Write down the Newton-Raphson formula for finding a numerical solution to the equation $e^{mx}+bx-a=0$. If $x_0=1$ find $x_1$.
\n\t\tIncluded in the Advice of this question are:
\n\t\t6 iterations of the method.
\n\t\tGraph of the NR process using jsxgraph. Also user interaction allowing change of starting value and its effect on the process.
\n\t\t", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\n\tConsider the following equation.
\n\t\\[\\simplify[std]{e^({m}x)+{b}x-{a}=0}\\]
\n\tFind the approximate solution in the range $0 \\le x \\le 1$ by using the Newton-Raphson method.
\n\tThe following diagram demonstrates the method.
\n\t$x_0$ is the starting value, you can slide it along the x-axis to see the effect of changing it.
\n\t \n\t{test(m,b,a,maxy)}
\n\t \n\t \n\t \n\t", "advice": "\n\ta)
\n\tRecall 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 $g(x)=0$
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}))}\\]
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})}\\]
(In your answers you would input $x$ rather than $x_n$.)
\n\tIn the following let $\\displaystyle f(x)=\\simplify[std]{ (({m} * x -1) * Exp({m} * x ) + {a}) / ({m} * Exp({m} * x ) + {b})}$
\n\tb)
\n\tIf $x_0=2$ then $x_1$ is simply given by:
\\[\\simplify[std]{x_1 = (({2*m} -1) * Exp({2*m}) + {a}) / ({m} * Exp({2*m}) + {b})}\\]
which to 4 decimal places is: $\\;\\;x_1= \\var{ans}$
\n\tWe find on running the iteration that the first six values are:
\n\t\\[\\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}\\]
\n\tSo the solution to the equation to four decimal places for $0 \\le x \\le 1$ is $x=\\var{precround(ans1,4)}$
\n\tHere we see the graph of $\\simplify{e^({m}*x)+{b}*x-{a}}$ and the first four successive approximations to the root:
\n\t{test(m,b,a,maxy)}
\n\t \n\tNote 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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\n\t\t\tUsing 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.
Please note that if you enter a function of the form $xe^{ax}$, then you must input it as $x*e^{ax}$.
\n\t\t\t$p(x_n)=\\;\\;$[[0]] In your answer use $x$ instead of $x_n$.
\n\t\t\t$g'(x_n)=\\;\\;$[[1]] In your answer use $x$ instead of $x_n$.
\n\t\t\tIf you have forgotten the Newton-Raphson formula you can click on Steps to see it. You will not lose any marks in doing so.
\n\t\t\t \n\t\t\t", "stepsPenalty": 0, "steps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "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 $g(x)=0$
If $x_0=2\\;\\;\\;$what is $x_1$ correct to $4$ decimal places?
\n\t\t\t$x_1=\\;\\;$ [[0]]
\n\t\t\tEnter your answer to 4 decimal places.
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\nSuppose $\\simplify{f(x) = {a}*x+{b}}$
\nfind $f(\\var{x0})$
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\n$\\simplify{h(x)={a3}x^2+{b3}x+{c3}}$
\nuse the Numbas syntas set(y,z)
If the roots of a quadratic $r$ and $s$, then the quadratic is $(x-r)(x-s)=x^2-(r+s)x+rs
\nIn this case, which of these statements are true?
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", "templateType": "anything", "group": "part d", "definition": "random(25..80)"}, "m": {"name": "m", "description": "this stands for minutes
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