Amicitas@lemmy.world to Technology@lemmy.worldEnglish · 2 months agoSecret calculator hack brings ChatGPT to the TI-84, enabling easy cheatingarstechnica.comexternal-linkmessage-square104fedilinkarrow-up1439arrow-down124 cross-posted to: [email protected][email protected]
arrow-up1415arrow-down1external-linkSecret calculator hack brings ChatGPT to the TI-84, enabling easy cheatingarstechnica.comAmicitas@lemmy.world to Technology@lemmy.worldEnglish · 2 months agomessage-square104fedilink cross-posted to: [email protected][email protected]
minus-squareVintageGenious@sh.itjust.workslinkfedilinkEnglisharrow-up5·2 months agoThe 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
The use of for makes sense.
k=0; for (i=0; i<n; i++) k=k+f(i);
is the same ask=\sum_{i=0}^{n-1} f(i)
and
k=1; for (i=0; i<n; i++) k=k*f(i);
is the same ask=\prod_{i=0}^{n-1} f(i)
In our case,
f(i)=1+r
andk=1; for (i=0; i<n; i++) k*(1+r);
is the same ask=\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