Talk:Padé approximant

where's epsilon?

The epsilon algorithm, the page refers to for getting the coefficients is missing. A search leads to nothing. I'm not from that field and thus cannot contribute the missing page. 84.226.41.90 17:52, 21 January 2007 (UTC)[reply]

Riemann-Padé Zeta function

(EDIT)

To study the resummation of divergent series, it can be useful to introduce the Padé o simply Rational zeta function as:

$\sum _{n=1}^{\infty }{\frac {[N/M]_{f}(n)}{n^{s}}}=\zeta _{Q}(s)$

so the zeta regularization value at s=0 it is equal to the 'sum' S, of the divergent series:

$S=\sum _{n=1}^{\infty }f(n)$

the functional equation for this Pade zeta as:

$\sum _{j=0}^{M}p_{j}\zeta _{Q}(s-j)=\sum _{j=0}^{N}q_{j}\zeta _{0}(s-j)$

here '0' means that the Pade is of order [0/0] and hence, we got the Riemann zeta function. and $Q=Q(n)=[N/M]_{f}(n)$

Sorry.. for my edition, perhaps now it looks clearer from the coefficients of the upper and lower Polynomials involving Padè approximant (If possible put it back) i thought i saw it into a book about Dirichlet series but can not remember its name.

The above section appeared in the article. It looks to me as it is valid, but unfortunately I have no idea what's meant. In the first formula, the numerator seems to be independent of n; and what is Q? In the last formula, the a_j and b_j is not defined. -- Jitse Niesen (talk) 13:28, 23 October 2007 (UTC)[reply]

Okay, it looks clearer now; thanks. I changed it a bit, hopefully without introducing any errors. -- Jitse Niesen (talk) 11:45, 26 October 2007 (UTC)[reply]

Introductory text

Consider the introductory text:

Padé approximant is the "best" approximation of a function by a rational function of given order - under this technique, the approximant's power series agrees with the power series of the function it is approximating.[1] The technique was developed by Henri Padé.

The external link "[1]" to "http://www.dattalo.com/technical/theory/sinewave.html" is inappropriate in a lead-in text. I have moved it into the references. --Михал Орела 14:26, 11 November 2010 (UTC) —Preceding unsigned comment added by MihalOrela (talk • contribs)

Slight correction: I have moved the said link into the External References section. --Михал Орела 14:38, 11 November 2010 (UTC) —Preceding unsigned comment added by MihalOrela (talk • contribs)

Definition

I have inserted a slightly more formal definition:
Given a function f and two integers m ≥ 0 and n ≥ 0, the Padé approximant of order [M/N] is the rational function

$R(x)={\frac {\sum _{m=0}^{M}p_{m}x^{m}}{1+\sum _{n=1}^{N}q_{n}x^{n}}}={\frac {p_{0}+p_{1}x+p_{2}x^{2}+\cdots +p_{M}x^{M}}{1+q_{1}x+q_{2}x^{2}+\cdots +q_{N}x^{N}}}$

--Михал Орела 14:04, 12 November 2010 (UTC) —Preceding unsigned comment added by MihalOrela (talk • contribs)

Now it is time to harmonize the notation used in this article with that of Padé table. Here in this article there are p's and q's. There, the more usual a's and b's are used. Since the Padé table article is already very well developed, I propose that harmonization of notation be undertaken here. Specifically,

$R(x)={\frac {\sum _{m=0}^{M}p_{m}x^{m}}{1+\sum _{n=1}^{N}q_{n}x^{n}}}={\frac {p_{0}+p_{1}x+p_{2}x^{2}+\cdots +p_{M}x^{M}}{1+q_{1}x+q_{2}x^{2}+\cdots +q_{N}x^{N}}}$

becomes

$R(x)={\frac {\sum _{m=0}^{M}a_{m}x^{m}}{1+\sum _{n=1}^{N}b_{n}x^{n}}}={\frac {a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{M}x^{M}}{1+b_{1}x+b_{2}x^{2}+\cdots +b_{N}x^{N}}}$

--Михал Орела 14:59, 12 November 2010 (UTC) —Preceding unsigned comment added by MihalOrela (talk • contribs)

Just fixed a bad editing error. --Михал Орела 15:06, 12 November 2010 (UTC)

References

In checking the Press et al. reference

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. Numerical recipes in C. Section 5.12., Cambridge University Press.

I noted that the given pdf link [2] no longer worked. In particular, I am not sure that the book is "freely" available electronically according to the usual copyright meaning. --Михал Орела 14:39, 12 November 2010 (UTC) —Preceding unsigned comment added by MihalOrela (talk • contribs)

I updated the Numerical Recipes reference to the current edition and fixed the link.

ServiceAT (talk) 01:03, 11 July 2011 (UTC)[reply]

method

In the absence of epsilon algorithm, a presentation of at least one way to generate a P.a. would help. If I understand correctly: given Taylor series coefficients $c_{i}$ , we have

$(\sum _{i=0}^{M+N}{c_{i}x^{i}})(\sum _{j=1}^{M}{d_{j}x^{j}})=\sum _{k=0}^{N}{n_{k}x^{k}}$

(with $d_{0}=1$ by convention), whence

$c_{0}=n_{0}$
$c_{1}=n_{1}-c_{0}d_{1}$
...
$c_{N}=n_{N}-c_{N-1}d_{1}-\ldots -c_{0}d_{N}$
$c_{N+1}=-c_{N}d_{1}-\ldots -c_{0}d_{N+1}$
...
$c_{N+M}=-c_{N+M-1}d_{1}-\ldots -c_{N}d_{M}$

which can be expressed with a matrix that I don't quite know how to format neatly ... but anyway, is that accurate? (Is it reasonable to assume that M ≥ N ?) —Tamfang (talk) 18:54, 12 November 2010 (UTC)[reply]

Hi,
the classical way to compute a Pade approximant is via the extended euclidean algorithm. The relation $R(x)=P(x)/Q(x)=f(x)\mod x^{M+N+1}$ is equivalent to the existence of some K(x) such that

P(x)=Q(x)f(x)+K(x)x^{M+N+1}

,

which can be interpreted as one step in the computation of the xgcd. Since the xgcd page is horrible: to compute the gcd of two polynomials a and b, one computes via long division the remainder sequence

r_{0}=a,\;r_{1}=b,\quad r_{k-1}=q_{k}r_{k}+r_{k+1}

with

\deg r_{k+1}<\deg r_{k}\,

,

and for the Bezout identities

u_{0}=1,\;v_{0}=0,\quad u_{1}=0,\;v_{1}=1,\quad u_{k+1}=u_{k}q_{k}-u_{k-1},\;v_{k+1}=v_{k}q_{k}-v_{k-1}

to obtain in each step the identity

r_{k}(x)=u_{k}(x)a(x)+v_{k}(x)b(x)

.

For the order (M/N) approximant, one thus carries out the extended euclidean algorithm for $a=f,\;b=x^{M+N+1}$ where f is replaced with its Taylor series up to degree M+N, until the step where $u_{k}$ has degree N. Then $P=r_{k},\;Q=u_{k}$ gives the Pade approximant. If one computes the full table of the xgcd, then one has obtained a(n anti-)diagonal of the Pade table. This method is mentioned in Bini/Pan: Polynomial and Matrix computations, but there should also be better sources.--LutzL (talk) 13:50, 13 November 2010 (UTC)[reply]

Thank you, that's very interesting! (Now I need to find or write a polynomial division function.) —Tamfang (talk) 21:47, 13 November 2010 (UTC)[reply]

Ah, there it is. —Tamfang (talk) 21:55, 13 November 2010 (UTC)[reply]

It appears to me that

q_{1}=0,\,r_{2}=f=r_{0}

... is that intentional? —Tamfang (talk) 17:41, 14 November 2010 (UTC)[reply]

Yes and no. The remark is correct and it was not intended, I just didn't have time to change it. Better start with descending degree,

r_{0}=a=x^{M+N+1},\;r_{1}=b=f

, but notice that with this swap now the denominator Q is taken from the v cofactor sequence.--LutzL (talk) 13:35, 15 November 2010 (UTC)[reply]

And of course you can omit the computation of the u cofactors since they are not used anywhere intermediately or in the result.--LutzL (talk) 13:40, 15 November 2010 (UTC)[reply]

Can anyone help to revert an edit

Found that the example for exp(x) was edited by some IP user in March, and becomes mathematically incorrect.

https://en.wikipedia.org/w/index.php?title=Pad%C3%A9_approximant&type=revision&diff=769403635&oldid=769402974

However it could not be simply "undo" as there is conflicts. Can anyone help to revert the mal-edit?

Billyauhk (talk) 04:30, 18 November 2017 (UTC)[reply]

Is the exp(x) example wrong?

The example for exp(x) here: https://en.wikipedia.org/wiki/Padé_approximant#Examples does not match the example in the Padé table here: https://en.wikipedia.org/wiki/Padé_table#An_example_–_the_exponential_function.

I believe the Padé table is correct and this page is wrong. I haven't edited a page before and wanted to get confirmation from someone else that there is actually an error first. — Preceding unsigned comment added by Aerojunkie (talk • contribs) 16:49, 15 March 2018 (UTC)[reply]

I have already noted the incorrect exp(x) Pade approximation above and identified the commit which make it wrong. Pls check my last section. Billyauhk (talk) 08:56, 3 September 2018 (UTC)[reply]

Simple example MatLab code

function [A,B] = paday(c,p,q)
% c is the Taylor Approximation, col vector
% p the number of terms in the numerator A, and q the number of terms in the denominator B
Temp = sparse(p+q+1, q+1);
for iq = 1:q+1
    temp(iq:length(c), iq)=c(1:length(c)-iq+1);
end
B = -Temp(p+2:p+q+1,2:q+1)\c(p+2:p+q+1);
A = Temp(1:p+1,iq+1)*(1;B)

QuentinUK (talk) 13:50, 9 December 2018 (UTC)[reply]

Pade approximation of erf(x) example is really bad

I just coded up that approximation. Return values are worse than single precision floating point for abs(x) > one sigma = sqrt(2)/2 and are completely unacceptable for abs(x) > two sigma.

In addition to some citations, these examples need error bounds or they should be removed. It is really morally wrong to waste people's time with bad algorithms. — Preceding unsigned comment added by 154.20.183.43 (talk) 22:08, 13 February 2019 (UTC)[reply]

Examples switch polynomial conventions

Many examples have highest order terms (say) first in the numerator, but bottom in the denominator. There is no consistency.

63.145.59.55 (talk) 17:17, 7 December 2020 (UTC)[reply]

Uniqueness

It says that the Padé approximant $[p/q]_{f}$ is unique, but from the definition it seems that for $f(x)=0$ both ${\frac {0}{1+x}}$ and ${\frac {0}{1+2x}}$ would go as $[1/1]_{f}$ . How to properly define uniqueness then? adamant.pwn — contrib/talk 11:44, 6 April 2022 (UTC)[reply]

On the other hand, I don't see any way for e. g.

[1/3]_{f}

to exist for

f(x)=x^{2}

. The way it's defined in the article,

[1/3]_{f}

will start from either

a_{0}

or

a_{1}x

, which can't match

x^{2}

in first

4

terms. adamant.pwn — contrib/talk 19:07, 6 April 2022 (UTC)[reply]

I came across the same problem. The point is that the Padé approximation does not always exist! This should be mentioned more explicitly in the article. Benji104 (talk) 17:11, 31 August 2023 (UTC)[reply]

Computation notation seems wrong

the c_k = f^(k)(0)… equation is a bit misleading since it makes it seem like the exponential function when really it seems it’s trying to denote the derivative number. I don’t know what the proper notation would be to denote that though and I haven’t found anything better. If this is a standard way of representing this then I see no issue but otherwise it could use a note addressing it. 2601:147:4701:EBC0:B113:AECF:59F6:7212 (talk) 11:15, 10 September 2024 (UTC)[reply]

This is probably the most common notation, see Derivative § Notation. An alternative would be to switch to D notation, replacing ⁠