Advanced Search

Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12
View unanswered posts
Maple 12 Questions and Posts Feed

how to simplify this logic?...

Asked by: ComputerUser 535
logic boolean
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: ComputerUser 535
solve
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: ComputerUser 535
sum summation
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: ComputerUser 535
dsolve
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: ComputerUser 535
series summation
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: ComputerUser 535
complexplot
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: ComputerUser 535
differential-geometry
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: ComputerUser 535
dsolve
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: ComputerUser 535
dsolve op indets
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: ComputerUser 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: ComputerUser 535
implicitplot3d
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: ComputerUser 535
pdetools
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: ComputerUser 535
detools
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

how to calculate probability of matrix in this cas...

Asked by: ComputerUser 535
May 11 2017
0  0
i count the number among group
but when the list a large such as over 1000 records, the count will be over 30,000
use which denominator to find probability?
is there any functions in maple for this case?
 
with(LinearAlgebra):
correlationlist1 := [[1,2,3],[1,3,5]....]:
PAB := Matrix(50):
for ii from 1 to nops(correlationlist) do
 for jj from 1 to nops(correlationlist[ii]) do
  for kk from 1 to nops(correlationlist) do
   for qq from 1 to nops(correlationlist[kk]) do
    if ii <> kk then
     #print("scan=",correlationlist2[kk],"kk=",kk,"qq=",qq,"row=",correlationlist[ii][jj],"column=",correlationlist[kk][qq]):
     PAB[correlationlist[ii][jj],correlationlist[kk][qq]] := PAB[correlationlist[ii][jj],correlationlist[kk][qq]] + 1: # group to group relations
    end if:
   od:
  od:
 od:
od:

How to use statistics correlate function with a li...

Asked by: ComputerUser 535
correlation
May 11 2017
0  0

If there is a list

[[1,2],[2,2],[3,3]...

how to use correlate function?

assume [1,2] and [2,1] count as 2 

when find correlation between 1and 2

and how about 

if [1,2] and [2,1] are different

first position as key

means 1 relate with 2 count as 1 case for 1 only

[2,1] count as 1 case for 2 only

 

First 12 13 14 15 16 17 18 Last Page 14 of 32