minus-squareVintageGenious@sh.itjust.workstoTechnology@lemmy.world•Secret calculator hack brings ChatGPT to the TI-84, enabling easy cheatinglinkfedilinkEnglisharrow-up5·7 days 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 linkfedilink
minus-squareVintageGenious@sh.itjust.workstoTechnology@lemmy.world•Secret calculator hack brings ChatGPT to the TI-84, enabling easy cheatinglinkfedilinkEnglisharrow-up4·7 days agoK•(1+r)^n linkfedilink
minus-squareVintageGenious@sh.itjust.worksOPtoFediverse@lemmy.world•Any android client that can do both Lemmy and Mastodon?linkfedilinkEnglisharrow-up2·30 days agoHow can I browse mastodon from a lemmy client? linkfedilink
VintageGenious@sh.itjust.works to Fediverse@lemmy.worldEnglish · 1 month agoAny android client that can do both Lemmy and Mastodon?plus-squaremessage-squaremessage-square10fedilinkarrow-up156arrow-down12
arrow-up154arrow-down1message-squareAny android client that can do both Lemmy and Mastodon?plus-squareVintageGenious@sh.itjust.works to Fediverse@lemmy.worldEnglish · 1 month agomessage-square10fedilink
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