-a)^{n}, ]

and the new series also converges for (|x-a|<R). The same principle holds for higher‑order derivatives and for integration (which introduces a constant of integration).

Thus, if we can start from a known series—most commonly the geometric series

[\frac{1}{1-x}= \sum_{n=0}^{\infty} x^n \qquad (|x|<1), ]

or its variants—we can obtain series for functions that are derivatives or integrals of (\frac{1}{1-x}) (or similar building blocks) by applying differentiation or integration term‑by‑term. ### General strategy

1. Identify a base function whose power series is known and whose derivative (or antiderivative) resembles the target function.
2. Write the known series in summation notation, explicitly stating its interval of convergence.
3. Differentiate (or integrate) the series term‑by‑term, adjusting indices as needed.
4. Simplify the resulting expression to match the desired function, possibly factoring out constants or re‑indexing.
5. State the final power‑series representation and its radius of convergence (which remains unchanged unless the operation introduces a singularity at the centre).

This method is especially useful for functions like (\frac{1}{(1-x)^2}), (\ln(1+x)), (\arctan x), and many rational combinations thereof.

Step‑by‑Step or Concept Breakdown

Below is a detailed walk‑through using the example (f(x)=\frac{1}{(1-x)^2}).

Step 1: Choose a known series

We know the geometric series

[ \frac{1}{1-x}= \sum_{n=0}^{\infty} x^n ,\qquad |x|<1 . ]

Step 2: Relate the target to a derivative

Observe that

[ \frac{d}{dx}\Bigl(\frac{1}{1-x}\Bigr)=\frac{1}{(1-x)^2}. ]

Thus, differentiating the known series will give us the series for (f(x)). ### Step 3: Differentiate term‑by‑term

Differentiate the right‑hand side:

[ \frac{d}{dx}\Bigl[\sum_{n=0}^{\infty} x^n\Bigr] =\sum_{n=0}^{\infty} \frac{d}{dx}\bigl(x^n\bigr) =\sum_{n=1}^{\infty} n x^{,n-1}. ]

(The (n=0) term vanishes because its derivative is zero.)

Step 4: Re‑index if desired

Let (m=n-1) (so (n=m+1)). Then

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

Renaming the dummy index back to (n) yields

[ \frac{1}{(1-x)^2}= \sum_{n=0}^{\infty} (n+1) x^{n},\qquad |x|<1 . ]

Step 5: Verify the radius of convergence

Since we only differentiated a series that converged for (|x|<1), the resulting series also converges for (|x|<1). No new singularities were introduced, so the radius remains (R=1).

This same pattern can be applied repeatedly: differentiating (\frac{1}{(1-x)^2}) yields (\frac{2}{(1-x)^3}), integrating (\frac{1}{1+x}) gives (\ln(1+x)), etc.

Real Examples

Example 1: Power series for (\ln(1+x))

We start from the geometric series for (\frac{1}{1+ x}):

[ \frac{1}{1+x}= \sum_{n=0}^{\infty} (-1)^n x^n ,\qquad |x|<1 . ]

Notice that [ \frac{d}{dx}\ln(1+x)=\frac{1}{1+x}. ]

Therefore, integrating the series term‑by‑term gives

[ \ln(1+x)= \int \frac{1}{1+x},dx = \int \sum_{n=0}^{\infty} (-1)^n x^n ,dx = \sum_{n=0}^{\infty} (-1)^n \int x^n ,dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}+C . ]

Since (\ln(1+0)=0), the constant (C=0). Re‑index with (k=n+1):

[ \boxed{\displaystyle \ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^{k}}{k}},\qquad |x|<1 . ]

(The series also converges at (x=1)