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Given two unit vectors $x, y\in S^n$, is there a way to obtain a 1st order aproximation at $x=y$ of the following function?:

$$ f(x, y) := \arccos(x^\top y) \frac{y-(x^\top y)x}{\lVert y- (x^\top y) x\rVert} $$

An alternative formulation may be the one resulting from defining $z := y - (x^\top y) x$, which leads to

$$ f(z) := \arccos(\sqrt{1-z^\top z}) \frac{z}{\lVert z\rVert} $$

Thus, the linearization point becomes $z=[0, 0, 0]^\top$. However, I have had no sucess in obtaining a Taylor approximation of $f(z)$.

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Let $x^Ty=\cos\theta$. Then $\|y-x\cos\theta\|=\sqrt{1-2\cos^2\theta+\cos^2\theta}=\sin\theta$ and only $y-x\cos\theta$ remains. As $\cos\theta\approx 1-\dfrac{\theta^2}2$, you can simplify to $y-x$.

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  • $\begingroup$ Thanks :) Just to check if I understand correctly: Is it right that the result comes from the following limit? $$ \lim_{\theta\to0} \theta \frac{y-x\cos(\theta)}{\sin(\theta)} = y-x $$ In that case, why can we "ignore" the $x$ and $y$ terms? $\endgroup$
    – abc
    Commented Sep 5 at 17:56
  • $\begingroup$ @abc: we don't ignore them !? $\endgroup$
    – Yves Daoust
    Commented Sep 6 at 6:58
  • $\begingroup$ It must be a misunderstanding from my side. I would greatly appreciate it if you could expand your answer a bit. What I think is blocking me is that, since $\theta$ depends on both $x$ and $y$, I don't understand why we can "disentangle" the terms involving $\theta$ and obtain the first-order approximation of $f$ by independently computing how these terms (those involving $\theta$) behave when $x = y$. $\endgroup$
    – abc
    Commented Sep 6 at 9:58
  • $\begingroup$ @abc: we don't need to care as the $\theta$s just vanish out of consideration. $\endgroup$
    – Yves Daoust
    Commented Sep 6 at 15:11

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