Numerical Method20785 marksNumerical: half-interval bisection methodHalf-Interval method and Convergence
How the half-interval method can be estimate a root of a non-linear equation? Find a real root of the following equation using the half-interval method to correct up to two decimal places. x2 - e-x - x = 1
How the half-interval method can be estimate a root of a non-linear equation? Find a real root of the following equation using the half-interval method to correct up to two decimal places. $x^2 - e^{-x} - x = 1$ [5]
- Equation: $x^2 - e^{-x} - x = 1$ - Rearranged: $f(x) = x^2 - e^{-x} - x - 1 = 0$ - Required accuracy: two decimal places The half-interval (bisection) method is based on the Intermediate Value Theorem: if $f(x)$ is continuous on $[a,b]$ and $f(a)\cdot f(b)<0$, then at least one real root lies i...