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Is there a closed form, or cleaner way of writing this function?


Does a closed form solution to this nonlinear ODE exist?Is there a closed form solution to this equation?Prove that the composition of two “closed form functions” is itself a “closed form function”?Areas where closed form solutions are of particular interestClassifying functions whose inverse do not have a closed formClosed form of planetary radial motion as time functionNonlinear ODE, closed form solution?Closed form for a series IIIs there a closed form for this “flowery” integral?Deriving the closed form of Gamma function using Euler-Chi function













1












$begingroup$


Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$frac{d^nf(x)}{dx^n}|_{x=0}=f(n)$$
What functions f could satisfy this equation? Do any functions of f have a closed form, or if not does it have a form that is just a normal ODE form?






ordinary-differential-equations derivatives closed-form







share|cite|improve this question









$endgroup$












  • $begingroup$
    From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
    $endgroup$
    – Henry Lee
    3 hours ago












  • $begingroup$
    @Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
    $endgroup$
    – tox123
    3 hours ago










  • $begingroup$
    They both mean the same thing and what you have showed is equal and shows what you mean
    $endgroup$
    – Henry Lee
    3 hours ago
















1












$begingroup$


Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$frac{d^nf(x)}{dx^n}|_{x=0}=f(n)$$
What functions f could satisfy this equation? Do any functions of f have a closed form, or if not does it have a form that is just a normal ODE form?






ordinary-differential-equations derivatives closed-form







share|cite|improve this question









$endgroup$












  • $begingroup$
    From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
    $endgroup$
    – Henry Lee
    3 hours ago












  • $begingroup$
    @Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
    $endgroup$
    – tox123
    3 hours ago










  • $begingroup$
    They both mean the same thing and what you have showed is equal and shows what you mean
    $endgroup$
    – Henry Lee
    3 hours ago














1












1








1





$begingroup$


Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$frac{d^nf(x)}{dx^n}|_{x=0}=f(n)$$
What functions f could satisfy this equation? Do any functions of f have a closed form, or if not does it have a form that is just a normal ODE form?






ordinary-differential-equations derivatives closed-form







share|cite|improve this question









$endgroup$




Given the following, and assuming that $f(x)$ is infinitely differentiable:
$$frac{d^nf(x)}{dx^n}|_{x=0}=f(n)$$
What functions f could satisfy this equation? Do any functions of f have a closed form, or if not does it have a form that is just a normal ODE form?






ordinary-differential-equations derivatives closed-form




ordinary-differential-equations derivatives closed-form






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









tox123tox123

572721




572721












  • $begingroup$
    From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
    $endgroup$
    – Henry Lee
    3 hours ago












  • $begingroup$
    @Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
    $endgroup$
    – tox123
    3 hours ago










  • $begingroup$
    They both mean the same thing and what you have showed is equal and shows what you mean
    $endgroup$
    – Henry Lee
    3 hours ago


















  • $begingroup$
    From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
    $endgroup$
    – Henry Lee
    3 hours ago












  • $begingroup$
    @Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
    $endgroup$
    – tox123
    3 hours ago










  • $begingroup$
    They both mean the same thing and what you have showed is equal and shows what you mean
    $endgroup$
    – Henry Lee
    3 hours ago
















$begingroup$
From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
$endgroup$
– Henry Lee
3 hours ago






$begingroup$
From what you've written it looks like: $$f(x=n)=f^{(n)}(0)$$ I've written $x=n$ to show that we are primarily defining $f$ in terms of $x$ which applies of the right. However you could also say: $$f(n)=f^{(n)}(0)$$
$endgroup$
– Henry Lee
3 hours ago














$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
3 hours ago




$begingroup$
@Henry Lee I used the notation I did only because I'm more familiar with it. Is the notation you suggest preferred, or the standard notation?
$endgroup$
– tox123
3 hours ago












$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
3 hours ago




$begingroup$
They both mean the same thing and what you have showed is equal and shows what you mean
$endgroup$
– Henry Lee
3 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $f(x)=a^{x+1}$, where $a$ satisfies $ln(a)=a$. Then $f^{(n)}(0)=ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}approx 0.318+1.337i$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is this exclusively the only answer or are there other functions that satisfy my conditions?
    $endgroup$
    – tox123
    3 hours ago










  • $begingroup$
    I can't see how $ln^n(a)a^1=a^{n+1}$...
    $endgroup$
    – Thehx
    3 hours ago






  • 1




    $begingroup$
    @Thehx: look at the definition of $a$.
    $endgroup$
    – Alex R.
    2 hours ago






  • 1




    $begingroup$
    oh god, this is beautiful.
    $endgroup$
    – Thehx
    2 hours ago



















0












$begingroup$

I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.



UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Let $f(x)=a^{x+1}$, where $a$ satisfies $ln(a)=a$. Then $f^{(n)}(0)=ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}approx 0.318+1.337i$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is this exclusively the only answer or are there other functions that satisfy my conditions?
      $endgroup$
      – tox123
      3 hours ago










    • $begingroup$
      I can't see how $ln^n(a)a^1=a^{n+1}$...
      $endgroup$
      – Thehx
      3 hours ago






    • 1




      $begingroup$
      @Thehx: look at the definition of $a$.
      $endgroup$
      – Alex R.
      2 hours ago






    • 1




      $begingroup$
      oh god, this is beautiful.
      $endgroup$
      – Thehx
      2 hours ago
















    4












    $begingroup$

    Let $f(x)=a^{x+1}$, where $a$ satisfies $ln(a)=a$. Then $f^{(n)}(0)=ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}approx 0.318+1.337i$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is this exclusively the only answer or are there other functions that satisfy my conditions?
      $endgroup$
      – tox123
      3 hours ago










    • $begingroup$
      I can't see how $ln^n(a)a^1=a^{n+1}$...
      $endgroup$
      – Thehx
      3 hours ago






    • 1




      $begingroup$
      @Thehx: look at the definition of $a$.
      $endgroup$
      – Alex R.
      2 hours ago






    • 1




      $begingroup$
      oh god, this is beautiful.
      $endgroup$
      – Thehx
      2 hours ago














    4












    4








    4





    $begingroup$

    Let $f(x)=a^{x+1}$, where $a$ satisfies $ln(a)=a$. Then $f^{(n)}(0)=ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}approx 0.318+1.337i$.






    share|cite|improve this answer











    $endgroup$



    Let $f(x)=a^{x+1}$, where $a$ satisfies $ln(a)=a$. Then $f^{(n)}(0)=ln^n(a) a^{1}=a^{n+1}=f(n)$, as desired. Note that $a$ will be a complex number here, explicitly in terms of Lambert’s W: $a=e^{-W(-1)}approx 0.318+1.337i$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 45 mins ago

























    answered 3 hours ago









    Alex R.Alex R.

    25.3k12454




    25.3k12454












    • $begingroup$
      Is this exclusively the only answer or are there other functions that satisfy my conditions?
      $endgroup$
      – tox123
      3 hours ago










    • $begingroup$
      I can't see how $ln^n(a)a^1=a^{n+1}$...
      $endgroup$
      – Thehx
      3 hours ago






    • 1




      $begingroup$
      @Thehx: look at the definition of $a$.
      $endgroup$
      – Alex R.
      2 hours ago






    • 1




      $begingroup$
      oh god, this is beautiful.
      $endgroup$
      – Thehx
      2 hours ago


















    • $begingroup$
      Is this exclusively the only answer or are there other functions that satisfy my conditions?
      $endgroup$
      – tox123
      3 hours ago










    • $begingroup$
      I can't see how $ln^n(a)a^1=a^{n+1}$...
      $endgroup$
      – Thehx
      3 hours ago






    • 1




      $begingroup$
      @Thehx: look at the definition of $a$.
      $endgroup$
      – Alex R.
      2 hours ago






    • 1




      $begingroup$
      oh god, this is beautiful.
      $endgroup$
      – Thehx
      2 hours ago
















    $begingroup$
    Is this exclusively the only answer or are there other functions that satisfy my conditions?
    $endgroup$
    – tox123
    3 hours ago




    $begingroup$
    Is this exclusively the only answer or are there other functions that satisfy my conditions?
    $endgroup$
    – tox123
    3 hours ago












    $begingroup$
    I can't see how $ln^n(a)a^1=a^{n+1}$...
    $endgroup$
    – Thehx
    3 hours ago




    $begingroup$
    I can't see how $ln^n(a)a^1=a^{n+1}$...
    $endgroup$
    – Thehx
    3 hours ago




    1




    1




    $begingroup$
    @Thehx: look at the definition of $a$.
    $endgroup$
    – Alex R.
    2 hours ago




    $begingroup$
    @Thehx: look at the definition of $a$.
    $endgroup$
    – Alex R.
    2 hours ago




    1




    1




    $begingroup$
    oh god, this is beautiful.
    $endgroup$
    – Thehx
    2 hours ago




    $begingroup$
    oh god, this is beautiful.
    $endgroup$
    – Thehx
    2 hours ago











    0












    $begingroup$

    I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.



    UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.



      UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.



        UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.






        share|cite|improve this answer











        $endgroup$



        I only see the trivial f(x)=0 solution so far. Couldn't yet prove there are no other solutions tho.



        UPD: alex-r provided an absolutely amazing non-trivial solution. Make sure to check it out.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 3 hours ago









        ThehxThehx

        687




        687






























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