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>> No.7683515 [View]
File: 923 KB, 1280x778, what is this all about.png [View same] [iqdb] [saucenao] [google]
7683515

I can try to point out the occurrence of -1/12 in a completely classical computation involving a smoothing process, that’s not analytic continuation. Here we replace the sum with one that eventually requires the basic Taylor series

[math]\dfrac {r^2} { \log(1+r)^2 } = 1 + r + \dfrac{1}{1! \, 2! \, 3!} r^2 + O(r^3)[/math]

and observe that 2!3! is 12.

As a precursor, consider the function [math]f(k):=k^2[/math].
It's value at the point k=43is 9 and it's value at the point k=4 is 16.

Now say we don't know which point k we really deal with, we only know that it's in the interval [3, 4]. A good estimate for the function evaluated on the unknown value should be the average

[math] \langle f(k) \rangle = \int_{3}^{4} \, f(k) \, dk = 12.333 = 9 + \frac {10} {3} [/math]

The continuous version of the evaluation differs by a term of [math] - \frac {10} {3} [/math].

Now the sum 0 + 1 + 2 + 3 + … equals the limit [math] \lim_{z \to 1} [/math] of the sum

[math] 0 + 1 \, z^1 + 2 \, z^2 + 3 \, z^3 + … [/math]

For z in (0,1) this can be computed

[math] \sum_{k=0}^\infty k \, z^k = z \dfrac {d} {dz} \sum_{k=0}^\infty z^k = z \dfrac {d} {dz} \dfrac {1} {1-z} = \dfrac{z} { (z-1)^2 } [/math]

As already knew and now can read off explicitly, this expression diverges for z to 1.
So let's consider the sum of smooth deviations around integers values. With

[math] \langle k\,z^k \rangle = z \dfrac {d} {dz} \langle z^k \rangle = z \dfrac {d} {dz} \langle e^{k \log(z) } \rangle = z \dfrac {d} {dz} \dfrac { z^{k'} } { \log(z) } |_{k}^{k+1} [/math].

we find the sum [math] \sum_{k=0}^n \langle k \, z^k \rangle[/math] involves canceling upper and lower bounds and we're left with [math] \dfrac {1} {\log(z)^2} [/math] plus terms suppressed by [math]z^n[/math].

Using the expansion of the log above, we find the difference

[math] \sum_{k=0}^\infty (k \, z^k - \langle k \, z^k \rangle ) = \dfrac {z} {(z-1)^2} - \dfrac {1} {\log(z)^2} = - \dfrac {1} {12} + O( (z-1)^1 )[/math]

>> No.7470713 [View]
File: 923 KB, 1280x778, what is this all about.png [View same] [iqdb] [saucenao] [google]
7470713

Well, are you aware of how to sketch elephants etc.?

en.wikipedia.org/wiki/Sketch_(mathematics)

http://ncatlab.org/nlab/show/Lawvere+theory

Can you express your thoughts in those terms or is this off? I don't know at which level you approach this.
Lawvere deals with universal algebra here, where afaik people can be more sparse with quantifiers.
An elaboration on the difference is given here
https://en.wikipedia.org/wiki/Universal_algebra#Examples

I recognize where you stole your image from.
Now the dependent type picture is the most straight forward to me, and there you have (constructive) Pi's and Sigmas. I general find the topos interpretations of those so complicated, and then mapping from an abstract theory to a concrete model would be even worse.
If that's related or helps: I tired to improve nLab exposition on the existential quantifier as dependent sum functor some time ago, and I made notes on all of those and tried to be more explicit here
axiomsofchoice.org/babbys_first_hott
axiomsofchoice.org/dependent_type_theory
axiomsofchoice.org/dependent_sum_functor
axiomsofchoice.org/dependent_product_functor

>The arithmetic hierarchy
Didn't see that coming.
Mhm, the ITT side where the quantifiers are those dependent type formers, all computational stuff should actually be covered by someone, no?

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