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        riddles >> hard >> Polynomial of Infinite Degree
        
(Message started by: Michael_Dagg on Dec 2^nd, 2005, 9:18pm)

Title: Polynomial of Infinite Degree
Post by Michael_Dagg on Dec 2^nd, 2005, 9:18pm

For n=1,2,3,..., find a formula for f such that

f(x) = (1-x^2)(1-x^2/4)(1-x^2/9)...(1-x^2/n^2)...

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 3^rd, 2005, 7:20am

[hide] 1/x!(-x)! [/hide], though one might find a nicer expression for it! ::)

Title: Re: Polynomial of Infinite Degree
Post by paul_h on Dec 3^rd, 2005, 10:29am

1/x!(-x)! - what? Problem don't say what x is anyway but x! requires integer x. This must be some kind of trick.

Title: Re: Polynomial of Infinite Degree
Post by paul_h on Dec 3^rd, 2005, 10:51am

That is more interesting that I thought it was.

For just n=1...4

f(x) = 1 - 205/144x^2 + 91/192x^4 - 5/96x^6 + 1/76x^8

But, when you factor f(x) you get (did it with Mathematica)

f(x) = 1/576 (x-1)(x+1)(x-2)(x+2)(x-3)(x+3)(x-4)(x+4)

I did this out to n=100 and it is obvious that if x is an integer f(x) = 0. But if x is not an integer it is another story.

Title: Re: Polynomial of Infinite Degree
Post by ace on Dec 3^rd, 2005, 11:54am

Duno the answer but pretty easy to see that it is an infinite degree poly whose zeros are s = {x: x <> 0, x integer}.

Funny thing is that since it is an infinite degree poly it has an infinite number of zeros which are all of the non-zero integers. As n->oo, the infinite size of s differs from the infinite size of Z because s has a hole in it around zero. Interesting. Where did this problem come from?

Title: Re: Polynomial of Infinite Degree
Post by towr on Dec 3^rd, 2005, 3:26pm

Drawing a few graphs might give an idea..

Title: Re: Polynomial of Infinite Degree
Post by Grimbal on Dec 3^rd, 2005, 4:25pm

If we just multiply that function by x, we get a function that has a zero for every integer in Z.
Now what was that function that has a zero for every multiple of a constant P: f(kP)=0?

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 3^rd, 2005, 5:22pm

on 12/03/05 at 10:29:45, paul_h wrote:

1/x!(-x)! - what? Problem don't say what x is anyway but x! requires integer x. This must be some kind of trick.

x! has a definition for non-integer x. In fact one means of defining it more generally looks an awful lot like that "infinite degree polynomial"...

As for what x can be, that itself is a good question. Perhaps z might be a better choice of variable. ;)

Title: Re: Polynomial of Infinite Degree
Post by JocK on Dec 4^th, 2005, 3:30am

f(x) = cos(pi x/2) cos( pi x/4) cos(pi x/8 ) .. cos(pi x/2ⁿ) ...

Anyone who knows a shorthand for this? ;D

Title: Re: Polynomial of Infinite Degree
Post by Michael_Dagg on Dec 4^th, 2005, 1:24pm

Nice - getting close.

Title: Re: Polynomial of Infinite Degree
Post by JocK on Dec 4^th, 2005, 1:38pm

on 12/04/05 at 13:24:52, Michael_Dagg wrote:

Nice - getting close.

Hmmm, it seems that:

Integral[-Infty,Infty] f(x-y) f(y) dy = f(x)

Or am I boarding a sinking ship..? ;D

Title: Re: Polynomial of Infinite Degree
Post by towr on Dec 4^th, 2005, 1:47pm

so [hide] sinc(pi x)[/hide]

Title: Re: Polynomial of Infinite Degree
Post by Michael_Dagg on Dec 4^th, 2005, 4:26pm

JocK, you have already given n-many life rafts for the ship.

Towr, my f is not zero at 0, but yours is 0 at zero. But, you are also close.

Edit: Actually I just noticed, towr, you said sinc(pi*x), I thought it read sin(pi*x) - I wondered how you might have missed f(0) = 1, but you didn't. My apologies, but sinc(pi*x) is it. That is,

sin(pi*x)/(pi*x) = (1-x^2)(1-x^2/2)(1-x^2/4)...(1-x^2/x^n)..., for n=1,2,3,..., x <> 0, but as sinc(x) is so defined we can include x=0. Actually, I could go on about how sinc() is defined and of course how it came about (due to Euler in fact) but I will let you research it.

One of the nice things now is to look at the right-hand side of f that JocK produced. This provides the n-many life rafts I mentioned.

Title: Re: Polynomial of Infinite Degree
Post by Michael_Dagg on Dec 4^th, 2005, 4:41pm

on 12/03/05 at 11:54:19, ace wrote:

. As n->oo, the infinite size of s differs from the infinite size of Z because s has a hole in it around zero. Interesting. Where did this problem come from?

That is a layman's observation regarding infinite sets (in this case Z and Z\{0}) and the countably of such sets (no disrespect intended) but it is a general response that is quite common. Consequently, it is not correct.

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 4^th, 2005, 7:46pm

As for where the problem came from, it is a fairly standard exercise (with a minor twist) in the study of entire functions.

Title: Re: Polynomial of Infinite Degree
Post by towr on Dec 5^th, 2005, 4:44am

Could anyone give a nice proof of why it's sinc(pi x)? Because the only way I can tell is from the graph.

(I've never done a study of entire functions either ;))

Title: Re: Polynomial of Infinite Degree
Post by Barukh on Dec 5^th, 2005, 9:21am

The following reasoning won’t pass the “rigor test” in modern mathematics, but that’s how Euler came to it, and it’s full of elegance.

If we have an 2n-degree polynomial P_2n(x) of the form a₀ - a₁x² + a₂x⁴ + … (-1)ⁿ a_nx²ⁿ with all a_i-s positive, then its roots are n pairs [pm]x₁, …, [pm]x_n. That is, P_2n(x) can be written in the form a₀(1-(x/x₁)²)…(1-(x/x_n)²).

Now, extend this reasoning to polynomials of infinite degree and consider the function sin(x), which as we know may be represented as such a polynomial. It has an infinite number of roots, all of which except one (0) come in pairs: [pm]pi, [pm]2pi, …, [pm]n pi, … Therefore, the function sin(x)/x – which has a₀ = 1 (http://www.ucl.ac.uk/Mathematics/geomath/level2/series/ser121.html) - “may be” written in the form
sin(x)/x = (1- (1-(x/pi)²) … (1-(x/n pi)²)…
Now substitute “pi x” instead of “x” and you will get the statement of the problem.

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 5^th, 2005, 6:19pm

It definitely won't pass the rigor test, for if g(z) is an entire function with g(0) = 1 and g(z) <> 0 for all z, then g(z)sin(z)/z has exactly the same roots, and the same coefficient a₀, and so by the same reasoning should also equal the product.

If g(z) were constrained to be a constant 1, this would be okay, but in fact many g(z) satisfy the condition. In particular if f(z) is an arbitrary entire function, then g(z) = e^zf(z) works.

Of course, the correct answer is sin(pi*x)/pi*x. But the question becomes, can you prove it?

P.S. For those who may not be aware of the terminology, an entire function is one which is defined and (complex) differentiable on the entire complex plane.

Title: Re: Polynomial of Infinite Degree
Post by Barukh on Dec 7^th, 2005, 1:25am

Let’s try from the other side:
sin(x) = (e^ix - e^-ix)/2i = lim_n->oo [(1+ix/n)ⁿ – (1-ix/n)ⁿ]
For an odd n, the expression under the limit is a polynomial of degree n, let’s say P_n(x). Its roots x₁, …, x_n are those satisfying (1+ix/n)ⁿ = (1-ix/n)ⁿ, or 1+ix/n = z^k(1-ix)/n, where z = e^(2[pi]i/n) is a primitive n-th root of unity. Therefore,
x_k = n/i * (z^k–1) / (z^k+1) = n tan(k pi/n), k = 0, …, n-1.
Therefore,
P_n(x)/x = [prod]_k=1,…,n-1 [1-x/n tan^-1(k pi/n)] = [prod]_{k = 1,…,(n-1)/2}[1-(x/n)²tan^-2(k pi/n)],
and for a fixed k, the last term has the limit (1-(x/k pi)²) when n -> oo.

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 7^th, 2005, 4:26pm

An interesting approach, though it falls short of full rigor yet.

Define P_m,n = [prod]_k=1^m 1 - (x/n)²tan^-2(k pi/n).

You have shown that lim_m lim_n P_m,n = f(x/pi), and that lim_n P_{(n-1)/2 , n} = sin(x)/x.

But you have not yet shown that the two limits are equal. I am sure you are aware that for arbitrary double sequences, this is not necessarily the case.

Title: Re: Polynomial of Infinite Degree
Post by Icarus on Dec 17^th, 2005, 9:01am

Some more information on what is going on here.

An entire function is a function f: C --> C, which is analytic (complex-differentiable) at every point on the complex plane. If a is any complex number, then f(z) = [sum]_n=0^oo b_n(a) (z - a)ⁿ for some coefficients b_n(a), where the series has an infinite radius of convergence (the radius of convergence of any analytic function is always the distance to its nearest non-removable singularity).

If f(a) = 0 and f is not identically 0, then b₀(a) = 0, and there must be some k such that for all n < k, b_n(a) = 0, while b_k(a) <> 0. k is called the order of the zero at a. The function f(z)/(z-a)^k has only a removable singularity at z = a, (which we always view has having been removed), and no longer has a zero at a.

Suppose two entire functions f, g have exactly the same zeros, and all those zeros have exactly the same order. Then f/g is an entire function with no zeros at all. Because it has no zeros, you can take a well-defined logarithm of it, h = ln(f/g), which is itself an entire function. So f(z) = e^h(z)g(z).

One consequence of the series representation of any non-zero analytic function about a point a, is that the zeros of analytic functions are always isolated. (f(z)/(z-a)^k is non-zero and continuous at a, where k is the order of the zero at a, and so must be non-zero in a neighborhood of a. Since (z-a)^k is only 0 at a itself, f(z) must also be non-zero in the deleted neighborhood). As a result, any limit point of the collection of zeros of f cannot be within the domain of f. So for an entire function, f can have only countably many zeros a_n. If the order of a root of f is k, we place it k times within the sequence.

Consider the product g(z) = [prod]_n=0^oo (1 - z/a_n). If the product converges, then g has exactly the same roots as f, with the same orders. So f(z) = e^h(z)g(z).

So for any entire function f(z), with root sequence {a_n},
f(z) = e^h(z) [prod]_n (1 - z/a_n)

where h(z) is an entire function, and assuming that the product converges (if it fails to converge, factors can be multiplied in to force it to do so, but this complicates matters more than I want to mention).

Now define P(z) = z [prod]_n=1^oo (1 - z²/n²). P(z) is easily seen to converge for all z, and P(z) has simple (first order) zeros at all integers (and no other zeros). P(z) is also easily seen to be analytic. sin(pi z) also has simple zeros at all integers, so it follows that sin(pi z) = e^h(z)P(z). It remains to identify h(z).

For this, consider the logarithmic derivative P'(z)/P(z) = 1/z + [sum]_n>0 (2z/(z²-n²) = 1/z + [sum]_n>0 (1/(z-n) + 1/(z+n)) = lim_k [sum]_n=-k^k 1/(z-n).

Taking the derivative again yields g(z) = -[sum]_n (z-n)^-2. d²/dz² (ln (sin(pi z))) = pi²/sin²(pi z). Consider the difference g(z) - pi²/sin²(pi z) = h"(z). Both g(z) and sin(pi z) are periodic with period 1, so h"(z) is as well. h"(z) is entire, since h was. lim_y->oo g(x+iy) - pi²/sin²(pi z) = 0 - 0 = 0. Thus h"(z) is bounded on the strip [0,1] + iR, and by periodicity, everywhere. By Liouville's theorem, h"(z) must be a constant, and h(z) = Az+B. Evaluation for specific points shows that h(z) = ln(pi), from which the identity

sin(pi z)/(pi z) = [prod](1-z²/n²)

follows.

(Disclaimer: this is not my argument - I copied the key elements out of a textbook.)

Title: Re: Polynomial of Infinite Degree
Post by JP05 on May 20^th, 2006, 3:31pm

Took me a while to see what M_D meant when he said JoCK has given n-many life rafts for the ship:

Thus, sin(pi *x)/(pi *x) = cos(pi x/2) cos( pi x/4) cos(pi x/8 ) .. cos(pi x/2n) ... , x <> 0.

Title: Re: Polynomial of Infinite Degree
Post by alien on May 21^st, 2006, 12:44pm

For n=1,2,3,..., there can be only this formula for f:

f(x) = (9-x^7)(7-x^2/4)(12-x^2/9)...(4-x^2/n^2)...(3-3+4)...(32+3-x^1/3)... thus, sin(pi *pim+x)/(pim *x+pi) = cos(pi x/2) cost( 100$) cos(pi x/8 ) .. cost(300$) ... , x <> 9+1

Alien children on my planet know this, as we would put this surly in the easy and not hard section.

Title: Re: Polynomial of Infinite Degree
Post by JP05 on May 21^st, 2006, 5:09pm

Alien, your formulas appear to make no sense at all.

Title: Re: Polynomial of Infinite Degree
Post by towr on May 22^nd, 2006, 12:57am

on 05/21/06 at 17:09:57, JP05 wrote:

Alien, your formulas appear to make no sense at all.

I'd go with your instincts on this.
Probably inebriation or willfull annoyance by our resident alien.

Title: Re: Polynomial of Infinite Degree
Post by alien on May 22^nd, 2006, 8:54am

I took a sip of your champagne, and that was enough to get intoxicated. It takes about two weeks for me to get sober, and then I will say the correct answer.