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[eqn]\displaystyle \zeta \left( \sum_{k=1}^{\infty} \frac{9}{10^k}

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[eqn]\displaystyle \zeta \left( \sum_{k=1}^{\infty} \frac{9}{10^k} \right) = \zeta(1) = \infty[/eqn] Therefore, [math]\displaystyle 0.999\ldots = 1[/math]
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I kek'd
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>0.999...=infinity so 0.9999...=1

lmfao anime damsel you are high AF right now
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>>8699108
>lmfao anime damsel you are high AF right now
nope, you are you fucking brainlet
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>>8699138
you just took an infinite series and then concluded that 0.999...=1 now go back to your shit gween tea cartoon.
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[eqn]\zeta(-4)=\zeta(-2)=0[/eqn]

Therefore, [math]-4=-2[/math]
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>>8699079
W-what are you trying to say?

are you seriously implying that f(x)=f(y) => x=y is true?
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>>8699108
>>8699144
fucking brainlets

>>8699525
yes. the only pole of the riemann zeta is 1
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>>8699535
Why should this be true though?
[math]\displaystyle \zeta \left( \sum_{k=1}^{\infty} \frac{9}{10^k} \right) = \zeta(1)[/math]

It is obviously true if your conclousion is true but that is a pretty awful proof then.
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>>8700105
this, the proof already assumes that 0.9999... = 1
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>>8700105
>>8700288
why don't you compute it?
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>>8700320
its not possible without the knowledge that 0.999..=1.
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>>8700105
>It is obviously true if
It's not true. The function isn't define at [math] z=1 [/math]. There is no valid equation involving the a value [math] \zeta(1) [/math].
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>>8700345
His proof goes 0.999...=1 therfore

\displaystyle \zeta \left( \sum_{k=1}^{\infty} \frac{9}{10^k} \right) = \zeta(1)

therfore 0.999...=1.

What is not obvious there, the middel part is irrelevant, his asumption is the same as his conclusion.
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>>8700349
>isn't defined
it's a meromorphic function with a pole of order 1 at 1, genius
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What if there is a 0.0.....001?
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>>8700370
What are you trying to say? Saying it's a pole is also just saying it's not defined there, plus saying it's well defined in a neighborhood of that point.
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>>8700380
no. there are different kinds of isolated singularities of holomorphic functions. a pole of order 1 is a very specific one. it means the function can be written as a power series at point 1 as

[math] \sum_{i=-1}^\infty a_i (x-1)^i [/math]

it's not that it's just "not defined" because it doesn't take a value in C, it literally takes infinity as a value at that point, an infinity with a very specific behavior.
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>>8699079
YOU WHORE!
TAKE THAT BACK
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>>8700374
>What if there is a 0.0.....001?
Excellent question.
What if there was a virtual 1?
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>>8699181
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>>8699079

Go count primes with Cocoa instead of doing this useless shit.
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>>8702375
What secret does Cocoa have about the Zeta function that she does not want to tell us? She counts primes too quick.
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>>8699079
>>8701449
>>8702375
Is there a reason that post-millenium anime girls are starting to look more and more like babies?
Thread posts: 24
Thread images: 3


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