Why Differentiate a Power Series?
A power series is an infinite polynomial-like expression. When it represents a function, differentiating its terms can produce a new power series for the derivative. This is useful when the derivative is simpler to represent than the original function, or when a known series can be turned into a series for a related function.
The key condition is that the input must be strictly inside the original series’ radius of convergence. The term-by-term rule does not automatically apply at an endpoint, even if the original series converges there.
The Term-by-Term Differentiation Rule
Consider a power series centered at \(a\):
Its radius of convergence \(R\) is the number such that the series converges when \(|x-a|<R\) and diverges when \(|x-a|>R\). Behavior at \(|x-a|=R\) is not determined by the radius alone. If \(R>0\), the function represented by the series can be differentiated term by term for \(|x-a|<R\):
The \(n=0\) term is the constant \(c_0\), whose derivative is zero, so the differentiated sum begins at \(n=1\). The derivative of each remaining term follows from the power rule and the chain rule: \(\frac{d}{dx}(x-a)^n=n(x-a)^{n-1}\).
Why the Rule Works
For a finite polynomial, differentiating term by term is just the sum rule. For an infinite series, we must also justify that taking a limit of partial sums is compatible with differentiation. The power-series condition supplies that justification on every smaller closed interval strictly inside the radius.
Fix \(r<R\), and consider \(|x-a|\le r\). Choose \(s\) with \(r<s<R\). Since the original series converges at distance \(s\), its terms \(c_n s^n\) are bounded: there is a constant \(M\) such that \(|c_n|s^n\le M\) for all \(n\).
For \(|x-a|\le r\), the absolute value of the \(n\)-th differentiated term is bounded by \(n|c_n|r^{n-1}\le \frac{M}{s}n(r/s)^{n-1}\). The bounding series converges because \(r/s<1\).
The differentiated partial sums therefore converge uniformly on this smaller interval to a function \(g\). If \(S_N\) is the \(N\)-th partial sum, then \(S_N(x)-S_N(a)=\int_a^x S_N'(t)\,dt\). Taking limits gives \(f(x)-f(a)=\int_a^x g(t)\,dt\), where \(f\) is the function represented by the original series.
The differentiated series converges uniformly on the smaller interval, so its sum \(g\) is continuous there. Thus \(f(x)-f(a)=\int_a^x g(t)\,dt\) implies \(f'(x)=g(x)\). Since \(r<R\) can be chosen to include any point strictly inside the radius, the rule holds throughout \(|x-a|<R\).
This argument also explains why the endpoint is different: no smaller interval of the form \(|x-a|\le r\) can include a point at distance exactly \(R\). The interior justification does not settle what happens there.
Worked Example: Differentiate the Geometric Series
Worked Example: Differentiate the Geometric Series
For \(|x|<1\), the geometric series is \(\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\). Differentiate each term to find a series for \(\frac{1}{(1-x)^2}\).
The original series is centered at \(0\), has coefficients \(c_n=1\), and has radius \(R=1\). Term-by-term differentiation is therefore valid for \(|x|<1\):
Differentiate the function represented by the original series:
Consequently, for \(|x|<1\),
Check: At \(x=0\), the series has value \(1\), and the rational function has value \(\frac{1}{(1-0)^2}=1\). Also, the direct derivative calculation above confirms that the represented function is the derivative of \(\frac{1}{1-x}\). The result is asserted for \(|x|<1\); it makes no endpoint claim.
Worked Example: Differentiate a Logarithmic Series
Worked Example: Differentiate a Logarithmic Series
For \(|x|<1\), use the known logarithmic series \(\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\) to obtain a power series for the derivative of \(\ln(1-x)\).
Write the series with its negative sign outside, then differentiate its terms. The factor \(n\) from differentiating \(x^n\) cancels the denominator \(n\):
Re-indexing with \(k=n-1\) gives \(-\sum_{k=0}^{\infty}x^k\). The geometric series then yields
Direct differentiation provides an independent check:
Check: Both the term-by-term result and the direct derivative equal \(-1/(1-x)\). At \(x=0\), the differentiated series begins with \(-1\), matching the direct derivative \(-1/(1-0)=-1\). The differentiation is justified for \(|x|<1\), the interior of the logarithmic series’ radius.
Repeated Differentiation and Rational Functions
The same rule can be applied again to a differentiated power series, provided the input remains strictly inside the radius. Each application lowers the exponent and multiplies the coefficient by the exponent then present. This can produce series for higher derivatives and related rational functions.
Worked Example: Differentiate Twice to Represent a Rational Function
Starting from \(\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) for \(|x|<1\), differentiate twice to obtain a series for \(\frac{2}{(1-x)^3}\).
The first differentiation gives
Differentiate this series again. The term with \(n=1\) is the constant \(1\) and contributes zero; for \(n\ge 2\), the power rule gives
Now differentiate the rational function from the first step:
Therefore, for \(|x|<1\),
Check: The \(n=2\) term is \(2(1)x^0=2\). At \(x=0\), the series therefore gives \(2\), which matches \(2/(1-0)^3=2\). The direct differentiation also verifies the rational expression and its positive sign.
The Endpoint Pitfall
A common error is to assume that if the original power series converges at an endpoint, then its differentiated series must converge there too. The term-by-term differentiation theorem does not say this.
Worked Example: Why Endpoint Convergence Is Not Guaranteed
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n^2}\). At \(x=1\), the original series becomes \(\sum_{n=1}^{\infty}\frac{1}{n^2}\), which converges. But formally differentiating the terms gives \(\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}\). At \(x=1\), this becomes \(\sum_{n=1}^{\infty}\frac{1}{n}\), which diverges.
Thus convergence of the original series at \(x=1\) does not justify differentiating there. The theorem guarantees term-by-term differentiation only when \(|x|<1\). At an endpoint, convergence of the differentiated series and any equality with the derivative require separate analysis.
The practical rule is simple: first identify the radius of convergence, then apply term-by-term differentiation only at points strictly inside it. Do not use endpoint convergence of the original series as a substitute for that condition.
What Differentiation Changes
For a series centered at \(a\), differentiation changes the coefficients and powers but preserves the radius of convergence. In the original series, the \(n\)-th term is \(c_n(x-a)^n\); in the differentiated series it is \(n c_n(x-a)^{n-1}\). The multiplier \(n\) does not alter the radius, though it can change convergence at a boundary point.
| Feature | Original series | Differentiated series |
|---|---|---|
| General term | \(c_n(x-a)^n\) | \(n c_n(x-a)^{n-1}\), for \(n\ge 1\) |
| Constant term | \(c_0\) | Derivative is zero |
| Interior validity | \(|x-a|<R\) | Term-by-term equality holds for \(|x-a|<R\) |
| Endpoint behavior | Must be checked at each endpoint | Must also be checked separately |
Common Mistakes to Avoid
- Keeping the same exponent: Differentiating \(c_n(x-a)^n\) gives \(n c_n(x-a)^{n-1}\), not \(n c_n(x-a)^n\).
- Forgetting the constant term: The \(n=0\) term is constant, so it disappears from the derivative series.
- Dropping a chain-rule sign: For example, \(\frac{d}{dx}(1-x)^{-1}=+(1-x)^{-2}\), because the negative from the power rule and the negative from differentiating \(1-x\) cancel.
- Including an endpoint without justification: The theorem guarantees the term-by-term result inside the radius, not at its boundary.
Check Your Understanding
Answer these questions using the term-by-term rule and its conditions.
- Differentiate \(\sum_{n=0}^{\infty}c_n(x-a)^n\). What happens to the \(n=0\) term?
- For what values of \(x\) does the differentiation theorem guarantee equality between the derivative and the differentiated series?
- Differentiate the geometric series to write a power series for \(1/(1-x)^2\). State the interval where this representation is guaranteed.
- Why does the series \(\sum_{n=1}^{\infty}x^n/n^2\) not provide automatic justification for term-by-term differentiation at \(x=1\)?
- Differentiate \(\ln(1-x)=-\sum_{n=1}^{\infty}x^n/n\) term by term and check the result by direct differentiation.
