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Question Number 112655 by mathdave last updated on 09/Sep/20
prove that   ∞!=(√(2π))
$${prove}\:{that}\: \\ $$$$\infty!=\sqrt{\mathrm{2}\pi} \\ $$
Commented by MJS_new last updated on 09/Sep/20
if you want people with less knowledge to  understand this, you should explain the  background of it. this would be helpful. show  us why the faculty of infinity has the same  value as the Gaussian integral
$$\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{people}\:\mathrm{with}\:\mathrm{less}\:\mathrm{knowledge}\:\mathrm{to} \\ $$$$\mathrm{understand}\:\mathrm{this},\:\mathrm{you}\:\mathrm{should}\:\mathrm{explain}\:\mathrm{the} \\ $$$$\mathrm{background}\:\mathrm{of}\:\mathrm{it}.\:\mathrm{this}\:\mathrm{would}\:\mathrm{be}\:\mathrm{helpful}.\:\mathrm{show} \\ $$$$\mathrm{us}\:\mathrm{why}\:\mathrm{the}\:\mathrm{faculty}\:\mathrm{of}\:\mathrm{infinity}\:\mathrm{has}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{value}\:\mathrm{as}\:\mathrm{the}\:\mathrm{Gaussian}\:\mathrm{integral} \\ $$
Commented by mathdave last updated on 09/Sep/20
alright sir
$${alright}\:{sir} \\ $$
Answered by mathdave last updated on 09/Sep/20
solution   since  η(s)=Σ_(k=1) ^∞ (−1)^(k+1) k^(−s) =(1−2^(1−s) )ζ(s)  η^′ (s)=Σ_(k=1) ^∞ (−1)^k ((lnk)/k^s )=ζ^′ (s)(1−2^(1−s) )+ζ(s)2ln2  η^′ (0)=Σ_(k=1) ^∞ (−1)^k lnk=−ln(1)+ln(2)−ln(3)+−−−−  η^′ (0)=ln((π/2))−η^′ (0)[walli′s product]  η^′ (0)=(1/2)ln((π/2))=ln((√(π/2))).........(1)  again  η^′ (0)=−ζ^′ (0)+ζ(0)2ln2=−[η^′ (0)−ln(π)]−ln2  2η^′ (0)=ln((π/2))  η^′ (0)=(1/2)ln((π/2))=ln((√(π/2))).........(2)  but ln((√(π/2)))=ln(Π_(k=1) k)−ln2  ln((√(π/2)))=ln(∞!)−ln2  ln(∞!)=ln2+ln((√(π/2)))  ln(∞!)=ln (2(√(π/2)))=ln((√(2π)))  ln(∞!)=ln((√(2π)))  ∵ ∞!=(√(2π))                   Q.E.D  by mathdave(09/09/2020)
$${solution}\: \\ $$$${since} \\ $$$$\eta\left({s}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {k}^{−{s}} =\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)\zeta\left({s}\right) \\ $$$$\eta^{'} \left({s}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{ln}{k}}{{k}^{{s}} }=\zeta^{'} \left({s}\right)\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)+\zeta\left({s}\right)\mathrm{2ln2} \\ $$$$\eta^{'} \left(\mathrm{0}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{ln}{k}=−\mathrm{ln}\left(\mathrm{1}\right)+\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{ln}\left(\mathrm{3}\right)+−−−− \\ $$$$\eta^{'} \left(\mathrm{0}\right)=\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)−\eta^{'} \left(\mathrm{0}\right)\left[{walli}'{s}\:{product}\right] \\ $$$$\eta^{'} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{ln}\left(\sqrt{\frac{\pi}{\mathrm{2}}}\right)………\left(\mathrm{1}\right) \\ $$$${again} \\ $$$$\eta^{'} \left(\mathrm{0}\right)=−\zeta^{'} \left(\mathrm{0}\right)+\zeta\left(\mathrm{0}\right)\mathrm{2ln2}=−\left[\eta^{'} \left(\mathrm{0}\right)−\mathrm{ln}\left(\pi\right)\right]−\mathrm{ln2} \\ $$$$\mathrm{2}\eta^{'} \left(\mathrm{0}\right)=\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\eta^{'} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{ln}\left(\sqrt{\frac{\pi}{\mathrm{2}}}\right)………\left(\mathrm{2}\right) \\ $$$${but}\:\mathrm{ln}\left(\sqrt{\frac{\pi}{\mathrm{2}}}\right)=\mathrm{ln}\left(\underset{{k}=\mathrm{1}} {\prod}{k}\right)−\mathrm{ln2} \\ $$$$\mathrm{ln}\left(\sqrt{\frac{\pi}{\mathrm{2}}}\right)=\mathrm{ln}\left(\infty!\right)−\mathrm{ln2} \\ $$$$\mathrm{ln}\left(\infty!\right)=\mathrm{ln2}+\mathrm{ln}\left(\sqrt{\frac{\pi}{\mathrm{2}}}\right) \\ $$$$\mathrm{ln}\left(\infty!\right)=\mathrm{ln}\:\left(\mathrm{2}\sqrt{\frac{\pi}{\mathrm{2}}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{2}\pi}\right) \\ $$$$\mathrm{ln}\left(\infty!\right)=\mathrm{ln}\left(\sqrt{\mathrm{2}\pi}\right) \\ $$$$\because\:\infty!=\sqrt{\mathrm{2}\pi}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\left(\mathrm{09}/\mathrm{09}/\mathrm{2020}\right) \\ $$$$ \\ $$
Commented by Her_Majesty last updated on 09/Sep/20
show that X^(⟨7⟩)  ⇒^!  (X−1)^(⟨−3⟩)   proof:  X^7 ⊂X^(⟨7∣−9⟩) ⊃(X+1)^(⟨−3∣−5∣=7⟩)   but (X+1)^(⟨−3∣−5∣=7⟩) =(X−5)^(⟨−3∣−5∣=7⟩)   and (X−5)^(⟨−3∣−5∣=7⟩) ⊂(X−6)^(⟨7∣−9⟩) ⊃(X−6)^(⟨7⟩)   but we know (x+u) ⇒^!  (x+v) ⇔ v= { ((u+5)),((u−7)) :}  (Beck−Runnel−Axiom)  ⇒ (x−6) ⇒^!  (x−1) ⇒  ⇒ (X−6)^(⟨7⟩)  ⇒^!  (X−1)^(⟨−3⟩)   but we have shown   (X−6)^(⟨7⟩) ⊂(X−6)^(⟨7∣−9⟩) ⊃(X−5)^(⟨−3∣−5∣−^(2o) 7⟩) =  =(X+1)^(⟨−3∣−5∣=7⟩) ⊂X^(⟨7∣−9⟩) ⊃X^(⟨7⟩)  ⇒  ⇒ X^(⟨7⟩)  ⇒^!  (X−1)^(⟨−3⟩)  q.e.d.    [b.t.w. because of the Axioms of Inversion  (X+u)^(⟨−3∣7⟩) =(X+v)^(⟨6⟩)  with v= { ((u+3)),((u−9)) :}  we also showed that X^(⟨7⟩)  ⇒^!  (X+2) which  is essential to Simon′s Approach]
$${show}\:{that}\:{X}^{\langle\mathrm{7}\rangle} \:\overset{!} {\Rightarrow}\:\left({X}−\mathrm{1}\right)^{\langle−\mathrm{3}\rangle} \\ $$$${proof}: \\ $$$${X}^{\mathrm{7}} \subset{X}^{\langle\mathrm{7}\mid−\mathrm{9}\rangle} \supset\left({X}+\mathrm{1}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid=\mathrm{7}\rangle} \\ $$$${but}\:\left({X}+\mathrm{1}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid=\mathrm{7}\rangle} =\left({X}−\mathrm{5}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid=\mathrm{7}\rangle} \\ $$$${and}\:\left({X}−\mathrm{5}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid=\mathrm{7}\rangle} \subset\left({X}−\mathrm{6}\right)^{\langle\mathrm{7}\mid−\mathrm{9}\rangle} \supset\left({X}−\mathrm{6}\right)^{\langle\mathrm{7}\rangle} \\ $$$${but}\:{we}\:{know}\:\left({x}+{u}\right)\:\overset{!} {\Rightarrow}\:\left({x}+{v}\right)\:\Leftrightarrow\:{v}=\begin{cases}{{u}+\mathrm{5}}\\{{u}−\mathrm{7}}\end{cases} \\ $$$$\left({Beck}−{Runnel}−{Axiom}\right) \\ $$$$\Rightarrow\:\left({x}−\mathrm{6}\right)\:\overset{!} {\Rightarrow}\:\left({x}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\Rightarrow\:\left({X}−\mathrm{6}\right)^{\langle\mathrm{7}\rangle} \:\overset{!} {\Rightarrow}\:\left({X}−\mathrm{1}\right)^{\langle−\mathrm{3}\rangle} \\ $$$${but}\:{we}\:{have}\:{shown}\: \\ $$$$\left({X}−\mathrm{6}\right)^{\langle\mathrm{7}\rangle} \subset\left({X}−\mathrm{6}\right)^{\langle\mathrm{7}\mid−\mathrm{9}\rangle} \supset\left({X}−\mathrm{5}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid\overset{\mathrm{2}{o}} {−}\mathrm{7}\rangle} = \\ $$$$=\left({X}+\mathrm{1}\right)^{\langle−\mathrm{3}\mid−\mathrm{5}\mid=\mathrm{7}\rangle} \subset{X}^{\langle\mathrm{7}\mid−\mathrm{9}\rangle} \supset{X}^{\langle\mathrm{7}\rangle} \:\Rightarrow \\ $$$$\Rightarrow\:{X}^{\langle\mathrm{7}\rangle} \:\overset{!} {\Rightarrow}\:\left({X}−\mathrm{1}\right)^{\langle−\mathrm{3}\rangle} \:{q}.{e}.{d}. \\ $$$$ \\ $$$$\left[{b}.{t}.{w}.\:{because}\:{of}\:{the}\:{Axioms}\:{of}\:{Inversion}\right. \\ $$$$\left({X}+{u}\right)^{\langle−\mathrm{3}\mid\mathrm{7}\rangle} =\left({X}+{v}\right)^{\langle\mathrm{6}\rangle} \:{with}\:{v}=\begin{cases}{{u}+\mathrm{3}}\\{{u}−\mathrm{9}}\end{cases} \\ $$$${we}\:{also}\:{showed}\:{that}\:{X}^{\langle\mathrm{7}\rangle} \:\overset{!} {\Rightarrow}\:\left({X}+\mathrm{2}\right)\:{which} \\ $$$$\left.{is}\:{essential}\:{to}\:{Simon}'{s}\:{Approach}\right] \\ $$
Commented by Her_Majesty last updated on 09/Sep/20
(1) find all possible sources of (X−1)^(⟨+5⟩)           [super easy]  (2) find possible sources of  (a) (X−3)^(⟨+5∣−7⟩)  and (b) (X+4)^(⟨−5∣−7⟩)   and their center angles [hard]
$$\left(\mathrm{1}\right)\:{find}\:{all}\:{possible}\:{sources}\:{of}\:\left({X}−\mathrm{1}\right)^{\langle+\mathrm{5}\rangle} \\ $$$$\:\:\:\:\:\:\:\:\left[{super}\:{easy}\right] \\ $$$$\left(\mathrm{2}\right)\:{find}\:{possible}\:{sources}\:{of} \\ $$$$\left({a}\right)\:\left({X}−\mathrm{3}\right)^{\langle+\mathrm{5}\mid−\mathrm{7}\rangle} \:{and}\:\left({b}\right)\:\left({X}+\mathrm{4}\right)^{\langle−\mathrm{5}\mid−\mathrm{7}\rangle} \\ $$$${and}\:{their}\:{center}\:{angles}\:\left[{hard}\right] \\ $$
Commented by MJS_new last updated on 09/Sep/20
what′s this???
$$\mathrm{what}'\mathrm{s}\:\mathrm{this}??? \\ $$
Commented by 1549442205PVT last updated on 09/Sep/20
Please,Sir Mathdave can explain    at tenth line  ln(√(π/2))=ln(Π_(k) k)−ln2  Π_k k is an infinite value while (√π) <2  (Math is beautiful because of its  precision)
$$\mathrm{Please},\mathrm{Sir}\:\mathrm{Mathdave}\:\mathrm{can}\:\mathrm{explain}\: \\ $$$$\:\mathrm{at}\:\mathrm{tenth}\:\mathrm{line} \\ $$$$\mathrm{ln}\sqrt{\frac{\pi}{\mathrm{2}}}=\mathrm{ln}\left(\underset{\mathrm{k}} {\Pi}\mathrm{k}\right)−\mathrm{ln2} \\ $$$$\underset{\mathrm{k}} {\prod}\mathrm{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{infinite}\:\mathrm{value}\:\mathrm{while}\:\sqrt{\pi}\:<\mathrm{2} \\ $$$$\left(\mathrm{Math}\:\mathrm{is}\:\mathrm{beautiful}\:\mathrm{because}\:\mathrm{of}\:\mathrm{its}\right. \\ $$$$\left.\mathrm{precision}\right) \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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