Approximating magnitude of sum of two vectors

Question

We have two vectors $\vec{v}$ and $\vec{u}.$ Knowing that $u\ll v,$ how can we approximate $|\vec{v}+\vec{u}|?$ Can we Taylor expand the norm of the sum?

Answer 1

If these vectors are elements of a finite-dimensional vector space with norm $$|u+v|=\sqrt{\sum (u_i+v_i)^2}$$ Then yes, $$|u+v|=\sqrt{\sum (u_i^2+2 u_i v_i+v_i^2)}$$ $$|u+v|=\sqrt{\sum u_i^2+2 \sum u_i v_i + \sum v_i^2}$$ $$|u+v|= |u|+\frac{1}{2|u|}\left(2 \sum u_i v_i + \sum v_i^2\right)+\cdots$$ $$|u+v|= |u|+\frac{1}{2|u|}\left(2 u \cdot v + |v|^2\right)+\cdots$$

Answer 2

Note that

$$|\vec{v}+\vec{u}|^2=(\vec{v}+\vec{u}) \cdot ( \vec{v} +\vec{u})=|\vec{v}|^2 + 2\vec{v}\cdot\vec{u} + |\vec{u}|^2 $$

thus for binomial expansion $x\to0 \quad(1+x)^a\approx 1+ax$

$$|\vec{v}+\vec{u}|= ( |\vec{v}|^2 + 2\vec{v}\cdot\vec{u} + |\vec{u}|^2)^\frac12= |\vec{v}|\cdot\left(1 + 2\frac{\vec{v}\cdot\vec{u}}{|\vec v|^2} + \frac{|\vec{u}|^2}{|\vec v|^2}\right)^\frac12 \approx |\vec{v}|\cdot \left(1 + \frac{\vec{v}\cdot\vec{u}}{|\vec v|^2} + \frac{|\vec{u}|^2}{2|\vec v|^2}\right)$$

Answer 3

The other answers here are actually incorrect, and mix different orders of the approximation. The expansion we are making is that $|\vec{u}| \gg |\vec{v}|$. Denoting the unit vectors and magnitudes as $$ \hat{u} \equiv \frac{\vec{u}}{u}, \,\,\, \hat{v} \equiv \frac{\vec{v}}{v}, $$ and writing $|\vec{u}+\vec{v}|$ explicitly, $$ |\vec{u}+\vec{v}| = \left[ u^2 + v^2 + 2 (\vec{u}\cdot\vec{v}) \right]^{1/2} $$ $$ = u \left[ 1 + 2\left(\frac{v}{u}\right)(\hat{u}\cdot\hat{v}) + \left(\frac{v}{u}\right)^2 \right]^{1/2}, $$ we can make a binomial expansion of $|\vec{u}+\vec{v}|$: $$ (1+x)^\alpha \approx 1 + \alpha x + \tfrac{1}{2} \alpha (\alpha-1)x^2 + \ldots $$ In this case, we have $$ \alpha \rightarrow \tfrac{1}{2}, \,\,\,\, x \rightarrow \left(\frac{v}{u}\right)2(\hat{u}\cdot\hat{v}) + \left(\frac{v}{u}\right)^2 $$ The place where previous answers go wrong is that they don't include the third term of the binomial expansion. But $x$ contains terms of both $O(v/u)$ and $O((v/u)^2)$, so the third term also contains terms of order $O((v/u)^2)$ and you need it for the approximation to be valid to that order. $$ |\vec{u}+\vec{v}| \approx u \left[ 1 + \tfrac{1}{2}\left( \left(\frac{v}{u}\right) 2 (\hat{u}\cdot\hat{v})+\left(\frac{v}{u}\right)^2\right) - \tfrac{1}{8}\left(\left(\frac{v}{u}\right)^2 4 (\hat{u}\cdot\hat{v})^2 + O((v/u)^3) \right)\right] $$ $$ \approx u \left[ 1 + \left(\frac{v}{u}\right)\left( \hat{u}\cdot\hat{v}\right) + \left(\frac{v}{u}\right)^2 \tfrac{1}{2}\left(1 - (\hat{u}\cdot\hat{v})^2 \right) \right] $$ Writing the answer this way makes the relative sizes of the various terms manifest, but I'll conclude by putting it in a slightly different form, to make contact with the previous answers.

To $O(v/u)$: $$ |\vec{u}+\vec{v}| \approx |\vec{u}| + \frac{1}{2 |\vec{u}|}(2 \vec{u}\cdot\vec{v}) + O((v/u)^2)$$ To $O((v/u)^2)$: $$ |\vec{u}+\vec{v}| \approx |\vec{u}| + \frac{1}{2 |\vec{u}|}\left(2 \vec{u}\cdot\vec{v} + |\vec{v}|^2 - \tfrac{1}{|\vec{u}|^2}(\vec{u}\cdot\vec{v})^2 \right) + O((v/u)^3)$$