Domain of validity for the asymptotic expansion of the Bessel function of the first kind
$begingroup$
A very short question here... I'd like to use an asymptotic approximation of a Bessel functions, which I've found given in several places as $J_nu(z)=sqrt{frac{2}{pi z}}cos(z-frac{1}{2}nu pi -frac{pi}{4})$ for $|arg(z)|<pi$ for large z.
My question is, doesn't this mean that the asymptotic form is valid for all nonnegative arguments z, since $|arg(z)|<pi$ means $-pi<arg(z)<pi$? Am I misunderstanding the references' conditions or is that all there is to it?
I don't know how to derive such an expansion myself, so I don't know where the condition came from. Is it correct that the asymptotic expression is valid for all large z away from the negative real line?
asymptotics bessel-functions
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
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add a comment |
$begingroup$
A very short question here... I'd like to use an asymptotic approximation of a Bessel functions, which I've found given in several places as $J_nu(z)=sqrt{frac{2}{pi z}}cos(z-frac{1}{2}nu pi -frac{pi}{4})$ for $|arg(z)|<pi$ for large z.
My question is, doesn't this mean that the asymptotic form is valid for all nonnegative arguments z, since $|arg(z)|<pi$ means $-pi<arg(z)<pi$? Am I misunderstanding the references' conditions or is that all there is to it?
I don't know how to derive such an expansion myself, so I don't know where the condition came from. Is it correct that the asymptotic expression is valid for all large z away from the negative real line?
asymptotics bessel-functions
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
$endgroup$
add a comment |
$begingroup$
A very short question here... I'd like to use an asymptotic approximation of a Bessel functions, which I've found given in several places as $J_nu(z)=sqrt{frac{2}{pi z}}cos(z-frac{1}{2}nu pi -frac{pi}{4})$ for $|arg(z)|<pi$ for large z.
My question is, doesn't this mean that the asymptotic form is valid for all nonnegative arguments z, since $|arg(z)|<pi$ means $-pi<arg(z)<pi$? Am I misunderstanding the references' conditions or is that all there is to it?
I don't know how to derive such an expansion myself, so I don't know where the condition came from. Is it correct that the asymptotic expression is valid for all large z away from the negative real line?
asymptotics bessel-functions
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
$endgroup$
A very short question here... I'd like to use an asymptotic approximation of a Bessel functions, which I've found given in several places as $J_nu(z)=sqrt{frac{2}{pi z}}cos(z-frac{1}{2}nu pi -frac{pi}{4})$ for $|arg(z)|<pi$ for large z.
My question is, doesn't this mean that the asymptotic form is valid for all nonnegative arguments z, since $|arg(z)|<pi$ means $-pi<arg(z)<pi$? Am I misunderstanding the references' conditions or is that all there is to it?
I don't know how to derive such an expansion myself, so I don't know where the condition came from. Is it correct that the asymptotic expression is valid for all large z away from the negative real line?
asymptotics bessel-functions
asymptotics bessel-functions
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
asked Dec 13 '18 at 22:54
BGreenBGreen
412211
412211
add a comment |
add a comment |
1 Answer
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From its expansion,
$$J_{nu}left(zright)=(tfrac{1}{2}z)^{nu}sum_{k=0}^{infty}(-1)^{k}frac{(%
tfrac{1}{4}z^{2})^{k}}{k!Gammaleft(nu+k+1right)}$$
is an analytic function of $z$, except for a branch point at $z=0$ when $nu$ is not an integer (see here). In this case, the given condition $left|operatorname{Arg}(z)right|<pi$ corresponds to the principal branch of $J_nu(z)$ (cut along the negative real axis).
If $nu$ is an integer, $J_nu(-z)=(-1)^nu J_nu(z)$, which allows to find the asymptotic expansion for $z<0$.
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
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$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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$begingroup$
From its expansion,
$$J_{nu}left(zright)=(tfrac{1}{2}z)^{nu}sum_{k=0}^{infty}(-1)^{k}frac{(%
tfrac{1}{4}z^{2})^{k}}{k!Gammaleft(nu+k+1right)}$$
is an analytic function of $z$, except for a branch point at $z=0$ when $nu$ is not an integer (see here). In this case, the given condition $left|operatorname{Arg}(z)right|<pi$ corresponds to the principal branch of $J_nu(z)$ (cut along the negative real axis).
If $nu$ is an integer, $J_nu(-z)=(-1)^nu J_nu(z)$, which allows to find the asymptotic expansion for $z<0$.
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
$endgroup$
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
add a comment |
$begingroup$
From its expansion,
$$J_{nu}left(zright)=(tfrac{1}{2}z)^{nu}sum_{k=0}^{infty}(-1)^{k}frac{(%
tfrac{1}{4}z^{2})^{k}}{k!Gammaleft(nu+k+1right)}$$
is an analytic function of $z$, except for a branch point at $z=0$ when $nu$ is not an integer (see here). In this case, the given condition $left|operatorname{Arg}(z)right|<pi$ corresponds to the principal branch of $J_nu(z)$ (cut along the negative real axis).
If $nu$ is an integer, $J_nu(-z)=(-1)^nu J_nu(z)$, which allows to find the asymptotic expansion for $z<0$.
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
$endgroup$
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
add a comment |
$begingroup$
From its expansion,
$$J_{nu}left(zright)=(tfrac{1}{2}z)^{nu}sum_{k=0}^{infty}(-1)^{k}frac{(%
tfrac{1}{4}z^{2})^{k}}{k!Gammaleft(nu+k+1right)}$$
is an analytic function of $z$, except for a branch point at $z=0$ when $nu$ is not an integer (see here). In this case, the given condition $left|operatorname{Arg}(z)right|<pi$ corresponds to the principal branch of $J_nu(z)$ (cut along the negative real axis).
If $nu$ is an integer, $J_nu(-z)=(-1)^nu J_nu(z)$, which allows to find the asymptotic expansion for $z<0$.
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
$endgroup$
From its expansion,
$$J_{nu}left(zright)=(tfrac{1}{2}z)^{nu}sum_{k=0}^{infty}(-1)^{k}frac{(%
tfrac{1}{4}z^{2})^{k}}{k!Gammaleft(nu+k+1right)}$$
is an analytic function of $z$, except for a branch point at $z=0$ when $nu$ is not an integer (see here). In this case, the given condition $left|operatorname{Arg}(z)right|<pi$ corresponds to the principal branch of $J_nu(z)$ (cut along the negative real axis).
If $nu$ is an integer, $J_nu(-z)=(-1)^nu J_nu(z)$, which allows to find the asymptotic expansion for $z<0$.
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
answered Dec 14 '18 at 15:21
Paul EntaPaul Enta
5,18111334
5,18111334
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
add a comment |
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
If I'm understanding correctly, then it really does just mean that the expansion is valid for all z not on the negative real line, correct?
$endgroup$
– BGreen
Dec 14 '18 at 17:04
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
$begingroup$
Yes, if $nu$ is not an integer. If it is, you can find a more general expression for the expansion.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:11
add a comment |
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