# Mortgage rate

uzless
Posers and Puzzles 24 Mar '10 21:39
1. 28 Mar '10 23:011 edit
Originally posted by geepamoogle
Anyone know how to derive the formula for interest rate given the principle amount, payments, and number of payments?

The base formula includes both "i" (interest rate per payment) and "(1+i)^-n"... I see no way to combine the two occurrences of i.

I do suppose you could zero in on the interest though, by plugging in rates and getting the answer closer for one of the variables closer and closer to the correct answer.
Could you give us the exact base formula you are using?
2. 29 Mar '10 01:32
M = P * i / [1 - (1+i)^-n]

M = monthly payment
P = Current principle
i = monthly interest rate (In this case, 5% / 12)
n = number of monthly payments
3. PBE6
Bananarama
29 Mar '10 17:502 edits
Originally posted by geepamoogle
M = P * i / [1 - (1+i)^-n]

M = monthly payment
P = Current principle
i = monthly interest rate (In this case, 5% / 12)
n = number of monthly payments
If we let x = 1+i, then the expression becomes:

M = P * (x-1) / [1 - x^n]

Rearranging, we have:

M/P * [1 - x^n] = (x-1)

Multiplying both sides by x^n and factoring out (x-1) from the left hand side, we have:

M/P * (x-1) * [x^n - x^(n-1) + x - 1] = x^n * (x-1)

Cancelling (x-1) from both sides, we're left with:

M/P [x^n - x^(n-1) + x -1] = x^n

Rearranging into polynomial form, we have:

(1-M/P) * x^n + x^(n-1) - x + 1 = 0

Now, the fundamental theorem of algebra states that "every non-constant single-variable polynomial with complex coefficients has at least one complex root." However, the Abel-Ruffini theorem also states that "there is no general solution in radicals to polynomial equations of degree five or higher."

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
http://en.wikipedia.org/wiki/Abel's_impossibility_theorem

So, the solution must exist, and a specific equation may exist to solve it, but there is no general formula for producing the result (provided n>4, which is usually the case for a mortgage ðŸ˜‰). I suppose this is just a long-winded way of saying approximation is the best method in this case.
4. 29 Mar '10 20:43
Originally posted by PBE6
If we let x = 1+i, then the expression becomes:

M = P * (x-1) / [1 - x^n]

Rearranging, we have:

M/P * [1 - x^n] = (x-1)

Multiplying both sides by x^n and factoring out (x-1) from the left hand side, we have:

M/P * (x-1) * [x^n - x^(n-1) + x - 1] = x^n * (x-1)

Cancelling (x-1) from both sides, we're left with:

M/P [x^n - x^(n-1) + x -1] = x^ ...[text shortened]... se this is just a long-winded way of saying approximation is the best method in this case.
M = P * i / [1 - (1+i)^-n]

if we let x = 1 + i

then shouldn't that be:

M = P * (x-1) / [1 - x^-n]

Or am I missing something?
5. 02 Apr '10 07:291 edit
Originally posted by iamatiger
M = P * i / [1 - (1+i)^-n]

if we let x = 1 + i

then shouldn't that be:

M = P * (x-1) / [1 - x^-n]

Or am I missing something?
Assuming I was on the right track:
M = P * (x-1) / [1 - x^-n]

multiply by [1-x^n] and expand
M - M/x^n = Px - P

Multiply by x^n
Mx^n -M = Px^(n+1) - Px^n

Rearrange
(M + P)x^n - Px^(n+1) = M

Divide by -P
x^(n+1) - (M/P + 1)x^n = -M/P

Which we will need some cunning way of solving