## MATH1013 Quizzes

Quiz 2: The Gregory-Dary method
Question 1 Questions
The function $f\left(x\right)={x}^{3}+3{x}^{2}+4x+1$ has a root between $x=0$ and $x=-1$. Which of the functions below are in the form $x=\varphi \left(x\right)$ and will lead to finding a solution to $f\left(x\right)=0$ using the Gregory-Dary method ?
 a) $\varphi \left(x\right)=-\frac{1}{4}\left({x}^{3}+3{x}^{2}+1\right)$ b) $\varphi \left(x\right)=-\left(3+\frac{4}{x}+\frac{1}{{x}^{2}}\right)$ c) $\varphi \left(x\right)=-\frac{1}{3}\left({x}^{2}+4+\frac{1}{x}\right)$ d) $\varphi \left(x\right)=-\frac{1}{4}\left({x}^{3}-3{x}^{2}-1\right)$

Choice (a) is correct!
${\varphi }^{\prime }\left(x\right)=-\frac{3}{4}{x}^{2}-\frac{3}{2}x$ so $|{\varphi }^{\prime }\left(x\right)|<1$ when $-1.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
The function $f\left(x\right)={x}^{3}+3{x}^{2}+4x+1$ has a root at $x=-0.3177$, correct to 4 decimal places. Using the Gregory-Dary method, how many steps are required to find this root, starting at $x=-0.5$ ?
 a) 3 b) 5 c) 7 d) 9

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
${x}_{1}=-\frac{1}{4}\left({{x}_{0}}^{3}+3{{x}_{0}}^{2}+1\right)$ where ${x}_{0}=-0.5$ ${x}_{1}=-0.4063\phantom{\rule{0.3em}{0ex}}$, ${x}_{2}=-0.3570\phantom{\rule{0.3em}{0ex}}$, ${x}_{3}=-0.3342\phantom{\rule{0.3em}{0ex}}$, ${x}_{4}=-0.3244\phantom{\rule{0.3em}{0ex}}$, ${x}_{5}=-0.3204$ ${x}_{6}=-0.3188\phantom{\rule{0.3em}{0ex}}$, ${x}_{7}=-0.3181\phantom{\rule{0.3em}{0ex}}$, ${x}_{8}=-0.3178\phantom{\rule{0.3em}{0ex}}$, ${x}_{9}=-0.3177\phantom{\rule{0.3em}{0ex}},{x}_{10}=-0.3177$.
Rearrange the equation $f\left(x\right)=3x-2-3sinx$ into the form $x=\varphi \left(x\right)$ and determine if the solution to $f\left(x\right)=0$ can be found using the Gregory-Dary iterative method.
 a) $x=\varphi \left(x\right)=sinx+2$ and the Gregory-Dary iterative method can be used since $|{\varphi }^{\prime }\left(x\right)|\le 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$ b) $x=\varphi \left(x\right)=sinx+\frac{2}{3}$ and the Gregory-Dary iterative method can be used since $|{\varphi }^{\prime }\left(x\right)|\le 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$ c) $x=\varphi \left(x\right)=sinx+2$ and the Gregory-Dary iterative method can be used since $|{\varphi }^{\prime }\left(x\right)|\ge 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$ d) $x=\varphi \left(x\right)=sinx+\frac{2}{3}$ and the Gregory-Dary iterative method can be used since $|{\varphi }^{\prime }\left(x\right)|\ge 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is correct!
If $x=\varphi \left(x\right)=sinx+\frac{2}{3}$ then $|{\varphi }^{\prime }\left(x\right)|=|cosx|\le 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Find the solution to $f\left(x\right)=3x-2-3sinx=0$ correct to 4 decimal places using the Gregory-Dary method.
 a) $x=1.6647$ b) $x=1.6621$ c) $x=1.6625$ d) The solution cannot be found since $|{\varphi }^{\prime }\left(x\right)|\le 1$ for all $x\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Using $x=\varphi \left(x\right)=sinx+\frac{2}{3}$ and starting at ${x}_{0}^{}=1$, we find ${x}_{1}=1.5081$, ${x}_{2}=1.6647$, ${x}_{3}=1.6623$, ${x}_{4}=1.6625$, ${x}_{5}=1.6625$.
Choice (d) is incorrect
Rearrange the equation $f\left(x\right)={x}^{2}-4x-2cos2x=0$ into the form $x=\varphi \left(x\right)$ and determine if the solution of $f\left(x\right)=0$ can be found using the Gregory-Dary iterative method for the root between $x=0.3$ and $x=0.6\phantom{\rule{0.3em}{0ex}}.$
 a) $x=\varphi \left(x\right)=\frac{1}{4}{x}^{2}+\frac{1}{2}cos2x$ b) $x=\varphi \left(x\right)=\sqrt{4x-2cos2x}$ c) $x=\varphi \left(x\right)=\frac{1}{2}arccos\left(2x-\frac{1}{2}{x}^{2}\right)$ d) No re-arrangements of the equation are possible to find a solution.

Choice (a) is incorrect
Choice (b) is correct!
Note that $|{\varphi }^{\prime }\left(x\right)|=|\frac{1}{2}x+cos2x|<1$ for $x$ between $x=0.3$ and $x=0.6\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Rearrange the equation $f\left(x\right)={x}^{2}-4x-2cos2x=0$ into the form $x=\varphi \left(x\right)$ and determine if the solution to $f\left(x\right)=0$ can be found using the Gregory-Dary iterative method for the root between $x=4$ and $x=5\phantom{\rule{0.3em}{0ex}}.$
 a) $x=\varphi \left(x\right)=\frac{1}{4}{x}^{2}+\frac{1}{2}cos2x$ b) $x=\varphi \left(x\right)=\sqrt{4x-2cos2x}$ c) $x=\varphi \left(x\right)=\frac{1}{2}arccos\left(2x-\frac{1}{2}{x}^{2}\right)$ d) No re-arrangements of the equation are possible to find a solution.

Choice (a) is incorrect
Choice (b) is correct!
Note that $\left|{\varphi }^{\prime }\left(x\right)\right|=\left|\frac{2-2sin2x}{\sqrt{4x-2cos2x}}\right|<1$ when $4
Choice (c) is incorrect
Choice (d) is incorrect
The function $f\left(x\right)={x}^{2}-4x+2$ has two roots, $x=\alpha$ and $x=\beta \phantom{\rule{0.3em}{0ex}}.$ Suppose $\alpha <\beta \phantom{\rule{0.3em}{0ex}}.$ Determine the functions ${\varphi }_{1}\left(x\right)$ and ${\varphi }_{2}\left(x\right)$ which would be suitable to find both $\alpha$ and $\beta$ using the Gregory-Dary method.
 a) ${\varphi }_{1}\left(x\right)=x=\frac{1}{4}\left({x}^{2}+2\right)$ is suitable to find both $\alpha$ and $\beta \phantom{\rule{0.3em}{0ex}}.$ b) ${\varphi }_{1}\left(x\right)=x=4-\frac{2}{x}$ is suitable to find both $\alpha$ and $\beta \phantom{\rule{0.3em}{0ex}}.$ c) ${\varphi }_{1}\left(x\right)=x=\frac{1}{4}\left({x}^{2}+2\right)$ is suitable to find $\alpha$ and${\varphi }_{2}\left(x\right)=x=4-\frac{2}{x}$ is suitable to find $\beta \phantom{\rule{0.3em}{0ex}}.$ d) ${\varphi }_{2}\left(x\right)=x=4-\frac{2}{x}$ is suitable to find $\alpha$ and${\varphi }_{1}\left(x\right)=x=\frac{1}{4}\left({x}^{2}+2\right)$ is suitable to find $\beta \phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${x}^{2}-4x+2=0$ has two roots $x=2±\sqrt{2}$ so $\alpha \approx 0.5858$ and $\beta \approx 3.4142\phantom{\rule{0.3em}{0ex}}.$
Consider ${\varphi }_{1}\left(x\right)=\frac{1}{4}\left({x}^{2}+2\right)$, then ${\varphi }_{1}^{\prime }\left(x\right)=\frac{1}{2}x$ and $|{\varphi }_{1}^{\prime }\left(x\right)|<1$ when $0 so ${\varphi }_{1}\left(x\right)$ is suitable to find $\alpha \phantom{\rule{0.3em}{0ex}}.$ Now consider ${\varphi }_{2}\left(x\right)=4-\frac{2}{x}$, then ${\varphi }_{2}^{\prime }\left(x\right)=\frac{2}{{x}^{2}}$ and $|{\varphi }_{2}^{\prime }\left(x\right)|<1$ when $x>\sqrt{2}$ so ${\varphi }_{2}\left(x\right)$ is suitable to find $\beta \phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
The function $f\left(x\right)=4x{e}^{2x}-2$ has a root between $x=0$ and $x=1\phantom{\rule{0.3em}{0ex}}.$ Which of the functions below are in the form $x=\varphi \left(x\right)$ and will lead to finding a solution of $f\left(x\right)=0$ using the Gregory-Dary method ?
 a) $x=\frac{1}{2}{e}^{-2x}$ b) $2x={e}^{-2x}$ c) $x=-\frac{1}{2}ln2x$ d) The function $|{\varphi }^{\prime }\left(x\right)|>1$ on $0 for all choices of $\varphi \left(x\right)\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
$\begin{array}{llll}\hfill \varphi \left(x\right)& =\frac{1}{2}{e}^{-2x}\phantom{\rule{1em}{0ex}}hence\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\varphi }^{\prime }\left(x\right)& =-{e}^{-2x}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}|{\varphi }^{\prime }\left(x\right)|<1\phantom{\rule{1em}{0ex}}on\phantom{\rule{1em}{0ex}}0
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Use the Gregory-Dary method to find the root of the function $f\left(x\right)=4x{e}^{2x}-2$ to 4 decimal places.
 a) $x=0.2835$ b) $x=0.2836$ c) $x=0.2837$ d) $x=0.2805$

Choice (a) is incorrect
Choice (b) is correct!
Iteration needs to be performed 15 times to obtain the solution correct to 4 decimal places and 19 for 5 decimal places.
Choice (c) is incorrect
Choice (d) is incorrect
The equation ${x}^{4}+3{x}^{2}-x-5=0$ has two solutions, $x=-1$ and a solution between $1$ and $2\phantom{\rule{0.3em}{0ex}}.$ Which of the following statements is correct ?
 a) The solution $x=-1$ can be found using the Gregory-Dary method with $x={x}^{4}+3{x}^{2}-5$ and starting at $x=-0.5$ but the other solution cannot be found. b) The solution $x=-1$ cannot be found using the Gregory-Dary method with $x={x}^{4}+3{x}^{2}-5$ but the other solution can be found starting at $x=1.5\phantom{\rule{0.3em}{0ex}}.$ c) Neither solution can be found using the Gregory-Dary method with $x={x}^{4}+3{x}^{2}-5$ since $|{\varphi }^{\prime }\left(x\right)|<1$ on the intervals containing the solutions. d) Neither solution can be found using the Gregory-Dary method with $x={x}^{4}+3{x}^{2}-5$ since $|{\varphi }^{\prime }\left(x\right)|>1$ on the intervals containing the solutions.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{llll}\hfill {\varphi }^{\prime }\left(x\right)& =4{x}^{3}+6x>1\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}11\phantom{\rule{1em}{0ex}}for\phantom{\rule{1em}{0ex}}-1.5 Note that it may be possible to calculate the roots using other methods, in particular the Newton-Raphson method.