proof of Faulhaber’s formula
Theorem 0.1.
If k∈N,2≤n∈Z, then
n-1∑m=1mk=1k+1k∑i=0(k+1i)Bink+1-i=∫n1bk(x)𝑑x |
where the Bi are the Bernoulli numbers and bi the Bernoulli polynomials
.
The exponential generating function for the Bernoulli numbers is
∞∑n=0Bnxnn!=xex-1 |
We develop an equation involving sums of Bernoulli numbers on one side, and a simple generating involving powers of e that gives us the appropriate sum of powers on the other side. Equating coefficients of powers of x then gives the result.
To get a generating function where the coefficient of xn/n! is ∑n-1m=1mk, we can use
n-1∑m=0emx | =n-1∑m=0∞∑k=0mkxkk! | ||
=∞∑k=0(n-1∑m=1mk)xkk! |
But this is also a geometric series, so
n-1∑k=0ekx | =1-enx1-ex | ||
=enx-1x⋅xex-1 | |||
=enx-1x∞∑l=0Blxll! | |||
=(∞∑k=0nk+1k+1⋅xkk!)(∞∑l=0Blxll!) | |||
=∞∑k=0(j∑i=01k-i+1(ki)Bink+1-i)xkk! |
Equating coefficients of xk/k! we get
n-1∑m=1mk | =k∑i=01k-i+1(ki)Bink+1-i | ||
=k∑i=0k!(k-i+1)i!(k-i)!Bink+1-i=1k+1k∑i=0(k+1i)Bink+1-i |
which proves the first equality.
If f(x) is a polynomial, write [xr]f(x) for the coefficient of xr in f(x). Then
[xr]bk(x)=1r[xr-1]b′k(x)=kr[xr-1]bk-1(x) |
and thus if r≤k, iterating, we get
[xr]bk(x)=(kr)[x0]bk-r(x)=(kr)Bk-r |
Then using the fact that b′k=kbk-1, we have
∫n1bk(x) | =1k+1(bk+1(n)-bk+1(1))=1k+1k+1∑r=0[xr]bk+1(x)(nr-1) | ||
=1k+1k+1∑r=0(k+1r)Bk+1-r(nr-1)=1k+1k+1∑r=1(k+1r)Bk+1-rnr |
Now reverse the order of summation (i.e. replace r by k+1-r) to get
∫n1bk(x)=1k+1k∑r=0(k+1k+1-r)Brnk+1-r=1k+1k∑r=0(k+1r)Brnk+1-r |
Title | proof of Faulhaber’s formula |
---|---|
Canonical name | ProofOfFaulhabersFormula |
Date of creation | 2013-03-22 18:43:50 |
Last modified on | 2013-03-22 18:43:50 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 4 |
Author | rm50 (10146) |
Entry type | Theorem |
Classification | msc 11B68 |