[Math] What does it mean that a function dominates over another function

algebra-precalculuslimits

Let $f,g$ be functions with Domain $\mathbb{R}$ and Range $\mathbb{C}$ What does it mean that $f$ dominates over $g$? Unfortunately I could not find any ressources online about this Topic.

But I remember there was a distinction between the behaviour at infinity and at $0$.

I understand why we call $f$ to be dominating over $g$ if

$\lim_{x\rightarrow x_0}f(x)=\infty$ and $\lim_{x\rightarrow x_0}g(x)=\infty$ and also $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=\infty$

Because the values for $f$ are bigger than those of $g$

However I don't understand why we also call $f$ to be dominating over $g$ if

If $\lim_{x\rightarrow x_0} f(x)=0$ and $\lim_{x\rightarrow x_0}g(x)=0$ and also $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=0$

Because the values of $f$ are smaller than those of $g$.

If we let $f(x)=x^2$ and $g(x)= x$

Hope someone can tell me why in both cases we would call $f$ to be dominating over $g$.

Best Answer

You've right that the second one is backwards. Whoever you got that from mixed it up.

The idea is that if we have a sum $$f(x)+g(x)$$ and factor out the “dominant” term $f(x)$, so that we get $$ f(x) \left( 1 + \frac{g(x)}{f(x)} \right) , $$ then we want $g(x)/f(x)$ to tend to zero, so that the whole parenthesis just tends to $1+0=1$. If not, it wouldn't be right to call $f$ the “dominant” term.

Saying that $f$ dominates over $g$ (as $x$ tends to whatever it tends to) means that the important contribution to the sum is the one that we get from $f(x)$, whose absolute value is large in comparison to the absolute value of $g(x)$. So regardless of whether $x \to \infty$ or $x \to x_0$, we say that $f(x)$ dominates over $g(x)$ if $$ \frac{g(x)}{f(x)} \to 0 . $$ (The values of $f(x)$ and $g(x)$ may both be “small”, or both “large”, or maybe one small and one large, but the quotient $g(x)/f(x)$ should be small.)

The distinction between $x \to \infty$ and $x \to 0$ that you seem to remember might be that for $x \to \infty$, higher powers $x^k$ dominate over lower powers, but as $x \to 0$ it's the other way around.

Related Question