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If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property

$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$

then is the following true?

$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$

This might look like homework but it isn't.

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I suppose you are interested in the asymptotics as $x \to \infty$? –  Aryabhata Nov 8 '10 at 22:48
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2 Answers 2

up vote 7 down vote accepted

Yes. One way to see this is to actually do the long division (like the kind you learned in elementary school)! Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.

Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$. So we have $$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$ since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$

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It is true. In the spirit of epsilon/delta, you are challenged to prove that $|f(x)-\frac{a}{b}| \lt \frac{M}{\sqrt{x}}$ for $x\gt x_0$ where your challenger gives M and you have to find an $x_0$ that works. But you get to challenge back saying the numerator should be within $\frac{N}{\sqrt{x}}$ of $a$ and similarly the denominator should be within $\frac{P}{\sqrt{x}}$ of $b$. So take $N=\frac{M}{2b}$ and $P=\frac{aM}{2b^2}$ and take the larger of the $x_0$'s that come back.

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Upvoted for a good explanation and an interesting narrative. Also, use of the epsilon delta proof here is probably more rigorous. –  AnonymousCoward Nov 9 '10 at 0:24
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@GottfriedLeibniz: Ross's answer is a good answer, but I would disagree that it's more rigorous. Big-O expressions have perfectly good (i.e., rigorously justified) algebraic rules that they follow. (See, for example, Chapter 9 of Concrete Mathematics.) My answer is based on those. –  Mike Spivey Nov 9 '10 at 0:37
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@Mike fair enough. As someone less familiar with asymptotics his also made more sense to me because it was more 'first principles' than your answer. –  AnonymousCoward Nov 9 '10 at 4:24
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@GottfriedLeibniz: That's what I suspected, which is partly why I answered your comment as I did. Big-O notation can seem quite mysterious at first, but if you work with it enough the mystery starts to disappear. That said, I still make mistakes. In fact, I even once made a Big-O mistake in a research paper that was subtle enough that none of the referees or editors caught it, either. Three years after the paper was published someone else wrote to the editor pointing out the mistake! –  Mike Spivey Nov 9 '10 at 4:56
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