Talk:Gaussian function

message: a,b and c are described in the intro paragraph, but d is not. Isn't d just the offset from 0?
The d parameter should not be part of this expression. Leaving it in there would mean that two thirds of the integrals below it are wrong. Its also not common, at least in physics, to have it there. http://hyperphysics.phy-astr.gsu.edu/hbase/math/gaufcn.html http://mathworld.wolfram.com/GaussianFunction.html 81.147.122.82 (talk) 09:00, 1 December 2014 (UTC)
thread_title:  d parameter 

message: Do we mean to say that 
:gaussian functions are eigenfunctions of the Fourier transform, 
or that 
:eigenfunctions of the Fourier transform are gaussian functions
or neither? -- Miguel
Not all eigenfunctions of the Fourier transform are Gaussian.  See Hermite polynomials. Michael Hardy 15:10, 30 Aug 2003 (UTC)
thread_title:  

message: Could someone add something about Gaussian functions being the ones with maximum entropy? I think this can also be related to the Heisenberg uncertainty principle since momentum and position are canonical conjugate variables.
:This article links to normal distribution, which I suspect already gives that information.  For non-normalized Gaussian functions, I'm not sure at this moment what the maximum-entropy statement would say. Michael Hardy 23:46, 27 Feb 2005 (UTC)
thread_title:  maximum entropy 

message: The function definition uses a,b,c as parameters, while the graph uses \mu,\sigma as parameters. What is the relation between the two sets of parameters?
--NeilenMarais 20:18, 24 May 2006 (UTC)
replies: text: Answering myself, it seems from looking at Gaussian function that a=\frac{1}{\sigma\sqrt{2\pi}}, b=\mu and c=\sqrt{2}\sigma. One could mention this, or perhaps even better, generate an image using the correct parameters. Opinions? --NeilenMarais 20:27, 24 May 2006 (UTC)
thread_title:  Definition of <math>\mu, \sigma</math> 

message: Would I be correct if i said that a gaussian function as such represents the values a variable can have .............that is to say ......it gives us a range of possible values of the variable , or it shows the region were the value of that variable lies ......
Is that what the Gaussian function does ... —The preceding unsigned comment was added by Hari krishnan07 (talk • contribs).
: Not directly. The gaussian function is the name for a function with specific properties e.g. as illustrated in the curves in the article. What you refer to is a probability distribution and can have the form of a gaussian Kghose 16:01, 16 December 2006 (UTC)
replies: text: I think not. This is related to Gaussian functions of course, but a true Gaussian function should not have the x^{2m} term in front. Oleg Alexandrov (talk) 18:41, 27 January 2007 (UTC)
thread_title:  Gaussian Function .. 

message: :Gaussian functions arise by applying the exponential function to a general quadratic function. The Gaussian functions are thus those functions whose algorithm is a quadratic function.
Did the author mean "logarithm" instead of "algorithm" here?
147.8.235.63 (talk) 10:18, 1 December 2008 (UTC)
thread_title:  Algorithm?  A misspelling? 

message: I'm confused... I see that the FFT of a Gaussian is Gaussian, but in a discrete implementation using scipy's fft to transform Gaussian functions, I get that σ -> N*5/(32σ) where N is the number of bins. This 5/32 seems like a weird magic number. What am I missing? —Ben FrantzDale (talk) 16:28, 2 April 2010 (UTC)
thread_title:  FFT? 

message: So, no mention of the fact that a linear-logarithmic Gaussian is a linear-linear parabola. I would imagine this would be considered a feature of note, but I don't know whether it is or not, or how to say that in a way that gives some substance to the fact. ᛭ LokiClock (talk) 06:21, 8 July 2010 (UTC)
thread_title:  No mention of parabola 

message: Because not everybody has access to Matlab due to its costs, I would like to see Python/Numpy code here instead. It is very similar, but nevertheless differs in some details. I could do the conversion myself, if people agree.
--maye (talk) 10:12, 27 October 2010 (UTC)
replies: text: I agree. There has been an attempt in the past to write this into policy, but the discussion was sidetracked and then ceased. The policies on media content do not merely disallow content they can't legally republish, but disallows content under non-commercial licenses, because anyone should be able to use all of the encyclopedia freely. At least in this case, you would be acting towards that end, and I see no reason why MATLAB has particular significance in this article. ᛭ LokiClock (talk) 14:10, 27 October 2010 (UTC)
replies: text: Most Matlab code can be run using Octave, which is free and open-source. Alhead (talk) 21:00, 7 December 2011 (UTC)
thread_title:  Matlab code better in Python? 

message: The article states:
:"Gaussian functions arise by applying the exponential function to a general quadratic function."
But how does applying the one to the other lead to that form for the quadratic? This should probably go in a History section. ᛭ LokiClock (talk) 00:47, 26 March 2011 (UTC)
thread_title:  Exponent's quadratic's form 

message: I guess they mean exactly the same thing, and normal distribution is the more formal name from the mathematical point of view. (Unsigned post.)
::Was suggestion to merge with Normal distribution.
:The Gaussian function does also have numerous applications outside the field of statistics, for example regarding solution of diffusion equations and Hermite functions and regarding feature detection in computer vision. If this article would be merged under normal distributions, these connections would be lost. Hence, I think that it is more appropriate to keep this article with appropriate cross referencing. Tpl (talk) 11:53, 8 June 2011 (UTC)
replies: text: Merge templates go on the articles not on the discussion page. New stuff on discussion page goes at the end. Have removed merge template as merge would not be good. Melcombe (talk) 14:44, 8 June 2011 (UTC)
thread_title:  Merging 

message: Can someone verify that the equation for 2-dimensional Gaussian elliptical is correct as currently stated:
:f(x,y) = A e^{- \left(a(x - x_o)^2 + 2b(x-x_o)(y-y_o) + c(y-y_o)^2 \right)}
Or should the second term in the exponent actually be negative like so:
:f(x,y) = A e^{- \left(a(x - x_o)^2 - 2b(x-x_o)(y-y_o) + c(y-y_o)^2 \right)}
I'm drawing from this page (wikipedia.org) and other sources for the bivariate gaussian pdf.  — Preceding unsigned comment added by 129.123.61.172 (talk) 21:04, 28 September 2011 (UTC)
replies: text: Well, it seems the only difference is the change of sign of the parameter b. Then it is simply the matter of definition. Bakken (talk) 10:52, 1 October 2011 (UTC)
thread_title:  Typo? 

message: An undefined variable called B keeps showing up inside the "Multivariate gaussian" paragraph. I have no idea if it is simply: B = A, or if there is something subtle going on here, but either we need to add a definition for B, or just remove it from the math.  — Preceding unsigned comment added by 2001:620:600:6000:ECDF:3A24:4CBA:7595 (talk) 11:47, 15 April 2013 (UTC)
replies: text: I believe you are right, I just let B = A in one place. But latter, there is a  B  that is defined as  B = A + A'  and I didn't yet figure out what it means. Bdmy (talk) 12:23, 15 April 2013 (UTC)

text: What's worse, I believe the definition of the Gaussian including the shift vector is wrong, or at least not explained enough. The definition used in the multivariate normal is something along the lines of n*e^{-(x-s)^T A(x-s)}, where n is the normalisation factor, A is the inverse of the covariance matrix and s is a shift (the mean for multivariate normal distributions). In the article it says e^{-x^T A x + s^T x}. According to my calculations, experiments and intuition those are not equivalent. For instance, with larger s, the value of the peak becomes larger, while this does not happen with the first definition. Also, if it's only shifted, the integral should not change. So imo, whatever it is, it is not a shifted gaussian. --128.130.118.54 (talk) 10:41, 24 October 2019 (UTC)
thread_title:  Multivariate gaussian 

message: Mathematically the same concept, except that the other page presents a two-argument version (i.e. where  is not a constant) and describes a single use case in more detail. QVVERTYVS (hm?) 19:51, 18 April 2014 (UTC)
thread_title:  Proposed merge with [[Radial basis function kernel]] 

message: RBF, while related to gaussians, is most identifiable as a type of kernel. It makes more sense to keep this separate since gaussians are a huge topic and someone is more likely to find this page while reading about different kernel types than applications of gaussian functions.  — Preceding unsigned comment added by 2602:306:371B:9920:4D9:8168:7CF6:D618 (talk) 00:30, 20 May 2014 (UTC)
thread_title:  Arguments Against 

message: In the definition of a Gaussian (as given in the lead section of the article):
:f(x) = a \exp{\left(- { \frac{(x-b)^2 }{ 2 c^2} } \right)}+d
there is a 2 in the denominator of the argument to the \exp function. Why is that 2 there? This is not explained in the article. The function would still be a Gaussian if the 2 would not be there, and it would have a simpler expression. I thought the 2 had something to do with that the integral from -\infty to +\infty would have a simpler expression, since a normal distribution function always should have the area 1, indicating that the integral of a Gaussian may be an interesting property. However, looking further down in the article, it turns out that the integral is
:\int_{-\infty}^\infty a e^{- { (x-b)^2 \over 2 c^2 } }\,dx=ac\cdot\sqrt{2\pi}.
so the 2 didn't simplify the expression of this integral either, just complicate it, since without the 2 in the denominator on the left hand side, there is no 2 in the square root on the right hand side either. So what does the 2 do there? It doesn't really feel like it belongs in the expression. —Kri (talk) 21:36, 12 October 2014 (UTC)
replies: text: No one knows?? WTF! I'm starting to think that it is someone who has just pulled a prank by putting that 2 there... —Kri (talk) 21:17, 22 October 2014 (UTC)
replies: text: Now I know why the 2 is there. It comes from the expression of the normal distribution, in which , or  which it is called there, is the standard deviation, so without the 2,  wouldn't equal to the standard deviation (although I saw now that the article already mentions that  is the standard deviation). So, mystery solved. —Kri (talk) 12:29, 30 October 2014 (UTC)
thread_title:  Why the 2 in the denominator? 

message: At the peak, when x=b, everything in the large parenthesis should be 0 ( x-b=0, so (x-b)^2=0, so -((x-b)^2)/(2c^2)=0 ).
This should result in: a EXP 0 = 1.0, regardless of the value of a.  But the result should be a, the peak value, by definition of a.
So what part of this am I not understanding?  The meaning of EXP?  Thanks.  — Preceding unsigned comment added by 67.249.?.? (talk) 22:24, 4 February 2015 (UTC)
thread_title:  Does the opening formula actually work properly?  

message: Can anyone provide the source of the equations listed under Meaning of parameters for the general equation, i.e. coefficients a, b and c? I have calculated them on my own and a couple of signs turned out different. I wonder if there's a textbook that contains the full derivation or the outcome. 
KenyaSong (talk) 14:45, 20 July 2018 (UTC)
thread_title:  Source for Meaning of parameters for the general equation 

message: Since the lower-case "gaussian" is pretty common now, I think adding this to the opening sentence is wise.
Spope3 (talk) 06:00, 30 March 2022 (UTC)
thread_title:  Should mention be made that "gaussian" is often not capitalized? 

message: Why is this called gaussian, since it already appeared in the works of de Moivre ? 
I think it would be useful to add some history part saying that this was not invented by Gauss. 2001:861:3008:37D0:34F7:B9ED:A9EB:BEF3 (talk) 20:30, 24 April 2023 (UTC)
thread_title:  History