minus-squareVintageGenious@sh.itjust.workstoTechnology@lemmy.world•Secret calculator hack brings ChatGPT to the TI-84, enabling easy cheatinglinkfedilinkEnglisharrow-up5·8 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-up4·8 days agoK•(1+r)^n linkfedilink
minus-squareVintageGenious@sh.itjust.workstoOpen Source@lemmy.ml•Seal: a fully featured YouTube downloader for Android based on yt-dlplinkfedilinkarrow-up2·28 days agoWorks well combined with revanced linkfedilink
VintageGenious@sh.itjust.works to Fediverse@lemmy.worldEnglish · 1 month agoAny android client that can do both Lemmy and Mastodon?plus-squaremessage-squaremessage-square11fedilinkarrow-up15arrow-down10
arrow-up15arrow-down1message-squareAny android client that can do both Lemmy and Mastodon?plus-squareVintageGenious@sh.itjust.works to Fediverse@lemmy.worldEnglish · 1 month agomessage-square11fedilink
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