Author |
Topic: Polynomial of Infinite Degree (Read 4463 times) |
|
Michael Dagg
Senior Riddler
Gender: 
Posts: 500
|
 |
Polynomial of Infinite Degree
« on: Dec 2nd, 2005, 9:18pm » |
 Quote  Modify
|
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)...
|
| « Last Edit: Dec 3rd, 2005, 12:07am by Michael Dagg » |
 IP Logged |
Regards,
Michael Dagg
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #1 on: Dec 3rd, 2005, 7:20am » |
Quote Modify
|
1/x!(-x)! , though one might find a nicer expression for it! 
|
| « Last Edit: Dec 3rd, 2005, 7:36am by Icarus » |
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
paul_h
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #2 on: Dec 3rd, 2005, 10:29am » |
Quote Modify
 Remove
|
1/x!(-x)! - what? Problem don't say what x is anyway but x! requires integer x. This must be some kind of trick.
|
|
IP Logged |
|
|
|
paul_h
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #3 on: Dec 3rd, 2005, 10:51am » |
Quote Modify
Remove
|
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.
|
|
IP Logged |
|
|
|
ace
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #4 on: Dec 3rd, 2005, 11:54am » |
Quote Modify
Remove
|
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?
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Polynomial of Infinite Degree
« Reply #5 on: Dec 3rd, 2005, 3:26pm » |
Quote Modify
|
Drawing a few graphs might give an idea..
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Polynomial of Infinite Degree
« Reply #6 on: Dec 3rd, 2005, 4:25pm » |
Quote Modify
|
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?
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #7 on: Dec 3rd, 2005, 5:22pm » |
Quote Modify
|
on Dec 3rd, 2005, 10:29am, 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. 
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
JocK
Uberpuzzler
Gender:
Posts: 877
|
|
Re: Polynomial of Infinite Degree
« Reply #8 on: Dec 4th, 2005, 3:30am » |
Quote Modify
|
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? 
|
| « Last Edit: Dec 4th, 2005, 3:56am by JocK » |
IP Logged |
solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Polynomial of Infinite Degree
« Reply #9 on: Dec 4th, 2005, 1:24pm » |
Quote Modify
|
Nice - getting close.
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
JocK
Uberpuzzler
Gender:
Posts: 877
|
|
Re: Polynomial of Infinite Degree
« Reply #10 on: Dec 4th, 2005, 1:38pm » |
Quote Modify
|
on Dec 4th, 2005, 1:24pm, Michael_Dagg wrote:
Hmmm, it seems that:
Integral[-Infty,Infty] f(x-y) f(y) dy = f(x)
Or am I boarding a sinking ship..?
|
| « Last Edit: Dec 4th, 2005, 1:44pm by JocK » |
IP Logged |
solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Polynomial of Infinite Degree
« Reply #12 on: Dec 4th, 2005, 4:26pm » |
Quote Modify
|
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.
|
| « Last Edit: Dec 4th, 2005, 6:19pm by Michael Dagg » |
IP Logged |
Regards,
Michael Dagg
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Polynomial of Infinite Degree
« Reply #13 on: Dec 4th, 2005, 4:41pm » |
Quote Modify
|
on Dec 3rd, 2005, 11:54am, 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.
|
|
IP Logged |
Regards,
Michael Dagg
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #14 on: Dec 4th, 2005, 7:46pm » |
Quote Modify
|
As for where the problem came from, it is a fairly standard exercise (with a minor twist) in the study of entire functions.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Polynomial of Infinite Degree
« Reply #15 on: Dec 5th, 2005, 4:44am » |
Quote Modify
|
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 )
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Polynomial of Infinite Degree
« Reply #16 on: Dec 5th, 2005, 9:21am » |
Quote Modify
|
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 P2n(x) of the form a0 - a1x2 + a2x4 + … (-1)n anx2n with all ai-s positive, then its roots are n pairs [pm]x1, …, [pm]xn. That is, P2n(x) can be written in the form a0(1-(x/x1)2)…(1-(x/xn)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 - “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.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #17 on: Dec 5th, 2005, 6:19pm » |
Quote Modify
|
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) = ezf(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.
|
| « Last Edit: Dec 5th, 2005, 6:20pm by Icarus » |
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Polynomial of Infinite Degree
« Reply #18 on: Dec 7th, 2005, 1:25am » |
Quote Modify
|
Let’s try from the other side:
sin(x) = (eix - e-ix)/2i = limn->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 Pn(x). Its roots x1, …, xn are those satisfying (1+ix/n)n = (1-ix/n)n, or 1+ix/n = zk(1-ix)/n, where z = e^(2[pi]i/n) is a primitive n-th root of unity. Therefore,
xk = n/i * (zk–1) / (zk+1) = n tan(k pi/n), k = 0, …, n-1.
Therefore,
Pn(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.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #19 on: Dec 7th, 2005, 4:26pm » |
Quote Modify
|
An interesting approach, though it falls short of full rigor yet.
Define Pm,n = [prod]k=1m 1 - (x/n)2tan-2(k pi/n).
You have shown that limm limn Pm,n = f(x/pi), and that limn 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.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Polynomial of Infinite Degree
« Reply #20 on: Dec 17th, 2005, 9:01am » |
Quote Modify
|
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=0oo bn(a) (z - a)n for some coefficients bn(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, bn(a) = 0, while bk(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) = eh(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 an. 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=0oo (1 - z/an). If the product converges, then g has exactly the same roots as f, with the same orders. So f(z) = eh(z)g(z).
So for any entire function f(z), with root sequence {an},
f(z) = eh(z) [prod]n (1 - z/an)
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=1oo (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) = eh(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)) = limk [sum]n=-kk 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. limy->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.)
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
JP05
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #21 on: May 20th, 2006, 3:31pm » |
Quote Modify
Remove
|
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.
|
|
IP Logged |
|
|
|
alien
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #22 on: May 21st, 2006, 12:44pm » |
Quote Modify
Remove
|
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.
|
|
IP Logged |
|
|
|
JP05
Guest
|
|
Re: Polynomial of Infinite Degree
« Reply #23 on: May 21st, 2006, 5:09pm » |
Quote Modify
Remove
|
Alien, your formulas appear to make no sense at all.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Polynomial of Infinite Degree
« Reply #24 on: May 22nd, 2006, 12:57am » |
Quote Modify
|
on May 21st, 2006, 5:09pm, 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.
|
| « Last Edit: May 22nd, 2006, 1:02am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print