# Directional derivative and non-appropriate solution?

• estro
In summary: I don't think so, as I can still use the same argument and just use the orthonormal basis.In summary, the directional derivative and "non-appropriate" solution is that there are n scalars a_{1},a_{2},...,a_{n} which 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).
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?

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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

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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.

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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?

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.

lanedance said:
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:
$$g(x,y,z)=f((x_0,y_0,z_0)+xu_1+yu_2+zu_3))$$
$$x=a_{2}u_{2}+a_{2}u_{2}+...+a_{n}u_{n}$$
$$f_x(x_0,y_0,z_0)=f_{a_{2}u_{2}+a_{2}u_{2}+...+a_{n}u_{n}}(x_0,y_0,z_0)=$$
$$g_{a_{1}x_{1}+a_{2}x_{2}+a_{n}x_{n}}(x_0,y_0,z_0)=\sum_{i=1}^n a_{i}g_{x_i}(x_0,y_0,z_0)=$$
$$\sum_{i=1}^n a_{i}f_{u_{i}}(x_0,y_0,z_0)$$

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estro said:
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
estro said:
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

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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
$$f_x(y) = \nabla_xf(y) = \nabla f(y) \bullet \frac{x}{|x|}$$

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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?

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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?

## What is a directional derivative?

A directional derivative is a measure of the rate of change of a function in a specific direction. It is calculated by taking the dot product of the gradient of the function and the unit vector in the desired direction.

## How is a directional derivative different from a regular derivative?

A directional derivative is a generalization of a regular derivative, which measures the rate of change in a specific direction. A regular derivative only measures the rate of change in the direction of the x-axis.

## What is a non-appropriate solution in relation to directional derivatives?

A non-appropriate solution is a situation where the directional derivative does not exist or is not defined. This can occur when the function is not differentiable or when the direction vector is not well-defined.

## How can non-appropriate solutions be avoided when using directional derivatives?

One way to avoid non-appropriate solutions is to ensure that the function is differentiable in the given direction. This can be done by checking for continuity and differentiability of the function at the given point.

## In what fields of science are directional derivatives commonly used?

Directional derivatives are commonly used in fields such as physics, engineering, and economics to measure rates of change in specific directions, such as in the direction of a force or a gradient. They are also used in optimization problems in mathematics and computer science.

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