Show that the equation \( x^3 - x - 1 = 0 \) has a root in the interval \( [1,2] \).
\( f(1)=-1<0 \), \( f(2)=5>0 \). Change of sign ⇒ root in \( (1,2) \).
Numerical Methods
10 exam-style questions on change of sign, Newton–Raphson, fixed-point iteration and numerical integration. Tap Show answer for full working.
3 Easy4 Medium3 Hard
Use the trapezium rule with 2 strips to estimate \( \displaystyle\int_0^2 (x^2+1)\, dx \).
\( h=1 \). \( \dfrac{h}{2}(y_0+2y_1+y_2)=\dfrac12(1+2\cdot2+5)=\dfrac12\cdot10=5 \). (Exact is \( 14/3\approx4.67 \).)
Starting with \( x_0=1 \), use the iteration \( x_{n+1}=\sqrt[3]{x_n+1} \) to find \( x_1 \) and \( x_2 \) to 3 d.p.
\( x_1=\sqrt[3]{2}\approx1.260 \), \( x_2=\sqrt[3]{2.260}\approx1.310 \).
Use the Newton–Raphson method with \( x_0=1 \) to find an approximation to the root of \( x^3-x-1=0 \), giving \( x_1 \) and \( x_2 \) to 3 d.p.
\( f'=3x^2-1 \). \( x_1=1-\dfrac{-1}{2}=1.5 \). \( x_2=1.5-\dfrac{1.5^3-1.5-1}{3(2.25)-1}=1.5-\dfrac{0.875}{5.75}\approx1.348 \).
Use the trapezium rule with 4 strips to estimate \( \displaystyle\int_0^1 e^{-x^2}\, dx \) to 3 d.p.
\( h=0.25 \). Values approx: 1, 0.939, 0.779, 0.570, 0.368. Estimate \( \dfrac{0.25}{2}(1+2(0.939+0.779+0.570)+0.368)\approx0.743 \).
Show that the equation \( x=\cos x \) has a root in \( (0,\pi/2) \) and use fixed-point iteration \( x_{n+1}=\cos x_n \) with \( x_0=0.5 \) to find the root to 2 d.p.
\( f(x)=x-\cos x \), \( f(0)=-1<0 \), \( f(\pi/2)>0 \). Iterating: converges to approx 0.74.
The equation \( x^3-3x+1=0 \) has a root near \( x=0.3 \). Use Newton–Raphson once to improve this approximation.
\( f'=3x^2-3 \). \( x_1=0.3-\dfrac{0.027-0.9+1}{0.27-3}=0.3-\dfrac{0.127}{-2.73}\approx0.347 \).
Explain why the Newton–Raphson method may fail to find a root, and give an example.
If \( f'(x_n)=0 \) (horizontal tangent), or if the iteration diverges / cycles. Example: \( f(x)=x^{1/3} \) near 0, or poor initial guess for \( \arctan x=0 \) etc.
Use the trapezium rule with \( n=4 \) to estimate \( \displaystyle\int_1^3 \dfrac{1}{x}\, dx \) and compare with the exact value \( \ln 3 \).
\( h=0.5 \). \( y \): 1, 2/3, 0.5, 0.4, 1/3. Estimate \( \dfrac{0.5}{2}(1+2(2/3+0.5+0.4)+1/3)\approx1.1167 \). Exact \( \ln3\approx1.0986 \). Overestimate.
The iterative formula \( x_{n+1}=\dfrac{1}{2}\left(x_n+\dfrac{2}{x_n}\right) \) is used to find \( \sqrt{2} \). Starting with \( x_0=1 \), find \( x_1 \), \( x_2 \) and \( x_3 \) to 4 d.p. and explain why the sequence converges to \( \sqrt{2} \).
\( x_1=1.5 \), \( x_2=1.4167 \), \( x_3=1.4142 \). Fixed point satisfies \( x=\dfrac12(x+2/x) \) ⇒ \( x^2=2 \). Newton for \( x^2-2=0 \).