The Product Rule from Cauchy Product Coefficients

Why the Product Rule Has a Coefficient Version

When two functions are represented by power series, their product is represented by a new power series whose coefficients are formed by convolution. Differentiating that product changes the coefficient at each degree. The product rule says that the same derivative can also be found by differentiating one factor at a time and adding the results.

This tutorial derives that agreement directly from the coefficients. The key is to keep track of the index shift caused by differentiation: the coefficient of degree $n$ in the derivative of a series comes from the original coefficient of degree $n+1$, not degree $n$.

Central idea: The product rule for power-series functions follows because the degree-$n$ coefficient of the derivative of their Cauchy product equals the sum of the degree-$n$ coefficients in $f'g$ and $fg'$.

Cauchy Product and Differentiation

Let two power series centered at $a$ be

f(x)=\sum_{n=0}^{\infty}a_n(x-a)^n, \qquad g(x)=\sum_{n=0}^{\infty}b_n(x-a)^n.

Suppose their radii of convergence are $R_f$ and $R_g$, and write $R=\min(R_f,R_g)$. For $|x-a|<R$, both series converge. Their product is given by the Cauchy product

f(x)g(x)=\sum_{n=0}^{\infty}c_n(x-a)^n, \qquad c_n=\sum_{k=0}^{n}a_kb_{n-k}.

The coefficient $c_n$ is a finite sum: it collects all pairs of terms whose degrees add to $n$. For instance, the degree-two coefficient is $a_0b_2+a_1b_1+a_2b_0$.

Termwise differentiation of a power series is valid at every point strictly inside its radius of convergence, and the differentiated series has the same radius. Thus, on $|x-a|<R$, the derivatives and products below are represented by convergent power series. The coefficient of degree $n$ in the derivative of the product is $(n+1)c_{n+1}$, because

\frac{d}{dx}\left(\sum_{m=0}^{\infty}c_m(x-a)^m\right) = \sum_{m=1}^{\infty}m c_m(x-a)^{m-1} = \sum_{n=0}^{\infty}(n+1)c_{n+1}(x-a)^n.

Deriving the Coefficient Identity

We now compare that derivative coefficient with the coefficients in $f'g$ and $fg'$. Differentiating the individual series gives

f'(x)=\sum_{k=0}^{\infty}(k+1)a_{k+1}(x-a)^k, \qquad g'(x)=\sum_{k=0}^{\infty}(k+1)b_{k+1}(x-a)^k.

By the Cauchy product formula, the degree-$n$ coefficient of $f'g$ is $\sum_{k=0}^{n}(k+1)a_{k+1}b_{n-k}$. The degree-$n$ coefficient of $fg'$ is $\sum_{k=0}^{n}a_k(n-k+1)b_{n-k+1}$. Their sum is therefore

\sum_{k=0}^{n}(k+1)a_{k+1}b_{n-k} + \sum_{k=0}^{n}a_k(n-k+1)b_{n-k+1}.

To see why this is $(n+1)c_{n+1}$, start from the definition of $c_{n+1}$. Each term $a_jb_{n+1-j}$ can be multiplied by $n+1$ by splitting that factor into $j+(n+1-j)$. The first part collects the terms from $f'g$, and the second part collects the terms from $fg'$.

\begin{aligned} (n+1)c_{n+1} &=(n+1)\sum_{j=0}^{n+1}a_jb_{n+1-j}\\ &=\sum_{j=0}^{n+1}j a_jb_{n+1-j} +\sum_{j=0}^{n+1}(n+1-j)a_jb_{n+1-j}\\ &=\sum_{k=0}^{n}(k+1)a_{k+1}b_{n-k} +\sum_{k=0}^{n}a_k(n-k+1)b_{n-k+1}. \end{aligned}

In the last line, the $j=0$ term in the first sum and the $j=n+1$ term in the second sum are zero. Reindexing the remaining terms gives exactly the two convolution coefficients above. This proves the coefficient identity.

Differentiate the product series.
The degree-$n$ coefficient becomes $(n+1)c_{n+1}$.

Differentiate each factor.
Use the Cauchy product to find the degree-$n$ coefficients in $f'g$ and $fg'$.

Match the coefficients.
Split the factor $n+1$ in each term of $c_{n+1}$ into $j+(n+1-j)$; the two resulting sums are precisely those from $f'g$ and $fg'$.

Theorem: If $f$ and $g$ are represented by power series centered at $a$ with radii $R_f$ and $R_g$, then for every $|x-a|<\min(R_f,R_g)$, $$ (fg)'(x)=f'(x)g(x)+f(x)g'(x). $$ The equality follows because the two sides have convergent power-series representations with the same coefficient at every degree.

This argument gives more than a formal resemblance. It proves equality of the functions on the common interval of convergence: the power series for each side converge there, and their coefficients agree degree by degree. No assertion about endpoints is needed; the conclusion is stated only at points strictly inside both radii.

Worked Example: Multiplying Two Polynomials

Let $t=x-0$, $f(t)=1+t+t^2$, and $g(t)=1+2t+3t^2$. First compute the Cauchy product coefficients:

\begin{aligned} c_0&=1\cdot1=1,\\ c_1&=1\cdot2+1\cdot1=3,\\ c_2&=1\cdot3+1\cdot2+1\cdot1=6,\\ c_3&=1\cdot3+1\cdot2=5,\\ c_4&=1\cdot3=3. \end{aligned}

Therefore $fg=1+3t+6t^2+5t^3+3t^4$, so direct differentiation gives

$$ (fg)'=3+12t+15t^2+12t^3. $$

Now calculate each side of the product rule. Since $f'=1+2t$ and $g'=2+6t$,

\begin{aligned} f'g&=(1+2t)(1+2t+3t^2)\\ &=1+4t+7t^2+6t^3,\\ fg'&=(1+t+t^2)(2+6t)\\ &=2+8t+8t^2+6t^3. \end{aligned}

Adding gives $f'g+fg'=3+12t+15t^2+12t^3$, which agrees with the derivative of the product.

The coefficient identity can also be checked degree by degree. For degree $0$, the coefficients of $f'g$ and $fg'$ are $1$ and $2$, respectively, so their sum is $3=(0+1)c_1$. For degree $1$, those coefficients are $4$ and $8$, so their sum is $12=2c_2$. For degree $2$, they are $7$ and $8$, so their sum is $15=3c_3$. For degree $3$, they are $6$ and $6$, so their sum is $12=4c_4$. These are exactly the coefficients obtained by differentiating the Cauchy product.

Worked Example: An Infinite Cauchy Product

For $|t|<1$, consider the power series

f(t)=\sum_{n=0}^{\infty}t^n, \qquad g(t)=\sum_{n=0}^{\infty}(n+1)t^n.

Their radii of convergence are both $1$, so the coefficient calculations and product rule apply on $|t|<1$. The product coefficient at degree $n$ is

c_n=\sum_{k=0}^{n}1\cdot(n-k+1) =1+2+\cdots+(n+1) =\frac{(n+1)(n+2)}{2}.

Thus the degree-$n$ coefficient after differentiating the product is $(n+1)c_{n+1}=\frac{(n+1)(n+2)(n+3)}{2}$. Check the first three degrees against the two differentiated products. We have $f'(t)=\sum_{m=0}^{\infty}(m+1)t^m$, while $g'(t)=\sum_{m=0}^{\infty}(m+1)(m+2)t^m$.

\begin{array}{c|c|c|c} n & [t^n](f'g) & [t^n](fg') & (n+1)c_{n+1}\\ \hline 0 & 1 & 2 & 3\\ 1 & 2+2=4 & 6+2=8 & 12\\ 2 & 3+4+3=10 & 12+6+2=20 & 30 \end{array}

For example, in degree $1$, the coefficient of $f'g$ is $1\cdot2+2\cdot1=4$, and the coefficient of $fg'$ is $1\cdot6+1\cdot2=8$. Their sum is $12=2c_2$, since $c_2=6$. In degree $2$, the corresponding sums are $1\cdot3+2\cdot2+3\cdot1=10$ and $1\cdot12+1\cdot6+1\cdot2=20$, adding to $30=3c_3$, since $c_3=10$. The degree-zero entries give $1+2=3=c_1$. Thus all three computed coefficients agree with differentiation of the product series.

Worked Example: Coefficient Cancellation

Let, for $|t|<1$,

f(t)=\sum_{n=0}^{\infty}t^n, \qquad g(t)=\sum_{n=0}^{\infty}(-1)^nt^n.

Here $a_k=1$ and $b_k=(-1)^k$. The Cauchy product coefficient is

c_n=\sum_{k=0}^{n}(-1)^{n-k}.

For even $n$, this alternating sum has an odd number of terms and equals $1$. For odd $n$, it has an even number of terms and equals $0$. Hence the product series is $1+t^2+t^4+\cdots$. Its derivative is $2t+4t^3+6t^5+\cdots$.

Check the first four derivative coefficients using the two products. Since $f'$ has coefficient $k+1$ at degree $k$, and $g'$ has coefficient $-(k+1)(-1)^k$ at degree $k$, the degree-zero coefficients are $1$ in $f'g$ and $-1$ in $fg'$, giving $0=c_1$. In degree one, the coefficients are $-1+2=1$ and $2-1=1$; their sum is $2=2c_2$. In degree two, they are $1-2+3=2$ and $-3+2-1=-2$; their sum is $0=3c_3$. In degree three, they are $-1+2-3+4=2$ and $4-3+2-1=2$; their sum is $4=4c_4$. These match the coefficients $0,2,0,4$ of the differentiated product.

This example shows why it is useful to compare coefficients carefully: convolution can create cancellations in the product even though neither original series has zero coefficients.

A Common Indexing Mistake

A frequent error is to say that the degree-$n$ coefficient in the derivative of $\sum c_n t^n$ is $n c_n$. Differentiation lowers each exponent by one, so that coefficient actually comes from the original term of degree $n+1$. The correct coefficient is $(n+1)c_{n+1}$.

For a concrete check, the polynomial example has $c_0=1$ and $c_1=3$. The constant term in the derivative is $1\cdot c_1=3$. The incorrect expression $0\cdot c_0$ would give $0$, contradicting the direct derivative of $1+3t+6t^2+5t^3+3t^4$. Keeping the shifted index visible prevents this error.

Takeaway: First write the differentiated series with a fresh degree index. Then the coefficient at degree $n$ is visibly $(n+1)c_{n+1}$, which can be compared with the two convolution sums.

When the Coefficient Argument Applies

The argument relies on the power series being used strictly inside their radii of convergence. There, the individual series, their differentiated series, and the Cauchy products used in the calculation all converge, and their coefficients represent the corresponding functions. The theorem therefore gives the ordinary derivative of the product on the common interior interval.

At an endpoint of a convergence interval, these interior convergence facts do not by themselves justify termwise differentiation or the same coefficient argument. The result above makes no endpoint claim. For the usual use of the product rule, the practical procedure is to identify the common open interval first, then perform the coefficient calculation there.

The coefficient proof matters because it connects two ways of organizing the same operation. Multiplication combines degrees by convolution; differentiation weights a term by its degree and shifts its exponent. Splitting the weight $n+1$ into the two contributing degree parts shows exactly why differentiating a product yields one term for each factor.

Check Your Understanding

Use the coefficient identities and convergence conditions developed above to answer these questions.

For two series centered at $a$, what is the Cauchy product coefficient $c_n$?
Why is the degree-$n$ coefficient of the derivative $(n+1)c_{n+1}$, rather than $nc_n$?
Write the degree-$n$ coefficient in $f'g$ in terms of $a_k$ and $b_k$.
What condition on $x$ ensures the coefficient proof applies to both power series?
In the cancellation example, calculate $c_5$ and state the coefficient of $t^4$ in the derivative of the product.

The Product Rule from Cauchy Product Coefficients

What You'll Learn

Why the Product Rule Has a Coefficient Version

Cauchy Product and Differentiation

Deriving the Coefficient Identity

Worked Example: Multiplying Two Polynomials

Worked Example: Multiplying Two Polynomials

Worked Example: An Infinite Cauchy Product

Worked Example: An Infinite Cauchy Product

Worked Example: Coefficient Cancellation

Worked Example: Coefficient Cancellation

A Common Indexing Mistake

When the Coefficient Argument Applies

Check Your Understanding

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