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 $$f(x,y) = \frac{2 \log(y) \sqrt{y ( x-y)}}{x \log(x)} $$ with $(x,y) \in D = \{(x,y)\mid 0 < y \leq x \le 1 \}$.

How would one show (or disprove) that $$ \forall \epsilon > 0\ \ \exists \delta>0 \ \ \forall (x,y) \in D \cap B_\delta(0) \quad f(x,y) < 1 + \epsilon $$ where $B_\delta(0)$ is the disk with radius $\delta$ and center $(0,0)$?

Please note that this is not just a bound on the limit of the function, as it is not well-defined. Choosing another coordinate $0 \le \theta \le 1$ with $y = \theta x$ we get $$ f(x, \theta x) = \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\ $$ such that $$ \lim_{x \to 0} f(x, \theta x) = 2 \sqrt{\theta (1-\theta)}. $$ So the limit is not well-defined as it depends on how the origin is approached.

Also, although the above limit $2 \sqrt{\theta (1-\theta)}$ is indeed bound by 1, it does not directly prove the original question, because $\theta$ is assumed constant with respect to $x$. More specifically, this limit shows (variable ranges omitted for clarity) $$ \forall \epsilon \ \ \forall \theta\ \ \exists \delta \ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon $$ while the original problem corresponds to the stronger (note the different quantifier ordering): $$ \forall \epsilon \ \ \exists \delta \ \ \forall \theta\ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon $$

share|improve this question

2 Answers 2

In polar coordinates: $$f(r,\theta) = \frac{2 \log(r\sin \theta) \sqrt{\sin \theta (\cos \theta-\sin \theta)}}{\cos \theta \log(r\cos \theta)}$$

You can show that the function: $$2\sqrt{\sin \theta (\cos \theta-\sin \theta)} \sec (\theta) <1$$ Simply by taking the derivative and showing that it obtains the maximum value $1$ at $\theta\to\tan^{-1}(1/2)$. So: $$f(r,\theta)< \frac{\log(r\sin \theta)}{\log(r\cos \theta)} = \frac{\log(r)+\log(\sin \theta)}{\log(r)+\log(\cos \theta)}\sim 1+\frac{\sin \theta - \cos \theta}{\log(r)}+O(\log^{-2}(r))$$ Clearly, for small enough $r\to\delta>0$, $$\epsilon = \frac{\sin \theta - \cos \theta}{\log(\delta)} > 0 $$.

share|improve this answer
    
Thx for the interest! But I think the reasoning is invalid. As $$\frac{\log(r \sin \theta)}{\log(r \cos \theta)} = \frac{\log y}{\log x}$$ the expression can e.g. stay equal to 2 by choosing $y=x^2$ –  veryltdbeard Feb 8 '13 at 17:53
    
@veryltdbeard - it can't, since $x>y$, and for small enough $r$ this is no longer true. –  nbubis Feb 8 '13 at 18:47
    
As $x \le 1$, $x^2 \le x$, so choosing $y=x^2$ is valid. –  veryltdbeard Feb 8 '13 at 20:37
    
I tried to add a clarification to the question pointing out the precise problem w.r.t taking limits. In case you're still interested, all help is still very welcome! –  veryltdbeard Feb 12 '13 at 11:34
up vote 0 down vote accepted

$$ \begin{eqnarray} f(x, \theta x) &=& \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\ &=& \frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{-\log(x)} + 2 \sqrt{\theta (1-\theta)} \\ &\le&\frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{- \log(x)} + 1 \\ &\le&\frac{V^*}{-\log(x)} + 1 \\ \end{eqnarray} $$ with $V^* = \max_{\theta \in [0,1]} (- 2 \log(\theta) \sqrt{\theta (1-\theta)}) \approx 1.3818$ (at $\theta^* \approx 0.1041$).

So $$ \begin{eqnarray} f(x, \theta x) < 1 + \epsilon & \Leftarrow & \frac{V^*}{-\log(x)} + 1 < 1 + \epsilon \\&\Leftrightarrow & x < e^{-V^*/\epsilon} \end{eqnarray} $$ So with $\delta(\epsilon)=e^{-V^*/\epsilon}$ we have shown that $$ \forall \epsilon > 0 \quad \forall x \in\ ]0, \delta(\epsilon)[\quad \forall \theta \in [0,1]\quad f(x, \theta x) < 1 + \epsilon. $$

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.