// Numbas version: exam_results_page_options {"name": "Newton-Raphson method #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "advice": "

\$$f(x)= x^3-\\simplify{{a}+{b}+{c}}x^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}x+\\simplify{-{a}*{b}*{c}}\$$

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The Newton-Raphson formula states:         \$$x_{n+1}=x_n-\\frac{f(x_n)}{f'(x_n)}\$$

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For this example tht gives:                        \$$x_{n+1}=x_n-\\frac{x_n^3-\\simplify{{a}+{b}+{c}}x_n^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}x_n+\\simplify{-{a}*{b}*{c}}}{3x_n^2-\\simplify{2*({a}+{b}+{c})}x_n+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}}\$$

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Take \$$x_0=\\var{x0}\$$

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\$$x_1=\\var{x0}-\\frac{(\\var{x0})^3-\\simplify{{a}+{b}+{c}}(\\var{x0})^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}(\\var{x0})+\\simplify{-{a}*{b}*{c}}}{3(\\var{x0})^2-\\simplify{2*({a}+{b}+{c})}(\\var{x0})+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}}\$$

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\$$x_1=\\var{x0}-\\frac{\\simplify{{x0}^3-({a}+{b}+{c})*{x0}^2+({a}*{b}+{a}*{c}+{b}*{c})*{x0}-{a}*{b}*{c}}}{\\simplify{(3)*{x0}^2-(2*({a}+{b}+{c}))*{x0}+({a}*{b}+{a}*{c}+{b}*{c})}}\$$

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\$$x_1=\\var{x1}\$$

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The second iteration:

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\$$x_2=\\var{x1}-\\frac{(\\var{x1})^3-\\simplify{{a}+{b}+{c}}(\\var{x1})^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}(\\var{x1})+\\simplify{-{a}*{b}*{c}}}{3(\\var{x1})^2-\\simplify{2*({a}+{b}+{c})}(\\var{x1})+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}}\$$

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\$$x_2=\\var{x1}-\\frac{\\simplify{{x1}^3-({a}+{b}+{c})*{x1}^2+({a}*{b}+{a}*{c}+{b}*{c})*{x1}-{a}*{b}*{c}}}{\\simplify{(3)*{x1}^2-(2*({a}+{b}+{c}))*{x1}+({a}*{b}+{a}*{c}+{b}*{c})}}\$$

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\$$x_2=\\var{x2}\$$

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The third iteration:

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\$$x_3=\\var{x2}-\\frac{(\\var{x2})^3-\\simplify{{a}+{b}+{c}}(\\var{x2})^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}(\\var{x2})+\\simplify{-{a}*{b}*{c}}}{3(\\var{x2})^2-\\simplify{2*({a}+{b}+{c})}(\\var{x2})+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}}\$$

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\$$x_3=\\var{x2}-\\frac{\\simplify{{x2}^3-({a}+{b}+{c})*{x2}^2+({a}*{b}+{a}*{c}+{b}*{c})*{x2}-{a}*{b}*{c}}}{\\simplify{(3)*{x2}^2-(2*({a}+{b}+{c}))*{x2}+({a}*{b}+{a}*{c}+{b}*{c})}}\$$

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\$$x_3=\\var{x3}\$$

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The fourth iteration:

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\$$x_4=\\var{x3}-\\frac{(\\var{x3})^3-\\simplify{{a}+{b}+{c}}(\\var{x3})^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}(\\var{x3})+\\simplify{-{a}*{b}*{c}}}{3(\\var{x3})^2-\\simplify{2*({a}+{b}+{c})}(\\var{x3})+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}}\$$

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\$$x_4=\\var{x3}-\\frac{\\simplify{{x3}^3-({a}+{b}+{c})*{x3}^2+({a}*{b}+{a}*{c}+{b}*{c})*{x3}-{a}*{b}*{c}}}{\\simplify{(3)*{x3}^2-(2*({a}+{b}+{c}))*{x3}+({a}*{b}+{a}*{c}+{b}*{c})}}\$$

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\$$x_4=\\var{x4}\$$

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Perform four iterations of the Newton-Raphson method on the function:

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\$$f(x)= x^3-\\simplify{{a}+{b}+{c}}x^2+\\simplify{{a}*{b}+{a}*{c}+{b}*{c}}x+\\simplify{-{a}*{b}*{c}}\$$

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taking  \$$x_0=\\var{x0}\$$  as your approximation.

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The first iteration, correct to three decimal places, gives:

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\$$x_1=\$$ [[0]]

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The second iteration, correct to three decimal places, gives:

\n

\$$x_2=\$$ [[0]]

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The third iteration, correct to three decimal places, gives:

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\$$x_3=\$$ [[0]]

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The fourth iteration, correct to three decimal places, gives:

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\$$x_4=\$$ [[0]]

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