This question is related to an answer I gave here:
So, please look at a simple worksheet containing only a few lines; the resuts are in the # comments.

restart;
evalf(frac(Pi^20));
#                              23.
restart;
printlevel:=40:
evalf(frac(Pi^20));
  ###  prinlevel stuff
#                              0.

And now the questions.
1. Why the first evalf(frac(Pi^20))  does not  call  `evalf/frac`?
     (the second does, trace(`evalf/frac`)  shows this  if inserted).
     Note that  `evalf/frac`(Pi^20)    returns  0.
2. Why evalf(frac(Pi^20))    depends on printlevel?
    Note that  if  printlevel is changed to 20 (say)  the result is again 23.
3. Why if we set interface(typesetting=standard)  in a fresh session
     the results are both 23?

I can't get a While do loop to work as expected.

For i from 2 while M[i,1]<>M[1,1] and i<25 do..   It doesn't catch row 15 where M[15,1] =M[1,1] but it does stop at i = 24 ok.
 

Download Test_While_do_loop.mw

I'm currently wondering about the cut I'm looking for in the following worksheet.

I evaluate it in 2 ways but get different answers. Any idea what the problem here is?

Thanks

Download CutErrorFunction.mw

restart; with(plots);
[animate, animate3d, animatecurve, arrow, changecoords, 

  complexplot, complexplot3d, conformal, conformal3d, 

  contourplot, contourplot3d, coordplot, coordplot3d, 

  densityplot, display, dualaxisplot, fieldplot, fieldplot3d, 

  gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, 

  interactive, interactiveparams, intersectplot, listcontplot, 

  listcontplot3d, listdensityplot, listplot, listplot3d, 

  loglogplot, logplot, matrixplot, multiple, odeplot, pareto, 

  plotcompare, pointplot, pointplot3d, polarplot, polygonplot, 

  polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, 

  semilogplot, setcolors, setoptions, setoptions3d, spacecurve, 

  sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]

fixedparameter1 := [n = .3, W[e] = .3, M = .2, gamma = 1, delta = -1, N[r] = .8, Pr = .72, Nb = .5, Nt = .5, Bi = 2, Pr = .72, Le = 5];
[n = 0.3, W[e] = 0.3, M = 0.2, gamma = 1, delta = -1, N[r] = 0.8, 

  Pr = 0.72, Nb = 0.5, Nt = 0.5, Bi = 2, Pr = 0.72, Le = 5]

eq1 := (1-n)*(diff(f(eta), eta, eta, eta))+f(eta)*(diff(f(eta), eta, eta))-M*(diff(f(eta), eta))+n*W[e]*(diff(f(eta), eta, eta, eta))*(diff(f(eta), eta, eta)) = 0;
        /  d   /  d   /  d         \\\
(1 - n) |----- |----- |----- f(eta)|||
        \ deta \ deta \ deta       ///

            /  d   /  d         \\     /  d         \
   + f(eta) |----- |----- f(eta)|| - M |----- f(eta)|
            \ deta \ deta       //     \ deta       /

            /  d   /  d   /  d         \\\ /  d   /  d         \\   
   + n W[e] |----- |----- |----- f(eta)||| |----- |----- f(eta)|| = 
            \ deta \ deta \ deta       /// \ deta \ deta       //   

  0
deq1; eval(eq1, fixedparameter1);
    /  d   /  d   /  d         \\\
0.7 |----- |----- |----- f(eta)|||
    \ deta \ deta \ deta       ///

            /  d   /  d         \\       /  d         \
   + f(eta) |----- |----- f(eta)|| - 0.2 |----- f(eta)|
            \ deta \ deta       //       \ deta       /

          /  d   /  d   /  d         \\\ /  d   /  d         \\   
   + 0.09 |----- |----- |----- f(eta)||| |----- |----- f(eta)|| = 
          \ deta \ deta \ deta       /// \ deta \ deta       //   

  0
eq2 := (1+(4/3)*N[r])*(diff(theta(eta), eta, eta))+Pr*f(eta)*(diff(theta(eta), eta))+Nb*(diff(phi(eta), eta))*(diff(theta(eta), eta))+Nt*(diff(theta(eta), eta))*(diff(theta(eta), eta)) = 0;
          /    4     \ /  d   /  d             \\
          |1 + - N[r]| |----- |----- theta(eta)||
          \    3     / \ deta \ deta           //

                         /  d             \
             + Pr f(eta) |----- theta(eta)|
                         \ deta           /

                  /  d           \ /  d             \
             + Nb |----- phi(eta)| |----- theta(eta)|
                  \ deta         / \ deta           /

                                    2    
                  /  d             \     
             + Nt |----- theta(eta)|  = 0
                  \ deta           /     
deq2; eval(eq2, fixedparameter1);
                      /  d   /  d             \\
          2.066666667 |----- |----- theta(eta)||
                      \ deta \ deta           //

                           /  d             \
             + 0.72 f(eta) |----- theta(eta)|
                           \ deta           /

                   /  d           \ /  d             \
             + 0.5 |----- phi(eta)| |----- theta(eta)|
                   \ deta         / \ deta           /

                                     2    
                   /  d             \     
             + 0.5 |----- theta(eta)|  = 0
                   \ deta           /     
eq3 := diff(phi(eta), eta, eta)+Pr*Le*f(eta)*(diff(phi(eta), eta))+Nt*(diff(theta(eta), eta, eta))/Nb = 0;
    /  d   /  d           \\                /  d           \
    |----- |----- phi(eta)|| + Pr Le f(eta) |----- phi(eta)|
    \ deta \ deta         //                \ deta         /

            /  d   /  d             \\    
         Nt |----- |----- theta(eta)||    
            \ deta \ deta           //    
       + ----------------------------- = 0
                      Nb                  
deq3 := eval(eq3, fixedparameter1);
    /  d   /  d           \\               /  d           \
    |----- |----- phi(eta)|| + 3.60 f(eta) |----- phi(eta)|
    \ deta \ deta         //               \ deta         /

                     /  d   /  d             \\    
       + 1.000000000 |----- |----- theta(eta)|| = 0
                     \ deta \ deta           //    
bcs1 := f(0) = 0, D(f)(0) = 1+gamma*(D@D)(F)(0)+delta*(D@D@D)(f)(0), D(f)(8) = 0;
 f(0) = 0, 

   D(f)(0) = 1 + gamma @@(D, 2)(F)(0) + delta @@(D, 3)(f)(0), 

   D(f)(8) = 0
bc1 := eval(bcs1, fixedparameter1);
   f(0) = 0, D(f)(0) = 1 + @@(D, 2)(F)(0) - @@(D, 3)(f)(0), 

     D(f)(8) = 0
bcs2 := D(theta)(0) = Bi*(theta(0)-1), theta(8) = 0;
         D(theta)(0) = Bi (theta(0) - 1), theta(8) = 0
bc2 := eval(bcs2, fixedparameter1);
           D(theta)(0) = 2 theta(0) - 2, theta(8) = 0
bcs3 := Nb*D(phi)(0)+Nt*D(theta)(0) = 0, Nb*D(phi)(0)+Nt*D(theta)(0) = 0, phi(8) = 0;
        Nb D(phi)(0) + Nt D(theta)(0) = 0, 

          Nb D(phi)(0) + Nt D(theta)(0) = 0, phi(8) = 0
bc3 := eval(bcs3, fixedparameter1);
       0.5 D(phi)(0) + 0.5 D(theta)(0) = 0, 

         0.5 D(phi)(0) + 0.5 D(theta)(0) = 0, phi(8) = 0
R := dsolve({bc1, bc2, bc3, deq1, deq2, deq3}, [f(eta), theta(eta), phi(eta)], numeric, output = listprocedure);
Error, (in dsolve/numeric/process_input) system must be entered as a set/list of expressions/equations

Hello! I am trying to make an if statement that is IF a bound is not equal to NULL, it does things, and if it IS equal to NULL, the bounds are set to zero. When a bound is null, they say 

bound1:=()

My first if statement will not work, please help!

bound1:=solve(tau(x)=(Intv||j)[1],x,useassumptions) assuming (Intv||i)[1]<=x<=(Intv||i)[2] ;  

bound2:=solve(tau(x)=(Intv||j)[2],x,useassumptions) assuming (Intv||i)[1]<=x<=(Intv||i)[2];

if bound1<>NULL;bound2<>NULL;  then

if bound1<=bound2   then  

lower:=bound1;  upper:=bound2  

else lower:=bound2;   upper:=bound1 end if;

else lower:=0; upper:=0 end if;

Let be given tetrahedron ABCD, where AB = BC = AC = a, AD = d, AD = e, CD = f. I know that, If the measure of angle of AB and CD equal to Pi/3, then we have d^2 - e^2 - a*f = 0. I tried:
ListTools[Categorize];
L := []; 
for a to 30 do for d to 30 do
for e to 30 do for f to 30 do
if abs(d-e) < a and a < d+e and abs(a-e) < d and d < a+e and abs(d-a) < e and e < d+a and abs(d-f) < a and a < d+f and abs(a-f) < d and d < a+f and abs(d-a) < f and f < d+a and abs(e-f) < a and a < e+f and abs(a-f) < e and e < a+f and abs(a-e) < a and a < a+e and -a*f+d^2-e^2 = 0 and igcd(a, d, e, f) = 1 and nops({a, d, e, f}) = 4
then L := [op(L), [a, d, e, f]] end if end do end do end do end do; 
nops(L); 
L;

Another way to find the length of edges of a tetrahedron knowing that the mesure angle of two opposite?

evalf[100](frac(exp(19*Pi)-19*Pi));

0.32853457802957784855876405976954586639886249604033514784046998713819112593

evalf[10](frac(exp(19*Pi)-19*Pi));

0.

Hello Friends

I have a critical problem that I wish to solve it with maple

suppose we have a list like following: y_obs=(2,4,8,7,9,52,35,478,52) and corresponding variance σy=(.2,.3,.5,.87,.1.2,.22,.78,.99,1.5)
we know y as the function of x described such as y_theoric=x+p and minimizing X is

X=Sigma [(y_theoric-y_obs)^2]/σy which includes the sum of nine numbers...

the question is:

How we can find p from likelihood function and plot general behavior of y versus of x through two above series?

for example this solution used in article under the names Hubble parameter data constraints on dark energy by Yun Chen and Bhatra Ratra (Physics Letters B)

Thank you

Hello everyone! I am currently solving on a basic coordinates points for my final year project. This is a part of my coding in maple.

ans := solve({eq5, eq6}, {P2, Q2});
             {P2 = 3.222860033, Q2 = 3.170614592}, 

               {P2 = 1.572224939, Q2 = 5.670614592}

 

I am finding the points of P2 and Q2. From there after solving for eq5 and 6 it will gives two points for P2 and two points for Q2. So how am i going to choose the points using maple coding without copy paste the answer?

Really appreciate if any of us can help me. Thank you in advanced :)

Consider for instance the following equation:

Eq:=(a-4)*exp(4*x)+(b+1)*exp(2*x)+(c-2)=0

How can I list the coefficients of the exponential functions and also solve the equation for the constant parameters 

a, b, and c?

I tried 

[coeffs(collect(lhs(EQ), exp), exp)] =~ 0;

but it did not work. Thank you for your help.

I mean 

restart;
 plots:-implicitplot(sqrt(b)*sqrt(1-4*p/b)-2*arctan(sqrt((9*p/b-22201/10000)/(9/4-9*p/b))) = 0, b = 0 .. 5,
 p = 0 .. 5, gridrefine = 2, rational);

I find the above result unsatisfactory.

Hi, my problem is that I have a set of variables stored in a list, then when I try to sum with differentiation inside the sum, Maple immediately tries to differentiate before summing, thus returning zero.

So I define a list coords := [t, r, theta, varphi], then call sum(diff(r^2, coords[k]), k = 1 .. 4), however Maple does the differentiation first, so it becomes 0 instead of 8r.

I attached the maple worksheet with what I did, on the first line I define the list with variables, on the second line I show that maple evaluates diff(r^2, coords[k]) to zero before doing the sum, where k is what is being summed over, on the third line I show that it copes fine if a specific element of the list is called, on the fourth line I show that summation over elements of the list is fine, and the last two lines show an example of the kind of thing I would like to do

Is there a way to make this work?

Dear Maple Primes,

could you, please, help me with numeric integration? I’m new in numeric integration and can’t reach desired precision of a result.
Here is the integral f(x_max) that I try to compute for different values of x_max from the interval 0.025..0.24 :

f:=(xmax)->Int(K*F*Int(G*F,x=x..xmax,method=integrationmethod),x=x0..xmax,method=integrationmethod)

where x₀ is lower limit of outer integral, x₀ := 0.025

and K, F and G are functions of x

K:=x-x0

F:=(a₁+a₂*x+a₃*x²+a₄*x⁵)/(b₁*x+b₂*x²+b₃*x⁶)

G:=exp(c₁+c₂*x+c₃*x⁷)

with

a₁:=8e3; a₂:=6e4; a₃:=3e4; a₄:=1.8e8;
b₁:=9.2e17; b₂:=1.1e18; b₃:=4.6e21;
c₁:=8.202046; c₂:=-12.31377; c₃:=-818043.42;

Please, notice, that G (as well as G*F) is a steeply decreasing function on the interval x = 0.025..0.24.

I get "a seemingly correct" result (that means that f increases as x_max intreases), when I try to plot f(x_max) for the following "guessed" options

Digits:=15
integrationmethod:=_d01akc
plot(f,0.21..0.24,color=black)

What is puzzling me is that I get a different "seemingly correct" result, when I modify the integral f by,
at fist, multiplying G by a constant (for example Const:=1e20; G:=Const*exp(c₁+c₂*x+c₃*x⁷) )
and, second, plotting the f divided by this constant:

plot(f/Const,0.21..0.24,color=red)

The following Figure presents the values of f plotted versus x_max with (red curve) and without (black curve) using of the constant Const:

Dear Primes, could you, please, comment on this difference? Because the only indicator that I have (from the analysis of G, F and K) is that f must be a monotonically (and stricktly) increasing function of x_max.

Please, find the maple worksheet in attachment.

Thank you in advance!
Maks

for_primes_numeric_integration_v02.mw

I want to find the maximize and minimize of the function
f:=x->(cos(x)+sqrt(3)*sin(x))/(cos(x)+sin(x)+2);
I tried 
minimize(f(x), x, location = 'true');
and
maximize(f(x), x, location = 'true');
But I didn't get the results.  How do I find the maximize and minimize of above funciton?

Hello,

     I have a very large expression I'm tring to average (by doing limit(1/L*int(<expr>,t=-L..L),L=infinty) ). However, given that this expression is so large (sum of 11k terms), I'm was looking for ways to speed up the calculation and use less RAM.

     I found that applying the lim@int seperately to each term of the sum helps. Also, since my integrand has various functions of other variables (eg. f(x), g(y), etc), Maple seems to go faster if I freeze those functions using frontend.

     However, the problem I'm running into is that, as I'm running through my for-loop to apply lim@int to each term, Maple starts running slower and slower, as well as taking more and more RAM. Additionally, it doesn't go to completion because it runs out of stack space.

     Is there an optimization trick to avoid this problem? I've tried garbage collecting after each iteration; that helps the RAM problem (at a cost of speed), but it still slows down over time.

      I've attached my code: MWE.maple. Fair warning: on a non-server computer, it is liable to run very slow.