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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.

joncodo
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1 Answers1

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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.

Ron Gordon
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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? – joncodo Jan 28 '13 at 01:07
  • 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 01:13
  • 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 :) – joncodo Jan 28 '13 at 01:14
  • 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 01:16
  • in the equation n is known as 365. A/P = 1.5 – joncodo Jan 28 '13 at 01:17
  • OK, observation: because $i>0$, $1+i>1$ so that $(1+i)^n$ will be very large, so large that we can neglect the constant term. I will provide an update. – Ron Gordon Jan 28 '13 at 01:22
  • In finance world, market participants won't solve this as a polynomial equation. One practical way is to first rewrite the equation as $i = \frac{A}{P}(1 - (1+i)^{-n})$ [#]. In normal market circumstances, $i$ is of the order of a few percents and duration $n$ is of the order of years. $e^{-i n}$ is usually a good approximation to $(1+i)^{-n}$ and $\frac{A}{P}(1 - e^{-i n})$ can be used as a first approximation of what $i$ should be. If more accurate value of $i$ is desired, one can plug it into R.H.S of [#] to crank out better approximation. – achille hui Jan 28 '13 at 01:28
  • @achillehui: thanks for the additional insight. I smartened up a bit and used a perturbational approach which should be roughly as accurate, albeit a little more messy. – Ron Gordon Jan 28 '13 at 01:45