Maple 12 Questions and Posts

how to guess the variables?...

Asked by: LeeHoYeung 535

July 09 2017

0 2

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?...

Asked by: LeeHoYeung 535

July 09 2017

0 2

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?...

Asked by: LeeHoYeung 535

July 03 2017

0 3

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 ?

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

Asked by: LeeHoYeung 535

June 22 2017

1 1

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...

Asked by: LeeHoYeung 535

June 17 2017

0 5

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?...

Asked by: LeeHoYeung 535

June 17 2017

0 2

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?...

Asked by: LeeHoYeung 535

June 17 2017

0 3

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?...

Asked by: LeeHoYeung 535

June 17 2017

0 2

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...

Asked by: LeeHoYeung 535

June 16 2017

0 1

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?...

Asked by: LeeHoYeung 535

June 16 2017

0 5

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...

Asked by: LeeHoYeung 535

June 16 2017

0 3

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?...

Asked by: LeeHoYeung 535

June 15 2017

0 6

F(exp(t)) = t

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

what is F ?

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

How to draw this graph?...

Asked by: LeeHoYeung 535

June 15 2017

0 0

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?...

Asked by: LeeHoYeung 535

June 10 2017

0 0

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) ?

how to buildsym in this case?...

Asked by: LeeHoYeung 535

June 08 2017

0 1

with(DEtools, buildsym, equinv, symtest):
ans := dsolve([eq2,eq3,eq4], Lie);

Error, (in dsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie]

ans := dsolve([eq2+eq3+eq4 = exp(t)], Lie);
Error, (in PDEtools/sdsolve) too many arguments; some or all of the following are wrong: [{a(t), b(t), c(t)}, Lie]

ans := dsolve([eq2,eq3,eq4]);
sym2 := buildsym(ans);
Error, (in buildsym) invalid input: `ODEtools/buildsym` expects its 1st argument, sol, to be of type {algebraic, algebraic = algebraic}, but received [{c(t) = ...}, {b(t) = ...}, {a(t) = ...)}]

PDEtools[declare](a(t), b(t), c(t), prime = t):
symgen(eq2+eq3+eq4=0);
                       a(t) will now be displayed as a
                       b(t) will now be displayed as b
                       c(t) will now be displayed as c
   derivatives with respect to t of functions of one variable will now be

displayed with 'symgen(....)'

update

if it can not do for 3 function a(t),b(t),c(t) system of differential equations

then

i change to use

eq2 := subs(b(t)=a(t),subs(c(t)=a(t),eq2));
eq3 := subs(b(t)=a(t),subs(c(t)=a(t),eq3));
eq4 := subs(b(t)=a(t),subs(c(t)=a(t),eq4));

with(DEtools, buildsym, equinv, symtest):
ans := dsolve(eq2 = 0, Lie);
buildsym(ans[1], a(t));
buildsym(ans[2], a(t));
buildsym(ans[3], a(t));

there are 3 answers, can i use one of it to recover the equation eq2 or eq3 or eq4?

ans := dsolve(eq3=0, Lie);
buildsym(ans[1], a(t));
sym2 := buildsym(ans[2], a(t));
buildsym(ans[3], a(t));

sym := [_xi=rhs(sym2[2]),_eta=rhs(sym2[1])];
ODE := equinv(sym, a(t));
eq3 - ODE;
sym := [_xi=rhs(sym2[1]),_eta=rhs(sym2[2])];
ODE := equinv(sym, a(t));
eq3 - ODE;

but ODE is not equal to original eq3

ans := dsolve(eq4=0, Lie);
buildsym(ans[1], a(t));
buildsym(ans[2], a(t));

ans := dsolve(eq2+eq3+eq4=0, Lie);
sym := buildsym(ans[1], a(t));
ODE := equinv(sym, a(t));
eq2+eq3+eq4 - ODE;

sym := buildsym(ans[2], a(t));
ODE := equinv(sym, a(t));
eq2+eq3+eq4 - ODE;

sym := buildsym(ans[3], a(t));
ODE := equinv(sym, a(t));
simplify(eq2+eq3+eq4 - - ODE);

can not recover the original result

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