Calculate the Mean mean of two numbers
$begingroup$
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
edited 40 mins ago
Jonathan Allan
53.5k535172
asked 8 hours ago
someonesomeone
557418
$endgroup$
add a comment |
$begingroup$
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
edited 40 mins ago
Jonathan Allan
53.5k535172
asked 8 hours ago
someonesomeone
557418
$endgroup$
$begingroup$
Sandbox link
$endgroup$
– someone
8 hours ago
$begingroup$
Somewhat related: Calculate the Arithmetic–Geometric Mean
$endgroup$
– user202729
2 hours ago
add a comment |
$begingroup$
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
edited 40 mins ago
Jonathan Allan
53.5k535172
asked 8 hours ago
someonesomeone
557418
$endgroup$
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
edited 40 mins ago
Jonathan Allan
53.5k535172
asked 8 hours ago
someonesomeone
557418
edited 40 mins ago
Jonathan Allan
53.5k535172
asked 8 hours ago
someonesomeone
557418
edited 40 mins ago
Jonathan Allan
53.5k535172
edited 40 mins ago
Jonathan Allan
53.5k535172
edited 40 mins ago
Jonathan Allan
53.5k535172
53.5k535172
asked 8 hours ago
someonesomeone
557418
asked 8 hours ago
someonesomeone
557418
asked 8 hours ago
someonesomeone
557418
557418
$begingroup$
Sandbox link
$endgroup$
– someone
8 hours ago
$begingroup$
Somewhat related: Calculate the Arithmetic–Geometric Mean
$endgroup$
– user202729
2 hours ago
add a comment |
$begingroup$
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
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Wolfram Language (Mathematica), 77 bytes
#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&
Try it online!
answered 7 hours ago
J42161217J42161217
13.6k21252
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
NeilNeil
82.2k745178
$endgroup$
add a comment |
$begingroup$
Jelly, 25 bytes
Æm
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ
Try it online!
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
$endgroup$
$begingroup$
I am fairly bad at Jelly, but could something similar toP*İLwork for the geometric mean?
$endgroup$
– someone
1 hour ago
$begingroup$
@someone it would need to beP*Lİ$so wouldn’t save bytes. It would mean I could bringÆmback 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
add a comment |
$begingroup$
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.
answered 2 hours ago
user202729user202729
14.1k12553
$endgroup$
add a comment |
$begingroup$
Jelly, 24 bytes
;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ
Try it online!
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Wolfram Language (Mathematica), 77 bytes
#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&
Try it online!
answered 7 hours ago
J42161217J42161217
13.6k21252
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 77 bytes
#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&
Try it online!
answered 7 hours ago
J42161217J42161217
13.6k21252
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 77 bytes
#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&
Try it online!
answered 7 hours ago
J42161217J42161217
13.6k21252
$endgroup$
Wolfram Language (Mathematica), 77 bytes
#&@@N@Nest[{Mean@#,GeometricMean@#,HarmonicMean@#,RootMeanSquare@#}&,{##},9]&
Try it online!
answered 7 hours ago
J42161217J42161217
13.6k21252
answered 7 hours ago
J42161217J42161217
13.6k21252
answered 7 hours ago
J42161217J42161217
13.6k21252
answered 7 hours ago
J42161217J42161217
13.6k21252
13.6k21252
add a comment |
add a comment |
$begingroup$
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.
answered 7 hours ago
NeilNeil
82.2k745178
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
NeilNeil
82.2k745178
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
NeilNeil
82.2k745178
$endgroup$
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.
answered 7 hours ago
NeilNeil
82.2k745178
answered 7 hours ago
NeilNeil
82.2k745178
answered 7 hours ago
NeilNeil
82.2k745178
answered 7 hours ago
NeilNeil
82.2k745178
82.2k745178
add a comment |
add a comment |
$begingroup$
Jelly, 25 bytes
Æm
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ
Try it online!
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
$endgroup$
$begingroup$
I am fairly bad at Jelly, but could something similar toP*İLwork for the geometric mean?
$endgroup$
– someone
1 hour ago
$begingroup$
@someone it would need to beP*Lİ$so wouldn’t save bytes. It would mean I could bringÆmback 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
add a comment |
$begingroup$
Jelly, 25 bytes
Æm
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ
Try it online!
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
$endgroup$
$begingroup$
I am fairly bad at Jelly, but could something similar toP*İLwork for the geometric mean?
$endgroup$
– someone
1 hour ago
$begingroup$
@someone it would need to beP*Lİ$so wouldn’t save bytes. It would mean I could bringÆmback 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
add a comment |
$begingroup$
Jelly, 25 bytes
Æm
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ
Try it online!
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
$endgroup$
Jelly, 25 bytes
Æm
Ç;ÆlÇÆeƊ;²Ç½Ɗ;İÇİƊµÐLḢ
Try it online!
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
edited 1 hour ago
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
answered 2 hours ago
Nick KennedyNick Kennedy
1,16649
1,16649
$begingroup$
I am fairly bad at Jelly, but could something similar toP*İLwork for the geometric mean?
$endgroup$
– someone
1 hour ago
$begingroup$
@someone it would need to beP*Lİ$so wouldn’t save bytes. It would mean I could bringÆmback 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
add a comment |
$begingroup$
I am fairly bad at Jelly, but could something similar toP*İLwork for the geometric mean?
$endgroup$
– someone
1 hour ago
$begingroup$
@someone it would need to beP*Lİ$so wouldn’t save bytes. It would mean I could bringÆmback 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
add a comment |
$begingroup$
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.
answered 2 hours ago
user202729user202729
14.1k12553
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
user202729user202729
14.1k12553
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
user202729user202729
14.1k12553
$endgroup$
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.
answered 2 hours ago
user202729user202729
14.1k12553
edited 1 hour ago
answered 2 hours ago
user202729user202729
14.1k12553
answered 2 hours ago
user202729user202729
14.1k12553
answered 2 hours ago
user202729user202729
14.1k12553
14.1k12553
add a comment |
add a comment |
$begingroup$
Jelly, 24 bytes
;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ
Try it online!
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
$endgroup$
add a comment |
$begingroup$
Jelly, 24 bytes
;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ
Try it online!
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
$endgroup$
add a comment |
$begingroup$
Jelly, 24 bytes
;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ
Try it online!
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
$endgroup$
Jelly, 24 bytes
;µ*Ɱ-;ؽ¤©Æm*İɼ;P½½ƊµÐLḢ
Try it online!
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
answered 49 mins ago
Erik the OutgolferErik the Outgolfer
32.9k429106
32.9k429106
add a comment |
add a comment |
If this is an answer to a challenge…
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Sandbox link
$endgroup$
– someone
8 hours ago
$begingroup$
Somewhat related: Calculate the Arithmetic–Geometric Mean
$endgroup$
– user202729
2 hours ago