this post was submitted on 20 Sep 2024
416 points (94.8% liked)
Technology
60052 readers
2809 users here now
This is a most excellent place for technology news and articles.
Our Rules
- Follow the lemmy.world rules.
- Only tech related content.
- Be excellent to each another!
- Mod approved content bots can post up to 10 articles per day.
- Threads asking for personal tech support may be deleted.
- Politics threads may be removed.
- No memes allowed as posts, OK to post as comments.
- Only approved bots from the list below, to ask if your bot can be added please contact us.
- Check for duplicates before posting, duplicates may be removed
Approved Bots
founded 2 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
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