this post was submitted on 16 Jul 2023
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[–] can@sh.itjust.works 3 points 1 year ago (2 children)

Yeah I don't really think that helps anyone that didn't understand the above example, sorry.

[–] Kecessa@sh.itjust.works 4 points 1 year ago

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!

[–] only_lurking@lemmy.world 1 points 1 year ago

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.