## Sunday, August 27, 2006

### The Basel Problem

Leonhard Euler first gained fame by his solution to what has been called the Basel Problem (named after Basel, Switzerland where the problem gained its notoriety). This is the resolution of the infinite sum:

∑ 1/(n2)

The problem had first been posed in 1644 by Pietro Mengoli. The problem was popularized by the celebrated mathematician Jakob Bernoulli in 1689. John Wallis calculated its value of its first three decimals (1.645...). The great Gottfriend von Leibnitz was unable to solve it. Johann Bernoulli, the younger brother of Jakob, also worked on it and failed to solve it.

By the 1730s the problem had gained a mystique as the greatest mathematicians of the age seemed unable to solve it. Then, in 1735, when he was 28, Euler announced that he had found a solution. At first, Euler succeeded in calculating the sum to six decimal places and then later, he reached his very insightful conclusion:

∑ 1/(n2) = π2/6

In today's blog, I will show the steps that Euler followed to arrive at this amazing result. This will not be a proof. Euler purposely made assumptions that enabled him to simplify the problem but which prevented his solution from rising to the level of a proof.

Solution: Euler's solution to the Basel Problem

∑ 1/(n2) = π2/6

(1) Using the Taylor expansion for sin x, we know that (see Lemma 3, here for details)

sin x = x - (x3/3!) + (x5/5!) - (x7/7!) + ...

(2) We know that sin x = 0 in the case where x = 0, ±π, ±2π, ±3π ... (see Lemma 1, here for details).

(3) Using the Fundamental Theorem of Algebra, we get:

x - (x3/3!) + (x5/5!) - (x7/7!) + ... = C(x)(x - π)(x + π)(x - 2π)(x + 2π)(x - 3π)(x + 3π)*... where C is a constant.

(4) Since for each value of π we have a pair: π and , we can multiply each of these pairs together to get:

x - (x3/3!) + (x5/5!) - (x7/7!) + ... = C(x)(x2 - π2)(x2 - 4π2)(x2 - 9π2)*...

(5) Since each of these x2 - nπ2 = 0, we can rearrange each to a form of (1 - x2/(nπ2) [See Corollary 1.1, here] so that we have:

x - (x3/3!) + (x5/5!) - (x7/7!) + ... = C(x)(1 - x22)(1 - x2/4π2)(1 - x2/9π2)*...

(6) Now, if we divide each side by x, we get:

(sin x)/x = 1 - (x2/3!) + (x4/5!) - (x6/7!) + ... = C(1 - x22)(1 - x2/4π2)(1 - x2/9π2)*...

(7) Now, since (sin x)/x = 1 as x approaches 0 (see Lemma 2, here), we see that C = 1

Here's why:

As x approaches 0, we have, C(1 - x22)(1 - x2/4π2)(1 - x2/9π2)*... = C(1 - 0)(1 - 0)(1 -0)*... = C

Since sin(x)/x = 1, we can see that C = 1.

(8) So, we are left now with:

1 - (x2/3!) + (x4/5!) - (x6/7!) + ... = (1 - x22)(1 - x2/4π2)(1 - x2/9π2)*...

(9) Now, in mathematics:

if we an identity (that is, an equation that is true for all cases of x):

a0 + a1x + a2x2 + ... anxn = b0 + b1x + b2x2 + ... + bnxn

then:

ai = bi

This means that we can take any power of x and the two sides must be equal.

(10) At this point, Euler took the factor x2 which gives us:

-(x2)/3! = -x22 - x2/4π2 - x2/9π2 - ...

(11) Dividing -x22 from both sides gives us:

π2/6 = 1 + 1/4 + 1/9 + ...

Euler would later publish a more thorough proof of his solution. See here for a proof that ∑ (1/n2) = π2/6.

References