A lemniscate is a polar curve of the form r^2=a^2*cos(2*theta) or r^2=a^2*sin(2*theta). I have just started using Maple and I wrote the following commands: 

> with(plots):
> polarplot(2*sqrt(cos(2*t)), axiscoordinates = cartesian, angularunit = radian, color = "Black");
 

But I am getting the following graph 

which is not satisfactory since some points are missing. I know that using the square root may have caused this, but I am not sure as to how should I resolve this issue. I used plus/minus symbol before the expression 2(cos(2t))^(0.5) but there was an error and the discontinuity still persisits. Kindly help me in plotting this curve. 

Thank you.

Hi everyone!

I would really appreciate if someone could give me a hand on telling me what is wrong with this problem! pdsolve gives the error: Error, (in pdsolve/sys/info) found functions with same name but depending on different arguments in the given DE system: {f(0, y), f(x, 0), f(x, y), (D[2](f))(0, y), (D[2](f))(x, 0)}.


Thanks in advance!!! 

Dear All,

I am plotting the following function using implicitplot command.:

plots[implicitplot3d]((17.31626331*M^3-(4*(z[1]-z[2])^2*M^2-1.171300684*(z[1]+z[2])^2)*(1.082266457-2*M)*(1.082266457-3*M))^2 = 4.598621420*(z[1]+z[2])^2*M*(1.082266457-2*M)^3*(4*(z[1]-z[2])^2*M^2-1.171300684*(z[1]+z[2])^2), M = 0 .. 1, z[1] = 0 .. 10, z[2] = -10 .. 0);

How can I extract data points from the plot obtained

with(DEtools):
rho := 0.1;
w0 := 2;
sys := {diff(x(t),t) = y(t),diff(y(t),t) = -2*rho*y - w0^2*x};
DEplot(sys, [x(t), y(t)], t = 0 .. 12, [[x(0) = 1, y(0) = 0]]);

i use flow above, would like to plot a circle move from right hand side to left hand side

and see how a circle influence the flow diagram in animation like weather diagram

I resolved the coefficients to a 2nd order diff eq of the form:ay''+by'+cy=f(t)

I have included the .mw file for convenience at the link at the bottom of the page.  I resolved the coefficients in 2 different ways & they do not concur.  The 1st approach used the LaPlace transform & partial fraction decomposition.  The coefficient results are given by equations # 14 & 15.  The 2nd approach used undetermined coefficients where I assumed the particular solution and then applied the initial conditions to resolve the coefficients pertaining to the homogeneous solution which are given in the results listed in equation #23.  Noted in the 1st case the coeff's are A₃ & A₄ and for the 2nd approach the coeff's are A₁ & A₂.  I have worked this numerous times & do not understand why they do not concur.  So I thought I should get some fresh eyes on the problem to find where I may have gone wrong.

Any new perspective will be greatly apprecieated.

I had trouble uploading the .mw file so I have included an alternative link to retrieve the file if the code contents is illegible or you cannot dowlad the file drectly from the weblink  Download coeffs_of_homogen_soln_discrepancy.mw.  You should be able to download from the alternative link below once you paste the link into your browser.  If you cannot & wish for me to provide the file in some other fashion respond with some specific instructions & I will attempt to get the file to you.

https://unl.box.com/s/dywe90wwpy0t4ilkuxshkivz2z26mud8

Thanks 4 any help you can provide.

Download coeffs_of_homogen_soln_discrepancy.mw

Download from_net.mw

I attempted to use convolution in AudioTools to convolve 2 vectors with arbitrary values, but could not since the operation is expecting numeric values.  Can this be done?    Is there not a convolution operastor in the LinearAlgebra package?  See attached:

convolutionWithAu[1].mw

After several attempts on this question,

Int(x*sqrt(2*x^4+3),x) with substitution u = sqrt(2)*x^2,

I don't seem to find the solution. Can you guys help me?

I have this function:

Z := (cos((1/2)*x)-I*sin((1/2)*x))*A0/r^(1/2)+(cos((1/2)*x)+I*sin((1/2)*x))*r^(1/2)*A1+(cos(3*x*(1/2))+I*sin(3*x*(1/2)))*r^(3/2)*A2;

I wanna to obtain another function Y which equal to

Y:=f1*(Re(Z)+Re(Z'))+f2*(Im(Z)+Im(Z'));

where: f1 and f2 are constant 

thanks :)

I am trying to have the output of DETOOLS as 3dpolarplot. As in the following example:

EF := {2*(diff(w[2](t), t)) = 10, diff(w[1](t), t) = sqrt(2/w[1](t)), diff(w[3](t), t) = 0}; with(DEtools); DEplot3d(EF, {w[1](t), w[2](t), w[3](t)}, t = 0 .. 100, [[w[1](0) = 1, w[2](0) = 0, w[3](0) = 0]], scene = [w[1](t), w[2](t), w[3](t)], stepsize = .1, orientation = [139, -106])

how can I get the output as a polarplot in 3d where, w[2] and w[3] have range 0..2*pi.

Please help in this respect asap.

Hello, I have a question.  I don't know why, but results of my calculations can't be saved in raschet document. This document excists, but there is no information in it! And I have an error with floating point format. How to solve that problems?

> restart;
> Digits := 5;
> NULL;
> NULL;
> NULL;
> NULL;
> NULL;
> ScS0 := P/(phi*f*kc*k0*deltad*Bm*etat);
> NULL;
> NULL;
> Sc := sqrt(ScS0);
> A := sqrt(Sc);
> B := A;
> NULL;
> mue := mu0*mur/(1+mur*dzet/lm);
> lm := 2*(LCA-A+(LC0+A))+dzet;
> NULL;
> LC0 := 3*A; LCA := .4*LC0; LD := .9*LC0;
> NULL;
> NULL;
> w1 := EE/(2*Pi*f*Bm*Sc);
> Lm := mue*w1^2*Sc/lm;
> ;
> I11 := sqrt((w2*Id/w1)^2+I0^2);
> ;
> NULL;
> ;
> h1 := sqrt(RAT*I11/deltad);
> ;
> h2 := sqrt(RAT*Id/deltad);
> NULL;
> A := .6;
> Ud := 35000;
> Id := 413;
> R := Ud^2/P;
> P := Ud*Id;
> P1 := P/eta;
> R1 := EE/I11;
> EE := 110000;
> I0 := EE/(2*Pi*f*Lm);
> w2 := w1*sqrt(P*R)/EE;
> mu0 := 4*Pi*10^(-7);
> mur := 1000;
> f := 50;
> k0 := .25;
> kc := .98;
> deltad := 0.3e7;
> Bm := 1.45;
> etat := .98;
> eta := .95;
> RAT := 1;
> dzet := 0.1e-3;
> phi := .5;
> W1 := evalf(w1);
324.55
> LLm := evalf(Lm);
13.407
> W2 := evalf(w2);
103.26
> evalf(lm);
7.2457
> evalf(LC0);
2.5877
> evalf(LCA);
1.0351
> Imax := evalf(I0);
26.117
> P1;
7
1.5215 10
> Rd := evalf(R);
84.746
> Bmm := evalf(mue*w1*I0/lm);
1.4500
> hâ := (.9*LC0*1000)/(w2+1)-4;

> evalf(hâ);

h￢
> Pred := Id/deltad;
> evalf(Pred);
0.00013767
> NULL;
> NULL;
> ll := hâ*(w2+1)+4*w2;
> NULL;
> a := am*nâ/nx;
> NULL;

> Pol := Vit*nâ;
> am := 5.1;
> am := 5.1;
> nâ := 4;
> evalf(a);
20.4
----
nx
> Vit := 35.19;
> evalf(Pol);
140.76
> plotnToka := Id/Pol;
> evalf(Id/Pol);
2.9341
> NULL;
> I1 := evalf(I11);
133.98
> NULL;
> evalf(mue);
0.0012395
> NULL;
> evalf(EE/I11);
821.02
> NULL;
> pr := "%";
"%"
> fd := fopen("C:\\Users\\Ñåìåí\\Desktop\\ÍÈÐ\\raschet4.ms", WRITE); fprintf(fd, "E=%g;Ud=%g;Imax=%g;P=%g;FR=%g;A=%g;B=%g;LC0=%g;LD=%g;LCA=%g;R=%g;BM=%g;", EE, Ud, Imax, P, f, A, B, LC0, LD, LCA, Rd, Bm); fprintf(fd, "\n %s P=%g;Id = %g;Bm=%g;I1=%g;Bmm=%g", pr, P, Id, Bm, I1, Bmm);
Error, (in fprintf) number expected for floating point format
Error, (in fprintf) number expected for floating point format
> fprintf(fd, "\n %s W1 = %g; W2 = %g; Lm=%g; Sc=%g; dzet=%g", pr, W1, W2, LLm, Sc, dzet);
Error, (in fprintf) file descriptor not in use
> fclose(fd);
Error, (in fclose) file descriptor not in use

OnesM:=Matrix(`%id`=119376536)

Anyone can solve this??

Many thanks!

Hi,

I have a first order differential eq. for some variable say $r(x)$, where $x$ is the independent variable.

After solving this differential equation numerically, I want to use its solution in other expression for $r(x)$ and plot the expession with $x$.

Please let me know how to do it.

Thanks in advance.

Probability_density_normalization.mw

In this code I'm trying to separately normalize two independent probability densities and then combine them to get the joint probability density that's normalized and then use it to calculate the probability that the two variables are equal. fD(x) is a Gaussian divided by x^2 and fA(x) is a Gaussian. The first problem occurs when I'm checking the normalization of the joint probability density by doing the double integral over all space for fD(x)*fA(y)dxdy, I get weird vanishing number when the parameter "hartree" takes a certain value, namely 27.211. If I change hartree to 27 or 1 or 2 it all worked, but 27.211 is not good. Also later when I do a single integral over all space for fD(x)*fA(x)dx to get the probability that these two are equal, I find the result is dependent on hartree. This hartree thing is a unit conversion in my physical problem and in principle should not interfere with either the normalization or the probability result at all. I suspect this is a coding bug but I can't find what it is. I'd appreciate any input.

Thank you very much!

Edit: I found out that the problem with the double integral normalization may have something to do with the discretization for numerical evaluation of the integral, since if I change the lower bound to 1/hartree and upper to 10/hartree then it's fine, however if I use lower bound at 1/hartree and upper at 5/hartree it doesn't work, although the distribution has no value between 5/hartree and 10/hartree. However after this is fixed I still have the problem with the single integral over all space for fD(x)*fA(x)dx changing with hartree. Well as a probability I would expect the integral to be bound between 0 and 1, but since it almost linearly depends on hartree, at hartree around 27 I would get the integral value to be about 25, which doesn't make sense. In fact, I now suspect it is not Maple, but my calculation of the probability of the two random variables taking the same value is wrong, I'd appreciate it very much if someone can confirm this.

I would have love to attach a document because I try pasting it but it is not allowed I want to integrate something of this nature;... I don't even get how to write anything here maybe because am using a phone.53e77f9f0cf21cc29fd9d4e8.pdf 

This is the paper i'm working on,

1) I couldn't get 11a and 11b on page 1918.

2) I don't know how to integrate 13b to 13e. Please somebody help my career I will never forget it.

My e-mail is foyt22@gmail.com