Does This Infinite Sum Equal 0 for all n?
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From: Leroy Quet on 3 Jul 2010 16:38 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
From: OwlHoot on 3 Jul 2010 17:49 On Jul 3, 9:38 pm, Leroy Quet <qqq...(a)mindspring.com> wrote: > 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 I'd try running it through Mathematica. The Home edition is only a couple of hundred dollars now. It's brilliant - I've spent the last week working through a good book called "A Student's Guide to Mathematica", by B Torrence and E Torrence (2nd ed, 2009), and I highly recommend it. P.S. If you do buy it, be sure to install it on a PC with up-to-date hardware and software, which you don't intend changing for some years, or decades preferably, because I gather that for copy protection reasons it is extremely finicky about being transplanted from one PC or OS to another! Cheers John Ramsden (jhnrmsdn(a)yahooo.co.uk) remove a letter to reply P.S. Sorry - Can't help with the series - It just looks too scary! But it's nice to see some of the regulars still posting here on sci.math.
From: Sue San on 3 Jul 2010 19:40 "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)!,
From: Leroy Quet on 4 Jul 2010 07:49 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. Thanks, Leroy Quet
From: Rob Johnson on 4 Jul 2010 13:30
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. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |