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>> No.4828839 [View]
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4828839

Just define the other side as the other side.

>> No.4201899 [DELETED]  [View]
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4201899

>>4201841
Well, imho if you consider how many people they have to train to get a "golden boy", then you must also consider the huge amount of foreigners in USA academia.

>>4201880
because I'm neither a racist, nor obsessed

Going further, you'd have to check and weight by the impact factor

http://en.wikipedia.org/wiki/Impact_factor

on all of these.

Anyway, the paper-system is like capitalism. In a direct way, but also in a methaphorical sense, in that it's pretty much terrible but, of course, better then the other options.

>> No.4145822 [DELETED]  [View]
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4145822

>>4145756
>mfw
choosing nasty bump functions will probably work, but thats really a cheaty way out ;)

here is my (non-piecewise) approach:

taylor expansion

<span class="math">f(x)=sum_{n=0}^\infty \frac{a_n}{n!} x^n[/spoiler]
now <span class="math">f^{(n)}(0)=a_n[/spoiler]. We want <span class="math">f^{(n)}(0)=f(n)[/spoiler], so <span class="math">a_n=sum_{k=0}^\infty \frac{a_k}{k!} n^k[/spoiler], i.e.
<span class="math">f(x)=sum_{n=0}^\infty \frac{1}{n!} (a_n=sum_{k=0}^\infty \frac{a_k}{k!} n^k) x^n[/spoiler].

for <span class="math">a_0[/spoiler] this will work anyway. since 0^n=0 or 1.
For a_1 we get
<span class="math">a_1=a_0+a_1+a_n=sum_{k=2}^\infty \frac{a_k}{k!} n^k[/spoiler].
Therefore

<span class="math">-a_0=a_n=sum_{k=2}^\infty \frac{a_k}{k!} n^k[/spoiler]
which is a restriction for a_2. This goes on, an you're always left with <span class="math">sum_{k={some number}}^\infty \frac{a_k}{k!} n^k[/spoiler]. This goes to 0 however, you you only need to formalize the procedure in a closed for. I'm too tired to do this

>> No.4120905 [View]
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4120905

If this sentence is true, then all results from all the other axioms in this framework lead to false results.

There goes your theory.
Curry's paradox in ya face, bitch.

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