Definition of Partial Derivative: Correct Form?

[Cyrus]

Which one is the correct definition? I have seen both being used in texts and it makes a world of difference which one is used when doing proofs.


$$\frac {\partial f}{\partial \bar{x} } = [ \frac {\partial f}{\partial x_1 }, \frac {\partial f}{\partial x_2 }, \frac {\partial f}{\partial x_n }]$$

OR:

$$\frac {\partial f}{\partial \bar{x} } = \left[\begin{array}{c} \frac {\partial f}{\partial x_1 } \\ \\ \frac {\partial f}{\partial x_2 }\\ \\ \frac {\partial f}{\partial x_n }\end{array}\right]$$

My book and wiki have the first definition, but I have seen some people use the second.

 

[christianjb]

They look like the same thing to me.

 

[Cyrus]

One is a column vector and one is a row vector.

 

[cristo]

What's the context here? Your expression looks very like grad(f).

 

[christianjb]

What's the context here? Your expression looks very like grad(f).

It's very nearly the same thing as writing grad.

f(r+dr)=grad[f(r)].dr
=dr df/dr
=dr. I df/dr

So grad[f(r)]=I df/dr

(I'm too lazy to latex the above, but all the r, dr are vectors)

 

[D H]

The notation $df/d\vec x$ is just a non-standard way to write $\nabla f$.

The standard representation of a displacement vector is as a column vector. With this representation, a displacement vector transforms from frame A to frame B as

$$\vec x_B = T_{A\to B} \, \vec x_A$$

The gradient of a scalar field transforms by a different set of rules. The gradient lives in the dual space of the displacement vector field. The gradient should thus be represented as a row vector. With the gradient expressed as a row vector, the gradient of a function transforms according to

$$\nabla f_B(x) = (\nabla f_A(x)) \, T_{A\to B}$$

 

[cristo]

The notation $df/d\vec x$ is just a non-standard way to write $\nabla f$.

That's what I thought; although the bar in the original post confused me!

The standard representation of a displacement vector is as a column vector. With this representation, a displacement vector transforms from frame A to frame B as

$$\vec x_B = T_{A\to B} \, \vec x_A$$

The gradient of a scalar field transforms by a different set of rules. The gradient lives in the dual space of the displacement vector field. The gradient should thus be represented as a row vector. With the gradient expressed as a row vector, the gradient of a function transforms according to

$$\nabla f_B(x) = (\nabla f_A(x)) \, T_{A\to B}$$

Of course, some people gloss over this and write all "vectors" as row vectors!

 

[Cyrus]

The bar means its a scalar differentiated W.R.T a vector.

Also, It never dawned on me to think of it as the gradient, but in fact that's exactly what it's doing (only if x is spatial coordinates). In that case, the correct definition is definition 2.

But I am asking which one is the correct definition? Why would you need context to answer my question?

 

[Mentz114]

According to post #6 by DH

The gradient should thus be represented as a row vector.

 

[pmb_phy]

Which one is the correct definition? I have seen both being used in texts and it makes a world of difference which one is used when doing proofs.


$$\frac {\partial f}{\partial \bar{x} } = [ \frac {\partial f}{\partial x_1 }, \frac {\partial f}{\partial x_2 }, \frac {\partial f}{\partial x_n }]$$

OR:

$$\frac {\partial f}{\partial \bar{x} } = \left[\begin{array}{c} \frac {\partial f}{\partial x_1 } \\ \\ \frac {\partial f}{\partial x_2 }\\ \\ \frac {\partial f}{\partial x_n }\end{array}\right]$$

My book and wiki have the first definition, but I have seen some people use the second.

The components of the first is a vector and as such is is layed out using a row matrix, which is the correct thing to do here. The second has components of a vector but is a column matrix of a 1-form which is the wrong thing to do. Only 1-forms are represented with the variables in the denuminator using subscripts and a change of sign for the spatial components.

Well ... it sounds good anyway. Does anyone agree?

Pete

 

[Cyrus]

According to post #6 by DH

Incorrect, the gradient is a column vector because my spatial coordinates is a column vector in R^n. So the gradient also has to be a column vector in R^n.

 

[Hurkyl]

Which one is the correct definition?

For better or worse -- the right answer to this question is "the one in the book." (Or "the one the instructor gave", or whatever is appropriate)

Also, for better or worse (worse, IMHO), it is common to formalize the mathematics so that no distinction is made between the two.


But if you are going to make a distinction (which I certainly advise) -- then $\partial f / \partial \vec{x}$ is going to be the opposite of what you use to denote $\vec{x}$. The most common convention is that $\vec{x}$ is a column, thus $\partial f / \partial \vec{x}$ is a row.



And, for maximum confusion, the notation $\nabla f$ is often used both to denote the gradient and to denote $\partial f / \partial \vec{x}$. (e.g. see Spivak's Differential Geometry)

 

[Cyrus]

Thats my problem. The book uses one way, my notes from class, which is from a different book, uses another way.

Here is a paper that basically says the same thing I've been saying in the footnote. But I wanted a second opinion from you guys.

http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.AppD.d/IFEM.AppD.pdf

Also note, the Jacobian is defined as the Transpose in my book of what's in the link above.

 

[Hurkyl]

Thats my problem. The book uses one way, my notes from class, which is from a different book, uses another way.

Here is a paper that basically says the same thing I've been saying in the footnote. But I wanted a second opinion from you guys.

http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.AppD.d/IFEM.AppD.pdf

Also note, the Jacobian is defined as the Transpose in my book of what's in the link above.

Ye gads! That's (IMHO) a very poor conventional choice -- so much that my first instinct would be to call it wrong! I would be a thousand times happier if x and y were row vectors, rather than the column vectors they were defined to be. (Or keep them as columns, but transpose everything else in that appendix)

 

[Cyrus]

Well, in Dynamics it makes more sense to define things as column vectors and not row vectors.

 

[christianjb]

I still maintain that

$\nabla f(\mathbf{r})=\mathbf{I}\frac{\partial f(\mathbf{r})}{\partial \mathbf{r}}$

Grad and df/dr are not completely equivalent.

 

[Cyrus]

I don't understand what your talking about christian. Can you explain what your saying in more detail?

 

[christianjb]

$f(\mathbf{r}+d\mathbf{r})=f(\mathbf{r})+\nabla f(\mathbf{r}).d\mathbf{r}$

Also

$f(\mathbf{r}+d\mathbf{r})=f(\mathbf{r})+\frac{\partial}{\partial \mathbf{r}} f(\mathbf{r})d\mathbf{r} <br /> = f(\mathbf{r})+\frac{\partial}{\partial \mathbf{r}} f(\mathbf{r})\mathbf{I}.d\mathbf{r}$

So

$\nabla f(\mathbf{r})=\mathbf{I}\frac{\partial f(\mathbf{r})}{\partial \mathbf{r}}$

 

[D H]

Incorrect, the gradient is a column vector because my spatial coordinates is a column vector in R^n. So the gradient also has to be a column vector in R^n.

If you represent spatial coordinates as a column vector (WHO DOES THIS GARBAGE?), you should represent the gradient as a row vector.

Another way to look at this is to view displacement vectors and the gradient of a scalar field as tensors. Displacement vectors are contravariant vectors, while gradients are covariant. They live in spaces dual to each other. It is best to represent the two vectors in different forms.

 

[Cyrus]

In fact, it is the standard in dynamics D_H. The configuration space will typically be a column in R^n.

 

[D H]

Christian, you implicitly chose a convention for $df/d\vec r$ by doing what you did and then used that choice to justify what you did.

It is better to write this as

$$f(\vec r+d\vec r) =<br /> f(\vec r) + d\vec r \cdot \frac{\partial}{\partial \vec r}f(\vec r)$$

and then it doesn't matter whether you represent spatial vectors as column vectors or as row vectors.

 

[D H]

In fact, it is the standard in dynamics D_H. The configuration space will typically be a column in R^n.

I'm an orientationally challenged idiot.

Read my post #6. I had it right there, where I said displacement vectors are typically represented as column vectors (and that gradients should be represented as row vectors).

Think of displacement vectors (and velocity vectors and acceleration vectors) as living in one space (e.g., column vectors); the gradient must necessarily live in the dual space (e.g., row vectors).

Suppose some units-challenged engineer gave you x, y, and z spatial coordinates in meters, meters, and furlongs. You naturally transform those goofy meters into furlongs and do all of your work in the furlongs/fortnight system. When you report your results back, you have to transform back to the mixed system. Different transformation rules apply for the displacement vectors and gradients your report. Displacement vectors transform as contravariant tensors, gradients as covariant tensors.

 

[christianjb]

Christian, you implicitly chose a convention for $df/d\vec r$ by doing what you did and then used that choice to justify what you did.

It is better to write this as

$$f(\vec r+d\vec r) =<br /> f(\vec r) + d\vec r \cdot \frac{\partial}{\partial \vec r}f(\vec r)$$

and then it doesn't matter whether you represent spatial vectors as column vectors or as row vectors.

Of course.

But my convention makes much more sense.

 

[Mentz114]

You naturally transform those goofy meters into
furlongs and do all of your work in the furlongs/fortnight system.

Aha - now I understand. A furlong is the length of a row (furrow) of beans. So we should use row vectors for position. I always work in roods per hour, so I don't have this problem.

 

[D H]

Of course.

But my convention makes much more sense.

Your convention involved dividing by a vector. That doesn't make much sense at all.

 

[christianjb]

Your convention involved dividing by a vector. That doesn't make much sense at all.

Hey- if it gives the right answer at the end of the manipulations then who cares?

You never end up with an answer that involves the division of a vector. It's just a matter of book-keeping.

Personally I don't like the convention that df/dr = grad f. It's better for the derivative to obey f(r+dr)-f(r)=dr df/dr, which this convention doesn't.

 

[D H]

The convention that $df/d\vec r$ is a $1\times n$ row vector and $d\vec r$ is a $n\times1$ column vector does obey the convention $f(\vec r+d\vec r) = f(\vec r) + df/d\vec r*d\vec r$. The product of a 1xn vector and a nx1 vector is a scalar.

 

[christianjb]

The convention that $df/d\vec r$ is a $1\times n$ row vector and $d\vec r$ is a $n\times1$ column vector does obey the convention $f(\vec r+d\vec r) = f(\vec r) + df/d\vec r*d\vec r$. The product of a 1xn vector and a nx1 vector is a scalar.

The inner product, yes.

 

[D H]

I wrote that as a matrix product, not the inner product.

 

[christianjb]

Now I'm confused.

"The product of a 1xn vector and a nx1 vector is a scalar"

That's the inner product right?