_Maxim_

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8 years, 238 days

MaplePrimes Activity


These are questions asked by _Maxim_

f := x -> x+1:
fsolve('f(x)', x);
                          -1.000000000

f := x -> x+2:
fsolve('f(x)', x);
                          -1.000000000

It's easy to avoid this, sure, but this can be really confusing. And I think it's wrong that Maple fails to notice that the cached result is no longer valid.

test := proc(x) print("x=", x); op(0, x) = f end proc:
applyrule(conditional(x::anything, _test(x)) = 0, f(1));
                           "x=", f(1)
                            "x=", 0
                               0

test := proc(x) print("x=", x); op(0, x) = g end proc:
applyrule(conditional(x::anything, _test(x)) = 0, f(1));
                               0

First, I think this is wrong. Second, how to make applyrule forget the (apparently) cached result? Third, if it doesn't like me using the underscore with my own function, is there a way to rewrite conditional without it?

EDIT: As Carl pointed out here, gc() can be used as a kind of a universal forget() function. The second applyrule correctly returns f(1) after gc().

1. The results are incorrect, because the integral diverges and the p.v. integral exists but is real-valued:

int(1/((1-16*t^2)*sqrt(1-t^2)*sqrt(1-4*t^2)), t = 0 .. 1/2);
                       (1/2)*EllipticPi(4, 1/2)

int(1/((1-16*t^2)*sqrt(1-t^2)*sqrt(1-4*t^2)), t = 0 .. 1/2, CauchyPrincipalValue);
                       (1/2)*EllipticPi(4, 1/2)

This integral converges, but the closed form is wrong:

int(1/((1-1/4*t^2)*sqrt(1-t^2)*sqrt(1-4*t^2)), t = 0 .. 3/2);
                       (1/2)*EllipticK(1/2)-(1/2)*EllipticF(2/3, 1/2)-(1/2)*EllipticPi(4, 1/2)+
                       (1/2)*EllipticPi(2/3, 4, 1/2)-(1/2*I)*EllipticK((1/2)*sqrt(3))-
                       (1/30*I)*EllipticPi(4/5, (1/2)*sqrt(3))+(1/2)*EllipticPi(1/16, 1/2)

2. The series expansion is wrong, because EllipticPi(1/4,16,2) is undefined:

series(EllipticPi(1/4+a, 16, 2), a = 0, 2);
                      EllipticPi(1/4, 16, 2)-(1/45)*sqrt(3)*sqrt(15)*ln(a)+O(a)

3. None of the methods listed on the evalf/Int page can handle this integral:

evalf(Int(1/((1-1/4*t^2)*sqrt(1-t^2)*sqrt(1-4*t^2)), t = 0 .. 3/2));

Although it can be done by splitting the integration range into 0..1/2, 1/2..1, 1..3/2.

None of the methods listed on the evalf/Int page can handle this computation:

evalf(Int(1/((1-1/4*t^2)*sqrt(1-t^2)*sqrt(1-4*t^2)), t = 1 .. 2-10^(-7), digits = 20));

4. Since there are values of the parameters for which the integral diverges, this probably needs a condition saying that 1/sqrt(n) and -1/sqrt(n) are not on the segment [0,z]:

FunctionAdvisor(integral_form, EllipticPi(z, n, k));
                      [EllipticPi(z, n, k) = Int(1/((-n*`_α1`^2+1)*sqrt(-`_α1`^2+1)*
                         sqrt(-k^2*`_α1`^2+1)), `_α1` = 0 .. z),
                       `with no restrictions on `(z, n, k)]

Besides, for z=-1-I, n=1/4, k=2, the lhs and the rhs do not agree:

evalf(subs({z = -1-I, n=1/4, k=2}, %[1]));
                      -.1413755772+1.748734618*I = -.1413755772-.7752517350*I

Either the integral representation is not supposed to be valid everywhere, or Maple computes EllipticPi incorrectly.

seq(i[1], i[1] = 1 .. 2);
i[1];
                              1, 2
                               2

I thought the note saying "The index variable i is NOT private to the seq invocation" referred only to parallel code.

applyrule((x::anything < y::anything) = 0, a < b);
applyrule('x::anything < y::anything' = 0, a < b);
applyrule(`<`(x::anything, y::anything) = 0, a < b);
applyrule('`<`(x::anything, y::anything)' = 0, a < b);
                             a < b
                             a < b
                             a < b
                               0

All quotes matter, no doubt, but why is `<` different from the infix "<" and why does it need uneval?

 

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