chapter 11 Special functions

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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
  0
1



1  x 
1
 x
xe


e
dx

.


2
0

0


 

0


2
23
2 x
3  x
Similarly,  x e dx  3 ,
x
e
dx

4


xe ax dx  
1

0


0
n  x
xe
dx 
n!

n 1



0
0

x n e  x dx  n!   1
2
3. Definition of the gamma function: recursion relation (p:
noninteger)

p 1  x
- Gamma function p   0 x e dx, p  0.

n    x n 1e  x dx  n  1!,
0

n  1   x n e  x dx  n!.
0

p x
- Recursion relation  p  1  0 x e dx  p!,
p  1.
 p  1  p p 
- Example 9 / 4  (5 / 4)5 / 4  5 / 41/ 41/ 4
so, 1/ 4  9 / 4  16 / 5.
3
4. The Gamma function of negative numbers
 p  
1
 p  1 ( p  0)
p
- Example
 0.3 
cf .
 p  
1
1
0.7 ,  1.3 
0.7 .
 0.3 1.3
 0.3
1
 p  1   as p  0.
p
- 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
 1/ 2  
(prove) 1 / 2   

0
1 / 2
2
  p 1  p  
 1

1 t
 y2
 y2
e dt   e 2 ydy  2  e dy.
0 y
0
t
 4

0


0
e 
 x2  y2
dxdy  4
 / 2  r 2
 
0
0
e
rdrd   .

.
sin p
5
6. Beta functions
B p, q    x p 1 1  x  dx,
1
q 1
0
 y
i ) B  p, q     
0
a
a
ii) B p, q   2 
 /2
0
iii ) B p, q   

0
p 1
p  0, q  0.
y

1



a


q 1
a
dy
1
q 1
 p  q 1  y p 1 a  y  dy. x  y / a 
0
a a
sin  2 p 1 cos 2 q 1 d .
y p 1dy
.
pq
1  y 
cf . B p, q   Bq, p 
x  sin  
2
x  y / 1  y 
6
7. Beta functions in terms of gamma functions
 p q 
 p  q 
B  p, q  
P rove)

 p    t
0
e dt  2  y
0
 p q   4 

0
 4

0



0
2 p 1  y 2
e
x 2 q 1 y 2 p 1e  x
 /2
2
2 p  2 q 1  r 2
e
dr
 /2
0

q   2 x 2 q 1e  x dx
dy,
 y2
2
0
dxdy
2 q 1
2 p 1  r




r
cos

r
sin

e
0
 4 r
0

p 1  t
2
rdrd
cos 2 q 1 sin  2 p 1 d  1  p  q   1 B p, q .
2
2
7
- Example
I 

0
x3dx
1  x 5
cf . B p, q   

0
y p 1dy
.
pq
1  y 
p  q  5, p  1  3  p  4, q  1.
41 3! 1
  .
5
4! 4
8
8. The simple pendulum
T
 
2
1 2 1
m v  m l
2
2
V   m glcos
L  T V 

1 2 2
m l   m glcos
2

d
m l2 2  m glsin   0 
dt
g
l
   sin  .
- Example 1 For small vibration,
sin   
g
l
   
 T
1

 2 l / g .
9
- Example 2
g
g
g
   sin      sin  or d   sin d :
l
l
l
1 2 g
  cos  const.
2
l
cf. elliptic integral
In case of 180 swings (-90 to +90)
  90  0  const. 0.
1 2 g
  cos ,
2
l

 /2
0
d
2g

cos ,
dt
l
d
2g

dt.
l
cos
d
2g T / 4
2g T

dt

 .

0
l
l
4
cos
 T 4
l  / 2 d
,

0
2g
cos
Beta function!!
Using computer, T  7.42 l / g .
10
9. The error function (useful in probability theory)
- Error function: erf x  
2



0
e  t dt.
2
- Standard model or Gaussian cumulative distribution function
1 x t 2 / 2
1 1
 x  
e
dt

 erf x / 2



2 2
2
 x  
1
1

2
2

x
et
2

/2




1
dt  erf x / 2 .
2
- Complementary error function
2  t 2 / 2
erfcx  
e
dt  1  erf x / 2 ,

x


2
 x 
erfc


2





x
et
2
/2

dt.
- in terms of the standard normal cumulative distribution function
erf x  2 x 2  1.
 
11
- Several useful facts
erf  x   erf  x 
erf   
2



0
e  t dt 
2
2 1 1
2 1
  
  1.
 2 2
 2



t4
2 
x3
x5
2



.  x  1
erf  x  
e
dt

1

t



dt

x









0
0
2!
3 5  2!

 



2
x
t 2
2
x
- Imaginary error function:erfix  
2

x
0
t2
e dt.
erf ix   ierfix 
12
10. Asymptotic series
erfcx   1  erf x  
2
 

x
t 2
e dt.
2
1 2 1 2 1 d  1 t 2 
Using e  t  te t  te t 
  e ,
t
t
t dt  2



x
e  t dt  
2

x


1 d  1 t 2 
1  1 t 2 
1  t 2  1 
  e dt    e   x   e   2 dt
t dt  2
t 2

x
 2
 t 
1  x 2 1  1 t 2

e   2 e dt.
2x
2 x t
13
 
 

Using 1 / t 2 e  t  1 / t 3 d / dt  12 e  t


x
2
2



d  1 t 2 
1  1 t 2 
1  t 2  3 
1  x 2 3  1 t 2
  e dt  3   e   x   e   4 dt  3 e  x 4 e dt.
dt  2
t  2
2x
2 t

x
 2
 t 

e 
1
1 3
1 3  5
erfcx   1  erf x  ~
1 2 

 .  x  1
2
3
2
2


x   2x
2x
2x

x2
   
- 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 ne n 2n
 p  1  p e
p p


1
1
p p

2p 1 



~
p
e
2p .
2

12
p
288
p


15
11. Elliptic integrals and functions
- Legendre forms:
- First kind : F  , k   
d

1  k sin 
2
0
- Second kind : E  , k   

0
2
0  k 1,
,
1  k 2 sin 2  d ,
0  k 1.
dt
0  k 1,
- Jacobi forms:
t  sin  ,
x  sin 
F  , k   

d

x
0
1  k sin 

1  k sin  d  
E  , k   
0
2
2
2
2
1 t
0
x
0
2
1 k t
2 2
1  k 2t 2
1 t
2
,
dt.
16
- Complete Elliptic integrals (=/2, x=sin=1):
1
d
dt
   /2
K or K k   F  , k   

,
2
2
2
2 t
0
0
2


1  k sin 
1 t 1 k t
2 t
1 1 k t
   /2
2
2
E or E k   E  , k   
1  k sin  d  
dt.
2
0
0
2


1 t
- Example 1

 /3
0
or


1  1 / 2 sin 2  d  E  , k   E  / 3, 1 2 ~ 0.964951
2

E x, k   E 3 / 2,1 / k




or E  , sin 1 k  E  / 3,  / 4
17
- Example 2

 /3
0
cf .
16  8 sin  d  4
2
 /3
0
2


1  1/ 2 sin 2  d
2
1  k 2 sin 2  d  E 1 , k   E 2 , k 
1
cf . F n   , k   2nK  F  , k 
E n   , k   2nE  E  , k .
18
- Example 4. Find arc length of an ellipse.
x  a sin  ,
y  b cos


ds2  dx2  dy2  a 2 cos2   b 2 sin 2  d 2 .
 ds  


a  a  b sin  d  a 
2
2
2
2
a 2  b2 2
1
sin  d .
2
a


a 2  b2
2
2
 elliptical integral of the second kind, k 
 e : eccentricity of ellipse
2
a


(using computer or tables)
19
- Example 5. Pendulum swing through large angles.
 2 
2g
cos  const.
l
 2 
2g
cos  cos .
l


0
d
2 g T
 

 2 K  sin  elliptic integral
l 4
2
cos  cos

T  4
l
l
 
 
2 K  sin   4
2 K  sin 
2g
2
g
2


20
For α

not too large sin 2 /2  12 , approximation by series.
2
2
2



l    1 

1
3
l



2
4
1 
T  4
1    sin
    sin
   2
 

g 2   2 
2 2 4
2
g  16


for small  ,
sin  / 2 ~  / 2
- For =30, this pendulum would get exactly out of phase with one
of very small amplitude in about 32 periods.
21
- Elliptic Functions
u
x
u
x
0
0
dt
1 t2
 sin 1 x
dt
1  t 2 1  k 2t 2
 sn 1 x (elliptic function) x  sn u .
cnu   1  x 2
dn u   1  k 2 x 2
d
sn u   cnu dn u 
du
22
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