proof of Faulhaber’s formula

proof of Faulhaber’s formula

Theorem 0.1

.

If

k

∈

N

,

2

≤

n

∈

Z

, then

∑

m

=

1

n

-

1

m

k

=

1

k

+

1

⁢

∑

i

=

0

k

(

k

+

1

i

)

⁢

B

i

⁢

n

k

+

1

-

i

=

∫

1

n

b

k

⁢

(

x

)

⁢

𝑑

x

where the

B

i

are the

Bernoulli numbers

and

b

i

the

Bernoulli polynomials

.

The

exponential generating function

for the Bernoulli numbers is

∑

n

=

0

∞

B

n

⁢

x

n

n

!

=

x

e

x

-

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

x

n

/

n

!

is

∑

m

=

1

n

-

1

m

k

, we can use

∑

m

=

0

n

-

1

e

m

⁢

x

=

∑

m

=

0

n

-

1

∑

k

=

0

∞

m

k

⁢

x

k

k

!

=

∑

k

=

0

∞

(

∑

m

=

1

n

-

1

m

k

)

⁢

x

k

k

!

But this is also a

geometric series

, so

∑

k

=

0

n

-

1

e

k

⁢

x

=

1

-

e

n

⁢

x

1

-

e

x

=

e

n

⁢

x

-

1

x

⋅

x

e

x

-

1

=

e

n

⁢

x

-

1

x

⁢

∑

l

=

0

∞

B

l

⁢

x

l

l

!

=

(

∑

k

=

0

∞

n

k

+

1

k

+

1

⋅

x

k

k

!

)

⁢

(

∑

l

=

0

∞

B

l

⁢

x

l

l

!

)

=

∑

k

=

0

∞

(

∑

i

=

0

j

1

k

-

i

+

1

⁢

(

k

i

)

⁢

B

i

⁢

n

k

+

1

-

i

)

⁢

x

k

k

!

Equating coefficients of

x

k

/

k

!

we get

∑

m

=

1

n

-

1

m

k

=

∑

i

=

0

k

1

k

-

i

+

1

⁢

(

k

i

)

⁢

B

i

⁢

n

k

+

1

-

i

=

∑

i

=

0

k

k

!

(

k

-

i

+

1

)

⁢

i

!

⁢

(

k

-

i

)

!

⁢

B

i

⁢

n

k

+

1

-

i

=

1

k

+

1

⁢

∑

i

=

0

k

(

k

+

1

i

)

⁢

B

i

⁢

n

k

+

1

-

i

which proves the first equality.

If

f

⁢

(

x

)

is a

polynomial

, write

[

x

r

]

⁢

f

⁢

(

x

)

for the coefficient of

x

r

in

f

⁢

(

x

)

. Then

[

x

r

]

⁢

b

k

⁢

(

x

)

=

1

r

⁢

[

x

r

-

1

]

⁢

b

k

′

⁢

(

x

)

=

k

r

⁢

[

x

r

-

1

]

⁢

b

k

-

1

⁢

(

x

)

and thus if

r

≤

k

, iterating, we get

[

x

r

]

⁢

b

k

⁢

(

x

)

=

(

k

r

)

⁢

[

x

0

]

⁢

b

k

-

r

⁢

(

x

)

=

(

k

r

)

⁢

B

k

-

r

Then using the fact that

b

k

′

=

k

⁢

b

k

-

1

, we have

∫

1

n

b

k

⁢

(

x

)

=

1

k

+

1

⁢

(

b

k

+

1

⁢

(

n

)

-

b

k

+

1

⁢

(

1

)

)

=

1

k

+

1

⁢

∑

r

=

0

k

+

1

[

x

r

]

⁢

b

k

+

1

⁢

(

x

)

⁢

(

n

r

-

1

)

=

1

k

+

1

⁢

∑

r

=

0

k

+

1

(

k

+

1

r

)

⁢

B

k

+

1

-

r

⁢

(

n

r

-

1

)

=

1

k

+

1

⁢

∑

r

=

1

k

+

1

(

k

+

1

r

)

⁢

B

k

+

1

-

r

⁢

n

r

Now reverse the order of

summation

(i.e. replace

r

by

k

+

1

-

r

) to get

∫

1

n

b

k

⁢

(

x

)

=

1

k

+

1

⁢

∑

r

=

0

k

(

k

+

1

k

+

1

-

r

)

⁢

B

r

⁢

n

k

+

1

-

r

=

1

k

+

1

⁢

∑

r

=

0

k

(

k

+

1

r

)

⁢

B

r

⁢

n

k

+

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

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