Talk:Laplace's method

message: The fr: link for this page points to the French article for 'Stationary Phase' -- not for 'Steepest Descent' -- I'm not sure how to fix this though as I don't know enough French to find the right article in French wikipedia. Help?
Zero sharp 23:21, 26 November 2006 (UTC)
thread_title: fr link points to wrong article

message: A question involving perhaps divergent series, let be:
 \int_{a}^{b} dx e^{Mf(x)}= g(M)(1+M^{-1}a1+a2M^{-2}+............)  
Laplace in admitted in general that this series in 1/M would be divergent if M isn't big but could be summable (in Borel or other sense??)? --85.85.100.144 22:12, 16 February 2007 (UTC)
replies: text: In general the series is an asymptotic series, and, as such, is often divergent, but sometimes summable. The theory for correctly dealing with such sums has a number of intricacies, but, for the intended use as an approximation, these are usually quite safe and accurate, especially when the sum is truncated at the smallest term.  i.e. once the terms start growing again, stop summing! linas (talk) 03:32, 31 January 2010 (UTC)
thread_title:  Asymptotic series 

message: It seems to me that example 2 is just a trick to promote the reference to Azevedo-Filho and Shachter. It is not a relevant example for an introductory page on the Laplace method. Both the reference and example 2 should be removed.  — Preceding unsigned comment added by Superhiggs (talk • contribs) 15:53, 24 September 2012 (UTC)
thread_title:  Self-promotional links 

message: Am i wrong or should the assumption be: f(x) is a twice differentiable function, where the second derivative is continuous?
Because the proof uses continuity of the second derivative, which is not given for many functions fulfilling the assumptions...  — Preceding unsigned comment added by 134.76.62.65 (talk) 17:28, 30 January 2013 (UTC)
thread_title:  Proof 

message: I saw this comment about the proof, which I am not knowledgeable enough to evaluate: "The biggest error is their usage of the limit; it's not even clear that this limit exists. One must either argue using limsups and liminfs or first somehow establish that the limit exists before proceeding with showing it's 1." Andrew Keenan Richardson (talk!) 17:09, 31 May 2013 (UTC)
thread_title:  Error in proof? 

message: Laplace's method basically says that a function which looks like a Gaussian has a similar integral as one.
I admit it's slightly more complicated than that: increasing the "M" does make it resemble a Gaussian more and more because function maximums usually resembles the -x^2 function more as you zoom in on them. And I admit there is a lot of bonus content at the bottom.
But how does this article make it all sound like scholarly science only high level mathematicians should understand? I'm betting that some portion of readers give up after not understanding "R = O\left((x-x_0)^3\right)." even though that has little to do Laplace's method. "What the heck is O?" is a certain thought to occur. Also, "Therefore, the function ƒ(x) may be approximated to quadratic order" is sure to add a tinge of confusion to some readers, although on its own that's endurable.
I still don't understand what
Importantly, the accuracy of the approximation can depend on the variable of integration, that is, what goes into h(x) and what stays in g(x).
meant.
The article seems written for someone who already knows everything about Laplace's method 135.0.167.2 (talk) 04:53, 27 November 2013 (UTC)
thread_title:  Why is it so hard to understand 

message: I have a question about the documents shown in this Laplace's method and the method of steepest descent. 
Both of these two documents use the n-dimensional vector " x " as " dx ".
My question is that: 
Does " dx " denote " \| dx \| " or " dx_1\wedge dx_2 \cdots \wedge dx_d "?
It looks like that you use " dx_1\wedge dx_2 \cdots \wedge dx_d " as your definition.
Ychsue (talk) 09:24, 14 May 2014 (UTC)
thread_title:  The question about the integration for d-dimensional case shown in section *Other formulations* 

message: I would just like to point out that the assumptions in Section 2 are fairly restrictive. A situation that appears often in practice, arguably as often as the one presented, is that f(x) attains its maximum at one of the endpoints of the interval, with a nonvanishing derivative. Then we will get another asymptotic equivalent, with a 1/M factor instead of 1/M^{1/2}. And for the most general situation, I remember it is treated in a good book (possibly in French) by Dieudonné.  — Preceding unsigned comment added by 64.18.85.10 (talk) 16:24, 24 July 2018 (UTC)
thread_title:  The "general theory" presented in Section 2 is actually not general 

message: I find the structure of this article very confusing. I'd expect it to define the subject (Laplace's method) or clearly have a section that does so. But it doesn't. To me, the method belongs in the summary and in a section that might lead the reader to believe that the section defines Laplace's method. 018 (talk) 02:27, 4 January 2019 (UTC)
thread_title:  definition 

message: In the "Other formulations" section the article reads, 
::By the way, although \mathbf{x} denotes a d-dimensional vector, the term  d\mathbf{x} denotes an infinitesimal volume here, i.e. d\mathbf{x} := dx_1dx_2\cdots dx_d.
Why is this important? also does "d\mathbf{x} := dx_1dx_2\cdots dx_d"? I know we write it that way, but this seems to imply that the product itself has some meaning. I think it's more like
:
\begin{align}
 & \int e^{M f(\mathbf{x})}\, d\mathbf{x} = \int dx_1 \int dx_2 \cdots \int dx_d e^{M f(x_1, x_2, \dots, x_d)}
\end{align}
but I don't think anyone who understood what a Hessian was would have this question. 018 (talk) 02:37, 4 January 2019 (UTC)
thread_title:  Confusing part in "Other formulations"