prove-that-2pi-

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

p r o v e t h a t

\infty! = \sqrt{2 π}

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

Commented by mathdave last updated on 09/Sep/20

a l r i g h t s i r

Answered by mathdave last updated on 09/Sep/20

s o l u t i o n

s i n c e

η (s) = \underset{k = 1}{\sum^{\infty}} {(- 1)}^{k + 1} k^{- s} = (1 - 2^{1 - s}) ζ (s)

η^{^{'}} (s) = \underset{k = 1}{\sum^{\infty}} {(- 1)}^{k} \frac{\ln k}{k^{s}} = ζ^{^{'}} (s) (1 - 2^{1 - s}) + ζ (s) 2 \ln 2

η^{^{'}} (0) = \underset{k = 1}{\sum^{\infty}} {(- 1)}^{k} \ln k = - \ln (1) + \ln (2) - \ln (3) + - - - -

η^{^{'}} (0) = \ln (\frac{π}{2}) - η^{^{'}} (0) [{w a l l i}^{'} s p r o d u c t]

η^{^{'}} (0) = \frac{1}{2} \ln (\frac{π}{2}) = \ln (\sqrt{\frac{π}{2}}) \dots \dots \dots (1)

a g a i n

η^{^{'}} (0) = - ζ^{^{'}} (0) + ζ (0) 2 \ln 2 = - [η^{^{'}} (0) - \ln (π)] - \ln 2

2 η^{^{'}} (0) = \ln (\frac{π}{2})

η^{^{'}} (0) = \frac{1}{2} \ln (\frac{π}{2}) = \ln (\sqrt{\frac{π}{2}}) \dots \dots \dots (2)

b u t \ln (\sqrt{\frac{π}{2}}) = \ln (\prod_{k = 1} k) - \ln 2

\ln (\sqrt{\frac{π}{2}}) = \ln (\infty!) - \ln 2

\ln (\infty!) = \ln 2 + \ln (\sqrt{\frac{π}{2}})

\ln (\infty!) = \ln (2 \sqrt{\frac{π}{2}}) = \ln (\sqrt{2 π})

\ln (\infty!) = \ln (\sqrt{2 π})

∵ \infty! = \sqrt{2 π} Q . E . D

b y m a t h d a v e (09 / 09 / 2020)

Commented by Her_Majesty last updated on 09/Sep/20

s h o w t h a t X^{⟨ 7 ⟩} \overset{!}{\Rightarrow} {(X - 1)}^{⟨ - 3 ⟩}

p r o o f :

X^{7} \subset X^{⟨ 7 ∣ - 9 ⟩} \supset {(X + 1)}^{⟨ - 3 ∣ - 5 ∣= 7 ⟩}

b u t {(X + 1)}^{⟨ - 3 ∣ - 5 ∣= 7 ⟩} = {(X - 5)}^{⟨ - 3 ∣ - 5 ∣= 7 ⟩}

a n d {(X - 5)}^{⟨ - 3 ∣ - 5 ∣= 7 ⟩} \subset {(X - 6)}^{⟨ 7 ∣ - 9 ⟩} \supset {(X - 6)}^{⟨ 7 ⟩}

b u t w e k n o w (x + u) \overset{!}{\Rightarrow} (x + v) \Leftrightarrow v = {\begin{cases} u + 5 \\ u - 7 \end{cases}

(B e c k - R u n n e l - A x i o m)

\Rightarrow (x - 6) \overset{!}{\Rightarrow} (x - 1) \Rightarrow

\Rightarrow {(X - 6)}^{⟨ 7 ⟩} \overset{!}{\Rightarrow} {(X - 1)}^{⟨ - 3 ⟩}

b u t w e h a v e s h o w n

{(X - 6)}^{⟨ 7 ⟩} \subset {(X - 6)}^{⟨ 7 ∣ - 9 ⟩} \supset {(X - 5)}^{⟨ - 3 ∣ - 5 ∣ \overset{2 o}{-} 7 ⟩} =

= {(X + 1)}^{⟨ - 3 ∣ - 5 ∣= 7 ⟩} \subset X^{⟨ 7 ∣ - 9 ⟩} \supset X^{⟨ 7 ⟩} \Rightarrow

\Rightarrow X^{⟨ 7 ⟩} \overset{!}{\Rightarrow} {(X - 1)}^{⟨ - 3 ⟩} q . e . d .

[b . t . w . b e c a u s e o f t h e A x i o m s o f I n v e r s i o n

{(X + u)}^{⟨ - 3 ∣ 7 ⟩} = {(X + v)}^{⟨ 6 ⟩} w i t h v = {\begin{cases} u + 3 \\ u - 9 \end{cases}

w e a l s o s h o w e d t h a t X^{⟨ 7 ⟩} \overset{!}{\Rightarrow} (X + 2) w h i c h

i s e s s e n t i a l t o {S i m o n}^{'} s A p p r o a c h]

Commented by Her_Majesty last updated on 09/Sep/20

(1) f i n d a l l p o s s i b l e s o u r c e s o f {(X - 1)}^{⟨ + 5 ⟩}

[s u p e r e a s y]

(2) f i n d p o s s i b l e s o u r c e s o f

(a) {(X - 3)}^{⟨ + 5 ∣ - 7 ⟩} a n d (b) {(X + 4)}^{⟨ - 5 ∣ - 7 ⟩}

a n d t h e i r c e n t e r a n g l e s [h a r d]

Commented by MJS_new last updated on 09/Sep/20

{what}^{'} s this ? ? ?

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

Please, Sir Mathdave can explain

at tenth line

\ln \sqrt{\frac{π}{2}} = \ln (\underset{k}{Π} k) - \ln 2

\prod_{k} k is an infinite value while \sqrt{π} < 2

(Math is beautiful because of its

precision)

Commented by Tawa11 last updated on 06/Sep/21

great sir

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