Solving $286.84 = \frac{15000\,x\,(1+x)^{60}}{(1+x)^{60}-1}$

Question

I am trying to solve this Basic APR calculation (via CalculatorSoup.com) with pen on paper.

$$286.84 = \frac{15000\,x\,(1+x)^{60}}{(1+x)^{60}-1}$$

What have I tried so far?

Am I doing anything wrong?

Answer 1

You can implement the Newton-Raphson method in $C\#$. There are many examples on the web.

Firstly let $a=286.84$ and $b=15000$, then we have the equation

\begin{eqnarray*} a=\frac{b\cdot i\cdot (1+i)^{60}}{ (1+i)^{60}-1} \\ &\texttt{Multiplying the equation by }(1+i)^{60}-1 \\ a\cdot (1+i)^{60}-a=b\cdot i\cdot (1+i)^{60} \\ & 1+i=q \\ a\cdot q^{60}-a=b\cdot (q-1)\cdot q^{60} \\ & \textrm{Multiplying out the brackets} \\ a\cdot q^{60}-a=b\cdot q^{61} -b\cdot q^{60} \\ & \textrm{Everything on the RHS} \\ 0=b\cdot q^{61}-(a+b)\cdot q^{60} +a \end{eqnarray*}

Thus we have to find the roots of the 61 degree polynomial $f(q)=b\cdot q^{61}-(a+b)\cdot q^{60} +a$. Next we have to find the first derivative: $$f'(q)=61\cdot b\cdot q^{60}-60\cdot (a+b)\cdot q^{59}$$

And the approximated value of $q$ at the n-th step is

$$q_{n+1} = q_n - \frac{f(q_n)}{f'(q_n)}$$

For verification I attach the table of the values which is made with Excel. The initial value $q_0$ is $1.01$, which is equivalent to $i=0.01=1\%$.

Update

I´ve used this compiler and this code for your problem. It works very well. If you have further questions feel free to ask. The output of the compiler is

$\color{white}{Math.Pow(Number1, Number2)}$