Preben Alsholm

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19 years, 313 days

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These are replies submitted by Preben Alsholm

@Carl Love In Maple 2023.2 it appears that the default in this case is adaptive=geometric.

@Carl Love I get:

With adaptive=false I get a much better result (not so with adaptive=geometric or adaptive=true):
This is what I get with adaptive = false:

@Carl Love Just a note: Your procedure doesn't handle Rouben Rostamian's example:

Include0View(plot(10 + 1/(x-1)^2, x=0..2, thickness= 3));

 

@Rouben Rostamian  My intention was to see if it is possible to include an option includeaxis1.
Erik Vestergaard seem to be looking for that kind of thing.
Should MapleSoft decide that an option like that makes sense, they obviously don't need my help!

The above isn't meant as a workaround.  I gave such one earlier.
# Incidentally I just simplified my code above, so that plot errors come directly from plot.

@Rouben Rostamian  You are right. This option must be restricted to bounded functions.
It could just ignore the option in this case.
Here is what must be a feeble attempt:

plot2:=proc(f::{algebraic,procedure},r::{range,name=range(realcons)},{includeaxis1::truefalse:=false}) local VW,p; 
      if not includeaxis1 then return plot(_passed) end if;
      p:=plot(f,r,_rest);
      VW:=min(0,Optimization:-Minimize(f,r)[1])..max(0,Optimization:-Maximize(f,r)[1]);
      if not VW::range(realcons) or hastype(VW,'infinity') then
         printf(cat("A vertical range could not be determined.","\n")); 
         printf(cat("Using includeaxis1= false instead.","\n"));        
         return p
      else 
         plots:-display(p,'view' = VW)
      end if
end proc:
plot2(sin(x),x=0..Pi,color=blue);
plot2(sin(x)+2,x=0..Pi,includeaxis1,color=blue);
plot2(10 + 1/(x-1)^2, x=0..2,includeaxis1,color=blue);
plot2(-10 - 1/(x-1)^2, x=0..2,includeaxis1);
plot2(x->6+sin(x)*x,0..2*Pi,includeaxis1,color=blue);
plot2(x->-10 - 1/(x-1)^2, 0..2,includeaxis1);
plot2(sin(x),0..2*Pi,includeaxis1); # error check
plot2(sin,x=0..2*Pi,includeaxis1);  # error check

# Somewhat revised: includes the procedural version.

@Carl Love Thanks Carl. I wasn't aware of that. I tried plotting sin(x) on -Pi..Pi with thickness=0.05.
First I thought there wasn't anything to see, it's that thin.
With a white background and color = white you obviously can't see anything regardless of thickness.

@erik10 An option to plot, something like 'includeaxis1' ::truefalse with default false.
To show that it should be possible I played a little below.
I prefer Optimization:-Maximize and Minimize to maximize and minimize. The former are more reliable.

restart;
P:=%plot(10 + sin(x), x=3*Pi..4*Pi,view=_VW);
f,r:=op(1..2,P);
VW:=min(0,Optimization:-Minimize(f,r)[1])..max(0,Optimization:-Maximize(f,r)[1]);
value(subs(_VW=VW,P));
############
P:=%plot(-10 + sin(x), x=3*Pi..4*Pi,view=_VW);
f,r:=op(1..2,P);
VW:=min(0,Optimization:-Minimize(f,r)[1])..max(0,Optimization:-Maximize(f,r)[1]);
value(subs(_VW=VW,P));
############################
P:=%plot(sin(x), x=3*Pi..4*Pi,view=_VW);
f,r:=op(1..2,P);
VW:=min(0,Optimization:-Minimize(f,r)[1])..max(0,Optimization:-Maximize(f,r)[1]);
value(subs(_VW=VW,P));

This is basically just for fun and certainly not a workaround.
 

@ Thanks, I get your point.
 

Int(exp(I*m*theta)*exp(-I*n*theta),theta=0..2*Pi)/(2*Pi);
F:=factor(combine(%));

value(F) assuming m=n; # 1
value(F) assuming posint; # 0,  The generic case

 

@Ronan Notice the underscore before rest. That is a clear signal that it means something to Maple.
 

p:=proc(x::algebraic) [x,_rest] end proc;
p(2);
p(2,4,fff=87);

Inside the procedure it stands for possible extra input to p that is not mentioned as a parameter in p.

We can define the otherwise neutral operator &sublist as an actual operator:
 

`&sublist` := proc(N::list,lis::list) 
    local find_element:=
     proc(e,L::list)
       local idx:=ListTools:-Search(e,L);
       if idx=0 then
         return [];
       elif idx=numelems(L) then
         return [L[idx]];
       else
         return L[idx+1..-1]
       end if
     end proc;
     local final_result::truefalse:=true;
     local item,r::list:=lis;
     for item in N do 
       r:=find_element(item,r);
       if numelems(r)=0 then
         final_result:=false;
         break;
       end if;
      end do;
   final_result;
end proc:

Then examples:
 

[7,2,4] &sublist [2,6,5,7,3,2,9,4];
[7,2,4] &sublist [2,6,5,7,3,2,9,4];
[7,2,4] &sublist [2,1,4,2,6,7,3,9];
[7,2,4] &sublist [7,4,2,7,2,4];
[7,2,4] &sublist [7,7,7,7,2,2,2,7,4];

 

@Kitonum Using nm's code I tried overloading subset to deal with this concept of sublist.
The idea is taking from the help page for overload. See the example "SetOperations".
 

ListOperations := module() option package;  export `subset`; 
   local find_element:=
     proc(e,L::list)
       local idx:=ListTools:-Search(e,L);
       if idx=0 then
         return [];
       elif idx=numelems(L) then
         return [L[idx]];
       else
         return L[idx+1..-1]
       end if
     end proc;
#########
    `subset` := proc(N::list,lis::list) option overload;
        local final_result::truefalse:=true;
        local item,r::list:=lis;
     for item in N do 
       r:=find_element(item,r);
       if numelems(r)=0 then
         final_result:=false;
         break;
       end if;
   end do;
   final_result;
end proc:
end module:

Here are examples:
 

with(ListOperations);
[7,2,4] subset [2,6,5,7,3,2,9,4];
[7,2,4] subset [2,1,4,2,6,7,3,9];
[7,2,4] subset [7,4,2,7,2,4];
[7,2,4] subset [7,7,7,7,2,2,2,7,4];
################## Now actual sets:
{7,2,4} subset {2,6,5,7,3,2,9,4};
{7,2,4} subset {2,1,4,2,6,7,3,9};
{7,2,4} subset {7,4,2,7,2,4};
{7,2,4} subset {7,7,7,7,2,2,2,7,4};
{a,b,c} subset {d,e,f,t};
{a,b,c} subset {d,e,b,f,c,t,a};

 

It is quite noticeable.

@mmcdara Yes, I do that routinely.
E.g. in a procedure using results from a dsolve/numeric/parameters defined outside the procedure.
 

Note 3 to my answer:
You can handle 0.005..0.995 if you split it:
 

res1:=evalf(Int(z(q1, q2), q1=0.1..0.9, q2=0.1..0.9));
  res2:=evalf(Int(z(q1, q2), q1=0.005...0.1, q2=0.005...0.1))+
        evalf(Int(z(q1, q2), q1=0.9...0.995, q2=0.9...0.995));
   res1+res2

J2() returns 1.742509218.

@mmcdara Thanks for clarifying the situation!

You are posting an image. That would require any responder to type those 6 equations himself.
Give us text we can copy and paste into Maple.
The syntax doesn't have to be perfect. I see one syntax error right away: You cannot use square brackets as if they were parentheses.

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