Proof involving homogeneous functions and chain rule

[hth]

Homework Statement

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

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

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

Homework Equations

The Attempt at a Solution

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

Then find dA/dt...

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

 

[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.

 

[HallsofIvy]

Homework Statement

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

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

You mean "f(tx₁, tx₂, tx₃,... tx_n)=t^s*f(x₁, x₂, x₃,... x_n) for all t"

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

Homework Equations

The Attempt at a Solution

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

Then find dA/dt...

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

 

[hth]

You mean "f(tx₁, tx₂, tx₃,... tx_n)=t^s*f(x₁, x₂, x₃,... x_n) for all t"

You're right. I apologize for my mistake.

 

[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,...,x_n).

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

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

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

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

 

[LCKurtz]

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,...,x_n).

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

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

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

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

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

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