# Calculate pi to 5 decimals [closed]

"Given that Pi can be estimated using the function $$\4 \times (1 - 1/3 + 1/5 - 1/7 + \cdots)\$$ with more terms giving greater accuracy, write a function that calculates Pi to an accuracy of 5 decimal places."

• Note, the estimation must be done by calculating the sequence given above.
• You should probably add some more rules, otherwise you will get answers like (python) p=lambda:3.14159
– Matt
Commented Aug 29, 2012 at 15:44
• Have you seen codegolf.stackexchange.com/questions/506/… , which is very similar? At the very least, trig functions should be banned for this problem because they allow for trivial solutions such as this QBASIC program: ?INT(4E5*ATN(1))/1E5 Commented Aug 29, 2012 at 17:30
• I think you should require that the algorithm be one of successive approximation: the longer you compute, the closer you get to pi. Commented Aug 29, 2012 at 22:59
• Dupe, but it's so old it's not here: stackoverflow.com/q/407518/12274
– J B
Commented Sep 1, 2012 at 21:42
• I’m voting to close this question because use of a specific algorithm is impossible to objectively validate Commented Jul 20, 2023 at 3:51

## JavaScript, 46 58 56 45 bytes

ES6 update: Turns out there's more features available to us now that five years have passed.

let f=(i=0,a=0)=>i>1e6?a:f(i+4,a+8/-~i/(i+3))


This version (45 bytes; yes, the let is required) works in ES6 strict mode in theory. In practice, you can run it in V8 (e.g. with node) with --use-strict --harmony-tailcalls; the Proper Tailcalls feature isn't widely implemented yet, alas. However, it's specified behaviour, so it should be fine.

If we want to stick to what's widely implemented, and not require strict-mode, we can simply use the ES6 fat-arrow syntax for functions but otherwise retain the same implementation as before (suggested by Brian H) for a cost of 48 bytes.

a=>{for(a=i=0;i<1e6;a+=8/++i/~-(i+=3));return a}


The choice of name for the single parameter doesn't really matter, but we might as well pick one of the names we use so as to minimise the global-scope pollution.

function(){for(a=i=0;i<1e6;a+=8/++i/~-(i+=3));return a}


This version is a function expression; add two characters (e.g. " f") if you want it named. This version clobbers the globals a and i; this could be prevented if we added "a,i" to the parameter list.

Makes use of a reformulated version of the algorithm in order to circumvent the need for subtraction.

 1/1 - 1/3  +   1/5 - 1/7   +    1/9 - 1/11  + ...
(3/3 - 1/3) + (7/35 - 5/35) + (11/99 - 9/99) + ...
2/3     +      2/35     +       2/99     + ...
2/(1*3)   +    2/(5*7)    +     2/(9*11)   + ...


Here's a "plain" version without this adjustment:

function(){for(a=0,i=1;i<1e6;i+=2)a+=[,4,,-4][i%4]/i;return a}


which clocks in at 64 62 characters.

Thanks to @ardnew for the suggestion to get rid of the 4* before the return.

### History

function(){for(a=i=0;i<1e6;a+=8/++i/~-(i+=3));return a}     // got rid of i+=4; restructured
// Old versions below.
function(){for(a=0,i=1;i<1e6;i+=4)a+=8/i/-~-~i;return a}    // got rid of 4*
function(){for(a=0,i=1;i<1e6;i+=4)a+=2/i/-~-~i;return 4*a}

• o.O very nice job, factoring out the subtraction. Commented Aug 30, 2012 at 17:01
• great work, but needs to be written as a proper function Commented Aug 30, 2012 at 17:55
• @ardnew: Thanks, I must've missed that detail when I read the problem description. I've updated it, and it's now a callable function expression (lambda); not sure if this is allowed or if it has to be given a name. If that's the case, it's just an additional two characters anyway. Commented Aug 30, 2012 at 18:11
• @FireFly you can also shave off 2 chars by changing a+=2/i/-~-~i;return 4*a to a+=8/i/-~-~i;return a Commented Aug 30, 2012 at 18:25
• @ardnew: oh, awesome; didn't think of that. :D Commented Aug 30, 2012 at 18:50

## Python 59 bytes

print reduce(lambda x,p:p/2*x/p+2*10**999,range(6637,1,-2))


This prints out 1000 digits; slightly more than the required 5. Instead of using the prescribed iteration, it uses this:

pi = 2 + 1/3*(2 + 2/5*(2 + 3/7*(2 + 4/9*(2 + 5/11*(2 + ...)))))


The 6637 (the innermost denominator) can be formulated as:

digits * 2*log2(10)

This implies a linear convergence. Each deeper iteration will produce one more binary bit of pi.

If, however, you insist on using the tan-1 identity, a similar convergence can be achieved, if you don't mind going about the problem slightly differently. Taking a look at the partial sums:

4.0, 2.66667, 3.46667, 2.89524, 3.33968, 2.97605, 3.28374, ...

it is apparent that each term jumps back and forth to either side of the convergence point; the series has alternating convergence. Additionally, each term is closer to the convergence point than the previous term was; it is absolutely monotonic with respect to its convergence point. The combination of these two properties implies that the arithmetic mean of any two neighboring terms is closer to the convergence point than either of the terms themselves. To give you a better idea of what I mean, consider the following image:

The outer series is the original, and the inner series is found by taking the average of each of the neighboring terms. A remarkable difference. But what's truly remarkable, is that this new series also has alternating convergence, and is absolutely monotonic with respect to its convergence point. That means that this process can be applied over and over again, ad nauseum.

Ok. But how?

Some formal definitions. Let P1(n) be the nth term of the first sequence, P2(n) be the nth term of the second sequence, and similarly Pk(n) the nth term of the kth sequence as defined above.

P1 = [P1(1), P1(2), P1(3), P1(4), P1(5), ...]

P2 = [(P1(1) +P1(2))/2, (P1(2) +P1(3))/2, (P1(3) +P1(4))/2, (P1(4) +P1(5))/2, ...]

P3 = [(P1(1) +2P1(2) +P1(3))/4, (P1(2) +2P1(3) +P1(4))/4, (P1(3) +2P1(4) +P1(5))/4, ...]

P4 = [(P1(1) +3P1(2) +3P1(3) +P1(4))/8, (P1(2) +3P1(3) +3P1(4) +P1(5))/8, ...]

Not surprisingly, these coefficients follow exactly the binomial coefficients, and can expressed as a single row of Pascal's Triangle. Since an arbitrary row of Pascal's Triangle is trivial to calculate, an arbitrarily 'deep' series can be found, simply by taking the first n partial sums, multiply each by the corresponding term in the kth row of Pascal's Triangle, and dividing by 2k-1.

In this way, full 32-bit floating point precision (~14 decimal places) can be achieved with just 36 iterations, at which point the partial sums haven't even converged on the second decimal place. This is obviously not golfed:

# used for pascal's triangle
t = 36; v = 1.0/(1<<t-1); e = 1
# used for the partial sums of pi
p = 4; d = 3; s = -4.0

x = 0
while t:
t -= 1
p += s/d; d += 2; s *= -1
x += p*v
v = v*t/e; e += 1

print "%.14f"%x


If you wanted arbitrary precision, this can be achieved with a little modification. Here once again calculating 1000 digits:

# used for pascal's triangle
f = t = 3318; v = 1; e = 1
# used for the partial sums of pi
p = 4096*10**999; d = 3; s = -p

x = 0
while t:
t -= 1
p += s/d; d += 2; s *= -1
x += p*v
v = v*t/e; e += 1

print x>>f+9


The initial value of p begins 210 larger, to counteract the integer division effects of s/d as d becomes larger, causing the last few digits to not converge. Notice here again that 3318 is also:

digits * log2(10)

The same number of iteratations as the first algorithm (halved because t decreases by 1 instead of 2 each iteration). Once again, this indicates a linear convergence: one binary bit of pi per iteration. In both cases, 3318 iterations are required to calculate 1000 digits of pi, as slightly better quota than 1 million iterations to calculate 5.

• That's much better than my solution: 4 * sum(1/(1+i*2) if not i%2 else -1/(1+i*2) for i in xrange(places*10**(places))) Commented Apr 30, 2014 at 22:41
• This is very similar to my approach, which happens to be a different form of yours. In mine, as k → ∞, f(-1,k) approaches your Euler-sum. Commented Sep 18, 2017 at 19:38
• Very cool; awesome analysis and explanation, thank you. Commented Jan 11, 2018 at 8:13
• Just a small thing. Didn't you mean after the P_1 = ..., P_2 = ..., P_3 = ..., P_4 = ..., "...multiply each by the corresponding term in the kth row of Pascal's Triangle, and dividing by 2^{k-1}.", instead of nth row and 2^{n-1}?. Commented Jan 11, 2018 at 9:36
• @jeremyradcliff I did, yes. Thanks for the correction. Commented Jan 11, 2018 at 11:07

## Java (67 chars)

float r(){float p=0,s=4,i=1E6f;while(--i>0)p+=(s=-s)/i--;return p;}


Note that this avoids loss of significance by adding the numbers up in the correct order.

• this is fully compliant C code too. if posted as C, you could change while(--i>0) to while(i--) and save 2 chars Commented Aug 30, 2012 at 18:13
• @ardnew, true, but with C there are much more interesting tricks to play... Commented Aug 30, 2012 at 19:03

# Mathematica 42 39 34 33 31 26 32

Archimedes' Approach 26 chars

N@#*Sin[180 Degree/#]&


This reaches the criterion when input is 822.

Question: Does anyone know how he computed the Sin of 180 degrees? I don't.

Leibniz' Approach (Gregory's series) 32 chars

This is the same function the problem poser gave as an example. It reaches the criterion in approximately one half million iterations.

N@4Sum[(-1)^k/(2k+1),{k,0,10^6}]


Madhava-Leibniz Approach 37 chars

This variation uses a few more characters but converges to criterion in only 9 iterations!

N@Sqrt@12 Sum[(-1/3)^k/(2k+1),{k,0,9}]

• those all compute it by the algorithm given in the problem definition? Commented Aug 30, 2012 at 17:00
• @acolyte Leibniz' approach (now the first one listed) is indeed the one mentioned in the description of the problem. It's very slow to converge. A slight variation on it (Madhava-Leibniz) converges very quickly. Commented Aug 30, 2012 at 20:07
• Sine of 180° is pretty easy. It's 180°/N that can get tricky outside of the usual suspects for N.
– J B
Commented Sep 5, 2012 at 21:35
• Please explain, @J.B. Tricky to measure? Commented Sep 5, 2012 at 23:02
• This entry should state "32" because only Leibniz' approach fulfils the requirements (counting the characters in the code as given, I get 34, but both spaces may be safely removed, giving indeed a length of 32). Commented Sep 19, 2017 at 19:15

## APL (14)

4×-/÷1-⍨2×⍳1e6


• 13, --/4÷1-2×⍳1e6 Commented Feb 24, 2014 at 22:29

## Haskell, 32

foldr(\k->(4/(2*k+1)-))0[0..8^7]


GHCi> foldr(\k->(4/(2*k+1)-))0[0..8^7]
3.141593130426724

Counting a function name it's

## 34

π=foldr(\k->(4/(2*k+1)-))0[0..8^7]


## R - 25 chars

sum(c(4,-4)/seq(1,1e6,2))


## C (GCC) (44 chars)

float p(i){return i<1E6?4./++i-p(++i):0;}


That's 41 chars, but it also has to be compiled with -O2 to get the optimiser to eliminate the tail recursion. This also relies on undefined behaviour with respect to the order in which the ++ are executed; thanks to ugoren for pointing this out. I've tested with gcc 4.4.3 under 64-bit Linux .

Note that unless the optimiser also reorders the sum, it will add from the smallest number, so it avoids loss of significance.

Call as p().

• Your recursive call is q(), not p(). And I don't think -O2 should be counted (but if you do count it, it's 4 chars because of the required space). Commented Sep 5, 2012 at 14:57
• Also: 1. gcc 4.1.1 doesn't optimize the recursion (and I don't see how it could), so the stack overflows. 2. it should be called as p(0). 3. Save a char by return++i.... 4. Two ++i makes undefined behavior. Commented Sep 5, 2012 at 15:03
• @ugoren, thanks for your comments. In order: q - that'll teach me to double-check after renaming. I think I'm following normal practice in counting -O2 as 3 chars, but we can open it up on meta if you want; meta.codegolf.stackexchange.com/questions/19 is the only relevant discussion I can find. I've added the version of gcc which I'm using, and which allows me to call it as p(). Saving the char stops the optimiser and gives segfault. I will clarify that I'm using undefined behaviour, as per meta.codegolf.stackexchange.com/questions/21 Commented Sep 5, 2012 at 21:51
• I added an answer to the meta question about flags. About p() - are you sure calling p() from any context would work? Or is it just what happened to be on the stack in your test? Commented Sep 6, 2012 at 7:20
• @ugoren, maybe I got lucky consistently. Even if I call it twice in a row, the second one still returns the correct value. gcc does seem to produce slightly different code for p() vs p(0), but I don't know what behaviour it documents and I'm not really a C programmer. Commented Sep 6, 2012 at 22:28

J, 26 chars

+/+/_2((4 _4)&%)>:+:i.100

Moved from 100 items of sequence to 1e6 items. Also now it's a code tagged and could be copypasted from browser to the console without errors.

+/+/_2((4 _4)&%)\>:+:i.1e6

• -/4%>:2*i.1e6 -- 13 characters. (Thanks to b_jonas in #jsoftware for making me realise that -/ works to compute a sum with alternating sign. [This is since all operators in J are of equal precedence and right-associative, so -/ 1 2 3 4 <=> 1 - (2 - (3 - 4)) <=> 1 - 2 + 3 - 4.]) Commented Aug 30, 2012 at 20:46
• that's neat and twice as awesome. Or even 2^10 more awesome!
– fftw
Commented Aug 31, 2012 at 1:16
• @FireFly that is beautiful Commented Sep 19, 2017 at 0:04

# Clojure - 79 chars

(fn [](* 4(apply +(map #(*(Math/pow -1 %1)(/ 1.0(+ 1 %1 %1)))(range 377000)))))


This creates a function of no arguments which will calculate a float which approximates pi correctly to five decimal places. Note that this does not bind the function to a name such as pi, so this code must either be evaluated in place with eval as (<code>) or bound to a name in which case the solution is

(defn p[](* 4(apply +(map #(*(Math/pow -1 %1)(/ 1.0(+ 1 %1 %1)))(range 377000)))))


for 82 chars

### About

(defn nth-term-of-pi [n] (* (Math/pow -1 n) (/ 1.0 (+ 1 n n))))
(defn pi [c] (* 4 (apply + (map nth-term-of-pi (range c)))))
(def  pi-accuracy-constant (loop [c 1000] (if (< (pi c) 3.14159) (recur (inc c)) c)))
; (pi pi-accuracy-constant) is then the value of pi to the accuracy of five decimal places


# Javascript - 33 Characters

p=x=>4*(1-(x&2))/x+(x>1?p(x-2):0)


Call p passing a positive odd number x and it will calculate Pi with (x-1)/2 terms.

## Ruby - 82 chars

def f(n,k=n)k>0?(f(n,k-1)+f(n+1,k-1))/2:n<0?0:f(n-1,0)+(-1)**n/(2*n+1.0)end;4*f(9)


Try it : https://repl.it/LQ8w

The approach uses the given series indirectly using a numerical acceleration approach. The resulting output is

pi ≈ 3.14159265161

vs.

pi = 3.14159265359

It starts with

f(n,0) = 1/1 - 1/3 + 1/5 - ... + ((-1)**n)/(2*n+1)


And then, since this is alternating, we can accelerate the convergence using

f(n,1) = (f(n,0) + f(n+1,0))/2


And it repeatedly applies this:

f(n,k) = (f(n,k-1) + f(n+1,k-1))/2


And for simplicity, f(n) = f(n,n).

## Ruby - 50 chars

If you don't mind running for a really long while, then you could simply use

def f(n)n<0?0:f(n-1)+(-1)**n/(2*n+1.0)end;4*f(1e7)


or

a=0;for k in 0..1e7 do a+=(-1)**k/(2*k+1.0)end;4*a


# VBA - 80 105 characters

From the VBE immediate window:

s=-1:x=1:i=3:While Left(4*x,7)<>3.14159:x=x+((1/i)*s):s=-s:i=i+2:Wend:Debug.?4*x


Using Left() instead of Round() saves a character but also makes it more accurate to the definition of the challenge. I'm getting my character count from Notepad++. I do see that other systems may count differently.

The algorithm below is easier to read:

Sub p
s=-1:x=1:i=3
While Left(4*x,7)<>3.14159
x=x+((1/i)*s)
s=-s
i=i+2
Wend
Debug.?4*x
Debug.?"Pi to 5 places solved in ";(i-1)/2;" steps."
End Sub


If you want to test this code and have Microsoft Excel, Word, Access, or Outlook installed (Windows only), press Alt+F11 to open the VBA IDE. Insert a new code module (Alt+I,M) and clear out Option Explicit if shown at the top. Then paste in the code and press F5 to run it. The results should appear in the Immediate Window (press Ctrl+G if you don't see it).

• Great solution, but you have some room for improvement. It looks like you have overcounted your solution, as is, I see 97 bytes of code. You can also drop the () from the sub call. If you are willing to make it an immediate window solution you can further knock it down to s=-1:x=1:i=3:While Round(4*x,5)<>3.14159:x=x+((1/i)*s):s=-s:i=i+2:Wend:?4*x for 75 bytes. Commented Oct 26, 2020 at 14:00
• Thanks for the tip. I wrote a minify algorithm for VBA and have updated it. I'm guessing that TIO counts CrLf different than Notepad++ but it's not something I am worried about. I had fun playing this round of code golf.
– Ben
Commented Oct 26, 2020 at 15:45

## C, 69 chars

float p,b;void main(a){b++<9e6?p+=a/b++,main(-a):printf("%f\n",4*p);}

• Run with no command line parameters (so a is initialized to 1).
• Must be compiled with optimization.
• void main is strange and non-standard, but makes things work. Without it, the recursion is implemented as a real call, leading to a stack overflow. An alternative is adding return.
• Two characters 4* can be saved, if running with three command line parameters.
• You could shorten that to int main(a) or even main(a), GCC only gives a warning. And it will give a warning for void main anyway, and maybe even because you have only one argument to main.
– user344
Commented Apr 30, 2014 at 14:32

## Python - 56 chars

Meh, My python-fu is not strong enough. I couldn't see any more shortcuts but maybe a more experienced golfer could find something to trim here?

t=s=0
k=i=1
while t<1e6:t,s,i,k=t+1,k*4./i+s,i+2,-k

• You could use Python 3 to save one byte for the float division (4. -> 4). In other news, I just found a case where Python 3 actually beats Python 2 in code golf!
– user344
Commented Apr 30, 2014 at 15:06

## Python (49)

print 4*sum((-1)**i/(2*i+1.)for i in range(9**6))

3.14159453527


# PHP - 56 55 chars

<?for($j=$i=-1;1e6>$j;){$p+=($i=-$i)/($j+=2);}echo$p*4;


I don't know that I can get it much smaller without breaking the algorithm rule.

• How about this for 45? <?for(;1e6>$j;)$p+=($i=-$i|4)/~-$j+=2;echo$p; Commented Sep 1, 2012 at 23:26
• I was trying to come up with that, but couldn't get the bitwise ops to work. Thanks for the suggestion! Commented Sep 18, 2012 at 20:21
• You can remove the last semicolon to save 1 character.
– user344
Commented Apr 30, 2014 at 14:28

## Perl - 43 39 chars

not sure the rules on anonymous subroutines, but here's another implementation using @FireFly's series construction

\$\>:#,_@>+\55+/:#^_"." v>p"~"/:"~"%08p"~"/00p:2\4%-*"(}" 8^90%"~":+2:+g90*+g80*< >*:**\/+>"~~"00g:"~"!|  Try it online! In case anyone is wondering, it's an elephant. # Husk, 16 bytes *_4ṁ\↑^9 9z*İ_İ1  Try it online! using series of only 6561 (9^4) terms (times-out on TIO for longer series). Output is a rational number expressed as a fraction: TIO header multiplies by 100,000 and rounds to nearest integer. *_4ṁ\↑^9 9z*İ_İ1 *_4 # multiply by -4: ṁ # sum of series of \ # reciprocals of ↑^9 9 # first 387420489 (9^9) terms of z* # pairwise products of İ_ # powers of -1 (-1,1,-1,1,...) and İ1 # odd numbers (1,3,5,7,9,...)  # jq, 34 characters [range(1;1e6;4)|1/.-1/(.+2)]|add*4  Sample run: bash-5.2$ jq -n '[range(1;1e6;4)|1/.-1/(.+2)]|add*4'
3.1415906535898936


Try it online!

## Java - 92 84 chars

I cannot beat by far Peter Taylor's result, but here is mine:

double d(){float n=0,k=0,x;while(n<9E5){x=1/(1+2*n++);k+=(n%2==0)?-x:x;}return 4*k;}


Ungolfed version:

double d() {
float n = 0, k = 0, x;
while (n < 9E5) {
x = 1 / (1 + 2 * n++);
k += (n % 2 == 0) ? -x : x;
}
return 4 * k;
}


Edit: Saved a few characters using ternary operator.

Ruby - 54 chars

def a()p=0;1000000.times{|i|p+=8/(4*i*(4*i+2))};p;end;


My first try on console

def a()i=1;p=0;while i<2**100 do p+=8/(i*(i+2));i+=4;end;p;end;


63 chars.

• You can save a byte by using def a; instead of def a().
– user344
Commented Apr 30, 2014 at 15:08
• Another one by removing the last semicolon.
– user344
Commented Apr 30, 2014 at 16:06

## Perl (76 chars)

$y=1e4;for$x(0..1e4-1){$y--while sqrt($x**2+$y**2)>1e4;$a+=$y}print 4*$a/1e8


(Result: 3.14159052)

Not the shortest possible solution, but maybe interesting. It's a geometrical one. I calculate the area under a circle.

I got another funny approach, but it's really slow. It counts the number of discrete points in a square that are below a quarter circle and calculates pi from it:

$i=shift;for$x(0..$i){for$y(0..$i){$h++if sqrt($x**2+$y**2)<$i}}print$h*4/$i**2  It expects the number of iterations as command line argument. Here you can see how run time relates to accuracy. ;) $ time perl -e '$i=shift;for$x(0..$i){for$y(0..$i){$h++if sqrt($x**2+$y**2)<$i}}print$h*4/$i**2' 100 3.1796 real 0m0.011s user 0m0.005s sys 0m0.003s$ time perl -e '$i=shift;for$x(0..$i){for$y(0..$i){$h++if sqrt($x**2+$y**2)<$i}}print$h*4/$i**2' 1000 3.14552 real 0m0.354s user 0m0.340s sys 0m0.004s$ time perl -e '$i=shift;for$x(0..$i){for$y(0..$i){$h++if sqrt($x**2+$y**2)<$i}}print$h*4/\$i**2' 10000
3.14199016
real    0m34.941s
user    0m33.757s
sys 0m0.097s


# k (25 chars)

4*+/%(i#1 -1)'1+2!i:1000000

Slightly shorter:

+/(i#4 -4)%1+2*!i:1000000

• Can save 6 bytes by doing -4*{y-x}/%1+2*!_1e6 (assuming a k where monadic % is inverse). Commented Oct 26, 2020 at 16:05
• Can trim another byte by doing {y-x}/-4%1+2*!_1e6 (and avoid the monadic % issue). Commented Oct 26, 2020 at 20:22

# CJam - 21

1e6{WI#4d*I2*)/}fI]:+


Pretty straightforward calculation of the given series.
CJam is http://sf.net/p/cjam

# Julia - 30 characters

sum(4./[1:4:1e6] - 4./[3:4:1e6])


# SQL, 253 bytes

DECLARE @B int=3, @A varchar(max), @C varchar(max)='1'
WHILE @B<100000
BEGIN
SELECT @C=@C+(select case when (@B-1)%4=0 then'+'else'-'end)+
(SELECT cast(cast(1.0/@B as decimal(9,8)) as varchar(max)))
SELECT @B=@B+2
END
EXECUTE('SELECT 4*('+@C+')')


I would provide a SQL Fiddle, but this goes too many loops deep finding the 1/3 1/5 1/7 etc. fractions and gives errors lol. However, if you change @B<100000 to 1000 then it runs (obviously not to the same number of digits of accuracy).

# GolfScript, 25 bytes

7.?,{2*)2.-1??./\/\-}*)4*


Try it online!

7.?                         # Push 7^7=823543, we just need an odd number bigger than 136120
,{               }*      # For every number k from 1 to 823542
2*)                    # 2*k+1
2.-1??./            # sqrt(2)/sqrt(2)=1.0 this number is just 1, but it is a float
\/          # 1.0 / (2*k+1)
\-        # Subtract the sequence from this number
)     # Add 1 because 1/1 was skipped
4*   # Multiply by 4


Output:

3.141593867855424


# Desmos, 30 bytes

4\sum_{n=0}^{9^6}(-1)^n/(1+2n)
`

Try it on Desmos

Pretty self-explanatory coding of the sequence. An array-based method, while it sounds like it should be more efficient, runs into Desmos' limitations on array length.