# need help with proof of simple theorem - linear algebra / analysis

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

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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.

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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.

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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