## how to dsolve for x(t) in this case?...

in the steps below, it is not fluent to do, and appear diff(1,t)

KineticEnergy := 1/2*m*diff(x(t), t)^2;
PotentialEnergy := subs(x=x(t),int((1/R^2)^2,x));
Action := KineticEnergy - PotentialEnergy;
AA := diff(Action,x(t)) - diff(diff(Action, diff(x(t),t)),t) = 0 <-------- Dsolve this
AA := eval(subs(diff(1,t)=0,diff(Action,x(t))) - Diff(subs(p=Diff(x(t),t),diff(subs(Diff(x(t),t)=p, Action), p)),t)) = 0
dsolve(AA, x(t));

Where R is constant

## how to dsolve for this equation?...

this equation is complicated

how to dsolve for this equation for function f ?

f(t,x,diff(x,t)) - f(t,x,p) - (diff(x,t)-p)*diff(f(t,x,p), p) = tan(t)

## how to solve this system?...

updated:
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));

restart:
with(Groebner):
p1 := vector([a*cos(q1), a*sin(q1)]);
p2 := evalm(p1 + b*vector([cos(q1+q2), sin(q1+q2)]));
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));
Pe := map(expand, P);
A := {cos(q1) = c1, sin(q1) =s1, cos(q2)=c2, sin(q2)=s2, cos(q3)=c3, sin(q3)=s3};
P := subs(A, op(Pe));
F1 := [x - P[1], y - P[2], s1^2+c1^2-1, s2^2+c2^2-1, s3^2+c3^2-1 ];
F2 := subs({a=1, b=1, c=1}, F1);

g2 := Basis(F2, plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[1], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[2], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[3], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[4], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[5], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[6], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[7], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[8], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[9], plex(c3, s3, c2, s2, c1, s1));

1, c1
2       2    2   2
16 y  + 16 x , s1  s2
2
8 x, c1 s2
2      2    2
2 y  + 2 x , s1  c2
2 x, c1 c2
3            2
2 x  - 2 x + 2 y  x, s2 c2
2
1, c2
2 x, s3
2, c3
originally i think
g2[1], g2[7], g2[9] have single variables c1, c2, c3 respectively
can be used to solve system

but without x and y, these equations can not be used
if choose leading term has x and y , but there is no single variable s1 or c1.

originally expect solve as follows
g2spec := subs({x=1, y=1/2}, [g2[3],g2[5],g2[6]]);
S1 := [solve([g2spec[1]])];
q1a := evalf(arccos(S1[1]));
q1b := evalf(arccos(S1[2]));
S2 := [solve(subs(s1=S1[1], g2spec[2])), solve(subs(s1=S1[2], g2spec[2])) ];
q2a := evalf(arccos(S2[1]));
q2b := evalf(arccos(S2[2]));
S3 := [solve(subs(s1=S2[1], g2spec[2])), solve(subs(s1=S2[2], g2spec[2])) ];
q2a := evalf(arccos(S3[1]));
q2b := evalf(arccos(S3[2]));

## Error, (in limit/dosubs) invalid input: `limit/dos...

f := -ln(-1-ln(exp(x)))+ln(-ln(exp(x)))-Ei(1, -1-ln(exp(x)))+Ei(1, -ln(exp(x)))
solve(limit(diff((subs(x=q, f)-f),h), h=0) = f, q);
limit(diff((subs(x=x*h, f)-f),h), h=0);
Error, (in limit/dosubs) invalid input: `limit/dosubs` uses a 3rd argument, newx, which is missing

guess an operator called Lee, Lee(f, x) = f

solve(limit(diff((subs(x=q, f)-f),h), h=0) = f, q);

suspect q = x*h or q=x*f

limit(diff((subs(x=x*h, f)-f),h), h=0);
Error, (in limit/dosubs) invalid input: `limit/dosubs` uses a 3rd argument, newx, which is missing

limit(diff((subs(x=f*h, f)-f),h), h=0);
Error, (in depends/internal) invalid input: `depends/internal` uses a 2nd argument, x, which is missing

## how to solve this system?...

sys1:=-.736349402144656384 = -1.332282598*10^12*(-.99999999999999966)^po1-1.332282598*10^12*(-.99999999999999966)^po2-.735533633151605248*Resid;

sys2:=.326676717828940144 = 1.331567176*10^12*(-.99999999999999966)^po1+1.331567176*10^12*(-.99999999999999966)^po2+.325144093024965720*Resid;

sys3:=.590327283775080036 = -1.072184073*10^9*(-.99999999999999966)^po1-1.072184073*10^9*(-.99999999999999966)^po2+.589610307487437146*Resid;

Minimize(sys1, {sys2,sys3},assume = nonnegative);

complex value encountered;

## how to calculate basis for this basis...

how to calculate basis <1,4,0>, <1,0,4> for eigenvalue 2;

how to calculate basis <1,0,1> for eigenvalue -1;

with(LinearAlgebra):
A := Matrix([[-2,1,1],[0,2,0],[-4,1,3]]);

sys1 := Eigenvalues(A)[1]*IdentityMatrix(3)-A;

sys1 := Eigenvalues(A)[2]*IdentityMatrix(3)-A;
sys1 := Eigenvalues(A)[3]*IdentityMatrix(3)-A;

B:=[<sys1[1,1],sys1[2,1],sys1[3,1]>,<sys1[1,2],sys1[2,2],sys1[3,2]>,<sys1[1,3],sys1[2,3],sys1[3,3]>,<0,0,0>];
LinearAlgebra:-Basis(B);

but not <1,4,0>, <1,0,4> for eigenvalue 2

## invalid input: LinearAlgebra:-Basis expects its 1s...

invalid input: LinearAlgebra:-Basis expects its 1st argument, V, to be of type {Vector, set(Vector), list(Vector)

A:=<<5,5,5>|<1,2,3>|<-5,1,2>>;
Basis(A);

## what are the equivalent maple command in python?...

1.op(0,Expr) , op(1,Expr)

2. indets(eq1,{string,name})

3. type(varlist[ii], function)

## how to find back the input if slope is this?...

sph := <R*cos(u)*cos(v)|R*sin(u)*cos(v)|R*sin(v)>;
GK(sph); #Gauss Curvature
MK(sph); #Mean Curvature

how to find sph if slope is tan(u) ?

## how to guess the variables?...

A. how to find xx1,xx2,xx3,yy1,yy2,yy3 that
Determinant(Matrix([[xx1,yy1,1],[xx2,yy2,1],[xx3,yy3,1]])) =(1/2)*aa*d*s*u+(1/2)*aa*d*s*a*t+(1/2)*d*v*u*t+(1/4)*d*v*a*t^2;

B. how to find x1,x2,x3,x4,y1,y2,y3,y4 that expand(
(x2 - x1)*(y4 - y3) - (y2 - y1)*(x4 - x3)) = (1/2)*d*s*aa*v+(1/2)*d*aa*v*u*t+(1/4)*d*aa*v*a*t^2+(1/2)*aa*d*s*u+(1/2)*aa*d*s*a*t+(1/2)*d*u^2*t+(3/4)*d*u*a*t^2+(1/4)*d*a^2*t^3;

## can this be a solution set?...

v=u+at                      (1)
s=u*t+1/2*a*t^2        (2)

below 3 equations, can substitute  (1)  into it to form (2)
s=1/2*(u+v)*t       (3)
v^2=u^2+2*a*s    (4)
s=v*t-1/2*a*t^2    (5)

can these 5 equations be considered as a solution set of solve function?

or

is only first 2 equations be a solution set?

if so, number of equations less than 5 variables, is there something missing?

## how to simplify this logic?...

would like to return K map of P1

Summation expression for logic only consider 1 but how about wildcard x ?

if consider wildcard x as 1 too, then will use below

source = [[0,0,1,0],[0,0,1,1],[0,1,0,1],[0,1,1,0],[0,1,1,1],[1,0,0,0],[1,0,0,1],[1,0,1,0],[1,0,1,1],[1,1,0,0],[1,1,0,1],[1,1,1,0],[1,1,1,1]];
i use Quine Mccluskey algorithm

got result below

wildcard is 5 or x
[[0, 5, 1, 5], [5, 0, 1, 5], [5, 5, 1, 0], [1, 0, 5, 5], [1, 5, 0, 5], [1, 5, 5, 0], [5, 5, 1, 1], [5, 1, 5, 1], [5, 1, 1, 5], [1, 5, 5, 1], [1, 5, 1, 5], [1, 1, 5, 5]]
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB

table1 = [[0,0,0,0],
[0,0,0,1],
[0,0,1,0],
[0,0,1,1],
[0,1,0,0],
[0,1,0,1],
[0,1,1,0],
[0,1,1,1],
[1,0,0,0],
[1,0,0,1],
[1,0,1,0],
[1,0,1,1],
[1,1,0,0],
[1,1,0,1],
[1,1,1,0],
[1,1,1,1]];

loand(lonot(tt[0]),tt[2])
loand(lonot(tt[1]),tt[2])
loand(lonot(tt[3]),tt[2])
loand(lonot(tt[1]),tt[0])
loand(lonot(tt[2]),tt[0])
loand(lonot(tt[3]),tt[0])
loand(tt[2],tt[3])
loand(tt[1],tt[3])
loand(tt[1],tt[2])
loand(tt[0],tt[3])
loand(tt[0],tt[2])
loand(tt[0],tt[1])

loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
def lonot(z):
if z == 1:
return 0
else:
return 1
def loand(a, b):
if a == 1 and b == 1:
return 1
else:
return 0
def loor(a, b):
if a == 0 and b == 0:
return 0
else:
return 1
#A'C + B'C + AB' + CD + A'BD
for tt in table1:
print loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));

finally i use python to verify
return
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1

seems correct if wildcard is 1 too, but
can boolean simplify function simplify this
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB

to

C + A + B.D   which is
P1 = D + Q0 + Q1.N in png file ?

## how to compute the most simplified result with sol...

solve(diff(-1/x,x) = (-1/x)^(b), b);

originally is 2, but it use ln(....) to express

if start from substitute, it seems need to replace manually.

solve(subs(a(x)=-1/x,diff(a(x),x) = (a(x))^(b)), b);

goal is to find b in equation below
solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)^(b), b);
(2*x+1)/(-1+x)^2-(2*(x^2+x+1))/(-1+x)^3 = ((x^2+x+1)/(-1+x)^2)^(b)

solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)*(b), b);

## is it possible to evaluate or how to evaluate this...

updated

after refer from

https://en.wikipedia.org/wiki/List_of_representations_of_e

exponential1 := sum((1/n!), n=0..infinity);
exponential1 is not a decimal number, it is exp(1)

hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x)), x=0..infinity);

hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x)), x=0..infinity);

how to evalute hoyeung1 or hoyeung2 as a decimal number?

how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc:

but i do not know whether sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x))*m^x, x=0..infinity) = hoyeung^x

can limit(1+(Int(exp(LambertW(1/(-1+x))*(-1+x)), x)))^x, x=infinity) = hoyeung^1 ?