For which real numbers k>0 is the function f:E^2->R given by f(x,y)=(x^2+y^2)^k differentiable?

Question

Advanced Calculus Problem

Answer 1

It would certainly help if we knew what you meant by E. In particular, assuming E is a subset of R, the answer will depend on whether E includes 0.

In general, ∂f/∂x = k(x^2+y^2)^(k-1) . 2x and ∂f/∂y = k(x^2+y^2)^(k-1) . 2y where these are defined. The only potentially problematic bit is (x^2 + y^2)^(k-1).

If E does not include 0, then x^2 + y^2 is always positive and so f is differentiable everywhere.

If E does include 0, then at the origin we will have (0)^(k-1) which is defined only for k>1.

Another possibility is that E is a subset of the complex line such that if x and y are in E, x^2 + y^2 is in R (so that f is indeed a function into R; obviously R is one such subset.) If we let x be in E, it would follow that x^2 + x^2 is in R, and hence x is either real or pure imaginary.

If we allow pure imaginary numbers in E, then on at least some of the domain of f, x and y will both be pure imaginary and x^2 + y^2 will be negative. Therefore f will only be defined when k is a rational number with an odd denominator in lowest-denominator form; in this case it will also be differentiable. If 0 is in E, then we get the further restriction (as before) that k > 1.

Answer 2

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