Given the partial derivatives, find the function or show it does not exist.

[Jormungandr]

Homework Statement

$f'_x = kx_k, k = 1, 2, ..., n$

The Attempt at a Solution

The partial should be f(sub)x(sub)k, as in, the partial derivative of $f$ with respect to $x_k$. I wasn't sure how to represent that using TeX.

I'm honestly at a complete loss here, because I'm not entirely sure what the subscripts refer to. I suppose if $k = 1$, then the derivative of $f$ with respect to $x_k$ $= 1x_1$, but that tells me absolutely nothing because I don't even know what that $x_1$ refers to...

Any help would be much appreciated!

 

[SammyS]

Homework Statement

$f'_x = kx_k, k = 1, 2, ..., n$

The Attempt at a Solution

The partial should be f(sub)x(sub)k, as in, the partial derivative of $f$ with respect to $x_k$. I wasn't sure how to represent that using TeX.

I'm honestly at a complete loss here, because I'm not entirely sure what the subscripts refer to. I suppose if $k = 1$, then the derivative of $f$ with respect to $x_k$ $= 1x_1$, but that tells me absolutely nothing because I don't even know what that $x_1$ refers to...

Any help would be much appreciated!

f_{x_k} → $\displaystyle f_{x_k}$

The subscripts mean that you have n variables.

Rather than using x, y, and z for three variable, you could use x₁, x₂, and x₃ .

What if n = 100 ? How else would you name all the variables ?

Added in Edit:

For instance, if we did use x, y, and z for x₁, x₂, and x₃, we would have:

$\displaystyle f_{x}=x$

$\displaystyle f_{y}=2y$

$\displaystyle f_{z}=3z\ .$

 

[STEMucator]

So you have some partial of f with respect to some x_k where :

f_xk = kx_k

For a varying constant k.

You want to find your original function f. What happens if you integrate here?

Note that _xk refers to an INDEPENDENT variable. We could have chosen y, z, u, etc.

 

[Jormungandr]

I realized that a few minutes after I posted it, my bad!

So we've got $f_{x_k}=kx_x$.

The function $f$ described by this set of partial derivatives should be in the form $f(x_1, x_2, ... , x_n)$, right? So I run into a bit of a snag when we integrate that partial.

If $f_{x_1} = x_1$, then
$f(x_1, x_2, ... , x_n)$ = $\frac{1}{2}x_1^2 + g(x_2, ... x_n)$

If this is correct, then moving on:

$f_{x_2}=g_{x_2}(x_2, x_3, ... , x_n)$

Implying:

$g_{x_2}(x_2, x_3, ... , x_n) = 2x_2$, right? And that:
$g(x_2, x_3, ..., x_n) = x_2^2 + h(x_3, x_4, ... , x_n)$

Well, I did this process over a few times and I'm reaching a definite pattern here, that shows:

$f(x_1, x_2, ... , x_n) = \frac{1}{2}x_1^2 + x_2^2 + \frac{3}{2}x_3^2 + 2x_3^2 + j(x_5, x_6, ... , x_n)$.

Which seems to imply that

$f(x_1, x_2, ... , x_n) = \frac{k}{2}x_k^2 +$ some function of g. Of course, I realize I would have arrived at this exact expression if I had integrated the beginning expression, I just wanted to see where this process would lead me.

Now, an issue arises with the function of g afterwards. Should it be:

$f(x_1, x_2, ... , x_n) = \frac{k}{2}x_k^2 + g(x_{k+1}, x_{k+2}, ... , x_n)$?

Have we arrived at the solution or is there some compact way to write the function of g? Or have I done it all wrong completely? :P Thanks!

 

[SammyS]

...

Well, I did this process over a few times and I'm reaching a definite pattern here, that shows:

$f(x_1, x_2, ... , x_n) = \frac{1}{2}x_1^2 + x_2^2 + \frac{3}{2}x_3^2 + 2x_4^2 + j(x_5, x_6, ... , x_n)$.
...

Have we arrived at the solution or is there some compact way to write the function of g? Or have I done it all wrong completely? :P Thanks!

It looks like you have:

$\displaystyle f(x_1, x_2, ... , x_n) = \frac{1}{2}\left((x_1)^2 + 2(x_2)^2 + 3(x_3)^2 + 4(x_4)^2 + \dots\right)$

Where does it end?

 

[HallsofIvy]

Look at the mixed partial derivatives. As long as the derivatives are continuous, mixed partial derivatives, with the same derivatives in different order, must be equal.

For example, if $f(x,y)= x^2+ 3xy+ y$, $f_x= 2x+ 3y$ and $f_y= 3x+ 1$ so that, differentiating $f_x$ with respect to y, $f_{xy}= 3$ and, differentiating $f_y$ with respect to x, $f_{yx}= 3$.

If we are told that $f_x= 3x+ 2y$ and $f_y= x- 2y$, we know that is impossible because that gives $f_{xy}= 2$ and $f_{yx}= 1$ which are not equal.