Hi,

I have a Matrix whose entries are polynomials in several formal parameters (the matrix is sparse and the polynomials are rather simple, though inverses of the parameters may also arise).
Then, when I compute the kernel with LinearAlgebra-NullSpace, maple naturally gives a basis of solutions over the same ring of polynomials.

Now for some reason there are some parameters that I don't want to see in the solutions (all but two of them, actually).

How can I compute the part of the kernel that lives in $\mathbb{Z}[a,a^{-1}, b,b^{-1}]$, i.e. that involves only the first two parameters?

Thank you,

NoThik

Edit : the coefficients of the polynomials are integers, and I expect the kernel elements to have integer coefficients as well.

Find the Fourier Series for the function f(x) defined as follows, and compare the graphs of some truncated Fourier series (try 1,2,3,5,6,30, terms) with the graph of f(x).

f(x)= min(|x|, pi/2), -pi less than or equal to x less than or equal to pi.

Also, let f(x) be periodic with a period of 2*pi. Thanks.

Hey, I have some data points:

I than entered:

X := Data[() .. (), 1];
Y := Data[() .. (), 2];
f := a+b*x^c+d*x^e;
Statistics:-NonlinearFit(f, X, Y, x);
 and when I entered Statistics:-NonlinearFit(f, X, Y, x);, It wouldn't output anything.

I am using maple 17 and I was trying to run a nested loop as shown below. I expected to get all combinations of i and j (9 components). But i and j always remain 3. What is the problem here? Can anyone suggest alternate solutions? I dont understand in which order maple increments i,j or k in each loop.

for k to 9 do

      for i to 3 do

          for j to 3 do

          a[k] := i, j

          end do

      end do

end do;

table([1=(3,3),2=(3,3),3=(3,3),4=(3,3),5=(3,3),6=(3,3),7=(3,3),9=(3,3),9=(3,3)

Thanks in advance

Hello,

I would like to make the operator L (factorization)

in the form

Thank you for your answer.

Greetings,

I have a differential equation:

an the solution:

I want to substitute

ekf := omega = sqrt(c/m);

algsubs(ekf, sol);
but sol does not change.

This works a bit but does not substitute everything in a way that you would expedect

This is the result:

Thanks in advance

I want to find the sum of the reciprocals of the factorials of the elements of in the sets [[0,2,3,0,0], [2,3,0,0,0], [1,1,2,2,0] .....]]  -----(1)

For example 1/(0!2!3!0!0!) + 1/(2!3!0!0!)+ 1/(1!1!2!2!0!)+........;----(2)

The lists are the partion(5) made into a list of 5 elements with 0s filling the partitions with less than 5 elemenst such as (2,2,1)=>(2,2,1,0,0)

All I need is help on how to make from (1) the sum (2) which is the sum of the reciprocals of the products of the factorials of each set. 

Hello,

I would like to choose the maximum interval which is here B3 not the maximum value.I used max which gave me the maximum value. Any ideas?

This is the output:

B1:=[2., 36.718220544331125]
B2:=[2., 35.08378362904457]
B3:=[2., 51.78712780854305]
datamax:= 51.78712780854305

how to get datamax:= [2., 51.78712780854305] instead of datamax:= 51.78712780854305?

This is my Maple code:

> restart;
> n := 3;
> for i to n do A[i] := RandomTools:-Generate(distribution(Uniform(.5, .75))) end do;
> eq := diff(X(t), t) = -S*X(t);
> ic[1] := X(0) = 150;
> for i to n do s[i] := dsolve({ic[1], subs(S = A[i], eq)}, X(t), range = 0 .. 10, numeric) end do;
> for i to n do B[i] := eval([t, X(t)], s[i](2)) end do;

datamax := max([seq(B[i], i = 1 .. n)]);

Thank you

Greetings to all.

Just a quick question, probably something obvious, please bear with me Your patience is appreciated.

As part of solving a posted problem at math.SE it became necessary for me to compute the bivariate generating function

sum(sum(binomial(2*q,n-1)*z^q/q!, q=0..infinity)*w^n,n=1..infinity)

The correct answer (in fact a simple computation) is

w exp(z(w+1)^2).

When asked to compute this sum with the command shown, Maple produces

hypergeom([], [], z) w

My question for you is, how does this represent the above answer that I computed manually? When I simplify the hypergeometric, Maple says

w exp(z)

which would appear to be the wrong value. Could someone enlighten me regarding this discrepancy?

BTW there seems to be no tag "hypergeometric" and I don't have enough points to create it.

Best regards, Marko Riedel

This is a problem of Russian in 2005: Solve inequality 

(3-x-sqrt(5-x^2))/(cos((2*x-7)/4)-cos((x-5)/4))>=0.

The problem can solve by hand. We only solve

(3-x-sqrt(5-x^2))/((2*x-7)/4-(x-5)/4) >=0.

I tried

eq1:=(3-x-sqrt(5-x^2))/((2*x-7)/4-(x-5)/4);

solve(eq1>=0,x);

I got

RealRange(1,Open(2)), RealRange(Open(2),5^(1/2)).

If I solve directly by

eq:=(3-x-sqrt(5-x^2))/(cos((2*x-7)/4)-cos((x-5)/4));

solve(eq>=0,x);

I don't get the exact answer. 

P.S. with Mathematica, I only use

Reduce[(3 - x - Sqrt[5 - x^2])/(Cos[(2 x - 7)/4] - Cos[(x - 5)/4]) >=
0, x, Reals]

I have an expression 

f:= x1 * x2 * x3^4;

I would like to do something like subs( (x1,x3) = (x3,x1), f) and get 

x1^4 * x2 * x3

back.

However this does not work. Is there some _lightweight_ way of doing it?

Hello everyone,

Here is my question,

restart:with(plots):

Eq:= diff(f(y),y$3)+f(y)*diff(f(y),y$2)-diff(f(y),y$1)^2-S*(diff(f(y),y$1)+y/2*diff(f(y),y$2))=0;

bc:=D(f)(0)=1,f(0)=0,(D@@2)(f)(b)=0, f(b)=S*b/2;

How we can plot b vs S?

Thanks

Hello,

I am solving eq1 and then plot it using odeplot. Then I am extracting data from odeplot using op[1,1]. From that I only care about  one point so I am using pt[1] := dt[1]([1, 2]); then I ploted it using pointplot. I run a do loop and at the end I am ploting all the points.

My code works fine but I am trying to have a line contacting each points. I tried  style=line but it did not work.

How can I have a line contacting each points on my final plot (which is display([pl[1], seq(allpl[k], k = 1 .. 3)])) ?

This is my code:

> restart; with(DEtools); with(plots);
> with(DEtools); with(plots);
> A := 0.2e-1; B := 10^(-5); k := 0;
> eq1 := diff(X(t), t) = -(A+B*X(t))*X(t);

> ic[1] := X(365*k) = 1000;
> s[1] := dsolve({eq1, ic[1]}, X(t), range = 0 .. .365, numeric);
> p[1] := odeplot(s[1], [[t, X(t)]], t = 0 .. .365);
> dt[1] := op([1, 1], p[1]);
> pt[1] := dt[1]([1, 2]);
> pl[1] := pointplot(pt[1], axes = boxed);
> for k to 3 do
tk := 365*k;
A := rhs(s[k](tk)[2]);
ic[k+1] := X(tk) = 500.*A;
s[k+1] := dsolve({eq1, ic[k+1]}, X(t), range = tk .. 2*tk, numeric);
p[k+1] := odeplot(s[k+1], [[t, X(t)]], t = tk .. 2*tk);
dt[k+1] := op([1, 1], p[k+1]);
pt[k+1] := dt[k+1]([1, 2]);
pl[k+1] := pointplot(pt[k+1], axes = boxed);
allpl[k] := display([pl[k+1]])
end do;
> display([pl[1], seq(allpl[k], k = 1 .. 3)]);

Thank you

I installed the shared library from http://www.maplesoft.com/applications/view.aspx?SID=1691

and it seems to have loaded ok. Now I can do

with(share);
See ?share and ?share,contents for information about the share library
                               []

When browsing help, I noticed it has a function called IntSolve() for solving linear integral equations. Maple itself has function called intsolve() which is build-in.

Since Maple is case sensitive, I assumed when I type IntSolve() then it will find the shared version of IntSolve() automatically and not call Maple's own intsolve. And that is indeed what happed. I typed an example from the help on share as you can see

eq1 := p(x)-1/2*Int(x*y*p(y),y=0....1)=5/6*x;
IntSolve(eq1,p(x));
                             7775  
                             ---- x
                             7776  
intsolve(eq1,p(x));
                            p(x) = x

Ok, So one of them is a little wrong :).

But suppose I want to explicitly add the share library there, in the call, as prefix, so that when I look at the code later, I can see which function I am calling. How should one do this? I tried

share:-IntSolve(eq1,p(x));
    Error, `share` does not evaluate to a module
share[IntSolve](eq1,p(x));

The strange thing, is that this worked

IntSolve[IntSolve](eq1,p(x));

Basically, all what I want is to way to add the share name to all the calls somewhere, so I know, when looking at the code, which call is from which library or package.

Any idea how to do this? So to avoid any possible name clash since the share library seems large.

I am using Maple 17.02

I am using 17.02. I looked at this question, http://www.mapleprimes.com/questions/35783-Inverse-Cosine-And-Sine-Transform but I do not understand the answer.

I have a function F(w), and want to find f(t) using inverse cosine forurier transform.

Here is F(w)

F:= w-> cos(w*L)*cos(w*x)/(w^6+w^2*k);

All parameters are positive.

How can I use the answer shown above, or another function to find f(t)?

I can offcourse simply apply the definition myself, and I did:

int(F(w)*cos(w*x),w=0..infinity) assuming positive;

but the result is

So it does not converge. And I just wanted to check using a build-in function to verify. Strange than Maple does not have these in its inttrans package.

with(inttrans);