Approximation to dot product of vectors I have two non-zero vectors $u,v \in \mathbb{R}^n$ such that $u \neq \lambda v$ for any $\lambda \in \mathbb{R}$. Let $\Delta=\frac{u^Tv}{\|u\|_2\|v\|_2}$. Now I have a vector $w \in \mathbb{R}^n$ such that $\|v-w\|_2 < \epsilon$ for some $\epsilon>0$. Let $\bar{\Delta}=\frac{u^Tw}{\|u\|_2\|w\|_2}$. Can we find a relation between $\bar{\Delta}$ and $\Delta$ ? In case if it helps, we can assume $u^Tv \neq 0$. I want an expression when $v \neq \lambda w$ for any $\lambda \in \mathbb{R}$.
 A: As you are probably already aware, your quantity $\Delta$ is just $\cos(\theta)$, where $\theta$ is the angle between $\vec u$ and $\vec v$. As such, it is probably more intuitive to look at this problem geometrically. If we look at the 2-dimensional subspace containing $\vec u$ and $\vec v$, and assume $\epsilon<|\vec v|$, we see something like this:

The shaded region corresponds to all the possible choices for $\vec v'=\vec v+\vec\epsilon$, where $\epsilon$ is a vector whose magnitude is at most $\epsilon$. We see that in two dimensions the extreme values of $\theta_{\vec u,\vec v}$ correspond to vectors contained in the same plane, tangent to the circle. It should be pretty clear that adding additional dimensions does not change these extreme values. Let $\Delta\theta$ be the maximal change in $\theta$, which is symmetric about the original value. If we assume $\epsilon<|\vec v|$, we see from the two right triangles in the figure that:
$$\sin(\Delta\theta)=\frac{\epsilon}{|\vec v|}$$
And thus, we get an upper and lower bound on $\theta'$. Of course, we need to take into account the restriction $0\le\theta'\le\pi$:
$$\max\left[0,\theta-\arcsin\left(\frac{\epsilon}{|\vec v|}\right)\right]\le\theta'\le\min\left[\pi,\theta+\arcsin\left(\frac{\epsilon}{|\vec v|}\right)\right]$$
Since $\cos$ in monotonic decreasing on $[0,\pi]$, these bounds will correspond to bounds on your parameter $\Delta=\cos(\theta)$. These can be obtained explicitly with a bit of triginometry, though accounting for all the cases of $\min$ and $\max$ is a bit tedious.
