I tried to toss Euler's solution to the Basel Problem onto the whiteboard this week, and didn't quite have it from memory.
The Basel Problem sought a solution to
∑ 1/k²
And what bits did I remember off the top of my head. Well, that it involved the Taylor Series for sin(x), a polynomial expansion based on the roots of sin(x), and equating like powers of x.
The problem that I had is that the polynomial expansion calls for a function P(x) such that P(0) = 1.
sin(0) != 1. Oops.
And deep down, it bugged me, because quite frankly if you are looking for a nice trigonometric function with many roots that passes through (0,1), you are generally talking about cos(x). Doesn't that work? And why didn't he use it?
Well, of course it does work, although the answer doesn't quite fall out. Without having done any extensive research of Euler, I'm guessing he was going for the style points.
The Taylor expansion of cos(x) is
cos(x) = 1 - x^2/2! + x^4/4! - ...
But we can also treat cos(x) as an analytic function with roots +π/2, -π/2, +3π/2, -3π/2... ±(2k+1)π/2
P(x)
= (1-x/(π/2))(1+x/(π/2))(1-x/(3π/2))(1+3/(3π/2))....
= (1-x²/(π/2)²)(1-x²/(3π/2)²)(1-x²/(5π/2)²)...
Collecting like terms, we get
1/2
= ∑ 1/((2k+1)²(π/2)²)
= ∑ 4/((2k+1)²(π)²)
π²/8
= ∑ 1/(2k+1)²
From here, there are two different approaches that you can take. The obvious one is to solve the infinite series in two pieces
∑ 1/k²
= ∑ 1/(2k)² + 1/(2k+1)²
= ∑ 1/(2k)² + ∑ 1/(2k+1)²
= ∑ (1/2²)(1/k²) + ∑ 1/(2k+1)²
= (1/4)∑ 1/k² + ∑ 1/(2k+1)²
= (4/3) ∑ 1/(2k+1)²
= (4/3) π²/8
= π²/6
An elegant alternative is to notice that every even square is some power of four times an odd square. Therefore
∑ 1/k² = ( ∑ 1/(2k+1)² ) * ( ∑ 1/4^k) = π²/8 * ( 1 / (1-1/4) ) = π²/8 * 4/3 = π²/6
Euler had several other solutions as well.
But at least now I have one way to solve this thing when I can't remember that Euler's idea of simple was to start from sin(x)/x.
December 20, 2003 10:25 PM
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