Difference between revisions of "BesselH0"

From TORI
Jump to: navigation, search
 
m (Text replacement - "\$([^\$]+)\$" to "\\(\1\\)")
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
[[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]] $=H_0$ 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==
  +
  +
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==
  
==References==

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

Retrieved from "https://mizugadro.mydns.jp/t/index.php?title=BesselH0&oldid=28005"