1

Can someone explain the following about big-O from the textbook to me? (I'm trying to catch up after missing classes due to illness.)

Show that $f(x) \in O(x^2)$ where $f(x) = 8x+9$. List the witnesses $C$ and $k$.

From what I have gathered I am looking at/for a distinct point witnessed by $C$/$k$ wherein there is a (lower bound?) point to where one of the functions ($f(x)$ or $g(x)$) is increasing at a faster rate for any given point $\geq C$?

The textbook talks testing/trying out different $C$ witnesses (ex. 1) to where I then solve for $8x+9\leq(1)x^2$ or (ex. 2) and solve for $8x+9\leq(2)x^2$, but I'm not sure where I go from there. If I'm letting $C=1$ then I get:

\begin{gather*} 8x+9\leq(1)x^2\\ 0\leq x^2-8x-9\,. \end{gather*}

Then from here I have tried the quadratic equation resulting in $(8 \pm \sqrt{100})/(2)$ or I saw where $0\leq x^2-8x-9$ could be factored to $(x+1)(x-9)$ so I know $-1$ and $9$ make zeroes on my number line, and I tried plugging in test points where I found everything less than $-1$ or greater than $9$ yields a positive, and everything between $-1$ and $9$ yields a negative, but from this point I am lost.

Can someone explain what I'm supposed to be doing and how to go about it?

Raphael
  • 72,336
  • 29
  • 179
  • 389
Analytic Lunatic
  • 119
  • 3
  • Maybe you could try complexity notation? Could you try calculating limit (f(x)/g(x)) instead of square root? – Evil Feb 18 '16 at 16:44
  • 2
    Questions like this abound on the site. Can someone link a close one/duplicate? – Raphael Feb 18 '16 at 19:41
  • You need to look at the definition of O. – Raphael Feb 18 '16 at 19:42
  • 1
    There are various definitions of big-O (they're all equivalent but they often differ in the details). Can you tell us what definition your textbook uses? Most of the definitions use two constants, often called $C$ and $k$ but they often use them to mean different things. WIthout knowing what definition you're using, it doesn't make sense to just talk about $C$ and $k$. – David Richerby Feb 18 '16 at 19:42

1 Answers1

2

I am going to guess what $C$ and $k$ mean in your context:

$f(x) = O(g(x))$ if and only if $\exists C, k \;\; \forall x \ge k \;\; (f(x) \le C|g(x)|)$.

You are asked to find constants $C$ and $k$ and prove that for all $x \ge k$ you have $8x+9 \le C x^2$.

I am going to take $C = 1$ and $k = 9$. We have $x \ge 9$ so $C g(x) = C x^2 = x^2 \ge 9 x = 8x + x \ge 8x + 9 = f(x)$.

Denis Pankratov
  • 1,483
  • 10
  • 16