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I'm pretty stuck on the following question

$f$ on $\mathbb{R}$ given by $xfy\Leftrightarrow (y(2x-3)-3x=y(x^2-2x)-5x^3)$ is a function.

Let $g$ be the restriction of $f$ to $\mathbb{Z}^+$, implying $g(n) = f(n),\,n \in\mathbb{Z}^+$

Determine $a\in R\,$, so $g\in \Theta(n^a)$

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Have you expressed $f(x)$ (by expressing $y$ in terms of $x$'s)? – Berci Oct 28 '12 at 22:12
In your last statement: "Determine $a \in R$ you mean $a \in \mathbb{R}$? – amWhy Oct 28 '12 at 22:22
@Berci: Could u eloborate it? amWhy Yes :) – Alek Oliver Oct 28 '12 at 22:30
y=(-5x^3+5x)/(-x^2+4x-3) is that what u mean :)? – Alek Oliver Oct 28 '12 at 22:37
If the question is solved, you indicate that by accepting an answer, not by changing the title to include the word solved. – Graphth Oct 30 '12 at 15:26
up vote 1 down vote accepted

Let's express $y=f(x)$: $$(x^2-2x-2x+3)y=5x^3-3x \implies y =\frac{5x^3-3x}{x^2-4x+3} $$ So, $a=3-2=1$ by the leading exponents.

Update: Formally, you have to prove that there are constants $A,B>0$ such that $An\le f(n)\le Bn\ $ (for $n\ge n_0$).

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Sadly there isnt no further explanation for it. So i simply have to look on the exponents, sounds kinda to simply but then again its only 1/5 of the solution – Alek Oliver Oct 28 '12 at 23:08
Ok. So, find 'Big Theta' in the following page, and use the definition given there: – Berci Oct 28 '12 at 23:36

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