1

Das Wallissche Produkt

kann wie folgt geschrieben werden:

2

= 2

4 n

2

4 n

1

n

1

-

=

2

A

Integral

das

betrachten

Wir

=

: m

sin x dx

m

0

2

2

gilt

Es

A

:

=

m

sin x

dx

sinx sin

x dx

m

=

m 1

-

0

0

Integr.

Part.

u(x)

:

= sin

m - 1x

)

u(x

,

= (m

-1)sin m-2 x cosx

v(x) = -

cosx

)

v(x = sinx

2

A =

-

m

[

sin m 1

- x ( cos x)] 2

-

0

(m- )1 sin

m-2 x osx

c

(-cosx) dx

0

2

A =

0

+

m

(

m - )

1 sin m-2 x cos2 x dx

0

2

A =

-

m

(

m - )

1 sin m-2x (1 sin2 x) dx

0

2

A =

-

m

(

m - )

1 (sin m-2x sin m x) dx

0

2

2

m-2

m

A =

-

m

(

m - )

1 sin

x dx

( m- )1 sin x dx

0

0

A =

m

(

m - )

1 A

m-2

(

-

m - )

1 A

m

---

2

m -1

m + 1

A

=

A

A

+

=

A

m

m-2

m 2

m

m

m + 2

Beweis mit Induktion:

2

2

A

= sin0x dx

= dx =

=

0

[ x]2

0

2

0

0

2

A

= sin1x dx =

=

1

[ -cosx]2

1

0

0

(2m- )1(2m- )3531

Gerade

A

:

=

2m

2m

(2m - 2)

6 4 2

2m

(2m - 2)6 4 2

Ungerade

A

:

+

=

2m 1

(2m+ )1(2m- )

1 5 3 1

A

+

-

-

2m 1

(2m 2m) (2m 2) (2m 2) (6 6) (4 4) (2 2)

=

A

+

-

-

2m

(2m )1 ( 2m )1 (2m )1 (5 5) (3 )3 (1 )

1

gilt

Es

sin

:

2m+2 x sin

2m 1

+ x sin

2m

x

für x

0,

m

,

2

2

2

2

sin2m+2

dx

sin2m 1+

dx

sin2mdx

0

0

0

A

+

A

+

A

2m 2

2m 1

2m

2m + 2 -1

2m + 1

A

:

Weiter

+

=

A

=

A

2m 2

2m

2m

2m + 2

2m + 2

A

2m

+

+1

A

2m

+

+1

2m 2

=

lim

2m 2

=

=

1

A

2m + 2

m

A

2m + 2

2m

2m

---

3

A

+

A

+

A

2m 2

2m 1

2m

A

A

A

+

+

2m 2 2m 1

2m

A

A

A

2m

2m

2m

A

A

A

+

+

lim

2m 2

lim

2m 1

lim

2m

m

A

m

A

m A

2m

2m

2m

A

+

1

lim

2m 1

1

m

A 2m

A

+

für

gilt

Also

lim

2m 1 =

1

m

A 2m

m

Für

gilt

:

A

+

-

-

2m 1

(2m 2m) (2m 2) (2m 2) (6 6) (4 4) (2 2)

lim

= lim

=

m

A

m

+

-

-

2m

(2m )1 ( 2m )1 (2m )1 (5 5) (3 3) (1 ) 1

1

2

(

2m 2m) (

2m - 2) (2m - 2) (6 6) (4 4) (2 2)

= 2

lim

m

(2m+ )1( 2m- )1 ( 2m- )1(2m-3)(75)(53)(3 )

1

(22)(44)(66)(2(n - )12(n - )1(2n2n)

= 2

lim

m

( 13)( 35) (2(n - )12(n - )1(2n - )1(2n + )1

2n 2n

=

2

n 1

= (2n + )

1 ( n

2 - )

1

---

4

Vorbemerkung für den Beweis der Stirlingschen Formel:

Die Trapezregel

Satz :

f

Sei

[ : ]

0,1

Funktion

erbare

differenzi

stetig

zweimal

eine

IR

:

ein

für

gilt

Dann

[ ]

1

,

0

:

1

1

f(x)dx

=

(f(0)+f(1)) 1

-

f ()

2

12

0

1

1

wobei

:

f ( = 1

)

x(1- f

x)

=

(x)dx R

:

12

2

0

f (x)

f (1)-

f (0) y = f(x) Für konvexe Funktionen:

f (

x) 0

0

1 x

1

2

g

Sei

=

(

x 1- x) x

x

1

Beweis:

=

-

g

,

=

g

,

x

-

= -

1

2

2

2

2

1

1

1

f (x dx

)

= -

dx

f(x)

(-1)

= - g

dx

f(x)

0

0

0

1

= - [

g

f(x)

(x)

]1 + f g

(x)

dx

(x)

0

0

1

= - [f(1)

g (1)

- f(0) g (0)

]+ f g

(x)

dx

(x)

0

1

1

=

[ f(0)

+ f(1)]

+ f g

(x)

dx

(x)

2

0

---

5

1

1

f (

x g

)

dx

(x)

=

[ f (x g(x)

)

]1 - f

dx

g(x)

(x)

0

0

0

1

= [ f

f

-

g(x)

(1)

g(0)

(0)

] - f

dx

g(x)

(x)

0

1

=

-

0

-

0

f

dx

g(x)

(x)

0

b

b

MWS

:

dx

g(x)

f(x)

= f(

) g(x)

a

a

1

= f

-

(

) g

dx

(x)

0

1

x2 x3

= f

-

(

)

-

4

6

0

1 1

1

= f

-

(

)

-

=

-

f

(

)

4

6

12

1

1

gilt

Es

:

dx

f(x)

=

[

f(0) + f(1)

] 1

-

f

(

)

2

12

0

---

6

Die Stirlingsche Formel

Die Stirlingsche Formel beschreibt das asymptotische Verhalten der Fakultät von natürlichen

Zahlen n (n!) für grosse n.

(Stirling, James: schottischer Mathematiker, 1692 ­ 1770)

Def .

a

Folgen

Zwei

:

b

und

(n

heissen

IN)

asymptotis

gleich

ch

(a

b

~

),

n

n

n

n

a

gilt

falls

lim

:

n =

,

1

n bn

(a

wobei

(b

und

)

müssen.

en

konvergier

nicht

)

n

n

Satz

Verhalten

che

asymptotis

das

hat

Fakultät

Die

:

n

n

~

n!

2 n

e

2

n! e

n

1

Daraus folgt für :

~

n

grosse

für

nn

2n

n = 1 : 3,694528

n = 10 :

3,1943835

n = 20 :

3,1767853

n = 30 :

3,1591127

n = 40 :

3,1546803

n = 50 :

3,1520848

n

:

Beweis:

1

g

Sei

(x :

)

=

(

x 1 - x)

IR

,

x

IR,

[ ]

0,1

2

:

f(x)

= ln x

k 1

+

1

dx

ln x

=

( k

ln + ln(k

+1)) 1

+

f

(

)

k

k +1

2

12

k

k

k

---

7

k

über

Summation

=

n

,

...

1,

-

1

n

n 1

-

1

n 1

-

dx

ln x

=

( k

ln + ln(k

+1)) 1

+

f (

)

k

k 1

= 2

12 k 1

1

=

1

1 n 1-

=

1

(ln

+

2

ln

+

2

ln

+

3

ln

+

3

ln

+

...

+

(n

ln

-

1) +

n)

ln

+

f (

)

2

12

k

k 1

=

1

n

n 1

-

=

(

n

ln

-

1

ln

)

1

+

k

ln

+

f (

)

2

k 1

=

12

k

k 1

=

n

dx

ln x

=

[ -x + ln x

x

]n = - n + n

ln

n

+1

1

1

1

n

n 1

-

-

n +

n

ln

n

+

1

=

(

n

ln

-

1

ln

)

1

+

k

ln

+

f (

)

2

k 1

=

12

k

k 1

=

n

1

1 n 1

-

k

ln

=

-

n

+ n

+

n

ln

+ 1

-

f (

)

k 1

=

2

12

k

k 1

=