1
$\begingroup$

In Multivariable Calculus, we can easily find the gradient of a scalar function (producing a scalar field) $f : \mathbb{R^n} \to \mathbb{R}$, and the gradient function would produce a vector field.

$$grad(f) = \vec{\nabla}(f) = \left< \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2} , ... , \frac{\partial f}{\partial x_n} \right> = \begin{bmatrix} \frac{\partial f}{\partial x_{1}} \\ \frac{\partial f}{\partial x_{2}} \\ ... \\ \frac{\partial f}{\partial x_{n}} \end{bmatrix}$$

Evaluating Vector Functions By Components

In Multivariable Calculus we learn that we can differentiate any vector function by taking the derivatives of its scalar components/functions, likewise we also learn that we can integrate any vector function by integrating each of its scalar components.

e.g.

Given a function $g : \mathbb{R^n} \to \mathbb{R^m}$, comprised of scalar functions $f_{i} : \mathbb{R^n} \to \mathbb{R} $

$${\vec{g}}{'}(t) = \left< {f_{1}}^{'}(t), {f_{2}}^{'}(t), ..., {f_{n}}^{'}(t)\right> = \begin{bmatrix} {f_{1}}^{'}(t) \\ {f_{2}}^{'}(t) \\ ... \\ {f_{n}}^{'}(t) \end{bmatrix}$$

$$\int\vec{g}(t) = \left< \int{f_{1}}^{'}(t) \ , \int{f_{2}}^{'}(t)\ , \ ..., \ \int{f_{n}}^{'}(t)\right> = \begin{bmatrix} \int{{f_{1}}^{'}(t)} \\ \int{f_{2}}^{'}(t) \\ ... \\ \int{f_{n}}^{'}(t) \end{bmatrix}$$

Can we do the same for the Del Operator?

Since we can differentiate an integrate any vector function, by taking the derivatives or integrals of its scalar components/functions, can we evaluate the gradient of a vector function by applying the Del Operator to each of it's scalar components to compute the gradient of each scalar function producing a scalar field. I realize that this would produce a Tensor field as a result.

Again given the same vector function $g : \mathbb{R^n} \to \mathbb{R^m}$, comprised of scalar functions $f_{i} : \mathbb{R^n} \to \mathbb{R} $, can we say the following :

$$ T = grad(\vec{g}) = \vec{\nabla}(\vec{g}) = \left< \vec{\nabla}(f_1), \vec{\nabla}(f_2), ..., \vec{\nabla}(f_n) \right> = \begin{bmatrix} \vec{\nabla}(f_1) \\ \vec{\nabla}(f_2) \\ ... \\ \vec{\nabla}(f_n) \end{bmatrix}$$

With $T$ denoting the tensor field outputted by taking the gradient of the vector field produced by the function $g$

Just to close off, I realize that a vector function, can take both vectors or scalars as inputs, and my question here only covers the case for scalar inputs to a vector function, however, extending this to vector inputs would be a fairly trivial task as we could just break up the vector inputs into its scalar components and then work from there, which we would now know how to do as that is covered within the scope of this question.

$\endgroup$
3
  • $\begingroup$ Also if I have left out any important cases of Vector/Scalar Functions that are important or should have been covered within the scope of the question, or if you have spotted any gaps in my understanding of Multivariable Calculus, please comment below. $\endgroup$
    – Perturbative
    Apr 26, 2016 at 0:58
  • $\begingroup$ Have a look at en.wikipedia.org/wiki/Jacobian_matrix_and_determinant $\endgroup$
    – John B
    Apr 26, 2016 at 1:02
  • $\begingroup$ Take a look at youtube.com/watch?v=p75-f0gN5c0 and see if that is a starting point that might appeal to you. As I point out frequently in comments, your confusion stems from what Hermann Weyl called "orgies of formalism". You have scalars, vectors, matrices, tensors, transposes, square brackets, triangular brackets, derivatives, gradients, little arrows above symbols. No wonder you find it confusing! But given the question you're asking, you might find Tensor Calculus cleansing. $\endgroup$
    – Lemma
    Apr 26, 2016 at 4:00

1 Answer 1

Reset to default
0
$\begingroup$

If $\vec{g}=\left[\begin{array}{c}f_1\\\vdots\\f_m\end{array}\right]$ then the derivative of $\vec{g}$ is the matrix
$$J\vec{g}=\left[\begin{array}{c}\nabla f_1\\\vdots\\\nabla f_m\end{array}\right],$$ which is an $m\times n$ - rectangular array.

In components, you would see it as $$J\vec{g}=\left[\dfrac{\partial f_i}{\partial x_j}\right],$$ where $i$ is for rows and $j$ is for columns, and where $x_1,...,x_n$ are the standard coordinate functions of $\Bbb R^n$.

$\endgroup$
4
  • $\begingroup$ But this doesn't seem to generalize the gradient properly. If we have a scalar valued function, i.e. $g = [f_1]$ in your example above, we usually write the gradient (like in the original post above) as $$\left[\begin{array}\\ \frac{\partial f_1}{\partial x_1} \\ \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_1}{\partial x_3} \\ \end{array}\right],$$ so that the index of the partial derivatives get incremented on each row. In your definition the partial definitions get increment on the columns so it doesn't seem to generalize the scalar case? $\endgroup$
    – ManUtdBloke
    May 8, 2018 at 21:03
  • 1
    $\begingroup$ it a matter of convention, but nobody writes the gradient in column form or... then how are you going to write laws as the chain's rule? and for that how do you going to explain this in matricial form? $\endgroup$
    – janmarqz
    May 8, 2018 at 23:50
  • $\begingroup$ I agree its a matter of convention. However don't we usually write vectors in column form, e.g. $Ax = b$ where $x$ and $b$ are vectors - as the gradient takes a scalar function to a vector I would expect that it would also be in column form? $\endgroup$
    – ManUtdBloke
    May 9, 2018 at 12:10
  • 1
    $\begingroup$ take a look: exemplifying with two dimension one uses $$\left(\begin{array}{c}x\\y\end{array}\right)\longmapsto\left(\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right)\left(\begin{array}{c}x\\y\end{array}\right),$$ so the gradient of $f$ at $p$ maps $$\Bbb R^2\to\Bbb R,$$ for a simple scalar $f:\Bbb R^2\to\Bbb R$. $\endgroup$
    – janmarqz
    May 9, 2018 at 14:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .