0
$\begingroup$

How does one Taylor expand a vector function of many variables? The question arises in the context of deriving the geodesic deviation in Newtonian gravity, where we need to subtract as follows, and the answer given was: $$\overline{g}(\overline{x}+\overline{N}, t) - \overline{g}(\overline{x}, t) = (\overline{N} \cdot \overline{\nabla})\overline{g} + \omicron(N^{2}).$$

I wanted to know how this obtained.

$\endgroup$
2
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Nov 27 '20 at 19:17
  • $\begingroup$ What is $\overline{\nabla}$ ? When it is just $\nabla_\overline{x}$, then the explanation is Taylor-Series. $\endgroup$ Dec 1 '20 at 18:00
0
$\begingroup$

The fact that $\overline{g}$ is a vector function is inconsequential here. Consider a component-wise expansion of $\overline{g}(\overline{x}+\overline{N}, t)$ as: $$g_{i}(\overline{x}+\overline{N}, t) = g_{i}(\overline{x}, t) + \sum_{j=1}^{n}(x_{j}+N_{j} - x_{j})\frac{\partial}{\partial N_{j}}g_{i}(\overline{x}+\overline{N}, t) + \omicron(N^{2}).$$

So, $$\overline{g}(\overline{x}+\overline{N}, t) = \overline{g}(\overline{x}, t) + \sum_{j=1}^{n}N_{j}\frac{\partial}{\partial N_{j}}\overline{g}(\overline{x}+\overline{N}, t) + \omicron(N^{2}).$$

The second term in the expansion can be interpreted as a dot product. It has a summation over $j$, which can be simplified by recognising that the summation over the partial derivatives is the gradient operator, while that over $N_{j}$ is the vector $\overline{N}$. So, the second term becomes: $\overline{N} \cdot \nabla$, and hence, the subtraction in your question yields the above dot product.

Hope this answers your question.

$\endgroup$
2
$\begingroup$

Analogous to the single-variable Taylor expansion $$ f(x) = f(a) + f'(a)(x-a)+O(x^2) $$ we have the multivariate expansion $$f(\mathbf{x})=f(\mathbf{a})+\nabla f(\mathbf{a})(\mathbf{x-a})+\ldots$$ Simply substitute in $\mathbf{x\rightarrow x+N}$ and $\mathbf{a \rightarrow x}$ into the second equation to obtain $$ f(\mathbf{x+N})=f(\mathbf{x})+(\nabla f(\mathbf{x}))\cdot \mathbf{N}+\ldots $$ and the result follows.

$\endgroup$
0
$\begingroup$

How does one Taylor expand a vector function of many variables?

$$\vec{f}=\vec{f}(\vec x)$$

taylor :

$$\vec{f}(\vec x+d\vec x)=\vec{f}(\vec x)+\frac {\partial \vec f}{\partial \vec x}\,d\vec{x}$$

where :

$$\frac {\partial \vec f}{\partial \vec x}\,d\vec{x}=(d\vec{x}\cdot \vec \nabla)\,\vec f(\vec x)$$

edit: for $$\vec f=\left[ \begin {array}{c} f_{{1}} \left( x_{{1}},x_{{2}} \right) \\ f_{{2}} \left( x_{{1}},x_{{2}} \right) \end {array} \right] $$ is: $$\frac {\partial \vec f}{\partial \vec x}= \left[ \begin {array}{cc} {\frac {\partial }{\partial x_{{1}}}}f_{{1} } \left( x_{{1}},x_{{2}} \right) &{\frac {\partial }{\partial x_{{2}}} }f_{{1}} \left( x_{{1}},x_{{2}} \right) \\ {\frac { \partial }{\partial x_{{1}}}}f_{{2}} \left( x_{{1}},x_{{2}} \right) &{ \frac {\partial }{\partial x_{{2}}}}f_{{2}} \left( x_{{1}},x_{{2}} \right) \end {array} \right] $$

the Jacobi matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy