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>> No.11001003 [View]
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11001003

>>11000887
Look Tooker, you can try and discredit the work if you want, and I try to be nice and answer you, even when I know most of your efforts are only to waste people's time, but I'm not going to sit around and engage in every one of your timesand expeditions, especially when you don't answer my disproofs of your "theorem."
Well, except for that last time when you admitted I found a flaw.
In any case, symbolic representations are different than evaluative ones, and (1-1)/(1-1) can be 1 symbolically in your little picture. But we don't even need to go that far.
Like you said,
>I'm not following you here.
And then you go on about "with the infinity symbol properly defined".
No where did I define infinity, you did. In fact, you can't prove those sums equal infinity, it's exactly what you're asking me to prove they aren't, and you know this. So your infinity isn't properly defined.
Then, like I so patiently explained, it's already known that they must converge for at least some b, as an infinite number of 1/2 + bi solutions are already known.
If you think that breaking the original sum into 2 is an illegal equivalence, prove it. Show me the cases where the original sum generates a,b pairs that the 2 do not, or vice versa.

>because it is notoriously impossible to cancel the non-convergence of infinity with the non convergence...
I don't have to prove that it does, it's enough that it could, which leaves open the possibility that it does, and indeed it does for some b.
>but actually, if you worked it out, you find it was impossible.
you already know it's impossible to prove for all b, or that it doesn't converge for some b, get out of here.
>there is, as caught by
that was you replying to yourself, dork.

>> No.10555103 [View]
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10555103

>>10555092
So, sleepy. You're not even trying.

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