Talk About Network

Google


Register and Login
Nick
Password
Register create new account Sign up is FREE and you can post replies, new topics, bookmark posts and more!
Recover lost password


Education > Math Recreational > Re: Root of Div...
Latest [ Topics | Posts ] Archive Post A New Topic Post a Reply
<< Topic < Post Post 2 of 2 Topic 2785 of 2912
Post > Topic >>

Re: Root of Divergent Series

by William Elliot <marsh@[EMAIL PROTECTED] > May 19, 2008 at 03:25 AM

On Sun, 18 May 2008, Jon G. wrote:

> |E| is the magnitude of vector E.
> all limits are as n approaches infinity
> ln e = 1
>
> The power series 1 + x + x^2 + x^3 + ... + x^n  diverges when
> x>=1 and has the root,
>
Huh?  1 + x + x^2 +..+ x^n
is not a power series and the notion of divergence doesn't apply.

1 + x + x^2 + x^3 + ...
	is a power series and converges	iff x in (-1,1)

The root of what?
	1 + x + x^2 +..+ x^n = 0
or
	1 + x + x^2 + x^3 + ...  = 0
?

(1 - x^(n+1))/(1 - x) = 0

x = any (n+1)-st root of 1 except 1
and has a real solution iff x = -1 and n is even.

or 1/(1 - x) = 0;  no solution

> x = lim ln[(n!(n+1)|E|^2 - n!e^2)/(n!e-(n+1))]
>
> where \
>
> |E|^2=(1/0!)^2 + (1/1!)^2 + (1/2!)^2 + (1/3!)^2 + ... + (1/n!)^2
>
Huh?  E is not a constant.  Do you mean E_n?

> Proof
>
> 1+lim{(1+2+3+...+n)ln[(n!(n+1)|E|^2 - n!e^2)/(n!e-(n+1))]=0
>
> lim ln[(n!(n+1)|E|^2 - n!e^2)/(n!e-(n+1))]=-lim 1/(1+2+3+...+n)=0
>
> lim [(n!(n+1)|E|^2 - n!e^2)/(n!e-(n+1))]=1
>
> lim 1/(n!e-(n+1)) = lim 1/(n!(n+1)|E|^2 - n!e^2)
>
> 0=0
>
> E.O.P.
>
E.O.P ?

Error Option Provided?
 




 2 Posts in Topic:
Root of Divergent Series
"Jon G." <jo  2008-05-18 05:54:10 
Re: Root of Divergent Series
William Elliot <marsh@  2008-05-19 03:25:46 

Post A Reply:
  Go here to Signup

AddThis Feed Button


About - Advertising - Contact - Frequently Asked Questions - Privacy Policy - Terms of Use - Signup

Contact
tan12V112 Fri Nov 21 7:51:59 CST 2008.