Newton Raphson method
NOTES ON STATISTICS, PROBABILITY and MATHEMATICS
Newton Raphson Method and Loan Interest:
Loan Mathematics:
We take a loan out from the bank: the principal \(P\) to be repaid in a fixed amount of time (let’s assume monthly payments \(n =360\), corresponding to monthly payments for \(30\) years) with interest compounded every month \(i = \text{nominal interest}/12\). Every month the payment is \(R\) (“rent”) dollars.
At time zero, we take the loan, and it starts accruing interest. After a month, the amount owed will be:
\[P \; (1 +i)\]
at which point we make the first payment. Now the balance will be left at
\[\text{balance owed}= P \; (1 +i) - R\]
on which interest will be accrued, so that by the end of the second month (and after making the second payment) the balance will be
\[\begin{align}\text{balance owed} &= \left [P\; (1 + i) - R\; \right] \;(1 + i) - R\\[2ex] &= P\;(1+i)^2-R(1+i)-R \end{align}\]
At the end of the third month the balance will be
\[\begin{align}\text{balance owed} &= \left[P\;(1+i)^2-R\;(1+i)-R\;\right]\;(1+i)-R\\[2ex] &= P\;(1+i)^3-R\;(1+i)^2-R\;(1+i)-R \end{align}\]
At the end of the \(n\) payments, the balance will be zero
\[\begin{align} 0 &= P\;(1+i)^n-R\;(1+i)^{n-1}-R\;(1+i)^{n-2}-\;\cdots\;-R\;(1+i)^0\\[2ex] &=P\;(1+i)^n - R\;\frac{1-(1+i)^n}{1-(1+i)}\\[2ex] &=P\;(1+i)^n - R\;\frac{(1+i)^n-1}{(1+i)-1}\tag 1\\[2ex] &=P\;(1+i)^n - R\;\frac{(1+i)^n-1}{i} \end{align}\]
because \(-R\;(1+i)^{n-1}-R\;(1+i)^{n-2}-\;\cdots\;-R\;(1+i)^0\) forms a geometrical series, and
\[\sum_{k=0}^{n-1}a\;r^k=a\;\frac{1-r^n}{1-r}\]
Rearranging,
\[\begin{align} R\;\frac{(1+i)^n-1}{i}&=P\;(1+i)^n\\[2ex] R\;\left((1+i)^n-1\right)&=P\;i\;(1+i)^n\\[2ex] R&=P\;i\;\frac{(1+i)^n}{\left((1+i)^n-1\right)}\tag 2 \end{align} \]
NEWTON-RAPHSON METHOD:
We want to find the zro values for Eq \((1)\), and we resort to the Newton-Raphson method, which solves for the \(x\) value at which \(F(x)=0\) based on a guess that this \(x\) value is close to an \(a\) value, wuch that
\[F'(a) = \lim_{a\to x}\; \frac{F(x)-F(a)}{x-a}\]
Assuming that the guess gets better and better, we can get rid of the \(\lim\), and express this formula as
\[F'(a) \approx \frac{F(x)-F(a)}{x-a}\]
Keeping in mind that \(F(x)\) is a zero,
\[x\approx a -\frac{F(a)}{F'(a)}\]
Since \(a\) can actually be far from \(x\) we update \(x\) as
\[x_{i+1} = x_i-\frac{F(x_i)}{F'(x_i)}\]
Going back to equation \((2),\) and taking into consideration that \(i\) is the variable to optimize,
\[F(x) = \frac{P\;x\;(1+x)^n}{\left((1+x)^n-1\right)}-R\]
Therefore,
\[\begin{align} F'(x)&=\frac{\partial \left(\frac{P x (x+1)^n}{(x+1)^n-1}-R\right)}{\partial x}\\[2ex] &=\frac{n P x (x+1)^{n-1}}{(x+1)^n-1}+\frac{P (x+1)^n}{(x+1)^n-1}-\frac{n P x (x+1)^{2 n-1}}{\left((x+1)^n-1\right)^2}\\[2ex] &=\frac{P (x+1)^{n-1} \left(x (x+1)^n+(x+1)^n-n x-x-1\right)}{\left((x+1)^n-1\right)^2} \end{align}\]
Let’s now give some values to the principal and the payment… Say we bought a new phone because we were eligible after a couple of years of using a the iFone 3.0. The new phone is the iFone 3.1, and costs \(P= \$ 700\). The salesman at AT&T tells you that you’ll pay \(R =\$ 35\) for \(n=24\) months. What is the interest rate being charged?
\[F(x) = \frac{700\;x\;(1+x)^{24}}{\left((1+x)^{24}-1\right)}-35\]
\[F'(x)=\frac{16800\, x (x+1)^{23}}{(x+1)^{24}-1}+\frac{P (x+1)^{24}}{(x+1)^{24}-1}-\frac{16800\, x (x+1)^{47}}{\left((x+1)^{24}-1\right)^2}\]
Plotting \(F(x),\)
would allow us to start the guessing game at \(0.02\), which corresponds to a nominal annual interest of \(0.02 \times 12=24\%.\) Otherwise we could just pretend that there is no compounding and guess \(\frac{R\times n}{P}=\frac{35\times 24}{700}= 1.2\implies i = 0.2/12= 0.016.\)
In this post there is a nice simplification that helps get a cleaner code expression, and a simpler derivative: instead of working with \(i=x,\) we set up \(x=1+\frac{i}{12},\) in which case, and going back to equation \((1),\)
\[\begin{align} &P\;(1+i)^n = R\;\frac{(1+i)^n-1}{(1+i)-1}\\[2ex] &P\,x^n\,(x-1)=R\,(x^n-1)\\[2ex] &F(x)=P\,x^n\,(x-1)-R\,(x^n-1)\\[2ex] &F(x)=P\,x^{n+1}\,-(P+R)\,x^n+R \end{align}\]
a nice polynomial, which derivative is
\[F'(x)=P\, (n+1)\, x^n-(P+R)\,n\,x^{n-1}\]
And the R code,
interest = function(loan, pmt, n){
F = function(x) (loan * x^(n+1) - (loan + pmt) * x^n + pmt)
F_prime = function(x) (loan * (n+1) * x^n - (loan + pmt) * n * x^(n-1))
guess = 1 + (((pmt * n / loan) - 1) / 12)
x = guess
while(abs(F(x)) > (1e-6)){
x = x - F(x)/ F_prime(x)
print(x)
}
paste("Yearly interest: ", round((12 * (x - 1) * 100),3)," percent")
}
interest(700,35,24)## [1] 1.01529
## [1] 1.015133
## [1] 1.015131## [1] "Yearly interest: 18.157 percent"Double-checking with a financial calculator:
700 PV
0 FV
-35 PMT
24 n
i
g 12x
Of course, this should also work out using the more convoluted formulas:
interest = function(loan, pmt, n){
F = function(x) (loan * x * (1 + x)^n)/((1+x)^n - 1) - pmt
F_prime = function(x) (loan * (x+1)^(n-1) * (x*(x+1)^n + (x+1)^n-n*x-x-1)) / ((x+1)^n - 1)^2
guess = 1 + (((pmt * n / loan) - 1) / 12)
x = guess
while(abs(F(x)) > (1e-6)){
x = x - F(x)/ F_prime(x)
print(x)
}
paste("Yearly interest: ", round((12 * x * 100),3)," percent")
}
interest(700,35,24)## [1] 0.04999943
## [1] 0.01812481
## [1] 0.01515984
## [1] 0.01513085
## [1] 0.01513084## [1] "Yearly interest: 18.157 percent"NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.