## 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 ?

## how to dsolve this case and evaluate the solution?...

sol := dsolve(diff(ln(y(x)),x) = y(x)^(1/(1-y(x))), y(x));
x-Intat(_a^(-(-2+_a)/(-1+_a)), _a = y(x))+_C1 = 0

the solution is not y(x) = , but y(x) at the right hand side

## how to find back the term in summation?...

Lee := (-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
sum(unknown, n=1..infinity) = Lee

how to find unknown?

## how to complex plot this function?...

complexpoint run a long time
there is no option numpoints in complexplot, how to fasten it?

Lee := (-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
complexplot(Lee, x = 0 .. 1);
Lee := Re(-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
plot(Lee, x = 0 .. 2, numpoints = 5);
Lee := Im(-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
plot(Lee, x = 0 .. 2, numpoints = 5);

## Is there geometric or statistical meaning for ln(d...

Is there geometric or statistical meaning for ln(dy/dx) = 0?

is there any feature in vector field plot when ln(dy/dx) = 0?

## how to evaluate or how to use these functions?...

Int(exp(LambertW(1/(-1+x))*(-1+x)), x)+1

x-Intat(1/exp((-1+_a)*LambertW(1/(-1+_a))), _a = y(x))-_C1 = 0

i use dsolve two equations, get two possible results,
how to evaluate these functions or how to use these functions?

## error when using variable...

mas := proc(f)
return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(x),x) = f))), x\$2));
end proc:
mas(exp(x));
mas(mas(exp(x)));
mas(x^2);
mas(x^2+x^3);

when i hard code x, there is no problem in above code.
but when i op to get variable x and run below, it do not have problem when run line by line, but it has problem when run in
procedure
Error, (in mas) invalid input: diff received exp(x), which is not valid for its 2nd argument

mas := proc(f)
local martin:
martin := op(f):
return ln(diff(rhs(subs(_C1=0,dsolve(diff(y(martin),martin) = f))), martin\$2));
end proc:

mas(exp(x));
mas(mas(exp(x)));
mas(x^2);
mas(x^2+x^3);

## How to find this function?...

F(exp(t)) = t

F(F(exp(t))) = 0

what is F ?

is it diff(ln(x),t) ?

## How to draw this graph?...

A system of algebraic equation

in terms of x, y, z

how draw 3 different circles to show the range of possible values for x, y and z respectively?

it may not be a circle

It may be 3 bounded area graph to show the range of x , y , z respectively

updated

like the graph in many examples in

algebraic and geometric ideas in the theory of discrete optimization

bound area have color