Cauchy Product of Series
Multiply two geometric series by diagonals and distinguish a truncated Cauchy sum from the product of two finite sums.
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Cauchy Product of Series
Given two geometric power series A(z) = Σi=0∞ (rz)i and B(z) = Σj=0∞ (sz)j, their Cauchy product groups terms along anti-diagonals of constant total degree n = i + j:
cn = Σk=0n rksn-k, with truncated Cauchy partial sum CN(z) = Σn=0N cnzn.
Whereas CN(z) sums the triangular region i + j ≤ N, the product of two finite degree-N partial sums (Σi=0N (rz)i)(Σj=0N (sz)j) sums the square grid 0 ≤ i, j ≤ N (including terms up to degree 2N).
Worked Example: Repeated Ratios r = 1/2, s = 1/2 at z = 1, N = 2
Compare the diagonal Cauchy partial sum against the square product of finite sums and the closed-form infinite product for r = 1/2, s = 1/2, z = 1, and N = 2:
- Diagonal coefficients:
c0 = 1,c1 = 1/2 + 1/2 = 1, andc2 = (1/2)2 + (1/2)(1/2) + (1/2)2 = 3/4 = 0.75. - Cauchy partial sum (
i + j ≤ 2):C2(1) = c0 + c1 + c2 = 1 + 1 + 3/4 = 11/4 = 2.75. - Product of finite sums (
0 ≤ i, j ≤ 2):(1 + 1/2 + 1/4)(1 + 1/2 + 1/4) = (7/4)2 = 49/16 = 3.0625, which exceedsC2by the three higher-degree corner terms of degrees 3 and 4. - Infinite product and Cauchy error:
1 / ((1 - 1/2)(1 - 1/2)) = 4, so the Cauchy truncation error is|11/4 - 4| = 5/4 = 1.25.
Interactive Experiment
Compare diagonal and square partial sums across sign cancellations in the Cauchy Product of Series interactive workspace:
- Start with the default Ratio r = 1/2, Ratio s = 1/2, Probe z = 1, and Degree N = 2. Verify Coefficient cN = 0.75, Cauchy partial sum = 2.75, and Product of finite sums = 3.0625.
- Set Ratio s = -1/2 and Degree N = 4 (keeping Ratio r = 1/2, Probe z = 1). Notice that odd anti-diagonals cancel to zero while even anti-diagonals give
c4 = 1/16 = 0.0625andC4 = 21/16 = 1.3125approaching4/3. - Set Ratio r = 0, Ratio s = 1/2, Probe z = 1/2, and Degree N = 2. Because the first series is the constant
1, the triangular and square regions contain the same nonzero column, so Cauchy partial sum and Product of finite sums both equal21/16 = 1.3125. - Test Degree N = 0 (where
C0 = 1differs from the infinite product4by3) and Probe z = 0 (where onlyc0 = 1contributes to the evaluated sum).
Limitations and Assumptions
By Mertens' theorem, the Cauchy product of two series converges to the product of their sums whenever both series converge and at least one converges absolutely. In this module, |rz| ≤ 1/2 < 1 and |sz| ≤ 1/2 < 1 hold across the entire declared parameter domain, guaranteeing absolute convergence of both geometric series. For conditionally convergent series outside the disk of absolute convergence, diagonal Cauchy summation can oscillate or diverge even when the individual series converge.
References
The definition of the Cauchy product of infinite series and Mertens' theorem on absolute convergence of series products are established in the following analysis text:
Experience it interactively
Adjust parameters, observe in real time, and build deep intuition with Riano’s interactive Cauchy Product of Series module.
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