Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f(n)$ and $g(n)$ be asymptotically nonnegative increasing functions. Show: $f(n) · g(n) = O((\max\{f(n), g(n)\})^2)$, using the definition of big-oh.

I can't quite figure this out, can someone help explain why this is true?

share|improve this question

3 Answers 3

We have $f(n)\leq \max\{f(n),g(n)\}$ and $g(n)\leq \max\{f(n),g(n)\}$. As $f(\cdot)$ and $g(\cdot)$ are non-negative, we get $$f(n)\cdot g(n)\leq \max\{f(n),g(n)\}^2,$$ what is wanted (take the constant equal to $1$ in the definition).

share|improve this answer

Another way, though with a worse constant:

Once $f$ and $g$ are positive, $f g =(f+g)^2-f^2-g^2 $, so

$\begin{align} |fg| &\le (f+g)^2+f^2+g^2\\ &\le (2\max(f, g))^2+2(\max(f,g))^2\\ &= 6(\max(f, g))^2\\ &= O(\max(f, g))^2\\ \end{align} $

share|improve this answer

Here is a proof that minimizes the number of 'surprises', i.e., one which I designed by just following the shape of the formulae.

Directly using the definition, I would calculate \begin{align} & f(n) \times g(n) = O((f(n) \text{ max } g(n))^2) \\ \equiv & \qquad \text{"definition of $\;O\;$"} \\ & \langle \exists k : k > 0 : \langle \forall_{l.e.} n :: k \times f(n) \times g(n) \le (f(n) \text{ max } g(n))^2 \rangle \rangle \\ \equiv & \qquad \text{"square of max = max of squares, since $\;f(n) \ge 0\;$ and $\;g(n) \ge 0\;$, for l.e. $\;n\;$"} \\ & \langle \exists k : k > 0 : \langle \forall_{l.e.} n :: k \times f(n) \times g(n) \le f(n)^2 \text{ max } g(n)^2 \rangle \rangle \\ \equiv & \qquad \text{"basic property of $\;\text{max}\;$"} \\ & \langle \exists k : k > 0 : \langle \forall_{l.e.} n :: k \times f(n) \times g(n) \le f(n)^2 \;\lor\; k \times f(n) \times g(n) \le g(n)^2 \rangle \rangle \\ \equiv & \qquad \text{"divide by $\;f(n)\;$, which is $\;\ge0\;$ for l.e. $\;n\;$; same for $\;g(n)\;$"} \\ & \langle \exists k : k > 0 : \langle \forall_{l.e.} n :: f(n) = 0 \;\lor\; k \times g(n) \le f(n) \\ & \phantom{\langle \exists k : k > 0 : \langle \forall_{l.e.} n ::} \lor\; g(n) = 0 \;\lor\; k \times f(n) \le g(n) \rangle \rangle \\ \Leftarrow & \qquad \text{"choose $\;k := 1\;$"} \\ & \langle \forall_{l.e.} n :: f(n) = 0 \;\lor\; g(n) \le f(n) \;\lor\; g(n) = 0 \;\lor\; f(n) \le g(n) \rangle \\ \equiv & \qquad \text{"real numbers are a total order, i.e., $\;x \le y \;\lor\; y \le x\;$"} \\ & \text{true} \end{align} Note that this proof only assumes that $\;f,g\;$ are eventually non-negative: we did nor need to assume that they are increasing.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.