1
$\begingroup$

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 $g\in \Theta(n^a)$

  • 0
    Noticed - thanks2012-10-30

1 Answers 1

1

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 $n\ge n_0$).

  • 0
    Ok. So, find 'Big Theta' in the following page, and use the definition given the$r$e: http://en.w$i$kipedia.org/wiki/Big_O_notation2012-10-28