Question

How can the following equation be simplified?
([{(235,750(1.0036)-1462.32)1.0036-1462.32}1.0036-1462.32]1.0036-1462.32)

Answer 1

hahah dan -.-

Answer 2

that is not an equation, you need an equal sign somewhere?

Answer 3

hmmm yea but im just tryna figure out if there is a faster way of computing this
Lets say i wanted to continue doing this for like 20 times

Answer 4

you can try distributing and see if a pattern emerges

Answer 5

ehhh but if you distribute than you have plenty calculations if its repeated like 20 or 30 times

Answer 6

{235,750(1.0036)-1462.32}1.0036-1462.32

Answer 7

\[(((a(b)-c)b-c)b-c)b-c\]

Answer 8

YEAAAAAA

Answer 9

\[(((abb-cb)-c)b-c)b-c\]

Answer 10

\[(abbc-cbb-cb-c)b-c\]

Answer 11

\[abbbc-bbbc-bbc-bc-c\]

Answer 12

\[ab^3c +\sum_{n=0}^{3}-(b^nc)\]

Answer 13

just not sure where the c came from on the first term abbbc

Answer 14

yea should be another b not c

Answer 15

Ok coooll
Thanks :)

Answer 16

http://www.wolframalpha.com/input/?i=%28%28%28a*b%E2%88%92c%29b%E2%88%92c%29b%E2%88%92c%29b%E2%88%92c

Answer 17

Here's the formula I got for payment we need to make each month if we have to pay off a debt d in t months at an interest i:

\[p=\frac{(1+i)^t *i*d}{(1+i)^t-1}\]

Answer 18

More simplified version
\[p=id+\frac{id}{(1+i)^t-1}\]
I guess I'll show the derivation with my reasoning idk

Answer 19

Ok I'm going to derive as best I can, but using new names: P=Payment, PV=Present Value, r=rate per period, n=number of periods.

However to derive this I'll need to introduce some intermediate stuff that will end up eventually going away. So I'll rename \(PV=d_1\) and this will decrease every time we make a payment. It's just our bank account's debt and eventually at the number of periods we have \(d_n=0\) since we want the final payment to be on the \(n th\) day so the final balance should be 0.

So throwing this together we get a recurrence relationship.

\(d_2=d_1(1+r)-P\)

This just says the new debt is the old debt plus the interest minus the payment you made.

More generally

\(d_{k+1}=d_k(1+r)-P\)

and the recurrence relation will have to end as

\(d_n=0\)

like we said earlier. I'll continue in the next post!

Answer 20

Let's plug the recurrence relation into itself to find the pattern, also let's set \(1+r = a\) to simplify the math cause I hate typing out extra garbage if I don't have to and we'll plug it back in at the end.

Plug in and shift around a little
\(d_2=d_1a-P\)
\(d_3=d_2a-P = (d_1a-P)a-P=d_1a^2-aP-P\)
\(d_4=d_3a-P = (d_2a-P)a-P=((d_1a-P)a-P)a-P=d_1a^3-a^2P-aP-P\)

I'll rewrite them with only the last part shown to make the pattern more clear:

\(d_2=d_1a-P\)
\(d_3=d_1a^2-aP-P\)
\(d_4=d_1a^3-a^2P-aP-P\)
\(d_5=d_1a^4-a^3P-a^2P-aP-P\)
And now I just sorta factor out the -P and write the geometric series in summation form
\[d_n=d_1a^{n-1} -P \sum_{s=0}^{n-2}a^s\]

Notice that these d's are special, \(d_n=0\) and also notice that \(d_1=PV\) so we can plug them in. We still have to simplify the geometric series, but I'll save that for next post, that's the final thing really.