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I'm taking a finance course, and I can't afford the financial calculator which can be used to solve this, so I would like to know how to solve this algebraically by hand (I don't care if it uses complex numbers or quadratic formula or anything tricky).

Note that I have a regular calculator that can grind out numbers, but not a financial calculator.

I tried and tried but wasn't able to isolate 'r' even into a quadratic form to solve it.

Trying to understand the process to duplicate it for any set of numbers...

The formula represents the effective monthly rate of a loan with a known principal value and interest rate of the loan, and the number of compounding periods.

The actual question is...

27000 = 2430 * { ( 1 - [1/(1+r)^12] ) / r }

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  • $\begingroup$ One cannot find an "algebraic" solution. If you know $x$, $y$, and $n$ you can fairly quickly home in on $r$, and find it to adequate accuracy. $\endgroup$ May 22 '14 at 2:33
  • $\begingroup$ Oh ok, thanks. If I know the numbers 'x' = 10, 'y' = 1, and 'n' = 12, how can we solve for 'r'. I tried and tried but wasn't able to isolate 'r' even into a quadratic form to solve it. $\endgroup$
    – Johnny
    May 22 '14 at 2:41
  • $\begingroup$ Please check your formula. Right now the way the parentheses work it looks as if you using $x=y\left(1-\frac{1}{r(1-r)^n}\right)$, which does not look right if it is intended to produce $10$ when $y=1$. $\endgroup$ May 22 '14 at 2:53
  • $\begingroup$ What is the formula supposed to represent? $\endgroup$
    – David K
    May 22 '14 at 3:18
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    $\begingroup$ Ahh, the formula is supposed to represent the effective monthly rate of a loan with a known principal value and interest rate of the loan, and the number of compounding periods. The formula is correct, Andre. I just triple-checked it. $\endgroup$
    – Johnny
    May 22 '14 at 3:29
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Let us consider the general case where the equation would write $$A = B \Big(1- \frac{1}{(1+r)^n} \Big) \frac{1}{r} $$ and we know that $r$ is a rather small number compared to $1$. We also know that $r \neq 0$ and $r>0$ (if your bank uses conditions which are different, please send me its name).

This being said, let us consider that the equation to solve is $$f(r)=B \Big(1- \frac{1}{(1+r)^n} \Big) -A r=0 $$ Developing $f(r)$ as a Taylor series around $r= 0$ gives $$f(r)=r (B n-A)-\frac{1}{2} r^2 (B n (n+1))+O\left(r^3\right)$$ and, so, a first estimate is given by $$r_0=\frac{2 (B n-A)}{B n(n+1)}$$ Applied to your numbers, this gives as estimate $r_0=\frac{4}{351}\simeq 0.0113960$ while the exact solution is $r=0.0120435$ which is not bad.

For polishing the solution, a typical method is Newton, which starting with an estimate $r_0$ will update the guess according to $$r_{n+1}=r_n-\frac {f(r_n)}{f'(r_n)}$$ Applied to your problem, the successive iterates will then be $0.0120803$, $0.0120436$, $0.0120435$ which is the solution.

Concerning the initial estimate, instead of stopping at the second order, we could have performed a third oder Taylor expansion and get $$f(r)= (B n-A)r-\frac{1}{2} (B n (n+1))r^2+\frac{1}{6} B n (n+1) (n+2) r^3+O\left(r^4\right)$$ and get as a firt estimate $$r_0=\frac{3 B n (n+1)-\sqrt{3} \sqrt{B n (n+1) (8 A (n+2)-B n (5 n+13))}}{2 B n (n+1) (n+2)}$$ which, for your numbers, would give $r_0=0.0120766$ which is quite good.

There are more sophisticated methods which would converge faster. If you are concerned, let me know.

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A simple, reasonably fast way to solve this numerically is to use the secant rule.

Say you have an equation $$ f(x) =0 $$ Say you have two estimates $x_0$ and $x_1$, with corresponding values $y_0 = f(x_0)$ and $y_1 = f(x_1)$. Hopefully they are near the root, ideally on both sides of it. But the method often works even starting from far away. A better estimate is given by: $$ x_2 = x_1 - \frac{y_1 (x_1 - x_0)}{y_1 - y_0} $$ If the result isn't near enough, reuse the point nearest the root or keep the pair straddling it, and repeat.

The explanation is that $x_2$ is where the straight line through $(x_0, y_0)$ and $(x_1, y_1)$ cuts the $x$ axis; if the graph of $f(x)$ isn't too wild, $x_2$ is much nearer the root than the starting points.

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