show that

\(\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)\) defines a smooth function \(\displaystyle f: S^{2}\to S^{2}\) where \(\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}\) so I am guessing it is a unit sphere.

How do we proceed with such problem?

f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

I started with computing these partial derivatives

\(\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)\)

\(\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)\)

\(\displaystyle \frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}\)

and these partial derivatives are all defined at any point from the domain of f

so I compute second order partial derivatives

\(\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}\)

\(\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}\)

\(\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}\)

and these 3 are also defined for all the points of the domain.

if I continue differentiating

\(\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0\) and

\(\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}\) which is defined for all points

What should I do next? can I conclude based on the calculations that f is certainly smooth?

Thanks in advance

\(\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)\) defines a smooth function \(\displaystyle f: S^{2}\to S^{2}\) where \(\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}\) so I am guessing it is a unit sphere.

How do we proceed with such problem?

f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

I started with computing these partial derivatives

\(\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)\)

\(\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)\)

\(\displaystyle \frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}\)

and these partial derivatives are all defined at any point from the domain of f

so I compute second order partial derivatives

\(\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}\)

\(\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}\)

\(\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}\)

and these 3 are also defined for all the points of the domain.

if I continue differentiating

\(\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0\) and

\(\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}\) which is defined for all points

What should I do next? can I conclude based on the calculations that f is certainly smooth?

Thanks in advance

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