How do I isolate for the interest with either the the future value or present value equations?

Question

I'm having troubles understanding how to isolate for the interest with either the future value or present value equations. Say we know what $K$ and $n$ are, and I'm trying to find $j$:

$$\frac{(1+j)^n-1}{j}=K$$

or

$$\frac{1-v^n}{j} = K$$

where $v = \frac {1}{1+j}$

Answer 1

Given that $w= j+1$, you have \begin{align} K &= \frac{w^n-1}{w -1}\\ \end{align}

We know that \begin{align} (w^n-1) &= (w-1)(w^{n-1}+w^{n-2}+\cdots+1) \end{align}

Therefore you can obtain the the value for $w$, and $j$ consequently, by finding the roots of the equation \begin{align} (w^{n-1}+w^{n-2}+\cdots+1) -K = 0 \end{align}

Doing it on paper I wouldn't expect $n\geq3$.

Answer 2

You have $\frac{(1+j)^{30} -1}{j}=41.487451$. Let $j+1=q \Rightarrow q-1=j$

And let $a=41.487451$

We get $\frac{q^{30} -1}{q-1}=a$

Multiplying both sides by $q-1$.

$q^{30} -1=aq-a$

Putting all terms on the RHS.

$q^{30}-aq -1+a=0$

Substituting back

$q^{30}-41.487451\cdot q +40.487451=0$

This equation can not be solved algebraically. You have to use an approximation method, for instance the Newton-Raphson method.

We know that j is greater than 0 and almost surely not bigger than $10\%=0.1$ since $j$ is an interest rate or something similar. And therfore q is between $1$ and $1.1$ A good initial value for q should be $1.05$