View Single Post
Old 2013-03-20, 01:05   #1
jinydu's Avatar
Dec 2003
Hopefully Near M48

6DE16 Posts
Default Can Euler's Basel Solution be Saved?

I'm thinking of giving a talk to an undergrad math club, and I had the idea of presenting Euler's solution to the Bessel Problem (1 + 1/4 + 1/9 + ... = pi^2/6) as the topic. While there are some very grave gaps, it a deliciously elegant solution, and it would be great if there were a way to repair those gaps. My (very rough draft of a) plan is to present Euler's solution as I read it, explain what the gaps are, and then try to fix them. While I know how to repair some of the problems, there are some I don't, which is why I'm posting this.

Anyway, here's Euler's solution:

1) Lemma 1: For all x, sin(x) = x - x^3/3! + x^5/5! - ...
Proof: Taken for granted.

2) Lemma 2: Let f be a complex polynomial with f(0) = 1 and distinct roots r_1, ... r_n. Then f(x) = (1 - x/r_1)(1- x/r_2)...(1-x/r_n)
Proof: Since f has n distinct roots, deg(f) = n. We have specified the values of f at n+1 distinct inputs; it is easy to check that there is at most one f of degree n which can satisfy those conditions. The given f satisfies those conditions.

3) Define sinc(x) = sin(x)/x when x =/= 0 and 1 when x = 0. Then sinc(x)= 1- x^2/3! + x^4/5! - ...

4) Clearly, sinc(0) = 1 and the roots of sinc(x) are just the roots of sin(x) with 0 removed (distinct because the roots of sin(x) are distinct), namely +/-npi for n = 0, 1, 2, etc.

5) Regard sinc(x) as an (infinite) polynomial. By Lemma 2, sinc(x) = (1-x/pi)(1+x/pi)(1-x/2pi)(1+x/2pi)...

6) So (1-x/pi)(1+x/pi)(1-x/2pi)(1+x/2pi)... = 1 - x^2/3! + x^4/5! - ...

7) (1-x^2/pi^2)(1-x^2/4pi^2)(1 - x^2/9pi^2)... = 1 - x^2/3! + x^4/5! - ...

8) Multiplying out the left-hand side and collecting like terms, we get
1 + (-1/pi^2 - 1/4pi^2 - 1/9pi^2 - ...)x^2 + O(x^4) = 1 - x^2/3! + x^4/5! - ...

9) Equate x^2 coefficients to get -1/pi^2 - 1/4pi^2 - 1/pi^2 = ... = -1/3!

10) Multiply both sides by -pi^2 to get the desired result.

--- Problems with the solution and my ideas for fixing them ---

4) We need to look for all complex zeroes of the function, not just the well-known ones on the real line. But this is easily fixed by using the formula sin(z) = e^(iz) - e^(-iz)/2i, setting the numerator equal to zero, splitting z into real and imaginary parts, and solving. It is quickly seen that Im(z) = 0, which brings us back to the well-known case.

5) This is the biggest problem. sinc(x) is of course an entire function, not a polynomial; so we can't legally just apply the lemma. To make matters much worse, the conditions in Lemma 2 aren't even enough to uniquely determine an entire function. Replacing sinc(x) with e^(h(x))sinc(x) for any entire function h with h(0) = 0 yields another function that satisfies exactly the same conditions; but is a very different function and of course will have a very different power series.

What I need is an analogue of Lemma 2 for entire functions that allows me to uniquely pin down the function using its roots, its value at the origin, and some extra data that needs to be easy to get for sinc(x).

7) Warning: Messing with an infinite product in such a nontrivial way is in general not ok. But in this case, it is ok assuming we have patched up the proof up till Line 6 because we know that the infinite product is convergent. In a convergent infinite product, multiplication is associative because the new sequence of partial products is just a subsequence of the original sequence of partial products.

8) Multiplying out an infinite product like that seems kind of fishy to me; can't think of a theorem to justify it off the top of my head. But I would think it's something that can be justified using standard undergrad analysis techniques. Am I right?


So in summary, the main thing I'm missing is a way to justify step 5. A justification for step 8 would also be nice.

Can it be done?

jinydu is offline   Reply With Quote