- 2071 Bhadra · 5 marks
Euler's theorem for cos⁻¹((x+y)/(√x+√y))
Short answer
Let z = cos u = (x+y)/(√x+√y). z is homogeneous of degree 1/2. By Euler’s theorem, x z_x + y z_y = (1/2)z. With z=cos u, this gives −sin u (x u_x + y u_y) = (1/2)cos u, so x u_x + y u_y = −(1/2) cot u.
If u = cos⁻¹((x+y)/(√x+√y)) then show that x ∂u/∂x + y ∂u/∂y = −(1/2) cot u.
Euler's theorem
If f(x, y) is homogeneous of degree n, then x ∂f/∂x + y ∂f/∂y = n f.
Proof
Given u = cos⁻¹((x + y)/(√x + √y)). Let
z = cos u = (x + y)/(x^(1/2) + y^(1/2)).
Replace x, y by tx, ty: 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
With z = cos u: ∂z/∂x = −sin u ∂u/∂x and ∂z/∂y = −sin u ∂u/∂y. So
−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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