This equation has a root in the range $0 \\lt x \\lt 1$.
\nUsing 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$p(x_n)=\\;\\;$[[0]] In your answer use $x$ instead of $x_n$.
\n$g'(x_n)=\\;\\;$[[1]] In your answer use $x$ instead of $x_n$.
\nIf 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 ", "type": "gapfill", "showCorrectAnswer": true, "gaps": [{"showpreview": true, "checkingaccuracy": 0.0001, "answersimplification": "std", "showCorrectAnswer": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "variableReplacements": [], "vsetrangepoints": 5, "type": "jme", "expectedvariablenames": [], "scripts": {}, "marks": 1, "checkingtype": "absdiff", "answer": "(((({m} * x) -1) * Exp(({m} * x))) + {a})", "vsetrange": [1, 1.5]}, {"showpreview": true, "checkingaccuracy": 0.001, "answersimplification": "std", "showCorrectAnswer": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "variableReplacements": [], "vsetrangepoints": 5, "type": "jme", "expectedvariablenames": [], "scripts": {}, "marks": 1, "checkingtype": "absdiff", "answer": "(({m} * Exp(({m} * x))) + {b})", "vsetrange": [0, 1]}], "variableReplacementStrategy": "originalfirst", "stepsPenalty": 0, "marks": 0, "steps": [{"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 $g(x)=0$
If $x_0=2\\;\\;\\;$what is $x_1$ correct to $4$ decimal places?
\n$x_1=\\;\\;$ [[0]]
\nEnter your answer to 4 decimal places.
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\nRecall 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 $x_n$.)
\nIn the following let $\\displaystyle f(x)=\\simplify[std]{ (({m} * x -1) * Exp({m} * x ) + {a}) / ({m} * Exp({m} * x ) + {b})}$
\nb)
\nIf $x_0=2$ 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: $\\;\\;x_1= \\var{ans}$
\nWe find on running the iteration that the first six values are:
\n\\[\\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}\\]
\nSo the solution to the equation to four decimal places for $0 \\le x \\le 1$ is $x=\\var{precround(ans1,4)}$
\nHere we see the graph of $\\simplify{e^({m}*x)+{b}*x-{a}}$ and the first four successive approximations to the root:
\n{test(m,b,a,maxy)}
\n\nNote 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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\nIncluded in the Advice of this question are:
\n6 iterations of the method.
\nGraph of the NR process using jsxgraph. Also user interaction allowing change of starting value and its effect on the process.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "Consider the following equation.
\n\\[\\simplify[std]{e^({m}x)+{b}x-{a}=0}\\]
\nFind the approximate solution in the range $0 \\le x \\le 1$ by using the Newton-Raphson method.
\nThe following diagram demonstrates the method.
\n$x_0$ is the starting value, you can slide it along the x-axis to see the effect of changing it.
\n \n{test(m,b,a,maxy)}
\n \n \n ", "name": "Maria's copy of Mario's copy of Interactive Newton-Raphson method", "functions": {"test": {"parameters": [["m", "number"], ["b", "number"], ["a", "number"], ["maxy", "number"]], "language": "javascript", "definition": "var div = Numbas.extensions.jsxgraph.makeBoard('600px','400px', {boundingbox:[0,maxy,3,-30], axis:false});\n var brd=div.board;\n // Initial function term\n var term = function(x) { return Math.exp(m*x)+b*x-a; };\n var graph = function(x) { return term(x); };\n // Recursion depth\n var steps = 4;\n // Start value\n var s = 2;\n \n //for (i = 0; i < steps; i++) {\n //document.write('