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I have the following complicated expression $E(n)$ and I am interested in finding a simple expression $S(n)$ such that $S(n) \leq E(n)$ for all sufficiently large $n$. Does anyone know how one can do this in maple?

$E(n) = $ $ 1/8\,\Biggl( 2\,\sqrt {2\,{n}^{2}-2\,n+1}\sqrt {2}n+\sqrt {2\,{n}^{2}- 2\,n+1}n\sqrt {{\frac {{n}^{2}+1}{{n}^{2}}}}-\sqrt {2\,{n}^{2}-2\,n+1} $ $ -\Biggl(18\,{n}^{4}-18\,{n}^{3}+13\,{n}^{2}+8\,\sqrt {2}{n}^{4}\sqrt { {\frac {{n}^{2}+1}{{n}^{2}}}}-8\,\sqrt {2}{n}^{3}\sqrt {{\frac {{n}^{2 }+1}{{n}^{2}}}}+4\,\sqrt {2}{n}^{2}\sqrt {{\frac {{n}^{2}+1}{{n}^{2}}} }-8\,\sqrt {2}{n}^{3}+ $ $8\,\sqrt {2}{n}^{2}-4\,\sqrt {2}n-4\,n+2-4\,{n}^ {3}\sqrt {{\frac {{n}^{2}+1}{{n}^{2}}}}+ $ $ 4\,{n}^{2}\sqrt {{\frac {{n}^{ 2}+1}{{n}^{2}}}}-2\,n\sqrt {{\frac {{n}^{2}+1}{{n}^{2}}}}+16\,\sqrt {2 \,{n}^{2}-2\,n+1}-16\,\sqrt {2\,{n}^{2}-2\,n+1}n\sqrt {{\frac {{n}^{2} +1}{{n}^{2}}}}\Biggr)^{-1/2}\Biggr) $ $ {\frac {1}{\sqrt {2\,{n}^{2}-2\,n+1}}} \left( -1+n\sqrt {{\frac {{n}^{2}+1}{{n}^{2}}}} \right) ^{-1} $

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    sorry I don't know how to add a line break to the expression2012-01-30

2 Answers 2

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asympt(E(n),n);

should give you an asymptotic expression for this as $n \to \infty$. (I would do this myself but I don''t feel like typing this monster into Maple myself, and Maple doesn't read LaTex very well). If e.g. you get something like $a n^p + b n^{p-1} + O(n^{p-2})$, then you might try $S(n) = a n^p + c n^{p-1}$ where $c < b$.

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I do not know how to do it in Mapple, but I am sure that you can do it "by hand" using some simple inequalities, like $ \sqrt{2}\,(n-1)\le\sqrt{2\,n^2-2\,n+1}=\sqrt{n^2+(n-1)^2}\le\sqrt{2}\,n $ and $ 1\le\sqrt{\frac{n^2+1}{n^2}}=\sqrt{1+\frac1{n^2}}\le1+\frac1{2\,n^2}\ . $ I am unable to read the rest of the expression, so I cannot give more inequalities.