2

I have the difference between the following log expressions and I am trying to bound the difference,

$$F= \log \left(1+ \left(2+\frac{1}{\sqrt{2}}\right)^2 x^2\right) - \log \left(1+ \left(1-\frac{1}{\sqrt{2}}\right)^2 x^2\right) $$

Can I say that $$F \leq \log\left(1+ \frac{ \left(2+\frac{1}{\sqrt{2}}\right)^2}{\left(1-\frac{1}{\sqrt{2}}\right)^2}\right)$$

Is this the tightest bound that exists?

Thanks in advance.

Henry
  • 1,204

2 Answers2

0

Set $\alpha=(2+\frac{1}{\sqrt{2}})^2$ and $\beta=(1-\frac{1}{\sqrt{2}})^2$, each positive, with $\alpha>\beta$.

We have $$F= \log\left(\frac{1+\alpha x^2}{1+\beta x^2}\right)$$ For all $x$, the fraction is in $[1,\frac{\alpha}{\beta})$. Further, as $x\to\infty$, the fraction gets closer and closer to the right endpoint. Hence the tightest possible bound for $F$ that will be valid for all $x$ is $$F\le \log \frac{\alpha}{\beta}=\log \alpha -\log \beta$$

If you don't like logs we can use the Taylor approximation $$\log x=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\frac{(x-1)^4}{4}+\cdots$$

We set $x=\frac{\alpha}{\beta}$. The series is alternating, so if we truncate after a negative term we get a succession of upper bounds: $$F\le \frac{\alpha}{\beta}-1$$ $$F\le \frac{\alpha}{\beta}-1-\frac{(\frac{\alpha}{\beta}-1)^2}{2}+\frac{(\frac{\alpha}{\beta}-1)^3}{3}$$ etc. Note that the first bound is already sharper than the one in the OP.

vadim123
  • 82,796
  • But for small $x$ i.e not $x \rightarrow \infty$, we can't say that the bound is $\log \frac{\alpha}{\beta}$ right? In this case it should be $1+ \log \frac{\alpha}{\beta}$ ? – Henry Dec 23 '14 at 16:55
  • If you want the bound to hold for all $x$, then you want $\log \frac{\alpha}{\beta}$. If you want the bound to hold for all $x\le M$, then you can use the slightly smaller bound $\log\frac{1+M^2\alpha}{1+M^2\beta}$. All of these are better (smaller) than $1+\log\frac{\alpha}{\beta}$. – vadim123 Dec 23 '14 at 17:05
  • Thanks, but isn't the fraction in $[1,\frac{1+\alpha}{1+\beta})$ instead of the fraction is in $[1,\frac{\alpha}{\beta})$? – Henry Dec 23 '14 at 19:58
  • No, the fraction is in $[1,\frac{1+\alpha x}{1+\beta x})$. As $x\to \infty$, the upper bound approaches $\frac{\alpha}{\beta}$. – vadim123 Dec 23 '14 at 21:42
0

$f(a) =\log (1+ax^2) - \ln a\Rightarrow f'(a) = \dfrac{x^2}{1+ax^2}- \dfrac{1}{a}< 0\Rightarrow \text{ since } b=\left(2+\dfrac{1}{\sqrt{2}}\right)^2 > \left(1-\dfrac{1}{\sqrt{2}}\right)^2=a\Rightarrow f(a) < f(b)\text{ and} \ln\left(\dfrac{b}{a}\right) \leq \dfrac{b}{a}-1 \Rightarrow \text{the inequality holds and the tighter bound is} \dfrac{b}{a}-1$

DeepSea
  • 77,651