Suppose $f(n)$ is $O(g(n))$ and $d(n)$ is $O(e(n))$. How do I use Big O to show that $f(n) + d(n)$ is $O(g(n) + e(n))$?
Suppose $f(n)$ is $O(g(n))$ and $d(n)$ is $O(e(n))$. How do i use Big O to show that $f (n)$ + $d(n)$ is $O(g(n) + e(n))$?
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statistics
asymptotics
computer-science
1 Answers
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In generally this is false (if we follow the definition given at Wikipedia). For example with $f(n)=g(n)=d(n)=-e(n)=n$ We have $f(n)=O(g(n))$ and $d(n)=O(e(n))$ but not $2n=f(n)+d(n)=O(g(n)+e(n))=O(0)$.
However, if we assume $g(n)$ and $e(n)$ are non-negative, then the claim follows: We are given that $|f(n)|\le c_1g(n)$ for all $n>N_1$ and $|d(n)|\le c_2e(n)$ for all $n>N_2$. Then $$|f(n)+d(n)|\le |f(n)|+|d(n)|\le c_1g(n)+c_2 e(n)\le\max\{c_1,c_2\}(g(n)+e(n))$$ for all $n>\max\{N_1,N_2\}$.
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0Hello. I know this is an old post, but I recently started an algorithm design class that has me completely confused. One of the class examples was like this and I am very confused about how you go about proving things like this. If you don't mind explaining the steps for this proof in more detail for me, I would be sincerely appreciative. Thanks! I just want to learn the nuances of big O as best as I can. – 2018-01-20