Numerical Methods: Newton, Bisection and Finite Differences

A lot of quant questions come down to an equation you cannot solve with algebra. The yield of a bond, the internal rate of return of a set of cash flows and the implied volatility of an option all have the same shape: find $x$ with $f(x) = 0$, where $f$ is easy to evaluate but has no inverse you can write down.

Bisection needs only continuity and a bracket. If $f(a)$ and $f(b)$ have opposite signs, the intermediate value theorem puts a root in $[a, b]$. Evaluate the midpoint, keep the half where the sign still changes, and repeat. Each step halves the bracket, so after $n$ steps the error is at most $(b - a)/2^n$.

Example: pay $100 today and receive $60 at the end of each of the next two years. The IRR solves

Since $f(0) = 20$ and $f(0.5) = -33.3$, a root lies in $[0, 0.5]$. The first midpoint $0.25$ gives $f = -13.6$, so the root is in $[0, 0.25]$. Next $0.125$ gives $f = +0.74$, so it is in $[0.125, 0.25]$. Four more steps leave the bracket $[12.5\%, 13.28\%]$. The exact answer is $13.066\%$.

Bisection is slow. Every step buys one binary digit, so shrinking a bracket of width $0.5$ to $10^{-6}$ takes $\lceil \log_2(0.5/10^{-6}) \rceil = 19$ steps. In return it is guaranteed to converge, and it never needs the derivative.