given-z-lim-n-n-z-z-1-z-n-n-z-proof-that-i-z-1-z-z-ii-wiertrass-definition-of-gamma-function-1-z-ze-z-m-1-1-z-m-e-z-m-is-euler-macheroni-con

Question Number 2891 by 123456 last updated on 29/Nov/15

given

Γ (z) = \lim_{n \to + \infty} \frac{n!}{z (z + 1) \cdot \cdot \cdot (z + n)} n^{z}

proof that

i)

Γ (z + 1) = z Γ (z)

i i) wiertrass definition of gamma function

\frac{1}{Γ (z)} = {z e}^{z γ} \underset{m = 1}{\prod^{+ \infty}} (1 + \frac{z}{m}) e^{- \frac{z}{m}}

γ is euler macheroni constant (Q 2818, Q 2783)

Commented by Filup last updated on 30/Nov/15

z (z + 1) (z + 2) \cdot \cdot \cdot (z + n) = \frac{(z + n)!}{(z - 1)!}

∴ Γ (z) = \lim_{n \to + \infty} \frac{n! \cdot (z - 1)!}{(z + n)!} \cdot n^{z}

Answered by Filup last updated on 30/Nov/15

Γ (z) = (z - 1) (z - 2) \cdot \cdot \cdot \times 3 \times 2 \times 1

Γ (z + 1) = z (z - 1) (z - 2) \cdot \cdot \cdot \times 3 \times 2 \times 1

Γ (z + 1) = z Γ (z)

I^{'} m sure you wanted a proof in terms of

the limit you gave above, but I am

unsure as to how to achieve the result

you most likely desired . Maybe you can

further prove by applying the limit ?

I shall update this post if I can . I m i g h t

have an idea to prove (i) in terms

of the limit

Commented by 123456 last updated on 30/Nov/15

the limit above is the most general definition

possible for Γ (analitic continuation of Γ)

its extend Γ to almost all C (with exeption

of negatives number and 0)

however its is so hard to use

Commented by Filup last updated on 30/Nov/15

I am trying a method of proof now .

If it leads nowhere, i^{'} ll delete my 2 nd post

Answered by Filup last updated on 30/Nov/15

P r o o f w i t h l i m i t

Γ (z) = \lim_{n \to \infty} \frac{n!}{z (z + 1) \cdot \cdot \cdot (z + n)} n^{z}

∴ Γ (z) = \lim_{n \to \infty} \frac{n! (z - 1)!}{(z + n)!} n^{z}

Show that Γ (z) = (z - 1) Γ (z - 1)

∴ \lim_{n \to \infty} \frac{n! (z - 1)!}{(z + n)!} n^{z} = (z - 1) (\lim_{n \to \infty} \frac{n! (z - 2)!}{(z + n - 1)!} n^{z - 1})

Let us assume Γ (z) = (z - 1) Γ (z - 1) . If this is

true then LHS = RHS

Γ (x) = (x - 1)!

\lim_{n \to \infty} \frac{n! \cdot Γ (z)}{Γ (z + n + 1)} n^{z} = (z - 1) (\lim_{n \to \infty} \frac{n! \cdot Γ (z - 1)}{Γ (z + n)} n^{z - 1})

∴ \lim_{n \to \infty} \frac{n! \cdot Γ (z)}{Γ (z + n + 1)} n^{z} = (z - 1) (\lim_{n \to \infty} \frac{n! \cdot \frac{Γ (z)}{z - 1}}{\frac{Γ (z + n + 1)}{z - 1}} n^{z - 1})

∴ \lim_{n \to \infty} \frac{n! \cdot Γ (z)}{Γ (z + n + 1)} n^{z} = (z - 1) (\lim_{n \to \infty} (\frac{n! \cdot Γ (z) \cdot (z - 1)}{(z - 1) \cdot Γ (z + n + 1)}) n^{z - 1})

∴ \lim_{n \to \infty} \frac{n! \cdot Γ (z)}{Γ (z + n + 1)} n^{z} = \lim_{n \to \infty} \frac{(z - 1) \cdot n! \cdot Γ (z)}{Γ (z + n + 1)} n^{z - 1}

Γ (z + n + 1) = (z + n)!

∴ \lim_{n \to \infty} \frac{n! \cdot Γ (z)}{(z + n)!} n^{z} = \lim_{n \to \infty} \frac{n! \cdot Γ {(z)}^{2}}{(z + n)! \cdot Γ (z - 1)} n^{z - 1}

∴ \lim_{n \to \infty} \frac{n!}{(z + n)!} n^{z} = \lim_{n \to \infty} \frac{n! \cdot Γ (z)}{(z + n)! \cdot Γ (z - 1)} n^{z - 1}

∴ \lim_{n \to \infty} \frac{n!}{(z + n)!} n^{z} = \lim_{n \to \infty} \frac{n! \cdot (z - 1)}{(z + n)!} n^{z - 1}

\lim_{x \to \infty} \frac{x!}{(x + a)!} = 1 (?)

∴ \lim_{n \to \infty} n^{z} = \lim_{n \to \infty} (z - 1) n^{z - 1}

TRUE

∴ Γ (z + 1) = z Γ (z)

please let me know of any errors

Commented by 123456 last updated on 30/Nov/15

Γ (z + 1) = z Γ (z)

Γ (z) = (z - 1) Γ (z - 1)

Commented by Filup last updated on 30/Nov/15

How should I apply this ?

Commented by 123456 last updated on 30/Nov/15

you have a typo in the 5 th line

Commented by Filup last updated on 30/Nov/15

T h a n k s! I^{'} l l r e r e a d e v e r y t h i n g!

I believe I have fixed everything

Commented by 123456 last updated on 30/Nov/15

hint : \frac{x}{x} = 1, \lim_{n \to \infty} \frac{n + a}{n + b} = 1

try to write Γ (z) and Γ (z + 1)

and compare each other to see what

you need to manipulate : D

ps : if you want i can post the proof for this

part, but try to do it

Commented by Filup last updated on 30/Nov/15

I^{'} ll let you know if i {can}^{'} t do it!

Commented by Filup last updated on 30/Nov/15

Am I able to remove like terms within

limits ? e . g . :

\lim_{x \to n} f (x) = \lim_{x \to n} g (x)

\Rightarrow \lim_{x \to n} \frac{f (x)}{g (x)} = 1

Commented by Filup last updated on 30/Nov/15

is this correct ?

Answered by Yozzi last updated on 30/Nov/15

(i) W e a r e g i v e n

Γ (z) = \lim_{n \to + \infty} \frac{n! n^{z}}{\underset{r = 0}{\prod^{n}} (z + r)} .

H e r e I a s s u m e z i s s u c h t h a t t h e

f u n c t i o n Γ (z) i s w e l l d e f i n e d . A

p o s s i b l e c o n s t r i c t i o n o n t h e d o m a i n

f o r Γ i s z > 0 . T h e r e f o r e w e h a v e

Γ (z + 1) = \lim_{n \to + \infty} \frac{n! n^{z + 1}}{(z + 1) (z + 2) \dots (z + n) (z + n + 1)}

Γ (z + 1) = \lim_{n \to + \infty} \frac{n! n^{z + 1} z}{{\underset{r = 0}{\prod^{n}} (z + r)} (z + 1 + n)}

Γ (z + 1) = \lim_{n \to + \infty} (\frac{n! n^{z}}{\underset{r = 0}{\prod^{n}} (z + r)}) \times \frac{z n}{z + 1 + n}

S i n c e \lim_{x \to + \infty} f (x) \times g (x) = (\lim_{x \to + \infty} f (x)) (\lim_{x \to + \infty} g (x)) i n

g e n e r a l, w e o b t a i n

Γ (z + 1) = Γ (z) \times \lim_{n \to + \infty} \frac{z n}{z + 1 + n} .

B u t, \frac{z n}{z + 1 + n} = \frac{z}{\frac{z}{n} + \frac{1}{n} + 1} .

\Rightarrow Γ (z + 1) = Γ (z) \lim_{n \to + \infty} \frac{z}{\frac{z}{n} + \frac{1}{n} + 1}

Γ (z + 1) = Γ (z) \times \frac{z}{\frac{z}{\infty} + \frac{1}{\infty} + 1}

Γ (z + 1) = Γ (z) \times \frac{z}{0 + 0 + 1}

Γ (z + 1) = z Γ (z) .

R e m a r k s :

(1) I f i t i s r e q u i r e d t o c a l c u l a t e t h e

v a l u e o f Γ (x) f o r x < 0 w e u s e t h e

r e c u r r e n c e r e l a t i o n Γ (x) = \frac{1}{x} Γ (x + 1)

u n t i l t h e p a r t Γ (x + 1) r e d u c e s f u l l y t o

s o m e n o n - n e g a t i v e f o r m .

E . g Γ (\frac{- 5}{2}) = \frac{- 2}{5} Γ (\frac{- 3}{2}) = \frac{- 2}{5} \times \frac{- 2}{3} Γ (\frac{- 1}{2})

Γ (- \frac{5}{2}) = \frac{4}{15} \times \frac{- 2}{1} Γ (\frac{1}{2})

S i n c e Γ (1 / 2) = \sqrt{π},

Γ (\frac{- 5}{2}) = \frac{- 8}{15} \sqrt{π} .

(2) F o r 0 < x < 1 w e h a v e

Γ (x) Γ (1 - x) = \frac{π}{s i n π z} .

S o, i f x = 1 / 2 w e d e r i v e t h e r e s u l t

Γ (0.5) = \sqrt{π} .

Γ (0.5) Γ (1 - 0.5) = \frac{π}{s i n 0.5 π}

{(Γ (0.5))}^{2} = \frac{π}{1} \Rightarrow Γ (0.5) = \pm \sqrt{π}

H o w e v e r, f o r x > 0 t h e l i m i t d e f i n i t i o n

f o r Γ (x) = \lim_{n \to + \infty} \frac{n! n^{x}}{\underset{r = 0}{\prod^{n}} (x + r)} i n d i c a t e s t h a t

Γ (x) i s n o n - n e g a t i v e .

\Rightarrow Γ (0.5) \neq - \sqrt{π} \Rightarrow Γ (0.5) = \sqrt{π} .

Commented by 123456 last updated on 02/Dec/15

the limit form of Γ function also tell

you that its have simple pole at

z \in {\dots, - 5, - 4, - 3, - 2, - 1, 0}

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