Difference between revisions of "Bernoulli B"

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$$B_n(x)=\displaystyle\sum_{k=0}^n {n \choose k} b_{n-k}x^k,$$
 
$$B_n(x)=\displaystyle\sum_{k=0}^n {n \choose k} b_{n-k}x^k,$$
 
where $b_k$ are [[Bernoulli numbers]].
 
where $b_k$ are [[Bernoulli numbers]].
 +
 +
$$B_0(x)=1$$
 +
$$B_1(x)=x-\dfrac{1}{2}$$
 +
$$B_2(x)=x^2-x+\dfrac{1}{6}$$
 +
$$B_3(x)=x^3-\dfrac{3x^2}{2}+\dfrac{x}{2}$$
 +
$$B_4(x)=x^4-2x^3+x^2-\dfrac{1}{30}$$
  
 
=Properties=
 
=Properties=

Revision as of 10:00, 23 March 2015

Bernoulli polynomials $B_n$ are orthogonal polynomials defined by the formula $$B_n(x)=\displaystyle\sum_{k=0}^n {n \choose k} b_{n-k}x^k,$$ where $b_k$ are Bernoulli numbers.

$$B_0(x)=1$$ $$B_1(x)=x-\dfrac{1}{2}$$ $$B_2(x)=x^2-x+\dfrac{1}{6}$$ $$B_3(x)=x^3-\dfrac{3x^2}{2}+\dfrac{x}{2}$$ $$B_4(x)=x^4-2x^3+x^2-\dfrac{1}{30}$$

Properties

Theorem: The following formula holds: $$\dfrac{te^{xt}}{e^t-1} = \displaystyle\sum_{k=0}^{\infty} B_k(x)\dfrac{t^k}{k!}.$$

Proof:

Theorem

The following formula holds: $$B_n(x)=-n \zeta(1-n,x),$$ where $B_n$ denotes the Bernoulli polynomial and $\zeta$ denotes the Hurwitz zeta function.

Proof

References

Orthogonal polynomials