A-Level Locating roots and iteration Practice Questions

Free A-Level Locating roots and iteration practice questions with full step-by-step worked solutions. Covers iteration, fixed-point, change-of-sign, root-location. Practise exam-style problems and check your method.

iterationfixed-pointchange-of-signroot-locationrearrangementfailure
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The iterative formula xn+1=cos⁡(x)x_{n+1}=\cos{\left(x \right)} is applied with x0=0.5000x_{0}=0.5000. Find the value of x1x_{1}, giving your answer to 4 decimal places.
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Worked solution

  1. Write down the iterative formula and the starting value

    xn+1=cos⁡(x),x0=0.5000x_{n+1}=\cos{\left(x \right)},\quad x_{0}=0.5000

    The sequence is generated by repeatedly applying g.

  2. Apply the iterative formula to find x_{1}

    x1=cos⁡(0.5000)≈0.8776x_{1}=\cos{\left(0.5000 \right)}\approx 0.8776

    Substitute the previous estimate and evaluate to 4 decimal places.

  3. State the value of x_{1}

    x1=0.8776x_{1}=0.8776

    This is the required iterate to 4 decimal places.

Answer
0.87760.8776
Question 2
2 markseasy
A fixed-point iteration is defined by xn+1=x2−2x_{n+1}=x^{2} - 2 with fixed point α\alpha. Using the value of g′(α)g'(\alpha), which statement best describes the behaviour of the iteration and its staircase or cobweb diagram?
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Worked solution

  1. Locate the fixed point alpha

    α≈2.0000\alpha\approx 2.0000

    The fixed point satisfies x=g(x).

  2. Differentiate g and evaluate the derivative at alpha

    g′(x)=2x,g′(α)≈4.0000g'(x)=2 x,\quad g'(\alpha)\approx 4.0000

    The size and sign of g'(alpha) fix the behaviour of the iteration.

  3. State the behaviour of the iteration near alpha

    ∣g′(α)∣>1\left|g'(\alpha)\right|>1

    This determines whether the staircase/cobweb converges or diverges.

Answer
The iteration diverges from the fixed point, and the diagram is a staircase (because g'(alpha) > 1).
Question 3
3 marksintermediate
A fixed-point iteration is defined by xn+1=cos⁡(x)x_{n+1}=\cos{\left(x \right)} with fixed point α\alpha. Using the value of g′(α)g'(\alpha), which statement best describes the behaviour of the iteration and its staircase or cobweb diagram?
Show worked solution

Worked solution

  1. Locate the fixed point alpha

    α≈0.7391\alpha\approx 0.7391

    The fixed point satisfies x=g(x).

  2. Differentiate g and evaluate the derivative at alpha

    g′(x)=−sin⁡(x),g′(α)≈−0.6736g'(x)=- \sin{\left(x \right)},\quad g'(\alpha)\approx -0.6736

    The size and sign of g'(alpha) fix the behaviour of the iteration.

  3. Apply the iterative formula to find x_{1}

    x1=cos⁡(0.50000)≈0.87758x_{1}=\cos{\left(0.50000 \right)}\approx 0.87758

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=cos⁡(0.87758)≈0.63901x_{2}=\cos{\left(0.87758 \right)}\approx 0.63901

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=cos⁡(0.63901)≈0.80269x_{3}=\cos{\left(0.63901 \right)}\approx 0.80269

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. State the behaviour of the iteration near alpha

    ∣g′(α)∣<1\left|g'(\alpha)\right|<1

    This determines whether the staircase/cobweb converges or diverges.

Answer
The iteration converges to the fixed point, and the diagram is a cobweb (because -1 < g'(alpha) < 0).
Question 4
5 markshard
A fixed-point iteration is defined by xn+1=x27+37x_{n+1}=\frac{x^{2}}{7} + \frac{3}{7} with fixed point α\alpha. Using the value of g′(α)g'(\alpha), which statement best describes the behaviour of the iteration and its staircase or cobweb diagram?
Show worked solution

Worked solution

  1. Locate the fixed point alpha

    α≈0.4586\alpha\approx 0.4586

    The fixed point satisfies x=g(x).

  2. Differentiate g and evaluate the derivative at alpha

    g′(x)=2x7,g′(α)≈0.1310g'(x)=\frac{2 x}{7},\quad g'(\alpha)\approx 0.1310

    The size and sign of g'(alpha) fix the behaviour of the iteration.

  3. Apply the iterative formula to find x_{1}

    x1=1.0000027+37≈0.57143x_{1}=\frac{1.00000^{2}}{7} + \frac{3}{7}\approx 0.57143

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=0.5714327+37≈0.47522x_{2}=\frac{0.57143^{2}}{7} + \frac{3}{7}\approx 0.47522

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=0.4752227+37≈0.46083x_{3}=\frac{0.47522^{2}}{7} + \frac{3}{7}\approx 0.46083

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. Apply the iterative formula to find x_{4}

    x4=0.4608327+37≈0.45891x_{4}=\frac{0.46083^{2}}{7} + \frac{3}{7}\approx 0.45891

    Substitute the previous estimate and evaluate to 5 decimal places.

  7. Apply the iterative formula to find x_{5}

    x5=0.4589127+37≈0.45866x_{5}=\frac{0.45891^{2}}{7} + \frac{3}{7}\approx 0.45866

    Substitute the previous estimate and evaluate to 5 decimal places.

  8. Apply the iterative formula to find x_{6}

    x6=0.4586627+37≈0.45862x_{6}=\frac{0.45866^{2}}{7} + \frac{3}{7}\approx 0.45862

    Substitute the previous estimate and evaluate to 5 decimal places.

  9. Apply the iterative formula to find x_{7}

    x7=0.4586227+37≈0.45862x_{7}=\frac{0.45862^{2}}{7} + \frac{3}{7}\approx 0.45862

    Substitute the previous estimate and evaluate to 5 decimal places.

  10. State the behaviour of the iteration near alpha

    ∣g′(α)∣<1\left|g'(\alpha)\right|<1

    This determines whether the staircase/cobweb converges or diverges.

Answer
The iteration converges to the fixed point, and the diagram is a staircase (because 0 < g'(alpha) < 1).
Question 5
8 markschallenging
A fixed-point iteration is defined by xn+1=1x+1x_{n+1}=\frac{1}{x + 1} with fixed point α\alpha. Using the value of g′(α)g'(\alpha), which statement best describes the behaviour of the iteration and its staircase or cobweb diagram?
Show worked solution

Worked solution

  1. Locate the fixed point alpha

    α≈0.6180\alpha\approx 0.6180

    The fixed point satisfies x=g(x).

  2. Differentiate g and evaluate the derivative at alpha

    g′(x)=−1(x+1)2,g′(α)≈−0.3820g'(x)=- \frac{1}{\left(x + 1\right)^{2}},\quad g'(\alpha)\approx -0.3820

    The size and sign of g'(alpha) fix the behaviour of the iteration.

  3. Apply the iterative formula to find x_{1}

    x1=11.00000+1≈0.50000x_{1}=\frac{1}{1.00000 + 1}\approx 0.50000

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=10.50000+1≈0.66667x_{2}=\frac{1}{0.50000 + 1}\approx 0.66667

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=10.66667+1≈0.60000x_{3}=\frac{1}{0.66667 + 1}\approx 0.60000

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. Apply the iterative formula to find x_{4}

    x4=10.60000+1≈0.62500x_{4}=\frac{1}{0.60000 + 1}\approx 0.62500

    Substitute the previous estimate and evaluate to 5 decimal places.

  7. Apply the iterative formula to find x_{5}

    x5=10.62500+1≈0.61538x_{5}=\frac{1}{0.62500 + 1}\approx 0.61538

    Substitute the previous estimate and evaluate to 5 decimal places.

  8. Apply the iterative formula to find x_{6}

    x6=10.61538+1≈0.61905x_{6}=\frac{1}{0.61538 + 1}\approx 0.61905

    Substitute the previous estimate and evaluate to 5 decimal places.

  9. Apply the iterative formula to find x_{7}

    x7=10.61905+1≈0.61765x_{7}=\frac{1}{0.61905 + 1}\approx 0.61765

    Substitute the previous estimate and evaluate to 5 decimal places.

  10. Apply the iterative formula to find x_{8}

    x8=10.61765+1≈0.61818x_{8}=\frac{1}{0.61765 + 1}\approx 0.61818

    Substitute the previous estimate and evaluate to 5 decimal places.

  11. Apply the iterative formula to find x_{9}

    x9=10.61818+1≈0.61798x_{9}=\frac{1}{0.61818 + 1}\approx 0.61798

    Substitute the previous estimate and evaluate to 5 decimal places.

  12. Apply the iterative formula to find x_{10}

    x10=10.61798+1≈0.61806x_{10}=\frac{1}{0.61798 + 1}\approx 0.61806

    Substitute the previous estimate and evaluate to 5 decimal places.

  13. Apply the iterative formula to find x_{11}

    x11=10.61806+1≈0.61803x_{11}=\frac{1}{0.61806 + 1}\approx 0.61803

    Substitute the previous estimate and evaluate to 5 decimal places.

  14. Apply the iterative formula to find x_{12}

    x12=10.61803+1≈0.61804x_{12}=\frac{1}{0.61803 + 1}\approx 0.61804

    Substitute the previous estimate and evaluate to 5 decimal places.

  15. State the behaviour of the iteration near alpha

    ∣g′(α)∣<1\left|g'(\alpha)\right|<1

    This determines whether the staircase/cobweb converges or diverges.

Answer
The iteration converges to the fixed point, and the diagram is a cobweb (because -1 < g'(alpha) < 0).

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