Axel Vogt

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16 years, 210 days
Munich, Germany

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These are replies submitted by Axel Vogt

Start with the inner integral. It will be a function of x.

@Christopher Tocci 

Essentially you want the integral Int(1/(theta*tan(theta)+c),theta = 0 .. Pi/2) or likewise theta/2 and up to Pi. It is computed really fast by Maple numerically. Your approximation needs a range for c. Note c = 0 means t=0 in the original question.

@daljit97 Modify the code to print that C in any case, given the x,y,z in the loop. You will see 2=0 or ... or 0=0. If the expression is true then it is printed.

Please use Maple syntax / code to ask your question

@Carl Love What is called "Quantile" seems to be a cumulative for the pdf.

off topic: I dislike questions without giving a context or reference, here it can not be judged (for me) whether it makes sense to treat it with Maple

@acer , thank you - I will try that (various versions and also old sheets)

@Nick_99 

roughly 1 second to compute the anti-derivative

 

NB: you need "Re( ... )" for plotting because 2F1 may have spurious numerical imaginaries - or use plot( [Re(...), Im(...)], ...)

NB2: do *not* use brackets like " [ ", use ordinary brackets like " ( " in your functions

@nm 

You are right, I made an error through copy + paste and will correct my answer, I have to add 800*t

Fails for me too (FF 83 portable, I have blocked some servers through DNS, but never observed a problem usinf Mapleprimes)

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/prove.mw .

 

 

@tomleslie I modified your solution
 

https://www.mapleprimes.com/questions/231147-Unevaluated-Integral

  restart; Digits:=10;

10

(1)

  eq1:=diff(f(y), y$4)+Uhs*diff(E(y),y$3)-(diff(f(y), y$2))+(diff(theta(y), y$1))= 0:

  eq2:=diff(theta(y), y$2)+(diff(f(y), y$2)+1)^2+1+diff(theta(y),y$2) = 0:

  E:=y->zeta*(cosh(k/2*(h1+h2-2*y)))/(cosh(k/2*(h1-h2))):

  bcs:= f(h1) = -(1/2)*(Q-1-d),
        f(h2) = (1/2)*(Q-1-d),
        (D(f))(h1) = -1,
        (D(f))(h2) = -1,
        theta(h1) = 0,
        theta(h2) = 1:

  epsilon1:=0.1:                    d:=1:                             omega:=Pi/6:
  h1:=-(1+epsilon1*sin(2*Pi*x)):    h2:=d+epsilon2*sin(2*Pi*x+omega): F:= Q-1-d:
  epsilon2:=0.5:                    x:=1:                             alpha:=Pi/6:

  d1 := subs( Uhs =-2,
              zeta=3,
              k=1,
              [eq1, eq2, bcs ]
            ):

d1_var:=eval(d1, [f=ff, Q=QQ]);

[diff(diff(diff(diff(ff(y), y), y), y), y)-3.52436433981067*sinh(-.125000000000000+y)-(diff(diff(ff(y), y), y))+diff(theta(y), y) = 0, 2*(diff(diff(theta(y), y), y))+(diff(diff(ff(y), y), y)+1)^2+1 = 0, ff(-1.) = -(1/2)*QQ+1, ff(1.25000000000000) = (1/2)*QQ-1, (D(ff))(-1.) = -1, (D(ff))(1.25000000000000) = -1, theta(-1.) = 0, theta(1.25000000000000) = 1]

(2)

 

F:=proc(Q)
local deq,sol,g;
deq:=eval(d1_var, QQ=Q);
sol:= dsolve(deq, numeric, output=listprocedure):
g:= unapply( rhs(sol[5])(z)-3.524364340*cosh(-0.1250000000 + rhs(sol[1])(z))-rhs(sol[3])(z)-1/2+rhs(sol[6])(z),z);
evalf(Int(g, 0..1, epsilon=1e-8));
end proc:

#F(0); # test

HFloat(-2.8641783992249827)

(3)

plot(F, -3 .. 3, numpoints=10);

 

 


 

Download MP_231147.mw

@Mo_Jalal 

Clearly it is never 0. But it can take any other value for x being a complex number: for that write g(x) = 1/w, any w not 0. Then you have cubic in x = w and you know it always has a solution.

combine                            ?

Plot the function - then you see what to do

and if not: it should be easy to define such a function (if - else and how to behave at the jump)

look up the help and find a sheet attached

MP_230629.mws

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