SO, A function $f(x)$ is given to me and but the initial intervals are not given. I need to find the root of the equation using Bisection method.
Sometimes when I randomly guess the initial interval which obeys $f(a)·f(b)<0$ (where $a$ and $b$ are initial intervals) the no of iterations gets ridiculously large, sometimes 10-15 iterations. Since, I'll only have 3 hours in examination, I don't want to waste too much time on a single question. I know that the closer to the actual root my initial guesses are the lesser the no of iterations will be required.
Is there any quick way to calculate the root using calculator. I have a casio fx-991ES. I have tried using the solve equation function but for 2nd and 3rd degree equations calculator gives 2 and 3 roots which is what it is supposed to do. But since I'll be finding only one root with bisection method, I am confused on which root to take.
Best Answer
It does defeat the task if you use other means to get at the solution. If you already know the solution, then there is no need to use the bisection method.
For polynomials, you can use some outer root radius estimate and calculate a function value table to find intervals with changing signs. There is no guarantee to find any such interval, for instance for the polynomial $f(x)=(x-\sqrt2)^2-10^{-8}$, which has two real roots, it is highly improbable to land in the small interval of negative values.
The bisection method is inherently slow. To reduce an initial interval length by a factor of $2^n$ always requires exactly $n$ steps. If you want faster convergence with a bracketed method use the Illinois variant of regula falsi, it is only modestly more complex than the bisection method.