## 20269 Reputation

15 years, 348 days

## It is there...

@vv  The volume of this tetrahedron is 21 (not 210) and it is on my list.

## ......

@mmcdara  I use Maple 2018.2

## Similar situation...

Here's another example with similar behavior:

```add(1,2,3); # Error
S:=1,2,3;

## diff, maximize...

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

## phaseamp...

@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)));
```

## Update...

@Randy233  See the update to my solution above.

## The same...

@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.

## Re...

@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);
```

## Re...

@AHSAN  See update to my answer.

## plot3d vs plots:-spacecurve...

@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]));
```

## Raising...

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

## OK...

@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.

## Re...

@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.

## Prefix notation...

Use the prefix notation for multiplication:

 (1)

 (2)