Sign of the derivative of the Hankel Function with purely imaginary arguments
Consider the Hankel function of the first kind $H^{(1)}_nu (z)$.
If I restrict $z$ to be a purely imaginary number of the form $ix$ where $x in mathbb{R}$ and $x>0$ and let $nu =0$. We can see that $iH^{(1)}_0 (ix) > 0$. Using the standard recursion relationship for the derivatives of the Hankel (and Bessel) functions, we see that $frac{partial}{partial x} big(iH^{(1)}_0 (ix)big) = H^{(1)}_1 (ix)$, and running some calculations, we see that $H^{(1)} _1 (ix) < 0$ for $x>0$.
If you continue taking the $x$ derivatives of $iH_0 ^{(1)} (ix)$, the sign of the derivatives alternates for $x>0$. My question is, how do I prove that for even $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) >0$ and for odd $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) <0$?
calculus complex-analysis bessel-functions
asked Nov 29 '18 at 16:43
user207526user207526
61
add a comment |
Consider the Hankel function of the first kind $H^{(1)}_nu (z)$.
If I restrict $z$ to be a purely imaginary number of the form $ix$ where $x in mathbb{R}$ and $x>0$ and let $nu =0$. We can see that $iH^{(1)}_0 (ix) > 0$. Using the standard recursion relationship for the derivatives of the Hankel (and Bessel) functions, we see that $frac{partial}{partial x} big(iH^{(1)}_0 (ix)big) = H^{(1)}_1 (ix)$, and running some calculations, we see that $H^{(1)} _1 (ix) < 0$ for $x>0$.
If you continue taking the $x$ derivatives of $iH_0 ^{(1)} (ix)$, the sign of the derivatives alternates for $x>0$. My question is, how do I prove that for even $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) >0$ and for odd $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) <0$?
calculus complex-analysis bessel-functions
asked Nov 29 '18 at 16:43
user207526user207526
61
add a comment |
Consider the Hankel function of the first kind $H^{(1)}_nu (z)$.
If I restrict $z$ to be a purely imaginary number of the form $ix$ where $x in mathbb{R}$ and $x>0$ and let $nu =0$. We can see that $iH^{(1)}_0 (ix) > 0$. Using the standard recursion relationship for the derivatives of the Hankel (and Bessel) functions, we see that $frac{partial}{partial x} big(iH^{(1)}_0 (ix)big) = H^{(1)}_1 (ix)$, and running some calculations, we see that $H^{(1)} _1 (ix) < 0$ for $x>0$.
If you continue taking the $x$ derivatives of $iH_0 ^{(1)} (ix)$, the sign of the derivatives alternates for $x>0$. My question is, how do I prove that for even $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) >0$ and for odd $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) <0$?
calculus complex-analysis bessel-functions
asked Nov 29 '18 at 16:43
user207526user207526
61
Consider the Hankel function of the first kind $H^{(1)}_nu (z)$.
If I restrict $z$ to be a purely imaginary number of the form $ix$ where $x in mathbb{R}$ and $x>0$ and let $nu =0$. We can see that $iH^{(1)}_0 (ix) > 0$. Using the standard recursion relationship for the derivatives of the Hankel (and Bessel) functions, we see that $frac{partial}{partial x} big(iH^{(1)}_0 (ix)big) = H^{(1)}_1 (ix)$, and running some calculations, we see that $H^{(1)} _1 (ix) < 0$ for $x>0$.
If you continue taking the $x$ derivatives of $iH_0 ^{(1)} (ix)$, the sign of the derivatives alternates for $x>0$. My question is, how do I prove that for even $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) >0$ and for odd $n$, $frac{rm d ^n}{rm d x^n} big(iH^{(1)}_0 (ix)big) <0$?
calculus complex-analysis bessel-functions
calculus complex-analysis bessel-functions
asked Nov 29 '18 at 16:43
user207526user207526
61
asked Nov 29 '18 at 16:43
user207526user207526
61
asked Nov 29 '18 at 16:43
user207526user207526
61
asked Nov 29 '18 at 16:43
user207526user207526
61
asked Nov 29 '18 at 16:43
user207526user207526
61
61
add a comment |
add a comment |
1 Answer
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We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$
begin{equation}
i{H^{(1)}_{0}}left(ixright)=frac{2}{pi}K_{0}left(xright)
end{equation}
Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function:
begin{align}
&K_{0}^{(k)}left(xright)=\
&frac{1}{2^{k}}left(e^{-ikpi}K_{-%
k}left(xright)+{kchoose1}e^{-i(k-2)pi}K_{-k+2}left(x%
right)+{kchoose2}e^{-i(k-4)pi}K_{-k+4}left(xright)+%
cdots+e^{ikpi}K_{+k}left(xright)right)
end{align}
All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As
begin{align}
frac{d^k}{dx^k}left[ i{H^{(1)}_{0}}left(ixright)right]&=frac{2}{pi}frac{d^k}{dx^k}left[K_{0}left(xright)right]
end{align}
Since $K_{-nu}(x)=K_nu(x)$ and $K_nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$
begin{equation}
i{H^{(1)}_{0}}left(ixright)=frac{2}{pi}K_{0}left(xright)
end{equation}
Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function:
begin{align}
&K_{0}^{(k)}left(xright)=\
&frac{1}{2^{k}}left(e^{-ikpi}K_{-%
k}left(xright)+{kchoose1}e^{-i(k-2)pi}K_{-k+2}left(x%
right)+{kchoose2}e^{-i(k-4)pi}K_{-k+4}left(xright)+%
cdots+e^{ikpi}K_{+k}left(xright)right)
end{align}
All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As
begin{align}
frac{d^k}{dx^k}left[ i{H^{(1)}_{0}}left(ixright)right]&=frac{2}{pi}frac{d^k}{dx^k}left[K_{0}left(xright)right]
end{align}
Since $K_{-nu}(x)=K_nu(x)$ and $K_nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
add a comment |
We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$
begin{equation}
i{H^{(1)}_{0}}left(ixright)=frac{2}{pi}K_{0}left(xright)
end{equation}
Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function:
begin{align}
&K_{0}^{(k)}left(xright)=\
&frac{1}{2^{k}}left(e^{-ikpi}K_{-%
k}left(xright)+{kchoose1}e^{-i(k-2)pi}K_{-k+2}left(x%
right)+{kchoose2}e^{-i(k-4)pi}K_{-k+4}left(xright)+%
cdots+e^{ikpi}K_{+k}left(xright)right)
end{align}
All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As
begin{align}
frac{d^k}{dx^k}left[ i{H^{(1)}_{0}}left(ixright)right]&=frac{2}{pi}frac{d^k}{dx^k}left[K_{0}left(xright)right]
end{align}
Since $K_{-nu}(x)=K_nu(x)$ and $K_nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
add a comment |
We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$
begin{equation}
i{H^{(1)}_{0}}left(ixright)=frac{2}{pi}K_{0}left(xright)
end{equation}
Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function:
begin{align}
&K_{0}^{(k)}left(xright)=\
&frac{1}{2^{k}}left(e^{-ikpi}K_{-%
k}left(xright)+{kchoose1}e^{-i(k-2)pi}K_{-k+2}left(x%
right)+{kchoose2}e^{-i(k-4)pi}K_{-k+4}left(xright)+%
cdots+e^{ikpi}K_{+k}left(xright)right)
end{align}
All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As
begin{align}
frac{d^k}{dx^k}left[ i{H^{(1)}_{0}}left(ixright)right]&=frac{2}{pi}frac{d^k}{dx^k}left[K_{0}left(xright)right]
end{align}
Since $K_{-nu}(x)=K_nu(x)$ and $K_nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
We can transform the Hankel function into a modified Bessel function using the correspondence (DLMF) for $x>0$
begin{equation}
i{H^{(1)}_{0}}left(ixright)=frac{2}{pi}K_{0}left(xright)
end{equation}
Now, the $k$-th derivative can be obtained from the expression (DLMF) of that of the modified Bessel function:
begin{align}
&K_{0}^{(k)}left(xright)=\
&frac{1}{2^{k}}left(e^{-ikpi}K_{-%
k}left(xright)+{kchoose1}e^{-i(k-2)pi}K_{-k+2}left(x%
right)+{kchoose2}e^{-i(k-4)pi}K_{-k+4}left(xright)+%
cdots+e^{ikpi}K_{+k}left(xright)right)
end{align}
All phase factors of the sum equal $-1$ if $k$ is odd and $1$ if $k$ is even. As
begin{align}
frac{d^k}{dx^k}left[ i{H^{(1)}_{0}}left(ixright)right]&=frac{2}{pi}frac{d^k}{dx^k}left[K_{0}left(xright)right]
end{align}
Since $K_{-nu}(x)=K_nu(x)$ and $K_nu(x)$ is positive, the sign of the $k$-th derivative is the same as $(-1)^k$ as expected.
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
answered Nov 29 '18 at 21:10
Paul EntaPaul Enta
4,28611129
4,28611129
add a comment |
add a comment |
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