Hi,

Thank you very much for your idea in previous discussion: this is the link.

http://www.mapleprimes.com/questions/202744-Calculs-Using-Maple

I asked how can I plot the phase portrait of this system

Sys1 := {diff(r(t),t) = r(t)^2*sin(theta(t)), diff(theta(t),t) = -r(t)^2*(-2*cos(theta(t))^2+1)};

I get this answer:

1) For me it's not necessary to give an initial condition to plot the pharse portrait.

2) What is the line in this phase portrait.

3) Here, I try a second method using Matlab code, I get:

and this is the code:

[r, theta] = meshgrid(0:pi/4:2*pi, 0:pi/4:2*pi);
rdot =r.^2.*sin(theta) ;
thetadot = r.^2.*(1-2*sin(theta).^2);
quiver(r,theta,rdot,thetadot)
xlabel('r')
ylabel('theta')

the arrow in the phase portrait are not the same.

Can someone give me more clear information about this problem.

Many thinks.

Hi, I'm trying to write a maple procedure to differentiate a function from first principles and so far I have

First_Principle := proc (f); for f do A := simplify((f(x+h)-f(x))/h); Limit(A, h = 0) = limit(A, h = 0) end do end proc

and from there I have no idea where to go. Does anybody have any ideas? Thanks!

How to solve the inequality

with Maple?

My attempts were the following.

Warning, solutions may have been lost

Of course, this works

f(1);
,

but one wishes to describe the solutions in the dependence on the parameter a. Unfortunately, both

and

produce wrong outputs(An SCR has been submitted by me.).

Egor has two parents, four grandparents, and so on.
Write an explicit formula and a recursive formula for the number of ancestors Egor has if we go back n generations.                

what would be the figure back to 25 generations ?

hi..i am a problem with solving following ....please help me ....thanks alot

dsys3 := {10*f2(x)+12*(diff(f1(x), x))+14*f3(x) = 0, 2*(diff(f1(x), x, x))+4*(diff(f2(x), x))+6*(diff(f3(x), x)) = 0, 16*(diff(f3(x), x, x, x, x))+19*(diff(f3(x), x, x))+22*(diff(f1(x), x))+25*f2(x)+27*f3(x)+29*f3(x)+31+32 = 0, f1(0) = 0, f1(1) = 0, f2(0) = 0, f2(1) = 0, f3(0) = 0, f3(1) = 0, ((D@@1)(f1))(0) = 0, ((D@@1)(f1))(1) = 0, ((D@@1)(f2))(0) = 0, ((D@@1)(f2))(1) = 0, ((D@@1)(f3))(0) = 0, ((D@@1)(f3))(1) = 0}; dsol5 := dsolve(dsys3, 'maxmesh' = 500, numeric, range = 0 .. 1, abserr = .1, output = listprocedure); fy3 := eval(f3(x), dsol5); fy2 := eval(f2(x), dsol5); fy1 := eval(f1(x), dsol5)

ERROR.mw

Hej everyone,

I am trying to run a Maple script for multiple parameters on our cluster, so far so good. I found that I can pass for example a:=1 and b:=2 to a Maple script using

maple -c "a:=1;" -c "b:=2;" < script.mpl

Is that the best/only way? For convenience I would like to read the parameters from a file, my first idea was to work with sed, so I tried

maple $(sed -n 1'p' data) < script.mpl

This should take the first line from data (which is -c "a:=1;" -c "b:=2;"), so that I would expect to get the same result.
Unfortunately, this is not the case, only the first parameter a:=1 is passed correctly and non of the further parameters.

I've appended the files, I used.

This may be not a Maple problem but rather me making a mistake using sed, but maybe you can see where I made a mistake.

Thanks in advance,

Sören

Download script.mw

The mw-file has to be exported as an mpl-file first. In my case data just contains one line: -c "a:=1;" -c "b:=2;" -c "c:=3;" (I couldn't upload either)

Hi,

I have a problem with a 1D loockup table with a Data set file. It seems that there are some problems with the file but I don't understand which.

I tried to use a .xlsx file and the result is this error (note that the table in excel has 17 rows, so at the 18 there is nothing):

Then I tried with a .csv file and I had this error:

I put below the MapleSim file, is not completed: just to have a look at the errors and the attached file, so don't consider other things.

srpingdamper.msim

the .xlsx attached files is the following, while i'm not able to upload the .csv but it's the same one just saved as .csv:

damperTR.xlsx

Thanks a lot.

Hello people in Mapleprimes,

I want to know how to use apply rule in the following problem.

Please tell me how to do about this.

applyrule(2*F(K,L)=F(2*K,2*L),2*F(K,L));

returns F(2*K,2*L), which is right.

But, what I want to do is to apply the rule with holding n as general variable of real number.

But, what I can get through 

applyrule(n*F(K,L)=F(n*K,n*L),2*F(K,L));

is only 2*F(K,L) not F(2*K,2*L).

What should I write additinally to the above to have Maple calcurate and produce the outcome as F(2*K,2*L), F(3*K,3*L) and so on?

Best wishes.

Taro

I'm brand new to Maple and was assigned a problem to modify a code provided by my professor to incorporate variable window size. However, I don't know where to begin or what I'm doing. I've attached the link to the problem below. Plz help.  NA1_Project_IDW_01.2014_Fall.pdf

Let (G, ·) be a group and X any set. Let F be the set of functions with domain

X and range G. Define a binary operation ∗ on F by (f ∗ g)(x) := f(x) · g(x). Is

prove that this is so.

Yes, (F, ∗) is a group.

prove it.

Exercise Prove that (-1)u = - u in any vector space. Note that (-1)u means the number -1 is multiplied to the vector u, and - u means the negative vector in the fourth property of the definition of vector spaces.

Answer

Exercise Prove that (a₁u₁ + a₂u₂) + (b₁u₁ + b₂u₂) = (a₁ + b₁)u₁ + (a₂ + b₂)u₂ in any vector space.

Answer

Exercise Give a detailed reason why, in any vector space,

• u + v = 0 ⇒ u = - v.

• 3u + 2v - 4w = 0 ⇒ v = - 3/2 u + 2w.

Hi, I've written some matrix computation in maple 18 . I wanna open this file in maple 16 ? how is it possible ? any help is appreciated in advance.

Hello, all.

I would like to plot a weighted graph.

Edge weights are floating point numbers (for example, 1,0; 2,0, 3,0, ...).

I tried:

>M := MATRIX([[0., 2.0, 1.0], [2.0, 0., 3.0], [1.0, 3.0, 0.]]);

> with(GraphTheory):
> G := Graph(undirected,weighted,M);

But it doesn't work.

Of course, I can take the edge weight and multiply it by 10 or 100.
May be somebody could provide a more simple solution.

Hello,

I have the following condition with two unknowns: t1 and t3:

-N*exp(Q1*alpha*eta*t1/(N*w))*exp(-(((N*w-z)*t1^2+((-N*w+z)*t3+2*Q1)*t1+(1/2)*t3*(t3*(N*w-z)-2*Q1))*alpha-2*N*w*C[max]*(t1-(1/2)*t3))*eta/(N*w))*S1*upsilon*w+N*exp(Q1*alpha*eta*t1/(N*w))*exp(-(((N*w-z)*t1^2+((-N*w+z)*t1+2*Q1)*t1+(1/2)*t1*((N*w-z)*t1-2*Q1))*alpha-2*N*w*C[max]*(t1-(1/2)*t1))*eta/(w*N))*S1*upsilon*w+K1^2*exp((1/2)*t1^2*alpha*eta*z/(N*w))*exp(-(1/2)*t1^2*alpha*eta)*exp(t1*eta*C[max])*alpha*eta*t1*z-K1^2*exp((1/2)*t1^2*alpha*eta*z/(N*w))*exp(-(1/2)*t1^2*alpha*eta)*exp(t1*eta*C[max])*alpha*eta*t3*z+K1*exp((1/2)*t1^2*alpha*eta*z/(N*w))*exp(-(1/2)*t1^2*alpha*eta)*exp(t1*eta*C[max])*S1*alpha*eta*t1*z-K1*exp((1/2)*t1^2*alpha*eta*z/(N*w))*exp(-(1/2)*t1^2*alpha*eta)*exp(t1*eta*C[max])*S1*alpha*eta*t3*z = 0

I know that this condition holds when t1=t3. Does there exist an additional solution for t1 and t3 which satisfies this condition?

Thanks,

Dmitry