Let $g: \mathbb R^{+} \times \mathbb R^{+}\to \mathbb R^{2}\setminus\{0\}$, $g(x,y):=(xy,\sqrt{x}/y)$


$f: \mathbb R^{2}\setminus\{0\}\to \mathbb R$, $f(x,y):=\ln(x^{2}y^{2})$

Find the Taylor Series of $h: f\circ g$ about $(x_{0},y_{0})$ to the second order.

My idea: After just having completed one-dimensional Taylor Series in our lecture, I would think that I could find the Taylor Polynomial of $f$ and then plug in my values $(g(x_{0},y_{0}))$ into the existing Taylor Polynomial of $f$ in order to get the $f \circ g$. This would be analogous to the one-dimensional Taylor Polynomial of a composite function (e.g. If I wanted to find the Taylor Polynomial of $e^{x^{3}-5x+1}$ about $0$, I would use the Taylor polynomial of $e^{x}$ about $0$ and then simply plug in $x^{3}-5x+1$( Correct me if I am wrong).

Why does this principle not work in multiple dimensions?

  • $\begingroup$ This still works as long as $f:\mathbb R^n→\mathbb R$ has (partial) derivatives of arbitrary orders at $g(x_0)$ and so does each component of $g:\mathbb R^m→\mathbb R^n$ at $x_0$. $\endgroup$ Jun 24, 2018 at 11:47
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    $\begingroup$ The principle will work, but it is counterproductive here: $f(g(x,y))=\ln(x^2y^2x/y^2)=\ln(x^3)=3\ln x$. Why suffer? $\endgroup$ Jun 29, 2018 at 9:18


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