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I have stumbled upon the following question: If f(x),g(x) > 1 for every x, prove/disprove the following:

f(x)+g(x)= O(f(x)*g(x))

I dont know how to start the proof, this is very basic for my level, please help.

shilovk
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Eiad Amer
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1 Answers1

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See https://en.wikipedia.org/wiki/Big_O_notation.

You basically need to show that there is a positive constant M and a number x0 such that:

abs(f(x)+g(x)) <= M*abs(f(x)*g(x))

for all x >= x0.

Which is trivial. I will omit this part, please do it yourself.

lexicore
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  • Should I divide the problem into cases? – Eiad Amer Apr 08 '18 at 14:41
  • @EiadAmer No. You should prove exactly what my comment says you should prove. Hint: divide both paths by `f(x)*g(x)`. What will you get? – lexicore Apr 08 '18 at 15:47
  • You get `abs(1/g(x) + 1/f(x)) <= M`. As `f(x) > 1`, `g(x) > 1` for every `x` the `1/f(x) < 1`, `1/g(x) < 1`. Meaning `abs(1/g(x) + 1/f(x)) < 2`, so you can choose any `M => 2`. – lexicore Apr 10 '18 at 05:31