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  That was not what I wanted. As I...

Pinetree 15
June 12 2018
0 0

 

That was not what I wanted. As I said, coefficients are long and ugly expressions in their own right, and I do not want to have to retype them both in arguments and in assumptions. Below is one grossly simplified example of what I have to handle (don't know why mtaylor() does not work while mtaylor(series()) does, but anyways ...). 

All that I want is to return that solution that has the form -b/(2*a)+sqrt((b/(2*a))^2-c/a).


 

``

``

mtaylor(series(exp(Int((1/2)*sqrt(x^2+4*p)/(arctanh((2*_z1*p-x)/sqrt(x^2+4*p))+arctanh(x/sqrt(x^2+4*p))), _z1 = 0 .. z)), x), [x], 3)

exp(int(p^(1/2)/arctanh(_z1*p^(1/2)), _z1 = 0 .. z))+exp(int(p^(1/2)/arctanh(_z1*p^(1/2)), _z1 = 0 .. z))*(int(-(1/2)*_z1^2*p/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))*x+(1/2)*exp(int(p^(1/2)/arctanh(_z1*p^(1/2)), _z1 = 0 .. z))*(2*(int((1/8)*(-arctanh(_z1*p^(1/2))*p^(3/2)*_z1^3+arctanh(_z1*p^(1/2))^2*_z1^4*p^2+2*_z1^4*p^2-arctanh(_z1*p^(1/2))*p^(1/2)*_z1-2*arctanh(_z1*p^(1/2))^2*_z1^2*p+arctanh(_z1*p^(1/2))^2)/(arctanh(_z1*p^(1/2))^3*(_z1^2*p-1)^2*p^(1/2)), _z1 = 0 .. z))+(int(-(1/2)*_z1^2*p/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))^2)*x^2

 (1)

"(->)"

[[x = 2*(p^(3/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))+(-p^3*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))^2+2*p^2*(int(_z1^3/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p^(5/2)*(int(_z1^4/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))-4*p^(5/2)*(int(_z1^4/(arctanh(_z1*p^(1/2))^3*(_z1^2*p-1)^2), _z1 = 0 .. z))+2*p*(int(_z1/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))+4*p^(3/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p^(1/2)*(int(1/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z)))^(1/2))/(p^(5/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))^2-p^(3/2)*(int(_z1^3/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))+p^2*(int(_z1^4/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))+2*p^2*(int(_z1^4/(arctanh(_z1*p^(1/2))^3*(_z1^2*p-1)^2), _z1 = 0 .. z))-p^(1/2)*(int(_z1/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p*(int(_z1^2/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))+int(1/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))], [x = -2*(-p^(3/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))+(-p^3*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))^2+2*p^2*(int(_z1^3/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p^(5/2)*(int(_z1^4/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))-4*p^(5/2)*(int(_z1^4/(arctanh(_z1*p^(1/2))^3*(_z1^2*p-1)^2), _z1 = 0 .. z))+2*p*(int(_z1/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))+4*p^(3/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p^(1/2)*(int(1/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z)))^(1/2))/(p^(5/2)*(int(_z1^2/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)), _z1 = 0 .. z))^2-p^(3/2)*(int(_z1^3/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))+p^2*(int(_z1^4/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))+2*p^2*(int(_z1^4/(arctanh(_z1*p^(1/2))^3*(_z1^2*p-1)^2), _z1 = 0 .. z))-p^(1/2)*(int(_z1/(arctanh(_z1*p^(1/2))^2*(_z1^2*p-1)^2), _z1 = 0 .. z))-2*p*(int(_z1^2/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))+int(1/(arctanh(_z1*p^(1/2))*(_z1^2*p-1)^2), _z1 = 0 .. z))]]

 (2)

``

``

``


 

Download quadratic-pick-plus.mw

But after a reboot?!?...

Pinetree 15
December 05 2017
0 0

This is getting weird. Re-running (executing using !!!) did not change a thing. Editing, saving, !!!ing did not change the errors.

But last night I had to reboot the computer for an update of some other software, and then when I relaunched Maple this morning and opened the SystemGoesWrong.mw file - the errors were still there, but everything was updated using !!! and the errors disappeared. 

What is going on? Is it so that "restart" does not really restart? Should I need to close Maple and reopen it?

 

And is this behaviour as expected?...

Pinetree 15
November 02 2017
0 0

Getting answers (thanks!) does not really stop me from asking ;)

As the integral is solvable, I fiddled around a bit with it. I tried the "implicit" option: 

restart; assume(nu > 2, beta > 0, lambda > 0, delta > 0, y >= 0);
ODEh := beta+y+(D(h))(y) = nu*((lambda+beta+y)^2-delta*h(y))^(1/2);
solimp := simplify(dsolve(ODEh, implicit));

Two puzzling elements in the output:

  • ln(negative). Why? There was a reason for the "assume".
  • The delta~+2 denominator. That does not look very simplified?

Afterwards, I played around with 
simplify(value(solimp))
simplify(value([copy+pasted the output after having fixed the delta~+2))
They do of course yield ugly output (and quite different formulae), but I don't immediately spot any ln(negative).

@_Maxim_ So in order to visually co...

Pinetree 15
November 02 2017
0 0

@_Maxim_ So in order to visually compare output, I use "rhs" in the second line. New document, paste 

restart; assume(nu > 2, beta > 0, lambda > 0, delta > 0, y >= 0);
rhs(simplify(dsolve(beta+y+(D(h))(y) = nu*((lambda+beta+y)^2-delta*h(y))^(1/2))));
applyrule(e::RootOf = 'applyop(simplify, 1, e)', %)

... but it makes no difference.

Then copy the RootOf parenthesis, paste it inside the parentheses of simplify() - and then I get a simplification. This one simplifies:

simplify(Intat((2*(2*_a*`δ`^2+`ν`*sqrt(-_a*`δ`^3-4*_a*`δ`^2-4*_a*`δ`+1)+8*_a*`δ`+8*_a+1))*(`δ`+2)*`λ`/(`ν`^2*_a*`δ`^3+4*_a^2*`δ`^4+4*`ν`^2*_a*`δ`^2+32*_a^2*`δ`^3+4*`ν`^2*_a*`δ`+96*_a^2*`δ`^2+128*_a^2*`δ`+4*_a*`δ`^2-`ν`^2+64*_a^2+16*_a*`δ`+16*_a+1), _a = _Z)*`δ`+2*`λ`*ln(-`β`*`δ`-y*`δ`-2*`β`-2*y-2*`λ`)+_C1*`δ`+2*Intat((2*(2*_a*`δ`^2+`ν`*sqrt(-_a*`δ`^3-4*_a*`δ`^2-4*_a*`δ`+1)+8*_a*`δ`+8*_a+1))*(`δ`+2)*`λ`/(`ν`^2*_a*`δ`^3+4*_a^2*`δ`^4+4*`ν`^2*_a*`δ`^2+32*_a^2*`δ`^3+4*`ν`^2*_a*`δ`+96*_a^2*`δ`^2+128*_a^2*`δ`+4*_a*`δ`^2-`ν`^2+64*_a^2+16*_a*`δ`+16*_a+1), _a = _Z)+2*_C1)

 

Maple 2016.2 (insert Friday 13th joke here) and ClumsyUser 2017.11 ...

@Carl Love  Everything is symbolic...

Pinetree 15
October 18 2017
0 0

@Carl Love 

Everything is symbolic. The only number is the "0" in the nonnegativity conditions. Oh, and indices. 

(And the problems are the same ... I don't think I messed up signs/parentheses?)

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