Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We can recursively define a sequence of polynomials by

$$P_0(x) := 1$$

and then with the definite integral

$$P_n(x) := \int_{c_n}^x P_{n-1}(t) ~\mathrm dt$$

where the $c_n$ are to be chosen so that

$$\int_0^1 P_n(t)~\mathrm dt = 0$$

so for $n = 1$ we have $P_1(x) = x - c_1$ so $c_1 = 1/2$. However already for $n=2$ it becomes difficult since $\int_{c_2}^x (t-1/2)~\mathrm dt = t^2/2 - t/2- c_2^2/2 + c_2/2$ and to have $\int_0^1 P_2(t)~\mathrm dt = 0$ there are two solutions for $c_2$ namely

$$c_2 = 1/2 \pm \sqrt 3/6.$$

So $P_2(x) = x^2/2 - x/2 + 1/12$. For $n=3$ we get $c_3 = 0$ and $P_3(x) = x^3/6 - x^2/4 + x/12$ however for $n=4$ we have $c_4=1/2 - 1/2\sqrt (1 - 2/15\sqrt 30)$ (and 3 other solutions) for $P_4(x)=x^4/24 - x^3/12 + x^2/24 - 1/720$. For $n=5$ again $c_5=0$ (however there are another 4 solutions). Is there some general Formula for the $c_n$ so we could have a shortcut for calculating the coefficients of $P_n(x)$? Also with an eye to fractional calculus it would be nice to know if for the Riemann-Liouville integral

$$\frac1{\Gamma(\alpha)} \int_c^x P(t) (x-t)^{\alpha-1}~\mathrm dt$$ and

$$\int_0^1 P(t)~\mathrm dt = 0$$

if there even is a solution for this - let alone if it would fit in the previously defined sequence. The dream result would of course be not only to have a formula for $c_n$ but a function $c(n)$ that continuously defines such a polynomial sequence. But that seems very remote since even in simple cases with for example $\alpha = 1/2$ and $c>0$ the integral produces complex values...

share|improve this question
    
Are you really interested in the $c$'s or do you simply want to have the polynomials $P_n(t)$? –  Fabian Jul 12 '11 at 4:47
    
well the bernoulli polynomials (or the ones defined here) might be interessting in their own right - but i am more interessted in generating them - for example is there some good way to define these polynomials for fractional index? I know for bernoulli polynomials which have the generating function $(te^{tx}/(e^t-1))$ and then you can define polynomials $(te^{\alpha tx}/(e^t-1))$ but i don't think thats consistent with the properties of the polynomials defined here... –  Peter Sheldrick Jul 12 '11 at 5:14
    
actually one generalization i saw wasn't $te^{\alpha tx}/(e^t-1)$ but $(1/(e^t-1))^\alpha e^{tx}$ these are called the 'Norlund Polynomials' - however i don't think they have the property $\int_0^1 P(t) ~\textrm{d}t = 1$ - so i would be happy about other ways to generalize the bernoulli polynomials (or the bernoulli numbers!) to fractional argument. –  Peter Sheldrick Jul 12 '11 at 5:38
add comment

2 Answers

up vote 5 down vote accepted

I can't comment, but your problem looks awfully like the method for generating the Bernoulli polynomials $B_n(x)$, except that instead of the relation $\frac{\mathrm d}{\mathrm dx}P_n(x)=P_{n-1}(x)$ that you have, the Bernoulli polynomials have the relation $\frac{\mathrm d}{\mathrm dx}B_n(x)=nB_{n-1}(x)$. That might be something to start with...


As Fabian seems to have settled the relationship of your polynomials to the more conventional Bernoulli polynomials, I'll address the question of generalizing the Bernoulli polynomials to arbitrary index; one way to this is the use of the exponential generating function

$$\frac{t\exp(zt)}{\exp(t)-1}=\sum_{n=0}^\infty \frac{B_n(z)}{n!}t^n$$

along with the Cauchy differentiation formula

$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\mathrm dz;$$

i.e., consider the contour integral

$$B_\alpha(z)=\frac{\Gamma(\alpha+1)}{2\pi i}\oint_\gamma \frac{t\exp(zt)}{\exp(t)-1} \frac{\mathrm dt}{t^{\alpha+1}},$$

where $\gamma$ is an anticlockwise contour that doesn't cross the negative real axis, and is within the circle with radius $2\pi$ (due to the restricted radius of convergence of the EGF). I'd say more, but it seems there is prior work by Butzer et al., and thus I'll just have to ask you to see that paper.

share|improve this answer
    
You're right - in fact $0!1 = B_0(x)$, $1!(x-1/2) = B_1(x)$, $2!(x^2/2-x/2+1/12) = B_2(x)$ so (probably) $n!P_n(x)=B_n(x)$ (i thought i saw this equation when reading about stirling polynomials - but then i couldn't find it again...). So finding the integration constant $c_n$ is a bit like calculating the bernoulli numbers - since those numbers are the constant terms in the bernoulli polynomials and that is what we fix when we determin $c_n$ - however what would the exact realation between $c_n$ and the nth bernoulli number be? The stuff i read didn't say much about this. –  Peter Sheldrick Jul 12 '11 at 5:00
    
@Peter: This and this may be of interest as well. –  gorilla Jul 12 '11 at 7:38
    
Thanks for the help! Btw i think the connection between the stirling and bernoulli polynomials i thought i read about but coudn't remember the reference was from the wikipedia page you linked! en.wikipedia.org/wiki/… --- and there the last bit under 'Bernoulli numbers as combinatorial objects.' –  Peter Sheldrick Jul 12 '11 at 9:59
    
I don't seem to recall linking to Wikipedia anywhere in my answer, @Peter... :) –  gorilla Jul 12 '11 at 10:06
    
ah yeah it was Fabian below this answer - sorry for mixup –  Peter Sheldrick Jul 12 '11 at 10:26
add comment

Don't know about the $c$'s. But finding the polynomials $P_n(t)$ is not so hard. As gorilla already noted $$\frac{d}{dx} P_n(t) = P_{n-1}(t). \qquad (1)$$ This determines the polynomials $P_n(t)$ together with the normalization $$\int_0^1 dt\, P_n(t) = 0 \qquad n\neq0 \qquad (2)$$ and the initial condition $P_0(t)=1$ uniquely (all the different $c$'s in you post which are the solutions of some polynomial equation lead to the same $P_n$!).

Every recursion step adds a degree to the polynomial and therefore we can parameterize the $n$-th polynomial as $$ P_n(t) = \sum_{j=0}^n \frac{a_{n-j}}{j!} t^j = \frac{a_0}{n!} t^n + \frac{a_1}{(n-1)!} t^{n-1}+ \dots\;.$$ With this ansatz Eq. (1) is automatically fulfilled. Equation (2) leads to the condition $$ \sum_{j=0}^n \frac{a_{n-j}}{(j+1)!} =0$$ with the solution $$a_n = -\sum_{j=1}^n \frac{a_{n-j}}{(j+1)!}.$$ This defines a recursion relation for the coefficients $a_n$ (with $a_0=1$ as initial condition).

The recursion relation has the solution $a_n = \frac{B_n}{n!}$ with $B_n$ the $n$-th Bernoulli number. The polynomial $P_n$ thus are given by $$P_n(t) = \sum_{j=0}^n \frac{B_{n-j}}{j!(n-j)!} t^j = \frac{B_n(t)}{n!},$$ with $B_n(t)$ the Bernoulli polynomials.

share|improve this answer
    
Thanks that is a very clear way to put it - however there are many ways to generate the bernoulli polynomials. For example the equation you use underneath $P_n(t)=...$ that relates the $a_n$ seems very 'linear'. So it shouldn't be suprising that you can generate the coefficients of the bernoulli polynomials with matrices for example. But really what i'm mainly interessted in is a way to define the bernuolli polynomials (or the ones i defined here which differ by a factor of $n!$ ($\Gamma(n+1)$ ?)) that lends itself to looking for such polynomials at fractional indices (a continous definition). –  Peter Sheldrick Jul 12 '11 at 5:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.