Tiny device installed inside TI-84 enables Wi-Fi Internet, access to AI chatbot.
  • linearchaos@lemmy.worldEnglish
    0·
    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.worksEnglish
      0·
      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