I have a question, about Bachman-Landau. How do you know when a function fits into some notation?

For example does fit into .

Or you just look and see, we had some equations in school but I cant' remember them all.

Thank you for your help.

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- Nov 16th 2012, 10:08 AMNforceBachman-Landau notation and functions
I have a question, about Bachman-Landau. How do you know when a function fits into some notation?

For example does fit into .

Or you just look and see, we had some equations in school but I cant' remember them all.

Thank you for your help. - Nov 16th 2012, 02:11 PMemakarovRe: Bachman-Landau notation and functions
- Nov 17th 2012, 01:43 AMNforceRe: Bachman-Landau notation and functions
- Nov 17th 2012, 03:43 AMemakarovRe: Bachman-Landau notation and functions
I mean that for every x, if x ≥ 5, then x² ≥ 25. Therefore, if x ≥ 5, then x² - 6x + 25 ≤ x² + 25 ≤ x² + x² = 2x².

- Nov 17th 2012, 04:31 AMNforceRe: Bachman-Landau notation and functions
Sorry but I still dont' understand, how did you get that , and then . From where?

Thank you for your help. - Nov 17th 2012, 04:56 AMemakarovRe: Bachman-Landau notation and functions
From where I got x² + 25 and x² + x² is a separate question from whether these inequalities are true. You don't have to understand why I chose these specific expressions, but you do have to understand that if you subtract 6x from x² + 25, where x >= 5 and therefore 6x is nonnegative, then you get a smaller value than x² + 25. Therefore, x² - 6x + 25 ≤ x² + 25. Further, you have to understand that x² ≥ 25 for x ≥ 5. Therefore, if you replace 25 with a bigger value x² in x² + 25, you get a bigger result. So, x² + 25 ≤ 2x².

- Nov 17th 2012, 07:08 AMNforceRe: Bachman-Landau notation and functions
Hmm, ok now I understand the ineqaulities, but why did you choose those terms? At every single expression there is one less term. So the purpose is to rewrite the function into inequalities.

At your first answer you said if deg(f()) ≤ deg(g()) is true, then the polynomial fits into O(x²), so 2 ≤ 2 is true, therefore it fits.

This is how I understand this. - Nov 17th 2012, 08:26 AMemakarovRe: Bachman-Landau notation and functions
You are right. The definition of big-O says that we must show that for some constants and , it is the case that for all . Therefore, the goal is to prove for some coefficient . One approach is to bound from above every term in the left-hand side by for some . Then you add them all together and obtain that the left-hand side is bounded from above by .

In this example, ; for and for . Adding these inequalities, we get . This holds when*and*, i.e., when . Therefore, we can take and and we get, as the definition of big-O demands, that for all . We could also say that for . Then for . There is some trade-off here: we got a smaller (1 instead of 5) but a significantly larger (26 instead of 2). If the polynomial were instead of , then, since for and for , we could say that for . Alternatively, since and for , we have for .

Still another way to prove is is to plot the graphs of the two functions and to note that eventually dominates . (If it did not, then we should plot , , etc. until it dominates the left-hand side). Then we can solve to get . Thus, we can take and .

This is correct, but this is a general theorem, and I am not sure you can provide the proof (though it is basically a generalization of the reasoning above). Whether this is sufficient depends on your course. It is possible that you need a way to show that is that you can fully justify.

For more information about the big-O notation see the external links in the Wikipedia article or search this forum. - Nov 17th 2012, 11:33 AMNforceRe: Bachman-Landau notation and functions
This is the best answer. Thank you. Another thing, what do you mean by "one function dominates another"? Does it mean that one function increases faster then another?

- Nov 17th 2012, 12:21 PMemakarovRe: Bachman-Landau notation and functions
Here by " eventually dominates " I mean that "eventually," i.e., for all starting from some point . That is, if is , then there exists a constant such that dominates (assuming that and are nonnegative). But, as far as I know, "dominates" does not have a universally accepted precise meaning in mathematics.