Prove f(n) + g(n) is O(max(f(n),g(n)))

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f(n) + g(n) <= 2* max{f(n),g(n)} 
(for each n>0, assume f(n),g(n) are none-negative functions)

Thus, for N=1, for all n>N: f(n) + g(n) <= 2*max{f(n),g(n)}, and we can say by definition of big O that f(n) + g(n) is in O(max{f(n),g(n)})

So basically, we use N=1, c=2 for the formal proof by definition.

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csnate
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Updated on November 18, 2020

Comments

  • csnate
    csnate over 3 years

    Hello I am having a bit of difficulty proving the following.

    f(n) + g(n) is O(max(f(n),g(n)))
    

    This makes logical sense, and by looking at this I can tell you that its correct but I'm having trouble coming up with a proof.

    Here is what I have so far:

    c * (max(f(n),g(n))) > f(n) + g(n) for n > N
    

    But I'm not sure how to pick a c and N to fit the definition because I don't know what f(n) and g(n) are.

    Any help is appreciated.

  • Savannah Madison
    Savannah Madison about 3 years
    how is c=2 in the above equation?