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