- 28 Mar '10 23:01 / 1 edit

Could you give us the exact base formula you are using?*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. - 29 Mar '10 17:50 / 2 edits

If we let x = 1+i, then the expression becomes:*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

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. - 29 Mar '10 20:43

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

if we let x = 1 + i

then shouldn't that be:

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

Or am I missing something? - 02 Apr '10 07:29 / 1 edit

Assuming I was on the right track:*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?

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