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Hi all.

I am using Maple2015.

I typed in as input y=x/sqrt(1-x^2).

I hit enter.  The output is:

 y=x/sqrt(1-x^2)

I know the 2 answers are equivalent.

My question is why did Maple swap 1-x^2 to -x^2+1???

Any advice to swap it back would be greatly appreciated.

How to find asymptotic behaviour of a function.

For example at infinity

sinh(x) behaves as 1/2*exp(x)

1/sinh(x)  behaves as 2*exp(-x)

exp(-x)*(exp(-x)+1) behaves as exp(-x)

so that it works with a more complex expression.

when solving a system of equations, I want to get rid of all the absolute functions.

 

for example, |y-2|=x,I don't want maple solve this equation directly...because maple may has difficulties when dealing with absolute values. Instead, I want to transform this equation by squaring the both sides at the same time which end up this equation: (y-2)^2=x^2.

 

The example I provide is kind of simple...what if there are multiple absolute term in the equations? Is there a general way to get what I need? Or Is it practical to use Maple to achive such thing?

 

thanks in advance.

When I use fsolve with equation 

-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0

I got only one solution.

fsolve(-x^2 + 2*x + 5 + (x^2 + 2*x - 1)* sqrt(2 - x^2)=0,x);

In fact, it have two reals solutions.  

I posted at here

http://mathematica.stackexchange.com/questions/83985/does-the-equation-have-two-roots/83991#83991

 

Hello Everyone

I have an expression which I wish to integrate. I would be grateful if you could please help me with it. I have uploaded the maple file for your refrence.

Thanks a lot for your time.

 

 

IntegrationExample.mw

 

restart;
pp:=-55471918776960000*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)+5350094400*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^2-129*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^4+2670899840*tanh((1/6450)*sqrt(41602500-cp^2)*Pi*x/cp)*sqrt(41602500-cp^2)*sqrt(10368400-cp^2);
Student[Calculus1]:-Roots(subs(x=8000,pp),cp=1..3220,numeric);
p1:=proc(v)
option hfloat;
local a;
a:=Student[Calculus1]:-Roots(subs(x=v,pp),cp=1..3220,numeric);
if nops(a)>=1 then seq([v,a[i]],i=1..nops(a));
end if;
end proc:
SS1:=[seq(p1(i),i=3500..20000,200)]:
plot(SS1,style=point,gridlines);

The final figure is different between maple12 and maple17.

On 17, unwanted points apprear.

is it a bug?

Hello,

I don't why I can no longer see my multibody model in the 3D construction mode.

I probably should desactive something.

I would like to observe my system with the kinematic constraints not enforced.

The idea is that my model is computing but the computation is very long. I should have done a mistake in my initial conditions.

Do you have an idea of what to do to reactivate/ show again the 3d contruction mode ?

Thank you for your help.

Dear Colleges

I have a problem with the following code. As you can see, procedure Q1 converges but I couldn't get the resutls from Q2.

I would be most grateful if you could help me on this problem.

 

Sincerely yours

Amir

 

restart;

Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)^2*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)^2*x,x$1);
eq1:=isolate(Eq1,diff(f(x),x,x,x));
eq2:=subs(g=f,isolate(Eq2,diff(g(x),x,x,x)));
EQ:=diff(f(x),x,x,x)=piecewise(x<c*0.1,rhs(eq1),rhs(eq2));
Eq11:=diff(theta(x),x$2)+pr*diff(theta(x),x$1)*f(x)+pr/prt*b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(theta(x),x$1)*x^2,x$1);
Eq22:=diff(g(x),x$2)+pr*diff(g(x),x$1)*f(x)+pr/prt*a^2*c*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(g(x),x$1)*x^1,x$1);
eq11:=isolate(Eq11,diff(theta(x),x,x));
eq22:=subs(g=theta,isolate(Eq22,diff(g(x),x,x)));
EQT:=diff(theta(x),x,x)=piecewise(x<c*0.1,rhs(eq11),rhs(eq22));
EQT1a:=eval(EQT,EQ):
EQT2:=eval(EQT1a,{f(x)=G0(x),diff(f(x),x)=G1(x),diff(f(x),x,x)=G2(x)}):
bd:=c;
a:=0.13:
b:=0.41:
pr:=1;
prt:=0.86;
reynolds:=12734151.135786774055543653356602;     #10^6;   #1.125*10^8:

c:=88.419896050808975395120916434619:
;
Q:=proc(pp2) local res,F0,F1,F2;
print(pp2);
if not type(pp2,numeric) then return 'procname(_passed)' end if:
res:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=pp2},numeric,output=listprocedure);
F0,F1,F2:=op(subs(subs(res),[f(x),diff(f(x),x),diff(f(x),x,x)])):
F1(bd)-1;
end proc;
fsolve(Q(pp2)=0,pp2=(0..1002));
se:=%;
res2:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=se},numeric,output=listprocedure):
G0,G1,G2:=op(subs(subs(res2),[f(x),diff(f(x),x),diff(f(x),x,x)])):
plots:-odeplot(res2,[seq([x,diff(f(x),[x$i])],i=1..1)],0..c);



Q2:=proc(rr2) local solT,T0,T1;
print(rr2);
if not type(rr2,numeric) then return 'procname(_passed)' end if:
solT:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-rr2},numeric,known=[G0,G1,G2],output=listprocedure):
T0,T1:=op(subs(subs(res),[theta(x),diff(theta(x),x)])):
T0(bd);
end proc;
fsolve(Q2(rr2)=0,rr2=(0..100));


shib:=%;
sol:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-shib},numeric,known=[G0,G1,G2],output=listprocedure):
plots:-odeplot(sol,[x,theta(x)],0..c);
#fsolve(Q2(pp3)=0,pp3=-2..2):

Amir

could you help me about maple
i try to calculating using chevypade rational approximating and the answer for cos(x) xe is(-.221091073962959*T(1, x-1)+.7710737338*T(0, x-1)-0.4212446689e-1*T(2, x-1))/(0.836360586596837e-1*T(1, x-1)+T(0, x-1)+0.3360079945e-1*T(2, x-1)) i can not to convert to rational form as x^^n .maple is not very friendship
Thanks


> restart;
> with*plots;
> Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2 = 0; 

> N := 1;

> blt := 10;
> Eq2 := (diff(theta(eta), eta, eta))/Pr+f(eta)*(diff(theta(eta), eta)) = 0; 
> bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(blt) = 0;
> bcs2 := (D(theta))(0) = -N*(1+theta(0)), theta(blt) = 0;
> L := [2.5, 3, 5, 7, 10];
> for k to 5 do R := dsolve(eval({Eq1, Eq2, bcs1, bcs2}, Pr = L[k]), [f(eta), theta(eta)], numeric, output = listprocedure); X1 || k := rhs(R[3]); X2 || k := rhs(R[4]); Y1 || k := rhs(R[5]); Y2 || k := -rhs(R[6]) end do;

 

 how I will draw the graph for Pr against theta   for Pr=2.5 until 7  taking rest of the parameter fix

 

Problem: I have two polynomials with arbitrary coefficients. I set them both to 0 and I used the 'map' , as well as 'coeffs' command to make the coefficients equal to 0.

Now for some reason, Maple does not print in order for one of the polynomials and it does for the other.

 

Note that 'order' refers to the coefficients attached to the powers of the variable.

 

Quick example: (this one actually works on Maple, but just not the one I have)

 

e1:= ax + (b + c)x^2 = 0

e2:= (c + d)x + (a + c)x^3 = 0

 

After applying map and coeff, one expects it to output

 

a = 0, (c + d) = 0, (b + c)=0, (a + c) = 0

 

instead I got

 

a = 0, (c + d) = 0, (a + c) = 0,(b + c)=0

 

Here is the problematic file

OutOfOrder.mw

I wish to evaluate the expression

knowing that

where a is a constant.  It is not hard to see, assuming enough differentiability,  that the expression evaluates to

I know how to do this when all the derivatives are expressed in terms of the diff() operator.  Here it is:

eq := diff(u(x,t),t) = a^2*diff(u(x,t),x,x);
expr := diff(u(x,t),t,t);
eval['recurse'](expr,[eq]);

However, I would prefer to do the computations when all derivatives are expressed in terms of the D operator but cannot get that to work.  What is the trick?

I've been having trouble figuring out some functions of Maple 18 specifically related to polar and cartesian integrals. My questions exactly would be; how does one go about translating a cartesian integral into a polar integral and just about most things related to polar and cartesian interactions/conversions.

 

For an example I have a double integral of 9xy^2 dxdy with bounds a= -sqrt(a^2-y^2) b= 0 c= 0, d= 1.

How exatcly do I use all of the available Maple functions to help me get to my answer?

 

Thanks in advance. Any help is appreciated.

according to help on timelimit

"Note: For efficiency reasons, the timelimit bound is ignored while in built-in routines."

Which is not very useful, since I want to limit  int() to some CPU time.

There are some integrals that can hang Maple easily. I'd like to set some CPU time on an int() and
have it terminate with error, but I am not able to find how to do that.

For example this

int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^(5/2),x)

Will hangs Maple.
Is there a trick some expert here can show to limit the CPU time on a build in operation?
May be some package or other command can do this?

I am only interested in int() now, but if it can work also on dsolve, that will be good.

thank you
ps. Mathematica supports putting time constraint on build-in commands. So I do
not see why Maple can't also do the same.

Hello. I have a question. If you can help me, i am pleasure.

Have nice day. :)

 

restart;
soru := proc(n,x)
local top::0;
for x from 2 to n do
top =top+(((x^(2+i))*top)^(1/(n+2-i)));
print(top);
od;
end proc;

 

soru(5,1);

Error, (in soru) illegal use of a formal parameter

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