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$$ \cos (\mathbf{v},\mathbf{w}) = \frac{{\langle\mathbf{v}, \mathbf{w}\rangle}} {{|| {\mathbf{v} } ||\;|| \mathbf{w}||}} $$

where $\langle\mathbf{v}, \mathbf{w}\rangle$ is the dot product, and $||\cdot||$ is the second norm.

Like this thread did, we can add a little $d\mathbf{v}$ to $\mathbf{v}$:

$$ \frac{{\langle\mathbf{v}+d\mathbf{v}, \mathbf{w}\rangle}}{{|| {\mathbf{v}+d\mathbf{v} } ||\;|| \mathbf{w}||}} $$

However, suppose $$ || {\mathbf{v}+d\mathbf{v} }|| = \sqrt{f(\mathbf{v})} $$ then, $$ f(\mathbf{v})=( \mathbf{v}+d\mathbf{v} )^T( \mathbf{v}+d\mathbf{v} )= \mathbf{v}^T \mathbf{v}+d\mathbf{v} ^T\mathbf{v}+\mathbf{v}^T d\mathbf{v}+d\mathbf{v}^T d\mathbf{v} \\=||\mathbf{v}||^2+||d\mathbf{v}||^2+d\mathbf{v} ^T\mathbf{v}+\mathbf{v}^T d\mathbf{v} $$ So basically, $$ || {\mathbf{v}+d\mathbf{v} }|| = \sqrt{||\mathbf{v}||^2+||d\mathbf{v}||^2+d\mathbf{v} ^T\mathbf{v}+\mathbf{v}^T d\mathbf{v}} $$ I don't see why this equals/approxiametely equals $$ || {\mathbf{v}+d\mathbf{v} }||\approx|| \mathbf{v} ||\left( {1 + \frac{\mathbf{v}} {{|| \mathbf{v} ||^2 }} \cdot d\mathbf{v}} \right) ? $$ Particularly, if $\mathbf{v}$ and $\mathbf{u}$ are complex, I think there is a problem, no?

Can anyone gieve some suggestions/hint here? Thanks a lot.

Community
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Nick X Tsui
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1 Answers1

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Factoring out $\|\vec{v}\|$ from inside the square root, and using symmetry of the dot product, gives you $$ \|\vec{v}+d\vec{v}\|=\|\vec{v}\|\sqrt{1+\frac{\|d\vec{v}\|^2}{\|\vec{v}\|^2}+2\frac{\vec{v}\cdot d\vec{v}}{\|\vec{v}\|^2}}. $$ Considering $\vec{v}$ as being itself fixed, and $d\vec{v}\to\vec{0}$, we can think about asymptotics. Note that $$ \sqrt{1+x}=1+\frac{x}{2}+O(x^2),\qquad x\to0, $$ so that $$ \sqrt{1+\frac{\|d\vec{v}\|^2}{\|\vec{v}\|^2}+2\frac{\vec{v}\cdot d\vec{v}}{\|\vec{v}\|^2}}=1+\frac{\vec{v}\cdot d\vec{v}}{\|\vec{v}\|^2}+O(\|d\vec{v}\|^2). $$

Nick Peterson
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  • Thakns Nick. I am wondering what if $v$ is complex? In that case $dv^Tv$ and $v^Tdv$ would not be equal, thus can not be combined to $ 2vdv$? – Nick X Tsui Mar 30 '17 at 16:01
  • True enough; off the top, it would seem that it would replace $\vec{v}\cdot d\vec{v}$ by its real part. – Nick Peterson Mar 30 '17 at 18:03