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2d
comment Non-standard numbers and exponential form of Zeta function
@mercio I highly appreciate your criticism and I found the mistake. I have corrected the formulas and pictures so now it should work well. $4(\omega_-)^2+0+(-1/2)+(-1/2)=4\tau^2$. The formulas for the exponentiation are now simplier in that they use simply Bernoulli numbers. And also it seems a more general formula works, $\operatorname{st}(\tau+y)^x=-x\zeta(1-x,1/2-y)$ although I should verify it. If so, the whole role of Hurwitz Zeta becomes clear.
2d
comment Non-standard numbers and exponential form of Zeta function
@mercio what do u mean?
2d
comment Non-standard numbers and exponential form of Zeta function
@mercio the number of Gaussian integers with positive real and imaginary parts not exceeding n (6th picture) is n(n+1)/2 which is partial sum of 1+2+3+4+... en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$
Now I fixed the problem. Sorry. In this form it is exactly what I want.
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@Daniel Fischer it came from there but modified. I just explained why there are no factorials.
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@michaelrccurtis it came from Maclauren's expansion for $1/(x + 1/2)$
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@Daniel Fischer well I would be OK with any of the summation methods, such as mean vanues etc.
2d
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@michaelrccurtis Bernoulli numbers.
2d
comment Non-standard numbers and exponential form of Zeta function
@Daniel Fischer maybe it is not totally exact, but it says that the objects are introduced either trough sets or series. The standard part is the Ramanujan's sum of the series. Each diverging series represents a non-standard number
2d
comment Non-standard numbers and exponential form of Zeta function
@Daniel Fischer all the following is exactly specifying what the objects are and how they behave
2d
comment Non-standard numbers and exponential form of Zeta function
@Asaf Karagila Particularly, the zeros of Zeta function are those numbers that satisfy $\operatorname{st}\omega_-^{1-z}=0$. This is quite an algebraic question.
2d
comment Non-standard numbers and exponential form of Zeta function
@Asaf Karagila I think the idea to represent Riemann zeta in exponential form may be useful. To me it is comparable with representing sine and cosine through exponents using complex numbers. At least to me this idea sheds light on the algebraic role of zeta function and Bernoulli numbers.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
"would seem to contradict the definition" - it would not if $\varepsilon^2=0$ as in dual/parabolic numbers. The given definition as I figured out leads exactly to the dual numbers system.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
Thank you for pointing out Euler's work. It seems this definition leads exactly to the system of dual numbers: en.wikipedia.org/wiki/Dual_number In dual numbers we have exactly $\exp(\varepsilon)=1+\varepsilon$ although the first form of the definition does not work because division by $\varepsilon$ is undefined.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
@Hurkyl well I think this question can be closed because in dual numbers en.wikipedia.org/wiki/Dual_number we have exactly that: $\exp(\varepsilon)=1+\varepsilon$
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
@Hurkyl what if to consider $\varepsilon$ not just an infinitesimal but as just some new algebraic element with this property?
Aug
23
comment Conventions adopted for extended reals
@mercio no, we get 6=5.
Aug
23
comment Conventions adopted for extended reals
@mercio wrong. if we define 0/0=0, and devide the both parts of $0\cdot 17=0$ by 0, we get 0=0, no contradiction.
Aug
23
comment Conventions adopted for extended reals
@mercio what contradictions appear if we define $0/0=0$?
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
@Hurkyl literally equal.