23.9 Euler Theorem
Consider a function f(x1, x2, x3,….,xn) that is homogeneous of degree r.
Let us rewrite f(kx1, kx2, kx3,….,kxn) as f(w1, w2, w3,….,wn) where wi = kxi for i = 1, 2,..,n.
By Definition 23.5.1 of the Total Differential we can write
df = f1 dw1 + f2 dw2 + ….+ fi dwi +…... ……+ fn dwn.
where each fi is the first order partial derivative with respect to wi.
If we now divide by dk we get the following result:
df = f1 dw1 + f2 dw2 + ….+ fi dwi +…... ……+ fn dwn.
dk dk dk dk dk
Since dwi = xi for each i = 1, 2, ….n we can refine the above result to
dk
df = f1 x1 + f2 x2 + ….+ fi xi +…... ……+ fn xn. *
dk
Recall that f(kx1, kx2, kx3,….,kxn) = kr f(x1, x2, x3,….,xn).
Thus df = r k r-1 f(x1, x2, x3,….,xn). **
dk
Also since each fi = f/wi it follows from the Chain Rule that
f/xi = f/wi wi/xi = k f/wi ***
Substituting ** into * we get
r kr-1 f(x1, x2, x3,….,xn) = f1 x1 + f2 x2 + ….+ fi xi +…... ……+ fn xn
Multiplying both sides by k we get
r kr f(x1, x2, x3,….,xn) = kf1 x1 + kf2 x2 + ….+ kfi xi +…... ……+ kfn xn
Substituting *** into the above equation we get
r kr f(x1, x2, x3,….,xn) = f/x1 x1 + f/x2 x2 + ….+ f/xi xi +…... ……+ f/xn xn
Hence the result in the next theorem.
23.9.1 Euler’s Theorem
Assume that f(x1, x2, x3,….,xn) is a homogeneous function of degree r.
Then r f(x1, x2, x3,….,xn) = f1(x) x1 + f2(x) x2 + ……….+ fi(x) xi +…..……+ fn(x) xn
where x = (x1, x2, x3,….,xn).
The right hand side is the sum of the products of the partial derivative fi(x) and the variable xi . The left hand side is simply r times the function f(x).
In the case of n = 2 variables, Euler’s Theorem for a homogeneous function f(x1, x2) of degree r is given by
r f(x1, x2) = f1(x) x1 + f2(x) x2