this post was submitted on 20 Sep 2024
413 points (94.8% liked)

Technology

59989 readers
2249 users here now

This is a most excellent place for technology news and articles.

Our Rules

  1. Follow the lemmy.world rules.
  2. Only tech related content.
  3. Be excellent to each another!
  4. Mod approved content bots can post up to 10 articles per day.
  5. Threads asking for personal tech support may be deleted.
  6. Politics threads may be removed.
  7. No memes allowed as posts, OK to post as comments.
  8. Only approved bots from the list below, to ask if your bot can be added please contact us.
  9. Check for duplicates before posting, duplicates may be removed

Approved Bots

founded 2 years ago
MODERATORS
L3s@lemmy.world
L3s@fry.gs
L4s@fry.gs
L4s@lemmy.world
enu@lemmy.world
413
Secret calculator hack brings ChatGPT to the TI-84, enabling easy cheating (arstechnica.com)
submitted 2 months ago by Amicitas@lemmy.world to c/technology@lemmy.world
104 comments fedilink hide all child comments
you are viewing a single comment's thread
view the rest of the comments
[–] VintageGenious@sh.itjust.works 5 points 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