Comment on Secret calculator hack brings ChatGPT to the TI-84, enabling easy cheating

• linearchaos@lemmy.world ⁨2⁩ ⁨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.

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  • VintageGenious@sh.itjust.works ⁨2⁩ ⁨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

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