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this code not answer:...

Asked by: bfathi 10
May 05 2015
0  8

Hi,

this code not answer:

> whit*evalf; tx := evalf((t1+t2)*(1/2));
> t1 := 326*40; t2 := 327*40;
> y1, y2 := eval([y1, y2], Ns(t1), Ns(t2));
>while (t1 < tx) and (tx < t2) do:
yx := rhs(Ns(tx)[c2]):
if yx > 0 then
y2 := yx: t2 := tx:
else
y1 := yx: t1 := tx:
fi;
od;
>Tx = evalf(tx);
> Hx := floor((1/3600)*tx):
> Mx := floor((tx-3600*Hx)*(1/60)):
> Sx := tx-3600*Hx-60*Mx:
> Hx, Mx, Sx;
> XS := [seq(X[i], i = 0 .. 328)]: YS := [seq(Y[i], i = 0 .. 328)]:
> VxS := [seq(Vx[i], i = 0 .. 328)]: VyS := [seq(Vy[i], i = 0 .. 328)]:
> save(G,Mz,Xs,VxS,VyS, 'orbit.sav'):

 can you help me?

Integrating an equation with unknown constants....

Asked by: AmrTawfik12 5
integration
May 05 2015
1  2

This might seem easy but i am a total beginner in Maple and i really need to know how to do this. If i have an expression like this:

X^2 + 3*y + 3*z + X^7

where (y) and (z) are constants of unknown values. I need to integrate the expression over an interval for example (from 1 to 10) so that the answer would still be in terms of (y) and (z).

Would anyone please help me and give me the right expression for writing this on Maple?

Thanks in advance.

How to solve Cauchy sequences...

Asked by: laporte bernard 45
May 05 2015
0  2

(a) Show that if {an} ∞ n=1 is Cauchy then {a 2 n} ∞ n=1 is also Cauchy. (b) Give an example of a Cauchy sequence {a 2 n} ∞ n=1 such that {an} ∞ n=1 is not Cauchy

how to solve an exercise with Maple involving the ...

Asked by: laporte bernard 45
limit
May 05 2015
0  6

Show that 2^3 + x ^2 − 3x + 2 is O(x ^3 ).

Covariant Derivative and Covariant Tensor and Cont...

Asked by: trace 75
May 05 2015
2  4

guys , i have a metric and i want to define a componenets of a tensor and then obtain its covariant derivative with respect to a metric, what is your idea ?

N_1=-A(r)^1/2 , A_2=A_3=A_4=0 , what is D_[nu] N_1 =?

 in general i want to define N[1]=-A(r)^(1/2) and N[2] = N[3]= N[3] = N[4] = 0 And define F[mu, nu] = 2*(D_[mu] N[nu]-D_[nu] N[mu]) And define Omega[mu, nu] = 2*(D_[mu] N[nu]+D_[nu] N[mu]) and compute expression F_[alpha, beta] F_[~alpha`, ~beta ] And N_[alpha] N_[~beta`] F_[ ~alpha, ~lambda ] Omega_[beta, lambda])

i have problem with this how to difine this tensorial terms and how to compute them.

Covariant.mw

 

thxxxx

How to compute direct exponentiation of the matrix...

Asked by: testht06 90
homework
May 04 2015
1  1

Hi EveryOne!

In the the answer of the question "How to find k^th root of the given matrix over finite field (at URL: http://www.mapleprimes.com/questions/203997-How-To-Find-Kth-Root-Of-The-Given-Matrix#comment215683). Carl Love helped to find k^th root of the given matrix M over GF(2m)/f(x).

Now, I need to compute direct exponentiation of the given matrix M in finite field. (If M = [mi,j], we say Direct Exponent (element-wise exponent matrix), Mdk of M is a matrix whose each element is the result of exponentiation of corresponding elements of M. If k=2, then we say Md2 is a direct square matrix of M )

Please help me!!! Thank you very much.

How to wrap code in a box for style...

Asked by: GPY 80
style block
May 04 2015
0  1

I've seen a few maple presentations where a block of code is wrapped in a box. Does anyone know how this is done?

 

Error, (in dsolve/numeric/process_input) unknown y...

Asked by: Mokusko 10
ode dsolve
May 04 2015
1  5

Hi, I'm trying to solve a system of equation and I keep getting this error. Could anyone help me figure out what I'm doing wrong?

My problem is:

> alpha := .3; G := 3.5; L := 6; f := 1.1;

for i to 50 do

I0 := x(z)+y[i](z); ICon := x(0) = 1, y[i](0) = 0;

for j to 50 do

i <> j;

d1 := diff(x(z), z) = -G*x(z)*y[i](z)/IC-alpha*x(z);

d2 := diff(y[i](z), z) = G*y[i](z)*y[j](z)/IC-alpha*y[i](z);

dsys := {d1, d2};

F := dsolve({ICon, op(dsys)}, [x(z), y[i](z)], numeric);

end do;

end do;
Error, (in dsolve/numeric/process_input) unknown y[2] present in ODE system is not a specified dependent variable or evaluatable procedure


 

Nullspace Calculation By Maple...

Asked by: Wtolrud 75
linear-algebra
May 04 2015
0  3

Hi,

When I calculate the nullspace of a Matrix my solution comes out in a different order than Maple.  So, the question is what steps does Maple use to calculate Nullspace, ColumnSpace, and eigenvalues.  All of these are calculated by Maple in a different order that when I calculate by hand.

What I meant to say is that the answer given by for nullspace is in a different order than if I were do the same calculation by Hand.   

A=<<[1,1,1,1],[1,2,3,4],[4,3,2,1]>>   this is what got calculating the Nullspace by hand <1,-2,1,0>,<2,-3,0,1>  when Maple does the calc. It returns the answer in the opposite order

 

Thanks

Bill

Plotting a function subject to an implicit constra...

Asked by: jschulzb 150
May 04 2015
2  3

Hello, 

     I an trying to plot a function of a single variable, which is an implicit function of another variable, i.e. I want to plot F(x(t)), given that x and t are related through the implicit constraint equation f(x,t) = 0. Is there any plot stuctures in Maple that would easily let me do this? I tried implicit plot but this seems insufficient. 

     As an example, consider plotting F = x + x^2 subject to f = x + sin(x) + ln (t) = 0. I could also write this as  a function subject o a differential constraint, as is f = diff(x(t), t) + 1/t + (diff(x(t), t))*cos(x(t)) = 0 and try to use some sort of implicit DE plotting routine. 

     Any ideas?

Thanks!

Open maple worksheet...

Asked by: Annonymouse 160
worksheet incomplete-question
May 04 2015
0  0

I am running Maple in a windows virtual machine, on a mac computer.

I have a number of worksheets on its disk

Windows advised me to run its error checking utility (chkdsk)

when I try and open them it gives me a number of options:

maple text

plain text 

and maple input

 

None of these are the same as the original files. What has happened? and how can i fix it?



How many different combinations of three numbers c...

Asked by: toandhsp 305
May 04 2015
0  5

I have a list L:={1,2,3,4,5,6,7,8}. I choose three elements from list L. How many different combinations of three numbers can be selected from L so that the numbers could represent the side lengths of a triangle?

Error, (in DEtools/convertsys) unable to convert t...

Asked by: patient 35
dsolve numeric
May 04 2015
0  20

Hi all,

Thanks for helping me to solve the problem below using Maple.

dsys := {(1-4*(diff(ln(v(z)), z)))*(diff(u(z), z))+((3/2)*z^{-1}-2*(3* z^{-1} *(diff(ln(v(z)), z))+2*(diff(ln(v(z)),z,z )))))*u(z) = 0, -z*(diff(v(z), z))-v(z)+v(z)^(1/2)*u(z) = 0, v(0) = 1, u(0) = 1, (D(v))(0) = 1/4, (D(u))(0) = 3/8}

When trying    sol := dsolve(dsys, numeric)

I got : Error, (in DEtools/convertsys) unable to convert to an explicit first-order system.

 

Note that the analytic solution for z<=0 is:

if z>-4   then  u(z)=(1+z/4)exp(z/8) and   v(z)=exp(z/4)

else u(z)=0  and v(z)= (-4/z)exp(-1)

Regards

Length of output exceed limit 1000000...

Asked by: Sofey 10
May 03 2015
0  3

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

 (1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

 (2)

``


Download consistency_expmle_4.mw

.

how to prove with maple...

Asked by: laporte bernard 45
student
May 03 2015
0  1


> 

"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

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