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.
\n\t", "rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "extensions": ["jsxgraph"], "variables": {"tol": {"name": "tol", "group": "Ungrouped variables", "definition": "0", "description": "", "templateType": "anything"}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "precround(tans,4)", "description": "", "templateType": "anything"}, "ans1": {"name": "ans1", "group": "Ungrouped variables", "definition": "precround(results[6],4)", "description": "", "templateType": "anything"}, "maxy": {"name": "maxy", "group": "Ungrouped variables", "definition": "ceil(e^(2*m)+2*b-a)+5", "description": "", "templateType": "anything"}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(1.5..2#0.1)", "description": "", "templateType": "anything"}, "tans": {"name": "tans", "group": "Ungrouped variables", "definition": "((2*m-1)*exp(2*m)+a)/(m*exp(2*m)+b)", "description": "", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(8..15)", "description": "", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random((a+1)..9)", "description": "", "templateType": "anything"}, "results": {"name": "results", "group": "Ungrouped variables", "definition": "nr(m,b,a,2,10,[])", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "m", "ans1", "ans", "b", "tol", "tans", "maxy", "results"], "variable_groups": [], "functions": {"test": {"parameters": [["m", "number"], ["b", "number"], ["a", "number"], ["maxy", "number"]], "type": "html", "language": "javascript", "definition": "\n\t\t\tvar div = Numbas.extensions.jsxgraph.makeBoard('600px','400px', {boundingbox:[0,maxy,3,-30], axis:false});\n\t\t\t var brd=div.board;\n\t\t\t // Initial function term\n\t\t\t var term = function(x) { return Math.exp(m*x)+b*x-a; };\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('This equation has a root in the range $0 \\lt x \\lt 1$.
\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.
\n\t\t\t", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "minValue": "ans-tol", "maxValue": "ans+tol", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}]}