1
$\begingroup$

Use summation notation to prove that $\vec\nabla (fg) = f \vec\nabla g + g\vec\nabla f$ where $f$ and $g$ are scalar functions.

So I'm assuming that I need to start by writing this in the summation notation. However, I am not familiar with how to write the gradient of a scalar in this notation. I know how to write dot product and cross, but how do we write using a scalar?

I think I'm confused on the idea of using a gradient on a scalar function? A scalar function is described as $(x_1,x_2,x_3)$, and the gradient is just the $(d/dx)$ operator on the function, so $(\frac{d}{dx} x_1, \frac{d}{dx} x_2, ..) $ Is this correct..?

$\endgroup$
  • $\begingroup$ I don't really see where you need to do any summation except for the one "+" in $\nabla(fg) = f\nabla g+ g \nabla f$. Concerning your edit: if $f=f(x_1,...,x_n)$ then $\nabla f = (\frac{df}{dx_1},...,\frac{df}{dx_n})^T$ $\endgroup$ – k1next Feb 19 '13 at 19:34
  • $\begingroup$ The method of proving this equation would be to essentially write out the left (or right side) of the equation in the notation, then doing simplifications to show that it is equivalent to the other side. I just don't know how to get started, on how to write the gradient of a just a scalar in the notation form. $\endgroup$ – julesverne Feb 19 '13 at 19:37
2
$\begingroup$

Let $[f]_i$ denote the $i$th component of the scalar function $f$. Then we need to prove that \begin{equation*} [\nabla(fg)]_i=[f\nabla g+g\nabla f]_i \end{equation*} But $\nabla f=(\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2},..., \frac{\partial f}{\partial x_n})$, so $[\nabla f]_i=\frac{\partial f}{\partial x_i}$. Using the product rule it follows that \begin{equation*} [\nabla(fg)]_i=\frac{\partial}{\partial x_i}(fg)=f\frac{\partial g}{\partial x_i}+g\frac{\partial f}{\partial x_i}=[f\nabla g+g\nabla f]_i \end{equation*} which is the result.

$\endgroup$
  • $\begingroup$ Perfect! I didn't realize there was no specific notation for the gradient, I was confusing myself with the dot product. Thank you so much! $\endgroup$ – julesverne Feb 19 '13 at 19:47
0
$\begingroup$

For what it's worth:

Let $f,g: \mathbb{R}^n\rightarrow \mathbb{R}$, thus: \begin{align} \nabla(fg)&= \begin{pmatrix} \partial_{x_1} \\ ..\\ \partial_{x_n}\end{pmatrix}(fg)\\&= \begin{pmatrix} \partial_{x_1} (fg)\\ ..\\ \partial_{x_n}(fg)\end{pmatrix} =\begin{pmatrix} g\partial_{x_1}f + f\partial_{x_1}g \\ ..\\ g\partial_{x_n}f + f\partial_{x_n}g \end{pmatrix} = f \begin{pmatrix} \partial_{x_1} g\\ ..\\ \partial_{x_n}g\end{pmatrix} +g \begin{pmatrix} \partial_{x_1} f\\ ..\\ \partial_{x_n}f\end{pmatrix} \\&= f \nabla g + g \nabla f \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.