Christian Wolinski

MaplePrimes Activity


These are answers submitted by Christian Wolinski

Could it be that you need a registered version of Maple?

How about:
plot(floor(x/3)*3, x=-5..5);

 

Simply compare these two codes:

restart;
P1 := 1007;
P2 := 1014;
P3 := 1014.1;
evalf(P2 - P1, 2);
evalf(P3 - P1, 2);
restart;
P1 := 1007.;
P2 := 1014.;
P3 := 1014.1;
evalf(P2 - P1, 2);
evalf(P3 - P1, 2);

The difference is obvious. No bugs here.

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For Your matrix you can use this:

F := (n, f) -> LinearAlgebra:-BandMatrix(map(f, [$1 .. n]), 0, n);
F(4, (n) -> d[n]);

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for i from 1 to n do proc(i, a, b) a[i]:=b[i] end(i, a ,b) od;

also

for i from 1 to n do proc(i) a[i]:=b[i] end(i) od;

provided a, b are not declared local.

Your formula appears to be:

fgP,fgQ:=unapply(gP,x,y), unapply(gQ,x,y);
(fgP(QS1,QS2)*fgQ(S1,-S2)/(fgP(PS1,PS2)*fgQ(S1,-S2))) mod p;

Notice the cancellation.

 

e_n_1b := n_1 = (-w^(2*sigma)*tau + w^sigma)*s*nsp_1/(w^sigma*tau^2 - w^(2*sigma)*tau + w^sigma - tau) + tau*(w^(sigma - 1)*tau - w^(2*sigma - 1))*s*nsp_2/(w^sigma*tau^2 - w^(2*sigma)*tau + w^sigma - tau);

wpowers := (A, e) -> frontend(convert, [subsindets(A, dependent(w)^anything, expand), parfrac, e], [{Non}(identical(w^sigma)), {}]);
wpowers(e_n_1b, w);
wpowers(e_n_1b, w^sigma);

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Edit: There was an error in my response, just like there was one in your code.
Correction:

PlotGraph := proc(func::anything, opts::(seq({identical(zoom) = [range, range]})))
   plots:-display(plot(func), ifelse(membertype(identical(zoom) = [range, range], [opts]), view = subs([opts], zoom), 'NULL'))
end proc;;
PlotGraph(x^2, zoom = [0 .. 1, -5 .. 5]);

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One variable, constant bounds:

a := solve({-1 < x, 0 < (2*x)/(x^2 - 1), x < 1}, {x});
solve(`and`(op(a)));

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See this example:

A := map((x -> ifelse( x<1/3 or x>2/3, x, 0)), LinearAlgebra:-RandomMatrix(25, 25, generator = rand(0 .. 1.0))):
P := plots[matrixplot](A, heights = histogram, transparency = 0.3, style = patchnogrid, shading = zhue, orientation=[30, 30, 0]):
subsindets(P, [[anything, anything, 0.]$4], (x -> NULL));

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The point at K=4*I..1+5*I that appears a solution is at:

{K = (1/2)*sqrt(nu*Pi*(-9*Pi*nu+(8*I)*h^2))/nu};

The above is a solution to cosh(Q)=0. cosh(Q) appears in the denominator.


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A := (4*x + 1)/(diff(f(y), y)) = a;
b := (3*(4*x + 1))/((diff(f(y), y))*(3*x + 1));
frontend(algsubs, [A, b, diff(f(y), y)], [{Non}(function), {}]); 

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Do You mean to do something like this:
"numer(L = 2/3)";
f := e -> lcm((denom@lhs, denom@rhs)(e))*e;
f(L = 2/3);


 

 

collect(Your_stuff_goes_here, s, factor); #normal might do also

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You can plot this directly by applying the transform to the original inequalities:

  T[0] := [abs(z) < 3, 1 < abs(z - 1)];
  T[1] := z = x + I*y;
  T[2] := evalc(subs(T[1], T[0])) assuming real;
  T[3] := w = radnormal(subs(x1 = 9/2 - 1/2*sqrt(45), x2 = 9/2 + 1/2*sqrt(45), ((3 + sqrt(5))*(z - x1))/(2*(z - x2))));
  T[4] := z = solve(T[3], z);
  T[5] := radnormal(subs(T[4], T[0]));
  T[6] := w = I*v + u;
  T[7] := factor(expand(subs(T[6], T[5]))) assuming real;
  plots[inequal](T[2], x = -4 .. 4, y = -4 .. 4);
  plots[inequal](T[7], u = -3/2 .. 3/2, v = -3/2 .. 3/2);

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