Directional derivative and non-appropriate solution?

[estro]

Directional derivative and "non-appropriate" solution?

Homework Statement

Given following normal vectors u_1,__2,...,u_n which are basis for R^N
Prove that there are n scalars which satisfy:
f_x(a)=\sum_{i=1}^na_if_{u_i}(a), given that f is differentiable at a\in R^n

Homework Equations

I know that if f is differentiable at a\in R^n for every t=(t_1,t_2,...,t_n) happens f_u=\sum_{i=1}^n f_{x_i}(a)t_i

The Attempt at a Solution

I changed from the standard basis [and it's cartesian coordinate system] to the new basis B=\{u_1,u_2,...,u_n\}
because B is basis i can write [x] as k_1u_1+k_2u_2+...+k_nu_n
and then:
f_x([a]_e)=f_{k_1u_1+k_2u_2+...+k_nu_n}([a]_B)=\sum_{i=1}^n f_{u_i}([a]_B)k_i
And the proof is finished.
What is bothers me is the fact that the "relevant equation" was proved using the standard basis, so I'm not sure that my solution is appropriate given how easy it becomes with such a trick.

Do you think my proof is valid?

 

[lanedance]

first, you need to write it like a proof to check if it is tight. What you have given so far seems to make a few jumps that you need to justify, but is in the right direction.

in particular your post does not describe a few things
- what the a_i are, are the components of a in the standard x basis?
- you final line has k_i not a_i as in the first line?
- how do you make the final leap
- x is referenced as both directional derivative direction, and "the standard" basis"

Also do you know (Einstein) summation notation? That will be very useful

so as b_j is basis you can write, the x_i in terms of the b_j as you implied, which is just matrix multiplication :
x_i = \sum_{j} k_{ij}b_j = k_{ij}b_j

Similarly you could use the to write a in terms of basis b_j and maybe not even have to go near the standard basis, but I'll leave it for you to clear up those points

 

[estro]

I know that the normal vectors: u_{1},u_{2},...,u_{n} are basis for R^n so there are n scalars k_{1},k_{2},...,k_{n} which satisfy x=\sum_{i=1}^n k_{i}u_{i}

Now I can write f_x([a]_E)=f_{k_{1}u_{1}+k_{2}u_{2}+...+k_{n}u_{n}}([a]_B) Can I do such a thing? Here I changed the cartesian coordinate system.
f_{k_{1}u_{1}+k_{2}u_{2}+...+k_{n}u_{n}}([a]_B)=\sum_{i=1}^n f_{u_i}([a]_B)k_i here I used the "relevant equation" which was already proved for me.

 

[lanedance]

how about first writing the question exactly as it was written..?

1st post, as i read it, has the a_i's related to the position 'a', whilst you replace the a_i's with k_i's in the final line of proof?

 

[estro]

You're right, instead of k_i I should have been using a_i
But is it really crucial?
Nevertheless I think my idea is wrong because the "relevant equation" on which my prove is based was proved using the standard basis.
So I'll try to think about another idea without changing the basis.

 

[estro]

how about first writing the question exactly as it was written..?

1st post, as i read it, has the a_i's related to the position 'a', whilst you replace the a_i's with k_i's in the final line of proof?

Lanedance, sorry only now I figured out what wrong with the question formulation, let me try again.

Given normal vectors u_{1},u_{2},...,u_{n} which are also basis for R^n, I need to show that there are n scalars a_{1},a_{2},...,a_{n} that satisfy: f_x(x_0,y_0,z_0)=\sum_{i=1}^n a_{i}f_{u_i}(x_0,y_0,z_0) given that f is differentiable at (x_0,y_0,z_0).

This is what I'm trying to do:
g(x,y,z)=f((x_0,y_0,z_0)+xu_1+yu_2+zu_3))
x=a_{2}u_{2}+a_{2}u_{2}+...+a_{n}u_{n}
f_x(x_0,y_0,z_0)=f_{a_{2}u_{2}+a_{2}u_{2}+...+a_{n}u_{n}}(x_0,y_0,z_0)=
g_{a_{1}x_{1}+a_{2}x_{2}+a_{n}x_{n}}(x_0,y_0,z_0)=\sum_{i=1}^n a_{i}g_{x_i}(x_0,y_0,z_0)=
\sum_{i=1}^n a_{i}f_{u_{i}}(x_0,y_0,z_0)

 

[lanedance]

You're right, instead of k_i I should have been using a_i
But is it really crucial?

its important as I'm trying to work out what you're attempting... which wasn't clear

Nevertheless I think my idea is wrong because the "relevant equation" on which my prove is based was proved using the standard basis.
So I'll try to think about another idea without changing the basis.

assuming a separate orthonormal basis isn't necessarily wrong either

 

[lanedance]

ok so, the notation is still a little confusing, so if needed let's call the frees variable y, with components y_i
y = (x, y, z) = (y_1,y_2,y_3)

now the direction you want to find the derivative in is x, express it in the u_i basis with scalars a_i
x = a_i u_i

now what is your definition of directional derivative? it we call the free variables y, something like
f_x(y) = \nabla_xf(y) = \nabla f(y) \bullet \frac{x}{|x|}

 

[estro]

I'm little bit confused with your last post.
As for the definition of directional derivative:
If the following limit exist then it is the directional derivative at x_0 in direction \vec u
\lim_{t\rightarrow 0} \frac {f(x_0+t\vec u)-f(x_0)} {t}=f_{\vec u} (x_0)

2 post above is my second try to prove, it is completely wrong?

 

[lanedance]

I guess its the leap
f_{x}(y) = f_{u_i k_i}(y) = f_{u_1 k_1 +...+u_n k_n}(y)= k_i f_{u_i}(y) = \sum_i k_i f_{u_i}(y)

can you justify that?