vv

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These are answers submitted by vv

The best method is probably using dsolve + fsolve, because fsolve + int(..., numeric) is too slow.

restart:
Digits:=15:
#eqList := ...
FS:=proc(a,f,t) 
    local u, ds, F;
    ds:=dsolve({diff(u(t),t) = f, u(-4)=0}, numeric, output=operator);
    F:=eval(u,ds);
    fsolve(F-a)
end proc: 
G:=e -> FS(op(1,e), op([2,1],e), op([2,2,1], e)):
map(G, eqList);

 [-4.00000000000000, -3.13995448471848, -2.11370056919619, 
  -1.52883183230736, -1.04840878317075, -0.706638188721254,
   -0.479577206139816, -0.315169354360080, -0.182432574734532, 
   -0.0625487819911634, 0.0891458203730475, 0.368690494498621, 
   1.50469145760957, 2.11552551065259, 2.46119669196488, 
   2.72088546517742, 2.96213514705916, 3.21307491249018, 
   3.59812548343646, 5.16349647450855, 6.56002211630892, 
   6.81413891022544, 6.66825093298453, 6.40577992538593, 
   6.19352683224088, 6.02201310955201, 5.95433562651451, 
   6.06761314123142, 6.32677648880848]

nops(eqList)=nops(%);
                            29 = 29

restart
L := [3*a0 + 2*a1, -a1/3 + a0/2 + 12*a3]:
x:=a3:

solve(select(has, L, x), x);

 

 

ex := x^4 + 3*(1+x^2)*f(x) + (x*x^2+x+1)*((D)(f)(x))^2 + (3*x+3)*(D@D)(f)(x):
eval( ex, f = (x -> (a1+a2*x+c*x^2+O(x^3)) ) ): 
series(%, x);
 #                   a2^2  + 3 a1 + 6 c + O(x)

Next time please post text, not pictures.

You have a polynomial function f in 2 variables and total degree 2, so
f(x,y) = a*x^2+b*y^2+c*x*y+p*x+q*y+r

You want that both partial functions f(., y0),  f(x0, .)   be concave [strictly].
This happens if and only if  a<0 and b<0.

Please note that f is concave (in both variables) iff a<0, b<0 and 4*a*b - c^2 > 0.

You post so many almost identical questions.
Here, the answer of your question why simplify does not work on this example? - MaplePrimes
applies ad litteram (but for x>0 now).

You have syntax erors. Use:

s := Maximize(TRC(tau1, lambda), C3 union C5, tau1 = 0 .. 1, lambda = 0 .. 1, assume = nonnegative);

 

What polynomials are congruent to u(x) modulo  p(x).

Answer: the equivalent class   < rem(u(x), p(x), x) > 
i.e.  rem(u(x), p(x), x) + A(x)*p(x),  where A(x) is an arbitrary polynomial,

Your example: 

rem(x^2+x+1, x+1, x) + A(x)*(x+1);  # for A(x)=x ==>  x^2+x+1

                               1 + A(x)*(x + 1)

 

In the last loop:
-  Insert a space between from and 1
and
-  replace sum with add.

Li is in the sense of Cauchy principal value, so, you must use:

int(1/ln(x),x=0..X, CPV);
#                           Ei(ln(X))

convert(%, Li);
#                             Li(X)

 

Just use := for assignments, instead of =.
So, 

n1 := 423; x := 16;   etc

 

Use

map(eval~, M);

 

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);

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

Example

restart;

F:=[x^2+y^2+z^2, x*y+y*z+z*x, x*y*z-1];

[x^2+y^2+z^2, x*y+x*z+y*z, x*y*z-1]

(1)

G,A := Groebner:-Basis(F, tdeg(x,y,z), output=extended);

[x*y+x*z+y*z, x^2+y^2+z^2, x*z^2+y*z^2+1, y^3+y^2*z+y*z^2+z^3-1, z^4-x-y-2*z, y^2*z^2+y+z], [[0, 1, 0], [1, 0, 0], [0, z, -1], [y+z, -x, 1], [z^2, -x*z-y*z, x+2*z+y], [0, y*z, -y-z]]

(2)

nops(G);

6

(3)

Matrix(A) . <F> = <G>: simplify(%);

(Vector(6, {(1) = (y+z)*x+y*z, (2) = x^2+y^2+z^2, (3) = 1+(x+y)*z^2, (4) = y^3+y^2*z+y*z^2+z^3-1, (5) = z^4-x-y-2*z, (6) = y^2*z^2+y+z})) = (Vector(6, {(1) = (y+z)*x+y*z, (2) = x^2+y^2+z^2, (3) = 1+(x+y)*z^2, (4) = y^3+y^2*z+y*z^2+z^3-1, (5) = z^4-x-y-2*z, (6) = y^2*z^2+y+z}))

(4)

#############

for i to nops(F) do
  Groebner[NormalForm](F[i], G, tdeg(x,y,z), R[i]);
od;

0

 

0

 

0

(5)

B:=Matrix([seq(R[i], i=1..nops(F))]):
B . <G> = <F>:  simplify(%);

(Vector(3, {(1) = x^2+y^2+z^2, (2) = (y+z)*x+y*z, (3) = x*y*z-1})) = (Vector(3, {(1) = x^2+y^2+z^2, (2) = (y+z)*x+y*z, (3) = x*y*z-1}))

(6)

 


Download groeb1vv.mw

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).

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