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Power Series Operations · Tutorial 757 of 998

The Termwise Differentiation Theorem

Learn why a power series may be differentiated term by term at every point strictly inside its radius of convergence.

Intermediate 10 min read

What You'll Learn

  • State the termwise differentiation theorem for a power series with center c and radius R.
  • Explain how uniform convergence on a closed interval inside the radius supports differentiation.
  • Follow a proof that the differentiated series represents the derivative of the original sum.
  • Differentiate geometric and sparse power series term by term.
  • Recognize why the theorem gives no automatic conclusion at an endpoint.

Why Termwise Differentiation Needs a Theorem

A power series is an infinite sum of polynomial terms. For a finite sum, differentiating term by term is immediate. For an infinite sum, however, convergence of the series alone does not automatically justify differentiating its terms. We need a result that identifies where the operation is valid and explains why.

The key condition is that the point lies strictly inside the radius of convergence. There, the original series and the series formed by differentiating its terms have enough uniform convergence on small closed intervals for the derivative of the sum to be established.

Definition: A power series centered at \(c\) is a series of the form \(\sum_{n=0}^{\infty}a_n(x-c)^n\), where \(a_n\) are real coefficients. Its radius of convergence \(R\) is the number such that the series converges absolutely when \(|x-c|<R\), diverges when \(|x-c|>R\), and may or may not converge at the endpoints when \(R\) is finite.

Here \(R\) may be finite or infinite. The theorem below concerns points satisfying \(|x-c|<R\), not the endpoints \(|x-c|=R\).

The Termwise Differentiation Theorem

Theorem: Suppose the power series \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) has radius of convergence \(R>0\). For every \(x\) with \(|x-c|<R\), the sum \(f\) is differentiable and $$ f'(x)=\sum_{n=1}^{\infty}n a_n(x-c)^{n-1}. $$ The differentiated series converges uniformly on every closed interval \([c-r,c+r]\) with \(0<r<R\). In particular, it represents the derivative throughout the open interval \((c-R,c+R)\).

The derivative of the \(n=0\) term is zero, which is why the differentiated series begins at \(n=1\). The theorem does not claim that the same formula works at an endpoint of the interval of convergence.

Why the Theorem Works

Fix any number \(r\) with \(0<r<R\), and choose \(s\) such that \(r<s<R\). The original series converges absolutely at \(x=c+s\), so the numerical series \(\sum_{n=0}^{\infty}|a_n|s^n\) converges. We use this convergence to control both the original series and its termwise derivative uniformly on \([c-r,c+r]\).

1
Control the original terms.
For \(|x-c|\le r\), we have \(|a_n(x-c)^n|\le |a_n|s^n\), since \(r<s\). The tails are therefore bounded by tails of the convergent numerical series \(\sum |a_n|s^n\). Thus the original power series converges uniformly on \([c-r,c+r]\).
2
Control the differentiated terms.
For \(n\ge1\) and \(|x-c|\le r\), the absolute value of the \(n\)th differentiated term is bounded as follows: $$ |n a_n(x-c)^{n-1}| \le |a_n|s^n\frac{n(r/s)^{n-1}}{s}. $$ The factor \(n(r/s)^{n-1}/s\) is bounded as \(n\) varies, because \(0<r/s<1\). Hence the differentiated terms are bounded by a constant times \(|a_n|s^n\). The comparison with the convergent numerical series proves uniform convergence of the differentiated series on this interval.
3
Pass from polynomial derivatives to the derivative of the sum.
Let \(P_N(x)=\sum_{n=0}^{N}a_n(x-c)^n\) be the \(N\)th partial sum. The polynomials \(P_N\) converge uniformly to \(f\), and their derivatives converge uniformly to a function \(g\), the sum of the differentiated series. Since each \(P_N'\) is continuous, their uniform limit \(g\) is continuous.

For completeness, the last step that identifies \(g\) as \(f'\) follows from the Mean Value Theorem. Fix an interior point \(x_0\), and take \(h\ne0\) small enough that the interval between \(x_0\) and \(x_0+h\) lies inside \([c-r,c+r]\). For each \(N\), the Mean Value Theorem applied to \(P_N\) gives a point \(\xi_N\) between \(x_0\) and \(x_0+h\) such that

$$ \frac{P_N(x_0+h)-P_N(x_0)}{h}=P_N'(\xi_N). $$

Uniform convergence of \(P_N\) and \(P_N'\) lets us pass to the limit in this equality. The left side tends to \(\bigl(f(x_0+h)-f(x_0)\bigr)/h\). On the right, uniform convergence of \(P_N'\) to \(g\) means the difference between \(P_N'(\xi_N)\) and \(g(\xi_N)\) tends to zero, regardless of the choice of \(\xi_N\). Consequently,

$$ \left|\frac{f(x_0+h)-f(x_0)}{h}-g(x_0)\right| \le \sup_{t\text{ between }x_0\text{ and }x_0+h}|g(t)-g(x_0)|. $$

Because \(g\) is continuous at \(x_0\), the right side tends to zero as \(h\to0\). Therefore the difference quotient for \(f\) tends to \(g(x_0)\). This proves that \(f'(x_0)=g(x_0)\), exactly the sum of the differentiated series. Since \(x_0\) was arbitrary inside the radius, the theorem holds throughout the open interval.

Worked Examples

Worked Example: Differentiating the Geometric Power Series

For \(|x|<1\), the geometric series has sum \(f(x)=\sum_{n=0}^{\infty}x^n=1/(1-x)\). Its center is \(0\) and its radius of convergence is \(1\). The theorem applies at every point in this open interval.

Differentiate each term and start the new sum at \(n=1\):

$$ f'(x)=\sum_{n=1}^{\infty}n x^{n-1}. $$

Since \(f(x)=1/(1-x)\), direct differentiation gives

$$ \frac{d}{dx}\left(\frac{1}{1-x}\right) =\frac{d}{dx}(1-x)^{-1} =-(1-x)^{-2}(-1) =\frac{1}{(1-x)^2}. $$

Thus, for \(|x|<1\),

$$ \sum_{n=1}^{\infty}n x^{n-1}=\frac{1}{(1-x)^2}. $$

Check: Differentiating the stated sum \(1/(1-x)\) gives \(1/(1-x)^2\), matching the termwise derivative. For example, at \(x=1/2\), the derivative is \(1/(1-1/2)^2=4\).

Worked Example: A Power Series with Center 2

Consider \(f(x)=\sum_{n=0}^{\infty}\bigl((x-2)/3\bigr)^n\). It is geometric with ratio \((x-2)/3\), so it converges when \(|x-2|<3\), and its radius is \(3\). Within that interval, its sum is

$$ f(x)=\frac{1}{1-(x-2)/3} =\frac{1}{(5-x)/3} =\frac{3}{5-x}. $$

Termwise differentiation gives

$$ f'(x)=\sum_{n=1}^{\infty}\frac{n(x-2)^{n-1}}{3^n}. $$

To check the result, differentiate the closed form:

$$ \frac{d}{dx}\left(\frac{3}{5-x}\right) =3(5-x)^{-2} =\frac{3}{(5-x)^2}. $$

Therefore, for \(|x-2|<3\), the differentiated series sums to \(3/(5-x)^2\). At \(x=1\), for instance, its value is \(3/(5-1)^2=3/16\). The series expression at that point is \(\sum_{n=1}^{\infty} n(-1)^{n-1}/3^n\), and the theorem identifies that sum with \(3/16\).

Check: Differentiating \(3/(5-x)\) gives \(3/(5-x)^2\), which agrees with the function represented by the termwise derivative.

Worked Example: Differentiating a Sparse Power Series

Consider

$$ f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x+1)^{2n}}{4^n}. $$

This is a geometric series in the quantity \(-(x+1)^2/4\). It converges when \((x+1)^2/4<1\), or \(|x+1|<2\), so its radius about the center \(-1\) is \(2\). On this open interval, its sum is

$$ f(x)=\frac{1}{1+(x+1)^2/4} =\frac{4}{4+(x+1)^2}. $$

Differentiate each term. The \(n=0\) term is constant, so its derivative is zero:

$$ f'(x)=\sum_{n=1}^{\infty}\frac{(-1)^n\,2n(x+1)^{2n-1}}{4^n}. $$

Differentiate the closed form to check the result:

$$ \frac{d}{dx}\left(\frac{4}{4+(x+1)^2}\right) =4\bigl(4+(x+1)^2\bigr)^{-2}\bigl(-2(x+1)\bigr) =\frac{-8(x+1)}{\bigl(4+(x+1)^2\bigr)^2}. $$

Hence the differentiated sparse series represents this derivative for \(|x+1|<2\). At \(x=0\), the value is \(-8/(4+1)^2=-8/25\). The series evaluated there is \(\sum_{n=1}^{\infty}(-1)^n2n/4^n\), so it too equals \(-8/25\).

Check: The derivative of the displayed closed form is \(-8(x+1)/(4+(x+1)^2)^2\), exactly the expression identified with the differentiated series.

A Common Mistake: Applying the Result at an Endpoint

The strict inequality \(|x-c|<R\) is essential. For example, the series \(\sum_{n=1}^{\infty}x^n/n^2\) converges at \(x=1\), because there it becomes \(\sum_{n=1}^{\infty}1/n^2\). A mistaken argument might say that its derivative at \(1\) is obtained by differentiating every term and substituting \(x=1\):

$$ \sum_{n=1}^{\infty}\frac{x^n}{n^2} \quad\stackrel{\text{incorrect at }x=1}{\longrightarrow}\quad \sum_{n=1}^{\infty}\frac{x^{n-1}}{n}, \qquad \sum_{n=1}^{\infty}\frac{1}{n}. $$

The final series diverges, and the termwise differentiation theorem says nothing about differentiability at \(x=1\): that point is an endpoint, not strictly inside the radius \(1\). Convergence of the original series at an endpoint does not make termwise differentiation there valid. The correction is to apply the theorem only for \(|x|<1\), and to analyze endpoint behavior separately if needed.

Why the Result Matters

Inside its radius of convergence, a power series is not only a way to represent a function; it also provides a series representation for the function's derivative. This makes it possible to obtain derivative formulas directly from coefficients and to study a function locally through its power series.

The theorem gives two closely related guarantees: the differentiated series converges at every point strictly inside the original radius, and its sum is the derivative of the original function there. Neither guarantee should be extended to boundary points without a separate argument.

Check Your Understanding

Use the theorem and its hypotheses to answer these questions.

  1. For a power series centered at \(c\) with radius \(R\), which points are covered by the termwise differentiation theorem?
  2. Why does the proof choose numbers \(r<s<R\) when controlling the differentiated terms?
  3. What is the termwise derivative of \(\sum_{n=0}^{\infty}a_n(x-c)^n\)? Why does its sum begin at \(n=1\)?
  4. What role does uniform convergence of the differentiated partial sums play in the proof?
  5. Does convergence of a power series at an endpoint by itself justify termwise differentiation there? Explain.

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