Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12


"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "

Proof:

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)]  in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."

Lemma:

"If [p/(q)]  in A with q  in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

(1)

"for all r  in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"

y > p/q

Now:

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

(2)

"a good choice for r, for instance r=2*p+1, we get from (1)"

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

(3)

thus:

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

(4)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

(5)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

(6)

"finally, supA<0"

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)

"so, for any x  in p/(q) in A, there exists y  in A such that y>x: in conclusion A does'nt admit a greatest element."

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):

`and`(p > q, p^2-2*q^2 >= 0)

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:

2*p^2-2*q^2 >= 1:

"for any r  in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:

"So, for any greatest element m of A, it exists m', greatest element of A such as:"

diff(m(x), x) < m(x):

"A does not admit a least upper bound."

``


Download rational_numbers.mw

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "

Proof:

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)]  in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."

Lemma:

"If [p/(q)]  in A with q  in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

(1)

"for all r  in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"

y > p/q

Now:

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

(2)

"a good choice for r, for instance r=2*p+1, we get from (1)"

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

(3)

thus:

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

(4)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

(5)

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

(6)

"finally, supA<0"

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)

"so, for any x  in p/(q) in A, there exists y  in A such that y>x: in conclusion A does'nt admit a greatest element."

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):

`and`(p > q, p^2-2*q^2 >= 0)

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:

2*p^2-2*q^2 >= 1:

"for any r  in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:

"So, for any greatest element m of A, it exists m', greatest element of A such as:"

diff(m(x), x) < m(x):

"A does not admit a least upper bound."

``


Download rational_numbers.mw

Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

I'd like to plot the following inequalities:

sqrt(x)<=1/sqrt(2)

1/sqrt(2)<sqrt(x)<=1/sqrt(2)

 

> restart;
> with(plots);
> Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-2*(diff(f(eta), eta))^2-M^2*(diff(f(eta), eta)) = 0;
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + f(eta) |----- |----- f(eta)||
\ deta \ deta \ deta /// \ deta \ deta //

2
/ d \ 2 / d \
- 2 |----- f(eta)| - M |----- f(eta)| = 0
\ deta / \ deta /
> Eq2 := 1+(4/3)*R*(diff(theta(eta), eta, eta))+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)) = 0;
4 / d / d \\
1 + - R |----- |----- theta(eta)||
3 \ deta \ deta //

/ / d \ / d \ \
+ Pr |f(eta) |----- theta(eta)| - |----- f(eta)| theta(eta)| = 0
\ \ deta / \ deta / /
> bcs1 := f(0) = S, (D(f))(0) = 1+L*G, (D(D(f)))(0) = .1, f(6) = 0;
f(0) = S, D(f)(0) = 1 + L G, @@(D, 2)(f)(0) = 0.1, f(6) = 0
> fixedparameter := [S = .1, M = .1];
[S = 0.1, M = 0.1]
> Eq3 := eval(Eq1, fixedparameter);
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + f(eta) |----- |----- f(eta)||
\ deta \ deta \ deta /// \ deta \ deta //

2
/ d \ / d \
- 2 |----- f(eta)| - 0.01 |----- f(eta)| = 0
\ deta / \ deta /
> fixedparameter := [R = .1, Pr = .7];
[R = 0.1, Pr = 0.7]
> Eq4 := eval(Eq2, fixedparameter);
/ d / d \\ / d \
1 + 0.1333333333 |----- |----- theta(eta)|| + 0.7 f(eta) |----- theta(eta)|
\ deta \ deta // \ deta /

/ d \
- 0.7 |----- f(eta)| theta(eta) = 0
\ deta /
> bcs2 := theta(0) = 1+T*B, (D(theta))(6) = B, theta(6) = 0;
theta(0) = 1 + T B, D(theta)(6) = B, theta(6) = 0

> T := .1; B := .1;
0.1
0.1
> L := [0., .1, .2, .3];
[0., 0.1, 0.2, 0.3]
> for k to 4 do R := dsolve(eval({Eq3, Eq4, bcs1, bcs2}, L = L[k]), [f(eta), theta(eta)], numeric, output = listprocedure); Y || k := rhs(R[2]); YL || k := rhs(R[3]) end do;
Error, (in dsolve/numeric/bvp/convertsys) too many boundary conditions: expected 6, got 7
> plot([YL || (1 .. 4)], 0 .. 6, 1 .. -.2, labels = [eta, diff(f(eta), eta)]);

 

 

 

       

 

 

                                                                          

A major-league pitcher releases a ball at a point 6 feet above the ground and 58 feet from home plate at a speed of 100 mi/hr ,  

If gravity had no effect, the ball would travel along a line and cross home plate 4 feet off the ground. Find the drop D caused by gravity.                                                                                                                                                      

 

NB: in this problem the angle alpha is the angle between the horizontal and the direction of the released ball. Since the ball is dropping, alpha will be negative.

I got a problem with a difficult ode,the commands are below.

restart;
sys := 1.*(diff(x(t), t, t)) = piecewise(b(t) = 1, 0, 1003.0-1000.*x(t)-30.*(diff(x(t), t))-25.*signum(diff(x(t), t)-.1)-.3*signum(diff(x(t), t))*exp(-2*abs(diff(x(t), t)))), x(0) = 1, (D(x))(0) = 0;
mu := 100;
stick := [diff(x(t), t) = .1, b(t) = piecewise((1000.-1000.*x(t))^2 < 10000, 1, 0)];
slip := [[0, 10000 < (1000.-1000.*x(t))^2], b(t) = 0];
sol:=dsolve({sys,b(0)=0},numeric,discrete_variables=[b(t)::float],events=[stick,slip],event_maxiter=1000000,output=listprocedure,maxfun=0,range=0..8);

any advice is appreciated.

How can I solve raychaudhuri equations numerically using GRtensor?

I want to use Maple (dsolve command) to solve this problem, how do I input the command? Part (a) is enought.

I have tried the command, but Maple says "Warning, for {u(t)} arbitrary the system is inconsistent".

Thank you very much for your help.

Further tags : implicit form, Matlab codegen for GPOPS purposes

Hi there,

I would like to introduce my self and ask you for an advice. 

I have a four linked robot that i would really like to control by the meaning of the optimal control theory. The arm has 4 actuators (giving torques U) in corrispondence of 4 rotational joints. P parameters describe the system.

I got ht edynamic system with the Lagrangian method and now I have 4 2nd order...

I'm using Maple to come up with equations, and then I'm substituting in variable names for functions which I can then output to Matlab.

I haven't had any problems until I ran into partial derivates described by the D[] operator.  For example, if i have a function f described by f(x(t),y(t)) and I differentiate it wrt t, I get 

D[1](f)(x(t),y(t))*(x(t)_dot) + D[2](f)(x(t),y(t))*(y(t)_dot)

I want to make a substitution for D[1](f)(x(t),y(t)) and D[2...

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