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I just wanted to ask whether one can get out of the expansion

(D-f(x))@@2(g)(x)=(D@@2)(g)(x)-(D((f(x))(g)))(x)-((f(x))(D(g)-(f(x))(g)))(x)

the intended result

(D@@2)(g)(x)-D(f*g)(x)-f(x)*D(g)(x)+f(x)^2*g(x)

The model of fixed-bed adsorption column

Fluid phase:

PDE:= diff(U(x, tau),tau)+ psi*Theta*diff(U(x, tau),x)-(1/Pe)*psi*Theta*diff(U(x, tau),$(x, 2))=-3*psi*xi*(U(x, tau)-Q/K);

 

IBC:={U(x, 0) = 0,U(0, tau) = 1+(1/Pe)*(D[1](U))(0, tau),(D[1](U))(1, tau)=0};

Particle:

PDE:= diff(Q(r, tau), tau) = diff(Q(r, tau), $(r, 2))+(2/r)*diff(Q(r, tau),r);

IBC:={Q(r, 0) = 0,(D[1](Q))(0, tau) = 0,(1/K)*(D[1](Q))(1, tau)=xi*(U-Q(1, tau)/K)};

Pe:=0.01:

psi:=6780:

Theta:=3.0:

xi:=10000:

I will really appreciate your help. Thanks in anticipation.

I'm just using it this way:

with(QDifferenceEquations):
QPochhammer(-1,5,10)

Error, (in QDifferenceEquations:-QPochhammer) wrong type of arguments

Am I doing something terribly wrong???

i dont know what`s wrong with it help me guys

Hello. I have a question. If you can help me, i am pleasure.

Have nice day. :)

 

Question: Enter a number after maple find(understand) to prime number or normal number.

 

restart;
> asal := proc(n)
> local m;
> if n<2 then
> m := s=n/2+1;
> print(m);
> else
> for i from 2 to n do
> if irem(n,i,0) then
> print("asal değil");
> if i = m then
> print("asaldir");
> end do;
> end if;
> end if;
> end proc;

 

Hi there, this is my first post in this forum. I hope you can help me with my problem :)

I've created a little program to solve a problem for my class. This is the first time, I wrote a program with Maple, so please don't laugh. 

I've posted the Maple code below:

 

restart; randomize():
with(plots): with(plottools): with(LinearAlgebra): with(Statistics): with(MTM):

s:={1=2, 3=2}:

 

Realisierung:=proc(np,volfrac)

local m, h, adresses, rndmAdrs, omega: omega:=Vector[row](np,s);

adresses:=Vector(np-nops(s));

h:=1;
for m from 1 to np do
if(omega[m]=0) then
adresses[h]:=m;
h:=h+1;
end if
od;

randomize():
rndmAdrs:=Shuffle(adresses);

for m from 1 to volfrac*np do
omega[rndmAdrs[m]]:=1
od:

for m from 1 to np -(volfrac*np+nops(s)) do
omega[rndmAdrs[m+int8(volfrac*np)]]:=2
od:

omega;

end:

 


RealMatrix:=proc(np,nr,volfrac)

local m, alle:

alle:=Matrix(nr,np); for m from 1 to nr do

randomize();

alle[m]:=Realisierung(np,volfrac); od;

alle;

end:

RealMatrix(5,5,0.4);

 

I used the Shuffle function from Statistics-package to randomly permutate the components in the Vector adresses. Which works fine. By setting Randomize(); I get randomly results vor this permutation, when I execute the whole worksheet.

But when the "Realisierung" procedure is called in the for loop of the "RealMatrix" procedure, always the same Permutation is calculated.

This leads to the fact, that every row in my matrix looks the same.
Like in the example

2 1 2 2 1
2 1 2 2 1
2 1 2 2 1
2 1 2 2 1
2 1 2 2 1

but the lines should be randmly like

2 1 2 2 1
2 2 2 1 1
2 1 2 1 2
2 1 2 2 1
.....

 

Obvioulsy the problem is in the randomize()-command. I tried to place it directly after "restart" and directly before the Shuffling is executed.
But I had no succes. 

Maybe one of you can help me?
Would be great!
Thanks

 

 

 

 

 

 

Hello,

please help check what's wrong with this code. I need the analytic solution and convert to Bessel but return error. Here is the worksheet ID_1.mw

Best regards.

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


 


"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

If a dosage Q units of a certain drug is administrated to an individual, then the amount remaining in the bloodstream at the end of t minutes is given by Q*exp^-ct, where c>0. Suppose this same dosage is given at successive T-minute intervals.

 

a) Show that the amount A(k) of the drug is given by A(k) = ∑n=0k-1 Q*exp(^-ncT).

b) Find an upper bound for the amount of the drug in the bloodsteam after any number of doses.

c) Find the smallest time between doses that will ensure that A(k) does not exceed a certain level M for M>Q.

hello everyone, I have problem in my project written by matlab to solving my equations with rang-kuta 4 order method. I send my project file.  please help me to run it correctly. thank you.

input Q_j n j V Z F_ve a_1 a_2 p_1 p_2 L A f m k c h t i;

for jj=1:n;

            s=Q_j(jj);

end

b=(-V)*Z;

o=(F_ve*A)*(((a_1^2)*p_1)-((a_2^2)*p_2))/L;

% RK4 method;

% md2z/dt2+cdz/dt+kz=f;

% dz_1/dt=z_2;

% mdz_2/dt+cdz_1/dt+z_1k=f;

% dz_2/dt=1/m(f-cz_2-kz_1);

f=0;

m=1;

k=s+b+o;

c=0;

h=0.5;

t=0:0.5:t;

z1=zeros (1,n);

z2=zeros (1,n);

z1 (1)=0;

z2 (1)=1;

% N=length (t);

f1=@(t,z1,z2)z2;

f2=@(t,z1,z2)f-(c*z2)-(k*z1);

for i=1:n-1;

    k1=f1 (t (i),z1 (i),z2 (i));

    m1=f2 (t (i),z1 (i),z2 (i));

    k2=f1 (t (i)+h/2,z1 (i)+0.5*k1*h,z2 (i)+0.5*m1*h);

    m2=f2 (t (i)+h/2,z1 (i)+0.5*k1*h,z2 (i)+0.5*m1*h);

    k3=f1 (t (i)+h/2,z1 (i)+0.5*k2*h,z2 (i)+0.5*m2*h);

    m3=f2 (t (i)+h/2,z1 (i)+0.5*k2*h,z2 (i)+0.5*m2*h);

    k4=f1 (t (i)+h,z1 (i)*k3*h,z2 (i)*m3*h);

    m4=f2 (t (i)+h,z1 (i)*k3*h,z2 (i)*m3*h);

    z1 (i+1)=z1 (i)+(h/6)*(k1+(2*k2)+(2*k3)+k4);

    z2 (i+1)=z2 (i)+(h/6)*(m1+(2*m2)+(2*m3)+m4);

end

plot (t,z1)

plot (t,z2)

plot (z,t)

 

 

 

 

Hi everyone!

I use two "ArrayTools:-Copy" commands to copy elements from Vector A to Vector B.

restart;
A := LinearAlgebra:-RandomVector(10); A := convert(A, Vector[row]);
B := Vector[row](10); ArrayTools:-Copy(2, A, 1, B, 1); ArrayTools:-Copy(2, A, 7, B, 7); B;

My question is: Could I obtain the same result using only one command?

P.S. I tried to use "ArrayTools:-BlockCopy", but didn't get correct result.

 

Hello,
I am looking here for a tutor, that can help me doing some Maple V (5) programming.
I am a mathematic student and we use Maple programming.
I will be happy to pay a small amount of money for each exercice you help me doing,

if anyone is interested, please contact me here.

(The procedures that we usually have to write are for example:
 Newton-Raphson Method, Chebyshev Polynomial,...  I don't think it is hard for you.
Thank you very time for your time and your help.

How to find the nth derivative of (logx)/x  and (e^x)logx by using leibenitz theorem....?

 

 

 

Hi all,

How to build an uneven ground by MapleSim?

I want simulation a car runnung on this uneven ground.

Is there any method , proposal , orientation, or any thing can help me

 

Thank you

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