Calculate the Mean mean of two numbers












3












$begingroup$


disclaimer: the Mean mean is made up by me



Define the arithmetic mean of $n$ numbers as
$$M_1(x_1,...,x_n)=frac{x_1+x_2+...+x_n}{n}$$
Define the geometric mean of $n$ numbers as
$$M_0(x_1,...,x_n)=root{n}of{x_1x_2...x_n}$$
Define the harmonic mean of $n$ numbers as
$$M_{-1}(x_1,...,x_n)=frac{n}{frac{1}{x_2} + frac{1}{x_2} + ... + frac{1}{x_n}}$$
Define the quadratic mean of $n$ numbers as
$$M_2(x_1,...,x_n)=rootof{frac{x_1^2+x_2^2+...+x_n^2}{n}}$$
The Mean mean ($M_M$) is defined as follows:
Define four sequences ($a_k, b_k, c_k, d_k$) as $$a_0=M_1(x_1,...,x_n),\b_0=M_0(x_1,...,x_n),\c_0=M_{-1}(x_1,...,x_n),\d_0=M_2(x_1,...,x_n),\
a_{k+1}=M_1(a_k,b_k,c_k,d_k),\b_{k+1}=M_0(a_k,b_k,c_k,d_k),\c_{k+1}=M_{-1}(a_k,b_k,c_k,d_k),\d_{k+1}=M_2(a_k,b_k,c_k,d_k)$$
All four sequences converge to the same number, $M_M(x_1,x_2,...,x_n)$.



Example



The Mean mean of 1 and 2 is calculated as follows: start with $$a_0 = (1+2)/2 = 1.5, b_0 = rootof{1 * 2} = rootof2 approx 1.4142,\ c_0 = frac2{frac1{1}+frac1{2}} = frac4{3} approx 1.3333, d_0 = rootof{frac{1^2+2^2}2} = rootof{frac5{2}} approx 1.5811.$$
Then
$$a_1 = frac{1.5+1.4142+1.3333+1.5811}4 approx 1.4571,\ b_1 = root^4of{1.5*1.4142*1.3333*1.5811} approx 1.4542,\ c_1 = frac4{frac1{1.5}+frac1{1.4142}+frac1{1.3333}+frac1{1.5811}} approx 1.4512,\ d_1 = rootof{frac{1.5^2+1.4142^2+1.3333^2+1.5811^2}4} approx 1.4601.$$
The further calculation of the sequences should be clear. It can be seen that they converge to the same number, approximately $1.45568889$.



Challenge



Given two positive real numbers, $a$ and $b$, calculate their Mean mean $M_M(a,b)$.



Test cases



1 1 => 1

1 2 => 1.45568889

100 200 => 145.568889

2.71 3.14 => 2.92103713

0.57 1.78 => 1.0848205

1.61 2.41 => 1.98965438

0.01 100 => 6.7483058


This is code-golf, so the shortest code wins!






code-golf math







share|improve this question











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    Somewhat related: Calculate the Arithmetic–Geometric Mean
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    – user202729
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3












$begingroup$


disclaimer: the Mean mean is made up by me



Define the arithmetic mean of $n$ numbers as
$$M_1(x_1,...,x_n)=frac{x_1+x_2+...+x_n}{n}$$
Define the geometric mean of $n$ numbers as
$$M_0(x_1,...,x_n)=root{n}of{x_1x_2...x_n}$$
Define the harmonic mean of $n$ numbers as
$$M_{-1}(x_1,...,x_n)=frac{n}{frac{1}{x_2} + frac{1}{x_2} + ... + frac{1}{x_n}}$$
Define the quadratic mean of $n$ numbers as
$$M_2(x_1,...,x_n)=rootof{frac{x_1^2+x_2^2+...+x_n^2}{n}}$$
The Mean mean ($M_M$) is defined as follows:
Define four sequences ($a_k, b_k, c_k, d_k$) as $$a_0=M_1(x_1,...,x_n),\b_0=M_0(x_1,...,x_n),\c_0=M_{-1}(x_1,...,x_n),\d_0=M_2(x_1,...,x_n),\
a_{k+1}=M_1(a_k,b_k,c_k,d_k),\b_{k+1}=M_0(a_k,b_k,c_k,d_k),\c_{k+1}=M_{-1}(a_k,b_k,c_k,d_k),\d_{k+1}=M_2(a_k,b_k,c_k,d_k)$$
All four sequences converge to the same number, $M_M(x_1,x_2,...,x_n)$.



Example



The Mean mean of 1 and 2 is calculated as follows: start with $$a_0 = (1+2)/2 = 1.5, b_0 = rootof{1 * 2} = rootof2 approx 1.4142,\ c_0 = frac2{frac1{1}+frac1{2}} = frac4{3} approx 1.3333, d_0 = rootof{frac{1^2+2^2}2} = rootof{frac5{2}} approx 1.5811.$$
Then
$$a_1 = frac{1.5+1.4142+1.3333+1.5811}4 approx 1.4571,\ b_1 = root^4of{1.5*1.4142*1.3333*1.5811} approx 1.4542,\ c_1 = frac4{frac1{1.5}+frac1{1.4142}+frac1{1.3333}+frac1{1.5811}} approx 1.4512,\ d_1 = rootof{frac{1.5^2+1.4142^2+1.3333^2+1.5811^2}4} approx 1.4601.$$
The further calculation of the sequences should be clear. It can be seen that they converge to the same number, approximately $1.45568889$.



Challenge



Given two positive real numbers, $a$ and $b$, calculate their Mean mean $M_M(a,b)$.



Test cases



1 1 => 1

1 2 => 1.45568889

100 200 => 145.568889

2.71 3.14 => 2.92103713

0.57 1.78 => 1.0848205

1.61 2.41 => 1.98965438

0.01 100 => 6.7483058


This is code-golf, so the shortest code wins!






code-golf math







share|improve this question











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  • $begingroup$
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    – someone
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    Somewhat related: Calculate the Arithmetic–Geometric Mean
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    – user202729
    2 hours ago














3












3








3





$begingroup$


disclaimer: the Mean mean is made up by me



Define the arithmetic mean of $n$ numbers as
$$M_1(x_1,...,x_n)=frac{x_1+x_2+...+x_n}{n}$$
Define the geometric mean of $n$ numbers as
$$M_0(x_1,...,x_n)=root{n}of{x_1x_2...x_n}$$
Define the harmonic mean of $n$ numbers as
$$M_{-1}(x_1,...,x_n)=frac{n}{frac{1}{x_2} + frac{1}{x_2} + ... + frac{1}{x_n}}$$
Define the quadratic mean of $n$ numbers as
$$M_2(x_1,...,x_n)=rootof{frac{x_1^2+x_2^2+...+x_n^2}{n}}$$
The Mean mean ($M_M$) is defined as follows:
Define four sequences ($a_k, b_k, c_k, d_k$) as $$a_0=M_1(x_1,...,x_n),\b_0=M_0(x_1,...,x_n),\c_0=M_{-1}(x_1,...,x_n),\d_0=M_2(x_1,...,x_n),\
a_{k+1}=M_1(a_k,b_k,c_k,d_k),\b_{k+1}=M_0(a_k,b_k,c_k,d_k),\c_{k+1}=M_{-1}(a_k,b_k,c_k,d_k),\d_{k+1}=M_2(a_k,b_k,c_k,d_k)$$
All four sequences converge to the same number, $M_M(x_1,x_2,...,x_n)$.



Example



The Mean mean of 1 and 2 is calculated as follows: start with $$a_0 = (1+2)/2 = 1.5, b_0 = rootof{1 * 2} = rootof2 approx 1.4142,\ c_0 = frac2{frac1{1}+frac1{2}} = frac4{3} approx 1.3333, d_0 = rootof{frac{1^2+2^2}2} = rootof{frac5{2}} approx 1.5811.$$
Then
$$a_1 = frac{1.5+1.4142+1.3333+1.5811}4 approx 1.4571,\ b_1 = root^4of{1.5*1.4142*1.3333*1.5811} approx 1.4542,\ c_1 = frac4{frac1{1.5}+frac1{1.4142}+frac1{1.3333}+frac1{1.5811}} approx 1.4512,\ d_1 = rootof{frac{1.5^2+1.4142^2+1.3333^2+1.5811^2}4} approx 1.4601.$$
The further calculation of the sequences should be clear. It can be seen that they converge to the same number, approximately $1.45568889$.



Challenge



Given two positive real numbers, $a$ and $b$, calculate their Mean mean $M_M(a,b)$.



Test cases



1 1 => 1

1 2 => 1.45568889

100 200 => 145.568889

2.71 3.14 => 2.92103713

0.57 1.78 => 1.0848205

1.61 2.41 => 1.98965438

0.01 100 => 6.7483058


This is code-golf, so the shortest code wins!






code-golf math







share|improve this question











$endgroup$




disclaimer: the Mean mean is made up by me



Define the arithmetic mean of $n$ numbers as
$$M_1(x_1,...,x_n)=frac{x_1+x_2+...+x_n}{n}$$
Define the geometric mean of $n$ numbers as
$$M_0(x_1,...,x_n)=root{n}of{x_1x_2...x_n}$$
Define the harmonic mean of $n$ numbers as
$$M_{-1}(x_1,...,x_n)=frac{n}{frac{1}{x_2} + frac{1}{x_2} + ... + frac{1}{x_n}}$$
Define the quadratic mean of $n$ numbers as
$$M_2(x_1,...,x_n)=rootof{frac{x_1^2+x_2^2+...+x_n^2}{n}}$$
The Mean mean ($M_M$) is defined as follows:
Define four sequences ($a_k, b_k, c_k, d_k$) as $$a_0=M_1(x_1,...,x_n),\b_0=M_0(x_1,...,x_n),\c_0=M_{-1}(x_1,...,x_n),\d_0=M_2(x_1,...,x_n),\
a_{k+1}=M_1(a_k,b_k,c_k,d_k),\b_{k+1}=M_0(a_k,b_k,c_k,d_k),\c_{k+1}=M_{-1}(a_k,b_k,c_k,d_k),\d_{k+1}=M_2(a_k,b_k,c_k,d_k)$$
All four sequences converge to the same number, $M_M(x_1,x_2,...,x_n)$.



Example



The Mean mean of 1 and 2 is calculated as follows: start with $$a_0 = (1+2)/2 = 1.5, b_0 = rootof{1 * 2} = rootof2 approx 1.4142,\ c_0 = frac2{frac1{1}+frac1{2}} = frac4{3} approx 1.3333, d_0 = rootof{frac{1^2+2^2}2} = rootof{frac5{2}} approx 1.5811.$$
Then
$$a_1 = frac{1.5+1.4142+1.3333+1.5811}4 approx 1.4571,\ b_1 = root^4of{1.5*1.4142*1.3333*1.5811} approx 1.4542,\ c_1 = frac4{frac1{1.5}+frac1{1.4142}+frac1{1.3333}+frac1{1.5811}} approx 1.4512,\ d_1 = rootof{frac{1.5^2+1.4142^2+1.3333^2+1.5811^2}4} approx 1.4601.$$
The further calculation of the sequences should be clear. It can be seen that they converge to the same number, approximately $1.45568889$.



Challenge



Given two positive real numbers, $a$ and $b$, calculate their Mean mean $M_M(a,b)$.



Test cases



1 1 => 1

1 2 => 1.45568889

100 200 => 145.568889

2.71 3.14 => 2.92103713

0.57 1.78 => 1.0848205

1.61 2.41 => 1.98965438

0.01 100 => 6.7483058


This is code-golf, so the shortest code wins!






code-golf math




code-golf math






share|improve this question















share|improve this question













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edited 40 mins ago









Jonathan Allan

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asked 8 hours ago









someonesomeone

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  • $begingroup$
    Sandbox link
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    – someone
    8 hours ago










  • $begingroup$
    Somewhat related: Calculate the Arithmetic–Geometric Mean
    $endgroup$
    – user202729
    2 hours ago


















  • $begingroup$
    Sandbox link
    $endgroup$
    – someone
    8 hours ago










  • $begingroup$
    Somewhat related: Calculate the Arithmetic–Geometric Mean
    $endgroup$
    – user202729
    2 hours ago
















$begingroup$
Sandbox link
$endgroup$
– someone
8 hours ago




$begingroup$
Sandbox link
$endgroup$
– someone
8 hours ago












$begingroup$
Somewhat related: Calculate the Arithmetic–Geometric Mean
$endgroup$
– user202729
2 hours ago




$begingroup$
Somewhat related: Calculate the Arithmetic–Geometric Mean
$endgroup$
– user202729
2 hours ago










5 Answers
5






active

oldest

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1












$begingroup$


Wolfram Language (Mathematica), 77 bytes



#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&


Try it online!






share|improve this answer









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    1












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    Charcoal, 40 bytes



    Ｗ‹⌊θ⌈θ≔⟦∕ΣθＬθＸΠθ∕¹Ｌθ∕ＬθΣ∕¹θ₂∕ΣＸθ²Ｌθ⟧θＩ⊟θ


    Try it online! Link is to verbose version of code. Takes input as an array of numbers. Explanation:



    Ｗ‹⌊θ⌈θ


    Repeat while the array contains different values...



    ≔⟦....⟧θ


    ... replace the array with a list of values:



    ∕ΣθＬθ


    ... the mean...



    ＸΠθ∕¹Ｌθ


    ... the geometric mean...



    ∕ＬθΣ∕¹θ


    ... the harmonic mean...



    ₂∕ΣＸθ²Ｌθ


    ... and the root mean square.



    Ｉ⊟θ


    Cast an element of the array to string and implicitly print it.






    share|improve this answer









    $endgroup$





















      0












      $begingroup$


      Jelly, 25 bytes



      Æm
      
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ


      Try it online!






      share|improve this answer











      $endgroup$













      • $begingroup$
        I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
        $endgroup$
        – someone
        1 hour ago












      • $begingroup$
        @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
        $endgroup$
        – Nick Kennedy
        1 hour ago



















      0












      $begingroup$


      J, 34 bytes



      (31 as an expression without the assignment to variable f)



      f=:1{(^.z,%z,*:z,[z=:(+/%#)&.:)^:_


      Try it online!



      For functions a and b, a &.: b ("a under b" (related challenge)) is equivalent to (b inv) a b -- apply b, then a, then inverse of b. In this case, geometric/harmonic/quadratic mean is the arithmetic mean "under" logarithm, inversion, and square respectively.






      share|improve this answer











      $endgroup$





















        0












        $begingroup$


        Jelly, 24 bytes



        ;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ


        Try it online!






        share|improve this answer









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          5 Answers
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          5 Answers
          5






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$


          Wolfram Language (Mathematica), 77 bytes



          #&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&


          Try it online!






          share|improve this answer









          $endgroup$


















            1












            $begingroup$


            Wolfram Language (Mathematica), 77 bytes



            #&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&


            Try it online!






            share|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$


              Wolfram Language (Mathematica), 77 bytes



              #&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&


              Try it online!






              share|improve this answer









              $endgroup$




              Wolfram Language (Mathematica), 77 bytes



              #&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&


              Try it online!







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 7 hours ago









              J42161217J42161217

              13.6k21252




              13.6k21252























                  1












                  $begingroup$


                  Charcoal, 40 bytes



                  Ｗ‹⌊θ⌈θ≔⟦∕ΣθＬθＸΠθ∕¹Ｌθ∕ＬθΣ∕¹θ₂∕ΣＸθ²Ｌθ⟧θＩ⊟θ


                  Try it online! Link is to verbose version of code. Takes input as an array of numbers. Explanation:



                  Ｗ‹⌊θ⌈θ


                  Repeat while the array contains different values...



                  ≔⟦....⟧θ


                  ... replace the array with a list of values:



                  ∕ΣθＬθ


                  ... the mean...



                  ＸΠθ∕¹Ｌθ


                  ... the geometric mean...



                  ∕ＬθΣ∕¹θ


                  ... the harmonic mean...



                  ₂∕ΣＸθ²Ｌθ


                  ... and the root mean square.



                  Ｉ⊟θ


                  Cast an element of the array to string and implicitly print it.






                  share|improve this answer









                  $endgroup$


















                    1












                    $begingroup$


                    Charcoal, 40 bytes



                    Ｗ‹⌊θ⌈θ≔⟦∕ΣθＬθＸΠθ∕¹Ｌθ∕ＬθΣ∕¹θ₂∕ΣＸθ²Ｌθ⟧θＩ⊟θ


                    Try it online! Link is to verbose version of code. Takes input as an array of numbers. Explanation:



                    Ｗ‹⌊θ⌈θ


                    Repeat while the array contains different values...



                    ≔⟦....⟧θ


                    ... replace the array with a list of values:



                    ∕ΣθＬθ


                    ... the mean...



                    ＸΠθ∕¹Ｌθ


                    ... the geometric mean...



                    ∕ＬθΣ∕¹θ


                    ... the harmonic mean...



                    ₂∕ΣＸθ²Ｌθ


                    ... and the root mean square.



                    Ｉ⊟θ


                    Cast an element of the array to string and implicitly print it.






                    share|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$


                      Charcoal, 40 bytes



                      Ｗ‹⌊θ⌈θ≔⟦∕ΣθＬθＸΠθ∕¹Ｌθ∕ＬθΣ∕¹θ₂∕ΣＸθ²Ｌθ⟧θＩ⊟θ


                      Try it online! Link is to verbose version of code. Takes input as an array of numbers. Explanation:



                      Ｗ‹⌊θ⌈θ


                      Repeat while the array contains different values...



                      ≔⟦....⟧θ


                      ... replace the array with a list of values:



                      ∕ΣθＬθ


                      ... the mean...



                      ＸΠθ∕¹Ｌθ


                      ... the geometric mean...



                      ∕ＬθΣ∕¹θ


                      ... the harmonic mean...



                      ₂∕ΣＸθ²Ｌθ


                      ... and the root mean square.



                      Ｉ⊟θ


                      Cast an element of the array to string and implicitly print it.






                      share|improve this answer









                      $endgroup$




                      Charcoal, 40 bytes



                      Ｗ‹⌊θ⌈θ≔⟦∕ΣθＬθＸΠθ∕¹Ｌθ∕ＬθΣ∕¹θ₂∕ΣＸθ²Ｌθ⟧θＩ⊟θ


                      Try it online! Link is to verbose version of code. Takes input as an array of numbers. Explanation:



                      Ｗ‹⌊θ⌈θ


                      Repeat while the array contains different values...



                      ≔⟦....⟧θ


                      ... replace the array with a list of values:



                      ∕ΣθＬθ


                      ... the mean...



                      ＸΠθ∕¹Ｌθ


                      ... the geometric mean...



                      ∕ＬθΣ∕¹θ


                      ... the harmonic mean...



                      ₂∕ΣＸθ²Ｌθ


                      ... and the root mean square.



                      Ｉ⊟θ


                      Cast an element of the array to string and implicitly print it.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 7 hours ago









                      NeilNeil

                      82.2k745178




                      82.2k745178























                          0












                          $begingroup$


                          Jelly, 25 bytes



                          Æm
                          
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ


                          Try it online!






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                            $endgroup$
                            – someone
                            1 hour ago












                          • $begingroup$
                            @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                            $endgroup$
                            – Nick Kennedy
                            1 hour ago
















                          0












                          $begingroup$


                          Jelly, 25 bytes



                          Æm
                          
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ


                          Try it online!






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                            $endgroup$
                            – someone
                            1 hour ago












                          • $begingroup$
                            @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                            $endgroup$
                            – Nick Kennedy
                            1 hour ago














                          0












                          0








                          0





                          $begingroup$


                          Jelly, 25 bytes



                          Æm
                          
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ


                          Try it online!






                          share|improve this answer











                          $endgroup$




                          Jelly, 25 bytes



                          Æm
                          
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ


                          Try it online!







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 1 hour ago

























                          answered 2 hours ago









                          Nick KennedyNick Kennedy

                          1,16649




                          1,16649












                          • $begingroup$
                            I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                            $endgroup$
                            – someone
                            1 hour ago












                          • $begingroup$
                            @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                            $endgroup$
                            – Nick Kennedy
                            1 hour ago


















                          • $begingroup$
                            I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                            $endgroup$
                            – someone
                            1 hour ago












                          • $begingroup$
                            @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                            $endgroup$
                            – Nick Kennedy
                            1 hour ago
















                          $begingroup$
                          I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                          $endgroup$
                          – someone
                          1 hour ago






                          $begingroup$
                          I am fairly bad at Jelly, but could something similar to P*İL work for the geometric mean?
                          $endgroup$
                          – someone
                          1 hour ago














                          $begingroup$
                          @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                          $endgroup$
                          – Nick Kennedy
                          1 hour ago




                          $begingroup$
                          @someone it would need to be P*Lİ$ so wouldn’t save bytes. It would mean I could bring Æm back down a line without costing bytes, but I quite like the fact each one currently has an arithmetic mean at its core.
                          $endgroup$
                          – Nick Kennedy
                          1 hour ago











                          0












                          $begingroup$


                          J, 34 bytes



                          (31 as an expression without the assignment to variable f)



                          f=:1{(^.z,%z,*:z,[z=:(+/%#)&.:)^:_


                          Try it online!



                          For functions a and b, a &.: b ("a under b" (related challenge)) is equivalent to (b inv) a b -- apply b, then a, then inverse of b. In this case, geometric/harmonic/quadratic mean is the arithmetic mean "under" logarithm, inversion, and square respectively.






                          share|improve this answer











                          $endgroup$


















                            0












                            $begingroup$


                            J, 34 bytes



                            (31 as an expression without the assignment to variable f)



                            f=:1{(^.z,%z,*:z,[z=:(+/%#)&.:)^:_


                            Try it online!



                            For functions a and b, a &.: b ("a under b" (related challenge)) is equivalent to (b inv) a b -- apply b, then a, then inverse of b. In this case, geometric/harmonic/quadratic mean is the arithmetic mean "under" logarithm, inversion, and square respectively.






                            share|improve this answer











                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$


                              J, 34 bytes



                              (31 as an expression without the assignment to variable f)



                              f=:1{(^.z,%z,*:z,[z=:(+/%#)&.:)^:_


                              Try it online!



                              For functions a and b, a &.: b ("a under b" (related challenge)) is equivalent to (b inv) a b -- apply b, then a, then inverse of b. In this case, geometric/harmonic/quadratic mean is the arithmetic mean "under" logarithm, inversion, and square respectively.






                              share|improve this answer











                              $endgroup$




                              J, 34 bytes



                              (31 as an expression without the assignment to variable f)



                              f=:1{(^.z,%z,*:z,[z=:(+/%#)&.:)^:_


                              Try it online!



                              For functions a and b, a &.: b ("a under b" (related challenge)) is equivalent to (b inv) a b -- apply b, then a, then inverse of b. In this case, geometric/harmonic/quadratic mean is the arithmetic mean "under" logarithm, inversion, and square respectively.







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 1 hour ago

























                              answered 2 hours ago









                              user202729user202729

                              14.1k12553




                              14.1k12553























                                  0












                                  $begingroup$


                                  Jelly, 24 bytes



                                  ;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ


                                  Try it online!






                                  share|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$


                                    Jelly, 24 bytes



                                    ;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ


                                    Try it online!






                                    share|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$


                                      Jelly, 24 bytes



                                      ;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ


                                      Try it online!






                                      share|improve this answer









                                      $endgroup$




                                      Jelly, 24 bytes



                                      ;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ


                                      Try it online!







                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 49 mins ago









                                      Erik the OutgolferErik the Outgolfer

                                      32.9k429106




                                      32.9k429106






























                                          draft saved

                                          draft discarded




















































                                          If this is an answer to a challenge…




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