The bisection method starts on an interval of length 3. Every step halves the interval. How many steps n are needed at least before the interval is shorter than 0.0001?
Numerical methods worksheet with answers
1.Easy
b − a = 3ε = 0.0001
n =
2.Easy
Use linear interpolation in the table to estimate y at x = 1.65.
x = 1.65
y =
3.Medium
The trapezoid rule is used for ∫02 x³ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 2]. Find the smallest number of strips n that guarantees an error below 0.0005.
f(x) = x³[a, b] = [0, 2]ε = 0.0005
n =
4.Medium
The trapezoid rule is used for ∫02 eˣ dx. The error is at most (b − a)³·M/(12n²) with M = max|f″| on [0, 2]. Find the smallest number of strips n that guarantees an error below 1 × 10⁻⁵.
f(x) = eˣ[a, b] = [0, 2]ε = 1 × 10⁻⁵
n =
5.Hard
Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = 1 + sin(ty), y(0) = 1. Find y₁ ≈ y(0.1).
y′ = 1 + sin(ty)y(0) = 1h = 0.1
y1 =
6.Hard
Use Euler's method with h = 0.25 for y′ = y − t², y(0) = 2 to estimate y5 ≈ y(1.25).
y′ = y − t²y(0) = 2h = 0.25
y5 =
Answer key
- 1. n = 15
- 2. y = 0.7296
- 3. n = 127
- 4. n = 702
- 5. y1 = 1.105
- 6. y5 = 5.55
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