## Sunday, October 29, 2006

### Bernoulli Numbers and the Riemann Zeta Function

Leonhard Euler used the Bernoulli numbers to generalize his solution to the Basel Problem. This is the problem that put Euler on the map mathematically.

The Bernoulli numbers are named after Jacob Bernoulli, the same Bernoulli who popularized the Basel Problem that Euler solved. Bernoulli had been unable to solve the Basel Problem but Euler later showed how the numbers he had identified could be used to provide a general solution to ζ(2s) = ∑ n-2s = 1/12s + 1/22s + ...

The content in today's blog is taken straight from Graham, Knuth, Patashnik's Concrete Mathematics.

The following are definitions for the hyperbolic functions. For those who would like a background on them, see here.

Definition 1: sinh z

sin h z = (ez - e-z)/2

Definition 2: cosh z

cosh z = (ez + e-z)/2

Definition 3: coth z

coth z = (cosh z)/(sinh z)

Now, we use these definitions in this lemma.

Lemma 1: z/(ez - 1) + (z/2) = (z/2) coth (z/2)

Proof:

(1) z/(ez - 1) + (z/2) = (2z + z[ez - 1])/(2[ez-1]) = (2z -z + z[ez])/(2[ez-1]) = (z/2)(ez+1)/(ez - 1)

(2) (z/2)(ez + 1)/(ez-1) = [e-(z/2)/e-(z/2)]* (z/2)(ez + 1)/(ez-1) =

= (z/2)[(ez/2 + e-z/2)/(ez/2 - e-z/2)] =

= (z/2) coth (z/2)

QED

Corollary 1.1: z coth z = ∑ (n ≥ 0) 4nB2n(z2n)/(2n!)

Proof:

(1) From Lemma 1 above:

z/(ez - 1) + (z/2) = (z/2) coth (z/2)

(2) Using a previous result (see Corollary, here), we also know that:

z/[ez - 1] + (z/2) = ∑ (n ≥ 0)B2nz(2n)/(2n)!

(3) Subsituting 2z for z gives us:

(2z)/(e2z - 1) + z = z coth z = ∑ (n ≥ 0)B2n(2z)(2n)/(2n)! =

= ∑ (n ≥ 1)(4)nB2n(z)(2n)/(2n)!

QED

Lemma 2: cot x = i coth ix

Proof:

(1) eix = i sin x + cos x [See Euler's Formula Theorem, here]

(2) cosh ix = (eix + e-ix)/2 = (isinx + cos x + isin(-x) + cos(-x))/2 = 2cos(x)/2 = cos(x)

(3) sinh ix = (eix - e-ix)/2 = (isin x + cos x - isin(-x) -cos(-x))/2 = 2isin(x)/2 = isin(x)

(4) i coth ix = i (cosh(ix)/sinh(ix)) = i ( cos(x)/isin(x)) = cot(x)

QED

Lemma 3: z cot z = ∑ (n ≥ 0) (-4)nB2n(z2n)/(2n)!

where Bi is a Bernoulli number [See Definition 1, here for definition of Bernoulli numbers]

Proof:

(1) From Corollary 1.1 above, we have:

z coth z = ∑ (n ≥ 0)B2n(2z)(2n)/(2n)!

(2) Using Lemma 2 above, we have:

z cot z = ∑ (n ≥ 0)B2n(2iz)(2n)/(2n)! =

= ∑ (n ≥ 1)(-4)nB2n(z)(2n)/(2n)!

QED

Lemma 4: cot z = (1/2n) ∑ (k=0, 2n-1) cot ([z + kπ]/2n)

Proof:

(1) I will use induction for this proof.

(2) For n=1, we have:

cot z = (1/2) cot(z/2) + (1/2) cot(z/2 + π/2)

(a) Since cot(z/2 + π/2) = - tan(z/2) [See Corollary 1.6, here], we have:

(1/2)cot(z/2) + (1/2)cot(z/2 + π/2) = (1/2)[cot(z/2) - tan(z/2)] = (1/2)[cos(z/2)/sin(z/2) - sin(z/2)/cos(z/2)] = (1/2)[cos2(z/2) - sin2(z/2)]/[sin(z/2)cos(z/2)]

(b) Since cos(z) = cos2(z/2) - sin2(z/2) [see Lemma 3, here], we have:

cot z = (1/2)[cos(z)]/[sin(z/2)cos(z/2)]

(c) Since sin(2x) = 2sin(x)cos(x) [See Lemma 2, here], we have:

cot z = (1/2)[cos(z)/[(1/2)sin(z)]] = cos(z)/sin(z) = cot z

(3) Assume that cot z = (1/2n) ∑ (k=0, 2n-1) cot ([z + kπ]/2n) for n ≥ 1.

(4) Using cot(2x) = (1/2)[cot(x) - tan(x)] [See Corollary 3.6, here], we have:

cot z = (1/2n) ∑ (k=0, 2n-1) cot ([z + kπ]/2n) =

= (1/2n+1) ∑ (k=0,2n-1) [ cot([z + kπ]/2n+1) - tan(z + kπ]/2n+1) ]

(5) Since -tan(x) = cot(x + π/2), we have:

cot z = (1/2n+1) ∑ (k=0,2n-1) [ cot([z + kπ]/2n+1) + cot(z + kπ]/2n+1 + π/2) ] =

(1/2n+1) ∑ (k=0,2n-1) [ cot([z + kπ]/2n+1) + cot(z + (k+ 2n)π]/2n+1) ] =

(1/2n+1) ∑ (k=0,2n-1) cot([z + kπ]/2n+1) + (1/2n+1) ∑ (k=0,2n-1) cot([z + (k + 2n)π]/2n+1) =

= (1/2n+1) ∑ (k=0,2n-1) cot([z + kπ]/2n+1) + (1/2n+1) ∑ (k=2n,2n+1-1) cot([z + kπ]/2n+1) =

= (1/2n+1) ∑ (k=0, 2n+1-1) cot ([z + kπ]/2n+1)

QED

Corollary 4.1: cot z = (1/2n) cot (z/2n) - (1/2n) tan (z/2n) + ∑ (k=1, 2n-1-1) (1/2n) (cot[(z + kπ)/2n] + cot[(z - kπ)/2n])

Proof:

(1) From Lemma 4, substituting (n-1) for (n), we get:

cot z = (1/2n-1) ∑ (k=0, 2n-1-1) cot ([z + kπ]/2n-1) =

= (1/2n-1) cot(z/2n-1) + (1/2n-1) ∑ (k=1, 2n-1-1) cot([z + kπ]/2n-1)

(2) Using cot(2x) = (1/2)[cot(x) - tan(x)] (see Corollary 3.6, here), we get:

(1/2n-1) cot(z/2n-1) = (1/2n)[cot(z/2n) - tan(z/2n)]

(3) We also get:

(1/2n-1) ∑ (k=1, 2n-1-1) cot([z + kπ]/2n-1) = (1/2n) ∑ (k=1, 2n-1-1) [cot([z + kπ]/2n) - tan([z+kπ]/2n)]

(4) To complete the proof, we just need to show that ∑ (k=1, 2n-1-1) -tan([z+kπ]/2n) = ∑ (k=1, 2n-1-1) cot([z - kπ]/2n)

(5) First, we note that -tan(z) = cot(z - π/2) since:

-tan(z) = cot(z + π/2) = cot(z - π + π/2) = cot(z - π/2) [See Corolary 1.6 and Corollary 1.5, here]

(6) This gives us that:

∑ (k=1, 2n-1-1) -tan([z + kπ]/2n) = ∑(k=1,2n-1-1) cot([z + kπ]/2n - π/2) =

= ∑ (k=1,2n-1-1) cot([z + (-2n-1 + k)π]/2n)

(7) Now we can see that -2n-1 + k = -1 when k=2n-1-1 and -2n-1 + k = -2n-1+1 when k = 1.

(8) Therefore:

∑ (k=1,2n-1-1) cot([z + (-2n-1 + k)π]/2n) = ∑ (k=1,2n-1-1) cot([z - kπ]/2n)

QED

Lemma 5: lim (z → 0) z cot z = 1

Proof:

(1) lim(z → 0) z cot z = lim(z → 0) cos(z)* lim(z → 0) (z/sin(z)) [By Product Rule for Limits, see Lemma 2, here]

(2) lim(z → 0) cos(z) = 1 [See Property 6, here if needed]

(3) lim (z → 0) (z/sin(z)) = lim(z → 0) 1/(sin(z)/z) [By the Reciprocal Law for Limits, see Lemma 6 here]

(4) lim (z → 0) (sin(z)/z) = 1 [See Lemma 2, here]

(5) Putting it all together gives us:

lim (z → 0) z cot z = lim(z → 0) cos(z) * 1/[lim(z → 0) sin(z)/z] = 1*(1/1) = 1

QED

Corollary 5.1: z cot z = 1 - 2 ∑ (k ≥ 1) z2/[k2π2 - z2]

Proof:

(1) Using Corollary 4.1 above, we have:

z cot z = (z/2n) cot (z/2n) - (z/2n) tan (z/2n) + ∑ (k=1, 2n-1-1) (z/2n) (cot[(z + kπ)/2n] + cot[(z - kπ)/2n])

(2) Since n can take any value, we can see that:

z cot z = lim (n → ∞) [ (z/2n) cot (z/2n) - (z/2n) tan (z/2n) + ∑ (k=1, 2n-1-1) (z/2n) (cot[(z + kπ)/2n] + cot[(z - kπ)/2n]) ]

(3) As n approaches , z/2n approaches 0, so that using Lemma 5 above we have:

lim(n → ∞) (z/2n)cot(z/2n) = lim (z → 0) (z) cot(z) = 1

(4) Likewise,

lim(n → ∞) -(z/2n)tan(z/2n) = lim(z → 0) -z*tan(z) = lim(z → 0) (-z) * sin(z)/cos(z) = (-z)*(0)*(1) = 0 [See Product Rule for Limits for details if needed, see Lemma 2, here]

(5) Now, (z/2n) = [(z + kπ)/2n]*[z/(z + kπ)] so that:

lim(n → ∞) (z/2n)*cot([z + kπ]/2n) =

lim(n → ∞) [z/(z + kπ)]*[(z+kπ)/2n]*cot([z + kπ]/2n)

(6) Let u = (z + kπ)/2n

(7) As n → ∞, u → 0, so that we have:

lim(n → ∞) (z/2n)*cot([z + kπ]/2n) =
lim(u → 0) [z/(z + kπ)]*u*cot(u) = [z/(z + kπ)]*1 = z/(z + kπ)

(8) Let v = (z - kπ)/2n

(9) As n → ∞, v → 0, so that we have:

lim(n → ∞) (z/2n)*cot([z - kπ]/2n) =
lim(v → 0) [z/(z - kπ)]*v*cot(v) = [z/(z - kπ)]*1 = z/(z - kπ)

(10) So, lim (n → ∞) [ (z/2n)cot([z + kπ]/2n) + (z/2n)cot([z - kπ]/2n) = z/(z + kπ) + z/(z - kπ) = [z*(z - kπ) + z(z + kπ)]/(z2 - k2π2) = (2z2)/(z2 - k2π2) [See Addition Rule for Limits, see Corollary 8.1, here]

(11) Finally,

lim (n → ∞) (z/2n) cot (z/2n) - (z/2n) tan (z/2n) + ∑ (k=1, 2n-1-1) (z/2n) (cot[(z + kπ)/2n] + cot[(z - kπ)/2n]) =

= 1 - 0 + ∑(k ≥ 1) 2z2/(z2 - k2π2) = 1 - 2*∑ (k ≥ 1) z2/(k2π2 - z2)

QED

Lemma 6: z2/[k2π2 - z2] = ∑ (i ≥ 1) z2i/[k2iπ2i]

Proof:

(1) 1/(1 - x) = 1 + x + x2 + ... [See Lemma 1, here]

(2) x/(1 - x) = x + x2 + x3 + ...

(3) Let x = (z2)/(k2π2)

(4) Then:

[(z2)/(k2π2)]/[1 - (z2)/(k2π2)] =

(z2)/[(k2π2](1 - (z2)/(k2π2)] =

= z2/(k2π2 - z2)

QED

Theorem: ζ(2n) = (-1)n-1[22n-1π2nB2n]/(2n)!

where:

ζ(s) = 1/1s + 1/2s + 1/3s + ...

Bi is a Bernoulli number

Proof:

(1) z cot z = 1 - 2 ∑ (k ≥ 1) z2/[k2π2 - z2] [See Lemma 5 above]

(2) Using Lemma 6 above, we have:

z cot z = 1 - 2∑ (k ≥ 1)[ z2/k2π2 + z4/k4π4 + z6/k6π6 + ... ]

(3) Since for each of term of this sum, k can take all integer values ≥ 1, we replace the ∑(k ≥ 1) with:

z cot z = 1 - 2(z2ζ(2)/π2 + z4ζ(4)/π4 + z6ζ(6)/π6 + ...

(4) From Lemma 3 above, we have:

z cot z = ∑ (n ≥ 0) (-4)nB2n(z2n)/(2n)! = B0 + ∑ (n ≥ 1) (-4)nB2n(z2n)/(2n)! =
= 1 + ∑ (n ≥ 1) (-4)nB2n(z2n)/(2n)!

(5) Equating the equations in step #3 and step #4 gives us:

- 2(z2ζ(2)/π2 + z4ζ(4)/π4 + z6ζ(6)/π6 + ... = ∑ (n ≥ 1) (-4)nB2n(z2n)/(2n)!

(6) This gives us for each term i ≥ 1:

-2z2*iζ(2*i)/π2*i = (-4)iB2i(z2i)/(2i)!

(7) Solving for ζ(2*i), this gives us:

ζ(2i) = (-4)iB2iπ2i/[(2i)!(-2)] =

= (-1)i-1(22i-1π2iB2i)/(2i)!

QED