How to input pi
1 316 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Anthony
le 20 Sep 2016
Commenté : Walter Roberson
le 1 Août 2023
How can i enter pi into an equation on matlab?
2 commentaires
Vignesh Shetty
le 6 Avr 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
Walter Roberson
le 16 Déc 2022
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π
Réponse acceptée
Geoff Hayes
le 20 Sep 2016
Modifié(e) : MathWorks Support Team
le 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)
1 commentaire
Plus de réponses (4)
Essam Aljahmi
le 31 Mai 2018
Modifié(e) : Walter Roberson
le 31 Mai 2018
28t2e−0.3466tcos(0.6πt+π3)ua(t).
5 commentaires
Image Analyst
le 20 Oct 2018
Attached is code to compute Ramanujan's formula for pi, voted the ugliest formula of all time.
.
Actually I think it's amazing that something analytical that complicated and with a variety of operations (addition, division, multiplication, factorial, square root, exponentiation, and summation) could create something as "simple" as pi.
Unfortunately it seems to get to within MATLAB's precision after just one iteration - I'd have like to see how it converges as afunction of iteration (summation term). (Hint: help would be appreciated.)
John D'Errico
le 28 Nov 2018
Modifié(e) : John D'Errico
le 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
le 20 Oct 2018
If you are constructing an equation using the symbolic toolbox use sym('pi')
3 commentaires
Steven Lord
le 22 Oct 2021
That's correct. There are four different conversion techniques the sym function uses to determine how to convert a number into a symbolic expression. The default is the 'r' flag which as the documentation states "converts floating-point numbers obtained by evaluating expressions of the form p/q, p*pi/q, sqrt(p), 2^q, and 10^q (for modest sized integers p and q) to the corresponding symbolic form."
The value returned by the pi function is "close enough" to p*pi/q (with p and q both equal to 1) for that conversion technique to recognize it as π. If you wanted the numeric value of the symbolic π to some number of decimal places use vpa.
p = sym(pi)
vpa(p, 30)
Dmitry Volkov
le 16 Déc 2022
Easy way:
format long
p = pi
1 commentaire
Walter Roberson
le 16 Déc 2022
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π
Voir également
Catégories
En savoir plus sur Elementary Math dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!