Good Morning,

This is the script that I have used to to an integration, as the command ''int'' didn't work because the coputation was really heavy.

p_old := 0;

g := 0; #index

E_tot_net_Point := 0;

for g to 2 do

 h := g+1;

 loopstart := SP_time[g];

 loopend := SP_time[h];

  for tt from loopstart by 0.01 to loopend do

   t := tt;

   P_tot_net_Fkt := t -> C_motore[g]*omega_motore[g]

if t≠SP_time[g] and t ≠ SP_time[h] then

 P_tot_net_Point := P_tot_net_Fkt(t);

 P_tot_net_Point_next := P_tot_net_Fkt(t+0.001);

 plt_P_tot_net_plt[t] := pointplot([t, P_tot_net_Point]);

 E_tot_net_Point := E_tot_net_Point+(P_tot_net_Point*0.01);

 plt_E_tot_net_plt[t] := pointplot([t, E_tot_net_Point]);

 p_old := P_tot_net_Point;

end if;      

  end do:

end do:

plots:-display(entries(plt_E_tot_net_plt, 'nolist'), title = "Energy", color = "Red", symbol = point, symbolsize = 20, gridlines = true, size = [1000, 500], labels = ['[s]', '[Watt*s]'], font = [axes, 12])


I have plotted the function as a series of points over the orizontal axis (t).

Now I would like to calculate the last value of the curve (which is itself a point) as a function of the orizontal axes t. How can I do it?

Something like E_tot_net_Point = f(t)

Thank you for the help

I have a maple homework on the collatz conjecture, but i am new to maple and does not know how to program it.

our professor gave us the first part, i tried a lot of different loop, but keep getting errors. Need HELP.

I need to test all integer m <= 60, i dont know how to set up a loop that can test all 60 numbers.

> restart;

> k := proc (x) options operator, arrow; piecewise(type(x, even), (1/2)*x, 3*x+1) end proc;
/ 1 \
x -> piecewise|type(x, even), - x, 3 x + 1|
\ 2 /
> 
> m := 3;
> k(m);
> for j to 200 while k(%) > 1 do k(%) end do;
> print(m, j+1);

I have a homework ask you to find the first string of (at least)10, 50, 100 consecutive composites. I have no idea how to use maple. HELP

all i can think of is 

>ithprime(i+1) - ithprime(i) = 10

>print(i+1, i)

and combine it with some loop

i dont know how to set up a loop

 need a lot of help 

I saw a presentation Calculus 1 -- I'm pretty sure it was Maple -- that showed how to set up a volume of solid of revolutionvisually rotated the region and looked at it from different points of view.

Is there someplace I could go to find out ho0w to do that.  I am new and in experienced in using Maple.

Thanks in advance.

Tim Wisecarver

Georgetown preparatory School

twisecarver@gprep.org

This post is my attempt to answer the question from   here : how to find all integer points (all points with integer coordinates) in the intersection of two cubes. The following procedure  IntegerPoints  solves a more general problem: it finds all the integer points of a bounded polyhedral region of arbitrary dimension, defined by a system of linear inequalities and / or equations.

Required parameters of the procedure: SN is a set or a list of linear inequalities and/or equations with any number of variables, the Var is the list of variables. The procedure returns the set of all integer points, satisfying the conditions  SN .

Code of the procedure:

restart;

IntegerPoints := proc (SN::{list, set}, Var::list)

local SN1, sn, n, Sol, k, i, s, S, R;

uses PolyhedralSets, SolveTools[Inequality];

SN1 := convert(evalf(SN), fraction);

for sn in SN1 do

if type(sn, `<`) then SN1 := subs(sn = (`<=`(op(sn))), SN1)

end if; end do;

if IsBounded(PolyhedralSet(SN1)) = false then error "The region should be bounded" end if;

n := nops(Var);

Sol := LinearMultivariateSystem(SN, Var);

if Sol = {} then return {} else

k := 0;

for s in Sol do if nops(indets(s[1])) = 1 then

S[0] := [[]];

for i to n do

S[i] := [seq(seq([op(j1), op(j2)], j2 = [isolve(eval(s[i], j1))]), j1 = S[i-1])] end do;

k := k+1; R[k] := op(S[n]);

end if; end do;

convert(R, set);

map(t->rhs~(t), %);

end if;

end proc:

Examples of use:

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z < 12}, [x, y, z]);

  {[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 1, 4], [1, 1, 5], [1, 1, 6], [1, 2, 1], [1, 2, 2], [1, 2, 3], [2, 1, 1], [2, 1, 2],

                                   [2, 1, 3], [2, 1, 4], [2, 2, 1], [3, 1, 1], [3, 1, 2]}

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z = 12}, [x, y, z]);

                                    {[1, 1, 7], [1, 2, 4], [1, 3, 1], [2, 1, 5], [2, 2, 2], [3, 1, 3], [4, 1, 1]}

IntegerPoints([x > 0, y > 0, z > 0, 2*x+3*y+z = 12, x+y+z <= 6], [x, y, z]);

                                                           {[1, 3, 1], [2, 2, 2], [4, 1, 1]}

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z < 12});  #  isolve fails with these examples

              Warning, solutions may have been lost

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z = 12});

              Warning, solutions may have been lost

In the following example (with a visualization) we find all integer point in the intersection of a square and a triangle:

S1 := {x > 0, y > 0, x < 13/2, y < 13/2}:

S2 := {y > (1/4)*x+1, y < 2*x, y+x < 12}:

S := IntegerPoints(`union`(S1, S2), [x, y]):

Region := plots[inequal](`union`(S1, S2), x = 0 .. 7, y = 0 .. 7, color = "LightGreen", nolines):

Points := plot([op(S)], style = point, color = red, symbol = solidcircle):

Square := plottools[curve]([[0, 0], [13/2, 0], [13/2, 13/2], [0, 13/2], [0, 0]], color = blue, thickness = 3):

Triangle := plottools[curve]([[4/7, 8/7], [4, 8], [44/5, 16/5], [4/7, 8/7]], color = blue, thickness = 3):

plots[display](Square, Triangle, Points, Region, scaling = constrained);

In the following example (with a visualization) we find all integer point in the intersection of two cubes. The second cube is obtained from the first cube by rotation with orthogonal matrix  A  and by a translation:

A := <1/3, 2/3, 2/3; -2/3, 2/3, -1/3; -2/3, -1/3, 2/3>:

f := unapply(A^(-1).<x+5, y-4, z-7>, x, y, z):

S1 := {x > 0, y > 0, z > 0, x < 6, y < 6, z < 6}:

S2 := eval(S1, {x = f(x, y, z)[1], y = f(x, y, z)[2], z = f(x, y, z)[3]}):

S := IntegerPoints(`union`(S1, S2), [x, y, z]);

Points := plots[pointplot3d](S, color = red, symbol = box):

Cube := plottools[cuboid]([0, 0, 0], [6, 6, 6], color = blue, linestyle = solid):

F := plottools[transform]((x, y, z)->convert(A.<x, y, z>+<-5, 4, 7>, list)):

plots[display](Cube,  F(Cube), Points, scaling = constrained, linestyle = solid, transparency = 0.7, orientation = [25, 75], axes = normal);

In the example below, all the ways to exchange $ 1 coins of 1, 5, 10, 25 and 50 cents, if the number of coins no more than 8, there is no pennies and there is at least one 50-cent coin:

IntegerPoints({x1 = 0, x2 >= 0, x3 >= 0, x4 >= 0, x5 >= 1,  x1+5*x2+10*x3+25*x4+50*x5 = 100, x1+x2+x3+x4+x5 <= 8}, [x1, x2, x3, x4, x5]);

nops(%);

                              {[0, 0, 0, 0, 2], [0, 0, 0, 2, 1], [0, 0, 5, 0, 1], [0, 1, 2, 1, 1], [0, 2, 4, 0, 1],

                                                 [0, 3, 1, 1, 1], [0, 4, 3, 0, 1], [0, 5, 0, 1, 1]}

                                                                                    8

Integer_points.mw

Addition: Below in my comments another procedure  IntegerPoints1  is presented that solves the same problem.

Hi so im trying to create a loop which allows me to find the first square after this number 
zahl:=1234567: could you guys tell me whats wrong and explain me why it is wrong, i suck at this :/

thank you so much 

zahl:=1234567:
 while sqrt(zahl)<> round(sqrt(zahl)) do  
   zahl:=zahl +1  
     if sqrt(zahl)<> round(sqrt(zahl))then  
  print(zahl):  
 end if:
 end do;

I assume that I'm not providing the correct input to the simplify command to get the simplification that I want.  In particular, for the following code:

assume(n, positive);
simplify(3^(-(1/2)*n)*2^((1/6)*n)-2^((2/3)*n)*6^(-(1/2)*n));
simplify(log(3^(-(1/2)*n)*2^((1/6)*n))-log(2^((2/3)*n)*6^(-(1/2)*n))); 

The expression should evaluate to 0.  However, the first expression does not simplify to 0 (it does not simplify at all in Maple) while the second expression simplifies to 0.

The simplification is fairly easy for the first expression by factoring 6 and combining terms; it seems like I'm not entering the command to simplify in this way.

Hai everyone

may i ask why solution have an error?

hope i have an answer

r

Download soret.mw

I want to build a custom component by the material’s stress-strain curve ,for the component signal input is the x(t)(displacement-time curve), then I can simulate the material’s acceleration-time curve，the stress-strain equation , parameter and the custom component I built has been given in the picture，but the x(t) is same as the s[rel](t),so how to solve this problem? Thanks first.

Hello,

I would like to build a periodic function based on the function x-->x².

For x belonging to [0,2], the function is defined like this :

And for the rest of the domain in R, the function should be periodic with a period of 2.

In other words, I would like to define a function which permits me to obtain this plot :

Sorry, for my picture which is very ugly, but it should enable to explain what I'm looking for.

1) Is a new piecewise function based on f function is the better solution to create this kind of periodic function ?

2) I didn't find or understand with the help how i can create "infinite" piecewise function. May you help me to define this kind of piecewise function ?

Thanks a lot for your help.

P.S: With the term "nested" in my title I only want to say that I want to apply another piecewise function (for example g) on a already created function (for example f). this term may be not very appropriated.

i have run higer order nonlinear ode bvp

but cant solve the error

please help me

Hello,

I would like to build this piecewise function :

I try this manner :

f:=x->piecewise(x=>0 and x <= 1, x^2,x > 1 and x<=2, (2-x)^2);

But, it doesn't work. I receive the following error message :

Error, `>` unexpected

Do you have some ideas about my mistake?

Thank you for your help.

Hello;

I need some help to compute the series approximation of the modulus and argument of hankel function for large x. The code display

 Error, (in asympt) unable to compute series

Thanks  for helping me.

#We define the hankel function as
#HankelH1(v,x) = BesselJ(v,x) + I*BesselY(v,x), where BesselJ and BesselY are bessel function.
#In this question the parameter "v " is  fixed. "

# Code

restart:
with(MTM):
HankelH1(v, x);
# The modulus of Hankel function
Mn:=x->abs(HankelH1(v, x));
thetan:=x->argument(HankelH1(v, x));
phin:=x->argument(diff(HankelH1(v, x),x));
dervthetan:=x->diff(thetan(x),x);
dervphin:=x->diff(phin(x),x);
# Compute series
series(thetan(x),x=infinity,7);
series(phin(x),x=infinity,7);
series(Nn(x),x=infinity, 7);
series(Mn(x),x=infinity,7);
# I define the following function
f:=x->(Nm(x)/Mn(x))*exp(I*(phin(x)-thetan(x)));

# Series approximation
series(f(x),x=infinity, 7);

Dear all;

Hello everybody, I need your help to dispaly some values obtained using my function f. When I run the code there is no results obtained. Many thanks.

restart:

# The vectors e(i) satify the folowing conditions
e(0)*e(1)=e(n-1) assuming  1<n;
e(0)*e(0)=e(2):
e(1)*e(1)=e(n-1) assuming  1<n: :
e(2)*e(1)=e(n) assuming  1<n:
#
for i from 1  to n-1 do
e(i)*e(0)=e(i+1);
end do:

# We define the function f
f:=e(0)->e(0)+(n-3)*e(1);
f:=e(1)->(n-2)*e(1);
for i from 2  to 3 do
f:=e(i)->(n+i-3)*e(i)+(i-1)*(n-3)*e(n-3+i);
end do:

for i from 4 to n do
f:=e(i)->(n+i-3)*e(i)
end do:

# We define the two vectors
x:=sum(alpha(k)*e(k),k=0..n);
y:=sum(beta(k)*e(k),k=0..n);

#Question : I would like to compute the following  but there is no display of the solution. 
(x*y);
f(x*y);
f(x);
f(y);
x*f(y);
f(x)*y;
f(x*y)- f(x)*y-x*f(y);