Technology

59374 readers

3794 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

founded 1 year ago

MODERATORS

416

submitted 1 month ago by [email protected] to c/[email protected]

104 comments fedilink hide all child comments

you are viewing a single comment's thread
view the rest of the comments

[–] [email protected] 4 points 1 month ago (1 children)

K•(1+r)^n

[–] [email protected] 1 points 1 month ago (1 children)

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.

[–] [email protected] 5 points 1 month 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