1585 Reputation

19 years, 261 days

MaplePrimes Activity

These are answers submitted by Christian Wolinski

d1 := LinearAlgebra:-Determinant(A):

or
d1 := LinearAlgebra:-Determinant(A):

Use fulldigits....

fsolve(SYS,VAR,fulldigits);

Lacking caution....

In Maple I would suggest care and caution when mixing trigs, exps with algebra.
You're solving this:

A:=(20 + 20*T + 2*T*(T + 1))*exp(-T) - 10*exp(-2*T) - 2*T - 10.0 = 0;
S:=(solve,((E,V)->V))(expand(A),exp(T));
plot([S],T=0..10,view=[DEFAULT,-1..5],axes=boxed,scaling=constrained,color=[grey,grey,red]);

solve(evalc(Im(S[1]))=0);

Cubic solutions....

This works better:

eq := a*x^3 + b*x^2 + c*x + d;
s := x1 + x2 + x3 = -b/a;
sp := x1*x2 + x1*x3 + x2*x3 = c/a;
p := x1*x2*x3 = -d/a;
eliminate({eq, s, p, sp}, {x});
sol := solve(%[2], {x1, x2, x3}, explicit):
#easier way to solve:
#sol := solve(identity(eq-a*(x-x1)*(x-x2)*(x-x3), x), {x1, x2, x3}, explicit=true);
# evaluate without assignment:
# eval( radnormal([sol], rationalized), [a=1, b=-5, c=6, d=-1]);
a := 1;
b := -5;
c := 6;
d := -1;

Simply option operator, arrow....

As demonstrated here:

f := proc(a)
option operator, arrow;
proc(b)
option operator, arrow;
a/b;
end;
end;

Use PolyhedralSets....

Try this:

with(PolyhedralSets):
V := [x, y, z];
p1 := x+z-1;
H1 := PolyhedralSet([p1 <= 0], V):
H2 := PolyhedralSet([p1 = 0], V):
H3 := PolyhedralSet([p1 >= 0], V):
C := PolyhedralSet([seq(n >= 0, n = V), seq(n <= 1, n = V)], V):
S := seq(h intersect C, h = [H1, H2, H3]);

ev:= edgeoptions = [color = "black", thickness=0], vertexoptions = [color = "red", symbolsize = 1]:
plots[display](
Plot(S[1], faceoptions = [color = pink, glossiness = 0.25, transparency=0.33], ev),
Plot(S[2], faceoptions = [color = aquamarine, glossiness = 0.25, transparency=0], ev),
Plot(S[3], faceoptions = [color = turquoise, glossiness = 0.25, transparency=0.33], ev),
view=[(-1..2)\$3], scaling = constrained, lightmodel = light3, axes = boxed, insequence=false
);

.

see ?interp...

y = interp(op(map2(map2, op, [1,2], [[-1, -4], [1, 0], [2, 0], [3, -4]])), x);

Without floats....

Converting into fractions we get:

Sum(binomial(196, j)*(1/2)^(321/10+j)*(Sum(binomial(109, l)*(-1)^(j+l)*(Sum((1/2)^((31/10)*h)*GAMMA(-41/10+h)*GAMMA(382/41+(10/41)*l+(10/41)*j+(31/41)*h)/(factorial(h)*(321/31+(10/31)*j+h)*GAMMA((10/41)*j+(10/41)*l+6321/410+(31/41)*h)), h = 0 .. infinity)), l = 0 .. 109)), j = 0 .. 196);

A guess....

I do not really understand what you are saying so my question follows:

We can indentify any point P using: P = [x, y, c]; where c is the color and (x,y) is the coordinate in the plane.
We have a rule: For all ( P[1] = [x[1], y[1], c[1]], P[2] = [x[2], y[2], c[2]] ) (c[1]=c[2]) implies not (y[1]=y[2]).
An observation: for every set of points satisfying above rule, the cardinality of (subset of points) having only particular (y, c) coordinates is not greater than 1.

You've given us a set of 63 points, call it set S. Which are you trying to do:
1) choose a subset of S that satisifies the rule
2) partition S into subsets satisfying the rule

[{x = 4, x = 5, x = 6, x = 7, x = 2, x = 3, x = 1}, {y = 1, c = 1}], [{x = 4, x = 5, x = 6, x = 7, x = 2, x = 3, x = 1}, {c = 2, y = 2}], [{x = 4, x = 5, x = 7, x = 1}, {c = 3, y = 3}], [{x = 4, x = 5, x = 7, x = 1}, {y = 4, c = 3}], [{x = 6, x = 2, x = 3}, {c = 4, y = 3}], [{x = 6, x = 2, x = 3}, {y = 4, c = 4}], [{x = 5, x = 7, x = 3, x = 1}, {c = 5, y = 5}], [{x = 5, x = 7, x = 3, x = 1}, {y = 6, c = 5}], [{x = 4, x = 6, x = 2}, {c = 6, y = 6}], [{x = 4, x = 6, x = 2}, {c = 6, y = 5}], [{x = 7, x = 2, x = 1}, {c = 7, y = 9}], [{x = 7, x = 2, x = 1}, {c = 7, y = 7}], [{x = 7, x = 2, x = 1}, {c = 7, y = 8}], [{x = 5, x = 3}, {c = 8, y = 7}], [{x = 5, x = 3}, {c = 8, y = 8}], [{x = 5, x = 3}, {c = 8, y = 9}], [{x = 4, x = 6}, {c = 9, y = 8}], [{x = 4, x = 6}, {c = 9, y = 7}], [{x = 4, x = 6}, {c = 9, y = 9}];

Linear transform....

My one bet is the OP is looking for this:

X := x, y;
V := x^4, x^3*y, x^2*y^2, x*y^3, y^4;
P := (-x+sqrt(3)*y)/2,(-sqrt(3)*x-y)/2;
VP := expand(unapply([V], X)(P));
map(proc(E) local C, W; C := coeffs(E, {x, y}, 'W'); subs( [W] =~ [C], ({V} minus {W}) =~ 0,  [V]) end, VP);

No need for recursion....

You can simply solve for J and use that, no need for recursion.

J=(n->(2*(n-2^(ilog[2](n)))+1));

Can you prove the formula?

Use diff and not D....

I would suspect the reason is you are not using diff and mtaylor does. So try this code:

restart;
f := (x,y) -> 1/(2 + x*y^2):
P := proc(x,y,x0,y0,N) local p,q; options operator, arrow;  add(1/factorial(n)*add(binomial(n,k)*subs(p=x-x0,q=y-y0,diff(f(p,q),[p\$(n-k),q\$k])*p^(n-k)*q^k), k=0..n), n=0..N): end;
P(x,y,x0,y0,30):

Exp....

Maple 5.4 says:

Int(ln(2^(1/2)*t^(1/2))*exp(-t),t = 0 .. infinity) = 1/2*ln(2)-1/2*gamma;

Combinat package....

This code creates a polynomial in x,y,z of degree 8:

f := proc(V, d) local n; map(mul, combinat[choose]([1\$d, op(map(`\$`, V, d))], d)) end;
L := f([x, y, z], 8);
add(c[i] * L[i], i = 1..nops(L));

Modify as needed.

Surd....

You're obviousy thinking of surd. Look at the following:
surd(x,3) assuming x, real;

it will give you:
signum(x)*abs(x)^(1/3);

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