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Homework Help: Finding the function, given the gradient.

  1. Mar 17, 2008 #1
    the gradient function is |x|^p-2 x
    and i need to find the function, which apparantly is 1/p |x|^p but i can't figure out how to show this.
    This is for a bigger problem where the function must be convex. and also p>1

    I tried, finding the derivative of 1/p |x|^p , but i don't get the gradient function.
    At first, I thought about this function: 1/p-1 |x|^p-1 where you can easily get the gradient function by taking the derivative but i was told that this is not a convex function.

    any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 17, 2008 #2
    Just to clarify, you're using the standard definition of the gradient as [tex]\nabla = \frac{\partial}{\partial x} \textbf{i} + \frac{\partial}{\partial y} \textbf{j} + \frac{\partial}{\partial z}\textbf{k}?[/tex]

    In other words, [tex]\nabla[/tex] is an operation on a scalar function which returns a vector function. The functions you've given me are both scalar. Is it possible you typed it in wrong?
     
    Last edited: Mar 17, 2008
  4. Mar 17, 2008 #3
    yes, i have the gradient f(x)= |x|^p-2 x, and i need to find f(x), in class, the definition of gradient is just the derivative w.r.t x of f(x)
    so i am asking why 1/p |x|^p is the answer because i don't see how you can use this, to find the gradient function |x|^p-2 x. so I thought the function was something else: 1/p-1 |x|^p-1, but i was told this function is not convex.
     
  5. Mar 17, 2008 #4
    So, if I'm reading this correctly, you're given [tex]f'(x)=|x|^p - 2x[/tex], and you're asked to find [tex]f(x)[/tex]. First of all, have you learned antiderivatives yet? Oh, and I'm pretty sure there's no possible way for [tex]f(x)=\frac{|x|^p}{p}[/tex] for [tex]x \neq 0[/tex].
     
    Last edited: Mar 17, 2008
  6. Mar 17, 2008 #5
    no the function is this: gradient f(x) = x |x|^(p-2) maybe this is more clear way to write it. and somehow get f(x) = 1/p |x|^p from it.
     
  7. Mar 17, 2008 #6
    Oh. I see. Try remembering that [tex]\frac{d}{dx}(|x|) = \texttt{sign}\ x = \frac{x}{|x|} = \frac{|x|}{x}[/tex]
     
  8. Mar 17, 2008 #7
    ok, should i be working from the gradient f(x) -> f(x) or vice versa.
    as well , i am getting confused.
    is this correct: to work from gradient f(x) -> f(x) we integrate. and f(x)-> gradient f(x) we differentiate.
    working from f(x) -> gradient.. i dont see how i can get gradient f(x).
    and going from gradient f(x) -> f(x) , i havn't a clue how to integrate that function
     
  9. Mar 17, 2008 #8
    yes, that's correct. Try working with the piecewise definition of [tex]|x|[/tex], i.e.

    [tex]|x| = \left{ \begin{cases} x, & x>0\\ -x, & x<0\end{cases} [/tex]
     
  10. Mar 17, 2008 #9
    so which way should i be working?
    gradient f(x) -> f(x)?
     
  11. Mar 17, 2008 #10
    can you tell me how to integrate this? or at least start, so i can get 1/p|x|^p , i need this small part for a bigger problem and this is making me stuck.

    i have thought about what you said about the piecewise, but that confuses me even more as i have to deal with not one but 2 functions now
     
  12. Mar 17, 2008 #11
    ok, so i can get 1/px^p for the x>0 case.
    but for the x<0 case:
    i am struggling
    i have,
    integ( (-x)^(p-2) x dx)
    can i write this as:
    = integ( (-1)^p (x)^(p-2) x dx )
    so,
    = (-1)^p integ (x^(p-2) x dx)
    which is just
    = (-1)^p 1/p x^p
    now how can i put the two together... to make x into |x|?
     
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