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How can i enter pi into an equation on matlab?

Vignesh Shetty
on 6 Apr 2020

Hi Anthony!

Its very easy to get the value of π. As π is a floating point number declare a long variable then assign 'pi' to that long variable you will get the value.

Eg:-

format long

p=pi

Geoff Hayes
on 20 Sep 2016

Edited: MathWorks Support Team
on 28 Nov 2018

Anthony - use pi which returns the floating-point number nearest the value of π. So in your code, you could do something like

sin(pi)

Essam Aljahmi
on 31 May 2018

Edited: Walter Roberson
on 31 May 2018

28t2e−0.3466tcos(0.6πt+π3)ua(t).

John D'Errico
on 28 Nov 2018

As I recall, these approximations tend to give a roughly fixed number of digits per term. I'll do it using HPF, but syms would also work.

DefaultNumberOfDigits 500

n = 10;

piterms = zeros(n+1,1,'hpf');

f = sqrt(hpf(2))*2/9801*hpf(factorial(0));

piterms(1) = f*1103;

hpf396 = hpf(396)^4;

for k = 1:n

hpfk = hpf(k);

f = f*(4*hpfk-3)*(4*hpfk-2)*(4*hpfk-1)*4/(hpfk^3)/hpf396;

piterms(k+1) = f*(1103 + 26390*hpfk);

end

piapprox = 1./cumsum(piterms);

pierror = double(hpf('pi') - piapprox))

pierror =

-7.6424e-08

-6.3954e-16

-5.6824e-24

-5.2389e-32

-4.9442e-40

-4.741e-48

-4.5989e-56

-4.5e-64

-4.4333e-72

-4.3915e-80

-4.3696e-88

So roughly 8 digits per term in this series. Resetting the default number of digits to used to 1000, then n=125, so a total of 126 terms in the series, we can pretty quickly get a 1000 digit approximation to pi:

pierror = hpf('pi') - piapprox(end + [-3:0])

pierror =

HPF array of size: 4 1

|1,1| -1.2060069282720814803655e-982

|2,1| -1.25042729756426e-990

|3,1| -1.296534e-998

|4,1| -8.e-1004

So as you see, it generates a very reliable 8 digits per term in the sum.

piapprox(end)

ans =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420199

hpf('pi')

ans =

3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420199

I also ran it for 100000 digits, so 12500 terms. It took a little more time, but was entirely possible to compute. I don't recall which similar approximation I used some time ago, but I once used it to compute 1 million or so digits of pi in HPF. HPF currently stores a half million digits as I recall.

As far as understanding how to derive that series, I would leave that to Ramanujan, and only hope he is listening on on this.

Walter Roberson
on 20 Oct 2018

If you are constructing an equation using the symbolic toolbox use sym('pi')

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