Factorial inverse equation












0












$begingroup$


I need to find a solution to a problem and I need to solve an equation. The following equation is the result of some manipulation.



I'm a little bit rusty and I can't see how to find a simple solution ( to be honest I can't find a solution at all ):



$e^{-A} cdot A^x = x! cdot K$



What I need is $x$. $A$ and $K$ are real, positive constant numbers an I'm expecting the solution would be real and positive. Does anyone has suggestions ?






factorial poisson-distribution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:08












  • $begingroup$
    Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:12










  • $begingroup$
    Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:14






  • 1




    $begingroup$
    It makes it almost certain that there is no solution.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 16:14
















0












$begingroup$


I need to find a solution to a problem and I need to solve an equation. The following equation is the result of some manipulation.



I'm a little bit rusty and I can't see how to find a simple solution ( to be honest I can't find a solution at all ):



$e^{-A} cdot A^x = x! cdot K$



What I need is $x$. $A$ and $K$ are real, positive constant numbers an I'm expecting the solution would be real and positive. Does anyone has suggestions ?






factorial poisson-distribution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:08












  • $begingroup$
    Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:12










  • $begingroup$
    Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:14






  • 1




    $begingroup$
    It makes it almost certain that there is no solution.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 16:14














0












0








0





$begingroup$


I need to find a solution to a problem and I need to solve an equation. The following equation is the result of some manipulation.



I'm a little bit rusty and I can't see how to find a simple solution ( to be honest I can't find a solution at all ):



$e^{-A} cdot A^x = x! cdot K$



What I need is $x$. $A$ and $K$ are real, positive constant numbers an I'm expecting the solution would be real and positive. Does anyone has suggestions ?






factorial poisson-distribution







share|cite|improve this question









$endgroup$




I need to find a solution to a problem and I need to solve an equation. The following equation is the result of some manipulation.



I'm a little bit rusty and I can't see how to find a simple solution ( to be honest I can't find a solution at all ):



$e^{-A} cdot A^x = x! cdot K$



What I need is $x$. $A$ and $K$ are real, positive constant numbers an I'm expecting the solution would be real and positive. Does anyone has suggestions ?






factorial poisson-distribution




factorial poisson-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '16 at 16:06









haster8558haster8558

43




43








  • 1




    $begingroup$
    So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:08












  • $begingroup$
    Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:12










  • $begingroup$
    Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:14






  • 1




    $begingroup$
    It makes it almost certain that there is no solution.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 16:14














  • 1




    $begingroup$
    So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:08












  • $begingroup$
    Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
    $endgroup$
    – Simply Beautiful Art
    Dec 19 '16 at 16:12










  • $begingroup$
    Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:14






  • 1




    $begingroup$
    It makes it almost certain that there is no solution.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 16:14








1




1




$begingroup$
So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
$endgroup$
– Simply Beautiful Art
Dec 19 '16 at 16:08






$begingroup$
So... are we using the Gamma function? Since it is very unlikely that $x$ is an integer...
$endgroup$
– Simply Beautiful Art
Dec 19 '16 at 16:08














$begingroup$
Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
$endgroup$
– Simply Beautiful Art
Dec 19 '16 at 16:12




$begingroup$
Note that if we restrict $x>0$, it is plenty likely that there may be no such solutions.
$endgroup$
– Simply Beautiful Art
Dec 19 '16 at 16:12












$begingroup$
Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
$endgroup$
– haster8558
Dec 19 '16 at 16:14




$begingroup$
Well, to be honest i don't know what the Gamma function is, but to be precise $x$ is a natural ( integer $>= 0$ ), because it rapresent the number of defect in an wafer of silicon. So it's reasonable to think to it as an integer bigger than zero. Does it help ?
$endgroup$
– haster8558
Dec 19 '16 at 16:14




1




1




$begingroup$
It makes it almost certain that there is no solution.
$endgroup$
– Robert Israel
Dec 19 '16 at 16:14




$begingroup$
It makes it almost certain that there is no solution.
$endgroup$
– Robert Israel
Dec 19 '16 at 16:14










3 Answers
3






active

oldest

votes


















1












$begingroup$

Before trying to solve this equation, I think it is better to get a sense of the possible solutions. Maybe this graph could help:



graph



As you see, the solution is not unique for some values of $A$ and $K$, while it doesn't have any solution for some positive values of $A$ and $K$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
    $endgroup$
    – haster8558
    Dec 19 '16 at 17:44










  • $begingroup$
    @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
    $endgroup$
    – polfosol
    Dec 19 '16 at 18:14










  • $begingroup$
    Ah right. I was in rush. It's definitely better. Thanks!!!
    $endgroup$
    – haster8558
    Dec 19 '16 at 18:38



















0












$begingroup$

$x!$ is not defined for arbitrary real positive $x$, just for nonnegative integers. You could replace it by $Gamma(x+1)$. But still your equation can't be solved in "closed form". Numerical methods might be tried.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:19










  • $begingroup$
    Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 20:36



















0












$begingroup$

Here's a possible solution - not sure if it will be of particularly good numerical stability. Take logs:



$$ ln{(e^{-A} A^x)} = ln{(x! K)} implies - A + x ln{A} = ln{x!} + ln{K}.$$



You might be able to solve this numerically if you use a an accurate enough approximation for $ln x!,$ using the Stirling series:



$$ ln x! approx x ln x - x + frac{1}{2} ln{2pi x} + frac{1}{12 x} - frac{1}{360 x^3} + dots$$



so a solution will be close to the solution of, say,



$$ x ln x - (1 + ln A)x+frac{1}{2} ln{2pi x} + (A + ln K) = 0.$$






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Before trying to solve this equation, I think it is better to get a sense of the possible solutions. Maybe this graph could help:



    graph



    As you see, the solution is not unique for some values of $A$ and $K$, while it doesn't have any solution for some positive values of $A$ and $K$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
      $endgroup$
      – haster8558
      Dec 19 '16 at 17:44










    • $begingroup$
      @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
      $endgroup$
      – polfosol
      Dec 19 '16 at 18:14










    • $begingroup$
      Ah right. I was in rush. It's definitely better. Thanks!!!
      $endgroup$
      – haster8558
      Dec 19 '16 at 18:38
















    1












    $begingroup$

    Before trying to solve this equation, I think it is better to get a sense of the possible solutions. Maybe this graph could help:



    graph



    As you see, the solution is not unique for some values of $A$ and $K$, while it doesn't have any solution for some positive values of $A$ and $K$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
      $endgroup$
      – haster8558
      Dec 19 '16 at 17:44










    • $begingroup$
      @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
      $endgroup$
      – polfosol
      Dec 19 '16 at 18:14










    • $begingroup$
      Ah right. I was in rush. It's definitely better. Thanks!!!
      $endgroup$
      – haster8558
      Dec 19 '16 at 18:38














    1












    1








    1





    $begingroup$

    Before trying to solve this equation, I think it is better to get a sense of the possible solutions. Maybe this graph could help:



    graph



    As you see, the solution is not unique for some values of $A$ and $K$, while it doesn't have any solution for some positive values of $A$ and $K$.






    share|cite|improve this answer









    $endgroup$



    Before trying to solve this equation, I think it is better to get a sense of the possible solutions. Maybe this graph could help:



    graph



    As you see, the solution is not unique for some values of $A$ and $K$, while it doesn't have any solution for some positive values of $A$ and $K$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 '16 at 17:14









    polfosolpolfosol

    5,55731945




    5,55731945












    • $begingroup$
      Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
      $endgroup$
      – haster8558
      Dec 19 '16 at 17:44










    • $begingroup$
      @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
      $endgroup$
      – polfosol
      Dec 19 '16 at 18:14










    • $begingroup$
      Ah right. I was in rush. It's definitely better. Thanks!!!
      $endgroup$
      – haster8558
      Dec 19 '16 at 18:38


















    • $begingroup$
      Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
      $endgroup$
      – haster8558
      Dec 19 '16 at 17:44










    • $begingroup$
      @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
      $endgroup$
      – polfosol
      Dec 19 '16 at 18:14










    • $begingroup$
      Ah right. I was in rush. It's definitely better. Thanks!!!
      $endgroup$
      – haster8558
      Dec 19 '16 at 18:38
















    $begingroup$
    Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
    $endgroup$
    – haster8558
    Dec 19 '16 at 17:44




    $begingroup$
    Looking this I got an idea. I can plot the function $e^{-A} cdot A^y = x! cdot K$ and then plot the equation $x = y$. The solution is in the cross. The only problem is to find the values a reasonable range for $x$ and $y$ .
    $endgroup$
    – haster8558
    Dec 19 '16 at 17:44












    $begingroup$
    @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
    $endgroup$
    – polfosol
    Dec 19 '16 at 18:14




    $begingroup$
    @haster8558 Why not just plot the function $e^{-A} cdot A^x - x! cdot K$ and then find its roots?
    $endgroup$
    – polfosol
    Dec 19 '16 at 18:14












    $begingroup$
    Ah right. I was in rush. It's definitely better. Thanks!!!
    $endgroup$
    – haster8558
    Dec 19 '16 at 18:38




    $begingroup$
    Ah right. I was in rush. It's definitely better. Thanks!!!
    $endgroup$
    – haster8558
    Dec 19 '16 at 18:38











    0












    $begingroup$

    $x!$ is not defined for arbitrary real positive $x$, just for nonnegative integers. You could replace it by $Gamma(x+1)$. But still your equation can't be solved in "closed form". Numerical methods might be tried.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
      $endgroup$
      – haster8558
      Dec 19 '16 at 16:19










    • $begingroup$
      Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
      $endgroup$
      – Robert Israel
      Dec 19 '16 at 20:36
















    0












    $begingroup$

    $x!$ is not defined for arbitrary real positive $x$, just for nonnegative integers. You could replace it by $Gamma(x+1)$. But still your equation can't be solved in "closed form". Numerical methods might be tried.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
      $endgroup$
      – haster8558
      Dec 19 '16 at 16:19










    • $begingroup$
      Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
      $endgroup$
      – Robert Israel
      Dec 19 '16 at 20:36














    0












    0








    0





    $begingroup$

    $x!$ is not defined for arbitrary real positive $x$, just for nonnegative integers. You could replace it by $Gamma(x+1)$. But still your equation can't be solved in "closed form". Numerical methods might be tried.






    share|cite|improve this answer









    $endgroup$



    $x!$ is not defined for arbitrary real positive $x$, just for nonnegative integers. You could replace it by $Gamma(x+1)$. But still your equation can't be solved in "closed form". Numerical methods might be tried.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 '16 at 16:13









    Robert IsraelRobert Israel

    320k23209459




    320k23209459












    • $begingroup$
      Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
      $endgroup$
      – haster8558
      Dec 19 '16 at 16:19










    • $begingroup$
      Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
      $endgroup$
      – Robert Israel
      Dec 19 '16 at 20:36


















    • $begingroup$
      Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
      $endgroup$
      – haster8558
      Dec 19 '16 at 16:19










    • $begingroup$
      Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
      $endgroup$
      – Robert Israel
      Dec 19 '16 at 20:36
















    $begingroup$
    Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:19




    $begingroup$
    Ok, so it isn't possible to get a solution in closed form but it has some solutions. Any idea how to find a numerical solution ? Any algorithm simple to implement ?
    $endgroup$
    – haster8558
    Dec 19 '16 at 16:19












    $begingroup$
    Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 20:36




    $begingroup$
    Newton's method should work. It might be better to work with the logarithms of both sides of the equation, to avoid numerical overflow.
    $endgroup$
    – Robert Israel
    Dec 19 '16 at 20:36











    0












    $begingroup$

    Here's a possible solution - not sure if it will be of particularly good numerical stability. Take logs:



    $$ ln{(e^{-A} A^x)} = ln{(x! K)} implies - A + x ln{A} = ln{x!} + ln{K}.$$



    You might be able to solve this numerically if you use a an accurate enough approximation for $ln x!,$ using the Stirling series:



    $$ ln x! approx x ln x - x + frac{1}{2} ln{2pi x} + frac{1}{12 x} - frac{1}{360 x^3} + dots$$



    so a solution will be close to the solution of, say,



    $$ x ln x - (1 + ln A)x+frac{1}{2} ln{2pi x} + (A + ln K) = 0.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here's a possible solution - not sure if it will be of particularly good numerical stability. Take logs:



      $$ ln{(e^{-A} A^x)} = ln{(x! K)} implies - A + x ln{A} = ln{x!} + ln{K}.$$



      You might be able to solve this numerically if you use a an accurate enough approximation for $ln x!,$ using the Stirling series:



      $$ ln x! approx x ln x - x + frac{1}{2} ln{2pi x} + frac{1}{12 x} - frac{1}{360 x^3} + dots$$



      so a solution will be close to the solution of, say,



      $$ x ln x - (1 + ln A)x+frac{1}{2} ln{2pi x} + (A + ln K) = 0.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here's a possible solution - not sure if it will be of particularly good numerical stability. Take logs:



        $$ ln{(e^{-A} A^x)} = ln{(x! K)} implies - A + x ln{A} = ln{x!} + ln{K}.$$



        You might be able to solve this numerically if you use a an accurate enough approximation for $ln x!,$ using the Stirling series:



        $$ ln x! approx x ln x - x + frac{1}{2} ln{2pi x} + frac{1}{12 x} - frac{1}{360 x^3} + dots$$



        so a solution will be close to the solution of, say,



        $$ x ln x - (1 + ln A)x+frac{1}{2} ln{2pi x} + (A + ln K) = 0.$$






        share|cite|improve this answer









        $endgroup$



        Here's a possible solution - not sure if it will be of particularly good numerical stability. Take logs:



        $$ ln{(e^{-A} A^x)} = ln{(x! K)} implies - A + x ln{A} = ln{x!} + ln{K}.$$



        You might be able to solve this numerically if you use a an accurate enough approximation for $ln x!,$ using the Stirling series:



        $$ ln x! approx x ln x - x + frac{1}{2} ln{2pi x} + frac{1}{12 x} - frac{1}{360 x^3} + dots$$



        so a solution will be close to the solution of, say,



        $$ x ln x - (1 + ln A)x+frac{1}{2} ln{2pi x} + (A + ln K) = 0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '16 at 17:05









        NitinNitin

        2,4441024




        2,4441024






























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