Iteration
Numerical methods
- A2
- Calculator allowed
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Show that the equation $x^3 + x - 3 = 0$ has a root in the interval $(1, 2)$.
3 marks
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Show that the equation $x^3 - 2x - 5 = 0$ can be rearranged into the form $x = \sqrt[3]{2x + 5}$.
3 marks
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Use the iteration $x_{n+1} = \sqrt[3]{2x_n + 5}$ with $x_0 = 2$ to find $x_1$, $x_2$ and $x_3$, each to $4$ decimal places.
4 marks
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The equation $x^3 + x - 3 = 0$ can be rearranged as $x = \sqrt[3]{3 - x}$. Explain why the iteration $x_{n+1} = \sqrt[3]{3 - x_n}$ would be expected to converge to the root near $x = 1.2$.
4 marks
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Use $x_0 = 1$ and $x_{n+1} = \sqrt[3]{3 - x_n}$ to find the root of $x^3 + x - 3 = 0$ to $3$ decimal places. Show your values of $x_1$, $x_2$ and $x_3$ to $4$ decimal places.
5 marks
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Show that $x^3 - 5x + 2 = 0$ has a root in the interval $(0, 1)$, and show that it can be rearranged as $x = \dfrac{x^3 + 2}{5}$.
5 marks
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Explain why the iteration $x_{n+1} = \dfrac{3}{x_n^2 + 1}$ for the equation $x^3 + x - 3 = 0$ would be expected to diverge near the root in $(1, 2)$.
6 marks
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Use the iteration $x_{n+1} = \dfrac{x_n^3 + 2}{5}$ with $x_0 = 0$ to find the root of $x^3 - 5x + 2 = 0$ that lies in $(0, 1)$, giving your answer to $3$ decimal places.
7 marks
Mark scheme
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Answer. There is a sign change, so there is a root in $(1, 2)$.
- $f(1) = -1$ and $f(2) = 7$.
- $f$ is continuous and changes sign, so a root lies between $1$ and $2$.
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Answer. The rearrangement is $x = \sqrt[3]{2x + 5}$.
- Add $2x + 5$ to both sides, or start from $x^3 = 2x + 5$.
- Take the cube root of both sides.
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Answer. $x_1 = 2.0801$, $x_2 = 2.0924$ and $x_3 = 2.0942$
- $x_1 = \sqrt[3]{9} = 2.0801$.
- $x_2 = 2.0924$.
- $x_3 = 2.0942$.
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Answer. Near the root, $|g'(x)| < 1$, so the iteration converges.
- Let $g(x) = (3 - x)^{1/3}$. Then $g'(x) = -\dfrac{1}{3}(3 - x)^{-2/3}$.
- Near $x = 1.2$, $|g'(x)|$ is about $0.2$, which is less than $1$.
- An iteration converges to a root when the absolute value of the derivative of the iterating function is less than $1$ there.
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Answer. The root is $1.213$ to $3$ decimal places.
- $x_1 = 1.2599$, $x_2 = 1.2028$ and $x_3 = 1.2158$, to $4$ decimal places.
- Further iterations settle at $1.213$ to $3$ decimal places.
- $f(1.213) < 0$ and $f(1.214) > 0$, which agrees with this rounding.
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Answer. There is a sign change on $(0, 1)$, and the rearrangement is $x = \dfrac{x^3 + 2}{5}$.
- $f(0) = 2$ and $f(1) = -2$.
- From $5x = x^3 + 2$, divide by $5$.
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Answer. Near the root, the absolute value of the derivative of the iterating function is greater than $1$, so the iteration diverges.
- The rearrangement $x(x^2 + 1) = 3$ gives $x = \dfrac{3}{x^2 + 1}$.
- $g'(x) = -\dfrac{6x}{(x^2 + 1)^2}$.
- Near $x = 1.2$, $|g'(x)| > 1$.
- The fixed-point iteration diverges when that absolute value is greater than $1$.
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Answer. $0.414$
- $x_1 = 0.4000$, $x_2 = 0.4128$, $x_3 = 0.4141$ and $x_4 = 0.4142$, to $4$ decimal places.
- The values settle at $0.414$ to $3$ decimal places.
- For $g(x) = \dfrac{x^3 + 2}{5}$, $|g'(x)| = \dfrac{3x^2}{5} < 1$ throughout $(0, 1)$, so the iteration converges.