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In [[mathematics]], the '''Bernoulli polynomials''' occur in the study of many [[special functions]] and in particular the [[Riemann zeta function]] and the [[Hurwitz zeta function]]. This is in large part because they are an [[Appell sequence]], i.e. a [[Sheffer sequence]] for the ordinary [[derivative]] operator. Unlike [[orthogonal polynomials]], the Bernoulli polynomials are remarkable in that the number of crossings of the ''x''-axis in the [[unit interval]] does not go up as the degree of the polynomials goes up. In the limit of large degree, the Bernoulli polynomials, appropriately scaled, approach the [[trigonometric function|sine and cosine functions]].
Those who are enduring hemorrhoids all agree it is quite painful. For those people even sitting is fairly big trouble. This healthy condition might affect on the career, technique of living, even your social and individual lifetime can very agonizing, when not fix this issue. Curing it is actually no problem nowadays, since there are plenty of hemorrhoid treatment methods available to take care of it for you. If you don't have the means to take surgery to promptly receive rid of piles or pricey painkillers to create the pain go away, then here are certain all-natural methods for this some natural hemorrhoid treatments we should try.<br><br>There are 2 issues associated with using lotions. The initially is that certain people experience burning sensations, sometimes thus bad that you need to discontinue the utilization of the cream. The second problem is that creams never treat the underlying problems which cause hemorrhoids; consequently lotions are a temporary [http://hemorrhoidtreatmentfix.com/hemorrhoid-relief hemorrhoid relief].<br><br>If the itch is driving you crazy, try extract of Echinacea on a cotton wool ball. Apply it straight to the area. This might bring we relief from this absolutely irritating symptom and ease the pain too. Echinacea can easily be bought at the local wellness food store.<br><br>A healthy digestion will also be a key for not wanting to undertake any hemorrhoid remedy. Good digestion ehances normal bowel movement. Exercising daily, strolling for about 20 to 30 minutes a day supports the digestive program.<br><br>Right today, there are a lot of hemorrhoid treatments. And yes, there are the painless hemorrhoid treatments also accessible. Examples of such as employ of petroleum jelly, the utilization of ointment phenylephrine or Preparation H, and even the easy use of soft cotton underwear. They are painless for with them we don't have to go below the knife.<br><br>Sitz Bath: This way is regarded as the most widespread techniques selected to relieve sufferers of the pain caused by hemorrhoids. A sitz bathtub is a tub filled with warm water, in the event you want we can add certain necessary oils to your bathtub water. You may need to soak the rectum to the warm water for at least 15 minutes. Do this three occasions a day plus it might greatly minimize the swelling and the pain of the the hemorrhoids.<br><br>Although these 7 steps are all great methods to heal hemorrhoids, they are no promise that you are able to completely heal the hemorrhoids or which they won't return. As a direction, when symptoms do not clear up completely, return in a limited days or deteriorate at several point inside medication, you ought to receive an appointment to visit a general specialist proper away.
[[Image:Bernoulli polynomials.svg|thumb|right|Bernoulli polynomials]]
 
==Representations==
 
The Bernoulli polynomials ''B''<sub>''n''</sub> admit a variety of different [[representation (mathematics)|representations]].  Which among them should be taken to be the definition may depend on one's purposes.
 
===Explicit formula===
 
:<math>B_n(x) = \sum_{k=0}^n {n \choose k} b_{n-k} x^k,</math>
 
for ''n'' ≥ 0, where ''b''<sub>''k''</sub> are the [[Bernoulli number]]s.
 
===Generating functions===
The [[generating function]] for the Bernoulli polynomials is
 
:<math>\frac{t e^{xt}}{e^t-1}= \sum_{n=0}^\infty B_n(x) \frac{t^n}{n!}.</math>
 
The generating function for the Euler polynomials is
:<math>\frac{2 e^{xt}}{e^t+1}= \sum_{n=0}^\infty E_n(x) \frac{t^n}{n!}.</math>
 
===Representation by a differential operator===
 
The Bernoulli polynomials are also given by
 
:<math>B_n(x)={D \over e^D -1} x^n</math>
 
where ''D'' = ''d''/''dx'' is differentiation with respect to ''x'' and the fraction is expanded as a [[formal power series]]. It follows that  
:<math>\int _a^x  B_n (u) ~du = \frac{B_{n+1}(x) - B_{n+1}(a)}{n+1}  ~.</math>
cf. [[#Integrals]] below.
 
===Representation by an integral operator===
 
The Bernoulli polynomials are the unique polynomials determined by
 
:<math>\int_x^{x+1} B_n(u)\,du = x^n.</math>
 
The [[integral transform]] 
 
:<math>(Tf)(x) = \int_x^{x+1} f(u)\,du</math>
 
on polynomials ''f'', simply amounts to  
:<math>
\begin{align}
(Tf)(x) = {e^D - 1 \over D}f(x) & {} = \sum_{n=0}^\infty {D^n \over (n+1)!}f(x) \\
& {} = f(x) + {f'(x) \over 2} + {f''(x) \over 6} + {f'''(x) \over 24} + \cdots  ~.
\end{align}
</math>
This can be used to produce the [[#Inversion]] formulas below.
 
==Another explicit formula==
 
An explicit formula for the Bernoulli polynomials is given by
 
:<math>B_m(x)=
\sum_{n=0}^m \frac{1}{n+1}
\sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m.</math>
 
Note the remarkable similarity to the globally convergent series expression for the [[Hurwitz zeta function]]. Indeed, one has
 
:<math>B_n(x) = -n \zeta(1-n,x)</math>
 
where ''ζ''(''s'',&nbsp;''q'') is the Hurwitz zeta; thus, in a certain sense, the Hurwitz zeta generalizes the Bernoulli polynomials to non-integer values of&nbsp;''n''.
 
The inner sum may be understood to be the ''n''th [[forward difference]] of ''x''<sup>''m''</sup>; that is,
 
:<math>\Delta^n x^m = \sum_{k=0}^n (-1)^{n-k} {n \choose k} (x+k)^m</math>
 
where Δ is the [[forward difference operator]]. Thus, one may write
 
:<math>B_m(x)= \sum_{n=0}^m \frac{(-1)^n}{n+1} \Delta^n x^m. </math>
 
This formula may be derived from an identity appearing above as follows. Since the forward difference operator Δ equals
:<math>\Delta = e^D - 1\,</math>
where ''D'' is differentiation with respect to ''x'', we have, from the [[Mercator series]]
 
:<math>{D \over e^D - 1} = {\log(\Delta + 1) \over \Delta} = \sum_{n=0}^\infty {(-\Delta)^n \over n+1}.</math>
 
As long as this operates on an ''m''th-degree polynomial such as ''x''<sup>''m''</sup>, one may let ''n'' go from 0 only up to&nbsp;''m''.
 
An integral representation for the Bernoulli polynomials is given by the [[Nörlund&ndash;Rice integral]], which follows from the expression as a finite difference.
 
An explicit formula for the Euler polynomials is given by
 
:<math>E_m(x)=
\sum_{n=0}^m \frac{1}{2^n}
\sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m\,.</math>
 
This may also be written in terms of the [[Euler number]]s ''E''<sub>''k''</sub> as
 
:<math>E_m(x)=
\sum_{k=0}^m {m \choose k} \frac{E_k}{2^k}
\left(x-\frac{1}{2}\right)^{m-k} \,.</math>
 
==Sums of ''p''th powers==
 
We have
 
:<math>\sum_{k=0}^{x} k^p = \frac{B_{p+1}(x+1)-B_{p+1}(0)}{p+1}.</math>
 
See [[Faulhaber's formula]] for more on this.
 
==The Bernoulli and Euler numbers==
The [[Bernoulli number]]s are given by <math>B_n=B_n(0).</math>
An alternate convention defines the Bernoulli numbers as <math>B_n=B_n(1)</math>. This definition gives B<sub>''n''</sub>&nbsp;=&nbsp;−''n''ζ(1&nbsp;−&nbsp;''n'') where for ''n''&nbsp;=&nbsp;0 and ''n''&nbsp;=&nbsp;1 the expression −''n''ζ(1&nbsp;−&nbsp;''n'') is to be understood as
lim<sub>''x''&nbsp;→&nbsp;''n''</sub>&nbsp;−''x''ζ(1&nbsp;−&nbsp;''x'').
The two conventions differ only for ''n''&nbsp;=&nbsp;1 since B<sub>1</sub>(1)&nbsp;=&nbsp;1/2&nbsp;=&nbsp;−B<sub>1</sub>(0).
 
The [[Euler number]]s are given by <math>E_n=2^nE_n(1/2).</math>
 
==Explicit expressions for low degrees==
The first few Bernoulli polynomials are:
:<math>B_0(x)=1\,</math>
:<math>B_1(x)=x-1/2\,</math>
:<math>B_2(x)=x^2-x+1/6\,</math>
:<math>B_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{2}x\,</math>
:<math>B_4(x)=x^4-2x^3+x^2-\frac{1}{30}\,</math>
:<math>B_5(x)=x^5-\frac{5}{2}x^4+\frac{5}{3}x^3-\frac{1}{6}x\,</math>
:<math>B_6(x)=x^6-3x^5+\frac{5}{2}x^4-\frac{1}{2}x^2+\frac{1}{42}.\,</math>
 
The first few Euler polynomials are
:<math>E_0(x)=1\,</math>
:<math>E_1(x)=x-1/2\,</math>
:<math>E_2(x)=x^2-x\,</math>
:<math>E_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{4}\,</math>
:<math>E_4(x)=x^4-2x^3+x\,</math>
:<math>E_5(x)=x^5-\frac{5}{2}x^4+\frac{5}{2}x^2-\frac{1}{2}\,</math>
:<math>E_6(x)=x^6-3x^5+5x^3-3x.\,</math>
 
==Maximum and minimum==
 
At higher ''n'', the amount of variation in ''B''<sub>''n''</sub>(''x'') between ''x''&nbsp;=&nbsp;0 and ''x''&nbsp;=&nbsp;1 gets large. For instance,
 
:<math>B_{16}(x)=x^{16}-8x^{15}+20x^{14}-\frac{182}{3}x^{12}+\frac{572}{3}x^{10}-429x^8+\frac{1820}{3}x^6
-\frac{1382}{3}x^4+140x^2-\frac{3617}{510}</math>
 
which shows that the value at ''x''&nbsp;=&nbsp;0 (and at ''x''&nbsp;=&nbsp;1) is −3617/510 ≈&nbsp;−7.09, while at ''x''&nbsp;=&nbsp;1/2, the value is 118518239/3342336 ≈&nbsp;+7.09. [[D.H. Lehmer]]<ref>D.H. Lehmer, "On the Maxima and Minima of Bernoulli Polynomials", ''[[American Mathematical Monthly]]'', volume 47, pages 533–538 (1940)</ref> showed that the maximum value of ''B''<sub>''n''</sub>(''x'') between 0 and 1 obeys
 
:<math>M_n < \frac{2n!}{(2\pi)^n}</math>
 
unless ''n'' is 2 modulo 4, in which case
 
:<math>M_n = \frac{2\zeta(n)n!}{(2\pi)^n}</math>
 
(where <math>\zeta(x)</math> is the [[Riemann zeta function]]), while the minimum obeys
 
:<math>m_n > \frac{-2n!}{(2\pi)^n}</math>
 
unless ''n'' is 0 modulo 4, in which case
 
:<math>m_n = \frac{-2\zeta(n)n!}{(2\pi)^n}.</math>
 
These limits are quite close to the actual maximum and minimum, and Lehmer gives more accurate limits as well.
 
==Differences and derivatives==
 
The Bernoulli and Euler polynomials obey many relations from [[umbral calculus]]:
 
:<math>\Delta B_n(x) = B_n(x+1)-B_n(x)=nx^{n-1},\,</math>
 
:<math>\Delta E_n(x) = E_n(x+1)-E_n(x)=2(x^n-E_n(x)).\,</math>
 
is the [[forward difference operator]]).
 
These [[polynomial sequence]]s are [[Appell sequence]]s:
 
:<math>B_n'(x)=nB_{n-1}(x),\,</math>
 
:<math>E_n'(x)=nE_{n-1}(x).\,</math>
 
===Translations===
 
:<math>B_n(x+y)=\sum_{k=0}^n {n \choose k} B_k(x) y^{n-k}</math>
 
:<math>E_n(x+y)=\sum_{k=0}^n {n \choose k} E_k(x) y^{n-k}</math>
 
These identities are also equivalent to saying that these polynomial sequences are [[Appell sequence]]s.  ([[Hermite polynomials]] are another example.)
 
===Symmetries===
 
:<math>B_n(1-x)=(-1)^nB_n(x),\quad n \ge 0,</math>
 
:<math>E_n(1-x)=(-1)^n E_n(x)\,</math>
 
:<math>(-1)^n B_n(-x) = B_n(x) + nx^{n-1}\,</math>
 
:<math>(-1)^n E_n(-x) = -E_n(x) + 2x^n\,</math>
 
[[Zhi-Wei Sun]] and Hao Pan <ref>{{cite journal |author1=Zhi-Wei Sun |author2=Hao Pan |journal=Acta Arithmetica |volume=125 |year=2006 |pages=21–39 |title=Identities concerning Bernoulli and Euler polynomials  |arxiv=math/0409035}}</ref> established the following surprising symmetry relation: If ''r''&nbsp;+&nbsp;''s''&nbsp;+&nbsp;''t''&nbsp;=&nbsp;''n'' and ''x''&nbsp;+&nbsp;''y''&nbsp;+&nbsp;''z''&nbsp;=&nbsp;1, then
 
:<math>r[s,t;x,y]_n+s[t,r;y,z]_n+t[r,s;z,x]_n=0,</math>
 
where
 
:<math>[s,t;x,y]_n=\sum_{k=0}^n(-1)^k{s \choose k}{t\choose {n-k}}
B_{n-k}(x)B_k(y).</math>
 
==Fourier series==
 
The [[Fourier series]] of the Bernoulli polynomials is also a [[Dirichlet series]], given by the expansion
 
:<math>B_n(x) = -\frac{n!}{(2\pi i)^n}\sum_{k\not=0 }\frac{e^{2\pi ikx}}{k^n}= -2 n! \sum_{k=1}^{\infty} \frac{\cos\left(2 k \pi x- \frac{n \pi} 2 \right)}{(2 k \pi)^n}.</math>
Note the simple large ''n'' limit to suitably scaled trigonometric functions.
 
This is a special case of the analogous form for the [[Hurwitz zeta function]]
 
:<math>B_n(x) = -\Gamma(n+1) \sum_{k=1}^\infty
\frac{ \exp (2\pi ikx) + e^{i\pi n} \exp (2\pi ik(1-x)) } { (2\pi ik)^n }. </math>
 
This expansion is valid only for 0&nbsp;≤&nbsp;''x''&nbsp;≤&nbsp;1 when ''n''&nbsp;≥&nbsp;2 and is valid for 0&nbsp;<&nbsp;''x''&nbsp;<&nbsp;1 when ''n''&nbsp;=&nbsp;1.
 
The Fourier series of the Euler polynomials may also be calculated.  Defining the functions
 
:<math>C_\nu(x) = \sum_{k=0}^\infty
\frac {\cos((2k+1)\pi x)} {(2k+1)^\nu}</math>
 
and
 
:<math>S_\nu(x) = \sum_{k=0}^\infty
\frac {\sin((2k+1)\pi x)} {(2k+1)^\nu}</math>
 
for <math>\nu > 1</math>, the Euler polynomial has the Fourier series
 
:<math>C_{2n}(x) = \frac{(-1)^n}{4(2n-1)!}
\pi^{2n} E_{2n-1} (x)</math>
 
and
 
:<math>S_{2n+1}(x) = \frac{(-1)^n}{4(2n)!}
\pi^{2n+1} E_{2n} (x).</math>
 
Note that the <math>C_\nu</math> and <math>S_\nu</math> are odd and even, respectively:
 
:<math>C_\nu(x) = -C_\nu(1-x)</math>
 
and
 
:<math>S_\nu(x) = S_\nu(1-x).</math>
 
They are related to the [[Legendre chi function]] <math>\chi_\nu</math> as
 
:<math>C_\nu(x) = \mbox{Re} \chi_\nu (e^{ix})</math>
 
and
 
:<math>S_\nu(x) = \mbox{Im} \chi_\nu (e^{ix}).</math>
 
==Inversion==
The Bernoulli and Euler polynomials may be inverted to express the [[monomial]] in terms of the polynomials.
 
Specifically, evidently from the above section on [[#Representation by an integral operator]], it follows that 
:<math>x^n = \frac {1}{n+1}
\sum_{k=0}^n {n+1 \choose k} B_k (x)
</math>
 
and
 
:<math>x^n = E_n (x) + \frac {1}{2}
\sum_{k=0}^{n-1} {n \choose k} E_k (x).
</math>
 
==Relation to falling factorial==
The Bernoulli polynomials may be expanded in terms of the [[falling factorial]] <math>(x)_k</math> as
 
:<math>B_{n+1}(x) =  B_{n+1} + \sum_{k=0}^n
\frac{n+1}{k+1}
\left\{ \begin{matrix} n \\ k \end{matrix} \right\}
(x)_{k+1} </math>
where <math>B_n=B_n(0)</math> and
 
:<math>\left\{ \begin{matrix} n \\ k \end{matrix} \right\} = S(n,k)</math>
 
denotes the [[Stirling number of the second kind]]. The above may be inverted to express the falling factorial in terms of the Bernoulli polynomials:
 
:<math>(x)_{n+1} = \sum_{k=0}^n
\frac{n+1}{k+1}
\left[ \begin{matrix} n \\ k \end{matrix} \right]
\left(B_{k+1}(x) - B_{k+1} \right) </math>
 
where
:<math>\left[ \begin{matrix} n \\ k \end{matrix} \right] = s(n,k)</math>
 
denotes the [[Stirling number of the first kind]].
 
==Multiplication theorems==
The [[multiplication theorem]]s were given by [[Joseph Ludwig Raabe]] in 1851:
 
:<math>B_n(mx)= m^{n-1} \sum_{k=0}^{m-1} B_n \left(x+\frac{k}{m}\right)</math>
 
:<math>E_n(mx)= m^n \sum_{k=0}^{m-1}
(-1)^k E_n \left(x+\frac{k}{m}\right)
\quad \mbox{ for } m=1,3,\dots</math>
 
:<math>E_n(mx)= \frac{-2}{n+1} m^n \sum_{k=0}^{m-1}
(-1)^k B_{n+1} \left(x+\frac{k}{m}\right)
\quad \mbox{ for } m=2,4,\dots</math>
 
==Integrals==
Indefinite integrals
:<math>\int_a^x B_n(t)\,dt =
\frac{B_{n+1}(x)-B_{n+1}(a)}{n+1}</math>
 
:<math>\int_a^x E_n(t)\,dt =
\frac{E_{n+1}(x)-E_{n+1}(a)}{n+1}</math>
 
Definite integrals
:<math>\int_0^1 B_n(t) B_m(t)\,dt =
(-1)^{n-1} \frac{m! n!}{(m+n)!} B_{n+m}
\quad \mbox { for } m,n \ge 1 </math>
 
:<math>\int_0^1 E_n(t) E_m(t)\,dt =
(-1)^{n} 4 (2^{m+n+2}-1)\frac{m! n!}{(m+n+2)!} B_{n+m+2}</math>
 
==Periodic Bernoulli polynomials==
A '''periodic Bernoulli polynomial''' ''P''<sub>''n''</sub>(''x'') is a Bernoulli polynomial evaluated at the [[fractional part]] of the argument ''x''.  These functions are used to provide the [[remainder term]] in the [[Euler–Maclaurin formula]] relating sums to integrals.  The first polynomial is a [[Sawtooth wave|sawtooth function]].
 
==References==
<references />
* Milton Abramowitz and Irene A. Stegun, eds. ''[[Abramowitz and Stegun|Handbook of Mathematical Functions]] with Formulas, Graphs, and Mathematical Tables'', (1972) Dover, New York. ''(See  [http://www.math.sfu.ca/~cbm/aands/page_804.htm Chapter 23])''
 
* {{Apostol IANT}} ''(See chapter 12.11)''
*{{dlmf|first=K. |last=Dilcher|id=24|title=Bernoulli and Euler Polynomials}}
 
* {{Cite journal | last1 = Cvijović | first1 = Djurdje | last2 = Klinowski | first2 = Jacek | year = 1995 | title = New formulae for the Bernoulli and Euler polynomials at rational arguments | url = | journal = Proceedings of the American Mathematical Society | volume = 123 | issue = | pages = 1527–1535 }}
 
* {{Cite journal | doi = 10.1007/s11139-007-9102-0 | last1 = Guillera | first1 = Jesus | last2 = Sondow | first2 = Jonathan | year = 2008 | title = Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent | arxiv = math.NT/0506319 | journal = The Ramanujan Journal | volume = 16 | issue = 3| pages = 247–270 }} ''(Reviews relationship to the Hurwitz zeta function and Lerch transcendent.)''
 
* {{cite book | author=Hugh L. Montgomery | authorlink=Hugh Montgomery (mathematician) | coauthors=[[Robert Charles Vaughan (mathematician)|Robert C. Vaughan]] | title=Multiplicative number theory I. Classical theory | series=Cambridge tracts in advanced mathematics | volume=97 | year=2007 | isbn=0-521-84903-9 | pages=495–519 | publisher=Cambridge Univ. Press | location=Cambridge }}
 
[[Category:Special functions]]
[[Category:Number theory]]
[[Category:Polynomials]]

Revision as of 19:23, 28 February 2014

Those who are enduring hemorrhoids all agree it is quite painful. For those people even sitting is fairly big trouble. This healthy condition might affect on the career, technique of living, even your social and individual lifetime can very agonizing, when not fix this issue. Curing it is actually no problem nowadays, since there are plenty of hemorrhoid treatment methods available to take care of it for you. If you don't have the means to take surgery to promptly receive rid of piles or pricey painkillers to create the pain go away, then here are certain all-natural methods for this some natural hemorrhoid treatments we should try.

There are 2 issues associated with using lotions. The initially is that certain people experience burning sensations, sometimes thus bad that you need to discontinue the utilization of the cream. The second problem is that creams never treat the underlying problems which cause hemorrhoids; consequently lotions are a temporary hemorrhoid relief.

If the itch is driving you crazy, try extract of Echinacea on a cotton wool ball. Apply it straight to the area. This might bring we relief from this absolutely irritating symptom and ease the pain too. Echinacea can easily be bought at the local wellness food store.

A healthy digestion will also be a key for not wanting to undertake any hemorrhoid remedy. Good digestion ehances normal bowel movement. Exercising daily, strolling for about 20 to 30 minutes a day supports the digestive program.

Right today, there are a lot of hemorrhoid treatments. And yes, there are the painless hemorrhoid treatments also accessible. Examples of such as employ of petroleum jelly, the utilization of ointment phenylephrine or Preparation H, and even the easy use of soft cotton underwear. They are painless for with them we don't have to go below the knife.

Sitz Bath: This way is regarded as the most widespread techniques selected to relieve sufferers of the pain caused by hemorrhoids. A sitz bathtub is a tub filled with warm water, in the event you want we can add certain necessary oils to your bathtub water. You may need to soak the rectum to the warm water for at least 15 minutes. Do this three occasions a day plus it might greatly minimize the swelling and the pain of the the hemorrhoids.

Although these 7 steps are all great methods to heal hemorrhoids, they are no promise that you are able to completely heal the hemorrhoids or which they won't return. As a direction, when symptoms do not clear up completely, return in a limited days or deteriorate at several point inside medication, you ought to receive an appointment to visit a general specialist proper away.