Bessel

Transcript of Bessel

What are Bessel Functions?Why do we care?Exp-arc explained

Results

Calculating Bessel Functions via the Exp-arcMethod

David Borwein1, Jonathan M. Borwein2, and O-Yeat Chan3

1Department of Mathematics, University of Western Ontario2Faculty of Computer Science, Dalhousie University

3Department of Mathematics and Statistics, Dalhousie University

Illinois Number Theory Fest, May 2007

D. Borwein, J. M. Borwein, O-Y. Chan Calculating Bessel Functions via the Exp-arc Method

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Outline

1 What are Bessel Functions?

2 Why do we care?

3 Exp-arc explained

4 Results


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Outline


2 Why do we care?

3 Exp-arc explained

4 Results


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The second order differential equation

z2y + zy + (z2 2)y = 0

is called Bessels Equation.

The ordinary Bessel function or order , or the Bessel functionof the first kind of order , denoted J(z), is a solution to thisdifferential equation.

J(z) can be represented as an ascending series

J(z) =k=0

(1)k (z/2)2k+k !(k + + 1)

.


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z2y + zy + (z2 2)y = 0




J(z) =k=0

(1)k (z/2)2k+k !(k + + 1)

.


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z2y + zy + (z2 2)y = 0




J(z) =k=0

(1)k (z/2)2k+k !(k + + 1)

.


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It is not difficult to show that for 6 Z, J(z) and J(z) arelinearly independent. Since Bessels Equation is second order,for non-integer this pair generates all the solutions.

When = n is an integer, Jn(z) is also given by the generatingfunction

ez2(t 1t ) =

n=

Jn(z)tn.

Replace t by t and we find that

Jn(z) = (1)nJn(z).


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ez2(t 1t ) =

n=

Jn(z)tn.


Jn(z) = (1)nJn(z).


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ez2(t 1t ) =

n=

Jn(z)tn.


Jn(z) = (1)nJn(z).


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So to obtain the second solution to Bessels Equation at integerorder, define the Bessel function of the second kind Yn(z),n Z, by

Yn(z) := limn

J(z) cos pi J(z)sin pi

.

For general , Y(z) is defined as above without the limit.

We also have the following ascending series representation forYn(z), n Z.

Yn(z) =1pi

(2( + log(z/2))Jn(z)

n1k=0

(n k 1)!(z/2)2knk !

k=0

(1)k (z/2)2k+n(Hk + Hk+n)k !(n + k)!

).


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Yn(z) := limn


.



Yn(z) =1pi

(2( + log(z/2))Jn(z)

n1k=0

(n k 1)!(z/2)2knk !

k=0

(1)k (z/2)2k+n(Hk + Hk+n)k !(n + k)!

).


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Yn(z) := limn


.



Yn(z) =1pi

(2( + log(z/2))Jn(z)

n1k=0

(n k 1)!(z/2)2knk !

k=0

(1)k (z/2)2k+n(Hk + Hk+n)k !(n + k)!

).


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In addition to the J and Y Bessel functions, there are also themodified Bessel functions I(z) and K(z), which are solutionsto the differential equation

z2y + zy (z2 + 2)y = 0.

I(z) is usually known as the Bessel function of imaginaryargument, and is related to J by

I(z) = epii/2J(iz) =k=0

(z/2)2k+

k !(k + + 1),

and K(z) is given by

K(z) =pi

2I(z) I(z)

sin pi.


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z2y + zy (z2 + 2)y = 0.



(z/2)2k+

k !(k + + 1),


K(z) =pi

2I(z) I(z)

sin pi.


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z2y + zy (z2 + 2)y = 0.



(z/2)2k+

k !(k + + 1),


K(z) =pi

2I(z) I(z)

sin pi.


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As z , we have the asymptotic expansion

J(z) (

2piz

)1/2(cos(z pi2 pi4 )

k=0

(1)k (12 )2k (12 + )2kk !(2z)2k

sin(z pi2 pi4 )k=0

(1)k (12 )2k+1(12 + )2k+1k !(2z)2k+1

),

as well as similar expressions for Y , I, and K .

Here the notation (a)k is the Pochhammer symbol given by

(a)k = a(a+ 1) . . . (a+ k 1).

Note that this is an asymptotic series, and diverges for fixed z.


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J(z) (

2piz


k=0

(1)k (12 )2k (12 + )2kk !(2z)2k

sin(z pi2 pi4 )k=0

(1)k (12 )2k+1(12 + )2k+1k !(2z)2k+1

),



(a)k = a(a+ 1) . . . (a+ k 1).



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J(z) (

2piz


k=0

(1)k (12 )2k (12 + )2kk !(2z)2k

sin(z pi2 pi4 )k=0

(1)k (12 )2k+1(12 + )2k+1k !(2z)2k+1

),



(a)k = a(a+ 1) . . . (a+ k 1).



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There exists, however, a Hadamard expansion that isconvergent:

I(z) =ez

( + 12)2piz

k=0

(12 )kk !(2z)k

2z0

t+k12etdt .

Note that the error behaves like N1/2 when the series istruncated after N terms.

For more information on the classical theory, see Watsonsbook, A Treatise on the Theory of Bessel Functions."


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I(z) =ez

( + 12)2piz

k=0

(12 )kk !(2z)k

2z0

t+k12etdt .




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I(z) =ez

( + 12)2piz

k=0

(12 )kk !(2z)k

2z0

t+k12etdt .




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Outline


2 Why do we care?

3 Exp-arc explained

4 Results


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Bessels Equation arises as a special case of LaplacesEquation with cylindrical symmetry.

Thus, Bessel functions occur often in the study of waves in twodimensions.

For example, the AMOEBA water tank uses Bessel functions tocompute the amplitudes at which to drive its pistons to formletters from standing waves.


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Applications to Number Theory

Hardy and the Circle Problem

Let r2(n) denote the number of representations of the positiveinteger n as a sum of two squares. The circle problem is todetermine the precise order of magnitude for the error term"P(x) defined by

0nx

r2(n) = pix + P(x),

where the prime on the summation sign on the left sideindicates that if x is an integer, only 12 r2(x) is counted.


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Applications to Number Theory

Hardy and the Circle Problem

Let r2(n) denote the number of representations of the positiveinteger n as a sum of two squares. The circle problem is todetermine the precise order of magnitude for the error term"P(x) defined by

0nx

r2(n) = pix + P(x),

where the prime on the summation sign on the left sideindicates that if x is an integer, only 12 r2(x) is counted.


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In 1915, Hardy proved that

0nx

r2(n) = pix +

n=1

r2(n)(xn

)1/2J1(2pi

nx).

This is equivalent to the following result of Berndt andZaharescu0nx

r2(n) = pix

+ 2xn=0

m=1

J1

(4pim(n + 14)x

)m(n + 14)

J1

(4pim(n + 34)x

)m(n + 34)

.


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In 1915, Hardy proved that

0nx

r2(n) = pix +

n=1

r2(n)(xn

)1/2J1(2pi

nx).

This is equivalent to the following result of Berndt andZaharescu0nx

r2(n) = pix

+ 2xn=0

m=1

J1

(4pim(n + 14)x

)m(n + 14)

J1

(4pim(n + 34)x

)m(n + 34)

.


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The BZ result is a corollary of an entry on page 335 ofRamanujans Lost Notebook.

That entry is one of a pair of equations involving adoubly-infinite series of Bessel functions. If we let

F (x) =

{[x ], if x is not an integer,x 12 , if x is an integer,

where, [x ] is the greatest integer less than or equal to x , and

Z(z) := Y(z) 2piK(z);

Then the two entries read


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F (x) =



Z(z) := Y(z) 2piK(z);



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F (x) =



Z(z) := Y(z) 2piK(z);



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For x > 0 and 0 < < 1,

n=1

F(xn

)sin(2pin) = pix

( 12

) 14 cot(pi)+12x

m=1

n=0

J1(4pim(n + )x

)m(n + )

J1(4pim(n + 1 )x

)m(n + 1 )

.


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Entry

For x > 0 and 0 < < 1,

n=1

F(xn

)cos(2pin) =

14 x log(2 sin(pi))

+12x

m=1

n=0

Z1(4pim(n + )x

)m(n + )

+Z1(4pim(n + 1 )x

)m(n + 1 )

.


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I wanted to evaluate a truncation of the right-hand side of thesecond entry to verify it against the left-hand side. As anexample, computing the right-hand side (with one of the sumstruncated at 50 terms, the other truncated at 1000 terms, for atotal of 50,000 summands) at 28-digit precision in PARI tooknearly 4 hours.

As a comparison, a related sum where Z1 is replaced by a sinewas evaluated in 11 minutes to 10 million summands.


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I wanted to evaluate a truncation of the right-hand side of thesecond entry to verify it against the left-hand side. As anexample, computing the right-hand side (with one of the sumstruncated at 50 terms, the other truncated at 1000 terms, for atotal of 50,000 summands) at 28-digit precision in PARI tooknearly 4 hours.

As a comparison, a related sum where Z1 is replaced by a sinewas evaluated in 11 minutes to 10 million summands.


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Outline


2 Why do we care?

3 Exp-arc explained

4 Results


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What is exp-arc?

In their recent paper to find asymptotics for Laguerrepolynomials with effective (explicit) error bounds, D. Borwein,J. M. Borwein, and R. Crandall were led to consider thefollowing integral

I(p,q) := pi/2pi/2

eiqep cosd.

A simple change of variable yields

I(p,q) = 4ep 1/20

cosh(2iq arcsin x)e2px21 x2 dx .


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What is exp-arc?


I(p,q) := pi/2pi/2

eiqep cosd.


I(p,q) = 4ep 1/20



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What is exp-arc?


I(p,q) := pi/2pi/2

eiqep cosd.


I(p,q) = 4ep 1/20



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This integral, therefore, reduces to an integral involvinge arcsin x , or what BBC calls an exp-arc integral.

They then exploit the fact that the exp-arc function has a verynice series expansion on (1,1), namely

e arcsin x = 1+k=1

ck ()xk

k !,

where

c2k+1() = k

j=1

(2 + (2j 1)2), c2k =k

j=1

(2 + (2j 2)2).


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This integral, therefore, reduces to an integral involvinge arcsin x , or what BBC calls an exp-arc integral.

They then exploit the fact that the exp-arc function has a verynice series expansion on (1,1), namely

e arcsin x = 1+k=1

ck ()xk

k !,

where

c2k+1() = k

j=1

(2 + (2j 1)2), c2k =k

j=1

(2 + (2j 2)2).


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Plugging this into the expression for I(p,q) and interchangingsummation and integration we obtain

I(p,q) = 4epk=0

gk (2iq)(2k)!

Bk (p), (1)

where

g0 := 1, gk () :=k

j=1

((2j 1)2 + 2

)for k 1,

and

Bk (p) := 1/20

x2ke2px2dx =

12k2

10

epuuk12du.


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I(p,q) = 4epk=0

gk (2iq)(2k)!

Bk (p), (1)

where

g0 := 1, gk () :=k

j=1

((2j 1)2 + 2

)for k 1,

and

Bk (p) := 1/20

x2ke2px2dx =

12k2

10

epuuk12du.


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I(p,q) = 4epk=0

gk (2iq)(2k)!

Bk (p), (1)

where

g0 := 1, gk () :=k

j=1

((2j 1)2 + 2

)for k 1,

and

Bk (p) := 1/20

x2ke2px2dx =

12k2

10

epuuk12du.


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Note that

gk and Bk are rapidly computable via recursion

The integral 10 e

puuk1/2du is uniformly bounded for allk > 0, and so the Bk decrease geometrically as 2k .gk ()/(2k)! are bounded for fixed .

Therefore, the series for I(p,q) is geometrically convergent.


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Note that


The integral 10 e




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Note that


The integral 10 e




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Note that


The integral 10 e




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Outline


2 Why do we care?

3 Exp-arc explained

4 Results


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Integral Representations

J(z) =1pi

pi0

cos(t z sin t)dt sin pipi

0

etz sinh tdt ,

Y(z) =1pi

pi0

sin(z sin t t)dt 1pi

0

(et + et cos pi)ez sinh tdt ,

I(z) =1pi

pi0

ez cos t cos t dt sin pipi

0

ez cosh ttdt ,

and

K(z) = 0

ez cosh t cosh t dt =12

ez cosh ttdt .


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A quick change of variable allows us to express the finiteintegrals in terms of I(p,q).

For integral order, the infinite integrals in J and I disappear dueto the sin pi. Thus we have

Jn(z) =12pi

(einpi/2I(iz,n) + einpi/2I(iz,n)

),

and

In(z) =12pi

(I(z,n) + epiinI(z,n)

).


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A quick change of variable allows us to express the finiteintegrals in terms of I(p,q).

For integral order, the infinite integrals in J and I disappear dueto the sin pi. Thus we have

Jn(z) =12pi

(einpi/2I(iz,n) + einpi/2I(iz,n)

),

and

In(z) =12pi

(I(z,n) + epiinI(z,n)

).


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To deal with the general case, we need to evaluate the infiniteintegrals. A change of variables plus integration by parts givesus

0etz sinh tdt =

1 z

0

ezse arcsinh sds.

This is an exp-arc integral, but our expansion is only valid on[0,1). So what should we do?


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To deal with the general case, we need to evaluate the infiniteintegrals. A change of variables plus integration by parts givesus

0etz sinh tdt =

1 z

0

ezse arcsinh sds.

This is an exp-arc integral, but our expansion is only valid on[0,1). So what should we do?


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Expand about a point other than zero!

For example, the expansion at infinity is

se arcsinh s =n=0

An()s2n

,

where A0() = 2 and for n 1,

An() =(1)n2( + n + 1)n1

22nn!,

This expansion is valid for |s| > 1. The coefficients satisfy

An = ( + 2n 2)( + 2n 1)4n(n + ) An1.


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Expand about a point other than zero!

For example, the expansion at infinity is

se arcsinh s =n=0

An()s2n

,

where A0() = 2 and for n 1,

An() =(1)n2( + n + 1)n1

22nn!,

This expansion is valid for |s| > 1. The coefficients satisfy

An = ( + 2n 2)( + 2n 1)4n(n + ) An1.


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How about at other points?

No nice formula for coefficients, but thats okay.

For fixed k , we have the expansion

e arcsinh(k+s) =n=0

an(k , )n!

sn

where for n 0,

an+2 =1

k2 + 1

((2 n2)an k(2n + 1)an+1

),

anda0 = (k +

k2 + 1) , a1 = a0

k2 + 1.


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e arcsinh(k+s) =n=0

an(k , )n!

sn

where for n 0,

an+2 =1

k2 + 1

((2 n2)an k(2n + 1)an+1

),

anda0 = (k +

k2 + 1) , a1 = a0

k2 + 1.


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e arcsinh(k+s) =n=0

an(k , )n!

sn

where for n 0,

an+2 =1

k2 + 1

((2 n2)an k(2n + 1)an+1

),

anda0 = (k +

k2 + 1) , a1 = a0

k2 + 1.


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Thus for any positive integer N, we have 0

ezse arcsinh sds =n=0

(an(0, )

n!n(z) + n(z)

Nk=1

ekzan(k , )

n!

+ An()Gn(N + 12 , z, )),

where

n(z) := 1/20

ezssnds = ez/2

2nz+

nzn1(z),

n(z) := 1/21/2

ezssnds =ez/2

(2)nz ez/2

2nz+

nzn1(z),


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and

Gn(N, z, ) :=eNz

N2n+1

0

eNzs(1+ s)2nds

=1

( + 2n 1)( + 2n 2)(eNz(2(n z 1) + )

N2n+1+ z2Gn1(N, z, )

).


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Key points:

All the summands are easily computable via recursionThe recursions only involve elementary operations. Theinitial conditions B0 (from I) and G0 each require oneevaluation of incomplete gamma, which can be done viacontinued fraction.Each series converges geometrically, like 2N (as opposedto the Hadamard expansion for I , which is like N)Can choose N large to avoid the AnGn sum if we want. Inthis way, we can pre-compute summands involving only and summands involving only z for one-z many- orone- many-z evaluations.


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Key points:



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Key points:



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Key points:



What are Bessel Functions?Why do we care?Exp-arc explainedResults

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