# AGM Series Hole 1: Calculate the Arithmetic–Geometric Mean

This question was inspired by this HNQ.

## About the series

This question is now part of a series about the AGM method. This first post in the series will be about actually calculating the AGM. You may treat this like any other code golf challenge, and answer it without worrying about the series at all. However, there is a leaderboard across all challenges.

## What is the Arithmetic–Geometric Mean

The Arithmetic–Geometric Mean of two numbers is defined as the number that repeatedly taking the arithmetic and geometric means converges to. Your task is to find this number after some n iterations.

## Clarifications

• You take three numbers, a, b, n in any reasonable format.
• For n iterations, take the arithmetic and geometric mean of a and b and set those to a and b.
• For two numbers a and b, the arithmetic mean is defined as (a + b) / 2.
• The geometric mean is defined as √(a * b).
• a and b should be approaching each other.
• Then, output both a and b.
• You don't have to worry about float imprecision and such.
• This is so shortest code in bytes wins!

## Test Cases

[0, [24, 6]] -> [24, 6]
[1, [24, 6]] -> [15.0, 12.0]
[2, [24, 6]] -> [13.5, 13.416407864998739]
[5, [24, 6]] -> [13.458171481725616, 13.458171481725616]
[10, [100, 50]] -> [72.83955155234534, 72.83955155234534]

The next one is 1/Gauss's Constant:
[10, [1, 1.41421356237]] -> [1.198140234734168, 1.1981402347341683]


## Leaderboard

Stolen from Martin's series.

The following snippet will generate a leaderboard across all challenges of the series.

To make sure that your answers show up, please start every answer with a headline, using the following Markdown template:

# Language Name, N bytes


where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes


/* Configuration */

var QUESTION_IDs = [66068]; // Obtain this from the url
// It will be like http://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!.FjwQBrX2KXuFkv6p2lChi_RjzM19";

/* App */

var answers = [], page = 1, currentQ = -1;

function answersUrl(index) {
return "http://api.stackexchange.com/2.2/questions/" +  QUESTION_IDs.join(";") + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function getAnswers() {
$.ajax({ url: answersUrl(page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); if (data.has_more) getAnswers(); else process(); } }); } getAnswers(); var SIZE_REG = /\d+(?=[^\d&]*(?:&lt;(?:s&gt;((?!&gt;).)*&lt;\/s&gt;|((?!&gt;).)+&gt;)[^\d&]*)*$)/;
var NUMBER_REG = /\d+/;
var LANGUAGE_REG = /^#*\s*([^\n,]+)(?=,)/;//

function shouldHaveHeading(a) {
var pass = false;
var lines = a.body_markdown.split("\n");
try {
pass |= /^#/.test(a.body_markdown);
pass |= ["-", "="]
.indexOf(lines[1][0]) > -1;
pass &= LANGUAGE_REG.test(a.body_markdown);
} catch (ex) {}
return pass;
}

function shouldHaveScore(a) {
var pass = false;
try {
pass |= SIZE_REG.test(a.body_markdown.split("\n")[0]);
} catch (ex) {}
if (!pass) console.log(a);
return pass;
}

function getAuthorName(a) {
return a.owner.display_name;
}

function getAuthorId(a) {
return a.owner.user_id;
}

function process() {
answers = answers.filter(shouldHaveScore)
.filter(shouldHaveHeading);
answers.sort(function (a, b) {
var aB = +(a.body_markdown.split("\n")[0].match(SIZE_REG) || [Infinity])[0],
bB = +(b.body_markdown.split("\n")[0].match(SIZE_REG) || [Infinity])[0];
return aB - bB
});

var users = {};
answers.forEach(function (a) {
var headline = a.body_markdown.split("\n")[0];
var question = QUESTION_IDs.indexOf(a.question_id);
var size = parseInt((headline.match(SIZE_REG)||[0])[0]);
var language = headline.match(LANGUAGE_REG)[1];
var user = getAuthorName(a);
var userId = getAuthorId(a);
if (!users[userId]) users[userId] = {name: user, nAnswer: 0, answers: []};
if (!users[userId].answers[question]) {
users[userId].answers[question] = {size: Infinity};
users[userId].nAnswer++;
}
if (users[userId].answers[question].size > size) {
users[userId].answers[question] = {size: size, link: a.share_link}
}
});

var sortedUsers = [];
for (var userId in users)
if (users.hasOwnProperty(userId)) {
var user = users[userId];
user.score = 0;
user.completedAll = true;
for (var i = 0; i < QUESTION_IDs.length; ++i) {
if (user.answers[i])
user.score += user.answers[i].size;
else
user.completedAll = false;
}
sortedUsers.push(user);
}

sortedUsers.sort(function (a, b) {
if (a.nAnswer > b.nAnswer) return -1;
if (b.nAnswer > a.nAnswer) return 1;
return a.score - b.score;
});

var place = 1;
for (var i = 0; i < sortedUsers.length; ++i) {
var user = sortedUsers[i];
var row = '<tr><td>'+ place++ +'.</td><td>'+user.name+'</td>';
for (var j = 0; j < QUESTION_IDs.length; ++j) {
var answer = user.answers[j];
if (answer)
row += '<td><a href="'+answer.link+'">'+answer.size+'</a></td>';
else
row += '<td class="missing"></td>';
}
row += '<td></td>';
if (user.completedAll)
row += '<td class="total">'+user.score+'</td>';
else
row += '<td class="total missing">'+user.score+'</td>';
row += '</tr>';
("#users").append(row); } } body { text-align: left !important} #leaderboard { width: 500px; } #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; } td.total { font-weight: bold; text-align: right; } td.missing { background: #bbbbbb; } <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="leaderboard"> <h2>Leaderboard</h2> <p> Missing scores are shown as grey cells. A grey total indicates that the user has not participated in all challenges and is not eligible for the overall victory yet. </p> <table class="_user-list"> <thead> <tr><td></td><td>User</td> <td><a href="http://codegolf.stackexchange.com/q/66068">#1</a></td> <td></td><td>Total</td> </tr> </thead> <tbody id="users"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> • Are the initial numbers positive integers? – xnor Dec 8, 2015 at 22:58 • "both a or b"—well, which one? Both, or either one? Dec 8, 2015 at 22:59 • @Doorknob -_- Its both. Dec 8, 2015 at 23:10 • @xnor no. Look at the last test-case. Dec 8, 2015 at 23:10 • Making this part of a series causes a kind of unfortunate situation. This is so simple that solutions are all going to look quite similar. And posting similar solutions in languages that were already used is generally frowned upon. I wrote my solution in about 2 minutes, but it's in a language that was already used, and it's the same length. If I follow typical posting etiquette, I won't be able to participate in the series. Dec 9, 2015 at 5:50 ## 30 Answers ## CJam, 16 bytes {{_2+2/@@*mq}*}


Takes input on the stack as a b n where a and b are doubles. Online demo

# Dyalog APL, 2221 15 bytes

.5∘(+.×,×.*⍨)⍣⎕


Takes (a,b) as right argument, and prompts for n:

(

+.× dot product of 0.5 and the right argument

, followed by

×.*⍨ "dot power" of the right argument and 0.5*

)⍣⎕ applied numeric-prompt times.

* "dot power" is like dot product, but using multiplication and power instead of plus and multiplication, as follows:

n
A ×.*⍨ B is BiA = B1AB2A
i=1

-3 bytes thanks to ngn.

Old version:

{((+/÷≢),.5*⍨×/)⍣⍺⊢⍵}


Takes n as left argument and a b as right argument.

⊢⍵ On the RightArg
(...)⍣⍺ recalculate LeftArg times
(+/÷≢) sum divided by tally
, followed by
.5*⍨×/ the square root of the product.

All the test cases:

      f←{((.5×+/),.5*⍨×/)⍣⍺⊢⍵}
0 1 2 5 10 10 f¨ (24 6)(24 6)(24 6)(24 6)(100 50)(1,2*.5)
┌────┬─────┬────────────────┬───────────────────────┬───────────────────────┬───────────────────────┐
│24 6│15 12│13.5 13.41640786│13.45817148 13.45817148│72.83955155 72.83955155│1.198140235 1.198140235│
└────┴─────┴────────────────┴───────────────────────┴───────────────────────┴───────────────────────┘

• Is f⍣⍺⊢⍵ or similar an idiom you use professionally? Dec 9, 2015 at 1:42
• @ThomasKwa Yes, see e.g. Of⍣core⊢TREE at miserver.dyalog.com (click the big "D" and scroll to line [266]).
– Adám
Dec 9, 2015 at 7:35

## TI-BASIC, 22 bytes

Input N
For(I,1,N
{mean(Ans),√(prod(Ans
End
Ans


Does exactly what the algorithm says. Takes N from the prompt, and A and B through Ans as a two-element list.

If N is 0, the For( loop is skipped entirely.

# JavaScript ES7, 48 43 bytes

-5 thanks to Downgoat!

f=(n,a,b)=>n?f(n-1,(a+b)/2,(a*b)**.5):[a,b]


Very simple recursive function.

• (a*b)**.5 is shorter than Math.sqrt(a*b). example Dec 8, 2015 at 23:47
• @Downgoat That's ES7, but meh. Dec 8, 2015 at 23:47

# MATLAB/Octave, 69 65 bytes

function [a,b]=r(a,b,n)
for i=1:n;j=(a+b)/2;b=(a*b)^.5;a=j;end

• You could do b=(a*b).^5 directly as you are not reusing b again in that iteration and save 4 bytes. Dec 9, 2015 at 12:19

# Jelly, non-competing

9 bytes This answer is non-competing, since it uses features that postdate the challenge.

SH;P½¥ðṛ¡


Try it online!

### How it works

SH;P½¥ðṛ¡    Input: x (vector) -- y (repetitions)

SH           Take the sum (S) of x and halve (H) the result.
P½        Take the product (P) of x and the square root (½) of the result.
¥       Combine the last two instructions in a dyadic chain.
;          Concatenate the results to the left and to the right.
ð      Push the preceding, variadic chain; begin a new, dyadic chain.
ṛ     Return the right argument (y).
¡    Repeat the pushed chain y times.


## C++, 108102 100 bytes

Thank you to @RetoKoradi and @AlexA for saving me 6 bytes.

This is non-competitive, because C++ is not a good golfing language. Did this for fun :)

#include<cmath>
std::string f(float a,float b,int n){return n==0?a+" "+b:f((a+b)/2,sqrt(a*b),n-1);}


This is a simple recursion function, very similar to the JS answer.

• You can get rid of the spaces after the commas. Also, using float instead of double is shorter. Dec 9, 2015 at 1:06
• You can also remove the space in the #include line. Dec 9, 2015 at 7:20
• Wow, I'm stupid not to notice that. Thanks! Dec 9, 2015 at 23:07
• I would consider f(float*s) which takes a pointer to 3 floats to be in a "reasonable format". Not sure if that actually makes it shorter.
– nwp
Dec 10, 2015 at 14:26

# K5, 15 bytes

Very literal:

{(+/x%2;%*/x)}/


In action:

 {(+/x%2;%*/x)}/[0; 24 6]
24 6
{(+/x%2;%*/x)}/[5; 24 6]
1.345817e1 1.345817e1


Unfortunately, this does not work in oK because that interpreter does not currently support projection (currying) of adverbs. Works in the real k5.

In oK, it would currently be necessary to wrap the definition in a lambda:

  {x{(+/x%2;%*/x)}/y}[5; 24 6]
13.4582 13.4582


# J, 18 13 bytes

-:@+/,%:@*/^:


Usage:

   agm =: -:@+/,%:@*/^:
5 agm 24 6
13.4582 13.4582

• Wow, this works. Conjunctions are weird. I would expect this expression to be an adverb (which it can be), but if presented with arguments it is also a function. Dec 9, 2015 at 16:32

## Seriously, 11 bytes

,p;π√@ækn


Hex Dump:

2c70603be3fb40916b606e


Try it online

Explanation:

,                    Read in the list as [n,a,b]
p                   pop list to yield: n [a,b]
      n          Push a quoted function and run it n times.
;                 Duplicate [a,b] pair
π√               Compute its product and square root it (GM)
@              Swap the other copy of the pair to the top
æ             Compute its mean.
k            Compile the stack back into a list.


# Japt, 24 bytes 25 33

Saved 9 7 bytes thank to @ETHproductions

Uo r@[VW]=[V+W /2(V*W q]


Takes advantage of ES6 destructuring.

Try it online

### Ungolfed && Explanation

Uo r@[VW]=[V+W /2(V*W q]

// Implicit: U: 1st input, V: 2nd input, W: 3rd input
Uo     // Range from 0 to 1st input
r@     // Loop over range
[V,W]=    // Set 2nd and 3rd input to...
[V+W /2,   // Add 2nd and 3rd inputs, divide by 2
(V*W q]    // Multiple 2nd and 3rd inputs, find square root
// Set's to the above respectively
// Implicit: return [V,W]

• Uo generates a range of numbers from 0 to U, so Uo m@[V,W]=[V+W /2,(V*W q] should work. (Untested) Dec 9, 2015 at 1:04
• Oh, and you shouldn't need the commas at all. :) Dec 9, 2015 at 1:14
• @ETHproductions thanks yet again! :) Dec 9, 2015 at 1:16
• Oh dear, this fails for any U other than 1, outputting each loop as it goes. Here's one that works properly: Uo £[VW]=[V+W /2(V*W q]};[VW] Dec 9, 2015 at 1:17
• @ETHproductions thanks, but using r seemed to work also Dec 9, 2015 at 1:20

# Matlab, 54 bytes

function x=f(x,n)
for k=1:n
x=[mean(x) prod(x)^.5];end


Example:

>> f([24 6], 2)
ans =
13.500000000000000  13.416407864998739


# Pyth, 12

u,.OG@*FG2EQ


Test Suite

### Explanation

u,.OG@*FG2EQ    ##  implicit: Q = eval(input())
u         EQ    ##  reduce eval(input()) times, starting with Q
##  the reduce lambda has G as the previous value and H as the next
.OG           ##  arithmetic mean of last pair
@*FG2      ##  geometric mean of last pair, uses *F to get the product of the list
##  and @...2 to get the square root of that
,              ##  join the two means into a two element list

• Forgot about @ and .O, but I didn't even know the new purpose of E.
– orlp
Dec 9, 2015 at 0:56
• @orlp ah, didn't see your post, my bad I would have just suggested these in the comments. And yeah, keeping track of all the changing stuff is a bit of a struggle :P Dec 9, 2015 at 1:15

# Minkolang v0.14, 23 bytes

Try it here!

$n[$d+2$:r*1Mi2%?!r]$N.
$n C get all input C [ ] C pop N; repeat inner N times C$d                   C duplicate stack [1,2] => [1,2,1,2] C
+                  C add top two elements C
2$: C divide by two C r C reverse stack (get the other two) C * C multiply them together C 1M C take square root C i2%?!r C reverse the stack if an odd step number C$N  C output stack
1M           C take square root C
i          C get step in for loop C


# Pyth, 15 bytes

u,^*FG.5csG2vzQ


# Python 3, 65 55 bytes

Thanks to mathmandan for pointing out a shorter version using the lambda operator.

f=lambda a,b,n:f((a+b)/2,(a*b)**.5,n-1)if n else(a,b)


## My original version:

def f(a,b,n):
if n:f((a+b)/2,(a*b)**.5,n-1)
else:print(a,b)


To my chagrin, a recursive function (a la the JavaScript and C++ answers) was shorter than a simple for loop.

• You can shorten this a little with lambda and the ternary if/else operator: f=lambda a,b,n:f((a+b)/2,(a*b)**.5,n-1)if n else(a,b) Dec 9, 2015 at 3:43
• No problem! (Also, I think this is 53 bytes.) Dec 9, 2015 at 14:21
• The .py file I have saved is listed as being 55 bytes. Is there a better way to calculate program size? Dec 11, 2015 at 22:18
• Sometimes people on this site copy and paste their code into mothereff.in/byte-counter . If you're wondering about the discrepancy, I would guess that Windows is inserting an unnecessary newline character at the end of your .py file (and Windows counts a newline as 2 bytes instead of 1). Either way, you don't have to count that last newline as part of your code for scoring purposes. If you do post a multiple-line entry, you should count 1 for each newline character, not 2, and not including any newline at the end of your last line of code. (As far as I understand the rules anyway!) Dec 12, 2015 at 0:53

# R, 66 bytes

f=function(a,b,n){while(n){x=(a+b)/2;b=(a*b)^.5;n=n-1;a=x};c(a,b)}


Usage:

> f(24,6,0)
[1] 24  6
> f(24,6,1)
[1] 15 12
> f(24,6,2)
[1] 13.50000 13.41641
> f(24,6,3)
[1] 13.45820 13.45814
> f(24,6,4)
[1] 13.45817 13.45817
> f(100,50,10)
[1] 72.83955 72.83955
> f(1,1.41421356237,10)
[1] 1.19814 1.19814

• You can remove the function name to save 2 bytes. Dec 9, 2015 at 17:30

# Mathematica, 31 30 bytes

Saved one byte thanks to Martin Büttner.

{+##/2,(1##)^.5}&@@#&~Nest~##&


Usage:

In[1]:= {+##/2,(1##)^.5}&@@#&~Nest~##&[{24, 6}, 5]

Out[1]= {13.4582, 13.4582}


# Lua, 62 bytes

n,a,b=...for i=1,n do a,b=(a+b)/2,math.sqrt(a*b)end print(a,b)


Uses command line arguments from ... to assign to n, a and b, a nifty trick I learned about Lua recently.

## Haskell, 40 bytes

(!!).iterate(\(a,b)->((a+b)/2,sqrt$a*b))  An anonymous function. Example usage: >> let f=(!!).iterate(\(a,b)->((a+b)/2,sqrt$a*b)) in f (1.0,1.41421356237) 10
(1.198140234734168,1.1981402347341683)


The lambda function (\(a,b)->((a+b)/2,sqrt$a*b)) takes the arithmetic and geometric mean on a tuple. This is iterated starting with the first input (a tuple), and then (!!) indexes the second input to specify the number of iterations. # Perl, 60 bytes perl -ape'F=($F[0]/2+$F[1]/2,sqrt$F[0]*$F[1])for 1..shift@F;$_="@F"'


N.B.: Per this meta post, I believe I've got the scoring correct. The actual code (between single quotes) is 58 characters, then I added +2 for a and p flags as that's the difference from the shortest invocation, perl -e'...'

### Vague complaints

I have this nagging feeling I'm missing an obvious improvement. I know, "welcome to code golf", but I mean more than usual I believe there's an easy opportunity to shorten this.

Early on, I had messed around with using $\ as the second term with some success, but the above approach ended up being 2 bytes shorter, even with the extra ap flags required. Similarly, avoiding the explicit $_ assignment would be nice, but the loop makes that difficult.

The shift@F bugs me, too; if I don't do it that way, though (or use @F=(0,...,...) instead, which doesn't save any bytes), there's an off-by-one error with the @F assignment.

echo 5 24 6 | perl -ape'F=($F[0]/2+$F[1]/2,sqrt$F[0]*$F[1])for 1..shift@F;$_="@F"'  ### Outputs 13.4581714817256 13.4581714817256  # Julia, 49 bytes (a,b,n)->(for i=1:n;a,b=(a+b)/2,√(a*b)end;(a,b))  Pretty direct iterative algorithm. Using the√ symbol and the multiple return saves a few bytes, but the for loop syntax costs a few. # Haskell, 47 Bytes f a b 0=(a,b) f a b n=f((a+b)/2)(sqrt$a*b)(n-1)

• you could save some bytes taking a b as a couple in f: f x 0=x;f(a,b)n=f((a+b)/2,sqrt$a*b)$n-1 Dec 9, 2015 at 7:34
• And define the function infix.
– xnor
Dec 9, 2015 at 10:34

# Julia, 42 bytes

f(a,b,n)=n>0?f((a+b)/2,(a*b)^.5,n-1):(a,b)


This is a recursive function f that accepts three numbers and returns a tuple.

Ungolfed:

function f(a::Real, b::Real, n::Integer)
if n > 0
# Recurse on the arithmetic and geometric means, decrementing n
return f((a + b) / 2, sqrt(a * b), n - 1)
else
# Return the pair
return (a, b)
end
end


# LabVIEW, 21 LabVIEW Primitives

Primitives counted as per this meta post.

pretty staightforward not much to explain.

# Python 2, 6261 62 bytes

def f(a,b,n):
while n:a,b=(a+b)/2.,(a*b)**.5;n-=1
print a,b

• The program should only print once, when it terminates. Dec 9, 2015 at 1:50
• My misunderstanding. Fixed. Dec 9, 2015 at 17:03

# CJam, 16 bytes

{{_:+2/\:*mq]}*}


This is an anonymous function. The input is a list with the two values (as doubles), followed by the iteration count. Try it online with I/O code for testing.

I wouldn't normally have posted this because @PeterTaylor posted an equally long CJam answer before I saw the question. But since this is advertised as the start of a series, I wanted to keep my options open in case the series is interesting.

While the length is the same as Peter's answer, the code is not. I chose a different input format by taking the two values in a list, where Peter used separate values. So while there's not much to it with either input format, the code looks quite different.

{     Start loop over number of iterations.
_     Copy the current pair of values.
:+    Reduce pair with + operator.
2/    Divide by 2.
\     Swap second copy of pair to top.
:*    Reduce pair with * operator.
mq    Calculate square root.
]     Wrap the two new values in a list for next iteration.
}*    End iteration loop.


# Perl 6,  53  47 bytes

{(($^a,$^b),->(\a,\b){((a+b)/2,sqrt(a*b))}...*)[$^n]} # 53 bytes  usage: # give it a name my &code = {(($^a,$^b),->(\a,\b){((a+b)/2,sqrt(a*b))}...*)[$^n]}

say code 100,50,10;          # (72.8395515523453 72.8395515523453)
say code 1,1.41421356237,10; # (1.19814023473417 1.19814023473417)


If I change the input from a,b,n to (a,b),n I can save a few bytes.

{($^l,->(\a,\b){((a+b)/2,sqrt(a*b))}...*)[$^n]} # 47 bytes


usage:

my &code = {($^l,->(\a,\b){((a+b)/2,sqrt(a*b))}...*)[$^n]}

say code (100,50),10;          # (72.8395515523453 72.8395515523453)
say code (1,1.41421356237),10; # (1.19814023473417 1.19814023473417)

say code (24,6),$_ for 0,1,2,5; # (24 6) # (15 12) # (13.5 13.4164078649987) # (13.4581714817256 13.4581714817256)  { ($^l,          # first 2 element tuple
->            # pointy block (lambda)
(\a,\b)     # take a single tuple, and give its 2 elements each a name
{
(           # create a 2 element tuple
(a+b)/2,  # arithmetic mean
sqrt(a*b) # geometric mean
)
} ... *       # create a lazy infinite sequence of tuples
)[ $^n ] # take the nth "tuple" from the outer sequence }  Really I would swap out the ... * with ... -> (\a,\b) { a =~= b }, then there would be no need for the $^n parameter.
( do not use == instead of =~=, or it may not stop )

my &code = {(\$^l,->(\a,\b){((a+b)/2,sqrt(a*b))}...->(\a,\b){a=~=b})[*-1]}

say code (24,6);           # (13.4581714817256 13.4581714817256)
say code (100,50);         # (72.8395515523453 72.8395515523453)
say code (1,1.41421356237) # (1.19814023473417 1.19814023473417)


# Prolog, 80 bytes

Code:

p(A,B,0):-write([A,B]).
p(A,B,N):-X is(A+B)/2,Y is sqrt(A*B),M is N-1,p(X,Y,M).


Example:

p(100,50,10).
[72.83955155234534, 72.83955155234534]


Try it online here

# Java, 10396 84 bytes

String f(int n,double a,double b){return n>0?f(n-1,(a+b)/2,Math.sqrt(a*b)):a+","+b;}


Verify all testcases.

Old version (96 bytes):

String f(int n,double a,double b){for(;n>0;a=(a+b)/2,b=Math.sqrt((b-2*a)*b))n--;return a+","+b;}


Old version (103 bytes):

String f(int n,double a,double b){double t;for(;n>0;t=(a+b)/2,b=Math.sqrt(a*b),a=t)n--;return a+","+b;}