Solution for equation with numerical method I'm solving my finances task with annuities and I am struggling in one moment. I already have done this: $$125a_{15,i}=1687$$ $$a_{15,i}=13.5$$ $$a_{15,i}=\frac{1-\left(\frac{1}{1+i}\right)^{15}}{i}$$ So: $$13.5i=1-\left(\frac{1}{1+i}\right)^{15}$$
And now I don't know how to solve this equation. I believe, that here should be used numerical method. For example: Newton method. But not see how to realize it in this case. 
 A: 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
$$i \approx \frac{74472569}{5529600000} =0.0134680$$
A: 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 $1.35\%$.
