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## The recent move

I disagree with gradient having been moved to gradient (calculus) (and making gradient into disambiguation) for three reasons:

1. It was done without prior consultation on the talk page.
2. The links pointing to gradient were not fixed to point to the new page.
3. There is little need for a disambiguation page.
4. It looks to me that the meaning of gradient as vector field is by far more dominant than the meaning of the gradient as slope. And either way, the second meaning is also mathematical, in spite of what the created disambiguation page at gradient seemed to imply.

gradient (calculus) is is now moved back to gradient. Discussion welcome. Oleg Alexandrov 23:41, 14 Jun 2005 (UTC)

I made a gradient (disambiguation) page, listing the two meanings. Oleg Alexandrov 23:52, 14 Jun 2005 (UTC)

## Jargon heavy?

Right now this article is far too jargon heavy and i can't make any sense of it. In lower levels of maths at least gradiant reffers to slope that is the m in y=mx+c or more generally the value of dy/dx. Plugwash 01:47, 20 July 2005 (UTC)

I added a sentence in the introduction saying that the gradient becomes the derivative in 1D. Now anybody could follow the derivative article to read more about that. Is that better, or you want more? :) Oleg Alexandrov 01:57, 20 July 2005 (UTC)
thats certainly better though i think we should get rid of the disabig and bring it all together in one article since the concepts are so closely related. Plugwash 18:13, 20 July 2005 (UTC)
I would not agree with that. In math the gradient is primarily exactly as described in the article (a vector field of partial derivatives), while what you mean by gradient is called slope. Also, most articles linkin to gradient do seem to mean the vector field definition. But this is an opinion. If you can do a good job at combining the things without being too biased towards the slope definition, then please go ahead. Oleg Alexandrov 18:38, 20 July 2005 (UTC)
mmm ok slope is a big article. lemme have a think about this. Plugwash 19:25, 20 July 2005 (UTC)
I think that maybe User:Oleg Alexandrov missed what User:Plugwash meant due to a more advanced understanding of the meaning that obscured previous understandings he may have had about the meaning that were perhaps even more similar to Plugwash'es own. Basically what I am saying is that when you are dealing with only one input and only one output variable, a single variable from the set of real numbers - gradient and slope are the same thing because the "field" is one dimensional anyways. Zaphraud (talk) 02:11, 7 April 2008 (UTC)
I surely understand that. But what I'm saying is that we have a dedicated slope article for the 1D case. Slope can be mentioned here, but this article is still primarily about the bigger concept of the gradient. Oleg Alexandrov (talk) 03:38, 7 April 2008 (UTC)

I don't agree with the recent changes. I believe the article now does a poor job at explaining what the gradient is. All this because there are two gradients, one is the vector field, the other is a number. They are of course related, via the dot product, but they are not the same. That's why I think it is better to have a disambiguation page explaining the similarities and differences, and keep this page only for the main meaning, which is the vector field. Oleg Alexandrov 00:54, 22 July 2005 (UTC)

You keep claiming that the vector calculus meaning is the main meaning yet almost everyone has heared and uses the word gradiant (at least here in britan, americans seem to prefer grade) yet only a comparitively small number will have even heared of vector calculus. Plugwash 02:15, 22 July 2005 (UTC)

Fair enough. At least now we agree that two meanings are involved. I understand that most people in Britan think of gradient as a slope, however, very few of the articles linking here mean that by gradient (I checked around 20 randomly chosen ones, maybe you are willing to do some research yourself and let me know what you think).

I believe one can write a short note mentioning the British usage, then link to grade (geography), which is the article describing the meaning as you want it.

Either way, I find the first paragraph in the text very clumsily written. Oleg Alexandrov 03:10, 22 July 2005 (UTC)

to me at least the vector calculus definition seems a strictening up of the laymans definition. That is it defines that the gradiant is measured in the direction where it is greatest and defines that the direction of greatest slope is stored as the angle part of the gradiant. Plugwash 13:02, 22 July 2005 (UTC)

Plugwash, good point. I still think however that the introduction is too clumsy now. I would like the first paragraph be split in two, as too many things are there. I will look at it later today. Oleg Alexandrov 16:06, 22 July 2005 (UTC)

Yeah, it's a little problematic that in British English, what we use "gradient" for "slope", and "gradient vector" for what is described in this article... Enochlau 11:00, 8 November 2005 (UTC)

One way to settle this discussion is to find some references (published textbooks) on the subject where gradient is defined one way or the other. Given these references, it should be straightforward to present two or three definitions (a vector field, slope, the general case for higher level mathematics) and discuss their relations. Right now, I only see some argumentation without any backup in the established literature.--KYN 17:56, 17 December 2005 (UTC)

Note that as taught in British high schools, "gradient" means the tangent of the angle of inclination, not the sine. I.e. grade (geography) describes an ambiguity that doesn't exist in practice in the British usage. Also, the concept is taught in terms of the derivative of a function, not as being specific to geography. -- David Hopwood 82.42.16.20 17:19, 19 December 2005 (UTC)

## Question

Shouldn't

${\displaystyle \nabla \phi ={\begin{pmatrix}{\frac {\partial \phi }{\partial x}},{\frac {\partial \phi }{\partial y}},{\frac {\partial \phi }{\partial z}}\end{pmatrix}}}$

be

${\displaystyle \nabla \phi ={\begin{pmatrix}{\frac {\partial \phi }{\partial x}},{\frac {\partial \phi }{\partial y}},{\frac {\partial \phi }{\partial z}}\end{pmatrix}}^{T}}$,

the gradient being (at least in engineering) usually thought as a column vector?

Andrea Censi 15:18, 22 March 2006 (UTC)

Does it matter that much if it is a row or column vector? I think this is a more general, vector calculus issue, and I would argue that it is not worth the trouble mentioning that a vector may be written both horizontally and vertically. Oleg Alexandrov (talk) 05:53, 23 March 2006 (UTC)
Well, the convention that it is used is that ${\displaystyle {\frac {\partial f}{\partial \mathbf {x} }}}$ and

${\displaystyle \nabla _{\mathbf {x} }f}$ are each other's transpose, such that

${\displaystyle {\frac {\partial f}{\partial x}}={\begin{bmatrix}{\frac {\partial f}{\partial {x_{1}}}}&\dots &{\frac {\partial f}{\partial {x_{n}}}}\end{bmatrix}}}$

is a row vector. In fact this is formally important because so you can write

${\displaystyle df={\frac {\partial f}{\partial x}}dx}$

and the multiplication is well defined ( first is a row, second is a column). And then it's better that the gradient being a column vector because it is of the same "kind" as ${\displaystyle x}$. So I propose to write the following definition for gradient:

${\displaystyle \nabla _{\mathbf {x} }f:=\left({\frac {\partial f}{\partial x}}\right)^{T}={\begin{bmatrix}{\frac {\partial f}{\partial {x_{1}}}}\\\vdots \\{\frac {\partial f}{\partial {x_{n}}}}\end{bmatrix}}}$

So it's not a question that an object is "written" horizontally or vertically, but in calculus a row vector and a column vector are really of different type. This is not an "aestetic" point, but having consistency in such matters really makes calculation easier. In fact, I see that you used the same convention in Gradient_descent

Andrea Censi 17:43, 23 March 2006 (UTC)

The distinction is between contravariant vectors and covariant vectors or forms. The fact that a contra- and a covector can be contracted to a scalar which can be represented by matrix multiplication of the components arranged in a column and a row matrix, does not imply a preference for representing a (contra)vector as either row or cloumn matrix. That is just convention, it doesn't mean anything. For me the gradient of a function f is ${\displaystyle \mathrm {d} f=\sum _{i}{\frac {\partial f}{\partial c^{i}}}\mathrm {d} c^{i}}$ where c is any set of curvilinear coordinates. Only if I suppress the choice of coordinates do I need to choose between listing the components ${\displaystyle \{{\frac {\partial f}{\partial c^{i}}}\}_{i}}$ as a row or column matrix. --MarSch 11:47, 11 April 2006 (UTC)

${\displaystyle \nabla f(\rho ,\phi ,\theta )={\begin{pmatrix}{\frac {\partial f}{\partial \rho }},{{\frac {1}{\rho }}{\frac {\partial f}{\partial \phi }}},{{\frac {1}{\rho \sin \phi }}{\frac {\partial f}{\partial \theta }}}\end{pmatrix}}}$

isn't consistent with the nomenclature of http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

I've switched ${\displaystyle \rho }$ and r, but the symbol usage in the page above isn't consistent with the definitions in Cylindrical_coordinates (where azimuthal is ${\displaystyle \theta }$) and Spherical coordinates (where azimuthal is ${\displaystyle \theta }$ and polar is ${\displaystyle \phi }$). Whosasking 15:06, 6 September 2007 (UTC)

I have always interpreted an expression on the form ${\displaystyle (a_{1},...,a_{N})}$ as merely being an N-tuple, i.e., it can be interepreted as either a column vector or row vector depending on the situation. (In most situations it really doesn't matter.) It's just a list of the components of any of such type of vectors. \\ Jens Persson (193.10.114.119 (talk) 09:39, 13 June 2008 (UTC))

Actually, it does matter. Strictly speaking, the gradient at a point is NOT a vector, and the general gradient function is NOT a vector field. The gradient at a point is a covector, and the general gradient is a one-form, i.e., a smooth section of the cotangent bundle. In euclidean space, the distinction between the tangent space and the cotangent space is manifested as row and column vectors. That said, in finite dimensions the tangent and cotangent spaces at a point (as well as the fibers of higher order covariant and contravariant tensor bundles) are canonically isomorphic, so for purposes of calculation it doesn't really matter. 74.192.193.21 (talk) 14:21, 4 March 2011 (UTC)

If you ask any mathematician to define gradient, the defintion that he gives will involve partial derivatives. The use of the word gradient to mean 'slope' results from the fact the slope is a special case of a gradient that is commonly encountered in lower level mathematics. Since mathematicians are forever generalizing, I believe that the gradient article should simply mention the word confusion and point out that the slope is a special case. This should take no more than two sentences. The article should describe the gradient in its full meaning. In that respect, the article is perhaps jargon-light as oppsed to jargon-heavy. --anon

That the slope is a special case is the second paragraph in the introduction. Should there be more than that? Oleg Alexandrov (talk) 23:40, 17 November 2005 (UTC)
And ask any photoshop user to define gradient you'll get a gradual blend. Needs a mention? ...dave souza, talk 16:06, 22 March 2006 (UTC)
I guess that could go in gradient (disambiguation), as this article seems to be about the vector calculus concept. Oleg Alexandrov (talk) 05:53, 23 March 2006 (UTC)
Which redirects to gradient. Is it worth a separate page? As an attempt, I'll just add a note here. The illustration shows nice circular and linear gradients..dave souza, talk 06:40, 23 March 2006 (UTC)
The redirect can be transformed in a disambig, if necessary.
The problem with the photoshop definition is that I don't see how it relates to the definition of gradient as slope, inclination. It seems to me a separate independent definition, and if this is so, it appears confusing inserted in the article where it is. That's when disambiguation pages are handy I think. Wonder what you think. Oleg Alexandrov (talk) 14:34, 23 March 2006 (UTC)
Good question, I've tried rephrasing the wording to clarify this, and also pointed out that gradient meaning road or surface slope is the usual usage in the UK. As this is the generic "gradient" page the various meanings should be outlined on it: of course it could be logical to move this page to Gradient (calculus) if it's to be confined to a particular mathematical meaning. ..dave souza, talk 18:39, 23 March 2006 (UTC)
I don't quite see how the "Photoshop gradient" implies the vector calculus gradient, so now there is a contradiction in paragraph 3. Oleg Alexandrov (talk) 04:34, 24 March 2006 (UTC)
Didn't mean to imply that, so have had another go at improving my phrasing. The difficulty is that vector calculus is over my head, but gradient is a term and concept I've often used in construction and imaging. Try thinking of photoshop and other graphics applications (vector as well as bitmap) as mathematical transformations on the values of pixels. This is what a transform layer in photoshop does: a gradient transform gradually changes values from one colour to another. Hope that helps. ..dave souza, talk 06:45, 24 March 2006 (UTC)
The "gradient" in Photoshop is a particularization. In fact, it means "produce an image whose gradient is fixed". For example, consider an image which goes from black to white from left to right. Let ${\displaystyle f(x,y)}$ be the gray value at each pixel x,y. The gradient of this image is then ${\displaystyle (1,0)}$ and is constant. If you set the gradient as ${\displaystyle (0,1)}$ then it's black at the top and white at the bottom. if you set the gradient as ${\displaystyle (1,1)}$ you get black at top-left and white at the bottom-right. So I think that this is the reason why this pattern was called "gradient" in photoshop. As for me, I'd mention that the other meanings are derived from the original meaning (calculus). Andrea Censi 10:26, 24 March 2006 (UTC)

Dear experts, how about a image gradient article, with pictures, references, and all that? And then we may link to it from gradient. Oleg Alexandrov (talk) 04:56, 25 March 2006 (UTC)

I agree Andrea Censi 15:49, 25 March 2006 (UTC)
Gradient is the logical page for disambiguating, and if a brief paragraph about the related graphics usage can't be accommodated then logically the calculus aspect should become Gradient (calculus), the graphics version could be Gradient (image) and the related term Grade also included on the disambig page. If you're desperate to keep this exotic page as is, then at the least it should have a For other uses disambiguation paragraph at the top. Mathematicians, please be logical. ...dave souza, talk 08:40, 26 March 2006 (UTC)
Yes, making multiple pages also seems a good solution. But please, Dave, please be clear to whom you are referring to as "mathematicians", because it could be an insult to some :-) Andrea Censi 09:31, 26 March 2006 (UTC)

## Question about wikibook on calculus

I discovered that this concept is well covered on a Wikibook http://en.wikibooks.org/wiki/Calculus:Multivariable_calculus#Gradient_vectors - what is the wikipedia policy in this case, just copy the material, summarize it or link to it? Andrea Censi 16:08, 26 March 2006 (UTC)

## Should gradient be a disambigutation page?

The current version of gradient is a bird of many feathers, part disambiguation page, and part the vector calculus gradient, which is a sorry mess.

There are two options now. Either

I would be for the second, as it appears to me that most pages linking here are about the vector gradient or related, and the "Photoshop gradient" needs forking to its own article. Oleg Alexandrov (talk) 17:17, 26 March 2006 (UTC)

I favor number 2. I think the main use of the word is the mathematical use, and other uses are just variations on a theme. -lethe talk + 17:25, 26 March 2006 (UTC)
Yes there is a need for Gradient (image processing), it not just photoshop a very common concept, worthy of more extensive treatment. Slope does a better job of providing a discription of the concept for non mathematicians and this page is better for mathematicians. In the light of this I favor the first so readers can find the right article for them quickly. --Salix alba (talk) 21:04, 26 March 2006 (UTC)
Naturally I favour the first, as a non mathematician searching for "gradient" to explain a road sign or description of a river gets hit by university level equations that are worse than useless. Nice to know that the graphics usage, which I thought a minor aside, is worthy of an article: that emphasises the need for the first port of call to provide disambiguation. The number of articles linking here was reduced by one when I noticed Khyber Pass which, as suspected, was referring to what some (but not all) in the US call a grade. Yosemite National Park and Inclined plane also seem to expect that usage, to name but two. Obviously there's effort involved in changing the mathematical links to gradient (calculus) or whatever, but it's worth it to make Wikipedia more accessible, and I'm willing to work on it. Note that gradient (disambiguation) currently redirects to gradient. ...dave souza, talk 22:54, 26 March 2006 (UTC)
Are you willing to fix the links to point in the appropriate places? If so, I would also agree with making this a disambig. Oleg Alexandrov (talk) 00:13, 27 March 2006 (UTC)
That's what I'm offering to do, though some help with it would be welcome. ..dave souza, talk 12:35, 27 March 2006 (UTC)

As far as I'm concerned, the revision to this page of 02:56, 4 April 2006 by Oleg Alexandrov, and splitting off image gradient for the graphics term, meets the need for disambiguation on this page. ..dave souza, talk 09:53, 4 April 2006 (UTC)

It is wrong for the term gradient to go to Gradient (calculus). The best example of the term gradient, that is nearly universal, are the winds: caused by the gradient, the potential, the "slope of the curve(line) between the H (High pressure) and the L (Low pressure). All the weatherPeople out there, will never stop referring to the Pressure "gradient" between the High and Low pressure centers.--- MMcAnnis,YumaAZMmcannis 21:43, 8 April 2006 (UTC)
I just got to this page by Climax community: ...."species distributed themselves along nutrient and other envirnmental gradients." I was going to wikify the term Gradient, instead I have to say all this stuff. (and Not wikify the term;Alas.) Gradient is not just a term for Calculus.--- MMcAnnis,YumaAZMmcannis 21:43, 8 April 2006 (UTC)
As I stated, i wanted to wikify Gradient, but was unable to for obvious reasons:
..."species distributed themselves along nutrient and other envirnmental gradients." ....i cannot wiki this over to a: gradient (mathematics); it makes no sense,...Alas. ---I am trying to state my case, without, anyone guessing what I am trying to say.--(See Climax community)--MMcAnnis////Mmcannis 02:25, 9 April 2006 (UTC)
But the mathematical usage is exactly the intended (only?) usage, but "along" should probably be changed to "according to".--MarSch 11:23, 11 April 2006 (UTC)

Nowhere in the talk page did I find a discussion as to the fact that the gradient of a scalar field at a point is equal normal to the field at that point. I believe this is a great use of (vector) gradient and ought to be incorporated. Shall I go ahead and do so? -unsigned

What do you mean by "normal to a scalar field"? All I know is normal to a surface. Oleg Alexandrov (talk) 03:23, 16 August 2006 (UTC)
I have no idea what he was talking about. But I thought of something similar that might be mentioned in the article. The gradient is perpendicular to the level set. --Spoon! 23:30, 31 August 2006 (UTC)

## The Gradient in other coordinate systems

I'm wondering if the expressions for the gradient in the other common 3D coordinate systems - such as spherical polar, and cylindrical coordinates - should be included in the article on the gradient? I think situations where these forms are convenient arise commonly enough that they should be at least mentioned and stated.

Added. This is basic reference material; no reason not to repeat it here. Whosasking 03:36, 9 April 2007 (UTC)

The gradient expressions in the alternate coordinate systems should include definitions of the basis vectors with respect to which the gradient is being written. For example, as it is listed now the gradient in cylindrical coordinates is given with respect to the vectors d/dρ, 1/ρ d/dθ, and d/dz. As a mathematician, it would have been more natural for me to express it in terms of d/dρ, d/dθ, and d/dz (the difference is the 1/ρ factor). I have in fact already had a student ask for clarification about this after reading this very article ... Holmansf (talk) 21:28, 11 March 2008 (UTC)

The same wonders me. Here, the gradient is defined as a vector, which components are subsequent partial derivatives. But is it also true for coordinate systems other than Cartesian? As far as I know, the components of the gradient in cylindrical and sperical coordinates are different, they're not simply the subsequent partial derivatives. So, can we say that? Is it general enough? SasQ —Preceding undated comment added 03:29, 19 January 2010 (UTC).

I see a lot of references in textbooks to gradients of vector quantities (which I don't understand, which is why I am here!). Could the gradient of a vector be defined and a couple of examples added?

Pagw 18:51, 22 May 2007 (UTC)

The gradient of a vector quantity is a second order tensor quantity. The gradient of a second order tensor is a 3rd order tensor. The gradient function raises the order. I agree, that this needs to be discussed within this page as well, it is also the reason why I am here. Does anyone know what the meaning of taking a vector field and dotting it with the gradient of that same vector field? —Preceding unsigned comment added by 67.39.207.253 (talk) 01:50, 12 December 2007 (UTC)

Well, seeing as the gradient of a vector field is orthogonal to that vector field, and the dot product of perpendicular vectors is zero, the result of this is zero. Hopefully that's what you mean by "meaning". siafu (talk) 01:42, 18 February 2008 (UTC)
Or, in math terms:
${\displaystyle {\overrightarrow {V}}\cdot \nabla {\overrightarrow {V}}=0}$
siafu (talk) 01:48, 18 February 2008 (UTC)

## No need for a disambiguation

I came across this page, and I agree with the previous comment. There is no need at all for a disambiguation for gradient.

Is this right? " Thus ${\displaystyle \nabla f}$ is a function from U to the space R " I think R should be Rn

## Total derivative

What do you think of my new article User:Saippuakauppias/Fréchet. Please post your comments to the articles talk site. --Saippuakauppias 17:13, 25 April 2008 (UTC)

## Chain Rule

Shouldn't the left hand side of the "more general" formula in the chain rule section be a gradient and not a Jacobian, as the function f takes a vector to a scalar (and also that the right hand side is a matrix times a vector)? If I'm wrong, then I believe this section needs some clarification.

18:50, 11 November 2008 (UTC) —Preceding unsigned comment added by 152.94.80.245 (talk)

I've got the same thought. If the gradient is the transposed Jacobian, it should be
${\displaystyle \nabla (f\circ g)(c)=(Dg(c))^{T}\nabla f(a)}$.
If the gradient is the Jacobian, it should be
${\displaystyle \nabla (f\circ g)(c)=\nabla f(a)Dg(c)}$. —Preceding unsigned comment added by 134.99.156.154 (talk) 09:05, 22 July 2010 (UTC)

## Gradient a vector or sum?

Dear people, At http://en.wikipedia.org/wiki/Gradient#Definition ${\displaystyle \nabla f}$ is a vector (of the derivatives in x,y and z coordinates). However, at http://en.wikipedia.org/wiki/Del#Gradient the Gradient is defined as the sum of these terms. Is the notation not consistent across the articles or am I missing something here? MrZap (talk) 10:08, 31 October 2009 (UTC)

It's not a mistake, but different notation. First notation specifies each component of the gradient vector, separated by a comma, and all that in parentheses [I saw round, rectangular ang angular parentheses]. The other notation specifies each component as a product of the component value [magnitude] [think of it as a coordinate, a number] and a unit vector in a direction of one of the coordinate axes [a versor]. All these component vectors, when ADDED vectorially, give the outcoming gradient vector. SasQ —Preceding undated comment added 03:21, 19 January 2010 (UTC).

## Gradient as a negative projection of the normal

I noticed one interesting thing: Suppose you have a "graph" of a function z = z(x,y), which is a surface over an x-y plane. Now, at every point of the surface you can nail a vector normal to the surface. Now observe the projection of this vector onto the x-y plane:
1. When the surface is flat [i.e. the z is constant], the normal will be pointing straight upward, and its projection will be a zero vector.
2. In places where the surface has positive slope, the normal will be leaning to the direction opposite to the direction of maximum increase - the more quicker the function increases, the more the normal vector leans in that opposite direction. And what with its projection? It will be a vector, which length depends on the maximum slope of the function in this point, and its direction will be opposite to the direction of the maximum increase. In other words, it'll point in the direction of fastest decrease [minimalization].
So, it's exactly opposite to the gradient vector!
Is it some well-known property or I'm the first one who noticed that? Because I searched over the Internet and haven't found any information about that. But if it's well-known, maybe it's worth to mention it here? SasQ

In 2-D it is well-known that the normal is perpendicular to the gradient. I don't know what your flat projection of the normal to a 3-D slope is called, but presumably the set of projections of unit normals completely determines the function z(x,y) up to a constant, i.e completely defines the shape of the surface (up to translation). I expect that there is a treatment of this somewhere, but I don't know where to find it. Dbfirs 12:00, 22 July 2010 (UTC)

## History

There is no discussion of history -- is this intentional? —Preceding unsigned comment added by 65.50.1.50 (talk) 18:50, 28 September 2010 (UTC)

What sort of history had you in mind? Can you find any history of gradient? As far as I know, the mathematics has been developed very slowly over many centuries, so there would not seem to be anything notable to mention, but if you can find some history, then I'm sure it could be included. Dbfirs 15:49, 5 October 2010 (UTC)

## "The example with the hill"

The third paragraph after the introduction asks the reader to recall "the example with the hill". This example must have gotten lost in the tubes somewhere... is anyone familiar enough with the article history to dive back in time and retrieve it? —Preceding unsigned comment added by 137.48.213.175 (talk) 13:27, 5 October 2010 (UTC)

Instead, I've removed the reference to a previous example, since the illustration stands on its own. Dbfirs 15:49, 5 October 2010 (UTC)

## “The gradient of a scalar field is a vector field” is wrong assumption

The gradient of a scalar field is a vector fieldhttp://en.wikipedia.org/wiki/Gradient is only assumption. The gradient of scalar function does not form the vector field. As well known from textbook Modern Geometry « at replacements of co-ordinates the function’s gradient will be transformed differently, than a vector»: hence the theory requiring such assumption must be false. Therefore classical representation in majority textbooks are wrong. Look please this university textbook:

or free book

Authors of this textbook – authoritative mathematicians:

I can present other and newer arguments on Web-pages. For example http://dxdy.ru/topic38647-30.html

Alexandr--94.27.93.132 (talk) 12:19, 29 December 2011 (UTC)

Definitions are not assumptions and cannot be wrong as such. Besides, Wikipedia follows the "classical representation in the majority of textbooks", whether right or wrong, so to speak - see wp:verifiability: "The threshold for inclusion in Wikipedia is verifiability, not truth—whether readers can check that material in Wikipedia has already been published by a reliable source, not whether editors think it is true." - DVdm (talk) 12:28, 29 December 2011 (UTC)

Dear DVdm! This problem already was discussed here http://en.wikipedia.org/wiki/User_talk:Dbfirs.“Dear Alexandr, :Wikipedia presents the majority viewpoint, but if there is disagreement amongst mathematicians, then this also should be mentioned.... Dbfirs 19:55, 27 December 2011 (UTC)”

From this citation follows that your arguments demand updating. Therefore let's try to formulate the text for editing of this article. Alexandr--188.163.97.5 (talk) 18:49, 20 January 2012 (UTC)

Yes, we do seem to have sufficient links to explain the controversy. Perhaps we could add a note to the disputed statement with links as above. Dbfirs 15:40, 21 January 2012 (UTC)
It's not really a controversy, but rather a matter of choosing an appropriate name in the context at hand. Anyway, I have added a sourced remark in the lead. - DVdm (talk) 16:35, 21 January 2012 (UTC)
Thanks. That's more than I was proposing, and should satisfy Alexandr's concerns. Dbfirs 16:53, 21 January 2012 (UTC)

Thanks!!! That's more than I hoped! However for full satisfaction of all readers it is necessary to add in your remark the citation from Dubrovin (page 15):

In the contexts of linear algebra and modern geometry the gradient is sometimes treated as a covector because «… under co-ordinate change the gradient of function transforms differently from a vector »[1]. Alexandr--94.27.94.176 (talk) 12:27, 22 January 2012 (UTC)

Let's keep the details out of the lead. This is explained later in the article. - DVdm (talk) 17:21, 22 January 2012 (UTC)

## Confusing too many technical terms

Somebody please put this article in layman terms or at least describe the technical terms without having to go >3 levels down.
--Jangirke (talk) 15:05, 3 December 2012 (UTC)

Could you please tell us which technical terms are causing a problem. A single click should suffice for explanations (though we cannot guarantee that absolutely everyone will understand the explanations at first reading). Dbfirs 18:24, 3 December 2012 (UTC)
I plan to add a diagram, showing how the gradient has the geometric interpretation of a vector normal to isosurfaces of a function, similar to those of linear functional. I know there are already good diagrams in the article, but one more in another perspective couldn't hurt... Maschen (talk) 11:44, 12 December 2012 (UTC)

## Gradients and critical points; example section improvements

I feel that the fact that critical points of a multivariable function can be found by setting the gradient of that function equal to the 0 vector or a vector of only 0 and DNE values. For example:

${\displaystyle \nabla f(x_{1},x_{2},\dots ,x_{n})=\langle {\frac {\partial f}{\partial x_{1}}}\mathbf {e} _{1},{\frac {\partial f}{\partial x_{2}}}\mathbf {e} _{2},\dots ,{\frac {\partial f}{\partial x_{n}}}\mathbf {e} _{n}\rangle ={\vec {0}}_{n}}$ or some vector of length n with only 0 and DNE values.

For example, the critical points of a two variable function that is differentiable at every point:

${\displaystyle \nabla f(x,y)}$ at${\displaystyle f(x_{0},y_{0})=\langle {\frac {\partial f(x_{0},y_{0})}{\partial x}}\mathbf {e} _{1},{\frac {\partial f(x_{0},y_{0})}{\partial y}}\mathbf {e} _{2}\rangle =\langle 0,0\rangle ={\vec {0}}_{2}={P}(x_{0},y_{0})}$ where ${\displaystyle P(x_{0},y_{0})}$ is the critical point at ${\displaystyle (x_{0},y_{0})}$

So it stands to say that when ${\displaystyle \nabla f(x_{1},x_{2},\dots ,x_{n})}$ at ${\displaystyle f(x_{0_{1}},x_{0_{2}},\dots ,x_{0_{n}})={\vec {0}}_{n}}$ then the values ${\displaystyle x_{0_{1}}}$ through ${\displaystyle x_{0_{n}}}$ make up a critical point of the function ${\displaystyle f(x_{1},x_{2},\dots ,x_{n})}$
I feel this is worth mentioning in the article, since this application of the gradient can give a geometric representation of multivariable optimization methods such as two-variable discriminants ${\displaystyle D=f_{xx}(x_{0},y_{0})f_{yy}(x_{0},y_{0})-f_{xy}(x_{0},y_{0})^{2}}$ as the geometric representation of that discriminant represented as a field of vectors around the <0, 0> gradient can help indicate if the critical point will be a local minimum, maximum, or saddle point.

Second, I feel like the examples in the first section past the article intro needs to be redone with examples that don't use a hatnoted page, specifically, Grade_(slope) in the hill examples. If no one is opposed to me providing different examples that don't use Grade_(slope), I may have some that won't have that possibly confusing factor. Many of the examples are also somewhat ineffective in illustrating applications of gradients, direction vectors, and gradients in partial differentials. I think it's important to see to it that this article distinguishes vector field representations of gradients from possibly mistakable concepts like Contour maps when using only vector field examples.
That said, I think a picture or explanation explaining that the gradient of a function is a line perpendicular to the lines of a contour map. This would be especially effective in some portion that distinguishes the differences between vector field representations and things like Slope fields and Contour maps. Since all these things are at least somewhat related, it is within reason to expect that there be an explanation of the association between these geometric representations of functions.

I may have time to edit these things in myself, but I'd like to get some sort of approval at least, first, since I am only in Calculus III as of now and have not gotten to some of the later components of gradients' geometric representations for functions outside of R^3 and R^4. Let me know if any of this sounds good. Penitence (talk) 17:02, 4 October 2013 (UTC)

## Directional derivative needs unit vector?

>Directional derivative needs unit vector? I'm sorry to waste your time if I'm wrong, but shouldn't "whose dot product with any vector v" be "whose dot product with any unit vector v"? Dave C (talk) 02:35, 23 October 2013 (UTC)dcromley

The article directional derivative defines the notion for any nonzero vector. To restrict the definition to unit vectors is a matter of choice and convention. D.Lazard (talk) 08:32, 23 October 2013 (UTC)

## Definition and Linear approximation

First of all, I'd like to point out I haven't learned about gradiens yet, so everything I say is according to my intuition and this page. So according to the definition, the gradian is defined to be the "unique vector field whose dot product with any vector v at each point x is the directional derivative of f along v". This sounds wrong since like limits of functions of several variables, gradien shouldn't (according to my intuition) be defined by directional derivatives, just like limits aren't defined by "directional limits" (if that's how they're called. I mean the value it approaches from every direction). Consider f(x,y)= sqrt(y^2+y) if y=x^2, 0 otherwise. According to the definition given above, the gradian of (0,0) should be (0,0), since the directional derivative is always 0 for any vector. But if that was true, then the linear approximation wouldn't be remotely true for any x and y for which y=x^2. Using the derivative analogy, this: ${\displaystyle \lim _{h\to 0}{\frac {\|f(x+h)-f(x)-\nabla f(x)\cdot h\|}{\|h\|}}=0}$ would no longer be true - since for h=(t,t^2) as t approaches zero, this limit is equal to 1, not 0.

Did I make a mistake, is my intuiotion wrong or is the definition wrong? — Preceding unsigned comment added by TheTranscendental (talkcontribs) 19:22, 29 March 2014 (UTC)

Your f is not differentiable. - DVdm (talk) 19:28, 29 March 2014 (UTC)

Ok, so this definition is true if f is differentiable, but one must first check if it is? Shouldn't the gradient not exist if f isn't differentiable? — Preceding unsigned comment added by TheTranscendental (talkcontribs) 20:06, 29 March 2014 (UTC)

## Where is the jar gone?

"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)

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)
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)
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)
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)

## Rewrite the section Gradient of a vector

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

${\displaystyle \nabla _{c}\mathbf {f} ^{b}=({\frac {\partial f^{\nu }}{\partial x^{\mu }}}+{\Gamma ^{\nu }}_{\mu \sigma }f^{\sigma })(\mathbf {e} ^{\mu })_{c}(\mathbf {e} _{\nu })^{b}.}$

Given the metric g, one obtains

${\displaystyle g^{ac}\nabla _{c}\mathbf {f} ^{b}=g^{\mu \rho }({\frac {\partial f^{\nu }}{\partial x^{\mu }}}+{\Gamma ^{\nu }}_{\mu \sigma }f^{\sigma })(\mathbf {e} _{\rho })^{a}(\mathbf {e} _{\nu })^{b}}$

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)

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)
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)

I think it is important to clarify whether this represents a covariant derivative, Frechet derivative, or Gateaux derivative. When I asked him, Maciej Zworski told me that the question "what is the gradient of a vector field" is ill-defined on account of the meaning of "gradient of a vector field". All I know is that yes, the gradient of a vector field is a tensor of type (0,2) as it's supposed to provide the "best linear approximation" of the vector field in question. As someone who has a difficult time understanding tensor math, I think it's also imperative that its expression in coordinates also be given, particularly the Euclidean case and the case of polar, cylindrical, and spherical coordinates. --Jasper Deng (talk) 07:03, 27 June 2015 (UTC)

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:

${\displaystyle \nabla f(x)={\frac {\partial f}{\partial q^{i}}}b^{i}={\frac {\partial f}{\partial q^{i}}}g^{ij}b_{j}}$

where ${\displaystyle f}$ is the function, ${\displaystyle x}$ represents the input into the function parameterized by the coordinate system, ${\displaystyle q^{i}}$ is the ith parameter of the coordinate system, ${\displaystyle b^{i}}$ is the ith basis vector in the contravariant (~inverse) coordinate system, ${\displaystyle b_{i}}$ is the ith basis vector in the (covariant) coordinate system, ${\displaystyle g^{ij}}$ 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: ${\displaystyle b_{1}={\vec {i}},b_{2}={\vec {j}},b_{3}={\vec {k}}}$

${\displaystyle g_{ij}=g^{ij}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}}$
So the gradient reduces to the familiar form of
${\displaystyle \nabla f(x)={\frac {\partial f}{\partial x^{m}}}{\frac {\partial x^{m}}{\partial b^{i}}}g^{ij}b_{i}={\frac {\partial f}{\partial x}}{\vec {i}}+{\frac {\partial f}{\partial y}}{\vec {j}}+{\frac {\partial f}{\partial z}}{\vec {k}}}$

Spherical:

${\displaystyle x=rsin(\theta )cos(\phi )}$
${\displaystyle y=rsin(\theta )sin(\phi )}$
${\displaystyle z=rcos(\theta )}$
${\displaystyle J={\begin{bmatrix}\cos \left(\phi \right)\sin \left(\theta \right)&r\,\cos \left(\phi \right)\cos \left(\theta \right)&r\,-\sin \left(\phi \right)\sin \left(\theta \right)\\\sin \left(\phi \right)\sin \left(\theta \right)&r\,\sin \left(\phi \right)\cos \left(\theta \right)&r\,\cos \left(\phi \right)\cos \left(\theta \right)\\\cos \left(\theta \right)&-r\,\sin \left(\theta \right)&0\end{bmatrix}}}$
${\displaystyle g_{ij}=J^{T}\,J={\begin{bmatrix}1&0&0\\0&r^{2}&0\\0&0&r^{2}\,\sin \left(\theta \right)^{2}\end{bmatrix}}}$
${\displaystyle g^{ij}={\begin{bmatrix}1&0&0\\0&r^{-2}&0\\0&0&r^{-2}\,\sin \left(\theta \right)^{-2}\end{bmatrix}}}$
${\displaystyle b_{1}=[\cos \left(\phi \right)\sin \left(\theta \right),\sin \left(\phi \right)\sin \left(\theta \right),\cos \left(\theta \right)]={\vec {r}}}$
${\displaystyle b_{2}=[r\,\cos \left(\phi \right)\cos \left(\theta \right),r\,\sin \left(\phi \right)\cos \left(\theta \right),-r\,\sin \left(\theta \right)]={\vec {\theta }}\,r}$
${\displaystyle b_{3}=[r\,-\sin \left(\phi \right)\sin \left(\theta \right),r\,\cos \left(\phi \right)\cos \left(\theta \right),0]={\vec {\phi }}\,r\,\sin \left(\theta \right)}$
${\displaystyle \nabla f(x)={\frac {\partial f}{\partial x^{m}}}{\frac {\partial x^{m}}{\partial b^{i}}}g^{ij}b_{i}={\frac {\partial f}{\partial r}}{\vec {r}}+{\frac {\partial f}{\partial \theta }}{\frac {\vec {\theta }}{r}}+{\frac {\partial f}{\partial \phi }}{\frac {\vec {\phi }}{r\sin \left(\theta \right)}}}$

Mouse7mouse9 22:53, 21 January 2015 (UTC) — Preceding unsigned comment added by Mouse7mouse9 (talkcontribs)

## Product Rule

The description for the product rule was a bit misleading.

If f and g are real-valued functions differentiable at a point aRn, then the product rule asserts that the product (fg)(x) = f(x)g(x) of the functions f and g is differentiable at a, and
${\displaystyle \nabla (fg)(a)=f(a)\nabla g(a)+g(a)\nabla f(a).}$

Initially, I assumed it had something to do with the derivative of ${\displaystyle \nabla (fg)}$. The little segment "(fg)(x) = f(x)g(x)" 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 aRn, then the product rule asserts that the product fg is differentiable at a, and
${\displaystyle \nabla (fg)(a)=f(a)\nabla g(a)+g(a)\nabla f(a).}$

Muntoo (talk) 01:03, 29 July 2015 (UTC)

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. 11:40, 29 July 2015 (UTC)