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dimitris
Posted: Thu Apr 26, 2007 8:20 am
Guest
Hello.

I have the folloiwng MeijerG function

In[11]:=
f = MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)];

where o,m are both real.

Can we write down f in terms of other special functions?

Thanks
Dimitris
Bhuvanesh
Posted: Thu Apr 26, 2007 9:11 am
Guest
I don't think it can be written in closed form in terms of anything else that is currently implemented in Mathematica/Maple. It's an entire function, though. How did you come across it?

Bhuvanesh,
Wolfram Research
dimitris
Posted: Fri Apr 27, 2007 11:02 am
Guest
A collegue of mine came across this function during the course of the
inversion of one Fourier Transform.

He suceeded in writing this function it terms of the sum of two
products; each of these products
was consist of BesselK*StruveL. He didn't give me more details. He put
me as a chalenge if I could
use a CAS in order to arrive at his formula (he checked his formula
numerically). I didn't
manage to find anything also, that's why the presence of this thread!

As regards your statement about the analyticity of this function, with
all of my respect to your
knowledge, are you 100% sure?

I don't know how you came to this conclusion. Mathematica can't get
neither the limit as o->0,
nor at infinity. But look the following plot (I know that plot can be
misleading sometimes, but I think here
is not the case)...

In[10]:f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

In[14]:(Limit[f[o, 1], o -> #1] & ) /@ {0, Infinity}

Out[14]{Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, o^2/4], o -> 0],
Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, o^2/4],
o -> Infinity]}

In[21]:Plot[{f[x, 1], -x^(-1) - 2*Pi}, {x, 0, 10}, Axes -> False, Frame ->
{True, True, False, False}]

I think that the function behaves as -1/x at zero and as -2*Pi.

Dimitris



Ο/Η Bhuvanesh έγραψε:
Quote:
I don't think it can be written in closed form in terms of anything else that is currently implemented in Mathematica/Maple. It's an entire function, though. How did you come across it?

Bhuvanesh,
Wolfram Research
dimitris
Posted: Fri Apr 27, 2007 12:12 pm
Guest
Quote:
I think that the function behaves as -1/x at zero and as -2*Pi.

....at infinity


Ο/Ξ— dimitris έγραψΡ:
Quote:
A collegue of mine came across this function during the course of the
inversion of one Fourier Transform.

He suceeded in writing this function it terms of the sum of two
products; each of these products
was consist of BesselK*StruveL. He didn't give me more details. He put
me as a chalenge if I could
use a CAS in order to arrive at his formula (he checked his formula
numerically). I didn't
manage to find anything also, that's why the presence of this thread!

As regards your statement about the analyticity of this function, with
all of my respect to your
knowledge, are you 100% sure?

I don't know how you came to this conclusion. Mathematica can't get
neither the limit as o->0,
nor at infinity. But look the following plot (I know that plot can be
misleading sometimes, but I think here
is not the case)...

In[10]:> f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

In[14]:> (Limit[f[o, 1], o -> #1] & ) /@ {0, Infinity}

Out[14]> {Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, o^2/4], o -> 0],
Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, o^2/4],
o -> Infinity]}

In[21]:> Plot[{f[x, 1], -x^(-1) - 2*Pi}, {x, 0, 10}, Axes -> False, Frame -
{True, True, False, False}]

I think that the function behaves as -1/x at zero and as -2*Pi.

Dimitris



Ï/Γ‡ Bhuvanesh ÝãñÑøΓ₯:
I don't think it can be written in closed form in terms of anything else that is currently implemented in Mathematica/Maple. It's an entire function, though. How did you come across it?

Bhuvanesh,
Wolfram Research
dimitris
Posted: Sat Apr 28, 2007 5:19 am
Guest
At last I succeeded!

Firstly, let

In[14]:f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

For m=1 here is a plot of this function

In[17]:Plot[{f[x, 1], -(1/x + 2*Pi)}, {x, 0, 10}, Axes -> False, Frame ->
True, PlotStyle -> {Red, Blue}]

Oberve that, (it seems...) the function behaves as -1/x as x->0+, and
tends to -2*Pi at infinity.
However, Mathematica fails to provide these limits.

In[18]:(Limit[f[x, 1], x -> #1] & ) /@ {0, Infinity}

Out[18]{Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, x^2/4], x -> 0],
Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, x^2/4],
x -> Infinity]}

Now, set

In[26]:g[y_] = f[x, m] /. x -> y*m

Out[26]MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, y^2/4]

Without loss of generality, assume y>0 (y is real) and take the
derivative of g[y].

In[27]:(Simplify[#1, y > 0] & )[D[g[y], y]]

Out[27](4*BesselK[1, y])/y

Take now the derivative of the numerator of the last expression.

In[30]:Expand[4*FullSimplify[D[BesselK[1, y], y]]]

Out[30]-4*BesselK[0, y] - (4*BesselK[1, y])/y

With this procedure, someone is able to see that the derivative of
g[y] (ie the
requested (normalized) MeijerG function), which is equal to
(4*BesselK[1, y])/y,
can be written as -4*(BesselK[0, y]+D[BesselK[1, y], y].

Hence the function g[y] will be given by the indefinite integral of
the last expression.
In other words

In[35]Surprisedo = FullSimplify[-4*(Integrate[BesselK[0, y], y] + BesselK[1, y])]

Out[35]-2*(Pi*y*BesselK[0, y]*StruveL[-1, y] + BesselK[1, y]*(2 +
Pi*y*StruveL[0, y]))

with

In[36]:Plot[oo, {y, 0, 10}]

In[37]:(Limit[oo, y -> #1] & ) /@ {0, Infinity}

Out[37]{-Infinity, -2*Pi}

and returning back to the original variable x=y/m, we finally get the
following simplified
expression for the MeijerG function

In[44]:mei = FullSimplify[oo /. y -> x/m]

Out[44]-((2*(Pi*x*BesselK[0, x/m]*StruveL[-1, x/m] + BesselK[1, x/m]*(2*m +
Pi*x*StruveL[0, x/m])))/m)

Dimitris

PS

A collegue of mine (Panagiotis Gourgiotis) came across this MeijerG
function during the course of the
inversion of one Fourier Transform.

He suceeded in writing this function as in Output[44]. He didn't give
me more details. He put
me as a chalenge if I could use a CAS in order to arrive at his
formula (he checked his formula
numerically).

However, having got "stick" I pushed him in order to give one or two
hints.
The step appeared at In[30], completely belongs to him! Hail Panos!


Ο/Η dimitris έγραψε:
Quote:
Hello.

I have the folloiwng MeijerG function

In[11]:> f = MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)];

where o,m are both real.

Can we write down f in terms of other special functions?

Thanks
Dimitris
A N Niel
Posted: Sat Apr 28, 2007 6:27 am
Guest
In article <1177755540.723527.220410@e65g2000hsc.googlegroups.com>,
dimitris <dimmechan@yahoo.com> wrote:

Quote:
At last I succeeded!

Firstly, let

In[14]:=
f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

here is the definition...

http://functions.wolfram.com/HypergeometricFunctions/MeijerG/02/0006/

but it doesn't seem to say what the contour is

Quote:

For m=1 here is a plot of this function

In[17]:=
Plot[{f[x, 1], -(1/x + 2*Pi)}, {x, 0, 10}, Axes -> False, Frame -
True, PlotStyle -> {Red, Blue}]

Oberve that, (it seems...) the function behaves as -1/x as x->0+, and
tends to -2*Pi at infinity.
However, Mathematica fails to provide these limits.

In[18]:=
(Limit[f[x, 1], x -> #1] & ) /@ {0, Infinity}

Out[18]=
{Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, x^2/4], x -> 0],
Limit[MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, x^2/4],
x -> Infinity]}

Now, set

In[26]:=
g[y_] = f[x, m] /. x -> y*m

Out[26]=
MeijerG[{{1}, {}}, {{-(1/2), 1/2}, {0}}, y^2/4]

Without loss of generality, assume y>0 (y is real) and take the
derivative of g[y].

In[27]:=
(Simplify[#1, y > 0] & )[D[g[y], y]]

Out[27]=
(4*BesselK[1, y])/y

Take now the derivative of the numerator of the last expression.

In[30]:=
Expand[4*FullSimplify[D[BesselK[1, y], y]]]

Out[30]=
-4*BesselK[0, y] - (4*BesselK[1, y])/y

With this procedure, someone is able to see that the derivative of
g[y] (ie the
requested (normalized) MeijerG function), which is equal to
(4*BesselK[1, y])/y,
can be written as -4*(BesselK[0, y]+D[BesselK[1, y], y].

Hence the function g[y] will be given by the indefinite integral of
the last expression.
In other words

In[35]:=
oo = FullSimplify[-4*(Integrate[BesselK[0, y], y] + BesselK[1, y])]

Out[35]=
-2*(Pi*y*BesselK[0, y]*StruveL[-1, y] + BesselK[1, y]*(2 +
Pi*y*StruveL[0, y]))

with

In[36]:=
Plot[oo, {y, 0, 10}]

In[37]:=
(Limit[oo, y -> #1] & ) /@ {0, Infinity}

Out[37]=
{-Infinity, -2*Pi}

and returning back to the original variable x=y/m, we finally get the
following simplified
expression for the MeijerG function

In[44]:=
mei = FullSimplify[oo /. y -> x/m]

Out[44]=
-((2*(Pi*x*BesselK[0, x/m]*StruveL[-1, x/m] + BesselK[1, x/m]*(2*m +
Pi*x*StruveL[0, x/m])))/m)

Dimitris

PS

A collegue of mine (Panagiotis Gourgiotis) came across this MeijerG
function during the course of the
inversion of one Fourier Transform.

He suceeded in writing this function as in Output[44]. He didn't give
me more details. He put
me as a chalenge if I could use a CAS in order to arrive at his
formula (he checked his formula
numerically).

However, having got "stick" I pushed him in order to give one or two
hints.
The step appeared at In[30], completely belongs to him! Hail Panos!


Ο/Η dimitris έγραψε:
Hello.

I have the folloiwng MeijerG function

In[11]:=
f = MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)];

where o,m are both real.

Can we write down f in terms of other special functions?

Thanks
Dimitris
A N Niel
Posted: Sat Apr 28, 2007 6:50 am
Guest
In article <280420070727472574%anniel@nym.alias.net.invalid>, A N Niel
<anniel@nym.alias.net.invalid> wrote:

Quote:
In article <1177755540.723527.220410@e65g2000hsc.googlegroups.com>,
dimitris <dimmechan@yahoo.com> wrote:

At last I succeeded!

Firstly, let

In[14]:=
f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

here is the definition...

http://functions.wolfram.com/HypergeometricFunctions/MeijerG/02/0006/

but it doesn't seem to say what the contour is



Here is a description including the contour

http://mathworld.wolfram.com/MeijerG-Function.html

from that description, it looks like you should try
Integrals and Series, Vol. 3: More Special Functions by Prudnikov et al
David W. Cantrell
Posted: Sun Apr 29, 2007 8:30 am
Guest
dimitris <dimmechan@yahoo.com> wrote:
Quote:
At last I succeeded!

Firstly, let

In[14]:=
f[o_, m_] := MeijerG[{{1}, {}}, {{-2^(-1), 1/2}, {0}}, o^2/(4*m^2)]

For m=1 here is a plot of this function

In[17]:=
Plot[{f[x, 1], -(1/x + 2*Pi)}, {x, 0, 10}, Axes -> False, Frame -
True, PlotStyle -> {Red, Blue}]

Oberve that, (it seems...) the function behaves as -1/x as x->0+, and
tends to -2*Pi at infinity.

Instead of the plot mentioned above, I suggest that you look at

Plot[{f[x, 1], -2*Sqrt[Pi^2 + 4/x^2]}, {x, 0, 10}, Axes -> False,
Frame -> True, PlotStyle -> {Red, Blue}]

That simple approximation has |relative error| < 0.022 for x >= 0.

David W. Cantrell
 
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