You could also define an alias for 1/alpha,

alias(alpha=(108+12*177^(1/2))^(1/3),1/alpha=1/(108+12*177^(1/2))^(1/3));
                                      1
                             alpha, -----
                                    alpha

> solve(x^3+2*x+1=0,x);

    alpha     4    alpha     2            1/2 /  alpha     4  \
  - ----- + -----, ----- - ----- + 1/2 I 3    |- ----- - -----|,
      6     alpha   12     alpha              \    6     alpha/

        alpha     2            1/2 /  alpha     4  \
        ----- - ----- - 1/2 I 3    |- ----- - -----|
         12     alpha              \    6     alpha/

Alec

Plotting the second football is slightly more tricky, but can be done as follows. First, I'll repeat the first construction,

with(Student:-LinearAlgebra):
P,Q:=<1,0,0>,<0,0,sqrt(2)>:
center:=(P+Q)/2:
e1,e2:=map(Normalize,NullSpace(Transpose(<P-Q>)))[]:
f1:=Normalize(P-Q):
f2:=e1*cos(alpha)+e2*sin(alpha):
cc:=center-f2/2:
a:=plot3d(cc+f1*cos(beta)+f2*sin(beta),
    alpha=0..2*Pi,beta=Pi/6..5*Pi/6,
    axes=normal,
    style=patchnogrid,
    color=red,
    lightmodel=light4,
    scaling=constrained,
    orientation=[-120,66],
    glossiness=0.85
):

Now, the second one,

 R,S:=<-1/2,sqrt(3)/2,0>,<0,0,-sqrt(2)>:
center1:=(R+S)/2:
g1,g2:=NullSpace(Transpose(<R-S>))[]:
g1:=solve((x*g1+g2).g2)*g1+g2:
g1,g2:=map(Normalize,[g1,g2])[]:
h1:=Normalize(R-S):
h2:=g1*cos(phi)+g2*sin(phi):
cc1:=center1-h2/2:
b:=plot3d(cc1+h1*cos(theta)+h2*sin(theta),
    phi=0..2*Pi,theta=Pi/6..5*Pi/6,
    axes=normal,
    style=patchnogrid,
    color=blue,
    lightmodel=light4,
    scaling=constrained,
    orientation=[-120,66],
    glossiness=0.85
):

Finally,

plots[display](a,b);

Alec

As usual in such cases, the answers should be rounded, and after that will satisfy the constraints.

Alec

If c is real and f(x)=x^3-c*x+16, then f(0)=16 and f(-infinity)=-infinity, so there always exists a negative real root. The sum of other 2 roots is equal to minus this negative real root, so it is positive which couldn't happen if they had negative real parts.

If c is not real, all 3 roots are not real, and their sum is 0, so again all 3 of them can not have negative real parts.

Alec

Another example,

with(Maplets):
with(Maplets:-Elements):
m := Maplet([
    ["Enter the function of x:", TextField['a']('value'='x^2')],
    Button("Diff and int",'onclick'=Action(
        Evaluate('d'='diff(a,x)'), Evaluate('i'='int(a,x)'))),	 
    ["Derivative:", TextField['d']()],
    ["Integral:", TextField['i']()]]):
Display(m):

See more examples in Maplet Examples in the wiki.

Alec

The following works,

int(5*x^(2/3)+3/(5*x^(2/5)), x);
                             (5/3)    (3/5)
                          3 x      + x

Alec

You could just use RiemannSum in Student:-Calculus1 package, see ?RightRiemannSum. For example,

Approx:=(f,a,b,N)->Student:-Calculus1:-RiemannSum(
    f,evalf(a..b),method=right,partition=N):

Alec

If you put Pi^2 instead of Pi in eq1, you would get the same answer.

Alec

GetBits in Bits package (in Maple 12.)

Alec

If I needed that for myself, I would, probably, use a parametric plot, which can be done with the same options as above, as

with(Student:-LinearAlgebra):
P,Q:=<1,0,0>,<0,0,sqrt(2)>:
center:=(P+Q)/2:
e1,e2:=map(Normalize,NullSpace(Transpose(<P-Q>)))[]:
f1:=Normalize(P-Q):
f2:=e1*cos(alpha)+e2*sin(alpha):
cc:=center-f2/2:
plot3d(cc+f1*cos(beta)+f2*sin(beta),
    alpha=0..2*Pi,beta=Pi/6..5*Pi/6,
    axes=normal,
    style=patchnogrid,
    color=red,
    lightmodel=light4,
    scaling=constrained,
    orientation=[-120,66],
    glossiness=0.85);

Alec

Or simple

P, Q, X := <1,0,0>, <0,0,sqrt(2)>, <x,y,z>:
eq:=cos(Student:-LinearAlgebra:-VectorAngle(X-P,X-Q))=-1/2:
plots[implicitplot3d](eq,
    x=-0.1..1.1,y=-0.5..0.5,z=-0.1..1.5,
    numpoints=10000, 
    axes=normal, 
    style=patchnogrid,
    color=red,
    lightmodel=light4,
    scaling=constrained,
    orientation=[-120,66],
    glossiness=0.85);

giving about the same picture as above.

Alec

a := eval(
    Student[Calculus1][SurfaceOfRevolution]
    (sqrt(1-x^2), x = -sqrt(3)/2 .. sqrt(3)/2, 
    distancefromaxis = 1/2, output = plot), 
    TITLE = NULL): 
b := plottools[rotate](
    a, 0, -arctan(sqrt(2)), 0): 
A := Student:-LinearAlgebra:-RotationMatrix(
    arctan(sqrt(2)), <0, 1, 0>): 
t := A.<sqrt(3)/2, 0, 1/2>; 
c := plottools[translate](b, 
    1/2-sqrt(6)/6, 0, sqrt(2)/2-sqrt(3)/6): 
plots[display](c, scaling = constrained);

Alec

It looks as if the 2 functions have different asymptotics, so their values shouldn't be very close for large x with constant P, Q, a, and b. Do you have a particular example?

Alec

Change the definion of Xbest to

Xbest := Sols;

i.e. get rid of [1] in Sols.

Alec

It is easy to do using deprecated student package,

with(student):
eval(slope(y=x^3+3*x-8),x=2);

                                  15

y:=x->x^3 -6*x^2 -34*x -9:
map(x->[x,y(x)],[solve](slope(y(x))=2))[];

                         [6, -213], [-2, 27]

eval(slope(x^5),x=-2);

                                  80

Note that the curve in b) is not a circle.

Alec