Talk:Gradient

message: "In mathematics, the gradient is a generalization of the usual concept of derivative to the functions of several variables."
The gradient is something like the rate at which something increases over an area.  The above sentence says something like, gradient is a non-specific reference to consequentials of non specifics, which isn't very specific, but is definitely a derivative consequential to specifics.  ~ R.T.G 13:33, 25 August 2014 (UTC)
replies: text: What does mean this comment in a non-jargon English, in particular, what are consequentials and specifics, and where are they defined? D.Lazard (talk) 13:56, 25 August 2014 (UTC)
replies: text: It means, using those words like that in a sentence makes it very difficult to understand without significant prior knowledge of the subject, and your response repeats and examples it beautifully in every sentiment, thankyou.  ~ R.T.G 15:14, 25 August 2014 (UTC)
replies: text: Perhaps it should say "In mathematics, the gradient is a generalization of the usual concept of derivative of a function in one dimension to a function in several dimensions"? In fact, as currently stated, it is not true; the gradient with arbitrary coordinates is not the same thing as a vector formed from the partial derivatives of a function of those coordinates, which the wording appears to suggest.  I'll change it accordingly.  —Quondum 06:29, 26 August 2014 (UTC)
replies: text: Your edit has improved the readability.  To be more helpful I'll have to take some time and actually figure it out.  But you are commended to approach a jargon request at all, kudos.  ~ R.T.G 22:45, 26 August 2014 (UTC)
thread_title:  Where is the jar gone? 

message: I have rewritten the section Gradient of a vector. The general definition of vector gradient is given by GTM 93, and I give the detailed reasoning as follows.
By Kozsul connection, one has
:\nabla_c \mathbf{f}^b = (\frac{\partial f^{\nu}}{\partial x^{\mu}}+{\Gamma^{\nu}}_{\mu\sigma}f^{\sigma})(\mathbf{e}^{\mu})_c(\mathbf{e}_{\nu})^b.
Given the metric g, one obtains
:g^{ac}\nabla_c \mathbf{f}^b = g^{\mu\rho}(\frac{\partial f^{\nu}}{\partial x^{\mu}}+{\Gamma^{\nu}}_{\mu\sigma}f^{\sigma})(\mathbf{e}_{\rho})^a(\mathbf{e}_{\nu})^b
in which the Greek symbols denote coordinates. By using English alphabet, one may get the results in the article.--IkamusumeFan (talk) 23:10, 26 October 2014 (UTC)
replies: text: This new version cannot been kept for several reasons. Firstly, it uses undefined notation, such that the g that appears in several formulas. Secondly, it is too technical, as it cannot be understood by most readers, even those who know the notion of gradient and use it frequently in the common case of Euclidean metric. Thirdly, your terminology is not standard and confusing: you use vector in the meaning of vector field. In standard terminology, "vector" means element of a vector space. If this vector space has a metric, this must been specified. If the vector depends on one or several variables, this must also been specified. Finally, when a special case (here, the gradient with respect of the Euclidean norm) is widely used, it must be explained before its generalization. Therefore, I'll revert your edits. Please, if you want to add or expand the generalizations of the simpler case, do it without destroying what is already there. D.Lazard (talk) 08:53, 27 October 2014 (UTC)
replies: text:  I disagree with the revert.  The issue of vector versus vector field is already a problem with that section independently of the recent edit, and should be fixed by normal editing.  The assumption of the exitence of a metric also has implicitly been assumed.  The proposed revision at least asserts flat out that there is a metric tensor, unlike the older version of the section.  (A metric tensor is always required to define the gradient, because it is needed to raise the index of the differential, even if all we care about are gradients of scalar fields.)  Also, it isn't any more technical than what is there already, which includes a formulation using Christoffel symbols for calculation in curvilinear coordinates.  Surely, the invariant formulation of this should also be mentioned.  So I have restored the edit, but moved it to the end of that section, with some tweaks.  Sławomir Biały (talk) 10:42, 27 October 2014 (UTC)
thread_title:  Rewrite the section Gradient of a vector 

message: According to tensor calculus, Tensors_in_curvilinear_coordinates#Gradient, the definition of gradient which produces the same vector from a scalar field independent of the coordinate system is:
\nabla f(x) = \frac{\partial f}{\partial q^i} b^i = \frac{\partial f}{\partial q^i} g^{i j} b_j 
where f is the function, x represents the input into the function parameterized by the coordinate system, q^i is the ith parameter of the coordinate system, b^i is the ith basis vector in the contravariant (~inverse) coordinate system, b_i is the ith basis vector in the (covariant) coordinate system, g^{i j} is the inverse metric tensor. 
I feel this is important to include on this wiki page as it says how to calculate the gradient in different coordinate systems. A few examples of the tensor / natural gradient include
Euclidean: b_1 = \vec{i}, b_2 = \vec{j}, b_3 = \vec{k}
:g_{i j} = g^{i j} = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
:So the gradient reduces to the familiar form of 
: \nabla f(x) = \frac{\partial f}{\partial x^m} \frac{\partial x^m}{\partial b^i} g^{i j} b_i = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k} 
Spherical: 
: x = r sin(\theta) cos(\phi)
: y = r sin(\theta) sin(\phi)
: z = r cos(\theta)
: J = \begin{bmatrix} 
\cos\left( \phi\right) \sin\left( \theta\right) & r\, \cos\left( \phi\right) \cos\left( \theta\right) & r\, -\sin\left( \phi\right) \sin\left( \theta\right) \\ 
\sin\left( \phi\right) \sin\left( \theta\right) & r\, \sin\left( \phi\right) \cos\left( \theta\right) & r\, \cos\left( \phi\right) \cos\left( \theta\right) \\
\cos\left( \theta\right) & -r \, \sin\left( \theta\right) & 0 \end{bmatrix}
: g_{i j} = J^T \, J = \begin{bmatrix}1&0&0\\0&r^2&0\\0&0&r^2 \, \sin\left( \theta\right)^2\end{bmatrix}
: g^{i j} = \begin{bmatrix}1&0&0\\0&r^{-2}&0\\0&0&r^{-2} \, \sin\left( \theta\right)^{-2}\end{bmatrix}
: b_1 = [\cos\left( \phi\right) \sin\left( \theta\right), \sin\left( \phi\right) \sin\left( \theta\right), \cos\left( \theta\right)] =
\vec{r} 
: b_2 = [r\, \cos\left( \phi\right) \cos\left( \theta\right),  r\, \sin\left( \phi\right) \cos\left( \theta\right), -r \, \sin\left( \theta\right) ] = \vec{\theta} \, r 
: b_3 = [r\, -\sin\left( \phi\right) \sin\left( \theta\right), r\, \cos\left( \phi\right) \cos\left( \theta\right), 0] = \vec{\phi} \, r \, \sin\left(\theta\right) 
: \nabla f(x) = \frac{\partial f}{\partial x^m} \frac{\partial x^m}{\partial b^i} g^{i j} b_i = \frac{\partial f}{\partial r} \vec{r} + \frac{\partial f}{\partial \theta} \frac{\vec{\theta}}{r} + \frac{\partial f}{\partial \phi} \frac{\vec{\phi}}{r \sin\left( \theta\right)} 
— Preceding unsigned comment added by Mouse7mouse9 (talk • contribs) 21:53, 21 January 2015 (UTC)
thread_title:  Natural Gradient (geometrically correct gradient) 

message: The description for the product rule was a bit misleading.
:If f and g are real-valued functions differentiable at a point , then the product rule asserts that the product  of the functions f and g is differentiable at a, and
::\nabla (fg)(a) = f(a)\nabla g(a) + g(a)\nabla f(a).
Initially, I assumed it had something to do with the derivative of \nabla (fg). The little segment "" confuses matters further and I think it should be removed or simplified because it is a bit obvious. My suggestion:
:If f and g are real-valued functions differentiable at a point , then the product rule asserts that the product  is differentiable at a, and
::\nabla (fg)(a) = f(a)\nabla g(a) + g(a)\nabla f(a).
Muntoo (talk) 01:03, 29 July 2015 (UTC)
replies: text: The new revision is an improvement because it is more concise.  I'm not sure I understand the source of confusion in the original wording, but I would support a revision like this if it clarifies things for some readers.  SławomirBiały 11:40, 29 July 2015 (UTC)
thread_title: Product Rule

message: I rewrote the first two paragraphs of the lead to make them mathematically accurate and supplied citations.  The wording was improper and sloppy.  There were no sources cited which is probably why the description was incorrect.  The next paragraph about the tangent space is sloppy and contains many intuitive but inaccurate words.  For example, what is the graph of a function of several variables?  I think the paragraph about the Jacobian should actually refer to the Jacobian matrix.  I am not familiar enough with tangent spaces, Jacobian matrices, and Banach spaces to rewrite the rest of the lead competently, but I hope someone knowledgeable will rewrite it properly and cite reliable sources for everything.—Anita5192 (talk) 16:20, 7 April 2019 (UTC)
thread_title:  Lead clean-up 

message: The gradient of a line I believe is difference in y divided by difference in x.
This surely is the single most common query people are asking when they want to find out what a gradient is.
Why on Earth am I reading about three dimension gradients already when the two dimensional case isn't yet discussed?
I do believe in fact the most definition is not present in this article.  I couldn't find it, at any rate.
This I have found is normal for Wikipedia math or physics articles.  They can only be understood by people who don't need to read them.
The search results from Bing are no better.  As far as I can tell, the entire first page of results has no page which says, simply, difference in y over difference in x.  — Preceding unsigned comment added by 176.58.238.150 (talk) 15:53, 9 April 2019 (UTC)
replies: text: The answer to your complaint is in the first line of the article (hatnote), where it is explicitly said that, for the gradient of a line, you have to go to the article Slope. Don't blame blame Wikipedia, if you are not able to read the beginning of an article. D.Lazard (talk) 16:09, 9 April 2019 (UTC)
thread_title:  Typical of wikipedia math/science articles 

message: I have added to the definition part of the article with the meaning of the word gradient outside the context of math.  — Preceding unsigned comment added by DJ Jerree Erickson (talk • contribs) 21:00, 4 December 2020 (UTC)
thread_title: Recent Changes

message: — Preceding unsigned comment added by 98.199.140.156 (talk) 21:00, 31 July 2021 (UTC)
replies: text: I don't understand the problem, the arrows point in the direction of the gradient. Pressure gradients move from low to high as well. It's the opposite direction that a particle would be pushed if you placed there, if that's what you mean; i.e. the force density would be given by \mathbf{f} = -\nabla p. ‑‑Volteer1 (talk) 17:22, 26 August 2021 (UTC)
replies: text: I didn't ask the question. An IP created this section with the question and no content other than the figure.—Anita5192 (talk) 17:46, 26 August 2021 (UTC)
replies: text: Oh, my bad, thanks for clarifying. I thought it was a bit strangely worded. ‑‑Volteer1 (talk) 17:52, 26 August 2021 (UTC)
thread_title: why is arrows moving from blue to red if concentration and pressure gradients move from high to low

message: I'm pretty sure that history section is wrong, or at least doesn't belong to this article. It says "The gradient was invented by Dan "the man" Lasker around 2016 where it quickly became one of the most popular trends in web development."  Gifah.28 (talk) 03:25, 8 December 2022 (UTC)
thread_title:  History section