Confusion regarding Kelvin functions
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I am trying to implement the following equation from this paper and having some troubles in the interpretation of $bei'$ and $ber'$. I understand from the definition of Kelvin functions that for integers n, $ber_n(x)$ can be defied, and usually, $bei(x)$ means $bei_0(x)$. I can compute it using Matlab function.
What is the usual interpretation of $bei'$ and $ber'$ ? is it $bei_1(x)$ and $ber_1(x)$ or is it the derivative? It appears that it is very common notation and not explained in the paper (I could not find anywhere else)
special-functions intuition matlab bessel-functions
asked Dec 5 '18 at 15:24
PojjPojj
347
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add a comment |
$begingroup$
I am trying to implement the following equation from this paper and having some troubles in the interpretation of $bei'$ and $ber'$. I understand from the definition of Kelvin functions that for integers n, $ber_n(x)$ can be defied, and usually, $bei(x)$ means $bei_0(x)$. I can compute it using Matlab function.
What is the usual interpretation of $bei'$ and $ber'$ ? is it $bei_1(x)$ and $ber_1(x)$ or is it the derivative? It appears that it is very common notation and not explained in the paper (I could not find anywhere else)
special-functions intuition matlab bessel-functions
asked Dec 5 '18 at 15:24
PojjPojj
347
$endgroup$
add a comment |
$begingroup$
I am trying to implement the following equation from this paper and having some troubles in the interpretation of $bei'$ and $ber'$. I understand from the definition of Kelvin functions that for integers n, $ber_n(x)$ can be defied, and usually, $bei(x)$ means $bei_0(x)$. I can compute it using Matlab function.
What is the usual interpretation of $bei'$ and $ber'$ ? is it $bei_1(x)$ and $ber_1(x)$ or is it the derivative? It appears that it is very common notation and not explained in the paper (I could not find anywhere else)
special-functions intuition matlab bessel-functions
asked Dec 5 '18 at 15:24
PojjPojj
347
$endgroup$
I am trying to implement the following equation from this paper and having some troubles in the interpretation of $bei'$ and $ber'$. I understand from the definition of Kelvin functions that for integers n, $ber_n(x)$ can be defied, and usually, $bei(x)$ means $bei_0(x)$. I can compute it using Matlab function.
What is the usual interpretation of $bei'$ and $ber'$ ? is it $bei_1(x)$ and $ber_1(x)$ or is it the derivative? It appears that it is very common notation and not explained in the paper (I could not find anywhere else)
special-functions intuition matlab bessel-functions
special-functions intuition matlab bessel-functions
asked Dec 5 '18 at 15:24
PojjPojj
347
asked Dec 5 '18 at 15:24
PojjPojj
347
edited Dec 7 '18 at 7:02
Pojj
asked Dec 5 '18 at 15:24
PojjPojj
347
asked Dec 5 '18 at 15:24
PojjPojj
347
asked Dec 5 '18 at 15:24
PojjPojj
347
347
add a comment |
add a comment |
1 Answer
1
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First of all: $operatorname{bei}(x)$ means $operatorname{bei}_0(x)$ and not $operatorname{ber}_0(x)$.
Second:
$operatorname{bei}'(x)$ is the derivative of $operatorname{bei}(x)$ and not $operatorname{bei}_1(x)$ but the derivatives can be expressed with the first order function, see https://dlmf.nist.gov/10.63.E3
$$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
$$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
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$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
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What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
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(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
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– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all: $operatorname{bei}(x)$ means $operatorname{bei}_0(x)$ and not $operatorname{ber}_0(x)$.
Second:
$operatorname{bei}'(x)$ is the derivative of $operatorname{bei}(x)$ and not $operatorname{bei}_1(x)$ but the derivatives can be expressed with the first order function, see https://dlmf.nist.gov/10.63.E3
$$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
$$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
$endgroup$
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
add a comment |
$begingroup$
First of all: $operatorname{bei}(x)$ means $operatorname{bei}_0(x)$ and not $operatorname{ber}_0(x)$.
Second:
$operatorname{bei}'(x)$ is the derivative of $operatorname{bei}(x)$ and not $operatorname{bei}_1(x)$ but the derivatives can be expressed with the first order function, see https://dlmf.nist.gov/10.63.E3
$$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
$$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
$endgroup$
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
add a comment |
$begingroup$
First of all: $operatorname{bei}(x)$ means $operatorname{bei}_0(x)$ and not $operatorname{ber}_0(x)$.
Second:
$operatorname{bei}'(x)$ is the derivative of $operatorname{bei}(x)$ and not $operatorname{bei}_1(x)$ but the derivatives can be expressed with the first order function, see https://dlmf.nist.gov/10.63.E3
$$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
$$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
$endgroup$
First of all: $operatorname{bei}(x)$ means $operatorname{bei}_0(x)$ and not $operatorname{ber}_0(x)$.
Second:
$operatorname{bei}'(x)$ is the derivative of $operatorname{bei}(x)$ and not $operatorname{bei}_1(x)$ but the derivatives can be expressed with the first order function, see https://dlmf.nist.gov/10.63.E3
$$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
$$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
answered Dec 6 '18 at 8:51
gammatestergammatester
16.8k21733
16.8k21733
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
add a comment |
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
Great! Thank you. Initial one was a typo. Do you know any book or a research paper which has the expansion of derivatives?
$endgroup$
– Pojj
Dec 7 '18 at 7:01
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
What kind of expansion? You can differentiate the expansions of the original functions, the section Basic properties shows some relations, the Wolfram Function Site 1 and 2 gives expansions in hypergeometric and other functions. There are polynomial approximations in Abramowitz/Stegun.
$endgroup$
– gammatester
Dec 7 '18 at 8:49
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
(continued) My own Pascal implementation uses the term-wise derivatives of the following expansions: Abramowitz/Stegun 9.9.10 if $x le 20$ and 9.10.1/2 for $x > 20$.
$endgroup$
– gammatester
Dec 7 '18 at 8:50
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
$begingroup$
Got it! thank you.
$endgroup$
– Pojj
Dec 7 '18 at 10:24
add a comment |
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