Calculating Coefficients After Differentiation

Why Track the Coefficients?

A power series records a function through its coefficients and powers. Differentiation changes both: the coefficient of a positive power is multiplied by its exponent, and that power drops by one. Keeping those two changes aligned is essential when writing the derivative as a new power series.

The index shift can be easy to overlook. In particular, the first term of the derivative is determined by the coefficient of the original linear term, not by the original constant term. Sparse series require the same care: each present power shifts down by one, while missing powers remain missing.

The Coefficient Rule

Suppose a power series centered at $c$ is written with its index starting at zero. Let $R$ be its radius of convergence. The termwise differentiation theorem states that at every point satisfying $|x-c|<R$, the function represented by the series is differentiable and its derivative is given by differentiating the terms.

Theorem: If $f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n$ has radius of convergence $R$, then for $|x-c|<R$, $$ f'(x)=\sum_{n=1}^{\infty}n a_n(x-c)^{n-1} =\sum_{k=0}^{\infty}(k+1)a_{k+1}(x-c)^k. $$ Thus the coefficient of $(x-c)^k$ in the derivative series is $(k+1)a_{k+1}$.

The first expression follows by differentiating each term: the derivative of $a_n(x-c)^n$ is $n a_n(x-c)^{n-1}$ when $n\geq 1$, while the constant term has derivative zero. The second expression puts the result into the standard form with powers $(x-c)^k$. Set $k=n-1$, so $n=k+1$; as $n$ runs from $1$ upward, $k$ runs from $0$ upward. The coefficient attached to the new power $k$ is therefore $(k+1)a_{k+1}$.

This is an index change, not an additional differentiation rule. It is often useful to name the new coefficients: if $f'(x)=\sum_{k=0}^{\infty}b_k(x-c)^k$, then $b_k=(k+1)a_{k+1}$. The derivative's constant coefficient is $b_0=a_1$, because only the original linear term contributes a constant after differentiation.

Worked Examples

Worked Example: Shift the Index and Evaluate Inside the Radius

Differentiate $f(x)=\sum_{n=0}^{\infty}\frac{(x-1)^n}{4^n}$, then evaluate the derivative at $x=2$.

Here $c=1$ and $a_n=1/4^n$. This geometric power series has radius $R=4$, so the differentiation rule applies at $x=2$, since $|2-1|=1<4$. Differentiate termwise and then let $k=n-1$:

\begin{aligned} f'(x) &=\sum_{n=1}^{\infty}\frac{n}{4^n}(x-1)^{n-1}\\ &=\sum_{k=0}^{\infty}\frac{k+1}{4^{k+1}}(x-1)^k. \end{aligned}

At $x=2$, each factor $(x-1)^k$ equals $1$, so

f'(2)=\sum_{k=0}^{\infty}\frac{k+1}{4^{k+1}}=\frac{4}{9}.

To check, sum the original geometric series as $f(x)=1/(1-(x-1)/4)$ for $|x-1|<4$. Its derivative is $f'(x)=\frac14(1-(x-1)/4)^{-2}$. At $x=2$, this is $\frac14(3/4)^{-2}=\frac14\cdot\frac{16}{9}=\frac49$, agreeing with the coefficient calculation.

Worked Example: A Zero Constant Term After Differentiation

Let $h(x)=\sum_{n=2}^{\infty}(x-1)^n$. Find the derivative series and evaluate it at $x=3/2$.

This is a geometric tail in $t=x-1$, convergent when $|t|<1$. Since the series begins at $n=2$, there is no original linear term. Differentiating and setting $k=n-1$ gives

\begin{aligned} h'(x) &=\sum_{n=2}^{\infty}n(x-1)^{n-1}\\ &=\sum_{k=1}^{\infty}(k+1)(x-1)^k. \end{aligned}

The new index starts at $k=1$, not $k=0$: the derivative has no constant term. At $x=3/2$, $x-1=1/2$, which is strictly inside the convergence interval, and

h'\!\left(\frac32\right) =\sum_{k=1}^{\infty}\frac{k+1}{2^k} =\sum_{k=1}^{\infty}\frac{k}{2^k} +\sum_{k=1}^{\infty}\frac{1}{2^k} =2+1=3.

For the check, the geometric tail sums to $h(x)=t^2/(1-t)$, where $t=x-1$. Differentiating this expression with respect to $x$ (and hence to $t$) gives $h'(x)=(2t-t^2)/(1-t)^2$. At $t=1/2$, this equals $(1-1/4)/(1/4)=3$. The missing constant term is also consistent with the original series having no term proportional to $x-1$.

Worked Example: Differentiate a Series with Sparse Exponents

Consider $p(x)=\sum_{n=0}^{\infty}\frac{(x-2)^{3n+2}}{5^n}$. Find its derivative series and evaluate at $x=3$.

The exponents in the original series are $2,5,8,\ldots$, so differentiate a general term using its actual exponent $3n+2$:

p'(x)=\sum_{n=0}^{\infty}\frac{3n+2}{5^n}(x-2)^{3n+1}.

The derivative exponents are $1,4,7,\ldots$. They remain sparse; the derivative does not acquire terms with exponents $0,2,3,5,\ldots$. The original series is geometric in $(x-2)^3/5$, so it converges when $|x-2|<\sqrt[3]{5}$. At $x=3$, $|x-2|=1<\sqrt[3]{5}$, and the derivative series becomes

\begin{aligned} p'(3) &=\sum_{n=0}^{\infty}\frac{3n+2}{5^n}\\ &=3\sum_{n=0}^{\infty}\frac{n}{5^n} +2\sum_{n=0}^{\infty}\frac{1}{5^n}\\ &=3\left(\frac{5}{16}\right) +2\left(\frac{5}{4}\right) =\frac{15}{16}+\frac{40}{16} =\frac{55}{16}. \end{aligned}

For a check, write $t=x-2$. Summing the geometric series gives $p(x)=t^2/(1-t^3/5)$. Differentiation yields

p'(x)=\frac{2t(1-t^3/5)+(3t^4/5)}{(1-t^3/5)^2}.

At $x=3$, $t=1$, so the numerator is $2(4/5)+3/5=11/5$ and the denominator is $(4/5)^2=16/25$. Therefore $p'(3)=(11/5)/(16/25)=55/16$, as obtained from the sparse coefficient series.

A Reliable Indexing Procedure

When asked for coefficients after differentiation, separate the operation into two tasks: differentiate each term correctly, then decide how to label the resulting powers. This avoids confusing the original index with the new exponent.

Record the original exponent.
For a term $a_n(x-c)^n$, the exponent is $n$. In a sparse series, it may instead be an expression such as $3n+2$.

Differentiate the term.
Multiply its coefficient by its exponent and reduce that exponent by one. A constant term contributes zero.

Reindex only if useful.
For consecutive powers, set the new power index equal to the old exponent minus one. For sparse powers, retaining the original index may make the pattern clearer.

Check the domain and the first term.
Use the differentiated series only at points strictly inside the original radius. Verify that its constant coefficient comes from the original linear term.

A Common Indexing Mistake

A frequent error is to multiply each coefficient by its exponent but leave the power unchanged. For example, differentiating $x^2+x^3$ incorrectly as $2x^2+3x^3$ applies the coefficient change but forgets the exponent change. The correct calculation is

\frac{d}{dx}(x^2+x^3)=2x+3x^2.

The powers have each decreased by one. If the result is then written using a new index $k$, the coefficient of $x^k$ comes from the original term with exponent $k+1$, so it is $(k+1)a_{k+1}$, not $k a_k$. In sparse series, do not try to force the result into a consecutive pattern: differentiate the exponents that actually occur.

Evaluating at an Interior Point

Coefficient calculations often lead to a numerical value for a derivative. The point must satisfy the condition in the differentiation theorem: if the original series has center $c$ and radius $R$, require $|x_0-c|<R$. Then the derivative series converges to $f'(x_0)$, and substitution into its terms is justified.

This strict interior condition matters. A power series can behave differently at a boundary point, and the termwise differentiation theorem alone does not give a boundary conclusion. Endpoint behavior requires separate analysis. For an interior point, however, the coefficient rule provides a direct route from the original coefficients to the derivative's value.

Careful indexing is useful beyond a single derivative. It makes explicit which original coefficients determine the new ones, keeps zero and sparse terms visible, and allows numerical evaluation without losing track of the series' center. The essential check is simple: every original term of positive degree contributes its exponent times its coefficient, attached to a power one lower.

Check Your Understanding

Use the coefficient rule and state any convergence condition needed for an evaluation.

If $q(x)=\sum_{n=0}^{\infty}a_n(x-3)^n$, what is the coefficient of $(x-3)^4$ in $q'(x)$?
For $r(x)=\sum_{n=3}^{\infty}(x+1)^n$, does $r'(x)$ have a constant term? Explain from the original exponents.
Differentiate $\sum_{n=0}^{\infty}(x-2)^{2n+1}/3^n$ and describe the exponents that occur in the derivative.
Why must a point where the derivative series is evaluated lie strictly inside the original radius of convergence?
What is the indexing error in assigning the coefficient $k a_k$ to the power $(x-c)^k$ in the derivative?