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};\n\t\t\t var graph = function(x) { return term(x); };\n\t\t\t // Recursion depth\n\t\t\t var steps = 4;\n\t\t\t // Start value\n\t\t\t var s = 2;\n\t\t\t \n\t\t\t //for (i = 0; i < steps; i++) {\n\t\t\t //document.write('x' + i + ' = ');\n\t\t\t //}\n\t\t\t \n\t\t\t var i;\n\t\t\t var ax = brd.create('axis', [[0,0], [1,0]], {strokeColor: 'black'});\n\t\t\t var ay = brd.create('axis', [[0,0], [0,1]], {strokeColor: 'black'});\n\t\t\t \n\t\t\t var g = brd.create('functiongraph', [function(x){return graph(x);}],{strokeWidth: 2, dash:0});\n\t\t\t var x = brd.create('glider',[s,0,ax], {name: 'x_{0}', strokeColor: 'magenta', fillColor: 'yellow'});\n\t\t\t \n\t\t\t newton(x, steps, brd);\t\n\t\t\t \n\t\t\t //function xval() {\n\t\t\t //for (i = 0; i < steps; i++)\n\t\t\t //document.getElementById('xv' + i).innerHTML = (brd.select('x_{' + i + '}').X()).toFixed(14);\n\t\t\t //}\n\t\t\t \n\t\t\t //brd.addHook(xval);\n\t\t\t \n\t\t\t function newton(p, i, board) {\t\n\t\t\t board.suspendUpdate();\t\n\t\t\t if(i>0) {\n\t\t\t var f = board.create('glider',[function(){return p.X();}, function(){return graph(p.X())},g], {name: '', style: 3, strokeColor: 'green', fillColor: 'yellow'});\n\t\t\t var l = board.create('line', [p,f],{strokeWidth: 0.5, dash: 1, straightFirst: false, straightLast: false, strokeColor: 'black'});\n\t\t\t var t = board.create('tangent',[f],{strokeWidth: 0.5, strokeColor: '#0080c0', dash: 0});\n\t\t\t var x = board.create('intersection',[ax,t,0],{name: 'x_{'+(steps-i+1) + '}', style: 4, strokeColor: 'magenta', fillColor: 'yellow'});\n\t\t\t newton(x,--i, board);\n\t\t\t }\n\t\t\t board.unsuspendUpdate(); \n\t\t\t \n\t\t\t \n\t\t\t }\t\n\t\t\t return div;\n\t\t\t", "type": "html", "parameters": [["m", "number"], ["b", "number"], ["a", "number"], ["maxy", "number"]], "language": "javascript"}, "funfornr": {"definition": "\n\t\t\tvar div = Numbas.extensions.jsxgraph.makeBoard('600px','200px',{boundingBox:[0,10,1,-8],grid:false});\n\t\t\t var board = div.board;\n\t\t\t var m = Numbas.jme.unwrapValue(scope.variables.m);\n\t\t\t var be = Numbas.jme.unwrapValue(scope.variables.b);\n\t\t\t var al = Numbas.jme.unwrapValue(scope.variables.a);\n\t\t\t var ans = Numbas.jme.unwrapValue(scope.variables.ans1)\n\t\t\t var a = board.create('point',[ans,0],'$A$');\n\t\t\t \n\t\t\t var graph = board.create('functiongraph',function(x){return Math.exp(m*x)+be*x-al});\n\t\t\t \n\t\t\t return div;\n\t\t\t \n\t\t\t", "type": "html", "parameters": [], "language": "javascript"}, "nr": {"definition": "if(n=5,l+[fun(m,b,a,c)]+[fun(m,b,a,fun(m,b,a,c))],nr(m,b,a,fun(m,b,a,c),n-1,l+[c]))", "type": "list", "parameters": [["m", "number"], ["b", "number"], ["a", "number"], ["c", "number"], ["n", "number"], ["l", "list"]], "language": "jme"}}, "parts": [{"steps": [{"showCorrectAnswer": true, "marks": 0, "variableReplacements": [], "checkvariablenames": false, "showpreview": true, "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 $g(x)=0$

", "checkingtype": "absdiff", "vsetrange": [0, 1], "type": "jme", "scripts": {}, "answer": "", "vsetrangepoints": 5, "showFeedbackIcon": true, "expectedvariablenames": [], "variableReplacementStrategy": "originalfirst", "checkingaccuracy": 0.001}], "type": "gapfill", "marks": 0, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "stepsPenalty": 0, "showFeedbackIcon": true, "prompt": "\n\t\t\t

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 $x_n$.

\n\t\t\t

$g'(x_n)=\\;\\;$[[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.

\n\t\t\t \n\t\t\t", "gaps": [{"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true, "vsetrangepoints": 5, "checkingtype": "absdiff", "vsetrange": [1, 1.5], "type": "jme", "scripts": {}, "answer": "(((({m} * x) -1) * Exp(({m} * x))) + {a})", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "expectedvariablenames": [], "answersimplification": "std", "checkingaccuracy": 0.0001}, {"showCorrectAnswer": true, "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true, "vsetrangepoints": 5, "checkingtype": "absdiff", "vsetrange": [0, 1], "type": "jme", "scripts": {}, "answer": "(({m} * Exp(({m} * x))) + {b})", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "expectedvariablenames": [], "answersimplification": "std", "checkingaccuracy": 0.001}], "showCorrectAnswer": true, "scripts": {}}, {"type": "gapfill", "marks": 0, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "prompt": "\n\t\t\t

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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\n\t

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

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\n\t

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\n\t", "advice": "\n\t

a)

\n\t

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 $g(x)=0$

\n\t

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}))}\\]

\n\t

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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\n\t

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

\n\t", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "\n\t\t

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.

\n\t\t"}, "rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Senida Krcic", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/755/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Senida Krcic", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/755/"}]}