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The Error Function Mathematical Physics R. I. Badran The Error Function Mathematical Physics The Error Function and Stirling’s Formula The Error Function: x 2 The curve of the Gaussian function y e is called the bell-shaped graph. The error function is defined as the area under part of this curve: x 2 2 erf (x) et dt 1. . 0 There are other definitions of error functions. These are closely related integrals to the above one. 2. a) The normal or Gaussian distribution function. x t2 1 1 1 x P(, x) e 2 dt erf ( ) 2 2 2 2 Proof: Put t 2u and proceed, you might reach a step of x 1 2 P(0, x) eu du P(,x) P(,0) P(0,x) , where 0 1 x P(0, x) erf ( ) Here you can prove that 2 2 . This can be done by using the definition of error function in (1). 0 u2 I I e du Now you need to find P(,0) where . To find this integral you have to put u=x first, then u= y and multiply the two resulting integrals. Make the change of variables to polar coordinate you get R. I. Badran The Error Function Mathematical Physics 0 2 2 I 2 er rdr d 0 From this latter integral you get 1 I P(,0) 2 and 2 . 1 1 x P(, x) erf ( ) 2 2 2 Q. E. D. x 2 t 1 2 1 x 2.b P(0, x) e dt erf ( ) 2 0 2 2 (as proved earlier in 2.a). 3. a) The complementary error function 2 2 erfc(x) et dt 1 erf (x) x Proof: Put t2= u and use the definition (1) of error function and the definition of (1/2). 2 x 2 t erfc( ) e 2 dt 3. b) 2 x Proof: (This will be left as an exercise for the students). 4. erf (x) 2P(0, x 2) 2P(, x 2) 1 Proof: (The student may take this as another exercise). R. I. Badran The Error Function Mathematical Physics Some useful properties of error function: 1. erf (x) erf (x) This means that the error function is an odd function. [solve problem 3 in section 9 chapter 11] 2. erf () 1. [This can be shown by putting the limit x equals and using the definition of (1/2)]. 2 x3 x5 erf (x) (x .......) 3. 3 5.2! when x1 Stirling’s Formula: This is an approximated formula for the factorial or Gamma function that can be used to simplify formulas involving factorials. n n n!~ n e 2n OR p p ( p 1) ~ p e 2p The sign ~, here, means asymptotic to. Derivation of Stirling's formula: ( p 1) p! x pexdx Starting with and rewrite the 0 integral as R. I. Badran The Error Function Mathematical Physics ( p 1) p! e pnxx dx . 0 Change the variable x p y p . The limits are y p for x0 and y for x. pn( p y p) p y p ( p 1) p! e pdy p Firstly we can write y y n( p y p) n[ p(1 )] np n(1 ) p p For large p, the second logarithm on the right hand side of the last equation can be expanded as follows: y y y 2 n(1 ) p p 2p [Hint: The binomial expansion used is: x 2 x3 x 4 n(1 x) x for 1 x 1]. 2 3 4 By substituting the expanded logarithm into the last integral, then we get: py py 2 pnp( ) p y p ( p 1) p!~ e p 2 p pdy p y 2 ( p 1) p!~ pe pnp p e 2 dy p The integral and its cofactor can be rewritten as: R. I. Badran The Error Function Mathematical Physics 2 2 y p y p p 2 2 ( p 1) p!~ p p e [ e dy e dy] . y2 Exercise: Show that e 2 dy 2 Using the last result mentioned above we get: 2 p y p p p p 2 ( p 1) p!~ 2p p e p p e e dy The integral in the last term tends to zero as p - p p ( py 1)y py!2~ p e 2p Q.E.D n(1 ) (This Stirling'sp p formula2p is a good approximation for large p) Note: When higher terms in the previous expansion of are considered, a better asymptotic expansion for the Gamma or factorial function can be obtained. i.e. 1 1 ( p 1) p!~ p pe p 2p(1 ) 12 p 288p2 (The student can try to prove that at home). .
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