Bernhard Riemann

Bernhard Riemann was born in the Kingdom of Hanover (now, Germany) on September 17, 1826. He was the second of six children. He was home schooled by his father, Lutheran minister, until he was 10. Through out his life, he remained very close to his family and very religious.

When he was 14, Riemann moved in with his grandmother and attended school in Lyceum. He entered the gymnasium in Luneburg in 1842. At the gymnasium, Riemann showed strong interest in mathematics. It is said that he read a 900 page book by Adrien Legendre on number theory in 6 days.

In 1846, Riemann enrolled in theology at the University of Gottingen. He continued to take classes in mathematics and later, after consulting with his father, changes his focus from theology to mathematics. At Gottingen, he was able to takes courses from the legendary mathematician Carl Friedrich Gauss.

Despite having Gauss on its faculty, the University of Gottingen was, at this time, secondary in mathematics to the University of Berlin. Riemann transferred there in 1847 and was able to attend courses in advanced mathematics given by Jakob Steiner, Carl Jacobi, Johann Dirichlet, and Ferdinand Eisenstein. These were exciting times for Riemann and he became particularly influenced by the theories of Dirichlet. It is said that at this time, Riemann built up what would become his general theory of complex variables.

Riemann returned to the University of Gottingein in 1849 to work on his Ph.D. thesis under the guidance of Gauss. Riemann was also greatly influenced by Wilhem Weber in theoretical physics and Johann Listing in topology.

Riemann's thesis was on what are today known as Riemann surfaces. In this groundbreaking work, Riemann used topology to analyze complex variables. The MacTutor biography describes this work as: "a strikingly original piece of work which examined the geometric properties of analytic functions, conformal mappings, and the connectivity of surfaces."

Today, Riemann's Ph.D. thesis is considered to be one of the most impressive that has ever been produced. Based on Gauss's recommendation, Riemann was offered a post at the University of Gottingen as a lecturer. To become a lecturer, he needed to achieve a post-doctoral degree called a Habilitation. To complete this degree, he needed to make a presentation on an advanced topic. He proposed three topics to Gauss and to his surprise, Gauss selected geometry. So, on June 10, 1854, Riemann presented a lecture on what is today known as Riemannian geometry which would later be the basis of Einstein's general theory of relativity. The lecture is today considered a classic.

The story goes that Riemann's lecture was so advanced for its time that only Gauss appreciated the depths of the ideas. While the other technical members of the audience listened politely, Gauss was greatly excited. The MacTutor biography quotes an eyewitness: "The lecture exceeded all his [Gauss's] expectations and greatly surprised him. Returning to the faculty meet, he [Gauss] spoke with the greatest phrase and rare enthusiasm to Wilhelm Weber about the depth of the thoughts that Riemann had presented."

When Gauss died in 1855, his replacement was Johann Dirichlet. In 1857, Riemann became a professor of mathematics. At this time, he wrote a paper on the theory of abelian functions. At the same time, Karl Weierstrass was working on the same topic. Riemann's paper was very advanced. Felix Klein writes: "It [Riemann's paper] contained so many unexpected, new concepts that Weierstrass withdrew his paper and in fact published no more."

In 1859, Dirichlet died and Riemann replaced him as the chair of mathematics. He was also admitted to the the Berlin Academy of Sciences. His nomination read (from MacTutor):

Prior to the appearance of his most recent work [Theory of abelian functions], Riemann was almost unknown to mathematicians. This circumstance excuses somewhat the necessity of a more detailed examination of his works as a basis of our presentation. We considered it our duty to turn the attention of the Academy to our colleague whom we recommend not as a young talent which gives great hope, but rather as a fully mature and independent investigator in our area of science, whose progress he in significant measure has promoted.

As a newly elected member of the Academy of Sciences, Riemann was expected to make a techincal presentation. Riemann's presentation unleashed what is today known as the Riemann Hypothesis. This is the most famous and most important open problem in number theory. It was identified by David Hilbert among his famous collection of 23 mathematical problems. The purpose of the paper was to outline a method for determining the number primes less than a given number.

At 36, he married Elise Koch and it seemed like his mathematical impact was only just beginning. Unfortunately, around this time, he got sick with what would later turn out to be tuberculosis. He continued to travel to Italy where he hoped a warmer environment would help his health and then returned to Gottingen. He died on July 20, 1866 in Italy.

Over all, Riemann's output was small but the influence of that output makes him one of the most influential mathematicians of the nineteenth century. He masterfully combined topics of topology, geometry, analysis, and number theory to show how each complemented the other areas. His impact on such a wide range of topics makes him one of the most important mathematicians of all time.

References

• "Bernhard Riemann", the MacTutor Biography
• John Derbyshire, Prime Obsession, Plume, 2004
  

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 = 1/1^2s + 1/2^2s + ...

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 = (e^z - e^-z)/2

Definition 2: cosh z

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

Definition 3: coth z

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

Now, we use these definitions in this lemma.

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

Proof:

(1) z/(e^z - 1) + (z/2) = (2z + z[e^z - 1])/(2[e^z-1]) = (2z -z + z[e^z])/(2[e^z-1]) = (z/2)(e^z+1)/(e^{z - 1})

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

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

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

QED

Corollary 1.1: z coth z = ∑ (n ≥ 0) 4ⁿB_2n(z²ⁿ)/(2n!)

Proof:

(1) From Lemma 1 above:

z/(e^z - 1) + (z/2) = (z/2) coth (z/2)

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

z/[e^z - 1] + (z/2) = ∑ (n ≥ 0)B_2nz⁽²ⁿ⁾/(2n)!

(3) Subsituting 2z for z gives us:

(2z)/(e^2z - 1) + z = z coth z = ∑ (n ≥ 0)B_2n(2z)⁽²ⁿ⁾/(2n)! =

= ∑ (n ≥ 1)(4)ⁿB_2n(z)⁽²ⁿ⁾/(2n)!

QED

Lemma 2: cot x = i coth ix

Proof:

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

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

(3) sinh ix = (e^ix - 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)ⁿB_2n(z²ⁿ)/(2n)!

where B_i 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)B_2n(2z)⁽²ⁿ⁾/(2n)!

(2) Using Lemma 2 above, we have:

z cot z = ∑ (n ≥ 0)B_2n(2iz)⁽²ⁿ⁾/(2n)! =

= ∑ (n ≥ 1)(-4)ⁿB_2n(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, 2^n-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 = (1/2^n-1) ∑ (k=0, 2^n-1-1) cot ([z + kπ]/2^n-1) =

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

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

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

(3) We also get:

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

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

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

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

(8) Therefore:

∑ (k=1,2^n-1-1) cot([z + (-2^n-1 + k)π]/2ⁿ) = ∑ (k=1,2^n-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, 2^n-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, 2^n-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, 2^n-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[2^2n-1π²ⁿB_2n]/(2n)!

where:

ζ(s) = 1/1^s + 1/2^s + 1/3^s + ...

B_i 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)ⁿB_2n(z²ⁿ)/(2n)! = B₀ + ∑ (n ≥ 1) (-4)ⁿB_2n(z²ⁿ)/(2n)! =
= 1 + ∑ (n ≥ 1) (-4)ⁿB_2n(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)ⁿB_2n(z²ⁿ)/(2n)!

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

-2z^2*iζ(2*i)/π^2*i = (-4)ⁱB_2i(z²ⁱ)/(2i)!

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

ζ(2i) = (-4)ⁱB_2iπ²ⁱ/[(2i)!(-2)] =

= (-1)^i-1(2^2i-1π²ⁱB_2i)/(2i)!

QED