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The formula is:

$$A = P\frac{i(i+1)^{n}}{(i + 1)^{n}- 1}$$

and I need to rearrange for $i$ because I don't know the interest. I can solve this if it were

Any help would be greatly appreciated.

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1 Answer 1

up vote 3 down vote accepted

Hint: $i = (1+i) - 1$, solve for $1+i$.

Rewrite the equation as

$$\frac{A}{P} = \frac{(1+i)-1}{1-(1+i)^{-n}}$$

Then multiply through the equation by the denominator of the RHS and rearrange to get

$$(1+i)^{n+1}-\left (1+\frac{A}{P} \right ) (1+i)^{n} + \frac{A}{P}=0$$

Now you can solve for $1+i$ as a (real, positive) root of this equation.

EDIT

When $n$ is large, in many cases $(1+i)^n$ will be much larger than $\frac{A}{P}$. We may then ignore that term will very little error and get as a very good approximation to the solution:

$$i \approx \frac{A}{P} $$

We may then find a first order approximation as follows: let $i=R+s$, where $R=\frac{A}{P}$. Then the equation takes the form:

$$s\left ( 1 + \frac{s}{1+R} \right )^n + R (1+R)^{-n}=0$$

To first order in $s$

$$i = R [1-(1+R)^{-n}]$$

Note that $(1+R)^{-n} \sim 10^{-64}$ with the current parameters, so there is no point in further approximation. But at least there is a general framework here.

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My n = 365 here. I got to where you are but How can i solve x^366 + x^365+x^364 = 0? –  Jonathan O Jan 28 '13 at 1:07
    
That equation would be easy, but we have, rather, $x^{366} + (a+1) x^{365} + a=0$. What are typical values for $a$? –  Ron Gordon Jan 28 '13 at 1:13
    
we know the fixed value for A. Lets say it is 0.5. What is the solution to that equation? You say it is easy but Im not so confident :) –  Jonathan O Jan 28 '13 at 1:14
    
No, I said your incorrect equation is easy. The one I posted is not. That's why I want a plausible number so I can see how hard this is. –  Ron Gordon Jan 28 '13 at 1:16
    
in the equation n is known as 365. A/P = 1.5 –  Jonathan O Jan 28 '13 at 1:17

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