Tricky Algebra: Determine Interest Rate For Compound Interest With Regular Payments

I am looking to find a formula that gives me the interest rate required to amass a certain wealth, given $$0$$ principle, but with regular payments of fixed value and a fixed interest rate.

In other words, if I invest some amount of momey every period, and interest is compounded once in every one of these periods, what was the interest rate if I end up with some amount after some time.

I can't seem to simplify the equation down in terms of interest rates.

Considering the variables:

$$A$$ = Final amount, $$P$$ = Payment amount, $$N$$ = Number of times interest is compounded per unit time, $$T$$ = duration, $$R$$ = Interest for time period.

It's easy to start with the compound interest formula (where $$p=Principle$$), which is $$A = p\left(1+\left(\frac{R}{N}\right)\right)^{NT}$$ and account for regular payments with the formula $$A = M \frac{\left(1+\frac{R}{N}\right)^{NT}-1}{R/N} \left(1+\frac{R}{N}\right)$$

Just to explain the formula, the final term $$(1+\frac{R}{N})$$ merely adds an extra compounding period. It has the effect that we are effectively depositing money at the start of the compounding period - without this term a total duration of t = 1/n (i.e the duration of a single compounding period) yields no interest at all.

The term $$((1+\frac{R}{N})^{NT}-1)$$ yields the total interest across the time period, and so we divide by the interest rate $$({\frac{R}{N}})$$ to get the return on investment per unit payment value. Multiply this by the value of our investment, $$M$$ and we have the total amount amassed.

Note it is a somewhat arbitrary choice that the payment period and interest compounding period align.

My issue is in the algebra to rearrange this formula in terms of R, so that R can be determined when given some A, M, N and T. How on earth do I write this equation in terms of R? Is there even a unique solution?

• Thanks @gt6989b, I have edited the formula. I was thinking numerical might be the way, but I wanted to see if there actually were an analytic solution which my poor algebra skills merely couldn't find. – Alan Feb 19 '20 at 20:02
• The regular payments formula still has some issues with the brackets ... – mjw Feb 19 '20 at 20:03
• @gt6989b, well, okay, we must have been editing at the same time? Can we undo my edits? – mjw Feb 19 '20 at 20:05
• @mjw :) np, i re-edited again, can you and the OP please validate this – gt6989b Feb 19 '20 at 20:06
• @gt6989b that's much better, thanks! – Alan Feb 19 '20 at 20:06

Considering the equation$$A = M \frac{\left(1+\frac{R}{N}\right)^{NT}-1}{\frac RN}\left(1+\frac{R}{N}\right)\tag1$$ define $$a=\frac AM \qquad, \qquad x=\frac RN\qquad,\qquad n=NT$$ to make the equation $$a=\frac{(1+x)^n-1} x(1+x) \tag 2$$ What we know is that $$x\ll 1$$; so, let us develop the rhs using the binomial theorem $$a=n+\sum_{k=1}^\infty \binom{n+1}{k+1}x^k$$ Transform it as simple Padé approximants which could be $$a=n\frac{ (n+5) x+6}{6-2 (n-1) x}\implies x=\frac{6 (a-n)}{2 a (n-1)+n (n+5)}$$

Let us try with $$a=500$$ and $$n=200$$. The above formula would give $$a=\frac{3}{400}=0.00750$$ while the exact solution would be $$0.00809$$; this is not so bad taking into account the huge value assigned to $$a$$.

Better, but at the price of a quadratic equation $$a=n\frac{60 +6 (n+13) x+ (n^2+3 n+20) x^2 } {60-24 (n-2) x+3 (n-2) (n-1) x^2 }$$ For the worked example, selecting the "reasonable" root, this would give $$x=0.00812$$

For sure, we could continue improving but this would be at the price of cubic or quartic equations which can be solved with radicals. To give you an idea, using cubic equations, we should get $$x=0.00809422$$ while the exact solution is $$0.00809450$$.

Play with that and, please, tell me how it works for your cases.

Edit

The first way described above is "neutral" in the sense that $$a$$ is set equal to the ratio of polynomials leading to linear, quadratic, cubic or quartic equations in $$x$$.

There is another way. Rewrite $$(2)$$ as $$(3)$$ $$\frac 1a=\frac{x}{(x+1) \left((x+1)^n-1\right)}\tag 3$$ Expand the rhs as a Taylor series centered at $$x=0$$ and use series reversion to get $$x=t+\frac {b_1} 6 t^2+\frac {b_2} {36} t^3+\frac {b_3} {1080} t^4+\frac {b_4} {6480} t^5+\frac {b_5} {90720} t^6+\frac {b_6} {2721600} t^7+O(t^{8})$$ where $$t=\frac{2(a-n)}{(n+1)a}$$. The coefficients are listed in the table below $$\left( \begin{array}{cc} k & b_k \\ 1 & n+5 \\ 2 & 2 n^2+11 n+23 \\ 3 & 22 n^3+153 n^2+402 n+503 \\ 4 & 52 n^4+428 n^3+1437 n^2+2438 n+2125 \\ 5 & 300 n^5+2836 n^4+11381 n^3+24879 n^2+30911 n+20413 \\ 6 & 3824 n^6+40692 n^5+188712 n^4+496259 n^3+799917 n^2+780417 n+411779 \end{array} \right)$$

Defining $$x_{(p)}=t+\sum_{k=0}^p \frac{b_k}{c_k}\,t^{k+1}$$ for the worked example we should get the following values $$\left( \begin{array}{cc} p & x_{(p)} \\ 0 & 0.0059701493 \\ 1 & 0.0071879409 \\ 2 & 0.0076739522 \\ 3 & 0.0078882746 \\ 4 & 0.0079897311 \\ 5 & 0.0080399577 \\ 6 & 0.0080655906 \\ 7 & 0.0080789649 \\ \cdots & \cdots \\ \infty &0.0080945103 \end{array} \right)$$

• Thanks Claude for the detailed answer. I'll need to spend some time with it to understand some of your magic. I'll get back with comments and/or accepted answer! – Alan Mar 4 '20 at 11:40
• +1 Magic indeed :) – gt6989b Mar 4 '20 at 14:29
• @gt6989b. Don't tell my wife that I am in love with Taylor (series) for more than 62 years ! No magic at all. Thanks for the kind word. Cheers :-) – Claude Leibovici Mar 4 '20 at 17:38

Even in the easy case of $$N=1$$, the expression comes down to $$A = M(1+R) \frac{\left(1+R\right)^T-1}{R},$$ which even for integer $$T \ge 4$$ is equivalent to looking for the roots of a quintic polynomials, which generally has no analytic solution in the radicals...