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

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]f'_x = kx_k, k = 1, 2, ..., n[/itex]

    3. The attempt at a solution

    The partial should be f(sub)x(sub)k, as in, the partial derivative of [itex]f[/itex] with respect to [itex]x_k[/itex]. 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 [itex]k = 1[/itex], then the derivative of [itex]f[/itex] with respect to [itex]x_k[/itex] [itex]= 1x_1[/itex], but that tells me absolutely nothing because I don't even know what that [itex]x_1[/itex] refers to...

    Any help would be much appreciated!
  2. jcsd
  3. Oct 15, 2012 #2


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    f_{x_k} → [itex]\displaystyle f_{x_k} [/itex]

    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:

    [itex]\displaystyle f_{x}=x [/itex]

    [itex]\displaystyle f_{y}=2y [/itex]

    [itex]\displaystyle f_{z}=3z\ . [/itex]
    Last edited: Oct 15, 2012
  4. Oct 15, 2012 #3


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

    fxk = kxk

    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.
  5. Oct 15, 2012 #4
    I realized that a few minutes after I posted it, my bad!

    So we've got [itex]f_{x_k}=kx_x[/itex].

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

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

    If this is correct, then moving on:

    [itex]f_{x_2}=g_{x_2}(x_2, x_3, ... , x_n)[/itex]


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

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

    [itex]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)[/itex].

    Which seems to imply that

    [itex]f(x_1, x_2, ... , x_n) = \frac{k}{2}x_k^2 +[/itex] 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:

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

    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!
  6. Oct 15, 2012 #5


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    It looks like you have:

    [itex]\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)[/itex]

    Where does it end?
  7. Oct 16, 2012 #6


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    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 [itex]f(x,y)= x^2+ 3xy+ y[/itex], [itex]f_x= 2x+ 3y[/itex] and [itex]f_y= 3x+ 1[/itex] so that, differentiating [itex]f_x[/itex] with respect to y, [itex]f_{xy}= 3[/itex] and, differentiating [itex]f_y[/itex] with respect to x, [itex]f_{yx}= 3[/itex].

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