chapter 11 Special functions

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Transcript chapter 11 Special functions

Mathematical methods in the physical sciences 3rd edition Mary L. Boas

Chapter 11 Special functions

Lecture 12 Gamma, beta, error, and elliptic 1

2. The factorial function (usually, n : integer)  0  

e ax dx

  1 

e

 

x

 0  1   0 

xe

ax dx

  1 

xe

 

x

0    0  Similarly,  0 

x

2

e

 

x dx

  2 3 ,

  0

1 

e

 

x dx

  1 2 .

 0 

x

3

e

 

x dx

 2  3  4  0 

x n e

 

x dx

 

n

!

n

 1   0 

x n e

x dx

n

!

  1

2

3. Definition of the gamma function: recursion relation ( p : noninteger) - Gamma function     0

x p

 1

e

x dx

,

p

 0 .

 

  0 

x n

 1

e

x dx

n

 1

!

, 

n

 1

  0 

x n e

x dx

n

!.

- Recursion relation 

p

 1

   0

x p e

x dx

p

 1

p

 

p

!

,

p

  1 .

- Example 

 

so

,  1 /  4 ( 5  / 4 )   9 /

      

 16 / 5 .

3

4. The Gamma function of negative numbers 

 

 1

p

p

 1

(

p

 0 ) - Example 

 0 .

3

  1 0 .

3 

 

, 

 1 .

3

     

.

cf

.

 

 1

p

p

 1

  as

p

 0 .

- Using the above relation, 1) Gamma( p = negative integers)  infinite.

2) For p < 0, the sign changes alternatively in the intervals between negative integers 4

5. Some important formulas involving gamma functions     ( prove )      0  1

t e

t dt

  0  1

y e

y

2 2

ydy

 2  0  

e y

2

dy

.

 2  4     0 0

e

 

x

2 

y

2 

dxdy

 4 0  / 2   0  

e r

2

rdrd

   .

 

  

1 

p

 sin  

p

.

5

6. Beta functions

B p

,   1 0

x p

 1

1 

x

q

 1

dx

,

p

 0 ,

q

 0 .

cf

.

B p

, 

q

,

i

)

ii

)

B

 

  0

a y a B

 

 2  0  / 2

p

 1 1

sin  2

a y q

 1

dy a

a

1

p

q

 1  0

a y p

 1

a

y

q

 1

dy

.

 1 cos 

2

q

 1

d

 .

x

 sin 2 

x

y

/

a

iii

)

B

,   0 

1

y

p

 1

dy y

p

q

.

x

y

/

1 

y

 

6

7. Beta functions in terms of gamma functions

B

  

    

p

 

q q

Prove )     0 

t p

 1

e

t dt

 2  0 

y

2

p

 1

e

y

2

dy

,  4     0 0

x

2

q

 1

y

2

p

 1

e

 

x

2 

y

2 

dxdy

  4    0 0  / 2

r

cos 

 

r

sin 

2

p

 1

e

r

2

rdrd

  2  0 

x

2

q

 1

e

x

2

dx

 4  0 

r

2

p

 2

q

 1

e

r

2

dr

 0  / 2

cos 

 

sin 

2

p

 1

d

  1 2 

p

q

 1 2

B

 

.

7

- Example

I

   0

1

x

3 

dx x

5

cf

.

B

 

   0

1

y

p

 1

dy y

p

q

.

p

q

 5 , 

   

 

 

1 

p

 1  3 3 !

4 !

 1 4 .

p

 4 ,

q

 1 .

8

8. The simple pendulum

T

 1 2

mv

2

V

 1 2

m l

  2  

mgl

cos 

L

T

V

 1 2

ml

2   2 

mgl

cos 

d dt

ml

2   2  

mgl

sin   0      

g l

sin  .

- Example 1 For small vibration, sin        

l g

 

T

  1  2 

l

/

g

.

9

- Example 2     

l g

sin         

l g

sin    1 2   2 

l g

cos   const .

cf.

elliptic

or

 

d

 integral   

l g

sin 

d

 : In case of 180   

  90 

 0  swings (-90  const.

 0 .

to +90  )  0 1 2   2 

l g

cos  ,  / 2

d

dt

 2

g l

cos  ,

d

 cos   2

g dt

.

l d

 cos   

T

 4 2

g l

 0

T

/ 4

dt

l

2

g

 0  / 2 2

g l

T

4 .

d

 cos  , Beta function!

!

Using computer,

T

 7 .

42

l

/

g

.

10

9. The error function (useful in probability theory) - Error function: erf  2   0 

e

t

2

dt

.

- Standard model or Gaussian cumulative distribution function   1 2   

x

e

t

2 / 2

dt

 1 2  1 2 erf 

x

/ 2    1 2  1 2   

x

e

t

2 / 2

dt

 1 2 erf 

x

/ 2 .

 - Complementary error function erfc  2  

x

 

e t

2 / 2

dt

 1  erf 

x

/ 2  , erfc

x

2  2  

x

 

e t

2 / 2

dt

.

- in terms of the standard normal cumulative distribution function erf

 

 2     1 .

11

- Several useful facts erf erf erf   erf   2   0  

e t

2

dt

 2   0

x e

t

2

dt

 2  1 2  1 2  2  1 2   1 .

2   0

x

  1 

t

2 

t

4 2 !

   

dt

 2   

x

x

3 3 

x

5 5  2 !

    .

x

 1  - Imaginary error function: erfi erf  2   ierfi  0

x e t

2

dt

.

12

10. Asymptotic series erfc  1  erf  2  

x

 

e t

2

dt

.

Using

e

t

2 

t

1

te

t

2 

t

1

te

t

2  1

t d dt

x

 

e t

2

dt

 

x

 1

t d dt

1 2

e

t

2

dt

 1

t

 1 2

x e

x

2  1 2 

x

t

2 1

e

t

2

dt

.

1 2

e

t

2   , 1 2

e

t

2

x

  

x

 1 2

e

t

2    1

t

2

dt

13

Using 

x

d dt

 

e

t

2 

 

d

/

dt

  1 2

e

t

2  1 2

e

t

2

dt

 1

t

3 1 2

e

t

2

x

  

x

 1 2

e

t

2    3

t

4

dt

 1 2

x

3

e

x

2  3 2 

x

t

4 1

e

t

2

dt

.

erfc  1  erf ~

x e

x

2  1  1 2

x

2  2 1

x

 2 3 2  1  3

x

 5 3     .

x

 1  - This series diverges for every x because of the factors in the numerator. For large enough x , the higher terms are fairly small and then negligible. For this reason, the first few terms give a good approximation. (asymptotic series) 14

11. Stirling’s formula - Stirling’s formula

n

!

~

n n e

n

2 

n

p

 1

p p e

p

2 

p

  1  1 12

p

 1 288

p

2     ~

p p e

p

2 

p

.

15

11. Elliptic integrals and functions - Legendre forms: First kind :

F

Second kind :

E

  0    0 

d

 1 

k

2 sin 2  , 0 

k

 1 , 1 

k

2 sin 2 

d

 , 0 

k

 1 .

- Jacobi forms:

t

 sin  ,

F

   0

x

 sin 

d

 1 

k

2 sin 2    0

x

1 

t

2

dt

1 

k

2

t

2 ,

E

  0  1 

k

2 sin 2 

d

   0

x

1 

k

2

t

2

dt

.

1 

t

2 0 

k

 1 , 16

- Complete Elliptic integrals (  =  /2, x =sin  =1):

K or K E or E

F

 2 ,

k

  0  / 2

d

 1 

k

2 sin 2    1 0 1 

t

2

dt

1 

k

2

t t

, 

E

 2 ,

k

  0  / 2 1 

k

2 sin 2 

d

   1 0 1 

k

2

t t dt

.

1 

t

2 - Example 1  0  

or

/ 3

E

1 

 

2 sin 2 

d

 

E

 3 / 2 , 1 / 

E k

 

or k

E

  / 3 , 1

E

  , sin  1

k

  2  ~ 0 .

964951

E

 / 3 ,  / 4

 17

- Example 2  0  / 3 16  8 sin 2 

d

  4  0  / 3 1 

 

2 sin 2 

d

 cf .

    1 2 1 

k

2 sin 2 

d

 

E

 1 ,

k

cf .

F

n

E

n

   ,

k

   ,

k

 2

nK

2

nE

 

F

k E

 

.

 2 ,

k

18

- Example 4. Find arc length of an ellipse.

x

ds

2

a

sin  , 

dx

2 

y dy

2  

b

a

cos  2 cos 2  

b

2 sin 2 

d

 2 .

ds

    elliptical

a

2 

a

2 

b

2

sin 2 

d

 

a

 integral of the second 1 

a

2 

a

2

b

2 sin 2 

d

 .

kind ,

k

2 

a

2 

a

2

b

2 

e

2 : eccentrici ty of ellipse   (using computer or tables) 19

- Example 5. Pendulum swing through large angles.

  2  2

g

cos 

l

 const .

   2  2

g

cos 

l

 cos 

.

0  cos 

d

  cos   2

g l T

 4 

T

  4

l

2

g

2

K

sin  2  4

l g

2

K

sin  2 2

K

sin  2

elliptic integral

20

For

α T

  4 not too large sin 2  1 2 , approximat ion by series.

l g

 2 2  1  1 2 sin

for small 2  2  ,  1 2 3 4 2 sin 4  2 sin  / 2 ~  / 2

   2 

l g

  1   2 16     - For  =30  , this pendulum would get exactly out of phase with one of very small amplitude in about 32 periods.

21

- Elliptic Functions

u

  0

x dt

1 

t

2  sin  1

x u

  0

x

1 

t

2

dt

1 

k

2

t

2  sn  1

x

( elliptic function) 

x

 sn

 

.

cn  1 

x

2 dn

d du

 sn  1 

k

2

x

2   cn 22