Faulhaberova formula

Faulhaberova formula predstavlja sumu:

$\sum _{k=1}^{n}k^{p}=1^{p}+2^{p}+3^{p}+\cdots +n^{p}$

Dobila je ime po nemačkom matematičaru Johanu Faulhaberu. Formula se može predstaviti preko Bernulijevih brojeva kao:

$\sum _{k=1}^{n}k^{m}={1 \over {m+1}}\sum _{k=0}^{m}{m+1 \choose {k}}B_{k}\;n^{m+1-k}$

Primeri

$1+2+3+\cdots +n={n(n+1) \over 2}={n^{2}+n \over 2}$

$1^{2}+2^{2}+3^{2}+\cdots +n^{2}={n(n+1)(2n+1) \over 6}={2n^{3}+3n^{2}+n \over 6}$

$1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left({n^{2}+n \over 2}\right)^{2}={n^{4}+2n^{3}+n^{2} \over 4}$

$1^{4}+2^{4}+3^{4}+\cdots +n^{4}={6n^{5}+15n^{4}+10n^{3}-n \over 30}$

$1^{5}+2^{5}+3^{5}+\cdots +n^{5}={2n^{6}+6n^{5}+5n^{4}-n^{2} \over 12}$

$1^{6}+2^{6}+3^{6}+\cdots +n^{6}={6n^{7}+21n^{6}+21n^{5}-7n^{3}+n \over 42}$

Dokaz

Definišemo li sumu

$S_{p}(n)=\sum _{k=1}^{n}k^{p}.$

Tada je:

$S_{p}(1)=1$

$S_{p}(n+1)=S_{p}(n)+(n+1)^{p}.$

Pokušajmo sada da $S_{p}(n)$ izrazimo u obliku polinoma:

$\sum _{k=0}^{\infty }a_{p,k}n^{k+1}$

Uvrstimo li to u drugi izraz u ovom poglavlju dobijamo:

$\sum _{k=0}^{\infty }a_{p,k}(n+1)^{k+1}=\sum _{k=0}^{\infty }a_{p,k}n^{k+1}+(n+1)^{p}$

Koristimo binomnu teoremu, pa sledi:

$\sum _{k=0}^{\infty }a_{p,k}\left[\left(\sum _{j=0}^{k+1}{\binom {k+1}{j}}n^{j}\right)-n^{k+1}\right]=(n+1)^{p}$

$\sum _{k=0}^{\infty }a_{p,k}\sum _{j=0}^{k}{\binom {k+1}{j}}n^{j}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.$

Dvostruku sumu na levoj strani preuredimo uzimajući u obzir j≤k:

$\sum _{j=0}^{\infty }n^{j}\sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.$

i konačno se dobija:

$\sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}={\binom {p}{j}}.$

Desna strana je jednaka nuli za j>p, pa je onda $a_{p,k}=0$ za k>p. Obe strane jednačine množimo sa j!, pa uz korišćenje Pohhamerovoga simbola vredi:

$\sum _{k=j}^{p}(k+1)_{j}a_{p,k}=(p)_{j}$

$\sum _{k=j}^{p}(k+1)_{t}(k+1-t)_{j-t}a_{p,k}=(p)_{t}(p-t)_{j-t}$

Supstitucijom k=k'+t i preuređenjem dobija se:

$\sum _{k'=j-t}^{p-t}(k'+1)_{j-t}\left[{\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}\right]=(p-t)_{j-t}.$

odnosno:

${\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}=a_{p-t,k'}.$

Za k'=0 je:

$a_{p,t}={\frac {(p)_{t}}{(t+1)_{t}}}a_{p-t,0}={\binom {p}{t}}{\frac {a_{p-t,0}}{t+1}}.$

a to upravo odgovara Bernulijevim brojevima, tako da konačno dobijamo:

$\sum _{k=1}^{n}k^{p}=\sum _{j=0}^{p}{\binom {p}{j}}{\frac {B_{p-j}}{j+1}}n^{j+1}$

Veza sa Bernulijevim polinomima

$\sum _{k=1}^{n}k^{p}={\frac {B_{p+1}(n+1)-B_{p+1}(0)}{p+1}},$

a tu su $B_{p}(x)$ Bernulijevi polinomi.

Faulhaberovi polinomi

Faulhaber je uočio da u slučaju neparnoga p suma

$1^{p}+2^{p}+3^{p}+\cdots +n^{p}$

predstavlja polinom od

$a=1+2+3+\cdots +n={\frac {n(n+1)}{2}}.$

Tako je npr:

$1^{3}+2^{3}+3^{3}+\cdots +n^{3}=a^{2};$

$1^{5}+2^{5}+3^{5}+\cdots +n^{5}={4a^{3}-a^{2} \over 3};$

$1^{7}+2^{7}+3^{7}+\cdots +n^{7}={12a^{4}-8a^{3}+2a^{2} \over 6};$

$1^{9}+2^{9}+3^{9}+\cdots +n^{9}={16a^{5}-20a^{4}+12a^{3}-3a^{2} \over 5};$

$1^{11}+2^{11}+3^{11}+\cdots +n^{11}={32a^{6}-64a^{5}+68a^{4}-40a^{3}+10a^{2} \over 6}.$

Literatura

• Faulhaberova formula
• Donald E. Knuth: Johann Faulhaber and Sums of Powers. Math. Comp. 61 (1993), no. 203, S. 277-294
• Abramowitz, Milton; Stegun, Irene A., eds. (1965), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover, ISBN 978-0486612720
• Bernulijevi brojevi