Author |
Topic: Alternating Sum (Read 1718 times) |
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender: 
Posts: 1948
|
I finally got around to working out the details:
For Re(a)>0, b,c>0, let
f(x) = eax-be^(cx).
Note that f(n) decays exponentially as n->-infinity, and even faster as n->infinity, so everything I say below should be true.
The Fourier transform F of f can be computed by substituting t=becx below:
F(w) =  f(x)e-2pi i xwdx
= ex(a-2pi i w)e-be^(cx)dx
= (t/b)(a-2pi i w)/c e-tdx/(ct)
= b(2pi i w-a)/c/c t(a-2pi i w)/c - 1 e-tdt
= b(2pi i w-a)/c/c Gamma((a-2pi iw)/c),
by the integral definition of Gamma(z) for Re(z)>0. By Poisson summation,
 f(n) = F(n),
where the sum is over all integers n. Fix z=e-b < 1, set c=log 2, and for A>1, set a=log(-A) = log(A) + pi i. Then
(-A)nz2^n = 1/log 2 Gamma((log A - (2n-1)pi i)/log 2)ex(log A - (2n-1)pi i)
Let S(z,A), T(z,A), denote the sum on the LHS above restricted to n>0, and n<0, respectively, and let U(x,A) denote the RHS. Now, we'll let A->1+.
The dominated convergence theorem, applied to the counting measure, says that if we have a series cn(A) depending on A, with cn(A) -> cn for each n, and |cn(A)|<bn for some bn with bn finite, then cn(A) -> cn.
Since |(-A)nz2^n| < bn=2nz2^n for 1<A<2, and n>0 bn converges (by the n-th root test, say), we have S(z,A) -> S(z) as A->1+.
Determining T(z) = limA T(z,A) is more tricky, since the series T(z,1) doesn't converge. For convenience set t=1/A < 1, r=1-z > 0, so that
T(z, A) = n>0 (-t)n(1-r)1/2^n
= n>0 (-t)n k>=0 (-r)k/k! (1/2n) (1/2n-1) ... (1/2n-(k-1))
= n>0 (-t)n + n>0 k>0 cn,k,r(t),
where cn,k,r(t) = -(-t)nrk/k! [((k-1)2n-1) ... (2n-1)]/2nk.
Since |cn,k,r(t)| < rk/2n, and rk/2n < r/(1-r), we have, for fixed z, as A,t -> 1,
T(z,A) = (-t)/(1+t) + n,k>0 cn,k,r(t)
-> -1/2 - n,k>0 (-1)n rk/k! [((k-1)2n-1) ... (2n-1)]/2nk
= -1/2 - r n>0 (-1/2)n - k>1 ck,r,
where
|ck,r| < n>0 |cn,k,r| < n>0 rk/(k2n) = rk/k,
so
T(z) = -1/2 + r/3 + g(r),
where
|g(r)| < k>1 rk/k < 1/2 k>1 rk = 1/2 r2/(1-r) < r2 = O((1-z)2),
uniformly for 1/2 < z < 1.
I'll skip the details, but an elementary argument from the product formula gives |Gamma(iy)| < 2-y^2/(2(y+1))/y, and this suffices to show that for fixed x,
U(x,A) -> U(x) := 1/log 2 Gamma(m pi i/log 2)ex m pi i,
where the sum is over odd m. Putting this all together, we have that
S(z) = -T(z) + U(x)
= 1/2 - (1-z)/3 + g(z) + 2/log 2 m>0 odd Re[ Gamma(m pi i/log 2)ex m pi i ],
where z = e-2^(-x), and g(z) = O((1-z)2).
Numerically, we have
U(x) = 0.00275 cos(pi (x+.48 )) + 10-9 cos(pi (x+.72)) + O(10-16).
Attached is a graph of
-S(z) + 1/2 - (1-z)/3 + 2/log 2 |Gamma(pi i/log 2)|cos(pi x + Arg[Gamma(pi i/log 2)])
(as a function of z). You can see the O((1-z)2) error term from T(z), as well as an oscillation of about 10-9, so you know I didn't just make all this up  .
|
| « Last Edit: Sep 12th, 2007, 7:06pm by Eigenray » |
 IP Logged |
|
|
|
Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261
|
 |
Re: Alternating Sum
« Reply #26 on: Aug 29th, 2006, 4:05pm » |
 Quote  Modify
|
on Aug 27th, 2006, 8:37am, THUDandBLUNDER wrote:
In spite of Eigenray's admirable display of mathematical erudition, this is not one of them.
You can still solve this for yourself by noting that S(x) > S(x4) and S(0.995) > 1/2
|
|
Eigenray just demonstrated that as x->1- the sum oscillates.. hence proving that it diverges.. (right?). Is that what you were trying to tell me using S(x) > S(x^4) and S(0.995) > 1/2 (our previously obtained limit). Is there a easier way to show this diverges? Needs more thought.. there must be something very basic I am missing here..
|
| « Last Edit: Aug 29th, 2006, 4:06pm by Sameer » |
IP Logged |
"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Alternating Sum
« Reply #27 on: Aug 29th, 2006, 6:15pm » |
Quote Modify
|
on Aug 29th, 2006, 4:05pm, Sameer wrote:
Is that what you were trying to tell me using S(x) > S(x^4) and S(0.995) > 1/2 (our previously obtained limit) |
|
Replacing x with t1/4 might be more suggestive...
|
|
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Alternating Sum
« Reply #28 on: Aug 30th, 2006, 1:35am » |
Quote Modify
|
Sameer, may be the following explanation will be helpful:
1. As shown by Icarus, if the limit exists, it equals 1/2.
2. Therefore, if we can find a sequence of numbers x0 < x1 < x2 < ... converging to 1-, for which S(xn) > 1/2 + constant, it will be shown that the limit doesn't exist.
3. Take x0 = 0.995, and xn = xn-11/4. We have:
a) xn is increasing and converges to 1-.
b) S(x0) > 1/2.
c) S(xn) is increasing, as indicated by T&B.
I've got 2 additional question:
1. Is there a way to predict the existance of numbers like 0.995 without resorting to numerical methods?
2. Consider the sequence S(xn) above. Does it have a limit, and if yes, what's its value?
Sorry, if Eigenray's post(s) already answer these questions.
|
| « Last Edit: Aug 30th, 2006, 1:38am by Barukh » |
IP Logged |
|
|
|
Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261
|
|
Re: Alternating Sum
« Reply #29 on: Aug 30th, 2006, 10:44am » |
Quote Modify
|
on Aug 30th, 2006, 1:35am, Barukh wrote:Sameer, may be the following explanation will be helpful:
2. Consider the sequence S(xn) above. Does it have a limit, and if yes, what's its value?
Sorry, if Eigenray's post(s) already answer these questions. |
|
Yes indeed Barukh that was the lines I was thinking as well.. so it ultimately turns to be the definition of limit that disproves the existence of limit!! 
For part 2. I am betting the limit tends to 1. Since for |x| < 1, x^(1/4) is increasing and it will tend to 1. For nth term will be x^(1/n) and as n-> infinity x^(1/n) will tend to 1.
|
|
IP Logged |
"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Alternating Sum
« Reply #30 on: Sep 1st, 2006, 4:01am » |
Quote Modify
|
on Aug 30th, 2006, 10:44am, Sameer wrote:| For part 2. I am betting the limit tends to 1. Since for |x| < 1, x^(1/4) is increasing and it will tend to 1. For nth term will be x^(1/n) and as n-> infinity x^(1/n) will tend to 1. |
|
I doubt this is the case. Remember, we are considering the sequence S(xn), and not the sequence xn.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print