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Euler's theorem and tan⁻¹ function

Short answer

Euler's theorem: if f is homogeneous of degree n, x ∂f/∂x + y ∂f/∂y = n f. Let z = tan u = (x³+y³)/(x+y); z is homogeneous of degree 2. Then x ∂z/∂x + y ∂z/∂y = 2z simplifies to x ∂u/∂x + y ∂u/∂y = sin 2u.

State Euler's Theorem for homogeneous function of two variables. If u = tan⁻¹((x³+y³)/(x+y)) then show that x ∂u/∂x + y ∂u/∂y = sin 2u.

Euler's theorem (1 mark)

If f(x, y) is homogeneous of degree n, then x ∂f/∂x + y ∂f/∂y = n f.

Proof (4 marks)

Given u = tan⁻¹((x³ + y³)/(x + y)), so

z = tan u = (x³ + y³)/(x + y).

Replacing x, y by tx, ty gives z(tx, ty) = t³(x³ + y³)/(t(x + y)) = t² z. So z is homogeneous of degree n = 2.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = 2z

As z = tan u: ∂z/∂x = sec²u ∂u/∂x, ∂z/∂y = sec²u ∂u/∂y. So

sec²u ( x ∂u/∂x + y ∂u/∂y ) = 2 tan u
x ∂u/∂x + y ∂u/∂y = 2 tan u / sec²u
                   = 2 tan u cos²u
                   = 2 sin u cos u
                   = sin 2u

Hence proved. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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