tomleslie

Well........

Answered: tomleslie 13876

December 17 2020

0 1

it depends a bit on how "sophisticated" you want to be!!!

After all, the position of the cursor after an error has been detected, is only a pretty good "guess" as to the location of the error: for example if you failed to terminate a 'for' loop correctly, then the cursor *might* be many lines after the point at which the loop should have been terminated!

The crudest way to "find" the cursor, is simply to realise that its position represents the point where anything you type will be inserted in the worksheet - so just type something obvious like 'TheCursorIsHere" - which you then *ought* to be able to locate just by reading the worksheet
The most sophisticated way to find a problem is to use the debugger and execute the code step-by-step. This may be "overkill" for your problem.
If you are using 1D input (NB this doesn't work with 2D input) then you can use the multiline comment characters (ie (* and *) ) to comment out sections of your code and perform a manual "binary search" to zoom in on where the problem might be. This approach is also pretty cude, but can be helpful in conjunction with (1) above.

Don't use smartplot!...

Answered: tomleslie 13876

December 15 2020

0 0

See the attached, whihc unfortunately won't display inline on this site:-(

(By the way the first equation in the original smartplot set looks a little doubtful compared with the others)

combPlot.mw

Simpler - and works(?)...

Answered: tomleslie 13876

December 15 2020

1 3

I didn't really understand whay you were trying to achieve with the statement

assign(seq(`print/_C`||k, k=0..10) = seq(c[k], k=0..10)):

but since it didn't seem "necessary" to achieve your objective, so I removed it, and edited the substituion statement to

S:=[seq(_C||k = c[k], k=0..10)]:

Both of your examples now appear to provide the required output - see the attached (which won't display inline - I seem to having the same problem CarlLove reproted yesterday)

toLatex.mw

More of an observation...

Answered: tomleslie 13876

December 11 2020

2 4

In your "toy" example

diff(u(x), x, x) = u(x), u(0) = 2, u(1) = 1

why would Maple need to return diff(u(x), x, x) explicitly? Since diff(u(x), x, x) = u(x), all it needs to return is u(x) - since these functions are obviously identical.

dsolve(..,numeric) will never return values for the "highest order derivatives" in an ODE system, because the ODE system itself will allow these to be computed in terms of lower order derivatives (which dsolve(...,numeric) will return).

There are cases where the highest-order derivatives cannot be (easily) isolated and hence computed from the information which dsolve(..,numeric) returns - this is usually where the "trick" of adding a subsidiary equation such as diff(u(x), x, x) = v(x) may be useful.

So just how difficult is your actual problem??

More of an observation...

Answered: tomleslie 13876

December 11 2020

0 0

It really doesn't matter very much whether you use a 'piecewise' or an 'if' statement - both will achieve the same thing.

Instinctively, I'd go with 'piecewise' (like Kitonum), if only because (particularly for more than two conditions) less typing of input will be involved!

See the attached where both methods are given

>

  restart;
  g:= (p, q) -> ((p - 200)^2 + q^2)*0.001/960000:
  f:= (r, s) -> 6.87e-03*(-0.5772 - ln(g(r, s)) + `if`( g(r, s) < 0.001,
                                                          0,
                                                          add(g(r, s)^j/(j*j!), j=1..12)
                                                       )
                         ):
  h:= (r, s) -> 6.87e-03*(-0.5772 - ln(g(r, s)) + piecewise
                                                  ( g(r, s) <= 0.001,
                                                     0 ,
                                                     add(g(r, s)^j/(j*j!), j=1..12)
                                                  )
                         ):
  plot3d( f(x,y), x=-1300..200, y=-1300..200);
  plot3d( h(x,y), x=-1300..200, y=-1300..200);

>

Download 2Cplot.mw

Check...

Answered: tomleslie 13876

December 10 2020

2 0

the attached

Note that the option scaling=constrained is a bad idea whenver the x-range and y-range differ noticeably, Also you should have 'uses plots' within the prpcedure which requires the odeplot command

Download odeplt.mw

my $0.02...

Answered: tomleslie 13876

December 10 2020

0 1

A straightforward double 'seq' seems to be about 75% faster.

Reorganisation of data (the final execution group) gains about a further factor of 2 - and would probably(?) "scale" better for larger problems

>

ans6 := [{alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 8], alpha[3, 6], alpha[3, 9]}, {alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 8], alpha[3, 3], alpha[3, 6]}, {alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 7], alpha[3, 6], alpha[3, 8]}, {alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 7], alpha[3, 3], alpha[3, 8]}, {alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 7], alpha[3, 3], alpha[3, 6]}, {alpha[1, 8], alpha[2, 2], alpha[2, 4], alpha[2, 5], alpha[3, 3], alpha[3, 6]}, {alpha[1, 8], alpha[2, 1], alpha[2, 2], alpha[2, 5], alpha[3, 3], alpha[3, 6]}, {alpha[1, 8], alpha[2, 0], alpha[2, 2], alpha[2, 5], alpha[3, 3], alpha[3, 6]}, {alpha[1, 5], alpha[2, 8], alpha[3, 6], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 5], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 5], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 7], alpha[3, 8]}];
ans7 := [{alpha[1, 8], alpha[2, 2], alpha[2, 5], alpha[2, 7], alpha[3, 3], alpha[3, 6], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 6], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 5], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 5], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 3], alpha[3, 4], alpha[3, 5], alpha[3, 6], alpha[3, 7]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 6], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 5], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 5], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 5], alpha[3, 6], alpha[3, 7], alpha[3, 8]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 4], alpha[3, 7], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 4], alpha[3, 6], alpha[3, 8], alpha[3, 9]}, {alpha[1, 5], alpha[2, 8], alpha[3, 2], alpha[3, 4], alpha[3, 6], alpha[3, 7], alpha[3, 9]}];

(1)

>	SubsetPairs:= (A::list(set), B::list(set))-> map(z-> [ListTools:-SearchAll](true, map(w-> z subset w, B)), A): CodeTools:-Usage(SubsetPairs(ans6, ans7),iterations=1000);

memory used=198.15KiB, alloc change=0 bytes, cpu time=2.23ms, real time=2.09ms, gc time=280.80us

(2)

>

#
# A bit quicker
#
  CodeTools:-Usage( seq
                    ( [ seq
                        ( `if`( ans6[j] subset ans7[i],
                                i,
                                NULL
                              ),
                          i=1..numelems(ans7)
                        )
                      ],
                      j=1..numelems(ans6)
                    ),
                    iterations=1000
                  );

memory used=63.75KiB, alloc change=0 bytes, cpu time=1.17ms, real time=1.17ms, gc time=0ns

(3)

>

#
# Even quicker - and (probably??) scales better
#
  getTab:=proc( L1, L2)
                local i,
                      j,
                      T:=table( [seq( j={}, j in`union`(L1[], L2[]) )]):
                for j from 1 to numelems(L1) do
                    for i in L1[j] do
                        T[i]:=`union`(T[i],{j});
                    od;
                od:
                return eval(T);
          end proc:
  T7:=CodeTools:-Usage(getTab(ans7, ans6),iterations=1000):
  CodeTools:-Usage( seq
                    ( `intersect`
                      ( seq
                        ( T7[j],
                          j in ans6[i]
                        )
                      ),
                     i=1..numelems(ans6)
                   ),
                   iterations=1000
                 );

memory used=72.00KiB, alloc change=0 bytes, cpu time=514.00us, real time=480.00us, gc time=78.00us
memory used=11.19KiB, alloc change=0 bytes, cpu time=94.00us, real time=92.00us, gc time=0ns

(4)

>

Download speedUP.mw

Cannot reproduce...

Answered: tomleslie 13876

December 10 2020

0 0

even in Maple 2015. See the attached

>	restart: interface(version);

(1)

>	Z := n -> op~(2, { allvalues(solve(LegendreP(n,x))) } );

(2)

>	Digits:=10: Z(17): numelems(%);

(3)

>	Z(18): numelems(%);

(4)

>	Digits:=15: Z(18): numelems(%);

(5)

>	Digits:=20: Z(18): numelems(%);

(6)

>	Zf := n -> op~(2, { allvalues(solve(evalf(LegendreP(n,x)))) } ); Z(18): numelems(%);

(7)

>

Download legZero.mw

Increase Digits...

Answered: tomleslie 13876

December 09 2020

1 1

see the attached.

I'm not sure why fsolve() is returning -0.1 as a solution when the function appears to be undefined at this value

Download fsol.mw

My $0.02...

Answered: tomleslie 13876

December 09 2020

2 1

see the attached

>	restart; with(DETools): randomize(): A:= a(A_5)^3-b; B := 3a(A_5)^2-3b/(A_5); C := (c_3)(A_5)-(c_4); u:=3y^2+2x^3B/A+3x^2C/A;

3*y^2+2*x^3*(3*a*A_5^2-3*b/A_5)/(A_5^3*a-b)+3*x^2*(A_5*c_3-c_4)/(A_5^3*a-b)

(1)

>

#
# Set all unknown parameters =10 (why not? they have to be
# set to something!!). Since the above set of parameter
# values is arbitrary probably best to let Maple pick the
# contours
#
  setPars:= indets(u, name) minus {x, y}=~10;
  plots:-contourplot( eval( u,
                            setPars
                          ),
                      x=-2.25..2.25,
                      y=-2.25..2.25,
                      axes=boxed,
                   # contours=[seq(-2.5..-1.3, 0.1)],
                      grid=[80,80],
                      coloring=["blue","red"]
                    );

${A_5 = 10, a = 10, b = 10, c_3 = 10, c_4 = 10}$

>

Download cPlot2.mw

Looks like a "bug"...

Answered: tomleslie 13876

December 08 2020

0 0

which occurred in Maple 2015 and was fixed in Maple 2015 - se the two attachments

Download m2015.mw

Download m2016.mw

Have you tried...

Answered: tomleslie 13876

December 07 2020

0 2

reading the error message which Maple provides - it is accurate and informative.

But since you can't/don't read the error message you are happy to sit around and wait while someone else reads it for you??

The attached shows my conclusions after reading the error message - basically the name 'M' is used in the definition of the system, but is nowhere specified - maybe it is supposed to be the same asa 'm'? Who knows???

(1+K)*(diff(diff(diff(f(eta), eta), eta), eta))+f(eta)*(diff(diff(f(eta), eta), eta))+K*(diff(g(eta), eta))-2*(diff(f(eta), eta))^2-M*(diff(f(eta), eta)) = 0

(1)

eq2 := (1+(1/2)*K)*(diff(g(eta), `$`(eta, 2)))+f(eta)*(diff(g(eta), eta))-3*g(eta)*(diff(f(eta), eta))-2*K*g(eta)-K*(diff(f(eta), `$`(eta, 2))) = 0

(1+(1/2)*K)*(diff(diff(g(eta), eta), eta))+f(eta)*(diff(g(eta), eta))-3*g(eta)*(diff(f(eta), eta))-2*K*g(eta)-K*(diff(diff(f(eta), eta), eta)) = 0

(2)

eq3 := (1+(4/3)*Rd)*(diff(theta(eta), `$`(eta, 2)))/Pr+f(eta)*(diff(theta(eta), eta))+Nb*(diff(chi(eta), eta))*(diff(theta(eta), eta))+Nt*(diff(theta(eta), eta))^2 = 0

(1+(4/3)*Rd)*(diff(diff(theta(eta), eta), eta))/Pr+f(eta)*(diff(theta(eta), eta))+Nb*(diff(chi(eta), eta))*(diff(theta(eta), eta))+Nt*(diff(theta(eta), eta))^2 = 0

(3)

diff(diff(chi(eta), eta), eta)+Sc*f(eta)*(diff(chi(eta), eta))+Nt*(diff(diff(theta(eta), eta), eta))/Nb = 0

(4)

(5)

(6)

>	# # Check the system to determine how many bcs/ics might # be required # `union`(indets~(eval([eq1, eq2, eq3, eq4, bcs], params))[]);

{M, eta, chi(eta), diff(chi(eta), eta), diff(diff(chi(eta), eta), eta), diff(diff(diff(f(eta), eta), eta), eta), diff(diff(f(eta), eta), eta), diff(diff(g(eta), eta), eta), diff(diff(theta(eta), eta), eta), diff(f(eta), eta), diff(g(eta), eta), diff(theta(eta), eta), f(eta), g(eta), theta(eta)}

(7)

>

#
# So from the output of the above command we can conclude
#
# 1. system is third order in f() so three bcs/ics on f() are necessary
# 2. system is second order in chi() so two bcs/ics on chi() are necessary
# 3. system is second order in g() so two bcs/ics on g() are necessary
# 4. system is second order in theta() so two bcs/ics on theta() are necessary
# 5. system contains the unknown name 'M' so one condition to
#    determine its value is necessary (possibly OP thinks that 'm'
#    and 'M' are the same?)
#
# Sum total
#
#    3+2+2+2+1=10 so 10 conditions are necessary
#
# and
#
  numelems([bcs]);
#
# returns 9. So 10 conditions required, only 9 given. This is never
# going to work
#

(8)

>

Download oneMoreCond.mw

As (almost) always...

Answered: tomleslie 13876

December 07 2020

0 0

there is more than one way to do this.

The "best" probably depends on how the created functions are subsequently going to be used.

One alternative is shown in the attached

>

  restart;
#
# Create a simple divide-by-5 function
#
  div5 := z -> 1/5*z;
#
# Create a 1-D array of "similar" functions, which
# depend on the array index
#
  f := Array(1 .. 5, i -> x -> sin(x + i));
#
# Some values of the above
#
  f[1](x);
  f[3](x);
  f[5](1);
#
# Apply the divide-by-5 function to each element of
# of the array 'f'
#
  g := div5~(f);
#
# Evaluate 'g' for varous index values and arguments
#
  g[1](x);
  g[3](x);
  g[5](1);

Array(%id = 36893488148141648820)

(1)

>

Download funcArr.mw

Well........

Answered: tomleslie 13876

December 05 2020

1 8

you have around a dozen parameters, none of whose values are given, so no definitive answer is possible.

The attached sets all of these parameters, other than x, y =1 (an arbirary choice) the plots the result

>

  restart:
  B4:=b1/b3:
  B5:=b4/b3:
  B:=(B4+(3*B5)/B4)/2:
  G2:=sin(b5*t+b6)^(2/3):
  X:=(x+c)/G2:
  Y1:=-exp(-(b7*y^2+b8*y+b9)):
  Y2:=-cot(b5*t+b6)/b5:
  Y:=Y1-Y2:
  G:=2*b5*cot(b5*t+b6):
  U:=(a*(x+c))/3-(2*a*c)/3-(((x+c)^2)*G)/18+(B5*Y+2*m2*tan(c7+m2*(X-B4*Y))+c8)/G2;
#
# Set all the parameters, other than 'x' and 'y' in the
# above expression to be 1 (why not OP doesn't specify!)-
# then plot the result with x,y varying from -20..20
#
# Note that varying numpoints can produce *visually* very
# "different" graphs - presumably because of the existence
# of "singularities" in the expression to be plotted
#
  plot3d( eval(U, (indets( U , name) minus {x,y})=~1),
          x=-20..20,
          y=-10..20,
          numpoints=10000
        );

(1/3)*a*(x+c)-(2/3)*a*c-(1/9)*(x+c)^2*b5*cot(b5*t+b6)+(b4*(-exp(-b7*y^2-b8*y-b9)+cot(b5*t+b6)/b5)/b3+2*m2*tan(c7+m2*(-b1*(-exp(-b7*y^2-b8*y-b9)+cot(b5*t+b6)/b5)/b3+(x+c)/sin(b5*t+b6)^(2/3)))+c8)/sin(b5*t+b6)^(2/3)

>

;

>

Download oddGraph.mw

Hmmm...

Answered: tomleslie 13876

December 05 2020

0 3

You can get a numerical solution with the DirectSearch add-on package, and the residuals seem "reasonable" - although with no knowledge of the overall "scale" of the system/variables, confidence in the obtained solution is not great