Are all fractional deriviatives/integrals of $e^x$ equal to $e^x$?












6














I have learned through calculus that the derivatives and the indefinite integrals of the exponential function are the same (at integer arguments) but was wondering if this holds true for fractional derivatives/integrals, such as $frac{1}{2}$, or do they turn into something monstrous?



I would test this in Wolfram Alpha but I cannot seem to get the entry correct for a fractional derivative/integral.






exponential-function fractional-calculus







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  • See fractional calculus (also here), differintegral, and fractional derivative
    – Lucian
    Jan 3 '14 at 19:33
















6














I have learned through calculus that the derivatives and the indefinite integrals of the exponential function are the same (at integer arguments) but was wondering if this holds true for fractional derivatives/integrals, such as $frac{1}{2}$, or do they turn into something monstrous?



I would test this in Wolfram Alpha but I cannot seem to get the entry correct for a fractional derivative/integral.






exponential-function fractional-calculus







share|cite|improve this question






















  • See fractional calculus (also here), differintegral, and fractional derivative
    – Lucian
    Jan 3 '14 at 19:33














6












6








6







I have learned through calculus that the derivatives and the indefinite integrals of the exponential function are the same (at integer arguments) but was wondering if this holds true for fractional derivatives/integrals, such as $frac{1}{2}$, or do they turn into something monstrous?



I would test this in Wolfram Alpha but I cannot seem to get the entry correct for a fractional derivative/integral.






exponential-function fractional-calculus







share|cite|improve this question













I have learned through calculus that the derivatives and the indefinite integrals of the exponential function are the same (at integer arguments) but was wondering if this holds true for fractional derivatives/integrals, such as $frac{1}{2}$, or do they turn into something monstrous?



I would test this in Wolfram Alpha but I cannot seem to get the entry correct for a fractional derivative/integral.






exponential-function fractional-calculus




exponential-function fractional-calculus






share|cite|improve this question













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asked Jan 3 '14 at 9:11









zerosofthezeta

2,48662746




2,48662746












  • See fractional calculus (also here), differintegral, and fractional derivative
    – Lucian
    Jan 3 '14 at 19:33


















  • See fractional calculus (also here), differintegral, and fractional derivative
    – Lucian
    Jan 3 '14 at 19:33
















See fractional calculus (also here), differintegral, and fractional derivative
– Lucian
Jan 3 '14 at 19:33




See fractional calculus (also here), differintegral, and fractional derivative
– Lucian
Jan 3 '14 at 19:33










3 Answers
3






active

oldest

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4














Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
Page 10, section 6, in the paper "The fractionnal derivation"
http://www.scribd.com/JJacquelin/documents






share|cite|improve this answer





















  • It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
    – Hans
    Jan 6 '14 at 17:35










  • @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
    – JJacquelin
    Jan 6 '14 at 18:17



















2














In fact, the fractional integals/dérivatives are based on the Riemann-Liouville operator in which the lower bound (a) of the integral plays an important role :



http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html



The conventional fractionnal calculus states a=0 :



http://mathworld.wolfram.com/FractionalCalculus.html



In this case, the fractional transform of the exponential function involves the Incomplete Gamma function as already said.
But in the case of a=-infinity, corresponding to the Weyl's operator, the Incomplete Gamma term fades and the exponential term remains alone. So, with the non-conventional definition based on the Weyl's operator, the transform of the exponential function is a simple exponential function.






share|cite|improve this answer





























    0














    This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:



    $$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$



    The same goes for the definition based on Fourier transform:



    $$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}omega^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$



    The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
      Page 10, section 6, in the paper "The fractionnal derivation"
      http://www.scribd.com/JJacquelin/documents






      share|cite|improve this answer





















      • It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
        – Hans
        Jan 6 '14 at 17:35










      • @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
        – JJacquelin
        Jan 6 '14 at 18:17
















      4














      Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
      Page 10, section 6, in the paper "The fractionnal derivation"
      http://www.scribd.com/JJacquelin/documents






      share|cite|improve this answer





















      • It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
        – Hans
        Jan 6 '14 at 17:35










      • @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
        – JJacquelin
        Jan 6 '14 at 18:17














      4












      4








      4






      Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
      Page 10, section 6, in the paper "The fractionnal derivation"
      http://www.scribd.com/JJacquelin/documents






      share|cite|improve this answer












      Not monstruous, but not very simple. The fractional integrals/dérivatives of the exponential function involves the Incomplete Gamma function :
      Page 10, section 6, in the paper "The fractionnal derivation"
      http://www.scribd.com/JJacquelin/documents







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 3 '14 at 10:33









      JJacquelin

      42.5k21750




      42.5k21750












      • It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
        – Hans
        Jan 6 '14 at 17:35










      • @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
        – JJacquelin
        Jan 6 '14 at 18:17


















      • It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
        – Hans
        Jan 6 '14 at 17:35










      • @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
        – JJacquelin
        Jan 6 '14 at 18:17
















      It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
      – Hans
      Jan 6 '14 at 17:35




      It appears to be a nice paper. Unfortunately it is in French, which I can not read. Do you have an English version? Thanks.
      – Hans
      Jan 6 '14 at 17:35












      @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
      – JJacquelin
      Jan 6 '14 at 18:17




      @ Hansen : The paper entitled "La dérivation fractionnaire, pp.1-6." is translated : "The Fractional Derivation, pp.7-12" Go directly to page 7.
      – JJacquelin
      Jan 6 '14 at 18:17











      2














      In fact, the fractional integals/dérivatives are based on the Riemann-Liouville operator in which the lower bound (a) of the integral plays an important role :



      http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html



      The conventional fractionnal calculus states a=0 :



      http://mathworld.wolfram.com/FractionalCalculus.html



      In this case, the fractional transform of the exponential function involves the Incomplete Gamma function as already said.
      But in the case of a=-infinity, corresponding to the Weyl's operator, the Incomplete Gamma term fades and the exponential term remains alone. So, with the non-conventional definition based on the Weyl's operator, the transform of the exponential function is a simple exponential function.






      share|cite|improve this answer


























        2














        In fact, the fractional integals/dérivatives are based on the Riemann-Liouville operator in which the lower bound (a) of the integral plays an important role :



        http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html



        The conventional fractionnal calculus states a=0 :



        http://mathworld.wolfram.com/FractionalCalculus.html



        In this case, the fractional transform of the exponential function involves the Incomplete Gamma function as already said.
        But in the case of a=-infinity, corresponding to the Weyl's operator, the Incomplete Gamma term fades and the exponential term remains alone. So, with the non-conventional definition based on the Weyl's operator, the transform of the exponential function is a simple exponential function.






        share|cite|improve this answer
























          2












          2








          2






          In fact, the fractional integals/dérivatives are based on the Riemann-Liouville operator in which the lower bound (a) of the integral plays an important role :



          http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html



          The conventional fractionnal calculus states a=0 :



          http://mathworld.wolfram.com/FractionalCalculus.html



          In this case, the fractional transform of the exponential function involves the Incomplete Gamma function as already said.
          But in the case of a=-infinity, corresponding to the Weyl's operator, the Incomplete Gamma term fades and the exponential term remains alone. So, with the non-conventional definition based on the Weyl's operator, the transform of the exponential function is a simple exponential function.






          share|cite|improve this answer












          In fact, the fractional integals/dérivatives are based on the Riemann-Liouville operator in which the lower bound (a) of the integral plays an important role :



          http://mathworld.wolfram.com/Riemann-LiouvilleOperator.html



          The conventional fractionnal calculus states a=0 :



          http://mathworld.wolfram.com/FractionalCalculus.html



          In this case, the fractional transform of the exponential function involves the Incomplete Gamma function as already said.
          But in the case of a=-infinity, corresponding to the Weyl's operator, the Incomplete Gamma term fades and the exponential term remains alone. So, with the non-conventional definition based on the Weyl's operator, the transform of the exponential function is a simple exponential function.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 '14 at 11:00









          JJacquelin

          42.5k21750




          42.5k21750























              0














              This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:



              $$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$



              The same goes for the definition based on Fourier transform:



              $$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}omega^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$



              The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.






              share|cite|improve this answer


























                0














                This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:



                $$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$



                The same goes for the definition based on Fourier transform:



                $$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}omega^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$



                The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.






                share|cite|improve this answer
























                  0












                  0








                  0






                  This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:



                  $$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$



                  The same goes for the definition based on Fourier transform:



                  $$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}omega^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$



                  The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.






                  share|cite|improve this answer












                  This depends on the definition of fractional derivatives but the definition based on Newton series interpolation over positive derivatives gives all fractional derivatives to be the same, $e^x$:



                  $$f^{(s)}(x)=sum_{m=0}^{infty} binom {s}m sum_{k=0}^mbinom mk(-1)^{m-k}f^{(k)}(x)$$



                  The same goes for the definition based on Fourier transform:



                  $$f^{(s)}(x)=frac{i}{2pi}int_{-infty}^{+infty} e^{- i omega x}omega^s int_{-infty}^{+infty}f(t)e^{iomega t}dt , domega$$



                  The both definitions coincide, and I consider them natural unlike other definitions that may produce a different result.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 at 23:09









                  Anixx

                  3,14512038




                  3,14512038






























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