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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

i'm stalled in my attempt to prove that $(1-a^2)^{\frac{1}{2}} - (1-b^2)^{\frac{1}{2}}$ goes to zero faster than $|A-B|$, where A,B are vectors in $\mathbb{R}^n$ with $a=|A| \leq 1$ and $b=|B|\leq 1$. Can anybody give me assistance? thanks if you can help

peace stm

share|cite|improve this question

Hint: consider a function $f(x) =\sqrt{1-x^2}$ for real $x$ and calculating first two derivatives show that $$ |f(0) - f(x)| = O(x^2). $$ If you need additional hints - please let me know.

share|cite|improve this answer
But $(f(0)-f(x))/x^2 \to 1/2$, so do you mean $O(x^2)$ instead of $o(x^2)$? – Antonio Vargas Mar 2 '12 at 8:19
@AntonioVargas: yes, thanks for fixing it. – Ilya Mar 2 '12 at 8:20
I think we need to be a little more careful. Say that $f(x)=1-2x^2$, then we still have $|f(0)-f(x)|=O(x^2)$, yet $f(|A|)-f(|B|)=2(|A|+|B|)(|A|-|B|)$ and $2(|A|+|B|)$ could be up to $4$ and $|A|-|B|$ could be up to $|A-B|$. – robjohn Mar 5 '12 at 13:19

Assuming $A \neq B$, use the reverse triangle inequality to get $|A - B| \geq |a-b|$, whence

$$\left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|}\right| \leq \left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{a-b}\right|.$$

Thus you can conclude, for example,

$$\limsup_{B \to A} \left|\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|}\right| \leq \frac{a}{\sqrt{1-a^2}}.\tag{1}$$

Now, consider how $B$ may approach $A$. If $A$ and $B$ are parallel with, say, $b < a$ then $|A-B| = a-b$, so that

$$\frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{|A-B|} = \frac{\sqrt{1-a^2} - \sqrt{1-b^2}}{a-b} \to -\frac{a}{\sqrt{1-a^2}}$$

as $B \to A$ since the expression you get is just the difference quotient for the function $\sqrt{1-x^2}$. In this case you can't say that the numerator goes to zero faster or slower than the denominator; in fact, they go to zero at the same rate in some sense (and that sense is known as big-$\Theta$).

And, if $a = b$ (that is, if $B$ approaches $A$ on the sphere of radius $a$) then the numerator is zero always.

By inequality $(1)$, every other path of approach must give you a behavior somewhere between these two. The numerator will always go to zero at least as fast as $|A-B|$, and in some cases--but not all--it will go to zero faster.

share|cite|improve this answer
good point. i was trying to prove the wrong thing. thanks for straightening me out about that. – sean mcilroy Mar 5 '12 at 7:06

Your Answer


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.