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1d
asked What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
1d
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
1d
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
1d
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.
1d
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.
1d
revised Non-standard numbers and exponential form of Zeta function
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1d
revised Non-standard numbers and exponential form of Zeta function
added 205 characters in body
1d
revised Non-standard numbers and exponential form of Zeta function
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2d
asked Non-standard numbers and exponential form of Zeta function
Aug
26
revised definite integral of $x^2e^{-x^2}$
this is not related to Euler's constant
Aug
26
suggested approved edit on definite integral of $x^2e^{-x^2}$
Aug
23
revised Conventions adopted for extended reals
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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.