@vas9pupkin I should've notced this before.... The command with does not work inside a procedure. So the IsDefinite is not being done at all in your last error. You should spell out the whole name in your procedures:
 LinearAlgebra:-IsDefinite(...). (There are some other options also, like the ?uses declaration.)

@vas9pupkin I should've notced this before.... The command with does not work inside a procedure. So the IsDefinite is not being done at all in your last error. You should spell out the whole name in your procedures:
 LinearAlgebra:-IsDefinite(...). (There are some other options also, like the ?uses declaration.)

There are numerous subtleties involved with using the < > constructors, especially regarding the distinction between row vectors and column vectors. It is understandable that one would prefer not to negotiate those subleties. So here's two other versions:

Matrix([a[x+m..y+m] $ m= 0..2]);

(which only works if m is unassigned) and

Matrix(['a[x+m..y+m]' $ 'm'= 0..2]);

(which works even if m is assigned).

There are numerous subtleties involved with using the < > constructors, especially regarding the distinction between row vectors and column vectors. It is understandable that one would prefer not to negotiate those subleties. So here's two other versions:

Matrix([a[x+m..y+m] $ m= 0..2]);

(which only works if m is unassigned) and

Matrix(['a[x+m..y+m]' $ 'm'= 0..2]);

(which works even if m is assigned).

@williamov Judging by your subsequent question "how to animate a vector", I don't think that it is a success. I think that you made roo into a big mess. You should use lists, not Vectors.

@williamov Judging by your subsequent question "how to animate a vector", I don't think that it is a success. I think that you made roo into a big mess. You should use lists, not Vectors.

What if the roots are sol = [3, 3, 1, 1+2*I, 1-2*I]? What do you want the result to be in that case? In other words, do the 3s need to be distinguished? If the imaginary parts are ignored, does the result still need to be distinguished from the real root 1?

I'd need to see the entire program/worksheet.

Vector(4, {(1) = ` 1 .. 99 `*Vector[column], (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

That's nonsense. Is that the contents of roo?

There's an animation toolbar that appears when you play an animation. Look for the icon labelled FPS. That's Frames Per Second. The lowest that you can set it is 1 FPS.

@Christopher2222 The osculating circle is the circle tangent to a curve which most closely approximates the curve at the point of tangency (among all tangent circles). The curvature is (by definition) the reciprocal of the radius of the osculating circle.

@K5ky:When I replace the equation with a name that points to it, this returns FAIL instead.Any ideas why?

NextZero takes an operator, not an expression, as its first argument.

@K5ky:When I replace the equation with a name that points to it, this returns FAIL instead.Any ideas why?

NextZero takes an operator, not an expression, as its first argument.

@williamov It's difficult to generalize that. It will work if you write a separate top-level assignment statement for each variable. But if you try to do it in a loop or procedure, or iterating over a list of names in a set or list, it gets tricky.

@williamov It's difficult to generalize that. It will work if you write a separate top-level assignment statement for each variable. But if you try to do it in a loop or procedure, or iterating over a list of names in a set or list, it gets tricky.

One reason that the absolute-value form might be preferred over the piecewise form is that the former makes the continuity obvious. One reason that the absolute-value form might be preferred over the list-of-points form is that most symbolic computation commands will not take the latter.

@whenyourestrange Actually, my command above can be simplified slightly: One of the two zero vectors (< 0 | 0 >), it doesn't matter which one, can be removed.