Bernoulli Numbers

Bernoulli Numbers: They are defined as the coefficient appearing in the following polynomial expansion of sums of integer powers.

$\displaystyle s_p(n) =\sum_{j=1}^{n-1}j^p = =\frac{B_{0}}{0 !} \frac{n^{p+1}}{p+1}+\frac{B_{1}}{1 !} n^{p}+\frac{B_{2}}{2 !} p n^{p-1}+\frac{B_{3}}{3 !} p(p-1) n^{p-2}+\ldots+\frac{B_{p}}{1 !} n$

$\displaystyle \begin{aligned} s_0(n) &= n-1\\ s_{1}(n) &=\frac{1}{2} n^{2}-\frac{1}{2} n & \\ s_{2}(n) &=\frac{1}{3} n^{3} -\frac{1}{2} n^{2} +\frac{1}{6} n \\ s_{3}(n) &=\frac{1}{4} n^{4} -\frac{1}{2} n^{3} +\frac{1}{4} n^{2} \\ s_{4}(n) &=\frac{1}{5} n^{5} -\frac{1}{2} n^{4} +\frac{1}{3} n^{3} \\ s_{5}(n) &=\frac{1}{6} n^{6} -\frac{1}{2} n^{5} +\frac{5}{12} n^{4}-\frac{1}{12} n^{2} \\ s_{6}(n) &=\frac{1}{7} n^{7} -\frac{1}{2} n^{6} +\frac{1}{2} n^{5}-\frac{1}{6} n^{3}+\frac{1}{42} n\\ s_{7}(n) & =\frac{1}{8} n^{8} -\frac{1}{2} n^{7}+\frac{7}{12} n^{6} -\frac{7}{24} n^{4} +\frac{1}{12} n^{2} & & \\ s_{8}(n) &=\frac{1}{9} n^{9} -\frac{1}{2} n^{8}+\frac{2}{3} n^{7} -\frac{7}{15} n^{5} +\frac{2}{9} n^{3} -\frac{1}{30} n \\ s_{9}(n) &=\frac{1}{10} n^{10}-\frac{1}{2} n^{9}+\frac{3}{4} n^{8} -\frac{7}{10} n^{6} +\frac{1}{2} n^{4} -\frac{3}{20} n^{2} \\ s_{10}(n) & =\frac{1}{11} n^{11}-\frac{1}{2} n^{10}+\frac{5}{6} n^{9} -n^{7} +n^{5} -\frac{1}{2} n^{3} +\frac{5}{66} n\\ \end{aligned}$

$\displaystyle B_{0}=1, B_{1}=-\frac{1}{2}, B_{2}=\frac{1}{6}, B_{3}=0, B_{4}-\frac{1}{30}, B_{5}=0, B_{6}=\frac{1}{42}, B_{7}=0$

$\displaystyle s_p(n+1)-s_p(n) =n^p$
$\displaystyle s_p(X+1)-s_p(X)=X^p$
$\displaystyle s_p'(X+1)-s_p'(X)=pX^{p-1}=ps_{p-1}(X+1)-ps_{p-1}(X)$

Therefore
$\displaystyle s_p'(X)= ps_{p-1}(X)+B_p$
${B_p}$ are the constants of integration (discrete integration!).
To get from ${s_{p-1}(n)}$ to ${s_p(n)}$ , just multiply with ${p}$ and integrate and add a term ${B_p n.}$
${B_p}$ can computed using the fact that ${s_p(1)=0.}$

$\displaystyle s_{m}(n)=\frac{1}{m+1} \sum_{k=0}^{m}\left(\begin{array}{c} m+1 \\ k \end{array}\right) B_{k} n^{m-k+1}$
$\displaystyle B_0=1, \sum_{k=0}^{n-1}\left(\begin{array}{l} n \\ k \end{array}\right) B_{k}=0$

Generating function:
$\displaystyle \frac{x}{e^{x}-1}=\sum_{k=0}^{\infty} \frac{B_{k} x^{k}}{k !}$
$\displaystyle B_{n}=\sum_{k=0}^{n} \frac{1}{k+1} \sum_{j=0}^{k}(-1)^{j}\left(\begin{array}{l} k \\ j \end{array}\right) j^{n}$

$\displaystyle (B+1)^n= B^n$

$\displaystyle \begin{aligned} &1=B_{0}\\ &\begin{array}{l} 0=B_{0}+2 B_{1} \\ 0=B_{0}+3 B_{1}+3 B_{2} \\ 0=B_{0}+4 B_{1}+6 B_{2}+4 B_{3} \end{array}\\ &0=B_{0}+5 B_{1}+10 B_{2}+10 B_{3}+5 B_{4} \end{aligned}$
$\displaystyle B_{n}+\left(\begin{array}{c} n \\ n-1 \end{array}\right) B_{n-1}+\cdots+\left(\begin{array}{c} n \\ 1 \end{array}\right) B_{1}+B_0=B_{n}$
$\displaystyle B^{n}+\left(\begin{array}{c} n \\ n-1 \end{array}\right) B^{n-1}+\cdots+\left(\begin{array}{c} n \\ 1 \end{array}\right) B^{1}+1=B^{n}$

Thus we can remember the relations by the formula ${ (B+1)^n= B^n}$ , where we interpret ${B^k}$ as ${B_k}$ when we expand the LHS.

Bernoulli Polynomials:

$\displaystyle \begin{aligned} &B_{0}(y)=1\\ &B_{1}(y)=y-\frac{1}{2}\\ &B_{2}(y)=y^{2}-y+\frac{1}{6}\\ &B_{3}(y)=y^{3}-\frac{3}{2} y^{2}+\frac{1}{2} y\\ &B_{4}(y)=y^{4}-2 y^{3}+y^{2}-\frac{1}{30}\\ &B_{5}(y)=y^{5}-\frac{5}{2} y^{4}+\frac{5}{3} y^{3}-\frac{1}{6} y\\ &B_{6}(y)=y^{6}-3 y^{5}+\frac{5}{2} y^{4}-\frac{1}{2} y^{2}+\frac{1}{42} \end{aligned}$

$\displaystyle \begin{aligned} &B_{0}(y)=1\\ &B_{k}^{\prime}(y)=k B_{k-1}(y)\\ &\int_{0}^{1} B_{k}(y) d y=0 \text { for } k \geq 1 \end{aligned}$
$\displaystyle s_{p}(n)=\sum_{k=0}^{n-1} k^{p}=\frac{1}{p+1}\left(B_{p+1}(n)-B_{p+1}\right)$

$\displaystyle B_{k}(y) = \sum_{n=0}^{k}\left(\begin{array}{l} k \\ n \end{array}\right) B_{n} y^{k-n}$
$\displaystyle B_{k}(x+y)=\sum_{n=0}^{k}\left(\begin{array}{l} k \\ n \end{array}\right) B_{n}(x) y^{k-n}$
$\displaystyle B_k(0)=B_k(1)=B_k$
$\displaystyle \frac{x e^{x y}}{e^{x}-1}=\sum_{k=0}^{\infty} \frac{B_{k}(y) x^{k}}{k !}$
$\displaystyle B_{n}(x)=\sum_{k=0}^{n} \frac{1}{k+1} \sum_{v=0}^{k}(-1)^{v}\left(\begin{array}{l} k \\ v \end{array}\right)(x+v)^{k}$
$\displaystyle B(1-x)=(-1)^{n} B(x)$

Growth, Bounds:

$\displaystyle \left|B_{2 k}\right| \sim 4\left(\frac{k}{\pi e}\right)^{2 k} \sqrt{\pi k}$
$\displaystyle \int_{0}^{1}\left|B_{n}(x)\right| d x \sim 8 \frac{n !}{(2 \pi)^{n+1}} <16 \frac{n !}{(2 \pi)^{n+1}}$
$\displaystyle \sup _{x \in[0,1]}\left|B_{2 n}(x)\right|=\left|b_{2 n}\right|$
$\displaystyle \sup _{x \in[0,1]}\left|B_{2 n+1}(x)\right| \leq \frac{2 n+1}{4}\left|b_{2 n}\right|$

Fourier Expansion:

$\displaystyle B_{1}(x)=-\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\sin (2 \pi k x)}{k}$
$\displaystyle B_{2 n}(x)=(-1)^{n+1} \frac{2(2 n) !}{(2 \pi)^{2 n}} \sum_{k=1}^{\infty} \frac{\cos (2 \pi k x)}{k^{2 n}}$
$\displaystyle B_{2 n+1}(x)=(-1)^{n+1} \frac{2(2 n+1) !}{(2 \pi)^{2 n+1}} \sum_{k=1}^{\infty} \frac{\sin (2 \pi k x)}{k^{2 n+1}}$

Zeta values:

$\displaystyle \zeta(2)=\frac{\pi^{2}}{6}, \quad \zeta(4)=\frac{\pi^{4}}{90}, \quad \zeta(6)=\frac{\pi^{6}}{945}, \quad \zeta(8)=\frac{\pi^{8}}{9450}$

$\displaystyle \zeta(2 k)=\sum_{n=1}^{\infty} \frac{1}{n^{2 k}}=\frac{4^{k}\left|B_{2 k}\right| \pi^{2 k}}{2(2 k) !}$

Consider $\displaystyle f(z)=\frac{z^{-2 m}}{e^{z}-1}=z^{-2 m-1} \frac{z}{e^{z}-1}=z^{-2 m-1} \sum_{k=0}^{\infty} \frac{B_{k}}{k !} z^{k}$
Take contour integral around a circle of of radius ${2\pi N +\epsilon}$ , by residue theorem we get
$\displaystyle \int_{|z|=2\pi N+\epsilon} f(z) d z=\frac{B_{2 m}}{(2 m) !}+ \sum_{n=-N, n \neq 0}^{N}(2 i \pi n)^{-2 m}$
$\displaystyle 0=\lim _{N \rightarrow \infty} \int_{|z|=2\pi N+\epsilon} f(z) d z=\frac{B_{2 m}}{(2 m) !}+(2 i \pi)^{-2 m} 2 \zeta(2 m)$
$\displaystyle \zeta(2 m)=\frac{B_{2 m}}{(2 m) !}(-1)^{m-1} \frac{1}{2}(2 \pi)^{2 m}$

Euler-Maclaurin Formula:

$\displaystyle \sum_{i=a}^{b-1} f(i)=\int_{a}^{b} f(x) d x+\left.\sum_{k=1}^{\infty} \frac{B_{k}}{k !} f^{(k-1)}(x)\right|_{a} ^{b}$

$\displaystyle \sum_{i=a}^{b-1} f(i)=\int_{a}^{b} f(x) d x+\left.\sum_{k=1}^{m} \frac{B_{k}}{k !} f^{(k-1)}(x)\right|_{a} ^{b}+ (-1)^{m+1} \int_{a}^{b} \frac{\left.B_{m}(y-| y\rfloor\right)}{m !} f^{(m)}(x) d x$

Clausen-von Staudt Theorem:

$\displaystyle \begin{aligned} B_{2}&=\frac{1}{6}=1-\frac{1}{2}-\frac{1}{3}\\ B_{4} &=-\frac{1}{30}=1 -\frac{1}{2}-\frac{1}{3}-\frac{1}{5} \\ B_{6} &=\frac{1}{42}=1 -\frac{1}{2}-\frac{1}{3}-\frac{1}{7} \\ B_{8} &=-\frac{1}{30}=1 -\frac{1}{2}-\frac{1}{3}-\frac{1}{5} \\ B_{10} &=\frac{5}{66}=1 -\frac{1}{2}-\frac{1}{3}-\frac{1}{11} \\ B_{12} &=-\frac{691}{2730}=1 -\frac{1}{2}-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{13} \\ B_{14} &=\frac{7}{6}=2 -\frac{1}{2}-\frac{1}{3} \\ B_{16} &=-\frac{3617}{510}=-6-\frac{1}{2}-\frac{1}{3}-\frac{1}{5}-\frac{1}{17} \end{aligned}$

$\displaystyle B_{2 k} \equiv-\sum_{(p-1) | 2 k} \frac{1}{p} \quad \bmod 1$

Proof:

${p B_{m}}$ is ${p-}$ integral.
${ 1^{m}+2^{m}+\cdots+(p-1)^{m} \equiv p B_{m} \quad(\bmod ~p)}$ for any even ${m \ge 0.}$
$\displaystyle s_m(p)=\sum_{k=0}^{m} B_{m-k}\left(\begin{array}{c}m \\ k\end{array}\right) \frac{p^{k+1}}{k+1}$
$\displaystyle =p B_{m}+\sum_{k=1}^{m} B_{m-k}\left(\begin{array}{l} m \\ k \end{array}\right) \frac{p^{k+1}}{k+1}$
$\displaystyle =p B_{m}+\sum_{k=1}^{m}\left(p B_{m-k}\right)\left(\begin{array}{l} m \\ k \end{array}\right) \frac{p^{k}}{k+1}$

By noting that ${\left(\begin{array}{l} m \\ k \end{array}\right) \frac{p^{k}}{k+1}}$ is an integer, we see that ${p B_{m}}$ is ${p-}$
integral by induction.

$\displaystyle \left(p B_{m-k}\right)\left(\begin{array}{l} m \\ k \end{array}\right) \frac{p^{k}}{k+1} \quad(\bmod ~p)$
$\displaystyle k \geq 2 \implies \frac{p^{k}}{k+1} \equiv 0~~(\bmod ~p)$
For ${m}$ even,
$\displaystyle k=1 \implies \left(\begin{array}{l} m \\ k \end{array}\right) \frac{p^{k}}{k+1}=\frac{m}{2} p \equiv 0~~(\bmod~ p)$

Therefore
$\displaystyle s_m(p)=p B_{m} (\bmod ~p)$
$\displaystyle 1^{m}+2^{m}+\cdots+(p-1)^{m} = \left\{\begin{array}{ll} -1, & \text { if } p-1 \mid m \\ 0, & \text { if } p-1\not | m \end{array}\right. (\bmod ~p)$

$\displaystyle \implies B_{2 k} \equiv-\sum_{(p-1) | 2 k} \frac{1}{p} \quad \bmod 1$

Fermat’s Last Theorem (Kummer): FLT is true for regular primes:
A regular prime ${p}$ is an integer such that class number of ${\mathbb{Q}\left(\zeta_{p}\right)}$ is relatively prime to ${p}$ and this is equivalent to it not dividing the numerator of ${B_{2}, B_{4}, B_{6}, \ldots B_{p-3}}$

$\displaystyle \begin{aligned} B_{0} &=1 \\ B_{1} &=-1 / 2 \\ B_{2} &=1 / 6 \\ B_{4} &=-1 / 30 \\ B_{6} &=1 / 42 \\ B_{8} &=-1 / 30 \\ B_{10} &=5 / 66 \\ B_{12} &=-691 / 2730 \\ B_{14} &=7 / 6 \\ B_{16} &=-3617 / 510 \\ B_{18} &=43867 / 798 \\ B_{20} &=-174611 / 330 \\ B_{22} &=854513 / 138 \\ B_{24} &=-236364091 / 2730\\ B_{26} &=8553103 / 6 \\ B_{28} &=-23749461029 / 870 \\ B_{30} &=8615841276005 / 14322 \\ B_{32} &=-7709321041217 / 510 \\ B_{34} &=2577687858367 / 6 \\ B_{36} &=-26315271553053477373 / 1919190 \\ B_{38} &=2929993913841559 / 6 \\ B_{40} &=-261082718496449122051 / 13530\\ B_{42}&=1520097643918070802691 / 1806\\ B_{44}&=-27833269579301024235023 / 690\\ B_{46}&=596451111593912163277961 / 282\\ B_{48}&=-5609403368997817686249127547 / 46410\\ B_{50}&=495057205241079648212477525 / 66\\ B_{52}&=-801165718135489957347924991853 / 1590\\ B_{54}&=29149963634884862421418123812691 / 798\\ B_{56}&=-2479392929313226753685415739663229 / 870\\ B_{58}&=84483613348880041862046775994036021 / 354\\ B_{60}&=-1215233140483755572040304994079820246041491 / 56786730\\ B_{62}&=12300585434086858541953039857403386151 / 6\\ B_{64}&=-106783830147866529886385444979142647942017 / 510\\ \end{aligned}$

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