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Length of output exceed limit 1000000...

Asked by: Sofey 10
May 03 2015
0  3

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

 (1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

 (2)

``


Download consistency_expmle_4.mw

.

how to prove with maple...

Asked by: laporte bernard 45
student
May 03 2015
0  1


> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

How can PDE be solved in terms of Bessel function?...

Asked by: emmantop 40
plot
May 02 2015
0  4

Good day everyone,

please how can one solve this pde in terms of Bessel function or any other analytic solution with the plot.

See the file ID.mw

Thanks.

how to copy and paste from maple 12 a quite big qu...

Asked by: laporte bernard 45
April 30 2015
0  4

Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

Maple V please help SOS urgently need help...

Asked by: c4meleon 30
April 30 2015
0  0

Hello,
I am looking here for a tutor, that can help me doing some Maple V (5) programming.
I am a mathematic student and we use Maple programming.
I will be happy to pay a small amount of money for each exercice you help me doing,

if anyone is interested, please contact me here.

(The procedures that we usually have to write are for example:
 Newton-Raphson Method, Chebyshev Polynomial,...  I don't think it is hard for you.
Thank you very time for your time and your help.

convert 2-D expression to 1-D...

Asked by: habex 15
2dmath worksheet editing
April 26 2015
2  1

Hello, I would like to customize the context menu in Maple 18 and looking for a way, to convert 2-D expressions to classical maple input using command(s) (same function as "2-D Math > Convert To > 1-D Math Input"). My main problem is, that using the standard procedures in the context menu I get the result of an expression, not the expression itself.

Can you give me an idea, how to do this? Thanks, Csaba

error appear in maple ...

Asked by: franky12 10
ode numeric
April 26 2015
0  1

@Carl Love Dear Sir i am trying to solve the system of nonlinear ODE equation with semiinfinite domain by using shooting method . but i am receiving following error 

"Error, (in isolate) cannot isolate for a function when it appears with different arguments"

how can i remove this error. i am unable to find the mistake. kindly help me 

Quantum Electrodynamics with Maple...

Asked by: Inacio Aprendanet 65
physics field lagrangian
April 24 2015
0  1

Hi!

 

I was trying to obtain the field equations for the QED Lagrangian and I was not sucessfull.

 

All my calculations are equal to 0.

 

Can someone give a hand?

 

QED.mw

 

Thanks a lot.

Maple 15 numerical evaluation of integrals cannot ...

Asked by: Denni 5
integral singularity
April 21 2015
1  6

I use Maple 15 to calculate some (nasty) integrals at my university. Because our university also offers a server on which I can run my Maple program, I would like to do that. (instead of occupying a workspace). But at the computer on my workspace the integrals are evaluated fine, but on the server the integrals are just returned with no numerical evaluation.

I constructed a MWE to look where it goes wrong. I set the printlevel to 25 so I could see what was going on. The MWE was suprisingly simple, on both machines (via ssh) I executed within maple:

evalf(Int(1/sqrt(x), x=0..2))

This of course would normally just give 2*sqrt(2). On my workplace-pc it worked fine and it found 2.828427125. The server just returned the integral. After looking at the steps, they where both exactly the same until the following part:

Workplace-PC:

         General_flags := {_NoNAG, _DEFAULT, _NoMultiple}

            NAG_methods := {_d01ajc, _d01akc, _d01amc}

                        Method := _DEFAULT

                          HFDigits := 15

                                       -12
                        HFeps := 0.1 10

                                            -9
                    HFeps := 0.5000000000 10

   oldEvents := overflow = default, division_by_zero = default

                         callNAG := true

                            fcns := {}

                  result := 2.82842712474618807

Server:

        General_flags := {_NoNAG, _DEFAULT, _NoMultiple}

           NAG_methods := {_d01ajc, _d01akc, _d01amc}

                       Method := _DEFAULT

                         HFDigits := 15

                                      -12
                       HFeps := 0.1 10

                                           -9
                   HFeps := 0.5000000000 10

   oldEvents := overflow = default, division_by_zero = default

                         callNAG := true

                           fcns := {}

       overflow = exception, division_by_zero = exception

It seems that the server has a problem with the singularity and thus throwing an exception, but I just don't get why. The Maple-versions are both the same.

Does somebody know what this could be?

Error trying to Solve unknown variable within a de...

Asked by: rbd66 10
solve units determinant
April 19 2015
1  2

Hi there,

 

I am trying to compute the following, and I am getting this error.

 

> A := map(convert, M, unit_free)*Unit('m'*(1/'s'^2));
(I had to put the Unit('m'*(1/'s'^2)) because the original units were kNm/s^2 (kg), and even though I simplified it, it's still using kNm/s^2, and leaves the m/s^2 for some reason when I try to remove the units. I tried simply changing the units on the original matrix, but the units menu has disappeared from the right-click menu!!)

> B := map(convert, K, unit_free);

 


Loading RealDomain;


solve;

Error, (in assuming) when calling 'Engine:-Dispatch'. Received: 'should not happen: Rename expects the input to contain unknown functions'

Why is it giving me this error? omega is an unknown variable that I am trying to solve for. I am going a modal analysis, so maybe there is a better way to find omega?

 

Any help appreciated!

 

Raquel

How do I copy and paste document maple to mathtype...

Asked by: minhhieuh2003 140
April 18 2015
1  1

I have document Maple.mw. I want to copy all text to mathtype. How do I copy and paste?

Link download: maple document

command "factors" doesn't work...

Asked by: alla 615
factorization
April 18 2015
0  1

Why does the following not work?

 

factors(x^4-1)

3D plotting of a piecewise...

Asked by: Shervin 20
April 18 2015
1  12

Hi

I have problem with 3Dplotting of this piecewise...can anyone help me please?!

since i know the output will be a value between 0 and 1 can i plot it for different values of Mu in this range to eliminate a dimension?

Mu(x1, x2, x3) = piecewise(`or`(`or`(x1 > 55, x2 > 10), y >= 2), 1, x1 < 35 and x2 <= 10 and x3 < 1, (1/10)*x2, `and`(`and`(x1 < 35 and x2 <= 10, x3 >= 1), x3 <= 2), max((1/10)*x2, x3-1), `and`(`and`(`and`(x1 >= 35, x1 <= 55), x2 <= 10), x3 <= 1), 1-((55-x1)*(1/20))((10-x2)*(1/10)), `and`(`and`(`and`(`and`(x1 >= 35, x1 <= 55), x2 <= 10), x3 >= 1), x3 <= 2), max(x3-1, 1-((55-x1)*(1/20))((10-x2)*(1/10))))

Thank you very much in advance...

I don not know why...

Asked by: wangwei 0
hypergeom
April 17 2015
0  0

Maple show "

Error, (in simplify/hypergeom/baseHG) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc"

after I finish the equations set and run it.

what's the reason?

procedure "proc"...

Asked by: Alger 524
April 16 2015
0  6

Dear all,

I have one probem in using Maple in comparison with Matlab.

Generally, when using Matlab, I write a long program with 500 lines (for example) and where I use some matricial calculations and other. When I want to repeat this program for some parameters (variables) many times, I transform the program to a procedure "Function" with just adding the command Function in the start of the program.

When using Maple, I have a difficulty to transform a long program which I want to repeat for some variables to a procedure "proc". The program uses packages like LinearAlgebra and other procedures.

Is there a simple way to transform a long Maple program to a procedure which should be repeated many times for some variables in Maple. 

I have especially problems with the commands local, global, use (for packages), etc.. and the warning related to the variables. It is not easy for a long program with many variables to write them in local or global variables.

Thank you.

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