// Numbas version: finer_feedback_settings {"name": "Maria's copy of Taylor series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Maria's copy of Taylor series (three terms)", "tags": [], "metadata": {"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 $x^2$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

You are asked to find the first 3 terms in the Taylor series at $x=\\var{c}$ for $f(x)=(\\simplify[all]{{a-b*c}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$.

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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} \\]

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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!.

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Do not input factorials or decimals in the Taylor series.

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