Difference between revisions of "BesselH0"

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m (Text replacement - "\$([^\$]+)\$" to "\\(\1\\)")
 
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[[File:BesselY0J0J1plotT060.png|500px|thumb|Real and imaginary parts of $H_0(x)$ for $x>0$ compared to [[BesselJ1]] (green)]]
+
[[File:BesselY0J0J1plotT060.png|500px|thumb|Real and imaginary parts of \(H_0(x)\) for \(x>0\) compared to [[BesselJ1]] (green)]]
[[File:Besselh0mapT100.png|500px|thumb|$u+\mathrm i v=H_0(x+\mathrm i y)$ ]]
+
[[File:Besselh0mapT100.png|500px|thumb|\(u+\mathrm i v=H_0(x+\mathrm i y)\) ]]
[[BesselH0]]$(z)=H_0(z)\!=$[[HankelH1]]$[0,z]$ is the [[Cylindric function]] H (called also the [[Hankel function]]) of zero order.
+
[[BesselH0]]\((z)=H_0(z)\!=\)[[HankelH1]]\([0,z]\) is the [[Cylindric function]] H (called also the [[Hankel function]]) of zero order.
   
BesselH0 is related with $J_0=$[[BesselJ0]] and $J_0=$[[BesselY0]] with simple relation
+
BesselH0 is related with \(J_0=\)[[BesselJ0]] and \(J_0=\)[[BesselY0]] with simple relation
: $H_0(z)=J_0(z)+\mathrm i Y_0(z)$
+
: \(H_0(z)=J_0(z)+\mathrm i Y_0(z)\)
   
In particular, for $x>0$, the relations $\Im(J_0(x))=0$ and $\Im(Y_0(x))=0$ hold, and, therefore,
+
In particular, for \(x>0\), the relations \(\Im(J_0(x))=0\) and \(\Im(Y_0(x))=0\) hold, and, therefore,
: $\Re(H_0(x))=J_0(x)$
+
: \(\Re(H_0(x))=J_0(x)\)
: $\Im(H_0(x))=Y_0(x)$
+
: \(\Im(H_0(x))=Y_0(x)\)
   
The [[explicit plot]] of real and imaginary parts of [[BesselH0]] versus real positive argument are shown in the upper right corner. <!-- in comparison with function $J_1=$[[BesselJ1]].!-->
+
The [[explicit plot]] of real and imaginary parts of [[BesselH0]] versus real positive argument are shown in the upper right corner. <!-- in comparison with function \(J_1=\)[[BesselJ1]].!-->
   
Below, the [[complex map]] of $H_0$ is plotted, $u+\mathrm i v=H_0(x+\mathrm i y)$.
+
Below, the [[complex map]] of \(H_0\) is plotted, \(u+\mathrm i v=H_0(x+\mathrm i y)\).
   
 
==Asymptotic expansions==
 
==Asymptotic expansions==
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does
 
does
   
$e^{i x} \left(1-\frac{i}{8 x}-\frac{9}{128 x^2}+\frac{75
+
\(e^{i x} \left(1-\frac{i}{8 x}-\frac{9}{128 x^2}+\frac{75
 
i}{1024 x^3}+\frac{3675}{32768 x^4}-\frac{59535
 
i}{1024 x^3}+\frac{3675}{32768 x^4}-\frac{59535
 
i}{262144
 
i}{262144
x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)$
+
x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)\)
   
 
===Square of BesselH0===
 
===Square of BesselH0===
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does
 
does
   
$e^{2 i x} \left(1-\frac{i}{4 x}-\frac{5}{32 x^2}+\frac{21
+
\(e^{2 i x} \left(1-\frac{i}{4 x}-\frac{5}{32 x^2}+\frac{21
 
i}{128 x^3}+\frac{507}{2048 x^4}-\frac{4035 i}{8192
 
i}{128 x^3}+\frac{507}{2048 x^4}-\frac{4035 i}{8192
x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)$
+
x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)\)
   
 
and
 
and
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does
 
does
   
$e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
+
\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\left(\frac{1}{x}\right)^{3/2}+\frac{381}{16384
 
\left(\frac{1}{x}\right)^{3/2}+\frac{381}{16384
 
x^2}-\frac{675 i
 
x^2}-\frac{675 i
 
\left(\frac{1}{x}\right)^{5/2}}{65536}+O\left(\left(\frac{1}{x}\right)^3\right)\right)
 
\left(\frac{1}{x}\right)^{5/2}}{65536}+O\left(\left(\frac{1}{x}\right)^3\right)\right)
  +
\)
$
 
   
 
At least one of these two asymptotic expansions seems to be wrong. Perhaps, the last one, because
 
At least one of these two asymptotic expansions seems to be wrong. Perhaps, the last one, because
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does
 
does
   
$e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
+
\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048
 
\left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048
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097152 x^3}+\frac{275625 i
 
097152 x^3}+\frac{275625 i
 
\left(\frac{1}{x}\right)^{7/2}}{16777216}+O\left(\left
 
\left(\frac{1}{x}\right)^{7/2}}{16777216}+O\left(\left
(\frac{1}{x}\right)^4\right)\right)$
+
(\frac{1}{x}\right)^4\right)\right)\)
   
 
Correcting the two terms (that showes deviation in the previous example); however, two new wrong terms are added. The control on the precision seems to be lost, because
 
Correcting the two terms (that showes deviation in the previous example); however, two new wrong terms are added. The control on the precision seems to be lost, because
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does
 
does
   
$e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
+
\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i
 
\left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048
 
\left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048
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\left(\frac{1}{x}\right)^{7/2}}{1048576}+\frac{4922662
 
\left(\frac{1}{x}\right)^{7/2}}{1048576}+\frac{4922662
 
5}{1073741824 x^4}-\frac{218791125 i
 
5}{1073741824 x^4}-\frac{218791125 i
\left(\frac{1}{x}\right)^{9/2}}{4294967296}+O\left(\left(\frac{1}{x}\right)^5\right)\right)$
+
\left(\frac{1}{x}\right)^{9/2}}{4294967296}+O\left(\left(\frac{1}{x}\right)^5\right)\right)\)
   
 
This phenomenon can be interpreted as bug in Mathematica [[Series]] routine.
 
This phenomenon can be interpreted as bug in Mathematica [[Series]] routine.

Latest revision as of 18:26, 30 July 2019

Real and imaginary parts of \(H_0(x)\) for \(x>0\) compared to BesselJ1 (green)
\(u+\mathrm i v=H_0(x+\mathrm i y)\)

BesselH0\((z)=H_0(z)\!=\)HankelH1\([0,z]\) is the Cylindric function H (called also the Hankel function) of zero order.

BesselH0 is related with \(J_0=\)BesselJ0 and \(J_0=\)BesselY0 with simple relation

\(H_0(z)=J_0(z)+\mathrm i Y_0(z)\)

In particular, for \(x>0\), the relations \(\Im(J_0(x))=0\) and \(\Im(Y_0(x))=0\) hold, and, therefore,

\(\Re(H_0(x))=J_0(x)\)
\(\Im(H_0(x))=Y_0(x)\)

The explicit plot of real and imaginary parts of BesselH0 versus real positive argument are shown in the upper right corner.

Below, the complex map of \(H_0\) is plotted, \(u+\mathrm i v=H_0(x+\mathrm i y)\).

Asymptotic expansions

TeXForm[Expand[Series[(HankelH1[0, x]) (Pi I x/2)^(1/2), {x, Infinity, 5}]]]

does

\(e^{i x} \left(1-\frac{i}{8 x}-\frac{9}{128 x^2}+\frac{75 i}{1024 x^3}+\frac{3675}{32768 x^4}-\frac{59535 i}{262144 x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)\)

Square of BesselH0

Some expansions of square of BesselH0 look counter-intuitive and require analysis. One example ix copy pasted below.

TeXForm[Expand[Series[(HankelH1[0, x])^2 Pi I x/2, {x, Infinity, 5}]]]

does

\(e^{2 i x} \left(1-\frac{i}{4 x}-\frac{5}{32 x^2}+\frac{21 i}{128 x^3}+\frac{507}{2048 x^4}-\frac{4035 i}{8192 x^5}+O\left(\left(\frac{1}{x}\right)^6\right)\right)\)

and

TeXForm[Expand[Series[(HankelH1[0, x^(1/2)])^2 Pi I x^(1/2)/2, {x, Infinity, 2}]]]

does

\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i \sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i \left(\frac{1}{x}\right)^{3/2}+\frac{381}{16384 x^2}-\frac{675 i \left(\frac{1}{x}\right)^{5/2}}{65536}+O\left(\left(\frac{1}{x}\right)^3\right)\right) \)

At least one of these two asymptotic expansions seems to be wrong. Perhaps, the last one, because

TeXForm[Expand[Series[(HankelH1[0, x^(1/2)])^2 Pi I x^(1/2)/2, {x, Infinity, 3}]]]

does

\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i \sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i \left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048 x^2}-\frac{5025 i \left(\frac{1}{x}\right)^{5/2}}{131072}-\frac{44325}{2 097152 x^3}+\frac{275625 i \left(\frac{1}{x}\right)^{7/2}}{16777216}+O\left(\left (\frac{1}{x}\right)^4\right)\right)\)

Correcting the two terms (that showes deviation in the previous example); however, two new wrong terms are added. The control on the precision seems to be lost, because

TeXForm[Expand[Series[(HankelH1[0, x^(1/2)])^2 Pi I x^(1/2)/2, {x, Infinity, 4}]]]

does

\(e^{2 i \sqrt{x}} \left(1-\frac{1}{4} i \sqrt{\frac{1}{x}}-\frac{5}{32 x}+\frac{21}{128} i \left(\frac{1}{x}\right)^{3/2}+\frac{507}{2048 x^2}-\frac{4035 i \left(\frac{1}{x}\right)^{5/2}}{8192}-\frac{163395}{20 97152 x^3}+\frac{50715 i \left(\frac{1}{x}\right)^{7/2}}{1048576}+\frac{4922662 5}{1073741824 x^4}-\frac{218791125 i \left(\frac{1}{x}\right)^{9/2}}{4294967296}+O\left(\left(\frac{1}{x}\right)^5\right)\right)\)

This phenomenon can be interpreted as bug in Mathematica Series routine.

References


http://mathworld.wolfram.com/BesselFunction.html

http://en.wikipedia.org/wiki/Bessel_function