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submitted 3 months ago by Amicitas@lemmy.world to c/technology@lemmy.world

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[-] VintageGenious@sh.itjust.works 4 points 3 months ago

K•(1+r)^n

[-] linearchaos@lemmy.world 1 points 3 months ago

I would just rebuild something in my head like this every time.

While i < n; k=k+(k*r); i++;

You'd think I could remember k(1+r)^n but when you posted, it looked as alien as it felt decades ago.

[-] VintageGenious@sh.itjust.works 5 points 3 months ago

The use of for makes sense.

k=0; for (i=0; i<n; i++) k=k+f(i); is the same as k=\sum_{i=0}^{n-1} f(i)

and

k=1; for (i=0; i<n; i++) k=k*f(i); is the same as k=\prod_{i=0}^{n-1} f(i)

In our case, f(i)=1+r and k=1; for (i=0; i<n; i++) k*(1+r); is the same as k=\prod_{i=0}^{n-1} (1+r) = (1+r)^n

All of that just to say that exponentiation is an iteration of multiplication, the same way that multiplication is an iteration of addition

this post was submitted on 20 Sep 2024

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