Solution for equation with numerical method

Parcly Taxel gave the answer.

Since we know that $i \ll 1$, we can have approximations of the solution expanding the function $$f(i)=1-(1+i)^{-15}-13.5i$$ as a Taylor series built around $i=0$ (or binomial expansion).

This would give $$f(i)=\frac{3 }{2}i-120 i^2+680 i^3+O\left(i^4\right)$$ Discarding the trivial $i=0$ and ignoring the higher order terms we are left with $$\frac{3 }{2}-120 i+680 i^2+O\left(i^3\right)$$ So,a first estimate is $i=\frac 1 {80} =0.0125$ but solving the quadratic will give $i=\frac{30-\sqrt{645}}{340} \approx 0.0135387$.

Starting from the simplest guess, Newton method should converge very fast $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0125000 \\ 1 & 0.0135440 \\ 2 & 0.0134691 \\ 3 & 0.0134687 \end{array} \right)$$

We could even do better using series reversion; this would give $$i=t+\frac{17 }{3}t^2+\frac{697}{18} t^3+\frac{38369 }{135}t^4+O\left(t^5\right)\qquad \text{where} \qquad t=\frac{1}{120} \left(\frac{3}{2}-f(i)\right)$$ Making $f(i)=0$, this would lead to $$i \approx \frac{74472569}{5529600000} =0.0134680$$

Rewrite the equation as $$1-(1+i)^{-15}-13.5i=0$$ Newton's method then tells us to iterate $$i\leftarrow i-\frac{1-(1+i)^{-15}-13.5i}{15(1+i)^{-16}-13.5}$$ where the numerator is the rewritten equation and the denominator is its analytic derivative. Starting from a reasonable value like $i=1$, we reach the correct result of $i=0.0134687$ quickly, or $1.35\%$.