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A loan of $A$ dollars is repaid by making $n$ equal monthly payments of $M$ dollars, starting a month after the loan is made. It can be shown that if the monthly interest rate is $r$, then
$A r=M\left(1-\frac{1}{(1+r)^{n}}\right) \text {. }$
A car loan of $\ 10000$ was repaid in 60 monthly payments of $\ 250$. Use the Newton Method to find the monthly interest rate correct to 4 significant figures.
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Even quite commonplace money calculations involve equations that cannot be solved by 'exact' formulas. Let $r$ be the interest rate. Then
$10000 r=250\left(1-\frac{1}{(1+r)^{60}}\right) .$
If we are going to work by hand, it is maybe worthwhile to simplify a bit to $f(r)=0$ where
$f(r)=40 r+\frac{1}{(1+r)^{60}}-1$

and therefore
$f^{\prime}(r)=40-\frac{60}{(1+r)^{61}} .$
The Newton Method iteration is now easy to write down. In raw form it is
$r_{n+1}=r_{n}-\frac{40 r_{n}+1 /\left(1+r_{n}\right)^{60}-1}{40-60 /\left(1+r_{n}\right)^{61}} .$
Compute. Particularly if we do the work by hand, it is helpful to make a good choice of $r_{0}$. If the interest rate were $2.5 \%$ a month, the monthly interest on $\ 10,000$ would be $\ 250$, and so with monthly payments of $\ 250$ we would never pay off the loan. So the monthly interest rate must have been substantially under $2.5 \%$. A bit of fooling around suggests taking $r_{0}=0.015$. We then find that $r_{1}=0.014411839$, $r_{2}=0.014394797$ and $r_{3}=0.01439477$. This suggests that to four significant figures the monthly interest rate is $1.439 \%$.
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