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Recall that the Newton-Raphson method is defined by:
\$x_{n+1}=x_n-\\frac{g(x_n)}{g'(x_n)}\$
where we would like to find the root of the equation $g(x)=0$

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This equation has a root in the range $0 \\lt x \\lt 1$.

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Using the Newton-Raphson formula, if $x_n$ is the $n$th estimate for this root, show that the next estimate can be written in the form \$x_{n+1}= \\frac{p(x_n)}{g'(x_n)}\$
Enter $p(x_n)$ and $g'(x_n)$ in the boxes below.

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Please note that if you enter a function of the form $xe^{ax}$, then you must input it as $x*e^{ax}$.

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$p(x_n)=\\;\\;$[[0]] In your answer use $x$ instead of $x_n$.

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$g'(x_n)=\\;\\;$[[1]] In your answer use $x$ instead of $x_n$.

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If you have forgotten the Newton-Raphson formula you can click on Steps to see it. You will not lose any marks in doing so.

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If $x_0=2\\;\\;\\;$what is $x_1$ correct to $4$ decimal places?

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$x_1=\\;\\;$ [[0]]

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Enter your answer to 4 decimal places.

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Consider the following equation.

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\$\\simplify[std]{e^({m}x)+{b}x-{a}=0}\$

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Find the approximate solution in the range $0 \\le x \\le 1$ by using the Newton-Raphson method.

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The following diagram demonstrates the method.

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$x_0$ is the starting value, you can slide it along the x-axis to see the effect of changing it.

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{test(m,b,a,maxy)}

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a)

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Recall that the Newton-Raphson method is defined by:
\$x_{n+1}=x_n-\\frac{g(x_n)}{g'(x_n)}\$
where we would like to find the root of the equation $g(x)=0$

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In this question we have:
\$\\simplify[std]{g(x) = Exp({m} * x) + {b} * x + { -a}} \\Rightarrow \\simplify[std]{g'(x) = {m}*Exp({m} * x) + {b}}\$
Substituting these expressions into the formula we have:
\$x_{n+1} =\\simplify[std]{ x_n -((Exp({m} * x_n) + {b} * x_n + { -a}) / ({m} * Exp({m} * x_n) + {b}))}\$

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which can be rearranged to give:
\$x_{n + 1} = \\simplify[std]{(({m} * x_n -1) * Exp({m} * x_n) + {a}) / ({m} * Exp({m} * x_n) + {b})}\$

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(In your answers you would input $x$ rather than $x_n$.)

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In the following let $\\displaystyle f(x)=\\simplify[std]{ (({m} * x -1) * Exp({m} * x ) + {a}) / ({m} * Exp({m} * x ) + {b})}$

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b)

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If $x_0=2$ then $x_1$ is simply given by:
\$\\simplify[std]{x_1 = (({2*m} -1) * Exp({2*m}) + {a}) / ({m} * Exp({2*m}) + {b})}\$

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which to 4 decimal places is: $\\;\\;x_1= \\var{ans}$

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We find on running the iteration that the first six values are:

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\\\begin{align}x_1&=f(x_0)=f(2)&=\\var{results[1]}\\\\x_2&=f(x_1)=f(\\var{results[1]})&=\\var{results[2]}\\\\x_3&=f(x_2)=f(\\var{results[2]})&=\\var{results[3]}\\\\x_4&=f(x_3)=f(\\var{results[3]})&=\\var{results[4]}\\\\x_5&=f(x_4)=f(\\var{results[4]})&=\\var{results[5]}\\\\x_6&=f(x_5)=f(\\var{results[5]})&=\\var{results[6]}\\end{align}\

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So the solution to the equation to four decimal places for $0 \\le x \\le 1$ is $x=\\var{precround(ans1,4)}$

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Here we see the graph of $\\simplify{e^({m}*x)+{b}*x-{a}}$ and the first four successive approximations to the root:

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{test(m,b,a,maxy)}

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Note that you can slide the first approximation $x_0$ along the x-axis to see the effect of changing the starting value.

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Write down the Newton-Raphson formula for finding a numerical solution to the equation $e^{mx}+bx-a=0$. If $x_0=1$ find $x_1$.

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Included in the Advice of this question are:

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6 iterations of the method.

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Graph of the NR process using jsxgraph. Also user interaction allowing change of starting value and its effect on the process.

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