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From: Leroy Quet on 5 Jul 2010 09:58


Rob Johnson wrote:
> In article <a06457a7-b0a4-4a57-9151-6bc251b118da(a)r27g2000yqb.googlegroups.com>,
> Leroy Quet <qqquet(a)mindspring.com> wrote:
> >Sue San wrote:
> >> "Leroy Quet" <qqquet(a)mindspring.com> wrote in message
> >> news:d2306d0a-fb62-454b-9e42-fcf82b05ca63(a)g19g2000yqc.googlegroups.com...
> >> > Let H(0,m) = 1/m for all positive integers m.
> >> > Let H(n,m) = sum{k=1 to m} H(n-1,k), for all positive integers n.
> >> > (So, H(1,m) = the mth harmonic number, 1 + 1/2 + 1/3 +... + 1/m.)
> >> >
> >> > Let, for all positive integers k, j(2k) = i, the squareroot of -1.
> >> > Let , for all positive integers k, j(2k-1) = 1.
> >> >
> >> > Does 0 =
> >> >
> >> > sum{k=1 to infinity}
> >> > H(n,k) e^(i*k*arctan(pi/2)) * (1 +pi^2 /4)^(k/2) * j(k) / (k+2n-1)!,
> >> >
> >> > for all nonnegative integers n, and where arctan(pi/2) is in radians
> >> > and is the value between 0 and pi/2?
> >> >
> >> > There are lots of places in my computations where I may have made at
> >> > least one error. But hopefully I didn't make any.
> >> >
> >> > Thanks,
> >> > Leroy Quet
> >> >
> >>
> >>
> >> The following is wrong, you are missing operators or ( ) or both;
> >> and what is arctan(pi/2) doing in there ?? it is a constant.
> >> Plase correct;
> >>
> >> H(n,k) e^(i*k*arctan(pi/2)) * (1 +pi^2 /4)^(k/2) * j(k) / (k+2n-1)!,
> >
> >
> >Yes, arctan(pi/2) is a constant.
> >
> >I checked this for n = 0, and it is WRONG. It seems, however, that a
> >simple fix would correct it. (My intent was valid, it seems.) But I do
> >not love this "result" enough to fix it now.
>
> I started working on this last night. I hadn't had any luck with
> proving it, and I was going to check out the n = 0 case numerically
> this morning. However, I came up with a couple of identities that
> might help with the "fixed" problem.
>
> i arctan(pi/2) 1 + i pi/2
> e = ----------------
> sqrt(1 + pi^2/4)
>
> so that the part of the original formula
>
> e^(i*k*arctan(pi/2)) * (1 +pi^2 /4)^(k/2)
>
> reduces to
>
> (1 + i pi/2)^k
>
> Also, a closed formula for j(k) is
>
> 1 + i k 1 - i
> j(k) = ----- - (-1) -----
> 2 2
>
> I wouldn't mind working on the fixed problem tonight, if you don't
> mind posting it.


Don't bother trying to fix it. I think my error, which may be one
error among a couple, was that I took e^1 to be 1 somewhere along the
line.

I started with the identities -- which themselves may be erroneous,
but I think they are correct:

sum{k>=1} H(m,k) cos(xk) * y^k /(k+m+n-1)!
=
exp(y cos(x)) *
sum{k>=1} H(n,k) (-1)^(k+1) * cos(xk +y sin(x)) /(k+m+n-1)!,

and

sum{k>=1} H(m,k) sin(xk) * y^k /(k+m+n-1)!
=
exp(y cos(x)) *
sum{k>=1} H(n,k) (-1)^(k+1) * sin(xk +y sin(x)) /(k+m+n-1)!,

for all nonnegative integers m and n.

Maybe something can be found from these identities, if they are true,
where, say,
y*sin(x) = pi/2. And where m = n. Maybe there is an interesting
special case there.

Thanks,
Leroy Quet


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