This is a follow up to one of my previous questions: Formula for integration bounds of recursively defined polynomial sequence. One aspect unfortunately wasn't answered at the time. That isn't a whine (in fact i was really pleased with the answer i got), i hold my hands up that the question wasn't specific enough - that's what this question is for.
Bernoulli polynomials have a neat characterization:
For $k \geq 1$
- $$\displaystyle \frac{\partial }{\partial x}B_k(x)=k\cdot B_{k-1}(x)$$
- $$\displaystyle \int_0^1 B_k(t) \, dt=\frac{1}{k+1}(B_{k+1}(1)-B_{k+1}(0))=0$$
Starting with $B_0(x)=1$ this generates all the polynomials. However, this is a two step process: first determine the indefinite integral $\tilde{B}_k = k\cdot \int B_k(t)\textrm{d}t$ and then determine the definite integral $\int_0^1\tilde{B}_k(t)\textrm{d}t$ to find out the integration constant that was unspecified in the indefinite integral so that the integral over $[0,1]$ is $0$.
It is possible to compress this into one integration with $$B_{k+1}(x) = (k+1) \int_{c_k}^x B_k(t)\textrm{d}t$$ where the $c_k$ determine the constant of integration, but they change with each $k$.
As in my other question it turns out that there are multiple possible $c_k$ for each $k$. In fact the number of possibilities is proportional to $k$ (since it involves solving a polynomial equation who's degree goes up with $k$).
Is there some formula that generates one $c_k$ for each $k$? Because then $B_{k+1}(x) = (k+1) \int_{F(k)}^x B_k(t)\textrm{d}t$ solves the problem.
Here is a list of the all the possible values for $c_k$ from $k=1..14$ and here are the corresponding polynomials (the possible $c_k$ are the roots of these polynomials), generated with octave, so please excuse the lack of formatting. For $k=5$ there is probably some numerical glitch (and others - also sometimes the root-finding fails to find all the roots).
EDIT: It turns out that these polynomials are just the Bernoulli polynomials again. So for odd $k$ the values $c_k=0,~0.5$ and $1$ work because those are always roots of odd Bernoulli polynomials. For even $k$ it is more tricky since the roots are more complicated, but those polynomials have roots that converge to $0.25$ and $0.75$. These are the roots of $\sin(2\pi x)$ and $\cos(2\pi x)$ in the interval $[0,1]$ that a scaled version of the Bernoulli polynomials converges to.
EDIT2: It is really obvious that this works, because if $r$ is a root of $B_{k+1}(x)$ then with property 1. we have: $$(k+1) \int_{r}^x B_k(t)\textrm{d}t = B_{k+1}(x)-B_{k+1}(r)=B_{k+1}(x).$$ If we integrate over $[0,x]$ this doesn't work because $B_k(0)$ isn't zero for even $k$ - in fact $B_k(0)$ are the Bernoulli numbers.
Consider the following Bernoulli polynomial identity for real variable $x$ and nonnegative integer $k$. $$ \displaystyle B_k(x)=\frac{1}{k+1}\frac{\partial B_{k+1}(x)}{\partial x} $$
Implement the dummy variable $u$ and integrate with respect to $u$ from $0$ to $x$. $$ \displaystyle B_k(u)du=\frac{1}{k+1}\partial B_{k+1}(u)\\ \displaystyle \int_0^x B_k(u) \, du=\frac{1}{k+1}(B_{k+1}(x)-B_{k+1}(0)) $$
Therefore: $$ \displaystyle B_{k+1}(x)=B_{k+1}(0)+(k+1) \int_0^x B_k(u) \, du $$
For example let $k=2$. $$ \displaystyle B_3(x)=B_3(0)+3 \int_0^x B_2(u) \, du\\ \displaystyle B_3(x)=0+3 \int_0^x \left(u^2-u+\frac{1}{6}\right) \, du\\ \displaystyle B_3(x)=x^3-\frac{3 x^2}{2}+\frac{x}{2} $$
The Digital Library of Mathematical Functions is a great resource for Bernoulli identities.