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Question Number 127641 by Study last updated on 31/Dec/20

Answered by Dwaipayan Shikari last updated on 31/Dec/20

$$1.2.3.4.5...=e^{-{\zeta }^{\prime }\left(0\right)}=e^{log\left(\sqrt{2\pi }\right)}=\sqrt{2\pi }$$

Commented by MJS_new last updated on 31/Dec/20

$$\mathrm{great}!\mathrm{from}\mathrm{now}\mathrm{on}\mathrm{these}\mathrm{shall}\mathrm{be}\mathrm{my}\mathrm{formulas}$$$$\mathrm{of}\mathrm{the}\mathrm{circle}$$$$p={\mathrm{e}}^{-2{\zeta }^{\prime }\left(0\right)}r$$$$A={\left(\frac{\mathrm{\infty }!\sqrt{2}r}{2}\right)}^2$$

Commented by Dwaipayan Shikari last updated on 31/Dec/20

$$Hahhahh!!$$$$1.2.3.4.5...={\mathit{e}}^{-{\mathit{\zeta }}^{{}^{\prime }}\left(0\right)}=\sqrt{2\pi }\left(Ramanujanproduct\right)$$$$\mathit{T}\mathit{h}\mathit{e}\mathit{q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{h}\mathit{a}\mathit{s}\mathit{n}\mathit{o}\mathit{m}\mathit{e}\mathit{a}\mathit{n}\mathit{i}\mathit{n}\mathit{g}\mathit{b}\mathit{u}\mathit{t}\mathit{a}\mathit{n}\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}$$$$$$😁

Commented by Dwaipayan Shikari last updated on 31/Dec/20

$$\mathcal{A}={((-\frac{1}{2})!r)}^2$$

Commented by MJS_new last updated on 31/Dec/20

$$yes$$

Commented by Study last updated on 31/Dec/20

$$isithasanotherfurmolasirteacher??$$

Commented by Dwaipayan Shikari last updated on 31/Dec/20

$$Haveagreatyearsir!$$$$\lfloor {\int }_0^1\frac{1}{\sqrt[2021]{x^{2020}-x^{2021}}}dx\rfloor =2021$$

Commented by Study last updated on 01/Jan/21

$$infinitebutnoknownnumbersowhy$$$$getafactorial?$$

Commented by Study last updated on 02/Jan/21

$$?????$$

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