# Finding the function, given the gradient.

1. Mar 17, 2008

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. Mar 17, 2008

### foxjwill

Just to clarify, you're using the standard definition of the gradient as $$\nabla = \frac{\partial}{\partial x} \textbf{i} + \frac{\partial}{\partial y} \textbf{j} + \frac{\partial}{\partial z}\textbf{k}?$$

In other words, $$\nabla$$ 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
3. Mar 17, 2008

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.

4. Mar 17, 2008

### foxjwill

So, if I'm reading this correctly, you're given $$f'(x)=|x|^p - 2x$$, and you're asked to find $$f(x)$$. First of all, have you learned antiderivatives yet? Oh, and I'm pretty sure there's no possible way for $$f(x)=\frac{|x|^p}{p}$$ for $$x \neq 0$$.

Last edited: Mar 17, 2008
5. Mar 17, 2008

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.

6. Mar 17, 2008

### foxjwill

Oh. I see. Try remembering that $$\frac{d}{dx}(|x|) = \texttt{sign}\ x = \frac{x}{|x|} = \frac{|x|}{x}$$

7. Mar 17, 2008

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

8. Mar 17, 2008

### foxjwill

yes, that's correct. Try working with the piecewise definition of $$|x|$$, i.e.

$$|x| = \left{ \begin{cases} x, & x>0\\ -x, & x<0\end{cases}$$

9. Mar 17, 2008

so which way should i be working?

10. Mar 17, 2008

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

11. Mar 17, 2008

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|?