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I have a matrix $A(\vec{v})$ whose entries are scalar functions of a vector $\vec{v}=(v_{x},v_{y})$. To be concrete I have something like

$A=\begin{pmatrix}f_{1}(\vec{v})& f_{2}(\vec{v})\\f_{3}(\vec{v}) &f_{4}(\vec{v}) \end{pmatrix}$

My goal is to Taylor expand the matrix $A$ around the point $\vec{v_{0}}$ at first oder in $|\vec{v}-\vec{v_{0}}|$.

I do that with:

Normal@Series[A[v], {vx ,v0x, 1}, {vy, v0y, 1}]//MatrixForm

The matrix I get is now a function of $\vec{v}-\vec{v_{0}}$ but I want to transform it in a matrix in $\vec{w}=\vec{v}-\vec{v_{0}}$. I have tried with

A[v]=A[v-v0]

but it doesn't work. I guess my question is how to change the variable of a function.

Also doing the expansion I don't get quadratic terms in $v_{x}$ $v_{y}$ (because it's first oder), but I get mixed terms like $v_{x}v_{y}$. Is there a way to get rid of those?

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    $\begingroup$ I think you want to adapt the accepted answer here. $\endgroup$
    – march
    Commented Apr 6, 2022 at 3:19

1 Answer 1

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You get the series expansion of A[{vx,vy}] by:

Normal@Series[A[{vx, vy}], {vx, v0x, 1}, {vy, v0y, 1}]/. Derivative[{1,1}][A] ->0

![enter image description here

and to replace v-v0 by w:

 Normal@Series[A[{vx, vy}], {vx, v0x, 1}, {vy, v0y, 1}]/. Derivative[{1,1}][A] ->0 /. {vx - v0x -> wx, vy - v0y -> wy}

![

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