1. ## Two questions :)

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!

2. I don't have much time, so my response is quick....

You have to use the identity: $\displaystyle \frac{df}{d\mathbf{n}} = \mathbf{n}^T \nabla f$

So we have:

$\displaystyle \left[\frac{df}{d\mathbf{a}} \frac{df}{d\mathbf{b}} \frac{df}{d\mathbf{c}}\right]^T = \mathbf{U}\nabla f$

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)

3. Thank you. However, I didn't understand anything from your post.
What's "T"? I couldn't see the logic of it.

Sorry, thanks.

BTW the second question (which wasn't answered anyway ) is not relavent anymore!

4. Originally Posted by aurora
Thank you. However, I didn't understand anything from your post.
What's "T"? I couldn't see the logic of it.

Sorry, thanks.

BTW the second question (which wasn't answered anyway ) is not relavent anymore!
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:

$\displaystyle \mathbf{a}^T \mathbf{b}$

So $\displaystyle \mathbf{n}^T \nabla f$ 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:
$\displaystyle \frac{df}{d\mathbf{n}} = \mathbf{n}^T \nabla f$
which is true for ANY vector n.

In particular for vectors a b and c we get three equations:
$\displaystyle \frac{df}{d\mathbf{a}} = \mathbf{a}^T \nabla f$
$\displaystyle \frac{df}{d\mathbf{b}} = \mathbf{b}^T \nabla f$
$\displaystyle \frac{df}{d\mathbf{c}} = \mathbf{c}^T \nabla f$

I guess you know that system of equations can be written as a one matrix equation. Thus we get:
$\displaystyle \left[ \begin{array}{c} \frac{df}{d\mathbf{a}} \\ \frac{df}{d\mathbf{b}} \\ \frac{df}{d\mathbf{c}} \end{array} \right] = \left [ \begin{array}{c} \mathbf{a}^T \\ \mathbf{b}^T \\\mathbf{c}^T \end{array} \right] \nabla f$

Or alternativly:$\displaystyle \left[\frac{df}{d\mathbf{a}} \frac{df}{d\mathbf{b}} \frac{df}{d\mathbf{c}}\right]^T = \mathbf{U}\nabla f$

Where $\displaystyle \mathbf{U} = \left [ \begin{array}{c} \mathbf{a}^T \\ \mathbf{b}^T \\\mathbf{c}^T \end{array} \right]$

U is unitary (which means $\displaystyle \mathbf{U}^T\mathbf{U} = \mathbf{I}$ $\displaystyle , \mathbf{I}$ - 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:
$\displaystyle \left[\frac{df}{d\mathbf{a}} \frac{df}{d\mathbf{b}} \frac{df}{d\mathbf{c}}\right] \left[\frac{df}{d\mathbf{a}} \frac{df}{d\mathbf{b}} \frac{df}{d\mathbf{c}}\right]^T = \left(\frac{df}{d\mathbf{a}}\right)^2 + \left(\frac{df}{d\mathbf{b}}\right)^2 + \left(\frac{df}{d\mathbf{c}}\right)^2$

If you take dot product of RHS the unitary matrix vanishes and you get:
$\displaystyle (\nabla f)^T \nabla f = \left[\frac{df}{dx} \frac{df}{dy} \frac{df}{dz}\right] \left[\frac{df}{dx} \frac{df}{dy} \frac{df}{dz}\right]^T = \left(\frac{df}{dx}\right)^2 + \left(\frac{df}{dy}\right)^2 + \left(\frac{df}{dz}\right)^2$

5. gotcha! Thank you very, very much!