Dear all,

here, I propose two methods for Adams Moulton Methos, but which one can I used.

The n-step Adams Moulton method to solve y'(x)=F(x,y(x)) is defined by the stencil

y(x+h)=y(x)+h *sum_{j=-1}^{n-1} beta_j F( x-j*h, y(x-j*h) ) + O(h^{n+2})

I want a procedure with single argument ''n'' that calculates and return the ''beta_i'' coefficients

I get two Methods. Which one correspond to my question please, and I don't understand the procedure proposed.

For me; the first give the iterative schemae used, but don't return the vector coefficients ( beta_i) and this methode method an interpolation of the function.

The second method, there is a function f, how this function is maded, and the same for the matrux A and the vector b...

the First Method:

> Adamsmoulton := proc (k::posint)

local P, t, f, y, n;

P := interp([t[n]+h, seq(t[n]-j*h, j = 0 .. k-2)], [f[n+1], seq(f[n-j], j = 0 .. k-2)], x);

y[n+1] = y[n]+simplify(int(P, x = t[n] .. t[n]+h))

end proc;

Second Method:

f:=proc(x,y) if x =0 and y=0 then 1 else x**y fi end;

n:=3; A:=matrix(n,n,(i,j)->f(1-j,1-i));

b:=vector(n, i->1/i);

linsolve(A,b);

I'm trying to run two statement sequences, one after the other, numerous times. I have the statement sequences:

>for j from 1 to N do

>S[j]:=V[j]+t;

>S[j]:=S[j]+3;

>end do:

>for j from 1 to N do

>if S[j]>99 then S[j]:=0

>end if:

>end do:

I can manage to run one of them multiple times, but when I try to encompass both of them within my 

>for counter from initial to final do statementsequence

end do:

it doesn't seem to work.

Thanks in advance

assume the word equation is

a_i *a_j - a_j *a_i = 0

how to find which permutation group is a_i and a_j

my understanding is to try all rotations

a book use underscript i and j

can i see them as upper script for i rotations which is shift i times to left for second row

and try all combination and composite them in two for loop? 

Whassup homies?

http://www.mathsisfun.com/puzzles/who-lives-in-the-city--solution.html

tried to solve this using C.Loves program, but didn't quite get their solution...

Who_Lives_in_the_Cit.mw

Vars:= [PN,Name, TV, Dest,Ages,Hair,Lives]:
PN:=[$1..5]:
Name:= [Bob, Keeley, Rachael, Eilish, Amy]:
TV:=[Simpsons, Coronation, Eastenders, Desperate, Neighbours]:
Dest:= [Fra, Aus, Eng, Afr,Ita]:
Ages:= [14, 21, 46, 52, 81]:
Hair:=[afro, long, straight, curly , bald]:
Lives:= [town, city, village, farm, youth]:
Con1:= Desperate=3: Con2:= Bob=1: Con3:= NextTo(Simpsons,youth,PN): Con4:= Succ(Afr,Rachael,PN): Con5:= village=52: Con6:= Aus=straight: Con7:= Afr=Desperate: Con8:= 14=5: Con9:= Amy=Eastenders: Con10:= Ita=long: Con11:= Keeley=village: Con12:= bald=46: Con13:= Eng=4: Con14:= NextTo(Desperate,Neighbours,PN): Con15:= NextTo(Coronation,afro,PN): Con16:= NextTo(Rachael,afro,PN): Con17:= 21=youth: Con18:= Coronation=long: Con19:= 81=farm: Con20:= Fra=town: Con21:= Eilish<>straight:

read "LogicProblem.mpl"; City:= LogicProblem(Vars): with(City);

we use modern computer algebra books

i) computer the GSO of (22,11,5),(13,6,3),(-5,-2,-1) belong to R^3.

ii)trace algorithm 16.10 on computer a reduced basis of the lattice in Z^3 spanned by the vectors form(i).

trace also the values of the d_i and of D, and compare the number of exchange steps to the theoretical upper bound from section 16.3

we use Modern Computer Algebra

let f=x^15-1 belong to Z[x]. take a nontrivial factorization f≡gh mod 2 with g,h belong to Z[x] monic and of degree at least 2. computer g*,h* belong to Z[x] such that   f≡g*h* mod 16 ,deg g*=deg g, g*≡g mod 2.

show your  intermediate. can  you guess some factors of f in Z[x]?

we use Modern Computer Algebra book  

trace algorithm 15.2 on factoring f=30x^5+39x^4+35x^3+25x^2+9x+2 belong to Z[x].choose the prime p=5003 in step.

I have this expression; f(a,c,n)=64a²n-16a⁴n-256c²n+32c⁴n-96a²c²+ 8a²c⁴+24a⁴c²-64a²n²+256a²-124a⁴+ 15a⁶+256n²+64a²c²n

such as "a" is between 0 and 2, "c" between 0 and 2 and "n" between 0 and 4.

How do I plot f(a,c,n) so that I can study its sign?

I have computed the infinite multiplication using Maple, $\Pi_{k=3}^{\infty} ( \cos (\frac{\pi}{k} ))$, as follows, but it resulted in 0! I wonder why this happened, although maple was using exact arithmetic.

    P := Product(cos(Pi/k), k = 3 .. infinity)   

    value(P)

Note that if I use floating-point instead, it gives me the right answer, 0.1149420449.

    evalf(P)
By the way, I did not expect such a situation! Exact arithmetic should be exact! I am multiplying non-zero numbers to each other, starting from $1/2$ to $1$ as $n \rightarrow \infty$. So it should not be zero! Why such a thing happened?

Dear all;

Special thanks for all the member who help me in Maple.

My last question is:

Write a maple procedure that solves for y(1) in the initial value problem y'(x)=f(y), y(0)=1

using a Numerical stencil based on the n^{th] order taylor series expansion of y.

The procedure arguments include an arbitrary function f, an integrer n, representing the accuracy of the taylor series expansion, and N representing the number of steps between x=0 and x=1.

Given a 2x2 matrix I am struggling to write a function that would return a list (a,b, a1, a2) of 2 complex numbers followed by 2 vectors such that the set of the 2 vectors is a basis for CxC and also Ab1=ab1, Ab2=Bb2 if these exist

Any ideas would be greatly appreciated

how to convert decimal to fraction without simplify

for example

convert(0.25, fraction)

expect 25/100, but not 1/4

Hi,

Please help me in solving system consist the three differential equations with three unknowns. I did already a few attempts, but I can not finish. Once in the final result was got RootOf and do not know what to do. I tried also numerically. I very very ask for some suggestions: ( 

with the boundary condition

parameters A,B,C are constans

1)

qa1 := A1*(diff(Tg(x), x, x))+A2*(diff(Tg(x), x))+(A3+A4)*Tg+A3*Tz+A4*Tw = 0;

eqa2 := B1*(diff(Tw(x), x, x))+B2*(diff(Tw(x), x))+(B3+B4)*Tw+B3*Tz+B4*Tg = 0;

eqa3 := C1*(diff(Tz(x), x, x))+(C3+C4)*Tz+C3*Tg+C4*Tw = 0

2)

On paper, the system of three equations with three unknowns I changed to system of two equations with two unknowns but still nothing. 

A[1] := 2, 

eqa1 := A[1]*C[1]*(diff(z(x), x, x, x, x))/C[3]+A[2]*C[1]*(diff(z(x), x, x, x))/C[3]+(A[1]*C[3]+A[1]*C[4]+A[3]*C[1]+A[4]*C[1])*(diff(z(x), x, x))/C[3]+(A[1]*C[3]+A[1]*C[4])*(diff(z(x), x))/C[3]+(A[3]*C[3]+A[3]*C[4]+A[4]*C[3]+A[4]*C[4]+A[3]*C[3])*z(x)/C[3]+A[1]*C[4]*(diff(y(x), x, x))/C[3]+A[2]*C[4]*(diff(y(x), x))/C[3]+(A[3]*C[4]+A[4]*C[4]+A[4]*C[3])*y(x)/C[3] = 0;

eqa2 := B[4]*C[1]*(diff(z(x), x, x))/C[3]+(B[4]*C[3]+B[4]*C[4]+B[3]*C[3])*z(x)/C[3]+B[1]*(diff(y(x), x, x))+B[2]*(diff(y(x), x))+(B[4]*C[3]+B[4]*C[4]+B[3]*C[3])*y(x)/C[3] = 0;

row := eqa1, eqa2;

sol := dsolve({row}, {y(x), z(x)});

Thank you very much for your help.

Ewa.

Dear all,

I need to compute the error, How to define the error between the exact and approximation.

                               d              
                              --- y(x) = -y(x)
                               dx             
                               y(x) = exp(-x)

I have a problem in this code, my goal is to compute the error between the approximate solution obtained by RK3 and Exact  and E ( approximation by RK3).

How to definie the error and prouve that the error is O(h^4)  ( with one step) and the global error is O(h^3).

Thank you  for helping me.

Hi,

I have a problem with dsolve in the following code

restart;
>
n:=20;
m:=1;
cc:=-200;
zzeta:=0.1;
sefr1:=0.3;
sefr:=0.2;
MM:=0;
lambda:=0.1;
Br:=1;
nn:=3;
>
>
#u(tau):=tau;
u(tau):=421.7129935*tau-2217.587728*tau^2+8897.376593*tau^3-27612.59182*tau^4+64248.00336*tau^5-1.083977605*10^5*tau^6-10.57029600-1.080951714*10^6*tau^13+7.999517316*10^5*tau^14-4.788741005*10^5*tau^15+2.309563748*10^5*tau^16+26511.11102*tau^18-5959.001794*tau^19+1.148523882*10^5*tau^7-95.23809524*tau^21+4.545454545*tau^22-9435.563781*tau^8-2.587683745*10^5*tau^9+6.473880128*10^5*tau^10+948.0272727*tau^20-88660.41892*tau^17-1.008692404*10^6*tau^11+1.175504242*10^6*tau^12;
>
>
B := 1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2*(1-zzeta)));
eq4 := 4*B*u(tau)-(1+zzeta)*(diff(tau*(diff(theta(tau), tau)), tau))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2;


theta(tau):=sum(p^ii*theta[ii](tau),ii=0..nn);
HH:= p*((4*(1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2-(1/2)*zzeta))))*u(tau)-(1+zzeta)*(diff(theta(tau), tau)+tau*(diff(theta(tau), tau, tau)))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2)+(1-p)*(diff(theta(tau),tau$2)):
eq5:=simplify(HH):
eq6:=collect(expand(eq5),p);

eq7:=
convert(series(collect(expand(eq5), p), p, nn+1), 'polynom');


for ii to nn do
ss[ii] := (coeff(eq7, p^ii)) ;
print (ii);
end do;

ss[0]:=diff(theta[0](tau), tau, tau);

icss[0]:=theta[0](zzeta/(2*(1-zzeta)))=0, D(theta[0])(1/(2*(1-zzeta)))=1;

dsolve({ss[0], icss[0]});
theta[0](tau):= rhs(%);


for ii to nn do
ss[ii]:=evalf[5](ss[ii]);
icss[ii]:=theta[ii](zzeta/(2*(1-zzeta)))=0, D(theta[ii])(1/(2*(1-zzeta)))=0;
dsolve({ss[ii], icss[ii]});
theta[ii](tau):=rhs(%);
end do;

I would be most grateful if you help me to find this problem.

Thanks for your attention in advance