this post was submitted on 20 Sep 2024
397 points (94.6% liked)
Technology
58180 readers
4479 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 1 year ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
K•(1+r)^n
I would just rebuild something in my head like this every time.
While i < n; k=k+(k*r); i++;
You'd think I could remember k(1+r)^n but when you posted, it looked as alien as it felt decades ago.
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+randk=1; for (i=0; i<n; i++) k*(1+r);is the same ask=\prod_{i=0}^{n-1} (1+r) = (1+r)^nAll of that just to say that exponentiation is an iteration of multiplication, the same way that multiplication is an iteration of addition