## 12915 Reputation

8 years, 356 days

## infinity...

The integral is + infinity for any real A,B.
Just evaluate:

integral assuming B>0;
and
integral assuming B<0;

Or, compute asympt(integrand, p, 1);

## You probably would have preferred...

```  lim   (x ln(x)) =   lim   ln(x) / (1/x) =    lim   (1/x) / (-1/x^2) = 0
x -> 0+             x -> 0+                  x -> 0+
```

## NormalForm extended...

Example

 > restart;
 > F:=[x^2+y^2+z^2, x*y+y*z+z*x, x*y*z-1];
 (1)
 > G,A := Groebner:-Basis(F, tdeg(x,y,z), output=extended);
 (2)
 > nops(G);
 (3)
 > Matrix(A) . = : simplify(%);
 (4)
 > #############
 > for i to nops(F) do   Groebner[NormalForm](F[i], G, tdeg(x,y,z), R[i]); od;
 (5)
 > B:=Matrix([seq(R[i], i=1..nops(F))]): B . = :  simplify(%);
 (6)
 >

## f := (x, y) -> piecewise([x, y] = [0...

```f := (x, y) -> piecewise([x, y] = [0, 0], 0, (x^3*y - x*y^3)/(y^2 + x^2)):
f(0,0);
#                               0

fx0:=limit((f(x,0)-f(0,0))/x, x=0);
#                            fx0 := 0

fy0:=limit((f(0,y)-f(0,0))/y, y=0);
#                            fy0 := 0

limit(( f(x,y)- fx0*x - fy0*y )/(x^2+y^2), [x=0,y=0]);
#                            -1/4 .. 1/4
```

==> f has partial derivatives (both are zero) at (0,0)

but it is not differentiable at (0,0).

## Corrected...

```restart;
phi := (p__1, p__2, q__1, q__2, xi) -> p__1*exp(q__1*xi) - p__2*exp(q__2*xi):
xi:=2:   # <----
for p__1 in [0,1,-1,I,-I] do
for p__2 in [0,1,-1,I,-I] do
for q__1 in [0,1,-1,I,-I] do
for q__2 in [0,1,-1,I,-I] do
result1 := evalf(phi(p__1, p__2, q__1, q__2, xi));
print('phi'(p__1, p__2, q__1, q__2, xi)=result1);
od od od od:
```

## . (dot)...

Don't use the dot (.) operator for usual multiplication; it is reserved for noncommutative multiplication  (usually dot product) . Use * instead.
For example,

simplify(v . ( 2/v + 1));
is not simplified if v is a name.

1. Use
or better:
because DEtools is a table-based package.

2.

ode:=y(x)*(2*x^2*y(x)^3+3)+x*(x^2*y(x)^3-1)*diff(y(x),x)=0;

## floats...

If possible, don't use floats in symbolic computation (here limit)

Compare:

```limit(n*(1/3 - 1/(3+1/n)),n=infinity);
#                               1/9
limit(n*(1/3. - 1/(3+1/n)),n=infinity);
#                        -Float(infinity)
```

So, replace in W:  3.6  with 36/10  etc.

## pdf...

Of course, just use the menu File > Page Setup...
or Ctrl+Shift+P
Then Export or Print to a pdf port.

## General form...

A is a diagonalizable matrix (the minimal polynomial being X^2 - X)

The possible diagonal forms are D1 and D2 below.

 > restart;
 > with(LinearAlgebra):
 > D1, D2 := DiagonalMatrix([0,0,1]), DiagonalMatrix([0,1,1]);
 (1)

The general form for a nonzero A is  T^(-1) . Dk . T,  where T is nonsingular and k in {1,2}.

Example:

 > T:=RandomMatrix(3, generator=rand(-3..3)): A:=T^(-1).D1.T;
 (2)
 > A^2-A;  # Check
 (3)
 >

## _EnvProbabilistic≔0...

 Try _EnvProbabilistic≔0; See:

Probabilistic Algorithm Control - Maple Help (maplesoft.com)

## Finite Differences...

```restart;
FD:=proc(f::procedure, i::nonnegint, h::algebraic)# Finite Difference
if i=0 then return f fi;
x -> ( FD(f,i-1,h)(x+h) - FD(f,i-1,h)(x) )
end proc:

f:= x -> x^5:
FD(f,3,h)(x): simplify(%);
seq( simplify(FD(f,i,1)(x)), i=0..6 );
```

150*h^5+180*h^4*x+60*h^3*x^2

x^5, 5*x^4+10*x^3+10*x^2+5*x+1, 20*x^3+60*x^2+70*x+30, 60*x^2+180*x+150, 120*x+240, 120, 0

## duplicate again...

Denote by p the polynomial.
Solving the equation p=0 wrt sigma and ploting the result
it is easy to conclude that p has a unique positive root in Z for any real sigma.
This root can be obtained by solve symbolically or numerically (for a numeric sigma).

 > p:=6125*Z^4 + 68644*Z^3*sigma - 219625*Z^3 + 255712*Z^2*sigma - 959250*Z^2 + 238144*Z*sigma - 1113500*Z - 245000;
 (1)
 > s:=solve(p,sigma);
 (2)
 > plot(s, Z=0 .. 60);
 > sol:=solve(eval(p, sigma=2), Z, explicit)[1]; # Take e.g.sigma=2
 (3)
 > evalf(sol)
 (4)
 >

## Useless examples...

```restart;