# Proof involving homogeneous functions and chain rule

1. Nov 20, 2009

### hth

1. The problem statement, all variables and given/known data

A function f is called homogeneous of degree s if it satisfies the equation

f(x1, x2, x3,..... xn)=t^s*f(x1, x2, x3,..... xn) for all t

Prove that the $$\sum$$ from i=1 to n of xi * df/dxi (x1, x2, x3,..... xn) = sf(x1, x2, x3,..... xn).

2. Relevant equations

3. The attempt at a solution

Take A=f(t1,t2,..,tn)=t^kf(t1,t2,..,tn)

Then find dA/dt...

I keep lost in the differentiation. Am I even on the right track? Any help is appreciated.

Last edited: Nov 20, 2009
2. Nov 20, 2009

### LCKurtz

Try something like this (3 variable example):

Let g(t) = f(tx,ty,tz) = t3f(x,y,z)

Now, using subscript notation for the partials of f with respect to its arguments, calculate g'(t) using the chain rule:

g'(t) = xf1(tx,ty,tz)+yf2(tx,ty,tz)+zf3(tx,ty,tz)=3t2f(x,y,z)

Now let x = u/t, y=v/t and z = w/t and see what happens. Should generalize to n variables.

3. Nov 21, 2009

### HallsofIvy

Staff Emeritus
You mean "f(tx1, tx2, tx3,..... txn)=t^s*f(x1, x2, x3,..... xn) for all t"

4. Nov 21, 2009

### hth

You're right. I apologize for my mistake.

5. Nov 21, 2009

### hth

Alright, here's my attempt.

f(tx1, tx2, tx3,..... txn)=t^s*f(x1, x2, x3,..... xn) for all t

Prove that the $$\sum$$ from i=1 to n of xi * df/dxi (x1, x2, x3,..... xn) = sf(x1, x2, x3,..... xn).

Proof.

Let f = f (x1, x2, x3,...,xn).

Then, by differentiating the function f(ty) = t^(s)f(y) by the chain rule,

$$\partial$$/$$\partial$$x1f(ty)d/dt(ty1) + ..... + $$\partial$$/$$\partial$$xnf(ty)d/dt(tyn) = st^(s-1) f(y).

So, y1$$\partial$$/$$\partial$$x1f(ty) + ..... + yn$$\partial$$/$$\partial$$xnf(ty) = st^(s-1)f(y).

This is the best I could come up with. I'm pretty confused, honestly.

6. Nov 21, 2009

### LCKurtz

Are you responding to my suggestion? It isn't clear to me. Don't call all the xi the same thing y. You want to start with

g(t) = f(tx1,tx2,...,txn) =tnf(x1, x2,..... xn)
and differentiate with respect to t similar to what you have shown. Then do what I did.