The numerator of equation (2) is an almost harmonic series. The function .... Now add and subtract a term which is equal to (1" ao )z2 in the numerator. =1". 1+. 1.
Paper 21 A method for inverting generating functions that are made up of the ratio of two harmonic series to find the survival probability and the first passage probability that occur in 2d random walk theory Ron Zamir © August 2015 (Revised April 2016) I.
Description of the problem
The generating function F(L, L;z) for the first passage probability for a symmetric random walk on a 2d square lattice, in which the walker starts at lattice point (x,y)=(L,L), moves to one of 4 adjacent lattice sites with
1 , and arrives for the first time at the origin at time n, is given by an expression which is the ratio of 4 ! expression was stated and derived in my unpublished paper 1 and is: two series. This probability
!
(1)
(2L)! 2L z Hypergeometric2F1[L + 12 , L + 12 ;2L + 1;z 2 ] 2 P(L, L;z) 2 (L!) . F(L, L;z) = = P(0,0;z) Hypergeometric2F1[ 12 , 12 ;1;z 2 ] 4L
In addition, this expression was shown in my unpublished paper 1 to be almost equal to
!
(2L)! 2L z 2 (L!) 2 4L
(2)
F(L, L;z) "
$ 4( L+k )*1 ' 2( L+k ) $ 1 ' & $&1 + L ') ) &1 * ) + ) 2k 1 1&% k( % 2(L + k) ( + z 2L , & z 1 2( L+k )*1 ) 2L+k+ # k=1 k $ 2$ 1 ' & 1 + 2L ' ) ) &1 * ) && ) k ( % (L + k) ( %% ( . 1 + $ 1 ' 2k 1 + , & )z # k=1 % k (
The denominator of equation (2) is a pure harmonic series with a term equal to 1 added on. The numerator of equation (2) is an almost harmonic series.
! The function
" 4( L+k )(1 % 2( L+k ) " 1 % $ "$1 + L %' ' $1 ( ' $# k& ' # 2(L + k) & (3) $ I call a cutoff function because it starts close to 1 2( L+k )(1 ' 2L+k+ 2" 1 % $ "1 + 2L % ' ' $1 ( ' $$ ' k & # (L + k) & ## &
0 for low values of k and then increases to a maximum value of 1 for higher values of k. Its effect is to lower (or cutoff) the value of the low k harmonic terms in the numerator and not to affect the high k harmonic terms.
!
2L
In the numerator, there is also a multiplicative factor of z which makes the starting coefficient of F(L, L;z) happen at a power of z that is equal to 2L. The physical meaning of this term is to not allow first passage probability until the walker has had enough time (n=2L) to reach the origin from its starting position of (x,y)=(L,L) which involves at least 2L steps to be made.
!
1
Since the numerator and denominator of F(L, L;z) are basically harmonic series with a cutoff function in the numerator, I thought it would be a good idea to form a series which is made up of the ratio of two generating functions that are pure harmonic, and then to see the effect of lowering the value of only one of the coefficients in the numerator. ! By doing this, I could see how the lowering of a single term in the numerator would affect the entire inverted series F(L, L;n) and the survival probability for the walker. II. Development of the method for the case in which the first term in the numerator of F(L, L;z) is ! lowered. I start by making all the terms of F(L, L;z) pure harmonic. So I choose F(L, L;z) to be: ! (4)
# 1 ) # 1 2k && 2L %1 + * % z ((z $ " $ k '' F(L, L;z) = ! k=1) = z 2L . # 1 # 1 2k && %1 + * % z (( $ " k=1 $ k ''
!
This generating function for F(L, L;z) implies that the particle reaches the origin with certainty (probability=1) when n=2L. ! This is not a physically reasonable generation function for F(L, L;z) . ! Things become much more interesting when the first term in the numerator is set to a o , which is chosen to be a number between 0 and 1. ! So F(L, L;z) becomes !
# 1 ) # 1 2k && 2L a + * % z ((z % o " k=1 $ k '' $ (5a) F(L, L;z) = . # 1 ) # 1 2k && %1 + * % z (( $ " k=1 $ k ''
!
Mathematica 10.1 can numerically invert equation (5a) if a value of L is chosen. For now, choose L=1. ! With this choice of L, equation (5a) becomes:
# 1 ) # 1 2k && 2 a + * % z ((z ) % o " k=1 $ k '' $ (5b) F(1,1;z) = = * (c2k z 2k ) . ) # 1 # 1 2k && k=1 %1 + * % z (( '' $ " k=1 $ k
!
2
The first eight series coefficients of the inversion of equation (5b) are shown below: c2 = a o
(1 " a o ) #
c4 =
c6 =
(1 " a o )("2 + # ) 2# 2
(1 " a o )(3 " 3# + # 2 ) c8 = 3# 3
(1 " a o )("12 + 18 # "11# 2 + 3# 3 ) c10 = 12 # 4
(1 " a o )(60 "120 # + 105 # 2 " 50 # 3 + 12 # 4 ) c12 = 60 # 5
c14 =
(1 " a o )("360 + 900 # "1020 # 2 + 675 # 3 " 274 # 4 + 60 # 5 ) 360 # 6
(1 " a o )(210 " 630 # + 875 # 2 " 735 # 3 + 406 # 4 "147 # 5 + 30 # 6 ) c16 = 210 # 7
To obtain numerical values for these coefficients, a o must be chosen. So choose a o to be 0 which is the lowest it can be. This eliminates the first term in the numerator of F(1,1;z) . With this choice of a, the values of the first 8 coefficients can be numerically calculated and are shown below. c2 = 0 ! ! ! c4 = 0.31831 c6 = 0.0578338 c8 = 0.0370336 c10 = 0.024811 c12 = 0.0184036 c14 = 0.0144489 c16 = 0.0117963
3
Now the first 40 numerical coefficients of F(1,1;z) with a o = 0 (which is F(1,1;n) ) are plotted on a log log plot. ! ! !
log[ F(1,1;n)]
log(n) After the anomalous first point, the data seems to be linear when it is plotted on a log log plot. However, we shall soon see that this data should not be modeled by a power law function. The survival probability S(1,1;n) is can be found by taking 1 – nth partial sum of the coefficients of F(1,1;z) when z is set to 1. As an equation, this is (6) S(1,1;n) = 1 " # F(1,1;z = 1) . nth partial sum
A graph of S(n), which is found by applying equation (6) to the numerically inverted F(1,1;z) with a o = 0 , is shown below. ! S(1,1;n)
!
4
n
III.
Finding asymptotic approximations to S(1,1;n) and F(1,1;n) when the first term of the F(1,1;z) series is reduced.
The method that I use to find the asymptotic approximations to S(1,1;n) and F(1,1;n) was originally ! stated in my unpublished paper 1. !
!
First form S(1,1;n) .
(6)
S(1,1;n) = 1 "
!
# F(1,1;z = 1)
!
nth partial sum
!
,$ 1 * $ 1 2k '' 2 / .& a o + + & z ))z 1 # k=1 % k (( 1 % . ! =1" + . $ 1 * $ 1 2k '' 1 nth partial sum. & z )) 1 .- &%1 + # + (( 0 z=1 k=1 % k
2
Now add and subtract a term which is equal to (1 " a o )z in the numerator.
,$ 1 * $ 1 2k '' 2 / 2 .&1 + + & z ))z " (1 " a o )z 1 (( % # k=1 % k 1 =1" + . ! * $ ' $ ' 1 1 1 2k nth partial sum. &1 + + & z )) .10 (( % # k=1 % k z=1
, / . 1 (1 " a o )z 2 2 . 1 . S(1,1;n) = 1 " + z " * $ ' $ ' 1 1 2k 1 nth partial sum . &1 + + & z ))1 .((0 z=1 % # k=1 % k
(7)
2
Consider now the nth partial sum of z : (8)
" [z 2 ]
nth partial sum
. z=1
! The nth partial sum of only the z 2 term in the brackets of equation (8) is always 1 because the coefficient 2 4 2 that multiplies z is 1. When the next partial sum is taken up to the z term, the z term still makes a 4 4 ! contribution of 1, and since there is no z term, the z term makes a contribution of 0. Adding these ! contributions together still equals 1. So " [z 2 ] = 1. !
!
Now apply this result to (7). !
nth partial sum
z=1
!
, ,! / / . . 1 1 2 (1 " a )z o 1 1 S(1,1;n) = 1 " .1 " + . . nth partial sum.$ 1 * $ 1 2k ''1 1 & z ))1 1 . .-&%1 + # + % k ((0 z=1 0 k=1
5
!
!
(9)
, / . 1 2 (1 " a )z o 1 S(1,1;n) = + . * $ ' $ ' 1 1 . 1 2k nth partial sum 1 + z & ) + & ) .-% # k=1 % k ((10 z=1
The next step is the key to the method for finding the approximation to S(1,1;n).
! Use the first two terms of the Taylor series expansion of 1 , which are 1 " x + O(x 2 ) , and apply this to 1+ x equation (9).
(10)
S(1,1;n) "
% 1 + % 1 2k ((0 2 (1 # a )z , / '1 # !, ' z **2 ! o ))1 z=1 & $ k=1 & k nth partial sum .
To achieve the nth partial sum of this series, the upper k index in the sum in brackets should be set equal
n "1. The highest power of z is then 2k max + 2 which when evaluated at ! 2 #n & n k max = "1 will give an upper z index of 2% "1( + 2 = n . This choice of k max will only allow terms up $2 ' 2 to a time = n when the nth partial sum is taken. ! ! ! Using this value of k max as the upper limit in the k sum is equivalent ! to taking the nth partial sum. ! Insert this result for k max into equation (10) and also let z-‐>1. to k max which should equal
%n ( %n ( %n ( % (/ % ( % ( !, ' #1* ' #1* ' #1* &2 ) &2 ) &2 ) . % ( 1 % ( % ( ' * ' * ' 1 1 1 1 1 1 * S(1,1;n) " .(1# a o )z 2 '1# + ' z 2k **1 = (1# a o )(1) 2 '1# + ' *(1) 2k * = (1# a o )'1# + ' **. ! ' $ k=1 & k )*1 ' $ k=1 & k ) * ' $ k=1 & k )* .& )0 z=1 & ) & )
Now transform this expression by using the opposite of the Taylor series expansion that was used above, now converting from 1 " x to
(11) So S(1,1;n) "
!
1 . 1+ x
(1 # a o )
%n ( % ( ' #1* ! ' 1 & 2 ) % 1 (* '1 + $ + '& k *)* k=1 ' * & )
=
(1 # a o ) % 1 % n (( '1 + HarmonicNumber' #1** & 2 )) & $
!
6
.
!
Now the fit from equation (11) is plotted along with the S(1,1;n) data that was graphed on the bottom of page 4.
The blue curve is the exact S(1,1;n) that was plotted on page 4 and which comes from using S(1,1;n) = 1 " # F(1,1;z = 1) (equation (6)), where F(1,1;z) is given by equation (5b). nth partial sum
The first few coefficients of F(1,1;z) obtained in this way are shown on the second half of page 3.
The yellow curve is the approximate S(1,1;n) that is found from equation (11).
!
S(1,1;n)
!
n
The agreement between these two curves is very good, the yellow being about 3% above the blue curve at n=80.
It is useful to note that when an arbitrary L is used so that F(L,L;z) is given by equation (5a), S(L,L;n) becomes the following using the steps that were outlined before on page 6.
, / . 1 2L (1 " a )z o (9b) S(L, L;n) = 1 + .$ * ' $ ' 1 1 2k 1 nth partial sum . & z ))1 .-&%1 + # + ((0 z=1 k=1 % k
!
Note that the only difference between equation (9) and (9b) is that in (9b) there is a z 2 numerator instead of z .
2L
in the
% 1 + % 1 2k ((0 2L (10b) S(L, L;n) " , /(1 # a o )z '1 # , '& z *)*2 ! & $ k=1 k )1 z=1 nth partial sum . ! Now the upper limit in the k sum should be k max . The total z exponent when k max is used should be n.
!
So 2L + 2k max = n .
!
7
!
So k max =
Now insert this value for k max into equation (10b). Inserting this value of k max as the upper limit of the sum in equation (10b) is equivalent to taking the nth partial sum of the series.
!
!
n " L . 2
, % .! ' 1 S(L, L;n) " .(1 # a o )z 2L 1 # '' $ .&
(/ % % 1 2k (*1 1 + '& z *)*1 = (1 # a o )''1 #! * ' $ k=1 k )10 z=1 &
n #L 2
( % 1 (* + '& *)* k=1 k * )
n #L 2
And finally
(11b) S(1,1;n) "
(1 # a o ) %n ( % ( ' #L * &2 ) % 1 (* ' 1 1 + ' ** + ' $ k=1 & k )* ' & )
=
(1 # a o ) % 1 %n (( 1 + HarmonicNumber # L ' ** ' &2 )) & $
.
! Now find an asymptotic fit to F(1,1;n) by using the relation
#S(1,1;n) . #n !
(12) F(1,1;n) = "
% ( ' * (1 " a o ) #' * ! =" + . % ( #n ' 1 n * '&-,1 + $ HarmonicNumber'& 2 "1*)0/*)
Mathematica 10.2 gives for this derivative the following:
$# 2 *$ n ' -' (1 " a o )& " HarmonicNumber,& "1),2/) )# n & , +% 2 ( .( % 6 (13) F(1,1;n) = , w here HarmonicNumber +%$ "1(',2. 2 * 2 $ 1 * n -' 2 # &1 + HarmonicNumber, "1/) + 2 .( % # n "1 2
1 . 2 k k=1
means #
!
!
So F(1,1;n) is not modeled by a power law fit in n. Instead it involves HarmonicNumber terms which become similar to ln when the HarmonicNumber argument becomes large. And so we now see why ! F(1,1;n) should not be modeled by a power law function. 8
Now the fit from equation (13) with L=1 is compared to the exact expression for F(1,1;n) that was graphed at the top of page 4. The blue curve is the exact F(1,1;n) that was plotted on page 4. The yellow curve is the approximate F(1,1;n) that is found from equation ! (13).
Log[F(1,1;n)]
The approximate (yellow) curve is too low by a factor that is just about equal to 2.
9
Log[n]
IV.
Finding asymptotic approximations to S(L, L,n) and F(L, L;n) when a single general term of the F(L, L;z) series is reduced.
Now consider the series
! (13.5) F(L, L;z) =
, 1 .1 + - "
! ! # 1 2k & 1 1 2m 1 z + * %$ z (' + a m k " m " k=1 # 1 + # 1 2k && %1 + * % z (( '' $ " k=1 $ k
m)1
# 1 2k &/ 2L % z (1z '0 k=m+1 $ k +
*
where 0 " a m < 1.
In this series, a middle term of index m has been lowered by a multiplicative ! constant a m that is between 0 and 1.
!
Now find an approximate expression for the survival probability S(L, L;n) using equation (6), which is
(6)
S(L, L;n) = 1 "
!
# F(L, L;z = 1) . nth partial sum
,, 1 ..1 + - # ! S(L, L;n) = 1 " * . nth partial sum . .-
! $ 1 2k ' 1 1 2m 1 z + * &% z )( + a m k # m # k=1 + $ 1 $ 1 2k '' &1 + * & z )) % # k=1 % k ((
m"1
$ 1 2k '/ 2L / * &% z )(1z 1 0 1 k=m+1 k 1 10 z=1 +
$ 1 1 2m ' 2L z )z in the numerator. %# m (
Now add and subtract a term which is equal to (1 " a m )&
, $ 1 1 2m ' 2L / (1 " a m )& z )z 1 . % ( 1 # m 2L (14) S(L, L;n) = 1 " . + .z !" $ 1 * $ 1 2k '' 1 nth partial sum . &1 + + & z )) 1 .(( 0 z=1 % # k=1 % k
There is no contribution from the bracketed partial sum in equation (14) until n=2L. So for n=0 to n=2(L-‐ 1), S(L, L;n) is 1-‐0=1.
! Starting at n=2L and going up to n=2m+2L, the partial sum from the z 2L term contributes 1 so that S(L, L;n) = 1 "1 = 0 .
!
For n>2m+2L, S(L, L;n) is given by
! , , $ 1 1 2m ' 2L / $ 1 1 2m ' 2L / (1" a ) z z (1" a ) z )z 1 ) 1 . . m & m & % ( % ( 1 # m # m 1 = + . (15) . ! S(L,L;n > 2m + 2L) = 1"1+ + . $ * * $ 1 $ 1 2k '' 1 1 $ 1 2k '' 1 nth partial sum. nth partial sum. & z )) 1 & z )) 1 .- &%1+ # + .- &%1+ # + (( 0 z=1 (( 0 z=1 k=1 % k k=1 % k
!
10
As was done before, use the approximation (16)
1 " 1 # x + O(x 2 ) to rewrite equation (15). 1+ x
% 1 1 2m ( 2L % 1 + % 1 2k ((0 S(L, L;n > 2m + 2L) " , /(1 # a m )' z *z '1 # , ' z **2 &$ m ) & $ k=1 & k ))1 z=1 nth partial sum. !
, $ 1 1 ' 2m+2L $ 1 * $ 1 2k ''/ = + .(1 " a m )& )z &1 " + & z ))1 . %# m( ((0 z=1 % # k=1 % k nth partial sum When n=2m+2L, the allowable value of k max will have to be 0 since n = 2m + 2L = 2m + 2L + 2k max . When n>2m+2L, the allowable value of k max will be such that n = 2m + 2L + 2k max , or
k max =
n " m " L . 2
!
!
!
!
So the upper index of the sum for S(L, L;n > 2m + 2L) should be adjusted to a value that is equal to k max . This is equivalent to taking the nth partial sum of the series.
, % 1 1 ( 2m+2L % 1 S(L, L;n > 2m + 2L) "! *z .(1 # a m )' '1 # &$ m) & $ -
% 1 2k ((/ z **1 ))0 k=1 k
kmax
+ '&
z=1
+ $ $ ' 1 1 1 2m+2L & = -(1 " a m )& 1" )z && # %# m( -, %
n "m"L 2
*
k=1
'. $ 1 2k ')0 & z ))0 . % k () (0/ z=1
Now set z=1.
+ $ $ 1 1 ' 2m+2L & 1 = -(1 " a m )& 1" )(1) && # %# m( -, %
n "m"L 2
*
k=1
'. + $ $ 1 2k ')0 $ 1 1 '& 1 & (1) ))0 = -(1 " a m )& ) 1" %k () % # m (&& # (0/ -, %
n "m"L 2
*
k=1
'. $ 1 ')0 & ))0. % k () (0/
Now use the opposite of the Taylor approximation (16), which is that 1 " x #
11
!
1 . This yields 1+ x
, / . 1 % ( 1 1 . (1 # a )' *1 m . &$ m) 1 . S(L, L;n > 2m + 2L) " . n % (1 #m#L .' 1 2 % 1 (*1 1 + ' **1 + .' $ ' k=1 & k )* .-& )10
+ . %1 1( (1 # a m )' * 0 & ) $ m 0 (17) S(L, L;n > 2m + 2L) " % ( + . 1 n 0 -,'&1 + $ HarmonicNumber-, 2 # m # L0/*)0/
Now take the negative derivative of S(L, L;n > 2m + 2L) with respect to n to find F(L, L;n > 2m + 2L) .
!
, / &1 1) (1 " a ) ( + . 1 !#S(L, L;n) m #. '% m* 1 F(L, L;n > 2m + 2L) = " $" ,n /)1 #n #n . & 1 .-('1 + % HarmonicNumber.- 2 " m " L10+*10
+ % 1 1 (% $ 2 +n .(. # HarmonicNumber- # m # L,20*0 - (1 # a m )' *' ,2 /)0 & ) $ m 6 & (18) F(L, L;n > 2m + 2L) " 2 0 % 1 +n .( 2 $ '1 + HarmonicNumber- # m # L0* 0 ,2 /) & $ , /
Now the approximate theoretical values for S(L, L;n > 2m + 2L) (equation (17)) and F(L, L;n > 2m + 2L) (equation (18)) are compared to the exact expressions for these functions which ! are found by numerically inverting the F(L, L;z) series given by equation (13.5) using the series coefficient command in Mathematica 10.1 to obtain F(L, L;n) , and then taking 1 – (the nth partial sum of the F(L, ! L;n) function) that results from the inversion to obtain S(L,L;n). For these calculations, L has ! been chosen to be 5, m has been chosen to be 10, and a10 has been chosen !of F(L, L;z) (equation (13.5)) is missing. to be 0 so the k=10 term in the numerator
!
! !
12
Here is the comparison for S(L, L;n > 2m + 2L) . The blue curve is the exact result that is obtained from taking 1 – (nth partial sum of F(L, L;n) ) which has been numerically inverted from F(L, L;z) that is given by equation (13.5). ! The yellow curve is the approximate theoretical value which is equation (17). ! S(L, L;n) !
!
The agreement for S(L,L;n) is excellent once again.
n
Here is the comparison for F(L, L;n > 2m + 2L) . The blue curve is the exact result that is obtained by inverting F(L, L;z) (equation 13.5) numerically. The yellow curve !is the approximate theoretical value which is equation (18). Log[F(L, L;n)] !
The theoretical (yellow) fit for F(L,L;n) is too low once again by a factor of about 2. 13
Log[n]
V. Approximating the full S(L, L;n) and F(L, L;n) functions when many series coefficients in F(L,L;z) are lowered. Now form the full F(L, L;z) function for a symmetric random walk in 2d that starts at (x,y)=(L,L) on a ! to approximate ! S(L, L;n) and F(L, L;n) . square lattice and try The exact expression for F(L, L;z) is given by equation (1), which is approximated well by equation (2). ! Equation (2) is of the form ! !
#! 1 ) # 1 & 2L 2k & % a o + * % a k z ((z '' " k=1 $ k $ where all the a coefficients are between 0 and 1. F(L, L;z) = # 1 ) # 1 2k && %1 + * % z (( $ " k=1 $ k ''
In section III (equation (11b) of this paper), the contribution to S(L, L;n) from the first lowered term in the numerator was shown to be (11b) S(L, L;n > 2L) contribution from first term "
(1 # a o )
. % 1 ! %n (( '1 + HarmonicNumber' # L ** &2 )) & $
In section IV (equation (17) of this paper), the contribution to S(L, L;n) from each of the other lowered terms in the numerator was shown to be ! (17)
S(L, L;n > 2m + 2L) contribution from other terms
+ . %1 1( (1 # a m )' * - ! 0 & ) $ m 0. "+n .(0 -% 1 -,'&1 + $ HarmonicNumber-, 2 # m # L0/*)0/
The a coefficients have the following form which is seen from equation (2):
!
ao =
(2L)! , 2 (L!) 2 4L
" 4( L+m )(1 % 2( L+m ) " % " % L 1 $ $1 + ' ' $1 ( ' $# m& ' # 2(L + m) & . am = $ 1 2( L+m )(1 ' 2L+m+ 2 " % 1 $ "1 + 2L % ' ' $1 ( ' $$ ' m& # (L + m) & ## &
14
!
The full expression for S(L, L;n) is found by adding all the contributions to S from equations (11b) and (17). This yields
S(L, L;n > 2L) " S(L, L;n > 2L) contribution from first term + S(L, L;n > 2m + 2L) contribution from other terms . ! $ ' *1 1(1 " a ) / ) m & m , (1 " a o ) +# m. & ) = +0 * 1 * $n ' m=1& $n ') 1 1 + HarmonicNumber " m " L ,1 + HarmonicNumber& " L)/ , & )(/.) &%+ # %2 (. %2 + # ( max
* # # 4( L+m )"1 && 2( L+m ) # & # & L 1 , % % %1 + ( ((/ %1 " ( , % %$ m' ((/ $ 2(L + m) ' 1 " , % % 1 (/ 2( L+m )"1 ( 2L+m+ 2 # & # & 2L 1 # , % % (2L)! & ((/ ( %1 " ( ((/ %1 " 4 L % % %$1 + m , 2( m' $ (L + m) ' # 1 1 &$ $ '' $ 2 (L!) ' / . = + 0,% ( # 1 *n -& m=1,$ ) m ' # 1 *n -& / %1 + HarmonicNumber, " L/( %1 + HarmonicNumber, " m " L/( / , +2 .' +2 .' $ ) $ ) , / , / , / ,+ /. max
The maximum value that m can be in the sum to the right is determined from the argument in the HarmonicNumber term. This argument must be at least 1. As m grows,
So the condition for finding m max is that
So m max =
n " L "1 . 2 !
n " m " L decreases. 2
n " m max " L = 1 . 2 !
!
A good approximation to S(L,L;n) is therefore
(19) * # # &&2( L+m ) # & 4( L+m )"1 ((/ 1 , % % #%1 + L &( %1 " ( , % %$ ((/ m' $ 2(L + m) ' 1 " , % % 1 (/ 2( L+m )"1 ( 2L+m + 2# & # & 2L 1 # , % (/ (2L)! & % ( n ( %1 " ( "L"1 ((/ %1 " 4 L % % %$1 + , 2( 2 m' $ (L + m) ' '' 1 1$ $ $ 2 (L!) ' / S(L, L;n > 2L) = + 0 , # *n -& m =1 , ) m # *n -& / 1 1 %1 + HarmonicNumber, " L/( %1 + HarmonicNumber, " m " L/( / , +2 .' +2 .' ) ) $ $ , / , / , / ,+ /.
15
Mathematica 10.2 could not find exact expressions for any of the sums at the right of equation (19) for general L, or for a fixed value of L. Now compare the results from equation (19) to the exact result for S(L,L;n) which is found by inverting F(L,L;z) that is given by equation (1) using the series command in Mathematica 10.1 and then applying equation (6), which is (6) S(L, L;n) = 1 " # F(L, L;z = 1) , to find S(L,L;n). nth partial sum
Below is shown this calculation for L=1. The blue curve is the exact S(L,L;n) using the inversion of equation (1). The yellow curve is the approximate solution, which is equation (19). ! S(1,1;n)
Next is shown the same comparison for L=10. S(10,10;n)
For both calculations, the approximate curve (yellow or red) starts under the exact curve (blue). Soon after, the approximate curve is above the blue curve and mimics its shape with about 3% error by n=150. 16
n
n
Now replace the sum at the right with an integral. This is an approximation.
Also rewrite the HarmonicNumber[x] terms using the approximation ln(x) + " , where " is Euler’s second constant which is about equal to 0.5772.
(20) + $ $ 4( L+m )#1 '' . !2( L+m ) $ ! 1 ' & & $&1 + L ') ))0 &1 # ) & &% ))0 m( % 2(L + m) ( &1 # & 1 ))0 2L+m + 2$ ' 2( L+m )#1 ))0 $ ' 2 L 1 $ & (2 L)! ' & n ) &1 # ) #L#1 & &1 + ))0. &1 # 4 L ) -$ dm ' &% 2 m( % (L + m) ( %% (( % 2 (L!) 2 ( 0 S(L, L;n) " + 2 -& ) $ $ '' $ +n '' % m ( . . 1 $ +n 1 0 &1 + & ln- # L 0 + 1 )) & * + & ln- # m # L 0 + 1 )) 0 / / * % ,2 (( % ,2 (( % % 0 0 0 -, 0/
Compare this result with the continuous result for S(R;T) that applies for a particle pulse that continuously diffuses, starting at a distance ro from the origin, to a central particle at the origin of radius a with an absorbing boundary condition at r=a. This formula is given in my unpublished paper 3 (equation (4)) and is ! (21)
$ 2 ' + d* "*!T [ J 0 (*R)Y 0 (* ) " J 0 (* )Y 0 (*R)] S(R;T) = "& ) , e . 2 2 %# ( 0 * J 0 (* ) + Y 0 (* ) 2
In this formula, R is a dimensionless distance and T is a dimensionless time. They are equal to
!
ro Dt and T = 2 , where ro is the initial distance of the center of the moving particle from the origin, a a ! particle, t is the dimensional time, and a is the radius of the D is ! the diffusion constant for the moving diffusing particle. J 0 and Y 0 are Bessel functions of the first kind. R=
! ! comparing the continuous equation for S(R;T) (equation (21)) to the lattice equation for S(L,L;n) When (equation (20)), ro should be set equal to ! 2 L . This is because when !the diffusing particle starts at
!
!
(x,y)=(L,L), it is a distance of
( )
( 2 ) L from the origin.
! It was !shown in my unpublished paper 20 that a good choice for a is 0.5, and a good choice for D is 1, ! so that the survival probability that is calculated from equation (21) and lattice results for S(L,L;n) are nearly the same. ! When finding a long time asymptotic result for S(R; T large) , only small values of " will contribute to "# 2T
the integral because of the exponential term e . Then the small argument approximations for the Bessel functions in (21) can be used since their ! argument is small. This is also done i! n my unpublished paper (3).
!
17
The small " expansions of J o (" ) and Y o (" ) are !
J o (" ) # 1 and Y!o (" ) #
2 %! % " ( ( ' ln' * + + * . $ & &2) )
So S(R;T) becomes:
!
% 2 , ,+ / / 2 , , +R / /( (1) ln + 2 # (1) . . 1 1 . ln. 1 + 2 1* ' % 2 ( 3 d+ #+ T & $ - - 2 0 0 $ - - 2 0 0) S(R;T large) " #' * 4 e 2 &$ ) 0 + , / 2 , ,+ / / (1) 2 + . . ln. 1 + 2 11 - $ - - 2 0 00 2
+ . $ 2 ' 0 &%" # ln ( R))( $ 2 ' 2 d* " * T 0 . = "& ) 3 e 20 %# ( 0 * + 2 + + * . .. --1 + - - ln- 0 + 1 00 00 , , # , , 2 / // / 2
-
(22)
S(R;T large) "
. 0
& ) ( + ln ( R)] [ d# $# T ( + . e 2 ( % 2 & &# ) ) + # + ( ln( + + , + + ( ' 4 ' '2* * * 2
Equation (22) for continuous diffusion is almost like equation (20) which approximates S(L,L;n) for lattice diffusion. ! In equation (22), " takes the place of m. There is a in equation (20).
d" dm term in equation (22), just as there is a term " m
"2 (20). !4 term takes the place of the " term in equation !
!enominator of equation (22), the In the d
The ln term in the denominator of equation (22) is raised to the second power, while the ln term in the denominator in equation (20) is raised to the first power. ! !
18
It is hard to see how the e
"# 2T
in equation (21) corresponds to the
# # && 2( L+m ) # & 4( L+m )"1 (( 1 % % #%1 + L &( %1 " ( % %$ m' (( $ 2(L + m) ' %1 " % ( term in equation (19) for any value of m. 1 !2L+m+ 2( L+m )"1 ( 2# 1 & % % # 2L & (( 1 + 1 " % ( % ( (( % %$ m' $ (L + m) ' '' $ $
However, as m becomes large, each of the terms in parentheses above do become exponential terms.
" % 2( L+m ) " % 4( L+m )(1 ' 1 $ "$1 + L %' $1 ( ' # e 2L e"2 & $# m& ' # 2(L + m) & In fact, when m is large, $ becomes % 2L "2 ( = 1 . 1 2( L+m )(1 ' 2L+m+ $e e ' 2" 1 % $ "1 + 2L % ' 1 ( ' $ ' $$ ' m& # (L + m) & ## &
Another interesting fact about equation (20) is that as m goes to !infinity, which means that n goes to infinity, equation (20) for S(L,L;n) for lattice diffusion goes to 0. !
*
This is seen because the first term in equation (20) has ,1 +
+
$n '1 HarmonicNumber& # L)/ in the %2 (. "
denominator. As n goes to infinity, the denominator goes to infinity and causes the first term to be 0. The second term in (20), which consists of a ! sum, also goes to 0 as m goes to infinity. This is because the
# # 4( L+m )"1 && 2( L+m ) # & # & L 1 % % %1 + ( (( %1 " ( % %$ m' (( $ 2(L + m) ' numerator is %1 " % ( which becomes (1-‐1) as m goes to infinity, 1 2( L+m )"1 ( 2L+m+ 2# 1 & % % # 2L & (( ( %1 " ( (( % % %$1 + m' $ (L + m) ' '' $ $ * $n 'and also because the denominator, which is , " + HarmonicNumber& # m # L)/ , becomes infinite as %2 (. + m g!oes to infinity making the second term zero also. S(L,L;n) should be 0 for large values ! of n because symmetric random walks in 2d are recurrent, so that the diffusing particle must eventually find the origin and then not survive anymore.
19
VI. Conclusion I have shown a method for approximating the Survival probability S(L,L;n) and the first passage probability F(L,L;n) for a single particle that symmetrically diffuses from a 2d lattice position (L,L) and travels to the origin (0,0). This method involves rewriting the generating function for the first passage probability F(L,L;z) as the ratio of two nearly harmonic series. By lowering a single term in the numerator of F(L,L;z) and showing how this can affect S(L,L,n) and F(L,L;n), I was able to create an approximate S(L,L;n) function by adding together the contributions from individual lowerings of the terms in the numerator of F(L,L;z). This yielded good results, especially for finding S(L,L;n). When the first term in the numerator of the series
# 1 ) # 1 2k && 2L % a o + * % z ((z " k=1 $ k '' $ (5a) F(L, L;z) = was lowered, the resulting S(L,L;n) and F(L,L;n) functions # 1 ) # 1 2k && %1 + * % z (( $ " k=1 $ k '' were
!
(11b) S(L, L;n > 2L) contribution from first term "
(1 # a o ) % 1 %n (( '1 + HarmonicNumber' # L ** &2 )) & $
and
!
$# 2 *$ n ' -' (1 " a o )& " HarmonicNumber,& " L ),2/) ( .( +% 2 % 6 (13b) F(1,1;n) = . 2 $ 1 *n -' 2 # &1 + HarmonicNumber, " L/) +2 .( % # When a general term was lowered in the numerator of the series
! (13.5) F(L, L;z) =
, 1 .1 + - "
F(L,L;n) functions were
m)1
#1 & 1 1 2m 1 z + * %$ z 2k (' + a m k " m " k=1 + # 1 # 1 2k && %1 + * % z (( '' $ " k=1 $ k
!
20
# 1 2k &/ 2L % z (1z '0 k=m+1 $ k +
*
, the resulting S(L,L;n) and
(17)
S(L, L;n > 2m + 2L) contribution from other terms
+ . %1 1( (1 # a m )' * 0 & ) $ m 0 "+n .(0 -% 1 -,'&1 + $ HarmonicNumber-, 2 # m # L0/*)0/
and
!
!
+ % 1 1 (% $ 2 +n .(. # HarmonicNumber- # m # L,20*0 - (1 # a m )' *' ,2 /)0 & $ m )& 6 (18) F(L, L;n > 2m + 2L) " . 2 0 % 1 +n .( 2 $ '1 + HarmonicNumber- # m # L0* 0 ,2 /) & $ , / Finally, when the complete series
# & 2L 1 ) #1 2k & a + a z % ( * % o (z k '' " k=1 $ k $ was considered, where F(L, L;z) = # 1 ) # 1 2k && %1 + * % z (( $ " k=1 $ k ''
ao =
(2L)! and 2 (L!) 2 4L
" 4( L+m )(1 % 2( L+m ) " % 1 $ "$1 + L %' ' $1 ( ' $# m& ' # 2(L + m) & , am = $ 1 2( L+m )(1 ' 2L+m+ 2" 1 % $ "1 + 2L % ' ' $1 ( ' $$ ' m& # (L + m) & ## & the resulting S(L,L;n) function was (19) * # # &&2( L+m ) # & 4( L+m )"1 ((/ 1 , % % #%1 + L &( %1 " ( , % %$ ((/ m' $ 2(L + m) ' 1 " , % % 1 (/ 2( L+m )"1 ( 2L+m + 2# & # & 2L 1 # , % (/ (2L)! & % ( n ( %1 " ( "L"1 ((/ %1 " 4 L % % %$1 + , 2( 2 ' m $ (L + m) ' '' 1 1 $ $ $ 2 (L!) ' / S(L, L;n > 2L) = + 0 , # # & *n -& * 1 ) m 1 n m =1 , / %1 + HarmonicNumber, " L/( %1 + HarmonicNumber, " m " L/( / , +2 .' +2 .' ) ) $ $ , / , / , / ,+ /.
.
21
A comparison was made between the form of the S(L,L;n) approximation for lattice diffusion, and the form of S(R;T) for continuous diffusion which had been previously found in my unpublished paper (3). The two long time expressions for S were seen to be similar in form and are (20)
!
+ $ $ 4( L+m )#1 '' . 2( L+m ) $ ' $ ' L 1 & & &1 + ) ))0 &1 # ) & &% ))0 m( % 2(L + m) ( &1 # & 1 )0 2( L+m )#1 ) 2L+m + 2 $ ' $ ' 2L 1 $ ' & & (2L)! ))0 n 1 + 1 # & ) & ) #L#1 & ))0 &1 # 4 L ) & -$ dm ' % % % 2 m( % (L + m) ( (( . % 2 (L!) 2 ( 0 S(L, L;n) " + 2 -& ) $ $ $ +n . '' . '' 1 $ +n 1 -% m ( 0 &1 + & ln- # L0 + 1 )) & * + & ln- # m # L0 + 1 )) 0 / (( / (( * % ,2 % ,2 % % 0 0 0 -, 0/ -
(22)
S(R;T) "
. 0
& ) ( + ln ( R)] [ d# $# T ( + e 2 . 2 ( # & &# ) ) + % + ( ( ln( + + , + + ' 4 ' '2* * * 2
The approximate S(L,L;n) (equation (20)) that was found in this paper is somewhat cumbersome for the ! use of calculation, and is not less complicated for computing than finding an exact result directly from equation (1). However, it is interesting to see the process by which the approximate S(L,L;n) is derived, and the similarities of the form of S(L,L;n) for lattice diffusion, and S(R;T) for continuous diffusion.
22
Appendix In this appendix is listed the Mathematica 10.4.1 code that was used to produce the graphs in this paper. Top Page 4: Graph of Log(F(1,1;n)) vs. Log(n)
ListLogLogPlot[ Table[ # 1 150 # 1 2*k && 2 %0 + * ) % * z (( * z '' " k=1 $ k $ {n,SeriesCoefficient[ ,{z,0,n}]}, # 1 150 # 1 2*k && %1 + * ) % * z (( '' $ " k=1 $ k {n,2,80,2}]] Bottom Page 4: Graph of S(1,1;n) vs. n
ListPlot[ Table[ $ ' $ 1 150 $ 1 2*k '' 2 &0 + * * & * z )) * z & ) % ( # k % ( k=1 )}, {n,&1 " Sum[SeriesCoefficient[ ,{z,0,k}],{k,4,n,2}] $ 1 150 $ 1 2*k '' & ) &1 + * * & * z )) & ) (( % # k=1 % k % ( {n,4,80,2}],PlotRange + {{0,82},{0,1}}]
23
Page 7: Graph of S(1,1;n) vs. n (Exact calculation and Approximate fit) ListPlot[
{Table[ $ '' 1 150 $ 1 0 + * * & * z 2*k )) * z 2 & (( # k=1 % k % {n,1 " Sum[SeriesCoefficient[ ,{z,0,k}],{k,2,n,2}]}, $ ' 1 150 $ 1 2*k ' &1 + * * & * z )) (( # k=1 % k % {n,2,80,2}], Table[ 1 {n, }, $ +n .' 1 &1 + * HarmonicNumber- " 10) ,2 /( # % {n,2,80,2}]},PlotRange 1 {0,1}]
Page 9: Graph of Log[F(1,1;n)] vs. Log[n] (Exact calculation and Approximate fit)
ListLogLogPlot[ {Table[ # 1 150 # 1 2*k && 2 %0 + * ) % * z (( * z '' " k=1 $ k $ {n,SeriesCoefficient[ ,{z,0,n}]}, # 1 150 # 1 2*k && %1 + * ) % * z (( '' $ " k=1 $ k {n,4,80,2}], Table[ # & +n . "2 % ( * HarmonicNumber- *1,20 , / 6 2 (}, {n,% 2 % # 1 + n .& ( % 2 * " * %1 + * HarmonicNumber- *10( ( , 2 /' ' $ " $ {n,4,80,2}]}]
24
Top Page 13: Graph of S(L,L;n) vs. n (Exact calculation and Approximate fit)
ListPlot[ {Table[ $ 1 9 $ 1 2*k ' 1 150 $ 1 2*k '' 2*5 &1+ * * & * z ) + 0 + * * & * z )) * z ( (( # k=11% k % # k=1% k {n,1" Sum[SeriesCoefficient[ ,{z,0,k}],{k,2,n,2}]}, $ 1 150 $ 1 2*k '' &1+ * * & * z )) (( % # k=1% k {n,2,80,2}], Table[ $ ' 1 1 & ) * )}, # 10 {n,& 2 &$ 1 +n .' ) & &1+ * HarmonicNumber- " 5 "100) ) ,2 /( ( %% #
{n,2,80,2}]},PlotRange 1 {{10,80},{".01,.04}}] Bottom Page 13: Graph of Log[F(L,L;n)] vs. Log[n] (Exact calculation and Approximate fit)
ListLogLogPlot[ {Table[ # 1 9 # 1 2*k & 1 150 # 1 2*k && 2*5 1 + * * z + 0 + * ) % * z (( * z % ( ) % ' '' " k=11$ k $ " k=1 $ k {n,SeriesCoefficient[ ,{z,0,n}]}, 300 # 1 # 1 2*k && %1 + * ) % * z (( '' $ " k=1 $ k {n,30,80,2}], Table[ # 1 1 #" 2 +n .& & * HarmonicNumber- *10 * 5,20( ( % * *% ,2 /' ( " 10 6 $ {n,% }, 2 % # 1 +n .& ( % 2 * " * %1 + * HarmonicNumber- *10 * 50( ( ,2 /' ' $ " $ {n,30,80,2}]}]
25
!
Top Page 16: Graph of S(L,L;n)] vs. n (Exact calculation and Approximate fit) (L=1) ListPlot[ {Table[ # (2 *1)! & 2*1 1 1 * z * Hypergeometric2F1[1 + ,1 + ,2 *1 + 1,z 2 ] % 4*1 2( $ 2 * (1!) ' 2 2 {n,1 " Sum[SeriesCoefficient[ ,{z,0,k}],{k,0,n,2}]}, 1 1 2 Hypergeometric2F1[ , ,1,z ] 2 2 {n,0,150,2}], Table[ # # # # && && 2*(1+m ) # & 4*(1+m )"1 (( (( 1 % % % % #%1 + 1 &( * %1 " ( % % % %$ m ' (( (( $ 2 * (1 + m) ' 1 " % % 1 % % ( (( 2*1+m+ 2*(1+m )"1 ( 2 # & # & # % 2 *1 1 % (2 *1)! & % ( (( % ( n * %1 " "1"1 ( ( (( (( %1 " 4*1 % % %1 + % 2( % 2 $ 1+ m' m ' '' 1 1 $ $ $ $ 2 * (1!) ' {n,% + 0 % * * ((}, # & # & * * m 1 n m=1 % ) % 1 + 1 * HarmonicNumber n "1 ( %1 + * HarmonicNumber, " m "1/( (( ,+ 2 /.(' % %$ ) % +2 .' ( $ ) % % (( % % (( %% %% (((( $ '' $ {n,6,150,2}]},PlotRange 1 {0,1.1}]
Bottom Page 16: Graph of S(L,L;n)] vs. n (Exact calculation and Approximate fit) (L=10) ListPlot[ {Table[ # (2 *10)! & 2*10 1 1 2 % 4*10 ( * z * Hypergeometric2F1[10 + ,10 + ,2 *10 + 1,z ] $ 2 * (10!) 2 ' 2 2 {n,1 " Sum[SeriesCoefficient[ ,{z,0,k}],{k,0,n,2}]}, 1 1 Hypergeometric2F1[ , ,1,z 2 ] 2 2 {n,0,500,40}], Table[ # # # # && && 2*(10+m ) # & 4*(10+m )"1 (( (( 1 % % % % #%1 + 10 &( * %1 " ( % % % %$ (( (( m' $ 2 * (10 + m) ' 1 " % % 1 % % ( (( 2*10+m+ 2*(10+m )"1 ( 2 # # % 1 & % (2 *10)! & % % #1 + 2 *10 & ( (( ( n * %1 " "10"1 ( ( (( (( %1 " 4*10 % %% % 2( % 2 $ 10 + m ' m ' '' 1 1 $ $$ $ 2 * (10!) ' {n,% + 0 % * * ((}, # & # & *n m 1 m=1 % ) % 1 + 1 * HarmonicNumber* n "10(( 1 + * HarmonicNumber " m "10 % ,+ 2 /.(' ,+ 2 /.(' % %$ ) % (( $ ) % % (( % % (( %% %% (((( $ '' $ {n,24,500,2}]},PlotRange 1 {0,1.1}]
26
For general L, the code for plotting S(L,L;n) and its approximate fit is: ListPlot[ {Table[ # (2 * L)! & 2*L 1 1 2 % 4*L ( * z * Hypergeometric2F1[L + , L + ,2 * L + 1,z ] $ 2 * (L!) 2 ' 2 2 {n,1 " Sum[SeriesCoefficient[ ,{z,0,k}],{k,0,n,2}]}, 1 1 Hypergeometric2F1[ , ,1,z 2 ] 2 2 {n,0,500,40}], Table[ # # # # && && 2*( L+m ) # & 4*( L+m )"1 (( (( 1 % % % % #%1 + L &( * %1 " ( % % % %$ m ' (( (( $ 2 * (L + m) ' 1 " % % 1 % % ( (( 2*L+m+ 2*( L+m )"1 ( 2 # # % 1 & % (2 * L)! & % % #1 + 2 * L & ( (( ( n * %1 " "10"1 ( ( (( (( %1 " 4*L ( % % 1 1 %$ %$ %$ 2 $ L + m' m ' '' $ 2 * (L!) 2 ' {n,% + 0 % * * ((}, # 1 *n -& % #1 + 1 * HarmonicNumber* n " L-& m=1 % ) m (( %1 + * HarmonicNumber, " m " L/( ,+ 2 /.(' % %$ ) % (( + . 2 $ ) ' % % (( % % (( %% %% (((( $ '' $ {n,2 * L + 4,500,2}]},PlotRange 1 {0,1.1}]
!
27
