Locating roots and iteration Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Locating roots and iteration questions. See exactly how to solve problems on iteration, fixed-point, change-of-sign, root-location.

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.

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
The iterative formula xn+1=1x+2x_{n+1}=\frac{1}{x + 2} is applied with x0=1.0000x_{0}=1.0000. Find the value of x1x_{1}, giving your answer to 4 decimal places.

Worked solution

  1. Write down the iterative formula and the starting value

    xn+1=1x+2,x0=1.0000x_{n+1}=\frac{1}{x + 2},\quad x_{0}=1.0000

    The sequence is generated by repeatedly applying g.

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

    x1=11.0000+2≈0.3333x_{1}=\frac{1}{1.0000 + 2}\approx 0.3333

    Substitute the previous estimate and evaluate to 4 decimal places.

  3. State the value of x_{1}

    x1=0.3333x_{1}=0.3333

    This is the required iterate to 4 decimal places.

Answer
0.33330.3333
Question 3
2 markseasy
The iterative formula xn+1=2x+3x_{n+1}=\sqrt{2 x + 3} is applied with x0=2.0000x_{0}=2.0000. Find the value of x1x_{1}, giving your answer to 4 decimal places.

Worked solution

  1. Write down the iterative formula and the starting value

    xn+1=2x+3,x0=2.0000x_{n+1}=\sqrt{2 x + 3},\quad x_{0}=2.0000

    The sequence is generated by repeatedly applying g.

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

    x1=2⋅2.0000+3≈2.6458x_{1}=\sqrt{2 \cdot 2.0000 + 3}\approx 2.6458

    Substitute the previous estimate and evaluate to 4 decimal places.

  3. State the value of x_{1}

    x1=2.6458x_{1}=2.6458

    This is the required iterate to 4 decimal places.

Answer
2.64582.6458
Question 4
2 markseasy
The iterative formula xn+1=x+63x_{n+1}=\sqrt[3]{x + 6} is applied with x0=1.0000x_{0}=1.0000. Find the value of x1x_{1}, giving your answer to 4 decimal places.

Worked solution

  1. Write down the iterative formula and the starting value

    xn+1=x+63,x0=1.0000x_{n+1}=\sqrt[3]{x + 6},\quad x_{0}=1.0000

    The sequence is generated by repeatedly applying g.

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

    x1=1.0000+63≈1.9129x_{1}=\sqrt[3]{1.0000 + 6}\approx 1.9129

    Substitute the previous estimate and evaluate to 4 decimal places.

  3. State the value of x_{1}

    x1=1.9129x_{1}=1.9129

    This is the required iterate to 4 decimal places.

Answer
1.91291.9129
Question 5
2 markseasy
The iterative formula xn+1=e−xx_{n+1}=e^{- x} is applied with x0=0.5000x_{0}=0.5000. Find the value of x1x_{1}, giving your answer to 4 decimal places.

Worked solution

  1. Write down the iterative formula and the starting value

    xn+1=e−x,x0=0.5000x_{n+1}=e^{- x},\quad x_{0}=0.5000

    The sequence is generated by repeatedly applying g.

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

    x1=e−0.5000≈0.6065x_{1}=e^{- 0.5000}\approx 0.6065

    Substitute the previous estimate and evaluate to 4 decimal places.

  3. State the value of x_{1}

    x1=0.6065x_{1}=0.6065

    This is the required iterate to 4 decimal places.

Answer
0.60650.6065

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