How to input pi

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Anthony
Anthony 2016-9-20
评论： Walter Roberson 2024-7-27
How can i enter pi into an equation on matlab?
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Vignesh Shetty
Vignesh Shetty 2020-4-6
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 2022-12-16
@Vignesh Shetty
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π

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Geoff Hayes
Geoff Hayes 2016-9-20
编辑：MathWorks Support Team 2018-11-28
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)
For more information, see https://www.mathworks.com/help/matlab/ref/pi.html.

更多回答（5 个）

Essam Aljahmi
Essam Aljahmi 2018-5-31
编辑：Walter Roberson 2018-5-31
28t2e−0.3466tcos(0.6πt+π3)ua(t).
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Image Analyst
Image Analyst 2018-10-20
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 2018-11-28
编辑：John D'Errico 2018-11-28
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.

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Walter Roberson
Walter Roberson 2018-10-20
If you are constructing an equation using the symbolic toolbox use sym('pi')
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James Emmanuelle Galvan
James Emmanuelle Galvan 2021-10-22
sym(pi) prints out "pi".
Steven Lord
Steven Lord 2021-10-22
Ran in:
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

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Dmitry Volkov
Dmitry Volkov 2022-12-16
Easy way:
format long
p = pi
  1 个评论
Walter Roberson
Walter Roberson 2022-12-16
@Dmitry Volkov
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π

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AKHIL TONY
AKHIL TONY 2023-8-1
using pi will give an approximate value
Meghpara
Meghpara 2024-7-27
it is easy to ge pi
in p=PI.

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