Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12

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) ?

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
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:

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

 

How to convert "123" into integer 123

i use sprintf %d still can not convert to integer

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