where there are a; + 1 options for each exponent x; because 0 ≤ x ≤ a,, and so the number of positive integer divisors is (a₁ + 1)(a₂ + 1)... (am + 1).
where there are a; + 1 options for each exponent x; because 0 ≤ x ≤ a,, and so the number of positive integer divisors is (a₁ + 1)(a₂ + 1)... (am + 1).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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