Find the first few terms of the Maclaurin and Taylor series of given functions.
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Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.
You are asked to find the first 3 terms in the MacLaurin series for $f(x)=(\\simplify[all]{{a}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$.
", "tags": ["approximate", "approximations", "calculus", "Calculus", "checked2015", "first three terms in a maclaurin series", "First three terms in MacLaurin series", "functions", "maclaurin series", "MacLaurin series", "mas1601", "MAS1601"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t20/06/2012:
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\n \t\t3/07/2012:
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Find the first 3 terms in the MacLaurin series for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "The first three terms in the MacLaurin series are given by $a+bx+cx^2$ where $\\displaystyle a=f(0),\\;\\;b=f'(0),\\;\\;c=\\frac{f''(0)}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(0)=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(0)=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(0)}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the MacLaurin series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2} \\]
Input the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.
\n
Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.
Do not input factorials or decimals in the Taylor series.
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", "tags": ["3 term Taylor series", "approximation", "approximations", "Calculus", "calculus", "checked2015", "function", "functions", "mas1601", "MAS1601", "series approximation", "series expansion", "Taylor series", "taylor series", "three term Taylor series"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n \t\t20/06/2012:
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\n \t\t3/07/2012:
Added tags.
\n \t\t9/07/2012:
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\n \t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Find the first 3 terms in the Taylor series at $x=c$ for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.
"}, "functions": {}, "advice": "The first three terms in the Taylor series are given by $\\simplify[all]{a+b(x-{c})+c(x-{c})^2}$ where $\\displaystyle a=f(\\var{c}),\\;\\;b=f'(\\var{c}),\\;\\;c=\\frac{f''(\\var{c})}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a-b*c}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a-b*c}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(\\var{c})=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(\\var{c})=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(\\var{c})}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the Taylor series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2} \\]