Kitonum

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This is a very simple task. See help on the commands  diff  and  maximize .

@Rouben Rostamian

 The simplification  alpha__max__1  with Maple:

alpha__max__1 := Pi - Pi*sqrt(29 - 4*sqrt(43)*cos(arctan((9*sqrt(191))/1121)/3) - 4*sqrt(3)*sqrt(43)*sin(arctan((9*sqrt(191))/1121)/3))/3;
e:=(1/3)*arctan(9*sqrt(191)*(1/1121)):
thaw(convert(subs(e=freeze(e), alpha__max__1), phaseamp, freeze(e)));

 

@Randy233  See the update to my solution above.

@srikantha087 It all depends on the size of the matrix, not how it was obtained. For your case I am using the function  i->0.5*i-0.75  that maps the set  {1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5}  onto the set  {0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0} . I took {1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5, 10.5, 11.5} instead of  {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}  so that the marks are in the middle of the base of the pillars.

@AHSAN  The line

p:=unapply(eval(p(x),%), x);

gives a standard way to define  p  as a function of  x  for later use.

To find  lambda  we only leave one condition  p(0)=0  and then express  lambda  through  k  using the second condition  p(1)=0 :

restart;
h := k - (k - 1)*x;
DE := diff(p(x), x) = 6/h^2 - 12*lambda/h^3;     
BC := p(0) = 0;                    
dsolve({BC, DE}, p(x));
eval(rhs(%),x=1);
solve(%,lambda);

 

@AHSAN  See update to my answer.

@Carl Love  Interestingly, plot3d also works for plotting spatial curves, but it doesn't respond to color input (everything is drawn in black):

restart;
with(plots):
f := (x, y) -> x^2 + y^2 - 12*x + 16*y;
display(plot3d(f(x, y), x = -9 .. 9, y = -9 .. 9), pointplot3d([[6, -8, f(6, -8)]], color = red, symbol = solidcircle, symbolsize = 18), view = [-4.2 .. 8.2, -8.2 .. 4.2, -100 .. 100], plot3d([cos(t), sin(t), 1 - 12*cos(t) + 15*sin(t)+1], t = 0 .. 2*Pi, color=red, orientation = [-15, 68, 5]));
                

                                  

@Carl Love  I just raised this ellipse up a little (by 1) so that it is entirely above the surface f(x,y)

@vv  I wrote   ln(2+5*t)  for  t>=0 . It is equivalent to  ln(piecewise(t = 0, 2, 0 < t, 2 + 5*t))  for  t>=0 .  Probably OP simply does not know that by default a piecewise function on unspecified intervals is considered equal to 0.  

@mbirkner  Yes you are right. This approach is only suitable for small numbers. Large numbers require a special custom procedure. I have no idea how this can be done yet.


Use the prefix notation for multiplication:

restart

M := Matrix(4, 4); for m from 0 to 4 do for n from 0 to 4 do M(m+1, n+1) := %factorial(2*m+3*n)/`%*`(%factorial(m+2*n+1), %factorial(m), %factorial(n)) end do end do; M

Matrix(%id = 18446745872818692814)

(1)

`~`[InertForm:-Display](M, inert = false)

Matrix(%id = 18446745872857841406)

(2)

 

``


 

Download How_to_display_factorial_unevaluated_ac_new1.mw

@acer  Thank you. You are right of course. I did not take into account that initially pairs are specified as sets, and when we get rid of  x  and   sets remain, and Maple automatically sorts them in ascending order, for example  {x=-12, y=-23} => {-23,-12} . I just thought that for a beginner, the approach using  seq  and  rhs is conceptually easier.

@tomleslie  Thanks. I didn't know that Maple accepts systems written in vector form. The help (Maple 2018) does not say about this.

@jalal  Nice! Only the shape of the heart needs to be slightly stretched horizontally.

@jalal 

Anim4 := animate(textplot, [[op(convert(Position(t)+(1.3/4)*Vitesse(t), list)), `#mover(mi("V"),mo("&rarr;"))`], align = [right, below], font = [times, bold, 18]], t = t0 .. t1, frames = 100):

 

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