Boundary conditions for Euler Maclaurin summation formula
John Washburn
value of the definite integral. Here is a link to the article on the
Euler-Maclaurin Summation formula on Mathworld,
http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html
The phrase of interest to me is: In certain cases, the last term tends
to 0 as n tends to infinity.
What are these cases?
I tracked down several of the references in the bibliography of the
above article as well as other papers and books. All present the same
basic form for the link between the sum and the integral.
Sum = integral + Bernoulli-Number Correction terms + integral
remainder term.
Where there are k correction terms and the remainder is given as an
integral of the 2k th derivative and the Bernoulli polynomial:
B_2k(x)/(2*k)!.
Knopp states that almost no functions allow you to ignore the integral
remainder term. Abramowitz and Stegun start with the infinite form.
This form has an infinite number of correction terms and no remainder
term. A & S give the impression that you can cut off the
Bernoulli-number correction terms anywhere it is convenient. The book
Mathematical Methods in Physics by Mathews and Walker uses operator
notation and thus, jumps directly to the infinite form and avoids the
integral remainder term altogether.
The several articles and books I have found are evenly split between
including and ignoring the integral remainder term. But, none of the
references I have describe when and how you can ignore the integral
remainder term.
My question is a clarification of these many papers and references.
What properties of f(x) and its derivatives allow you to pass to the
infinite form and ignore the integral remainder term? Is smoothness
sufficient? Is simple boundedness sufficient, given B_2k(x)/(2k!)
tends to 0 as k tends to infinity?
Thanks for your help on this matter.
John Washburn
Peter Spellucci
In article <cirqc6$48k$1...@dizzy.math.ohio-state.edu>,
John Washburn <jo...@WashburnResearch.org> writes:
>The Euler MacLaurin summation formula links the values of a sum to the
>value of the definite integral. Here is a link to the article on the
>Euler-Maclaurin Summation formula on Mathworld,
>http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html
>
>The phrase of interest to me is: In certain cases, the last term tends
>to 0 as n tends to infinity.
>
>What are these cases?
>
>I tracked down several of the references in the bibliography of the
>above article as well as other papers and books. All present the same
>basic form for the link between the sum and the integral.
>
>Sum = integral + Bernoulli-Number Correction terms + integral
>remainder term.
>
>Where there are k correction terms and the remainder is given as an
>integral of the 2k th derivative and the Bernoulli polynomial:
>B_2k(x)/(2*k)!.
you mean
R_k = \int_0^n (B_{2k}-\hat B_{2k}(t))/((2k!))f^{(2k)}(t) dt ?
= f^({2k})(\xi)n B_{2k}/((2k!))
B_{2k} the Bernoulli numbers with
| B_{2k}/(2k!) | <= 4/((2\pi)^{2k})
Hence you need a function whose supremum of |f^{2k}| on [0,n]
goes to zero as k to infinity for n fixed.
(if I not mistaken, the entire functions on |C do that?)
hth
peter
Peter Spellucci
this was a little bit too strong:
since we have the factor 1/(2*pi)^2k it suffices that the derivatives
are bounded for all x or grow slower than this factor.
but these are not all entire functions, no, this is false,
for example sin(7x) doesn't the job.
sorry for answering too fast
peter
John Washburn
>
>
>In article <cirqc6$48k$1...@dizzy.math.ohio-state.edu>,
> John Washburn <jo...@WashburnResearch.org> writes:
> >The Euler MacLaurin summation formula links the values of a sum to
the
> >value of the definite integral. Here is a link to the article on
the
> >Euler-Maclaurin Summation formula on Mathworld,
href="http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html">http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html</a>
> >
> >The phrase of interest to me is: In certain cases, the last term
tends
> >to 0 as n tends to infinity.
> >
> >What are these cases?
> >
> >I tracked down several of the references in the bibliography of the
> >above article as well as other papers and books. All present the
same
> >basic form for the link between the sum and the integral.
> >
> >Sum = integral + Bernoulli-Number Correction terms + integral
> >remainder term.
> >
> >Where there are k correction terms and the remainder is given as an
> >integral of the 2k th derivative and the Bernoulli polynomial:
> >B_2k(x)/(2*k)!.
>you mean
> R_k = \int_0^n (B_{2k}-\hat B_{2k}(t))/((2k!))f^{(2k)}(t) dt ?
> = f^({2k})(\xi)n B_{2k}/((2k!))
>B_{2k} the Bernoulli numbers with
> | B_{2k}/(2k!) | <= 4/((2\pi)^{2k})
Just t be clear: Is the bound here on the Bernoulli number,
B_{2k}/((2k!)), the Bernoulli polynomial, B_{2k}(t)/((2k!)) or both,
because |B_{2k}(t)/((2k!))| <= |B_{2k}/((2k!))|?
Peter Spellucci
In article <cisi4g$bti$1...@news.ks.uiuc.edu>,
polynomial
>> = f^({2k})(\xi)n B_{2k}/((2k!))
>>B_{2k} the Bernoulli numbers with
>> | B_{2k}/(2k!) | <= 4/((2\pi)^{2k})
>
>Just t be clear: Is the bound here on the Bernoulli number,
>B_{2k}/((2k!)), the Bernoulli polynomial, B_{2k}(t)/((2k!)) or both,
>because |B_{2k}(t)/((2k!))| <= |B_{2k}/((2k!))|?
>
the bound concerns the Bernoulli numbers and I applied the
intermediate value theorem for integrals.
but it is also valid for that \hat B_{2k}(t) since
|\hat B_{2k}(t)| <= B_{2k} = B_{2k}(0)= \hat B_{2k}(0) for k=1,2,3,.. and all
real x
and
\hat B_k(x) = B_k(x-floor(x)) ,
B_k(x) being the ordinary Bernoulli polynomial.
the \hat B_k can be written as
\hat B_k(x) = B_k(x) -k \sum_{i=0}^{n-1}( (x-i)_{+})^{k-1} for x in [0,n]
>>Hence you need a function whose supremum of |f^{2k}| on [0,n]
>>goes to zero as k to infinity for n fixed.
or at least does grow slower than (2\pi)^{2k}
>>(if I not mistaken, the entire functions on |C do that?)
(false as stated in my second posting)
here I myself have a question:
If f is an entire function on the complex plane, is it true that for every
compact real interval [a,b] there exists a constant C(f,a,b) such that
max_{x in [a,b]} |f^{k}(x)| <= C(f,a,b)^k
(geometric growth of magnitude of the derivatives with order ?)
Robert Israel
Peter Spellucci <spel...@fb04373.mathematik.tu-darmstadt.de> wrote:
> If f is an entire function on the complex plane, is it true that for every
> compact real interval [a,b] there exists a constant C(f,a,b) such that
> max_{x in [a,b]} |f^{k}(x)| <= C(f,a,b)^k
> (geometric growth of magnitude of the derivatives with order ?)
No. For example, f(z) = sum_{n=0}^infinity z^n/sqrt(n!) is entire and has
f^(k)(0) = sqrt(n!) which is not O(c^k) for any c.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
John Washburn
series, so the summation is over all integers. This question arose in
trying to find the value of:
sum(sin(Pi*n/P)*sin(2*Pi*x*(n/P+beta))/(n*(n/P+beta)),
n=-infinity..infinity);
For n=0, the value of the limit as n-->0 is used.
P, x and beta are parameters. P is a prime number > 7, beta is a real
parameter <> 0, and x is real with x > 1.
The Euler-MacLaurin formula would seem to indicate the sum over all
integers is equal to the integral over the whole of the real line.
This is because all the derivatives exist, are continuous, and are
able to be integrated. Further, at the “endpoints”, every derivative
tends to 0 as abs(n) tends to infinity. If the integral remainder
term can be dismissed then
sum(sin(Pi*n/P)*sin(2*Pi*x*(n/P+beta))/(n*(n/P+beta)),
n=-infinity..infinity)
=
int(sin(Pi*n/P)*sin(2*Pi*x*(n/P+beta))/(n*(n/P+beta)),
n=-infinity..infinity)
This seemed hard to believe and is what prompted my question. I did
not think I could properly dismiss the integral remainder term since
the limits of integration is the whole of the real line. The
remainder term would have to be dismissed only after careful analysis
of the limit processes involved.
On the other hand:
int(sin(2*Pi*n/P*x)/(Pi*n), n=-infinity..infinity) = 1
for all values of x and P.
Thus:
int(sin(2*Pi*n/P*x)/(Pi*n), n=-infinity..infinity) = 1 =
sum(sin(2*Pi*n/P*x)/(Pi*n), n=-infinity..infinity);
1 = sum(sin(2*Pi*n/P*x)/(Pi*n), n=-infinity..infinity) = (2*x/P) +
2*sum(sin(2*Pi*n/P*x)/(Pi*n), n=1..infinity)
1 = (2*x/P) + 2*sum(sin(2*Pi*n/P*x)/(Pi*n), n=1..infinity)
(1 / 2) - (x/P) = sum(sin(2*Pi*n/P*x)/(Pi*n), n=1..infinity)
This is the correct. f(t) = (1 / 2) - (x/P) has a Fourier series
representation of sum(sin(2*Pi*n/P*x)/(Pi*n), n=1..infinity) for 0 < x
< P.
On Thu, 23 Sep 2004 13:30:02 +0000 (UTC), Peter Spellucci wrote:
>
>In article <cisi4g$bti$1...@news.ks.uiuc.edu>,
> John Washburn <jo...@washburnResearch.org> writes:
> >On 22 Sep 2004 09:54:46 -0400, Peter Spellucci wrote:
> >>
> >>
> >>In article <cirqc6$48k$1...@dizzy.math.ohio-state.edu>,
> >> John Washburn <jo...@WashburnResearch.org> writes:
> >> >The Euler MacLaurin summation formula links the values of a sum
to
> >the
> >> >value of the definite integral. Here is a link to the article
on
> >the
> >> >Euler-Maclaurin Summation formula on Mathworld,
> >> ><a
>
>href="http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html"><a
href="http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html">http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html</a></a>