August 4th, 2012, 04:47 AM  #1 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  A challenge to the MMF members
Prove that using calculus. Balarka . 
August 4th, 2012, 05:04 AM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: A challenge to the MMF members 
August 4th, 2012, 05:54 AM  #3  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A challenge to the MMF members Quote:
But the Dirichlet Series for and can be regularized by this calculation: implies And so, This is the algebraic calculation.  
August 4th, 2012, 05:58 AM  #4 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A challenge to the MMF members
If you are interested, I could post my method which uses differential equations involving trigonometric functions.

August 4th, 2012, 06:29 AM  #5 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A challenge to the MMF members
This is actually the famous Grandi's series. My calculations: Satisfies the differential equation: . Now put f(x) = sin(x). A wellknown solution to is And by (1), a particular solution is So, Now put x = 0. eliminate all sin(0) = 0. substitute cos(0) = 1: 
August 4th, 2012, 06:43 AM  #6 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A challenge to the MMF members
Another way to handle the Grandi's series is to use the relation: put z = 0, by the analytic continuation of the zeta function. 
August 4th, 2012, 01:12 PM  #7 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: A challenge to the MMF members
The article says you can rearrange the terms to arrive at any integer solution. I wonder if you can rearrange the terms to arrive at ? Using this order... the sum alternates between 0 and 1 now you regroup like this.. again the sum INSIDE parentheses alternates between 0 and 1, so this second series with different grouping will be or the 1/2 you got is the AVERAGE of 0 and 1 because you used a variable S to represent an indeterminate value, but variables are supposed to represent unknown determinate (convergent) values, then we can be sure that legal algebraic operations may lead to a correct result. 
August 4th, 2012, 09:12 PM  #8  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: A challenge to the MMF members Quote:
But what about the analytic continuation of the zeta function? It clearly says that  
August 5th, 2012, 12:35 AM  #9 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: A challenge to the MMF members
Well, I don't understand too much about analytic continuation of the zeta function. i don't trust the result By definition... now if you put in s = 0 you should get which is clearly 1+1+1+ ... > divergent how do they get 1/2 out of this? they must be using another definition of zeta that i do not understand... 
August 5th, 2012, 12:40 AM  #10  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: A challenge to the MMF members Quote:
 

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