Asymptotic Notation: f(N)+g(N)=O(h(N))? f(N)*g(N)=O(h(N))?

[needhelp83]

Suppose f(N) = O(h(N)) and g(N) = O(h(N))

1. Is f(N) + g(N) = O(h(N))
2. Is f(N) * g(N) = O(h(N))

Justify answers

I am totally lost on these questions. ANY help would be greatly appreciated.

 

[wildman]

This is the Big-O Notation. It is involved in comparing the growth of one number-theoretic function with that of another. There is a sequence of these functions. For instance:

$$f_2 = log_2 n$$
.
.
.
$$f_{7.2} = n^2$$
$$f_{7.3} = n^3$$
.
.
.
$$f_{7.k} = n^k$$
.
.

Now answer these questions:

Does adding $$n^2$$ to $$n^2$$ change the level in the sequence?
How about multiplying? (different constants don't count)

 

[piercas]

Asymptotic notation is a mathematical tool used to describe the behavior of functions as the input size, denoted by N, approaches infinity. In this context, the statement "f(N) + g(N) = O(h(N))" means that the sum of the functions f(N) and g(N) will not grow faster than h(N) as N gets larger. Similarly, the statement "f(N) * g(N) = O(h(N))" means that the product of f(N) and g(N) will not grow faster than h(N) as N gets larger.

In order to determine if these statements are true, we need to understand the definition of big O notation. In general, a function f(N) is said to be O(g(N)) if there exists a constant C and a value N0 such that for all values of N greater than or equal to N0, f(N) is less than or equal to Cg(N). In other words, f(N) is bounded by a constant multiple of g(N) for sufficiently large values of N.

Now, let's apply this definition to the statements in question.

1. Is f(N) + g(N) = O(h(N))?

To answer this, we need to show that there exists a constant C and a value N0 such that for all values of N greater than or equal to N0, f(N) + g(N) is less than or equal to Ch(N). This is true because if f(N) = O(h(N)) and g(N) = O(h(N)), then there exists constants C1 and C2 and values N1 and N2 such that for all values of N greater than or equal to N1, f(N) is less than or equal to C1h(N) and for all values of N greater than or equal to N2, g(N) is less than or equal to C2h(N). Therefore, for all values of N greater than or equal to max(N1, N2), we have f(N) + g(N) is less than or equal to C1h(N) + C2h(N) = (C1 + C2)h(N), where C = C1 + C2 and N0 = max(N1, N2). This satisfies the definition of big O notation, hence f(N) + g(N) = O(h(N)) is true.

2. Is f(N