How to  compute the recurrence relation and I find the problem when the summation of U because appear noise term self-canceling and I can not find the nth component of U?Mixed_volterra_-Fredholm_(278)_Ex(8.17).mw

Download Mixed_volterra_-Fredholm_(278)_Ex(8.17).mwMixed_volterra_-Fredholm_(278)_Ex(8.17).mw

thank you for helping:)

Hello,

I have an non coupled non linear oscillator.

I notice that, if I try to plot for a time too big, my plot doesn't converge anymore and didn't keep an elliptic trajectory. In other words, the plot didn't stay in the limit cycle.

Do you know why, if tmax is too big, the solution is no longer stable ? Do you have ideas so that I can keep a stable limit cycle even if I increase tmax ?

My code is the following :

r:=sqrt((x(t)/a)^2+(z(t)/b)^2);
eqx:=diff(x(t),t)=alpha*(1-r^2)*x(t)+w*a/b*z(t);
eqz:=diff(z(t),t)=beta*(1-r^2)*z(t)-w*b/a*x(t);
EqSys:=[eqx,eqz];

params := alpha=1, beta=1, a=0.4, b=0.2, w=1;

EqSys := eval([eqx,eqz], [params]);
xmax := 0.8; zmax := 0.4;
tmax := 400;
ic:=[x(0)=0.4, z(0)=0];
DEplot(EqSys, [x(t),z(t)], t= 0..tmax, [ic],linecolor=black, thickness=1,x(t)=-xmax..xmax, z(t)=-zmax..zmax, scaling=constrained,arrows=none);

Thanks a lot for your help.

How to evaluate The Abel integral has the form I can not compute this

> restart;

> f := proc (x) options operator, arrow; (4/3)*x^(3/2) end proc; k := proc (x, t) options operator, arrow; 1/sqrt(x-t) end proc;

> int((4/3)*t^(3/2)/sqrt(x-t), t = 0 .. x);

Thank you :)

> restart;> f := proc (x) options operator, arrow; (4/3)*x^(3/2) end proc; k := proc (x, t) options operator, arrow; 1/sqrt(x-t) end proc;> int((4/3)*t^(3/2)/sqrt(x-t), t = 0 .. x);

I want to plot the curve of a pythagorean quadruple. Any suggestions? I think I have to use space curve. I never liked the options for color on those though. I'm guessing it's just a single curve based on what I know about these things. I have no idea what to expect but I'm sure it will please me no matter what.

X=2*m*p/(m^2+n^2+p^2)

y=2*n*p/(m^2+n^2+p^2)

z=(p^2-(m^2+n^2))/(m^2+n^2+p^2)

using FDM or FEM rather than dsolve?

ode.docxode.docx

Hello everyone! I'm currently trying to plot the effective potential for Sun-Jupiter system, to show the lagrangian points in this system. I've converted to a system of units where G=1, m_sun+m_jupiter=1 and R=1, whereby I get the following equation describing the effective potential of a third mass m in the system:

U(x,y)=-1 * 0.999047/abs(sqrt((x-(-0.000953))^2 +y^2 ))-1 * 0.000953/abs(sqrt((x-0.999047)^2 +y^2 ))

However, when I plot this in Maple, I don't get a contour plot that matches up with what I would [expect](https://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Lagrange_points2.svg/2000px-Lagrange_points2.svg.png), but [this](http://imgur.com/oZLE3gV). Is there possibly anyone here who could point out a mistake I've done, or give me some advice? Thanks in advance!

Occasionally I have need to import 1D Maple expressions into Mathematica, because I believe mma has superior simplify functionality (sorry Maple). Mathematica has its own syntax which (of course) differs to maple.  eg sin(x) in maple is Sin[x] in mma ; ln(x)=Log[x] ;....... I'm aware of the package MmaTranslator, but I think it only converts mma to maple.

so i need some code pls.

(i have bolded the functions which need to be converted)

eg: i/p:

(1/365)*(365*B21*EP*ln(1+i)*(1+i)^B21*hr*kw*p-365*EP*hr*kw*p*(1+i)^(-1+B21)*i-365*EP*hr*kw*p*(1+i)^(-1+B21)+365*EP*hr*kw*p+SC*i)/(hr*kw*(EP*ln(1+i)*(1+i)^B21*p-FIT*i*p+cos(omega)))

o/p:

(1/365)*(365*B21*EP*Log[1+i]*(1+i)^B21*hr*kw*p-365*EP*hr*kw*p*(1+i)^(-1+B21)*i-365*EP*hr*kw*p*(1+i)^(-1+B21)+365*EP*hr*kw*p+SC*i)/(hr*kw*(EP*Log[1+i]*(1+i)^B21*p-FIT*i*p+Cos[omega]))

I ran into a problem with define that I traced to the 'linear' option in the define statement, and I would like to know whether this is my lack of understanding define, or a bug.

I want to define a function called avg(parm) (which I never will need to use explicitly but  I need to be able to manipulate). This function has the property of being linear, and terms not depending on x can be pulled out. Also, avg(parm)=parm unless parm depends on x. (Yes, it stands for a kind of averaging, but that is not relevant here).

So I programmed it in this way:

define(avg,'linear','conditional'(avg(a::algebraic) = a, _type(a, freeof(x))),\
'conditional'(avg(a::(nonunit(algebraic))*b::(nonunit(algebraic))) = a*avg(b), _type(a, freeof(x))));

For the heck of it I could not get it to work. Basically, none of the equations in the definition were ever used, except that constant numeric factors were pulled out, as they should.

Eventually I found that the 'linear' option was the culprit. Once I got rid of that, the definition of avg did what I wanted it to do (like avg(a*x) => a*avg(x); avg(a) => a). The reason a constant numeric factor was pulled out before is because the 'linear' option does that; it never gets to my definitions (one can see that when printing avg).

Am I missing something fundamental here, or is define not working right?

For the record, I did this in Maple 15 on a PPC Mac. I don't have my Maple 2015 handy to test on that version.

Any insights?

Mac Dude

Hi,

I'm null in statistic. I'm doing a calculation to calculate a set of parameter. By example:

I have to calculate 5 parameters x1,x2,x3,x4,x5 from 7 equations f1,f2,f3,f4,f5,f6,f7. Because it is difficult to calculate directly 5 parameters, I used chisquare to minimize the difference between experimental and theorical data. Then, I can get the results. After that, I used these 5 parameters to back-calculate the data using 7 equations above. My question is about how to calculate the error bar (or standard error) of the back-calculate datato add to the plot.

Thank you  for helping me,

Best regards.

Download transpose_doubt_1.mwtranspose_doubt_1.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

I don't know if I am doing it right or not because Maple isn't solving the equation. The equation solution is x=2^(1/2).

Here I let a picture of which equation I want to solve and Maple's response.

In Maple 15 it seems that plottools:-transform only accepts this form of conditional statement:  

`if`(conditional expression, true expression, false expression).

Is there any way to have plottools:-transform process more than one condition? Do later versions of Maple permit this?

Hi, Im now trying to run my code. But it took like years to even getting the results. may I know any solutions on how to get faster results? Because I have run this code for almost 4 hours yet there is still 'Evaluating...' at the corner left. And when I tried to stop the program, it will stop at 'R1...'.

Digits := 18;
with(plots):n:=1.4: mu(eta):=(diff(U(eta),eta)^(2)+diff(V(eta),eta)^(2))^((n-1)/(2)):
Eqn1 := 2*U(eta)+(1-n)*eta*(diff(U(eta), eta))/(n+1)+diff(W(eta), eta) = 0;
Eqn2 := U(eta)^2-(V(eta)+1)^2+(W(eta)+(1-n)*eta*U(eta)/(n+1))*(diff(U(eta), eta))-mu(eta)*(diff(U(eta), eta, eta))-(diff(U(eta), eta))*(diff(mu(eta), eta)) = 0;
Eqn3 := 2*U(eta)*(V(eta)+1)+(W(eta)+(1-n)*eta*U(eta)/(n+1))*(diff(V(eta), eta))-mu(eta)*(diff(V(eta), eta, eta))-(diff(V(eta), eta))*(diff(mu(eta), eta)) = 0;
bcs1 := U(0) = 0, V(0) = 0, W(0) = 0;
bcs2 := U(4) = 0, V(4) = -1;
R1 := dsolve({Eqn1, Eqn2, Eqn3, bcs1, bcs2}, {U(eta), V(eta), W(eta)}, initmesh = 30000, output = listprocedure, numeric);
Warning, computation interrupted
for l from 0 by 2 to 4 do R1(l) end do;
plot1 := odeplot(R1, [eta, U(eta)], 0 .. 4, numpoints = 2000, color = red);

Thankyou in advance :)

Hi there, I am a new user of Maple. I am trying to use intsolve for a course project requiring solving an integral equation. I saw Maple has the intsolve function, then I bought it. However, it didn't seem to work for me.

I typed s(x) and r(y) in and try to solve this problem.

It took 20 minutes; it's still evaluating... what should I do?

I am currently trying to solve a geometric problem where I have to calculate angles in two connected four bar linkages parallel to a serial chain of rotatory joints (closed-loop kinematic chain).

The angle is calculated with

 > alpha:=arctan(exp_y, exp_x):

The expressions exp_y and exp_x contain long products of sines and cosines of 6 other time-dependant angles, square roots of these products, constant geometric lengths (not time-dependant) and constant geometric angles (not time-dependant).

The lenghts are already assumed positive

 > assume (l1>0): # similar for all lengths l2, l3, ...

The time dependant angles are defined as

> qJ_t := Matrix(6, 1, [qJ1(t), qJ2(t), qJ3(t), qJ4(t), qJ5(t), qJ6(t)]): # generalized coordinates of the system in the sense of technical mechanics

Other assumptions are not set, since the angles can be positive as well as negative.

Calculating this expression takes up to two days on a fast computer. In my opinion this takes much too long compared to other calculations with similar amount of variables (more complex robotic structures).Also, the arctan function does not "calculate" a result, it just writes down "arctan(...)".

Is there a way to speed up this calculation e.g. by using more assumptions?

On the arctan help page, the examples suggest that Maple is trying to already simplify the solution e.g. by drawing Pi out of the solution.