Gamma and betta function harsh shah
12 likes6,503 views
The document defines the gamma function and some of its key properties. It provides two definitions of the gamma function: an improper integral definition and an alternate definition involving an integral of x2n-1 from 0 to infinity. It then proves the alternate definition and discusses some important properties, including: Γ(n+1) = nΓ(n); Γ(1/2) = π; and examples of evaluating gamma functions for various values of n.
![Proof of the property :
1] Γ n+1 = 𝑒−𝑥∞
0
. 𝑥 𝑛
𝑑𝑥,
Inter changing by parts ,
Γ n+1 = | 𝑒−𝑥
. 𝑥 𝑛
|0
∞
- 𝑒−𝑥∞
0
. 𝑥 𝑛−1
𝑑𝑥
= 𝑛 𝑒−𝑥∞
0
. 𝑥 𝑛−1
𝑑𝑥
=n Γ n
Hence Γn+1 = n Γ n proved.
This is known as recurrence reduction formula for gamma function.](https://image.slidesharecdn.com/gammaandbettafunction-harshshah-140314021821-phpapp02/85/Gamma-and-betta-function-harsh-shah-5-320.jpg)
![ This shows that the region of integration is the first quadrant.
Draw the elementary radius vector in the region which starts from
the pole extend up to ∞.
Limits of r r=0 to r → ∞
Limits of 𝜃 𝜃 =0 to 𝜃 →
𝜋
2
.
Γ
1
2
Γ
1
2
= 4 𝑒−𝑟2∞
0
𝜋
2
0
. 𝑟𝑑𝑟𝑑𝜃
= 4 𝑑𝜃 (
−1
2
) 𝑒−𝑟2∞
0
𝜋
2
0
. (−2𝑟) 𝑑𝑟
=
4
−2
|𝜃|0
𝜋
2
|𝑒−𝑥2
|0
∞
[∵ 𝑒 𝑓 𝑟 𝑓′ 𝑟 𝜃𝑟 = 𝑒 𝑓 𝑟 ]
= -2.
𝜋
2
(0 − 1)
=𝜋
Γ
1
2
= 𝜋](https://image.slidesharecdn.com/gammaandbettafunction-harshshah-140314021821-phpapp02/85/Gamma-and-betta-function-harsh-shah-8-320.jpg)
Gamma and betta function harsh shah
- 1. GAMMA AND BETTA FUNCTION GROUP - 1 ENROLLMENT NO: 1-12
- 2. Gamma function : Gamma function is define by the improper interval 𝑒−𝑥 ∞ 0 𝑥 𝑛−1 𝑑𝑥 , 𝑛 > 0 And it is denoted by Γ n. Alternate form of gamma function , Γ n = 𝑒−𝑥2∞ 0 𝑥2𝑛−1 𝑑𝑥
- 3. Proof by definition Γ n = 𝑒−𝑥∞ 0 . 𝑥 𝑛−1 𝑑𝑥 Let x=𝑡2 and 𝑑𝑥 = 2tdt Γ n= 𝑒−𝑡2∞ 0 . 𝑡2𝑛−2 𝑑𝑡 =2 𝑒−𝑡2∞ 0 . 𝑡2𝑡−1 𝑑𝑡 Now changing the variable t to x, Γ n=2 𝑒−𝑥2∞ 0 𝑥2𝑛−1 𝑑𝑥
- 4. Proprieties of GAMMA function : Γ n+1 = n Γ n Γ1 2 = Π
- 5. Proof of the property : 1] Γ n+1 = 𝑒−𝑥∞ 0 . 𝑥 𝑛 𝑑𝑥, Inter changing by parts , Γ n+1 = | 𝑒−𝑥 . 𝑥 𝑛 |0 ∞ - 𝑒−𝑥∞ 0 . 𝑥 𝑛−1 𝑑𝑥 = 𝑛 𝑒−𝑥∞ 0 . 𝑥 𝑛−1 𝑑𝑥 =n Γ n Hence Γn+1 = n Γ n proved. This is known as recurrence reduction formula for gamma function.
- 6. NOTE : Γ n+1 = n! if n is a positive integer Γn+1 = n Γn if n is a real number . Γn = Γ 𝑛+1 𝑛 if n is negative fraction. Γn Γ1-n = Π sin 𝑛 Π
- 7. (2) Γ 1 2 = Π PROOF : By alternate form of gamma function , Γ 1 2 = 2 𝑒−𝑥2∞ 0 . 𝑥2( 1 2 )−1 𝑑𝑥 = 𝑒−𝑥2 . 𝑑𝑥 ∞ 0 Γ 1 2 Γ 1 2 = 𝑒−𝑥2 𝑑𝑥 ∞ 0 . 𝑒−𝑦2 . 𝑑𝑦 ∞ 0 changing to polar coordinates x= 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 ∴ 𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑎𝑑𝜃 Limits of x x=0 to x → ∞ Limits of y y=0 to y → ∞
- 8. This shows that the region of integration is the first quadrant. Draw the elementary radius vector in the region which starts from the pole extend up to ∞. Limits of r r=0 to r → ∞ Limits of 𝜃 𝜃 =0 to 𝜃 → 𝜋 2 . Γ 1 2 Γ 1 2 = 4 𝑒−𝑟2∞ 0 𝜋 2 0 . 𝑟𝑑𝑟𝑑𝜃 = 4 𝑑𝜃 ( −1 2 ) 𝑒−𝑟2∞ 0 𝜋 2 0 . (−2𝑟) 𝑑𝑟 = 4 −2 |𝜃|0 𝜋 2 |𝑒−𝑥2 |0 ∞ [∵ 𝑒 𝑓 𝑟 𝑓′ 𝑟 𝜃𝑟 = 𝑒 𝑓 𝑟 ] = -2. 𝜋 2 (0 − 1) =𝜋 Γ 1 2 = 𝜋
- 9. Examples : Find the value of Γ 𝟓 𝟐 Γn = 𝛤𝑛+1 𝑛 Γ −5 2 = 𝛤 −5 2 +1 5 2 = −2 5 Γ −3 2 = −2 5 Γ −3 2 +1 −3 2 = 4 15 Γ −1 2 = 4 15 Γ −1 2 +1 −1 2 = −8 15 Γ 1 2 = - −8 𝜋 15 ∵ 𝑓𝑟𝑜𝑚 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
- 10. THANK YOU