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I am new to Mathematica. I would like to understand if this output of $$\text{Series}\left[\left\{u v,\frac{u^2}{2}+w^2,\log \left(\frac{1}{u^2+1}\right)\right\},\{u,0,1\},\{v,0,1\},\{w,0,1\}\right]$$ is correct, because "by hand" I found that the jacobian matrix is $$J=\left( \begin{array}{ccc} v & u & 0 \\ u & 0 & 2 w \\ -\frac{2 u}{u^2+1} & 0 & 0 \\ \end{array} \right)$$ and $$\text{Normal}\left[\text{Series}\left[\left\{u v,\frac{u^2}{2}+w^2,\log \left(\frac{1}{u^2+1}\right)\right\},\{u,0,1\},\{v,0,1\},\{w,0,1\}\right]\right]$$ We get as output $\{u v,0,0\}$ but it should be $\{0,0,0\}$. Thanks.

\[Delta][u_, v_, w_] := {u*v, 1/2 u^2 + w^2, 
  Log[1/(u^2 + 1)]}

Normal[
 Series[{u*v, 1/2 u^2 + w^2, Log[1/(u^2 + 1)]}, {u, 0, 1}, {v, 0, 
   1}, {w, 0, 1}]]

Out[124]= {u v, 0, 0}
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    $\begingroup$ Please directly post your Mathematica code instead of the LaTeX formula. $\endgroup$
    – cvgmt
    Dec 16 '20 at 8:39
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I think Series is not suitable for multiple vector value function. So I recommend to create the Taylor expand by hand.

δ[u_, v_, w_] := {u*v, 1/2 u^2 + w^2, Log[1/(u^2 + 1)]}
D[δ[u, v, w], {{u, v, w}, 1}]
D[δ[u, v, w], {{u, v, w}, 2}]
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Mathematica must know the "smallness" of u,v,w !

Assuming same order try

Normal[Series[{u*v, 1/2 u^2 + w^2, Log[1/(u^2 + 1)]} /. {u -> eps u, v -> eps v, w -> eps w}
, {eps, 0, 2}]] /. eps -> 1
(*{u v, u^2/2 + w^2, -u^2}*)
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