Taylor Series in Fractional Calculus












2














I recently studied fractional calculus, namely the possibility to define fractional derivatives of some functions, like



$$frac{text{d}^{1/2}}{text{d}x^{1/2}} f(x) ~~~~~~~~~~~~~ frac{text{d}^{2/3}}{text{d}x^{2/3}} f(x)$$



and so on.



Now the question that came up into my mind is: if such a construction is possible, can we built " new " Taylor series for well known function in order to take into account (some) fractional derivatives too?



I know the first problems that would arise would be: how could we take the whole possible derivatives of order between $0$ and $1$? They are infinite. And Yeah, that could be a really huge problem..



Are there any example of Fractional Taylor Series?



P.s. I've already read other similar questions, but they are too arid and I didn't find any good answer yet..






derivatives taylor-expansion fractional-calculus







share|cite|improve this question






















  • How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
    – Simply Beautiful Art
    Feb 27 '16 at 0:54
















2














I recently studied fractional calculus, namely the possibility to define fractional derivatives of some functions, like



$$frac{text{d}^{1/2}}{text{d}x^{1/2}} f(x) ~~~~~~~~~~~~~ frac{text{d}^{2/3}}{text{d}x^{2/3}} f(x)$$



and so on.



Now the question that came up into my mind is: if such a construction is possible, can we built " new " Taylor series for well known function in order to take into account (some) fractional derivatives too?



I know the first problems that would arise would be: how could we take the whole possible derivatives of order between $0$ and $1$? They are infinite. And Yeah, that could be a really huge problem..



Are there any example of Fractional Taylor Series?



P.s. I've already read other similar questions, but they are too arid and I didn't find any good answer yet..






derivatives taylor-expansion fractional-calculus







share|cite|improve this question






















  • How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
    – Simply Beautiful Art
    Feb 27 '16 at 0:54














2












2








2


1





I recently studied fractional calculus, namely the possibility to define fractional derivatives of some functions, like



$$frac{text{d}^{1/2}}{text{d}x^{1/2}} f(x) ~~~~~~~~~~~~~ frac{text{d}^{2/3}}{text{d}x^{2/3}} f(x)$$



and so on.



Now the question that came up into my mind is: if such a construction is possible, can we built " new " Taylor series for well known function in order to take into account (some) fractional derivatives too?



I know the first problems that would arise would be: how could we take the whole possible derivatives of order between $0$ and $1$? They are infinite. And Yeah, that could be a really huge problem..



Are there any example of Fractional Taylor Series?



P.s. I've already read other similar questions, but they are too arid and I didn't find any good answer yet..






derivatives taylor-expansion fractional-calculus







share|cite|improve this question













I recently studied fractional calculus, namely the possibility to define fractional derivatives of some functions, like



$$frac{text{d}^{1/2}}{text{d}x^{1/2}} f(x) ~~~~~~~~~~~~~ frac{text{d}^{2/3}}{text{d}x^{2/3}} f(x)$$



and so on.



Now the question that came up into my mind is: if such a construction is possible, can we built " new " Taylor series for well known function in order to take into account (some) fractional derivatives too?



I know the first problems that would arise would be: how could we take the whole possible derivatives of order between $0$ and $1$? They are infinite. And Yeah, that could be a really huge problem..



Are there any example of Fractional Taylor Series?



P.s. I've already read other similar questions, but they are too arid and I didn't find any good answer yet..






derivatives taylor-expansion fractional-calculus




derivatives taylor-expansion fractional-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 '16 at 15:05









Von Neumann

16.3k72543




16.3k72543












  • How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
    – Simply Beautiful Art
    Feb 27 '16 at 0:54


















  • How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
    – Simply Beautiful Art
    Feb 27 '16 at 0:54
















How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
– Simply Beautiful Art
Feb 27 '16 at 0:54




How/Why does the Taylor Series expansion work? Does the Taylor Series expansion always work? Why? Try to apply these ideas on Fractional Calculus (though I don't see it working, due to one of the theories of Fractional Calculus being that a fractional derivative may be independent of close values of the original function, ie they don't work like derivatives, you can approximate the derivative by looking at the graph. Conversely, I imagine it hard to approximate a graph by looking at the fractional derivative.)
– Simply Beautiful Art
Feb 27 '16 at 0:54










3 Answers
3






active

oldest

votes


















3














Yes:



Series expansion in fractional calculus and fractional differential equations



Fractional calculus and the Taylor-Riemann series



Examples: see p.7-9 of the first link and p.18 of second link.






share|cite|improve this answer

















  • 1




    very interesting reference, thanks (+1)
    – G Cab
    Dec 9 at 0:51



















1














Doing a few calculations, I find the following must be:



$$f^{(n)}(x)=sum_{k=-infty}^inftyfrac{f^{(k)}(a)x^{k-n}}{Gamma(k-n+1)}$$



where we define ${1overGamma(-n)}=0$ for $ninmathbb N$



The regular Taylor series doesn't hold by induction, but this one does.



You could probably adjust $k$ to fit what you want.



(note, it fails to converge for most $f(x)$)






share|cite|improve this answer































    1














    An example of the formula for Caputo fractional derivative at zero is
    begin{equation}
    D_0^{alpha}f(x) = sum_{n=k}^{infty}frac{f^{(n)}(0)cdot x^{n-alpha}}{Gamma(n+1-alpha)} text{ where } k=text{rounded up } alpha.
    end{equation}






    share|cite|improve this answer























    • Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
      – hardmath
      Dec 9 at 1:05










    • This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
      – speepyjoe
      Dec 10 at 16:21











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Yes:



    Series expansion in fractional calculus and fractional differential equations



    Fractional calculus and the Taylor-Riemann series



    Examples: see p.7-9 of the first link and p.18 of second link.






    share|cite|improve this answer

















    • 1




      very interesting reference, thanks (+1)
      – G Cab
      Dec 9 at 0:51
















    3














    Yes:



    Series expansion in fractional calculus and fractional differential equations



    Fractional calculus and the Taylor-Riemann series



    Examples: see p.7-9 of the first link and p.18 of second link.






    share|cite|improve this answer

















    • 1




      very interesting reference, thanks (+1)
      – G Cab
      Dec 9 at 0:51














    3












    3








    3






    Yes:



    Series expansion in fractional calculus and fractional differential equations



    Fractional calculus and the Taylor-Riemann series



    Examples: see p.7-9 of the first link and p.18 of second link.






    share|cite|improve this answer












    Yes:



    Series expansion in fractional calculus and fractional differential equations



    Fractional calculus and the Taylor-Riemann series



    Examples: see p.7-9 of the first link and p.18 of second link.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 '16 at 15:15









    Martín-Blas Pérez Pinilla

    34.1k42771




    34.1k42771








    • 1




      very interesting reference, thanks (+1)
      – G Cab
      Dec 9 at 0:51














    • 1




      very interesting reference, thanks (+1)
      – G Cab
      Dec 9 at 0:51








    1




    1




    very interesting reference, thanks (+1)
    – G Cab
    Dec 9 at 0:51




    very interesting reference, thanks (+1)
    – G Cab
    Dec 9 at 0:51











    1














    Doing a few calculations, I find the following must be:



    $$f^{(n)}(x)=sum_{k=-infty}^inftyfrac{f^{(k)}(a)x^{k-n}}{Gamma(k-n+1)}$$



    where we define ${1overGamma(-n)}=0$ for $ninmathbb N$



    The regular Taylor series doesn't hold by induction, but this one does.



    You could probably adjust $k$ to fit what you want.



    (note, it fails to converge for most $f(x)$)






    share|cite|improve this answer




























      1














      Doing a few calculations, I find the following must be:



      $$f^{(n)}(x)=sum_{k=-infty}^inftyfrac{f^{(k)}(a)x^{k-n}}{Gamma(k-n+1)}$$



      where we define ${1overGamma(-n)}=0$ for $ninmathbb N$



      The regular Taylor series doesn't hold by induction, but this one does.



      You could probably adjust $k$ to fit what you want.



      (note, it fails to converge for most $f(x)$)






      share|cite|improve this answer


























        1












        1








        1






        Doing a few calculations, I find the following must be:



        $$f^{(n)}(x)=sum_{k=-infty}^inftyfrac{f^{(k)}(a)x^{k-n}}{Gamma(k-n+1)}$$



        where we define ${1overGamma(-n)}=0$ for $ninmathbb N$



        The regular Taylor series doesn't hold by induction, but this one does.



        You could probably adjust $k$ to fit what you want.



        (note, it fails to converge for most $f(x)$)






        share|cite|improve this answer














        Doing a few calculations, I find the following must be:



        $$f^{(n)}(x)=sum_{k=-infty}^inftyfrac{f^{(k)}(a)x^{k-n}}{Gamma(k-n+1)}$$



        where we define ${1overGamma(-n)}=0$ for $ninmathbb N$



        The regular Taylor series doesn't hold by induction, but this one does.



        You could probably adjust $k$ to fit what you want.



        (note, it fails to converge for most $f(x)$)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jun 16 '16 at 15:52

























        answered Jun 16 '16 at 15:18









        Simply Beautiful Art

        50.2k578181




        50.2k578181























            1














            An example of the formula for Caputo fractional derivative at zero is
            begin{equation}
            D_0^{alpha}f(x) = sum_{n=k}^{infty}frac{f^{(n)}(0)cdot x^{n-alpha}}{Gamma(n+1-alpha)} text{ where } k=text{rounded up } alpha.
            end{equation}






            share|cite|improve this answer























            • Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
              – hardmath
              Dec 9 at 1:05










            • This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
              – speepyjoe
              Dec 10 at 16:21
















            1














            An example of the formula for Caputo fractional derivative at zero is
            begin{equation}
            D_0^{alpha}f(x) = sum_{n=k}^{infty}frac{f^{(n)}(0)cdot x^{n-alpha}}{Gamma(n+1-alpha)} text{ where } k=text{rounded up } alpha.
            end{equation}






            share|cite|improve this answer























            • Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
              – hardmath
              Dec 9 at 1:05










            • This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
              – speepyjoe
              Dec 10 at 16:21














            1












            1








            1






            An example of the formula for Caputo fractional derivative at zero is
            begin{equation}
            D_0^{alpha}f(x) = sum_{n=k}^{infty}frac{f^{(n)}(0)cdot x^{n-alpha}}{Gamma(n+1-alpha)} text{ where } k=text{rounded up } alpha.
            end{equation}






            share|cite|improve this answer














            An example of the formula for Caputo fractional derivative at zero is
            begin{equation}
            D_0^{alpha}f(x) = sum_{n=k}^{infty}frac{f^{(n)}(0)cdot x^{n-alpha}}{Gamma(n+1-alpha)} text{ where } k=text{rounded up } alpha.
            end{equation}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 11 at 16:05

























            answered Dec 8 at 23:48









            speepyjoe

            112




            112












            • Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
              – hardmath
              Dec 9 at 1:05










            • This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
              – speepyjoe
              Dec 10 at 16:21


















            • Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
              – hardmath
              Dec 9 at 1:05










            • This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
              – speepyjoe
              Dec 10 at 16:21
















            Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
            – hardmath
            Dec 9 at 1:05




            Your post is quite terse, and this is has undesirable consequences. For one thing the statement is conclusory, without reasoned mathematical argument to support it. Even more undesirable is the lack of evident connection to the Question's posed problem ("can we buil[d] new Taylor series for well known function[s]").
            – hardmath
            Dec 9 at 1:05












            This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
            – speepyjoe
            Dec 10 at 16:21




            This is just an example of fractional Taylor series as was requested. I can prove the formula but that will take a couple of pages. Too long for this forum, I think.
            – speepyjoe
            Dec 10 at 16:21


















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