I am studying big O notation from this book.
The deffinition of big O notation is:
We say that f (x) is O(g(x)) if there are constants C and k such that |f (x)| ≤ C|g(x)| whenever x > k.
Now here is the first example:
EXAMPLE 1 Show that f (x) = x^2 + 2x + 1 is O(x^2).
Solution: We observe that we can readily estimate the size of f (x) when x > 1 because x 1. It follows that
0 ≤ x^2 + 2x + 1 ≤ x^2 + 2x^2 + x^2 = 4x^2
whenever x > 1. Consequently, we can take C = 4 and k = 1 as witnesses to show that f (x) is O(x^2). That is, f (x) = x^2 + 2x + 1 1. (Note that it is not necessary to use absolute values here because all functions in these equalities are positive when x is positive.)
I honestly don't know how they got c = 4, looks like they jump straight to the equation manipulation and my algebra skills are pretty weak. However, I found another way through [The accepted answer to this question])What is an easy way for finding C and N when proving the Big-Oh of an Algorithm?) that says to add all coefficients to find c if k = 1. So x^2+2x+1 = 1+2+1 = 4.
Now for k = 2, I'm completely lost:
Alternatively, we can estimate the size of f (x) when x > 2. When x > 2, we have 2x ≤ x^2 and 1 ≤ x^2. Consequently, if x > 2, we have
0 ≤ x^2 + 2x + 1 ≤ x^2 + x^2 + x^2 = 3x^2.
It follows that C = 3 and k = 2 are also witnesses to the relation f (x) is O(x^2).
Can anyone explain what is happening? What method are they using?
First alternative:
C=4?
The C=4 come from the inequalities
0 ≤ x^2 + 2x + 1 ≤ x^2 + 2x^2 + x^2 = 4x^2 = C*x^2, with C=4 (+)
The second inequality in (+) is true for all x greater than 1, since, term by term
2x < 2x^2, given x>1
1 < x^2, given x>1
k = 1? From above, we've shown that (+) holds as long as x is larger than 1, i.e.
(+) is true given x > k, with k=1
Second alternative:
k=2?
By the statement, we want to study f(x) for x larger than 2, i.e.
Study f(x) for x > k, k=2
Given x > 2, it's apparent that
0 ≤ x^2 + 2x + 1 ≤ x^2 + x^2 + x^2 = 3x^2 = C*x^2, with C=3 (++)
since, for x>2, we have
2x = x^2 given x=2 ==> 2x < x^2 given x>2
for x=2, 1 < x^2 = 4, so 1 < x^2 for all x>2
Both examples show that f(x) is O(x^2). By using your constants C and k, recall that then Big-O notation for f(x) can be summarized as something along the lines
... we can say that f(x) is O(g(x)) if we can find a constant C such that |f(x)| is less than C|g(x)| or all x larger than k, i.e., for all x>k. (*)
This, by no means, implies that we need to find a unique set of (C, k) to prove that some f(x) is some O(g(x)), just some set (C, k) such that (*) above holds.
See e.g. the following link for some reference on how to specify the asymptotic behaviour of a function: https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-o-notation