How to input pi

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Anthony 2016-9-20

How can i enter pi into an equation on matlab?
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Walter Roberson 2022-12-16
That is what @Geoff Hayes suggested years before. But it does not enter π into the calculation, only an approximation of π

采纳的回答

Geoff Hayes 2016-9-20

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)
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Walter Roberson 2022-12-16
Also see sinpi and cospi

更多回答（3 个）

Walter Roberson 2018-10-20
If you are constructing an equation using the symbolic toolbox use sym('pi')
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Steven Lord 2021-10-22
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

Essam Aljahmi 2018-5-31

28t2e0.3466tcos(0.6πt+π3)ua(t).
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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.

Dmitry Volkov 2022-12-16
Easy way:
format long
p = pi
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Walter Roberson 2022-12-16
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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