Multiply Two Maclaurin Series With a Convolution Sum

When you multiply two Maclaurin series with a convolution sum, the coefficient of xⁿ is a finite sum of products of earlier coefficients. If f(x) = Σ a_nx^{n and g(x) = Σ b_nxn, then h(x) = f(x)g(x) = Σ c_nxn where c_n = Σ_k=0n a_kb_n-k. This formula, known as the Cauchy product, is a finite summation for each resulting power. The sum runs from k=0 to n, giving n+1 terms per coefficient. No infinite loops or limits appear inside the coefficient itself. You compute each c_n directly from the sequence of a_k and b_k.}

Work the standard example: let f(x) = e^x and g(x) = e^x. Their Maclaurin series are both Σ xⁿ/n!. So a_n = 1/n! and b_n = 1/n!. For n=2, c₂ = a₀b₂ + a₁b₁ + a₂b₀. That is (1/0!)(1/2!) + (1/1!)(1/1!) + (1/2!)(1/0!) = 1/2 + 1 + 1/2 = 2. The product series should be e^2x, whose Maclaurin series is Σ (2ⁿ/n!)xⁿ. For n=2, 2²/2! = 4/2 = 2. The convolution sum gives 2, matching the known expansion exactly. Now compute c₃ the same way: c₃ = a₀b₃ + a₁b₂ + a₂b₁ + a₃b₀ = 1/6 + 1/2 + 1/2 + 1/6 = 8/6 = 4/3. The series for e^2x gives 2³/3! = 8/6 = 4/3. The pattern holds for every n because the convolution of the two sequences {1/n!} with itself is {2ⁿ/n!}.

Use this method whenever you have two known expansions and need the product series without recomputing derivatives. For n=5, the convolution sum has six terms. A series sum calculator handles the arithmetic for larger n, but the formula stays the same finite sum. The key is that each c_n depends only on a₀ through a_n and b₀ through b_n, never on terms beyond n.

Step through the calculation for a different pair. Take f(x) = e^x and g(x) = 1/(1-x) = Σ xⁿ for |x|<1. Here a_n = 1/n! and b_n = 1. For n=2, c₂ = a₀b₂ + a₁b₁ + a₂b₀ = 1*1 + 1*1 + (1/2)*1 = 1 + 1 + 0.5 = 2.5. The product h(x) = e^x/(1-x). Its series coefficients are not a standard table entry, but the convolution sum gives them term by term. For n=3, c₃ = a₀b₃ + a₁b₂ + a₂b₁ + a₃b₀ = 1 + 1 + 1/2 + 1/6 = 2.6667. The product series converges for |x| < 1, the smaller radius between e^x (all real x) and 1/(1-x) (|x|<1). The convolution sum works within that common radius. Outside it, the series may diverge even if the formula still produces numbers.

The reason the convolution sum is a finite summation comes from how polynomials behave. When you multiply two finite-degree polynomials, the coefficient of xⁿ is a sum over all pairs of monomials whose exponents add to n. An infinite series is a limit of polynomials, but for any fixed n, only finitely many terms from each series can contribute to xⁿ because any term a_kx^k with k > n would produce an exponent larger than n when multiplied by any b_jx^j with j ≥ 0. So the sum is always finite, exactly n+1 terms.

Each term in the sum is a product of one coefficient from the first series and one from the second. The indices sum to n. For n=2, we have (a₀,b₂), (a₁,b₁), (a₂,b₀). There is no term like a₃b_-1 because the series start at n=0. The formula works for any two Maclaurin series regardless of their radii, as long as both are centered at 0.