Staff Versus Members! (Current record: 200)

[MrRazot]

0
my evil plan is to continue posting 0's on this thread 

[White Wolf Guardian]

1

See, meaningless threats always work.

[MrRazot]

0
curses... you mods are obviously too smart and cunning for the likes of me

[Nikolai the Lynx]

1 Indeed.

[Kaya]

zero

[Nikolai the Lynx]

1 Well thank you.

[Midnight Madness]

2

Hooray, discussing my characters' histories with people.

[Nikolai the Lynx]

3 Sweet, that is always fun.

[Midnight Madness]

4

Especially considering that many of them find their origins in my Grade School years. Particularly Sytex.

[MrRazot]

0
that is indeed an apparent trend

[anoni]

1.

I'll be pretty impressed if you can do this, it's a finance question.

Gary gets a loan of $100,000. Gary repays the loan with monthly repayments of $M and the interest on the balance is calculated by 6% p.a. calculated monthly.
The balance at the end of n months is given by...
Bn = 100,000(rⁿ) - M(1 + r + r² + .... + r^{n - 1})

Where r = 1.005.
i) If the interest was repaid in 20 years (240 months), what is his monthly repayment?
ii) If he has a monthly repayment of $2,000 how long will it take, to the nearest month, to pay back the loan?
iii) Is the payment method i) or ii) better? How much money does he save using the better method?

I'll answer this tomorrow

i) Ok first we need to simplify the equation so that it is equal to
0 = 100,000(rⁿ) - M(rⁿ - 1) / (r - 1)
Rearranging in terms of M gives us
M = 100,000(rⁿ) * (r - 1) / (rⁿ - 1)
Where n = 240; r = 1.005
M = 716.4310585...
Therefore; in order to pay the loan back you would need 240 monthly repayments of $716.43

ii) Ok now we need to find n, we know r and M
0 = 100,000(rⁿ) - M(rⁿ - 1) / (r - 1)
0 = 100,000(rⁿ) - (Mrⁿ - M) / (r - 1)
0 = 100,000(rⁿ)(r - 1) - Mrⁿ + M
-M = rⁿ(100,000(r - 1) - M)
-M / (100,000(r - 1) - M) = rⁿ
ln(-M / (100,000(r - 1) - M)) = nln(r)
n = ln(-M / (100,000(r - 1) - M)) / ln(r)
Knowing that M = $2,000; r = 1.005.
n = 57.68013596...
n = 58 months.
It would take 58 months to repay that loan with a $2,000 monthly repayment

iii) This one is fairly easy, as the person only pays his monthly repayments it's just comparing 240 x 716.43 and 58 x 2,000
the first one will be more, so method ii) will be the cheaper method
He will save $(240 x 716.43 - 58 x 2,000)
By picking ii) over i) you will save (using the exact values of ii) and i))
$56,583.18c

[MrRazot]

0
i have a formula that i created for finding the amount of time it takes to achieve the final amount in a financial situation
for example 12000 = 10000(1+0.04)^n


the very simple formula is:
log(A/P) / log(1+i) = n


so:
log(12000/10000) / log(1+0.04) = 4.648 (rounded off to the third decimal place)


so it would take 4 years and 7 months (0.648 x 12) to get to that ammount


your stuff however is just too damn confusing for me

[anoni]

1.
That formula is only finding the amount of stuff you'd have to pay at one time.
I made some interesting formula's for sets and recursive functions.

[MrRazot]

0
i know lol
but you know what i find fun and easy?
patterns


determine the formula for the general term of this sequence
-1;6;17;32;51

[anoni]

1.
Ok lets see, first it doesn't seem to be an arithmetic sequence so let look at the gaps between each number
7 - 11 - 15 - 19
So it is arithmetic where the rate increases by 4 each step.

General arithmetic sequence formula is An = a + (n - 1)r
However for this sequence it's actually
An = 7 + 4(n - 1) + 7 + 4(n - 2) + ... + 7 - 1
Factorizing we get
An = -1 + 7(n - 1) + 4(0 + 1 + 2 + ... + n - 2)
The bracket being multiplied by 4 is also an arithmetic series, it will follow the arithmetic equation (n / 2)(a + l) where l is the last term, in this case n - 2
An = -1 + 7n - 7 + (4(n -1) / 2)(n - 2)
An = -1 + 7n - 7 + (2n - 2)(n - 2)
An = -1 + 7n - 7 + 2n² - 4n - 2n + 4
An = 2n² + n - 4

That's our equation to get the nth term,
An = 2n² + n - 4
Lets see if it works, for the first term n = 1
A1 = 2(1)² + 1 - 4
A1 = -1
seems to work, what about the 5th term or 51
A5 = 2(5)² + 5 - 4
A5 = 50 + 1
A5 = 51
Lastly, the 6th term as predicted by the formula should be...
A6 = 74

This was actually doing it a stranger way, where n = T where T is the term, usually T = n + 1
If T = n + 1 then the formula would be...
An = 2n² + 5n - 1
That is, this equation works where if you accept that for the first term n = 0 (rather than n = 1) and the second term n = 1 (rather than n = 2).

For fun, the 10,000th term (where T = n) should be 200,009,999.
Finding the pattern was easy, getting the general term was a bit more strange but it was fun! Give me MOAR!