Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 112655 by mathdave last updated on 09/Sep/20

$$provethat$$$$\mathrm{\infty }!=\sqrt{2\pi }$$

Commented by MJS_new last updated on 09/Sep/20

$$\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{people}\mathrm{with}\mathrm{less}\mathrm{knowledge}\mathrm{to}$$$$\mathrm{understand}\mathrm{this},\mathrm{you}\mathrm{should}\mathrm{explain}\mathrm{the}$$$$\mathrm{background}\mathrm{of}\mathrm{it}.\mathrm{this}\mathrm{would}\mathrm{be}\mathrm{helpful}.\mathrm{show}$$$$\mathrm{us}\mathrm{why}\mathrm{the}\mathrm{faculty}\mathrm{of}\mathrm{infinity}\mathrm{has}\mathrm{the}\mathrm{same}$$$$\mathrm{value}\mathrm{as}\mathrm{the}\mathrm{Gaussian}\mathrm{integral}$$

Commented by mathdave last updated on 09/Sep/20

$$alrightsir$$

Answered by mathdave last updated on 09/Sep/20

$$solution$$$$since$$$$\eta \left(s\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{k+1}{k}^{-s}=(1-2^{1-s})\zeta \left(s\right)$$$${\eta }^{{}^{\prime }}\left(s\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{\ln k}{k^s}={\zeta }^{{}^{\prime }}\left(s\right)(1-2^{1-s})+\zeta \left(s\right)2\ln 2$$$${\eta }^{{}^{\prime }}\left(0\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}{(-1)}^k\ln k=-\ln \left(1\right)+\ln \left(2\right)-\ln \left(3\right)+----$$$${\eta }^{{}^{\prime }}\left(0\right)=\ln \left(\frac{\pi }{2}\right)-{\eta }^{{}^{\prime }}\left(0\right)\left[{walli}^{\prime }sproduct\right]$$$${\eta }^{{}^{\prime }}\left(0\right)=\frac{1}{2}\ln \left(\frac{\pi }{2}\right)=\ln \left(\sqrt{\frac{\pi }{2}}\right).........\left(1\right)$$$$again$$$${\eta }^{{}^{\prime }}\left(0\right)=-{\zeta }^{{}^{\prime }}\left(0\right)+\zeta \left(0\right)2\ln 2=-[{\eta }^{{}^{\prime }}\left(0\right)-\ln \left(\pi \right)]-\ln 2$$$$2{\eta }^{{}^{\prime }}\left(0\right)=\ln \left(\frac{\pi }{2}\right)$$$${\eta }^{{}^{\prime }}\left(0\right)=\frac{1}{2}\ln \left(\frac{\pi }{2}\right)=\ln \left(\sqrt{\frac{\pi }{2}}\right).........\left(2\right)$$$$but\ln \left(\sqrt{\frac{\pi }{2}}\right)=\ln \left(\prod _{k=1}k\right)-\ln 2$$$$\ln \left(\sqrt{\frac{\pi }{2}}\right)=\ln (\mathrm{\infty }!)-\ln 2$$$$\ln (\mathrm{\infty }!)=\ln 2+\ln \left(\sqrt{\frac{\pi }{2}}\right)$$$$\ln (\mathrm{\infty }!)=\ln \left(2\sqrt{\frac{\pi }{2}}\right)=\ln \left(\sqrt{2\pi }\right)$$$$\ln (\mathrm{\infty }!)=\ln \left(\sqrt{2\pi }\right)$$$$\because \mathrm{\infty }!=\sqrt{2\pi }Q.E.D$$$$bymathdave\left(09/09/2020\right)$$$$$$

Commented by Her_Majesty last updated on 09/Sep/20

$$showthat{X}^{⟨7⟩}\stackrel{!}{\Rightarrow }{(X-1)}^{⟨-3⟩}$$$$proof:$$$$X^7\subset X^{⟨7\mid -9⟩}\supset {(X+1)}^{⟨-3\mid -5\mid =7⟩}$$$$but{(X+1)}^{⟨-3\mid -5\mid =7⟩}={(X-5)}^{⟨-3\mid -5\mid =7⟩}$$$$and{(X-5)}^{⟨-3\mid -5\mid =7⟩}\subset {(X-6)}^{⟨7\mid -9⟩}\supset {(X-6)}^{⟨7⟩}$$$$butweknow(x+u)\stackrel{!}{\Rightarrow }(x+v)\iff v=\{\begin{array}{l}u+5\\ u-7\end{array}$$$$(Beck-Runnel-Axiom)$$$$\Rightarrow (x-6)\stackrel{!}{\Rightarrow }(x-1)\Rightarrow$$$$\Rightarrow {(X-6)}^{⟨7⟩}\stackrel{!}{\Rightarrow }{(X-1)}^{⟨-3⟩}$$$$butwehaveshown$$$${(X-6)}^{⟨7⟩}\subset {(X-6)}^{⟨7\mid -9⟩}\supset {(X-5)}^{⟨-3\mid -5\mid \stackrel{2o}{-}7⟩}=$$$$={(X+1)}^{⟨-3\mid -5\mid =7⟩}\subset X^{⟨7\mid -9⟩}\supset X^{⟨7⟩}\Rightarrow$$$$\Rightarrow X^{⟨7⟩}\stackrel{!}{\Rightarrow }{(X-1)}^{⟨-3⟩}q.e.d.$$$$$$$$[b.t.w.becauseoftheAxiomsofInversion$$$${(X+u)}^{⟨-3\mid 7⟩}={(X+v)}^{⟨6⟩}withv=\{\begin{array}{l}u+3\\ u-9\end{array}$$$$wealsoshowedthat{X}^{⟨7⟩}\stackrel{!}{\Rightarrow }(X+2)which$$$$isessentialto{Simon}^{\prime }sApproach]$$

Commented by Her_Majesty last updated on 09/Sep/20

$$\left(1\right)findallpossiblesourcesof{(X-1)}^{⟨+5⟩}$$$$\left[supereasy\right]$$$$\left(2\right)findpossiblesourcesof$$$$\left(a\right){(X-3)}^{⟨+5\mid -7⟩}and\left(b\right){(X+4)}^{⟨-5\mid -7⟩}$$$$andtheircenterangles\left[hard\right]$$

Commented by MJS_new last updated on 09/Sep/20

$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{this}???$$

Commented by 1549442205PVT last updated on 09/Sep/20

$$\mathrm{Please},\mathrm{Sir}\mathrm{Mathdave}\mathrm{can}\mathrm{explain}$$$$\mathrm{at}\mathrm{tenth}\mathrm{line}$$$$\ln \sqrt{\frac{\pi }{2}}=\ln \left(\underset{\mathrm{k}}{\mathrm{\Pi }}\mathrm{k}\right)-\ln 2$$$$\prod _{\mathrm{k}}\mathrm{k}\mathrm{is}\mathrm{an}\mathrm{infinite}\mathrm{value}\mathrm{while}\sqrt{\pi }<2$$$$(\mathrm{Math}\mathrm{is}\mathrm{beautiful}\mathrm{because}\mathrm{of}\mathrm{its}$$$$\mathrm{precision})$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com