Repeated Differentiation and Falling Factorials

Repeated Differentiation of a Power Series

A power series can be differentiated more than once, provided we remain strictly inside its radius of convergence. Each differentiation lowers the exponent of every surviving term by one and multiplies its coefficient by the exponent that was present. Repeating this process produces a product of consecutive integers called a falling factorial.

This product gives a direct formula for the \(m\)th derivative. It also explains a useful fact about evaluating at the center: every term vanishes except the term whose original degree is exactly \(m\).

Definition: For a nonnegative integer \(m\), the falling factorial of \(n\) of order \(m\) is \(n_{\underline m}=n(n-1)\cdots(n-m+1)\) when \(m\geq 1\), and \(n_{\underline 0}=1\). If \(n<m\), then the \(m\)th derivative of \((x-c)^n\) is zero.

The Repeated-Differentiation Formula

Let a power series centered at \(c\) be

f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n

and suppose its radius of convergence is \(R>0\). The termwise differentiation theorem states that inside the open interval \(|x-c|<R\), the series can be differentiated term by term; the resulting derivative series has the same radius of convergence. Applying that theorem repeatedly gives the formula below for every nonnegative integer \(m\).

Theorem: For \(|x-c|<R\) and any nonnegative integer \(m\),

f^{(m)}(x) = \sum_{n=m}^{\infty} n_{\underline m}\,a_n(x-c)^{n-m} = \sum_{n=m}^{\infty} \frac{n!}{(n-m)!}\,a_n(x-c)^{n-m}.

The sum begins at \(n=m\) because terms with \(n<m\) are polynomials of degree less than \(m\), so their \(m\)th derivatives are zero. For \(m=0\), the formula is the original power series.

The restriction \(|x-c|<R\) matters. The formula is guaranteed throughout the open interval of convergence, but it does not by itself assert convergence or differentiability at either endpoint. Endpoint behavior requires separate analysis.

Why the Falling Factorial Appears

Consider one term of the power series. Differentiating it once multiplies it by its exponent and lowers that exponent by one. Repeating this produces the successive factors \(n\), \(n-1\), and so on.

\begin{aligned} \frac{d}{dx}(x-c)^n &=n(x-c)^{n-1},\\ \frac{d^2}{dx^2}(x-c)^n &=n(n-1)(x-c)^{n-2},\\ \frac{d^3}{dx^3}(x-c)^n &=n(n-1)(n-2)(x-c)^{n-3}. \end{aligned}

For example, the third derivative of a quadratic term is zero: after two differentiations it is constant, and one more differentiation gives zero. This is why it is correct to start the \(m\)th-derivative sum at \(n=m\), rather than to retain terms with negative exponents.

Key takeaway: The coefficient multiplier for the \(m\)th derivative of a degree-\(n\) term is \(n_{\underline m}\). It is not \(n^m\): the factors decrease by one at each differentiation.

Derivation by Induction

The repeated-differentiation formula follows by induction on \(m\). The case \(m=0\) is the original series. For \(m=1\), it is the termwise differentiation theorem applied inside \(|x-c|<R\).

Now suppose the formula holds for some \(m\geq 1\). The \(m\)th derivative is itself a power series convergent inside the same radius, so it can be differentiated term by term there. Differentiating a term with \(n>m\) gives

\frac{d}{dx}\left(n_{\underline m}a_n(x-c)^{n-m}\right) = n_{\underline m}(n-m)a_n(x-c)^{n-m-1} = n_{\underline{m+1}}a_n(x-c)^{n-(m+1)}.

The term with \(n=m\) is constant in the \(m\)th derivative series, so its next derivative is zero. Consequently, the new series starts at \(n=m+1\), and every remaining coefficient has the falling-factorial multiplier of order \(m+1\). This proves the formula for \(m+1\), completing the induction.

What Happens at the Center?

Set \(x=c\) in the formula. For every term with \(n>m\), the factor \((x-c)^{n-m}\) becomes zero. The term with \(n=m\) has exponent zero and remains. Thus exactly one term contributes.

Theorem: If \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) has radius of convergence \(R>0\), then for every nonnegative integer \(m\),

f^{(m)}(c)=m!a_m.

This result is the center-evaluation case of the repeated-differentiation formula. It is also a useful check on computations: the coefficient of \((x-c)^m\), and no other coefficient, determines the \(m\)th derivative at \(c\).

Worked Example: A Second Derivative at the Center

Consider \(f(x)=\sum_{n=0}^{\infty}\frac{(x-1)^n}{n+1}\), which has radius of convergence \(1\). Find \(f''(1)\) using repeated differentiation.

Here the center is \(c=1\), and the coefficient of \((x-1)^n\) is \(a_n=\frac{1}{n+1}\). For \(m=2\), the formula gives

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

At \(x=1\), every term with \(n>2\) contains a positive power of \(x-1\) and vanishes. The \(n=2\) term is

f''(1)=2(1)\frac{1}{3}(1-1)^0=\frac{2}{3}.

Check using the original series: its first terms are \(1+\frac12(x-1)+\frac13(x-1)^2+\cdots\). The constant and linear terms have zero second derivative, while the quadratic term has second derivative \(2\cdot\frac13=\frac23\). This agrees with the formula.

Reindexing a Derivative Series

The formula is often easiest to use as written, with the original index \(n\). Reindexing can also make the new series easier to read. After \(m\) differentiations, setting \(j=n-m\) turns the exponent into \(j\), while the coefficient must be rewritten using \(n=j+m\). Do not change the exponent without making the corresponding change in the coefficient and summation index.

Worked Example: A Third Derivative of an Exponential-Type Series

Let \(f(x)=\sum_{n=0}^{\infty}\frac{2^n}{n!}(x+1)^n\). This series has infinite radius of convergence. Find its third derivative as a power series and evaluate it at the center.

Here \(c=-1\), and \(a_n=2^n/n!\). The third-derivative formula is

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

Since \(n!=n(n-1)(n-2)(n-3)!\), the coefficient simplifies to \(2^n/(n-3)!\). Let \(j=n-3\). Then \(n=j+3\), so

f'''(x) = \sum_{j=0}^{\infty}\frac{2^{j+3}}{j!}(x+1)^j.

At the center \(x=-1\), every term with \(j>0\) vanishes. The \(j=0\) term is \(2^3/0!=8\), so \(f'''(-1)=8\).

Check from the original series: its degree-three term is \(\frac{2^3}{3!}(x+1)^3=\frac43(x+1)^3\). Its third derivative is \(3!\cdot\frac43=8\); lower-degree terms differentiate to zero, and higher-degree terms contribute zero at \(x=-1\). The result checks.

Sparse Series and Missing Coefficients

A power series may have many zero coefficients. The center formula still applies without modification: if the coefficient \(a_m\) is zero, then the \(m\)th derivative at the center is zero. A sparse series makes the distinction between “terms present” and “terms that survive at the center” especially clear.

Worked Example: A Fourth Derivative of a Sparse Series

Consider

g(x)=\sum_{j=0}^{\infty}\frac{(x+2)^{2j}}{(j+1)4^{2j}}.

The center is \(-2\). Expanding the first terms gives

g(x)=1+\frac{(x+2)^2}{32}+\frac{(x+2)^4}{768}+\cdots.

There is no cubic term, and the coefficient of the fourth-degree term is \(1/768\). Therefore the fourth derivative at the center is

g^{(4)}(-2)=4!\frac{1}{768}=\frac{24}{768}=\frac{1}{32}.

To check, the constant and quadratic terms have fourth derivative zero. The fourth-degree term differentiates to the constant \(4!\cdot\frac{1}{768}=\frac{1}{32}\), and terms of degree greater than four still contain a positive power of \(x+2\), so they vanish at \(x=-2\). The coefficient formula gives the same result: the coefficient \(a_4\) is \(1/768\), and \(4!a_4=1/32\).

A Common Mistake: Using the Same Factor Repeatedly

When differentiating a degree-three term three times, a common error is to multiply its coefficient by \(3^3=27\). That uses the original exponent at every step. But the exponent decreases after each differentiation.

\begin{aligned} \frac{d}{dx}\bigl(a_3(x-c)^3\bigr) &=3a_3(x-c)^2,\\ \frac{d^2}{dx^2}\bigl(a_3(x-c)^3\bigr) &=3\cdot2a_3(x-c),\\ \frac{d^3}{dx^3}\bigl(a_3(x-c)^3\bigr) &=3\cdot2\cdot1a_3=6a_3. \end{aligned}

The incorrect calculation \(3^3a_3=27a_3\) fails because it does not account for the exponent falling from \(3\) to \(2\) to \(1\). The falling factorial \(3_{\underline 3}=3\cdot2\cdot1=6\) records the correct factors.

Using the Formula Effectively

For a derivative away from the center, retain the full series formula and ensure that the point lies strictly inside the radius of convergence. For a derivative at the center, use \(f^{(m)}(c)=m!a_m\) directly; this avoids differentiating every term separately. In either case, the lower-degree terms disappear after enough differentiations, and terms of higher degree vanish at the center.

Identify the center and radius.
Confirm that the point where the derivative is evaluated satisfies \(|x-c|<R\).

Choose the derivative order.
For order \(m\), use the multiplier \(n_{\underline m}\) and begin the sum at \(n=m\).

Evaluate or reindex carefully.
At \(x=c\), only \(a_m\) contributes. Away from \(c\), keep the powers and coefficients aligned if you change indices.

Check Your Understanding

Use the falling-factorial formula and the center-evaluation rule to answer these questions.

Write the formula for the fourth derivative of \(\sum_{n=0}^{\infty}a_n(x-c)^n\) inside its radius of convergence.
For \(p(x)=\sum_{n=0}^{\infty}a_n(x-2)^n\), what is \(p^{(5)}(2)\)?
Why does the \(m\)th-derivative series begin at \(n=m\)?
For \(q(x)=1+3(x+1)^2+2(x+1)^4+\cdots\), find \(q^{(4)}(-1)\), assuming the displayed terms give its coefficients through degree four.
What is the error in multiplying the coefficient of \((x-c)^4\) by \(4^3\) when finding its third derivative?