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







share|cite|improve this question











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    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 7:02







      Pojj

















      asked Dec 5 '18 at 15:24









      PojjPojj

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          0












          $begingroup$

          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



          $$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
          $$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











          Your 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



          $$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
          $$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          0












          $begingroup$

          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



          $$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
          $$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          0












          0








          0





          $begingroup$

          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



          $$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
          $$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$






          share|cite|improve this answer









          $endgroup$



          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



          $$operatorname{ber}'(x) = frac{1}{sqrt{2}}Big( operatorname{ber}_1(x) + operatorname{bei}_1(x)Big)$$
          $$operatorname{bei}'(x) = frac{1}{sqrt{2}}Big( operatorname{bei}_1(x) - operatorname{ber}_1(x)Big)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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