How to input pi

3.633 visualizaciones (últimos 30 días)
Anthony
Anthony el 20 de Sept. de 2016
Comentada: Walter Roberson el 1 de Ag. de 2023
How can i enter pi into an equation on matlab?
  2 comentarios
Vignesh Shetty
Vignesh Shetty el 6 de Abr. de 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
Walter Roberson el 16 de Dic. de 2022
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π

Iniciar sesión para comentar.

Respuesta aceptada

Geoff Hayes
Geoff Hayes el 20 de Sept. de 2016
Editada: MathWorks Support Team el 28 de Nov. de 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)

Más respuestas (4)

Essam Aljahmi
Essam Aljahmi el 31 de Mayo de 2018
Editada: Walter Roberson el 31 de Mayo de 2018
28t2e0.3466tcos(0.6πt+π3)ua(t).
  5 comentarios
Image Analyst
Image Analyst el 20 de Oct. de 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
John D'Errico el 28 de Nov. de 2018
Editada: John D'Errico el 28 de Nov. de 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.

Iniciar sesión para comentar.


Walter Roberson
Walter Roberson el 20 de Oct. de 2018
If you are constructing an equation using the symbolic toolbox use sym('pi')
  3 comentarios
James Emmanuelle Galvan
James Emmanuelle Galvan el 22 de Oct. de 2021
sym(pi) prints out "pi".
Steven Lord
Steven Lord el 22 de Oct. de 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)
p = 
π
vpa(p, 30)
ans = 
3.14159265358979323846264338328

Iniciar sesión para comentar.


Dmitry Volkov
Dmitry Volkov el 16 de Dic. de 2022
Easy way:
format long
p = pi
  1 comentario
Walter Roberson
Walter Roberson el 16 de Dic. de 2022
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π

Iniciar sesión para comentar.


AKHIL TONY
AKHIL TONY el 1 de Ag. de 2023
using pi will give an approximate value
  1 comentario
Walter Roberson
Walter Roberson el 1 de Ag. de 2023
Yes, multiple people pointed that out years ago

Iniciar sesión para comentar.

Categorías

Más información sobre Elementary Math en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by