# Simple Theorem

1. Feb 28, 2006

### ak416

Well actuallly 2 thms. They have to do with homogeneous functions. f(tx1,...,txn) = t^k * f(x1,...,xn). Now how do you show A) d/dx1 f(tx1,...,txn) = t^k-1 * d/dx1 f(x1,...,xn) and B) kt^(k-1)*f(x1,...,xn) = x1*d/dx1 f(tx1,...,xn) + xn*d/dxn f(x1,...,xn)

A) In the book They say that differentiating the first equation (definition of homogeneous function of degree k) by its first argument yields: d/dx1 f(tx1,...,txn) * t = t^k d/dx1 f(x1,...,xn) from which A easily follows. But how do they get this? I know you have to apply the chain rule somehow but im not sure exactly...
B)Same as A, i end up with expressions like d/d(tx1) x1 which intuitively seems like t but im not sure.

2. Mar 1, 2006

### HallsofIvy

Staff Emeritus
First rule: faced with a complicated problem, try a few simple examples. Suppose f(x,y)= xy. Clearly, f(tx,ty)= (tx)(ty)= t2xy so this is a homogeneous function. f(tx,ty)x= fx(tx,ty)(tx)x= tfx(tx,ty). But since f(tx,ty)= t2xy that is also equal to fx(tx,ty)= t2y.

Since x1 is multiplied by t, differentiating f(tx1,...) with respect to x1 is just (chain rule) the derivative of f times the derivative of txw which is t: t df()/dx1. But f(tx1, ...)= tnf(x1,...) so that derivative is equal to tnfx(x1,...). Divide both sides by t now.
What you've written for B is incorrect. It should be
kt^(k-1)*f(x1,...,xn) = x1*d/dx1 f(tx1,...,xn)+ ... + xn*d/dxn f(x1,...,xn). Do you see the difference? You just have x1 and xn terms on the right but it is in fact a sum through all the variables.

Try f(x,y,z)= xyz. Then f(tx,ty,tz)= t2xyz.
fx= t2yz so xfx= t2xyz, yfy= t2xyz, zfz= t2xyz and their sum is 3t2 xyz= ktk-1 f(x,y,z) with k= 3.