The bisection method starts on an interval of length 4. Every step halves the interval. How many steps n are needed at least before the interval is shorter than 0.0001?
Find n
Show hints
- After n steps the interval has length (b − a)/2ⁿ.
- Solve 4/2ⁿ < 0.0001, so 2ⁿ > 40000.
- n > log₂(40000) = 15.29
Show the answer and the step-by-step solution
Root finding and interpolation: bisection keeps an interval [a, b] on which f changes sign; each step takes the midpoint c = (a + b)/2 and keeps the half where the sign still changes, so after n steps the interval is (b − a)/2ⁿ wide. Newton–Raphson follows the tangent line to the axis: xn+1 = xn − f(xn)/f′(xn); near a simple root it roughly doubles the number of correct digits per step. Linear interpolation between two table points: y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁).