do normies actually buy this shit?

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# do normies actually buy this shit?

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do normies actually buy this shit?

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Guy single-handedly fricked numberphile https://www.youtube.com/watch?v=YuIIjLr6vUA

Ramanujan was based though

>Ramanujan was based though

how's the weather in Calcutta?

pretty good.

what an outdated video

https://arxiv.org/abs/2401.10981

Pretty compelling

>debunk video

>says all the statements made were wrong

>the very first statement made by himself is wrong

You can't say that sum of natural numbers is equal to infinity - infinity is not a number.

He explains later in the video how when infinity is involved, the meaning of the equals sign will change depending on the context

No, he doesn't.

Series is divergent, therefore has no value.

Unless you're accepting different more generalized definitions - but then you may as well accept the one where 1+ 2 + 3 + ... = -1/12

as you should

Normies do not understand anything about an infinite series anon, and most don't know much about finite ones either.

>muh analytic continuation

This is what "complex number" ideologists unironically believe.

that an Indian man can be intelligent?

Obvious wrong, but because a mathematician did it, it's right. And it also developed a new sub branch of maths lmao

You are too moronic to understand what is going on there.

you are too moronic to realise that the math we have is entirely man made.

so yes it is susceptible to mistakes.

there's proof 1=-1,does it make it true? hell no.

>there's proof 1=-1,does it make it true? hell no.

1) no there isnt.

2) if we couldnt ever distinguish between true and false statements in math, then math would be pointless. you arent allowed to say that some proofs are technically correct but dont count because you dont like them. either state that the proof is erroneous or accept the conclusion.

e^(iπ) + 1 = 0

e^(iπ) + e^(iπ)^2 = 0

-e^(iπ)^2 = e^(iπ)

1 = -1

e^(iπ)^2 ≠ 1

>-1^2 ≠ 1

lmfao dumb

-(e^(iπ))^2 = -(e^(iπ)^2) = -1(e^(iπ)^2) = (-(1)e^(iπ)^2) = (-e^(iπ)^2) = -e^(iπ)^2 = 1

Based. Nobody should be injective or boosted.

>>-1^2 ≠ 1

>lmfao dumb

Except that you are the dumb one.

e^iπ^2 = e^(i^2*π^2) = e^(-π^2)

>-(e^(iπ))^2 = -(e^(iπ)^2) = -1(e^(iπ)^2) = (-(1)e^(iπ)^2) = (-e^(iπ)^2) = -e^(iπ)^2 = 1

Again, you're falling victim to your own ambiguous notation. Written properly, your chain of equalities looks like this:

>-(e^(iπ))^2 = -1 * (e^(iπ))^2 ?= (-1 * e^(iπ))^2 = (-(e^(iπ*~~^2 = 1

And clearly the ?= step is incorrect since -(a^2) is not equal to (-a)^2 in general. No more (You)s until you're honest about your intended order of operations.

-1^2 = -1

moron

>-e^(iπ)^2 = e^(iπ)

>1 = -1

I think you're just confusing yourself by excluding parentheses. (e^(iπ))^2 is 1 (which is a fact you use yourself), so both sides of this equation are -1, there's no contradiction.

welp, you fricked up the proof. ignoring that, the flaw is that f(x)=x^2 is not an injective function.

Srinivāsa Aiyangār Rāmānujan (ஸ்ரீனிவாஸ ஐயங்கார் ராமானுஜன்) was kino.

It doesn't equal -1/12. If you redefine basic algebra so you can assign non-convergent values to symbols then yes, but all of math collapses into madness if you permit this.

But it works

way to show your outdated knowledge

Wtf are you even asking? This is no different than the proofs that 1=2 or whatever. It's just your average "using correct math with incorrect assumptions returns contradictory results" thought experiment. I know this is a shitpost thread, but I don't understand what there is to "buy" about this.

here is something new for you to learn

you morons do realize that math is made up right?