this post was submitted on 16 Jul 2023
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Yeah I don't really think that helps anyone that didn't understand the above example, sorry.
Yeah they might as well have written it in assembly... Some people are just not very good at understanding that others don't have their knowledge/ease of understanding certain things, especially people who are very good at what they do, the ability to simplify is as much a skill as understanding complex concepts!
For the case that n = 0 (before the first run of the loop), x(0) = 1.
For the first actual case, n = 1. X(1) = x(0)*3*n = 1*3*1 = 3.
For the next case, n = 2. X(2) = x(1)*3*n = 3*3*2 = 18.
For the next case, n = 3. X(3) = x(2)*3*n = 18*3*3 = 162.
For the next and last case, n = 4. X(4) = 162*3*4 which I'm not computing. The computer value of x(4) is the value of the product loop.
If that doesn't help, I could try helping again to rephrase, but I'm not sure what else to add.