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First 3 terms = ?
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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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 $x^2$.

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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 $x^2$.

", "licence": "Creative Commons Attribution 4.0 International", "notes": "\n \t\t

20/06/2012:

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Added !collectNumbers to some rules so that polynomials presented in standard order.

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3/07/2012:

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9/07/2012:

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Improved display of first line in Advice.

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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} \$

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