Author |
Topic: harmonic (Read 4622 times) |
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: harmonic
« Reply #51 on: Jan 27th, 2008, 5:48pm » |
Quote Modify
|
To be honest, nobody cares if srn347 and sm347 are/were the same person. Whether you and srn347 are the same person (and I am 99% sure you are) is much more pertinent.
|
|
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: harmonic
« Reply #52 on: Jan 27th, 2008, 7:02pm » |
Quote Modify
|
In any case, temporary and srn347 both certainly tend to sully any thread they touch. Getting back on topic, it was still to be shown that Sr = 1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r) + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r) + 1/(4r+1) + .... = 1/r [ log 2 + 2 tan ] ~ log(2r/) + , where the sum is over = k/2r, k= r-1, r-3, r-5, ..., k 1. In fact it looks like Sr ~ log(2r/) + + log(2)/r - 0.137078/r2 Any takers?
|
« Last Edit: Jan 27th, 2008, 7:26pm by Eigenray » |
IP Logged |
|
|
|
temporary
Full Member
Posts: 255
|
|
Re: harmonic
« Reply #53 on: Jan 27th, 2008, 7:31pm » |
Quote Modify
|
on Jan 27th, 2008, 5:48pm, Obob wrote:To be honest, nobody cares if srn347 and sm347 are/were the same person. Whether you and srn347 are the same person (and I am 99% sure you are) is much more pertinent. |
| Wasn't srn347 ip banned also, or is that just a myth. Also, if what you are saying is true, then 1/1 + 1/2 + ... + 1/r - 1/(r+1) - 1/(r+2) - ... - 1/(2r) + 1/(2r+1) + ... + 1/(3r) - 1/(3r+1) - ... - 1/(4r) + 1/(4r+1) + .... ~ log(2r/) + + log(2)/r - 0.137078/r2
|
|
IP Logged |
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
|
|
|
|