Math

[Anonymous No.4711621]

0 = 0
0 + 0 + 0 + 0 + ... = 0
(1 - 1) + (1 - 1) + (1 - 1) + ... = 0
1 - 1 + 1 - 1 + 1 - 1 + ... = 0
1 + (-1 + 1) + (-1 + 1) + ... = 0
1 + 0 + 0 + 0 + ... = 0

1 = 0

Problem mathfags?

[Anonymous No.4711627]

Not with poorly formed mathematics, no no problem at all.

[Anonymous No.4711644]

Yes there is a problem, you're missing a -1 in the second to last step.
But that was good try indeed, regards.

[Anonymous No.4711648]

>>4711627
> Can't figure out what's wrong.
> Criticizes notation instead.

[Anonymous No.4711654]

>>4711644
It's an infinite sequence, there's no end.

[Anonymous No.4711657]

>>4711648
>no absolute convergence

[Anonymous No.4711672]

>1 + (-1 + 1) + (-1 + 1) + ... = 0

So, 1 plus a series of zero is suddenly zero?

[Anonymous No.4711674]

>implying convergence

[Anonymous No.4711679]

>>4711654
exactly why OP is wrong :)

[Anonymous No.4711697]

>>4711621
OP just pulled a (+1) out of his ass and added it to the sequence.

[Anonymous No.4711698]

>>4711672
Logically arrived from a true statement:
0 + 0 + 0 + 0 + ... = 0
Unless there's a flaw in the logical steps. It would be nice for you to point out which.

[Anonymous No.4711704]

>>4711698
see >>4711657

[Anonymous No.4711711]

2/10

[Anonymous No.4711715]

>>4711704
0 + 0 + 0 + 0 + ... does not converge to 0?

[Anonymous No.4711730]

>>4711715
it does

problem comes when associativity is assumed.

(1 - 1) + (1 - 1) + (1 - 1) + ... = 0
1 - 1 + 1 - 1 + 1 - 1 + ... = 0
1 + (-1 + 1) + (-1 + 1) + ... = 0

not allowed unless absolute convergence

[Anonymous No.4711741]

>>4711715

It sure does, but 1 + (0) + (0)... does not.

[Anonymous No.4711744]

>>4711730
>it does
>hur 0*infinity = 0 hur.

[Anonymous No.4711750]

>>4711744
>thinking a limit of a series is a simple multiplication
>hurr

[Anonymous No.4711762]

>>4711744

infinite series of addition =/= multiplication

[Anonymous No.4711772]

>0 + 0 + 0 + 0 + ... = 0
Prove this OP. I'll believe you when you've summed an infinite number of zeros in your head.

[Anonymous No.4711791]

<div class="math">1+1+1+1=\sum_{i=1} \frac{1}{i^0} = \zeta \left( 0 \right) = -\frac{1}{2} </div>
Problem Riemann hypothe-fags?

[Anonymous No.4711805]

<div class="math"> 1+1+1+1 + \ldots </div> <div class="math"> \sum_{i=1}^{\infty} \frac{1}{i^0} </div> <div class="math"> \zeta \left( 0 \right) = -\frac{1}{2} </div>
Problem Riemann hypothe-fags?

[Anonymous No.4711811]

>>4711805
But...but...

[Anonymous No.4711819]

Suppose the series converges to a real number S.
<div class="math"> S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots [\eqn]
<div class="math"> 1 - S = 1 - \left( 1 - 1 \right) - \left( 1 - 1 \right) - left( 1 - 1 \right) +-\left( 1 - 1 \right) - \ldots [\eqn]
1 - S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots
1-S = S
Therefore <span class="math"> S = \frac{1}{2} [/spoiler]</div></div>

[Anonymous No.4711823]

Suppose the series converges to a real number S.
<div class="math"> S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots </div>
<div class="math"> 1 - S = 1 - \left( 1 - 1 \right) - \left( 1 - 1 \right) - left( 1 - 1 \right) +-\left( 1 - 1 \right) - \ldots </div>
<div class="math"> 1 - S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots </div>
<div class="math"> 1-S = S</div>
Therefore <span class="math"> S = \frac{1}{2} [/spoiler]

[Anonymous No.4711827]

Suppose the series converges to a real number S.
<div class="math"> S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + \left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots </div>
<div class="math"> 1 - S = 1 - \left( 1 - 1 \right) - \left( 1 - 1 \right) - \left( 1 - 1 \right) - \left( 1 - 1 \right) - \ldots </div>
<div class="math"> 1 - S = \left( 1 - 1 \right) + \left( 1 - 1 \right) + \left( 1 - 1 \right) + \left( 1 - 1 \right) + \ldots </div>
<div class="math"> 1-S = S</div>
Therefore <span class="math"> S = \frac{1}{2} [/spoiler]

[Anonymous No.4711833]

>>4711805
<div class="math"> 1+2+3+4+\ldots = \zeta \left( -1 \right) = - \frac{1}{12} </div>

[Anonymous No.4711849]

0 = 0
0 + 0 + 0 + 0 + ... = 0
(1 - 1) + (1 - 1) + (1 - 1) + ... = 0
1 - 1 + 1 - 1 + 1 - 1 + ... = 0
1 + (-1 + 1) + (-1 + 1) + ... ( -1 goes here fucktard)= 0
1 + 0 + 0 + 0 + ... = 0

1 = 0

This has got to be the lamest troll math I've seen

[Anonymous No.4711852]

>>4711827

Shouldn't

1 − S = 1 − (1 − 1) − (1 − 1) − (1 − 1) − (1 − 1) −...

be

1 − S = 1 − (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) +...

?

[Anonymous No.4711867]

>>4711852
You have to subtract each term in the series when you subtract S. So there is a minus before each term. Then you just distribute through the negative and rearrange the parentheses.