Why Product Coefficients Require Care
A coefficient in a product of power series is built from all pairs of terms whose powers add to the target power. This is easy to overlook: the coefficient of degree \(n\) is generally not obtained by multiplying the two coefficients of degree \(n\), nor by multiplying terms with matching indices.
The useful feature is that only finitely many pairs can contribute to any one specified coefficient. We can therefore calculate that coefficient directly, without expanding an entire product or finding a closed form for it.
Write both series about the same center \(c\), and use \(t=x-c\) to keep track of powers:
When both series converge absolutely at \(x\), their product is represented by the Cauchy product. The coefficient at degree \(n\) is a finite sum, called a convolution sum.
The formula applies to power series with a common center, at points where both series converge absolutely. For power series, this condition is satisfied at every point strictly inside both radii of convergence. The coefficient formula itself involves only the finitely many indices \(0,1,\ldots,n\).
How to Read the Convolution Sum
To find \(\gamma_n\), list the possible ways to split the target exponent \(n\) into two nonnegative exponents. For each split \(k+(n-k)=n\), multiply the corresponding coefficients \(a_k b_{n-k}\), then add the products. A compact layout is often helpful:
The first factor’s index increases from \(0\) to \(n\), while the second factor’s index decreases from \(n\) to \(0\). Every possible split appears exactly once. A coefficient that is absent from a displayed series is treated as zero; this convention makes the formula work for sparse series as well as dense ones.
Worked Example: A Coefficient from Two Dense Series
Suppose, about the center \(c\), the coefficients through degree \(3\) are \[ A(x)=1+2t-3t^2+t^3+\cdots,\qquad B(x)=4-t+2t^2+5t^3+\cdots, \] where \(t=x-c\). Find the coefficient of \(t^3\) in the product.
For degree \(3\), the convolution sum has four terms: \[ \gamma_3=a_0b_3+a_1b_2+a_2b_1+a_3b_0. \] Insert the coefficients from the two series: \[ \gamma_3=(1)(5)+(2)(2)+(-3)(-1)+(1)(4). \] Compute each contribution: \(1\cdot5=5\), \(2\cdot2=4\), \((-3)(-1)=3\), and \(1\cdot4=4\). Therefore \[ \gamma_3=5+4+3+4=16. \]
Check by multiplying only the displayed terms that can affect degree \(3\): \(1\cdot5t^3\), \(2t\cdot2t^2\), \((-3t^2)(-t)\), and \(t^3\cdot4\) contribute \(5t^3+4t^3+3t^3+4t^3=16t^3\). Higher-degree terms cannot contribute to degree \(3\), so the coefficient is indeed \(16\).
Why the Sum Is Finite
The coefficient formula can be understood by multiplying finite truncations. Fix a nonnegative integer \(n\) and retain terms only through degree \(n\): \[ A_n(t)=\sum_{k=0}^{n}a_kt^k, \qquad B_n(t)=\sum_{j=0}^{n}b_jt^j. \] In their product, a term \(a_kb_jt^{k+j}\) contributes to degree \(n\) precisely when \(k+j=n\). Since \(k\) and \(j\) are nonnegative integers, the possible pairs are exactly \((0,n),(1,n-1),\ldots,(n,0)\). Adding their contributions gives the convolution formula.
Terms of either original series with exponent greater than \(n\) cannot affect the coefficient of \(t^n\): multiplying them by a term with a nonnegative exponent produces a power greater than \(n\). This also explains why a coefficient calculation can be completed from finite information, even when both series continue indefinitely.
For convergent power series, the Cauchy product theorem additionally justifies that this coefficient calculation represents the actual product function wherever both series converge absolutely. The finite index argument explains the coefficient rule; absolute convergence is the condition that permits the product of the infinite series to be represented by the resulting series.
Using Sparsity to Reduce the Work
A sparse series has many zero coefficients. The convolution formula still runs from \(k=0\) to \(k=n\), but there is no need to calculate terms that contain a zero. First identify which exponents actually occur in each factor, then keep only pairs whose exponents sum to the target degree.
Worked Example: A Sparse Factor
Let \[ A(x)=2+3t^2-t^5,\qquad B(x)=1-2t+4t^3+5t^5, \] where \(t=x-c\). Find the coefficient of \(t^5\) in \(A(x)B(x)\).
The nonzero exponents in \(A\) are \(0,2,5\). For a degree-\(5\) product term, the matching exponent in \(B\) must be \(5,3,0\), respectively. Thus the only possible contributions are \[ a_0b_5+a_2b_3+a_5b_0 =(2)(5)+(3)(4)+(-1)(1). \] Evaluating gives \(10+12-1=21\). The coefficient of \(t^5\) is \(21\).
Check by multiplying the terms that can reach degree \(5\): \(2\cdot5t^5=10t^5\), \(3t^2\cdot4t^3=12t^5\), and \((-t^5)\cdot1=-t^5\). Their sum is \(21t^5\), as found. For comparison, the coefficient of \(t^4\) is zero: the exponents \(0,2,5\) in \(A\) would require exponents \(4,2,-1\) in \(B\), and none gives a nonzero contribution.
This example illustrates two different reasons a term in a convolution may contribute nothing. A required coefficient might be zero because that power is absent; alternatively, the proposed exponent split might be impossible, as with a negative required exponent. In either case, the corresponding contribution is zero.
Products with Vanishing Initial Coefficients
Sometimes each factor begins with one or more zero coefficients. It helps to identify the first exponent present in each factor before calculating a target coefficient. If the first nonzero term of one factor has degree \(r\), and that of the other has degree \(s\), every product term has degree at least \(r+s\). Consequently, all coefficients below degree \(r+s\) vanish.
Worked Example: Finding the First Possible Nonzero Coefficient
Consider \[ A(x)=t^2(1+2t+3t^2+\cdots),\qquad B(x)=t^3(2-t+4t^2+\cdots). \] Find the coefficients of degrees \(4,5,6,\) and \(7\) in the product, and explain why the earlier coefficients vanish.
The first factor has no terms below degree \(2\); the second has no terms below degree \(3\). A product term must therefore have degree at least \(2+3=5\). There can be no contribution to degree \(4\), so \(\gamma_4=0\).
For degree \(5\), the only available split is \(2+3\). The coefficients are \(1\) and \(2\), giving \(\gamma_5=(1)(2)=2\).
For degree \(6\), the possible splits are \(2+4\) and \(3+3\). The corresponding coefficients are \(1,-1\) and \(2,2\), so \[ \gamma_6=(1)(-1)+(2)(2)=-1+4=3. \]
For degree \(7\), the possible splits are \(2+5\), \(3+4\), and \(4+3\). The coefficients are \(1,4\), \(2,-1\), and \(3,2\), respectively. Hence \[ \gamma_7=(1)(4)+(2)(-1)+(3)(2)=4-2+6=8. \]
Check by multiplying the terms through the needed degrees: \[ A(x)B(x) =t^5(1+2t+3t^2+\cdots)(2-t+4t^2+\cdots). \] The coefficients of the first three powers inside the parentheses are \(2\), \(-1+4=3\), and \(4-2+6=8\). Thus the product begins \(2t^5+3t^6+8t^7+\cdots\); it has no \(t^4\) term, in agreement with the convolution calculations.
The same formula handles vanishing caused by cancellation rather than by a lack of possible exponent pairs. For example, if the nonzero contributions to one coefficient are \(6\) and \(-6\), their sum is zero. Therefore, a zero product coefficient may mean either that no nonzero pair of terms can reach that degree or that available contributions cancel.
A Common Indexing Mistake
A frequent error is to pair coefficients with the same index. That would use \(a_kb_k\), but the powers in such a product add to \(2k\), not to a fixed target degree in general. The correct pairing for degree \(n\) is \(a_kb_{n-k}\).
For the coefficient of \(t\), the valid exponent splits are \(0+1\) and \(1+0\). Thus \[ \gamma_1=a_0b_1+a_1b_0=(1)(-1)+(2)(4)=-1+8=7. \] Direct multiplication confirms that the degree-one terms are \(-t\) from \(1\cdot(-t)\) and \(8t\) from \(2t\cdot4\), which combine to \(7t\). Always check that the two exponents in each proposed contribution add to the target degree.
A Reliable Calculation Procedure
For a specified product coefficient, organize the work before doing arithmetic:
Write down the exponent \(n\) whose coefficient is required.
Use \(k+(n-k)=n\), with both exponents nonnegative.
For sparse factors, retain only pairs for which both coefficients are nonzero.
Compute each pair’s contribution with its sign, then sum the results.
Confirm that every contributing pair of powers adds to \(n\), and verify the result by multiplying just the relevant terms.
Computing coefficients this way is useful whenever a product is needed only to a particular order. It avoids unnecessary expansion, makes sparse factors manageable, and reveals structural zeros before any arithmetic is done. The central discipline is simple: preserve the exponent sum and include every valid pair exactly once.
Check Your Understanding
Use finite convolution sums, and show which exponent pairs contribute.
- If \(A(x)=1+3t+2t^2\) and \(B(x)=2-t+4t^2\), what is the coefficient of \(t^2\) in \(A(x)B(x)\)?
- If the only nonzero exponents in \(A\) are \(1\) and \(4\), and the only nonzero exponents in \(B\) are \(0\) and \(3\), which pairs can contribute to degree \(4\)?
- Suppose \(A(x)\) begins at degree \(3\) and \(B(x)\) begins at degree \(2\). Which product coefficients must vanish before degree \(5\)?
- For \(A(x)=1-t+t^2\) and \(B(x)=1+t\), calculate the coefficient of \(t^2\). What causes the result?
- In a degree-\(n\) convolution sum, why can no coefficient with index greater than \(n\) contribute?