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I'm trying to solve the following problem:

Consider the function $f: S^2 \to \Bbb{R}$ given by $f(x,y,z)=x^{2023}+y^{2023}+z^{2023}$. Show that $df_p\neq 0$ for all $p\in f^{-1}(0)$.

I'm thinking about this:

  • Differentiate $f$ with respect to $x,y,z$ to obtain $f_x=2023 x^{2022}, f_y=2023 y^{2022}, f_z=2023 z^{2022}$.

  • Taking the standard parametrization of the unit sphere: $s(u,v)=(\cos(u) \sin(v),\sin(u) \sin (v), \cos (v))$.

  • We compute:

$$f_x(s(u,v))=2023 \cos ^{2022}(u) \sin ^{2022}(v)\\ f_y(s(u,v))=2023 \sin ^{2022}(u) \sin ^{2022}(v)\\f_z(s(u,v))=2023 \cos ^{2022}(v)$$

Now we verify by cases if it is possible that $(f_x(s(u,v)),f_y(s(u,v)),f_z(s(u,v)))=0$. Would this work for this problem?

Red Banana
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1 Answers1

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Yes, it works: $f_z=0$ first gives $\sin v=\pm1,$ and then $f_x=f_y=0$ give $\cos u=\sin u=0,$ which has no solution.

But a simpler method (without spherical coordinates) is: if $$2023 x^{2022}=2023 y^{2022}=2023 z^{2022}=0$$then $$x=y=z=0,$$ which is not consistent with $(x,y,z)\in S^2.$

Anne Bauval
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