Taylor series expansion for $P_0 \, r \, [1−(1+r)^{−N}]^{-1}$

Question

I am trying to estimate APR of a compound interest loan with a fixed payment.

For a loan where:$P_0$ represents the total amount borrowed from the lender, $r$ represents the monthly interest rate, $N$ is the total amount of terms/periods.

The equation for payment ($C$) is as follows:

$$C = P_0 \, r \, [1−(1+r)^{−N}]^{-1}$$

I want to solve for $r$ using a Taylor series. Can you someone help me with it and also provide the proof?

Thanks a lot!

Answer 1

There is no closed-form solution for $r$. Equation 7 of this link gives a power series solution.

I derived the following Taylor series for $C$:

$$ C=\frac{P_0}{N}\left[1 + \frac{N+1}{2}r + \frac{(N-1)(N+1)}{12}\frac{r^2}{2!}+\cdots\right] $$

If you solve this for $r$

$$ r\approx\left(\frac{3}{N-1}\right)\left[-1+\sqrt{1+\frac{4}{3}\frac{N-1}{N+1}\left(\frac{NC}{P_0}-1\right)}\right] $$

I haven't extensively tested this but is seems to get within 1% of the true interest rate for parameters in the consumer mortgage regime (e.g. 4%, 360 month).

The Taylor series will be most accurate at interest rates close to zero. To improve the accuracy you add a third term to the Taylor series. You could also explore curve fitting some real data in the region of interest; (the goal is to improve accuracy in the region of interest at the expense of accuracy near $r=0$). So you might try fitting an equation like:

$$ C=\frac{P_0}{N}\left[A_0 + A_1\frac{N+1}{2}r + A_2\frac{(N-1)(N+1)}{12}\frac{r^2}{2!}\right] $$

getting $A_0, A_1, A_2$ by least squares, then solving the quadratic for $r$.

Hope that helps.

Answer 2

If you have a look here, you will notice that Cantrell had the idea of developing the rhs of equation $$a=r \, [1−(1+r)^{−n}]^{-1}$$ as a Taylor series around $r=0$ and then used series reversion to get series expansion to get expressions looking like $$r=y\left(1+\sum_{k=1}^n b_k\, y^k\right)\qquad \text{where}\qquad y=\frac{2 (a n-1)}{n+1}$$ One could notice that $y$ is the first iterate of Newton method.

The first coefficients being $$b_1=-\frac{n-1}{6}\qquad b_2=\frac{(n-1) (2 n+1)}{36}\qquad b_3= -\frac{(n-1) (2 n+1) (11 n+7)}{1080}$$ $$b_4=\frac{(n-1) (2 n+1)^2 (13 n+11)}{6480}\qquad b_5=-\frac{(n-1) (2 n+1) (3 n+1) (50 n^2+89n+41)}{90720}$$ We can make shorter expressions transforming the series expansion as a Padé approximant and write, for example, $$r=y\,\frac{1+\sum_{k=1}^3 c_k\, y^k }{1+\sum_{k=1}^3 d_k\, y^k }$$ where $$c_1=\frac{-8 n^3-129 n^2-213 n+26}{12 \left(n^2-20 n+1\right)}\qquad c_2=\frac{-90 n^4-355 n^3-942 n^2-705 n+148}{168 \left(n^2-20 n+1\right)}$$ $$c_3=\frac{-194 n^5-776 n^4-1463 n^3-1867 n^2-844 n+284}{3780 \left(n^2-20 n+1\right)}$$ $$d_1=\frac{-2 n^3-57 n^2-57 n+8}{4 \left(n^2-20 n+1\right)}\qquad d_2=\frac{-170 n^4-823 n^3-1560 n^2-508 n+145}{252 \left(n^2-20 n+1\right)}$$ $$d_3=-\frac{(2 n+1) \left(874 n^4+1559 n^3+3336 n^2+1139 n-428\right)}{15120 \left(n^2-20 n+1\right)}$$

For illustration purposes, let us try with a principal of $250000$, monthly payments of $1500$ and a term of $360$ months. The last expressions yields to $$r=\frac{10989249624307166280122}{2195658966408547872140675}\approx 0.00500499$$ while the exact solution (obtained using Newton method) would give $$r\approx 0.00500583$$ corresponding to a very small relative error $\frac{\Delta r} r=0.017$%

Edit

For given $r$ and $n$ values, $a$ is exactly computed so $y$ and the approximated value proposed above. For illustration purposes, I produced the following table $$\left( \begin{array}{ccc} r_{exact} & n & r_{approx} \\ 0.0025 & 120 & 0.00250000001 \\ 0.0025 & 240 & 0.00250000388 \\ 0.0025 & 360 & 0.00249996731 \\ & & \\ 0.0050 & 120 & 0.00500001927 \\ 0.0050 & 240 & 0.00499985193 \\ 0.0050 & 360 & 0.00499916940 \\ & & \\ 0.0075 & 120 & 0.00749993315 \\ 0.0075 & 240 & 0.00749878217 \\ 0.0075 & 360 & 0.00749326614 \\ & & \\ 0.0100 & 120 & 0.00999973818 \\ 0.0100 & 240 & 0.00999452070 \\ 0.0100 & 360 & 0.00997350925 \end{array} \right)$$