Find the fixed interest rate of a regular monthly deposit with a certain duration and fixed end capital.

Question

I have a bit unusual calculation to do and I am struggling with an analytical solution.

I have a regular monthly deposit $r$ with a certain duration $n$ and fixed end capital $X$. I want to find the fixed interest rate $z$.

By putting $q = 1 + \frac{z}{100}$, I get the following equation to solve for $q$:

$$ r \cdot q \cdot \frac{q^n-1}{q-1}=X $$

Obviously $r \in \mathbb{R}^*$, $X \in \mathbb{R}^*$, $n \in \mathbb{N}^*$, and $q \in \mathbb{R}^*$ (actually $0.9 < q < 1.1$).

I have no problem solving this numerically, but I'm looking for the analytical solution (or at least a very very good approximation).

Answer 1

There is no general analytical solution for large $n$. In effect you are trying to solve an $n$ degree polynomial equation $$q+q^2+\cdots +q^n= \tfrac{X}{r}$$

This is easy enough when $n=1$: you get $q= \tfrac{X}{r}$ as you would expect

It is not much harder when $n=2$: you get $q=\pm \frac12 \sqrt{1+\tfrac{4X}{r}} -\frac12$ and in practice you want the positive square-root

But for $n=3$ and $n=4$ you are solving a cubic or a quartic; you get complicated answers which may involve complex arithmetic: see Wolfram Alpha's solutions when $n=3$ and when $n=4$

That is as far as you can go in aiming for a general closed-form solution: Galois theory suggests that except in special cases, there is not such solution for quintics or higher degree polynomials. So numerical approaches are the best way to go, and in practice also for the $n\ge 3 $ cases

Answer 2

There are many things we could do because $q$ is close to $1$. However, for more simplity, I prefer to use $q=1+z$; so, we have
$$\frac r X = \frac{q-1}{q(q^n-1)}=\frac z{(z+1) \left((z+1)^n-1\right)}\tag 1$$ Let $\color{blue}{k=\frac r X }$ and use methods similar to Newton but of much higher order (have a look here).

In order to fit in the page I limited to order $6$ (this is very high order but I could continue ad infinitum) and I wrote $z_6$ as $$z_6=\frac {\sum_{m=0}^6 a_m\, k^m } {\sum_{m=0}^6 b_m\, k^m } $$ and the coefficients are reported below $$\left( \begin{array}{cc} m & a_m \\ 0 & 24 (n-5) (n-4) (n-3) (n-2) (n-1) \\ 1 & 6 (n-2) (n-1) n \left(95 n^3-118 n^2-1079 n+2014\right) \\ 2 & 2 (n-1) n^2 \left(313 n^4+4746 n^3-5609 n^2-30954 n+43888\right) \\ 3 & -4 (n-1) n^3 \left(281 n^4+42 n^3-13453 n^2-12138 n+44276\right) \\ 4 & -12 (n-1) n^4 \left(12 n^4+644 n^3+2609 n^2-4676 n-17453\right) \\ 5 & 2 n^5 \left(26 n^5+268 n^4-4615 n^3-27725 n^2-6211 n+68497\right) \\ 6 &-2 n^6 \left(2 n^5-2 n^4-355 n^3+355 n^2+11153 n+19087\right) \end{array} \right)$$ $$\left( \begin{array}{cc} m & b_m \\ 0 & 3 (n-6) (n-5) (n-4) (n-3) (n-2) (n-1) \\ 1 & 2 (n-3) (n-2) (n-1) n \left(73 n^3-157 n^2-604 n+1516\right) \\ 2 & (n-2) (n-1) n^2 \left(590 n^4+2617 n^3-7952 n^2-14425 n+33354\right) \\ 3 & 4 (n-1) n^3 \left(66 n^5+1501 n^4+3930 n^3-12065 n^2-18036 n+34324\right) \\ 4 & -(n-1) n^4 \left(62 n^5-456 n^4-17851 n^3-69126 n^2+3317 n+168510\right) \\ 5 & 2 (n-1) n^5 (n+8) \left(2 n^4-112 n^3+247 n^2+4480 n+7359\right) \\ 6 & 7 n^6 (n+7) \left(2 n^4-16 n^3-33 n^2+376 n+751\right) \end{array} \right)$$

For illustration purposes, let us try for $X=100\, r$, that is to say $k=\frac 1 {100}$, and try for different $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 12 & +0.3044256119 & +0.3048502749 \\ 24 & +0.1017554837 & +0.1017680157 \\ 36 & +0.0497192682 & +0.0497202620 \\ 48 & +0.0274883517 & +0.0274884336 \\ 60 & +0.0156874365 & +0.0156874415 \\ 72 & +0.0086040173 & +0.0086040173 \\ 84 & +0.0040002749 & +0.0040002749 \\ 96 & +0.0008365056 & +0.0008365056 \\ 108 & -0.0014293462 & -0.0014293462 \\ 120 & -0.0031045139 & -0.0031045139 \\ 132 & -0.0043745620 & -0.0043745620 \\ 144 & -0.0053573204 & -0.0053573204 \\ 156 & -0.0061306994 & -0.0061306994 \\ 168 & -0.0067479655 & -0.0067479566 \\ 180 & -0.0072465748 & -0.0072465500 \\ 192 & -0.0076535041 & -0.0076534458 \\ 204 & -0.0079885900 & -0.0079884685 \\ 216 & -0.0082666844 & -0.0082664544 \\ 228 & -0.0084990861 & -0.0084986823 \\ 240 & -0.0086945140 & -0.0086938466 \\ 252 & -0.0088597823 & -0.0088587326 \\ 264 & -0.0090002789 & -0.0089986950 \\ 276 & -0.0091203088 & -0.0091180012 \\ 288 & -0.0092233457 & -0.0092200830 \\ 300 & -0.0093122178 & -0.0093077227 \\ 312 & -0.0093892477 & -0.0093831926 \\ 324 & -0.0094563578 & -0.0094483613 \\ 336 & -0.0095151519 & -0.0095047746 \\ 348 & -0.0095669776 & -0.0095537185 \\ 360 & -0.0096129758 & -0.0095962653 \end{array} \right)$$

Edit

Another thing which could be done is to Taylor expand $(1)$ and use series reversion. This would lead to $$z=\sum_{m=1}^p \frac {c_m} {d_m} t^m \qquad \text{where} \qquad t=\frac{2(1- k n)}{n+1}$$ where the $d_m$ make the sequence $$\{1,6,36,1080,6480,90720,2721600\}$$ which is not in $OEIS$ and the $c_m$ are given in the next table $$ \left( \begin{array}{cc} 1 & 1 \\ 2 & n+5 \\ 3 & 2 n^2+11 n+23 \\ 4 & 22 n^3+153 n^2+402 n+503 \\ 5 & 52 n^4+428 n^3+1437 n^2+2438 n+2125 \\ 6 & 300 n^5+2836 n^4+11381 n^3+24879 n^2+30911 n+20413 \\ 7 & 3824 n^6+40692 n^5+188712 n^4+496259 n^3+799917 n^2+780417 n+411779 \end{array} \right) $$ but we should much more terms to reach the same accuracy as with the Padé like approximant.