The Newton-Raphson Method for finding a correct monthly interest rate

Question

I am very new to this topic and just started to learn about this method. Trying to understand this method intuitively.

In this document I found and interesting real life question:

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 $$Ar=M\left(1-\frac1{(1+r)^n}\right).$$ A car loan of $10000$ dollars was repaid in $60$ monthly payments of $250$ dollars. Use the Newton Method to find the monthly interest rate correct to $4$ significant figures.

Can somebody please explain intuitively why do we use this method in real life? If I understood correctly so far, we can make a guess and then find a very close number to the real answer, using this method.

Appreciate your time and other interesting examples, ideally with a code example in R/Python. Thanks!

Answer 1

For intuition, you might be interested in reading Why does Newton's method work? and Math Insight.

As for this specific example, we cannot find a nice closed-form solution, so we are stuck using numerical methods instead and will choose Newton's Method, but many other root finding methods will work.

Some of the goals of numerical methods are to be stable and have as fast as convergence as possible for each iteration of the algorithm. In this problem, we are given:

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 $$Ar=M\left(1-\dfrac1{(1+r)^n}\right).$$ A car loan of $10000$ dollars was repaid in $60$ monthly payments of $250$ dollars. Use the Newton Method to find the monthly interest rate correct to $4$ significant figures.

So, we know

$$Ar = M\left(1-\dfrac1{(1+r)^n}\right) \rightarrow 10000 r = 250 \left(1-\dfrac1{(1+r)^{60}}\right)$$

We want to solve this function for $r$, but there is no closed form solution, so Newton's Method it is. For the Newton iteration step below, we can write our function as

$$f(r) = 40 r + \dfrac1{(r+1)^{60}} - 1$$

Taking the derivative, we have

$$f'(r) = 40 - \dfrac{60}{(r+1)^{61}}$$

The Newton iteration is given by $x_{n+1} = x_n - \dfrac{f(x)}{f'(x)}$, so we have

$$r_{n+1} = r_n - \dfrac{40 r_n +\dfrac{1}{(r_n+1)^{60}}-1}{40-\dfrac{60}{(r_n+1)^{61}}} = $$

Now, we choose a starting value, say, $r_0 = 1$, and keep our fingers crossed and have the iteration

• $r_0 = 1.0000000$
• $r_1 = 0.0250000$
• $r_2 = 0.0164861$
• $r_3 = 0.0145644$
• $r_4 = 0.0143962$
• $r_5 = 0.0143948$
• $r_6 = 0.0143948$

So, we find that the monthly interest rate is $1.439 \%$ to four significant figures.

Answer 2

This is probably too long for the comment section since @Moo already gave a good answer.

$$Ar=M\left(1-\frac1{(1+r)^n}\right)$$ which means that you look for the zero of function $$f(r)=\frac 1 r\left(1-\frac1{(1+r)^n}\right)-\frac A M$$ The part which depends or $r$ varies quite fast and it is far away from linearity; this implies first a "good" guess and second, probably more itarations than necessary.

At the opposite, consider the "reciprocal" function $$g(x)=\frac{r}{1-\frac{1}{(1+r)^{n}}}-\frac MA $$ It is much more linear and this is good news for any root-finding method.

Using an extremely limited Taylor expansion around $r=0$, we should have $$\frac{r}{1-\frac{1}{(1+r)^{n}}}=\frac{1}{n}+\frac{(n+1) r}{2 n}+O\left(r^2\right)$$ and, ignoring the higher order terms $$r_0=\frac{2 (M n-A)}{A (n+1)}$$ For your case $(A=10000,M=250,n=60)$ this leads to $r_0=\frac{1}{61}\approx 0.0164$ and Newton method will converge very fast. Continuing working with function approximation, you can generate a better initial guess, namely $$r_0=\frac{6 (M n- A)}{2 A (n+2)+M (n-1) n}$$ which, for your case, would give $r_0=\frac{6}{425}\approx 0.0141$ which is still closer to the solution given by @Moo.