A young boy (13 years old), son of friends of mine, is already very dedicated to mathemetics. He told me that, in the classical formula $$A=P\frac{i \,(i+1)^n}{(i+1)^n-1}$$ using his calculator he was able to compute any number knowing the three other except $i$ for which he needs to use a trial and error method. Since we use to "play mathemetics" (as he says) together, he asked me how he could do it simply.
For sure, I explained what are the problems and how the equation in $i$ could be easily solved by a numerical method such as Newton. This was probably looking too complex and he asked me if a "good" approximation could be given to him. But, deceptively, he asked for a formula which could be used on paper without any calculator.
Accepting the challenge, I worked the equation $$r=\frac{i\,(i+1)^n}{(i+1)^n-1}$$ for which I built two approximations.
The first one by Taylor series (around $i=0$) limited to first order $$r=\frac{i\, (n+1)}{2 n}+\frac{1}{n}\implies i_1=\frac{2 (n r-1)}{n+1}$$ which is an overestimate (Darboux theorem since this corresponds to the first iteration of Newton method).
The second one building the simplest Pade approximant (around $i=0$) $$r=\frac{\frac{1}{n}+\frac{ (n+2)}{3 n}i}{1-\frac{n-1}{6} i }\implies i_2=\frac{6 (n r-1)}{2(n+2)+n(n-1)r}$$ which is, as one could expect, much better than the first one but which corresponds to an underestimate (which I cannot prove).
Trying to improve that, using generated data ($0 \leq i\leq 0.01$ and $1 \leq n \leq 240$) and using linear regression, I finally proposed (rounding the coefficients) $$i_3\approx \frac 14 i_1+\frac 34 i_2$$ which is not too bad.
For example, using $r=0.01$ and $n=180$, the above formulas give $i_1=0.008840$, $i_2=0.006995$, $i_3=0.007456$ while the exact solution is $i=0.007299$.
My question is : does any one know a simple and accurate formula which could be proposed to this young man (remember : pen and paper only) ?
I must confess that I am now stuck (if he would have accepted a formula to be used with a calculator, I could have proposed a nice expression from a slightly more complex Pade approximant but this would require solving an ugly looking quadratic equation). For the conditions used above, it would lead to the estimate $0.007304$.
I was about to give up with this problem when an idea came to my mind.
The approximate solution must not require solving even quadratic equations; so, if the approximation is a Padé approximant, it should be of degree $1$ in numerator and degree $m$ in denominator but, at the same time, the function to be approximated, around $i=0$, has to be $$\frac{i\,(i+1)^n}{(i+1)^n-1}-r$$ In such a way I should need to solve for $i$ equation $$\frac{1-n\, r}{n}+b_m\, i=0$$ which is simple except that coefficients $b_m$ become very quickly quite ugly.
For $m=2$, introducing as an intermediate variable $$y=\frac{2 (n r-1)}{n+1}$$ (which is $i_1$ in the question), I arrived to $$i\approx\frac{2y\left(6+(n-1)y\right)}{12+(n-1)(4-y)y}$$ which is, for sure, not as good as the solution given by a $[2,2]$ Padé approximant but better than the $i_3$.
Repeating the calculation for $n=180$ and $r=0.01$, this leads to $i=0.007319$ while the exact value is $i=0.007299$.
Now, I give up !
Edit (almost three years later)
In an naswer to this question, I reported results obtained meanwhile for the approximation $$i=y\,\frac{\sum_{k=0}^3 c_k\, y^k }{\sum_{k=0}^3 d_k\, y^k }\qquad \text{where}\qquad y=\frac{2 ( nr-1)}{n+1}$$ with $$\left( \begin{array}{cc} \color{red} {k} & \color{red} {c_k} \\ 0 & 15120 \left(n^2-20 n+1\right) \\ 1 & -1260 (n+2) \left(8 n^2+113 n-13\right) \\ 2 & -90 \left(90 n^4+355 n^3+942 n^2+705 n-148\right) \\ 3 & -4 (n+2) \left(194 n^4+388 n^3+687 n^2+493 n-142\right) \end{array} \right)$$ and $$\left( \begin{array}{cc} \color{red} {k} & \color{red} {d_k} \\ 0 & 15120 \left(n^2-20 n+1\right) \\ 1 & -3780 \left(2 n^3+57 n^2+57 n-8\right) \\ 2 & -60 \left(170 n^4+823 n^3+1560 n^2+508 n-145\right) \\ 3 & -(2 n+1) \left(874 n^4+1559 n^3+3336 n^2+1139 n-428\right) \end{array} \right)$$ Repeating the calculation for $n=180$ and $r=0.01$, this leads to $$i=\frac{338717216186170264}{46405051549088067895}\approx 0.00729915$$ while the exact value is $i\approx0.00729945$.