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 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 $1-1+1-1+1-1+1-1+...= \frac{1}{2}$ 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 $\sum_{k=0}^{n}(-1)^{k}=\frac{(-1)^n+1}{2}\neq \frac{1}{2}\;\;\forall \;\;n\in\mathbb{N}$
August 4th, 2012, 05:54 AM   #3
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Re: A challenge to the MMF members

Quote:
 Originally Posted by ZardoZ $\sum_{k=0}^{n}(-1)^{k}=\frac{(-1)^n+1}{2}\neq \frac{1}{2}\;\;\forall \;\;n\in\mathbb{N}$
Actually, yes you are right.

But the Dirichlet Series for $a_n= (-1)^n$ and  can be regularized by this calculation:

$1-1+1-1+1-1+...= S$

$1-(1-1+1-1+1-1+...)=S$

$1-S= S$ implies $2S= 1$

And so,

$S= \frac{1}{2}$

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: $y= f(x) + f'(x) + f'#39;(x) + . . . \;\;\;\;\; (1)$ Satisfies the differential equation: $y' - y + f(x) = 0$. Now put f(x) = sin(x). A well-known solution to $y' - y + \sin(x) = 0$ is $y= \frac{1}{2} \cdot $\sin(x) + \cos(x)$$ And by (1), a particular solution is $y= \sin(x) + \cos(x) - \sin(x) - \cos(x) + \sin(x) + \cos(x) - \sin(x) - \cos(x) + . . .$ So, $y= \sin(x) + \cos(x) - \sin(x) - \cos(x) + \sin(x) + \cos(x) - \sin(x) - \cos(x)+ . . . = \frac{1}{2} \cdot $\sin(x) + \cos(x)$$ Now put x = 0. eliminate all sin(0) = 0. substitute cos(0) = 1: $1-1+1-1+1-1+...= \frac{1}{2} \cdot 1 = \frac{1}{2}$ $\text{QED}$
 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: $\eta(z)= $$1-2^{1-z}$$\zeta(z)$ put z = 0, $\eta(0)= -1 \cdot \zeta(0) = -1 \cdot \frac{-1}{2} = \frac{1}{2}$ 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: 234 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 $sqrt{2}$ ? Using this order... $1-1+1-1+1-...$ the sum alternates between 0 and 1 now you regroup like this.. $1-(1-1+1-1+1-...)$ again the sum INSIDE parentheses alternates between 0 and 1, so this second series with different grouping will be $1-0=1$ or $1-1=0$ 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
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Re: A challenge to the MMF members

Quote:
 Originally Posted by agentredlum The article says you can re-arrange the terms to arrive at any integer solution, i wonder if you can re-arrange the terms to arrive at $sqrt{2}$ ? Using this order... $1-1+1-1+1-...$ the sum alternates between 0 and 1 now you regroup like this.. $1-(1-1+1-1+1-...)$ again the sum INSIDE parenthesis alternates between 0 and 1 so this second series with different grouping will be $1-0=1$ or $1-1=0$ 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.
I know that Grandi's formulation is rather questionable.
But what about the analytic continuation of the zeta function? It clearly says that $\zeta(0)= \frac{-1}{2}$

 August 5th, 2012, 12:35 AM #9 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 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 $\zeta(0)=-\frac{1}{2}$ By definition... $\zeta(s)= \sum_{n=1}^{\infty} \frac{1}{n^s}$ now if you put in s = 0 you should get $\zeta(0)= \sum_{n = 1}^{\infty} \frac{1}{n^0}$ which is clearly 1+1+1+ ... -> $\infty\$ 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
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Re: A challenge to the MMF members

Quote:
 Originally Posted by agentredlum they must be using another definition of zeta that i do not understand...
[color=#000000]I have seen such strange proofs of such results in Ramanujan's notebooks, but to tell you the truth my knowledge of mathematics is not so advanced. I like to keep my mathematics to Earthly grounds. [/color]

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