Find the smallest integer value of N such that f(x)=O(x^N).
a) f(x) = x^7 - 150x^6 + 3x^3 + 111
b) f(x) = (x^3 + 2x^2) / (x^4 + x^2 + 1)
c) f(x) = ⌈x⌉
d) f(x) = ⌊x⌋
See whether this article gives you what you need.
http://math.stackexchange.com/questions/1661986/finding-the-least-integer-n-for-a-function-big-o-of-another-function
Oh, hello there! Let's find the smallest integer value of N for each function, shall we?
a) f(x) = x^7 - 150x^6 + 3x^3 + 111
Hmm, the highest power of x in this function is x^7. So, the value of N for f(x) = O(x^N) is N = 7.
b) f(x) = (x^3 + 2x^2) / (x^4 + x^2 + 1)
Let's simplify the function a bit. When we divide x^3 by x^4, we get 1/x. And when we divide 2x^2 by x^4, we get 2/x^2. So, the simplified version is f(x) = (1/x + 2/x^2) / (1 + 1/x^2 + 1/x^4).
Now, as x approaches infinity, all the terms with x in the denominator become less and less significant. Therefore, the highest power we have left is 1/x^4. So, the value of N for f(x) = O(x^N) is N = -4.
c) f(x) = ⌈x⌉ (the ceiling function)
Well, this function just rounds up x to the nearest integer. No matter how large x gets, the function will always stay the same, as it can only take integer values. Therefore, N can be any positive integer you'd like, as f(x) doesn't grow with x.
d) f(x) = ⌊x⌋ (the floor function)
Similarly to the previous function, this function just rounds down x to the nearest integer. It also doesn't grow with x, so N can be any positive integer as well.
I hope that answers your question with a touch of humor! If you have any more questions, feel free to ask!
To find the smallest integer value of N such that f(x) = O(x^N), we need to determine the highest power of x, denoted by N, such that the function f(x) can be bounded by a polynomial of the form x^N. In other words, we are looking for the smallest N value such that f(x) grows at a rate slower than or equal to x^N.
Let's analyze each function given:
a) f(x) = x^7 - 150x^6 + 3x^3 + 111
In order to determine the highest power of x that dominates the growth of this function, we ignore the lower power terms (3x^3 + 111) and focus on the highest power term, which is x^7. Therefore, the smallest integer N for this function is 7, since the function can be bounded by x^7.
b) f(x) = (x^3 + 2x^2) / (x^4 + x^2 + 1)
We can simplify this function by dividing both the numerator and denominator by x^4. This yields f(x) = (1/x + 2/x^2) / (1 + 1/x^2 + 1/x^4). As x approaches infinity, all the terms involving x in the denominator become negligible. Therefore, the highest power of x in the simplified function is 0. Hence, the smallest integer N for this function is 0.
c) f(x) = ⌈x⌉
The ceiling function ⌈x⌉ rounds up any given value x to the nearest integer. Since the function f(x) is already an integer, we can say that it can be bounded by a constant value. Any constant value can be represented as x^0, where N=0. Therefore, the smallest integer N for this function is 0.
d) f(x) = ⌊x⌋
The floor function ⌊x⌋ rounds down any given value x to the nearest integer. Similar to the ceiling function, the function f(x) is already an integer, and thus it can be bounded by a constant value. So, the smallest integer value of N for this function is also 0.
In summary, the smallest integer value of N for all the given functions is 0.