Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12
read example
 
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) ?
 
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;

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?
 

https://drive.google.com/file/d/0Bxs_ao6uuBDUNmd5ZVJtX29GT3c/view?usp=sharing
https://drive.google.com/file/d/0Bxs_ao6uuBDUSUdHSTcwSEQtS3M/view?usp=sharing

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 ?

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);
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 ?

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

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?
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(dy/dx) = 0?

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

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?
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);

F(exp(t)) = t

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

what is F ?

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

 

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

is it possible to change ODE to PDE?

the ODE has diff(a(t),t) and diff(b(t),t)

how to convert to diff(t, a), diff(t, b) ?

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