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I am trying to decide if each of these two statements are true or false:

  • $n\log(n)+m\log(m)=O(n\log(m)+m\log(n))$

  • $n\log(m)+m\log(n)=O(n\log(n)+m\log(m))$

I've tried using some properties of logarithms so as to take these two statements to these inequalities:

  • For the first one, the statement is true if and only if there exist $c$ and $(n_0,m_0)$ such that for all $(n,m)>(n_0,m_0)$ (which means $n>n_0$ and $m>m_0$)

$$n^n* m^m \leq c(m^n* n^m)$$ $${n}^{n-m}\leq cm^{n-m} $$

If I take $m=m_0+1$ and $n=(n_0+1)(m_0+1)c$ then $(n,m)>(n_0,m_0)$ but $$n^{n-m} > cm^{n-m}$$

From here it follows that the first statement is false.

  • As for the second one, the statement is true if and only if there exist $c, (n_0,m_0)$ such that for all $(n,m)>(n_0,m_0)$ $$m^n*n^m \leq c(n^n*m^m)$$ $$n^{m-n} \leq cm^{m-n}$$

I got stuck at this part, I would appreciate some help to prove or disprove the second statement and also to know if I've done the first part correctly. Thanks in advance.

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Since the functions being compared are functions of two variables, $m$ and $n$ it seems the answer would depend entirely upon whether $m$ and $n$ are of the same order.

Let $m=nk$. Then $k=\dfrac{m}{n}$.

The question of whether

$$ n\log(n)+m\log(m)=O(n\log(m)+m\log(n)) $$

becomes a question of whether

$$ n\log(n)+nk(\,\log(n)+\log(k)\,)=O(\,n\log(n)+n\log(k)+nk\log(n\,) $$ which reduces to the question whether

$$ k\cdot(n\log(n))=O(n\log(n)) $$

And since $k=\dfrac{m}{n}$ this reduces to the question whether

$$ m\log(n)=O(n\log(n)) $$

Next consider the second quesion. Is

$$ n\log(m)+m\log(n)=O(n\log(n)+m\log(m)) $$

Taking the same approach as in the first question, this reduces to the question of whether

$$ n\log(n)=O(m\log(n)) $$

So it comes down to whether or not $m$ and $n$ are of the same order.

So, for example, if $m=O(x^3)$ and $n=O(x^2)$ then the two functions are not of the same order. But if $m=O(x^2)$ and $n=O(x^2)$ then the two functions are of the same order.

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