Can this power tower function be optimized to perform faster?

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Take the following function defined in Mathematica:



Itr[x_, p_, n_] := x^Nest[Power[p, #] &, 1, n];


Evaluating this function even for small values of p results in very slow and processor-intensive evaluation. Is there an equivalent expression that will evaluate more efficiently and quickly?



For example, Itr[3, 1/2, 25] // N takes nearly 3 minutes to evaluate on a 2016 MacBook Pro.






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    up vote
    2
    down vote

    favorite












    Take the following function defined in Mathematica:



    Itr[x_, p_, n_] := x^Nest[Power[p, #] &, 1, n];


    Evaluating this function even for small values of p results in very slow and processor-intensive evaluation. Is there an equivalent expression that will evaluate more efficiently and quickly?



    For example, Itr[3, 1/2, 25] // N takes nearly 3 minutes to evaluate on a 2016 MacBook Pro.






    performance-tuning evaluation







    share|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Take the following function defined in Mathematica:



      Itr[x_, p_, n_] := x^Nest[Power[p, #] &, 1, n];


      Evaluating this function even for small values of p results in very slow and processor-intensive evaluation. Is there an equivalent expression that will evaluate more efficiently and quickly?



      For example, Itr[3, 1/2, 25] // N takes nearly 3 minutes to evaluate on a 2016 MacBook Pro.






      performance-tuning evaluation







      share|improve this question













      Take the following function defined in Mathematica:



      Itr[x_, p_, n_] := x^Nest[Power[p, #] &, 1, n];


      Evaluating this function even for small values of p results in very slow and processor-intensive evaluation. Is there an equivalent expression that will evaluate more efficiently and quickly?



      For example, Itr[3, 1/2, 25] // N takes nearly 3 minutes to evaluate on a 2016 MacBook Pro.






      performance-tuning evaluation




      performance-tuning evaluation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 hours ago









      LBushkin

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      21629




















          1 Answer
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          Yes, use floating point numbers right from the start: 



           Itr[3., 0.5, 50000000]


          takes about one second on my machine.



          You can also perform this in higher precision, but that will take longer; for example the following computation with 100-digit precision needs also about one second on my laptop:



           Itr[3.`100, 0.5`100, 500000];


          Assuming that the nested powers converge quickly, one can also use FixedPoint in order to iterate as long as it is needed:



           f[x_, p_] := x^FixedPoint[Power[p, #] &, 1];





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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            Yes, use floating point numbers right from the start: 



             Itr[3., 0.5, 50000000]


            takes about one second on my machine.



            You can also perform this in higher precision, but that will take longer; for example the following computation with 100-digit precision needs also about one second on my laptop:



             Itr[3.`100, 0.5`100, 500000];


            Assuming that the nested powers converge quickly, one can also use FixedPoint in order to iterate as long as it is needed:



             f[x_, p_] := x^FixedPoint[Power[p, #] &, 1];





            share|improve this answer


























              up vote
              4
              down vote



              accepted










              Yes, use floating point numbers right from the start: 



               Itr[3., 0.5, 50000000]


              takes about one second on my machine.



              You can also perform this in higher precision, but that will take longer; for example the following computation with 100-digit precision needs also about one second on my laptop:



               Itr[3.`100, 0.5`100, 500000];


              Assuming that the nested powers converge quickly, one can also use FixedPoint in order to iterate as long as it is needed:



               f[x_, p_] := x^FixedPoint[Power[p, #] &, 1];





              share|improve this answer
























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                Yes, use floating point numbers right from the start: 



                 Itr[3., 0.5, 50000000]


                takes about one second on my machine.



                You can also perform this in higher precision, but that will take longer; for example the following computation with 100-digit precision needs also about one second on my laptop:



                 Itr[3.`100, 0.5`100, 500000];


                Assuming that the nested powers converge quickly, one can also use FixedPoint in order to iterate as long as it is needed:



                 f[x_, p_] := x^FixedPoint[Power[p, #] &, 1];





                share|improve this answer














                Yes, use floating point numbers right from the start: 



                 Itr[3., 0.5, 50000000]


                takes about one second on my machine.



                You can also perform this in higher precision, but that will take longer; for example the following computation with 100-digit precision needs also about one second on my laptop:



                 Itr[3.`100, 0.5`100, 500000];


                Assuming that the nested powers converge quickly, one can also use FixedPoint in order to iterate as long as it is needed:



                 f[x_, p_] := x^FixedPoint[Power[p, #] &, 1];






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                Henrik Schumacher

                39k254114




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