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Euler's theorem statement with cos⁻¹ example

Short answer

Euler’s theorem: If f(tx,ty)=tⁿf(x,y) with continuous first partials, then x f_x + y f_y = n f. Here z=cos u=(x+y)/(√x+√y) is homogeneous of degree 1/2, so x z_x+y z_y=z/2. Since z_x=−sin u u_x, similar for y, we get −sin u (x u_x+y u_y)=(1/2)cos u, giving x ∂u/∂x+y ∂u/∂y=−(1/2)cot u.

State Euler's theorem for homogeneous function of two variables. If u = cos⁻¹((x+y)/(√x+√y)), then prove that x ∂u/∂x + y ∂u/∂y = −(1/2) cot u.

Euler's theorem (1 mark)

If f(x, y) is a homogeneous function of degree n, i.e. f(tx, ty) = tⁿ f(x, y), with continuous first partial derivatives, then

x ∂f/∂x + y ∂f/∂y = n f.

Proof (4 marks)

Given u = cos⁻¹((x + y)/(√x + √y)). Let

z = cos u = (x + y)/(x^(1/2) + y^(1/2)).

Then z(tx, ty) = t(x + y)/(t^(1/2)(√x + √y)) = t^(1/2) z, so z is homogeneous of degree n = 1/2.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = (1/2) z

Since z = cos u: ∂z/∂x = −sin u ∂u/∂x, ∂z/∂y = −sin u ∂u/∂y. Therefore

−sin u ( x ∂u/∂x + y ∂u/∂y ) = (1/2) cos u
x ∂u/∂x + y ∂u/∂y = −(1/2) cos u / sin u = −(1/2) cot u

Hence proved. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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