I don't have much time, so my response is quick....
You have to use the identity:
So we have:
Where U is unitary because base is orthonormal
If you take dot product of LHS and RHS you will get the right answer (because U is unitary)
1) a, b and c are in R^3, and they form an orthogonal base.
I need to prove that, for every diffrential function in "p", f: R^3-->R:
(df/d(a) (p))^2 + (df/d(b) (p))^2 + (df/d(c) (p))^2 = (df/dx (p))^2 + (df/dy (p))^2 + (df/dz (p))^2
Note that these are the directional derivatives. a, b and c are vectors of course, and they are all calculated in point "p", where f is diffrential.
I'm playing with it, but I don't get anything... looks kinda easy, but I dunno.
2) This looks even easier, but I'm just not seeing it:
prove that if f is a homogeneous function of order k (=> f(tx) = t^kf(x) ), then the partial derivatives of f are homogenous of order k-1.
I'll keep trying those two, but I'm having so much deadlines, I figured I better not waste time and write here... usually takes some time 'till I get a response anyway (not that I complain!!!)
If I make any progress with these I'll let you know.
THANK YOU VERY MUCH!!!! You are all wonderful to help like that!
T is matrix transpose. If you have to vectors a b (which are matrices with one column and 3 rows) you can write dot product as:
So is dot product of vector n and gradient f (which is vector too).
I hope you know that directional derivatives are dot product of the vector and gradient. Hence the equation:
which is true for ANY vector n.
In particular for vectors a b and c we get three equations:
I guess you know that system of equations can be written as a one matrix equation. Thus we get:
Or alternativly:
Where
U is unitary (which means - matrix identity) because vectors a b an d c are ORTHONORMAL (not orthogonal as you said)!
Now if you take this equation and do the dot product of LHS you get:
If you take dot product of RHS the unitary matrix vanishes and you get: