firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old. 

ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A,
"triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

                             [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

                            ( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

 Please and thank you for any help in advance <3

Hello,

Maple needs 827 characters to write a equation of a straight line.
Is that true or what am I doing wrong?   

Can anybody help me or give a direction to handle with such problems?

Putting
  assume(2<alpha , alpha<=4);about (alpha);
before it does not help either.

f:=-(6*(3*alpha^2*(alpha-1+sqrt(alpha^2-3*alpha+2))^4/(alpha-1)^4-12*alpha^2*(alpha-1+sqrt(alpha^2-3*alpha+2))^3/(alpha-1)^3-6*alpha*(alpha-1+sqrt(alpha^2-3*alpha+2))^4/(alpha-1)^4+16*alpha^2*(alpha-1+sqrt(alpha^2-3*alpha+2))^2/(alpha-1)^2+24*alpha*(alpha-1+sqrt(alpha^2-3*alpha+2))^3/(alpha-1)^3+3*(alpha-1+sqrt(alpha^2-3*alpha+2))^4/(alpha-1)^4-8*alpha^2*(alpha-1+sqrt(alpha^2-3*alpha+2))/(alpha-1)-24*alpha*(alpha-1+sqrt(alpha^2-3*alpha+2))^2/(alpha-1)^2-12*(alpha-1+sqrt(alpha^2-3*alpha+2))^3/(alpha-1)^3+8*(alpha-1+sqrt(alpha^2-3*alpha+2))^2/(alpha-1)^2+8*alpha+(8*(alpha-1+sqrt(alpha^2-3*alpha+2)))/(alpha-1)-7))/(alpha*(alpha-1+sqrt(alpha^2-3*alpha+2))^2/(alpha-1)^2-2*alpha*(alpha-1+sqrt(alpha^2-3*alpha+2))/(alpha-1)-(alpha-1+sqrt(alpha^2-3*alpha+2))^2/(alpha-1)^2+(2*(alpha-1+sqrt(alpha^2-3*alpha+2)))/(alpha-1)-1)^4;

Hello,

• Is there a simple way to find the domain for the real solutions of f(x)?

• And is there a way to let maple get the part of f(x) with the sqrt?
   (not by typing it by hand as I dit below)

• Is there a way to write the summary of the found domains in one line?

Thanks for your help. 





restart:
# How to find the Domain for real solutions for x?
f(x):=(x-1+sqrt(x^2-3*x+2))/(x-1);
discont_for_x=discont(f(x),x);
# x<>+1 (because the de denom=0 is not allowed)
denom(f(x))=0;
x={solve(denom(f(x))=0,x)};
# x<=1 union  2<=x (because the part under the sqrt must be >=0 to give Real solutions)
sqrt(x^2-3*x+2);
0<=x^2-3*x+2;
x=solve(0<=x^2-3*x+2,x);

Hi all

Please, I'd like to clarify some basic points about performing computations in Maple 18. Up to now I have been doing some numeric calcs using matrices normally composed of 10 to 15 columns and 200 to 600 rows.

When doing the calcs with a matrix of ~200 rows, it is just ok but as the number of rows increase the calculation speed reduces significantly.

The data contained in each row is calculated within a loop cycle (i.e.: for i from 1 to 200 do ......).

The number of rows is controlled by a slider so when I drag the slider the calcs are automatically updated and the results shown graphically.

As I said it is too slow so I don't know if I should be looking into the option of doing calcs symbolically first? However you can't use symbolic notation when working with matrices. I still got a lot of calcs to do but prefer not to continue as it will only get slower so better to see if I can optimise speed.

Any comment is trully appreciated.

Regards

Cesar

m 18 cubacuhre integration fail, intergrand unknown

Download stovrlp2.mw

m 18 cubacuhre integration fail, intergrand unknown

Hi all,
I need to find min_a(w) then a=-1,0,1 and

w := piecewise((x-a)/t <= -10, 202*a-200*x-(202001/101)*t, (x-a)/t <= -910/101, 3*a-x-(1011/101)*t, (x-a)/t <= 910/101, 2*a-t, (x-a)/t <= 10, a+x-(1011/101)*t, 10 < (x-a)/t, -198*a+200*x-(202001/101)*t)

Also need to plot it, i think it should look like this 

Thanks

I am getting a complex solution for a physical quantity, how can I plot it. Should I plot the real part or modulus?

solution is,

C1*sqrt(4*tau^2*M^2*lambda^2+Q^2)*LegendreP((1/2*I)*sqrt(7)-1/2, I*sqrt(7), (2*I)*lambda*M*tau/Q)+C2*sqrt(4*tau^2*M^2*lambda^2+Q^2)*LegendreQ((1/2*I)*sqrt(7)-1/2, I*sqrt(7), (2*I)*lambda*M*tau/Q)

where C1 and C2 are integration constants.

Please i don't want to convert from list to set then apply set minus and convert back because it loses its order and repititions.

so i want something that can for example do the following:

list_minus([1,2,4,6,2,1,3,6,2],[7,4,2,5,2]) = [1,6,1,3,6,2]

so it removes repititions from left and order is retained.

Hi all

I dont know why some Z1 appears on the screen and the code does not converge.
please help me
thanks alooooot

restart;
n:=3;
nn:=3;
m:=1;
BB:=1;
BINF:=5:
pr:=7;
digits:=10;
>
eq1:=diff(f(tau),tau$3)+((3/5)*f(tau)*diff(f(tau),tau$2))-(1/5)*(diff(f(tau),tau$1))^2+((2/5)*tau*diff(h(tau),tau$1))-((2/5)*h(tau))-BB*diff(f(tau),tau$1)=0;
eq11:=(1/pr)*diff(h(tau),tau$3)+(3/5)*f(tau)*diff(h(tau),tau$2)=0;

h(tau):=sum(p^i*h[i](tau),i=0..nn);
f(tau):=sum(p^i*f[i](tau),i=0..n);

H1:= p*(diff(f(tau),tau$3)+((3/5)*f(tau)*diff(f(tau),tau$2))-(1/5)*(diff(f(tau),tau$1))^2+((2/5)*tau*diff(h(tau),tau$1))-((2/5)*h(tau))-BB*diff(f(tau),tau$1))+(1-p)*(diff(f(tau),tau$3)):
H11:= p*((1/pr)*diff(h(tau),tau$3)+(3/5)*f(tau)*diff(h(tau),tau$2))+(1-p)*(diff(h(tau),tau$3)):
>
eq2:=simplify(H1):
eq22:=simplify(H11):
eq3:=collect(expand(eq2),p):
eq33:=collect(expand(eq22),p):
eq4:=
convert(series(collect(expand(eq2), p), p, n+1), 'polynom');
eq44:=
convert(series(collect(expand(eq22), p), p, n+1), 'polynom');
for i to n do
s[i] := coeff(eq4, p^i) ;
print (i);
end do:
for i to nn do
ss[i] := coeff(eq44, p^i) ;
print (i);
end do:
s[0]:=diff(f[0](tau), tau$3);
ss[0]:=diff(h[0](tau), tau$3);
ics[0]:=f[0](0)=0, D(f[0])(0)=0, D(f[0])(BINF)=0;
icss[0]:=h[0](BINF)=0, D(h[0])(0)=1, D(h[0])(BINF)=0;

dsolve({s[0], ics[0]});
f[0](tau):= rhs(%);
#dsolve({ss[0], icss[0]});
h[0](tau):= -exp(-tau); #;rhs(%);

>
>
for i to n do
f[ii-1](tau):=convert(series(f[ii-1](tau), tau, nn+1), 'polynom');
h[ii-1](tau):=convert(series(h[ii-1](tau), tau, nn+1), 'polynom');
s[i]:=simplify((s[i]));
ics[i]:=f[i](0)=0, D(f[i])(0)=0, D(f[i])(BINF)=0;
dsolve({s[i], ics[i]});
f[i](tau):=rhs(%);
ss[i]:=(ss[i]);
icss[i]:=h[i](BINF)=0, D(h[i])(0)=0, D(h[i])(BINF)=0;
dsolve({ss[i], icss[i]});
h[i](tau):=rhs(%);

end do;

f(tau):=sum((f[j])(tau),j=0..n);
with(numapprox):

plot(diff(f(tau),tau),tau=0..5,color=blue,style=point,symbol=circle,symbolsize=7,labels=["tau","velocity"]);
plot(pade(diff(f(tau),tau), tau, [7, 7]),tau=0..5,color=blue,style=point,symbol=circle,symbolsize=7,labels=["tau","velocity"]);

x*(y+z)/x

convert any polynomial to Divide(Mult(x, Plus(y+z)), x)

Hi can you please help me? I want to write a program in Maple code, using the Newton-Raphson method, to solve the equation tan(x) −x-1=0 .  the approximate root is 1.5.
I tried doing this using a while loop to compare the last and current iterates, but something is not working.
so can you please help me do that?

Thanks

y = -x+sqrt(-5*x)

how to plot this in 2d or 3d graph

I'm trying to plot the direction field of the second order differential equation x''=x'-cos(x) using dfieldplot: 

> with(DEtools); with(plots);
> f1 := (x, y) options operator, arrow; diff(x(t), t)-cos(x(t)) end proc;
/ d \
(x, y) -> |--- x(t)| - cos(x(t))
\ dt /
> dfieldplot([diff(x(t), t) = y(t), diff(y(t), t) = f1(x(t), y(t))], [x(t), y(t)], t = -2 .. 2, x = -2 .. 2, y = -2 .. 2);
Error, (in DEtools/dfieldplot) cannot produce plot, non-autonomous DE(s) require initial conditions.
>

The error I'm getting says I need initial conditions, but I wasn't provided with any. Is there another way to plot this? Sorry if this is dumb question, but I've only ever plotted first order equations.

The equation

x^7+14*x^4+35*x^3+14*x^2+7*x+88 = 0

has the unique real root

x = (1+sqrt(2))^(1/7)+(1-sqrt(2))^(1/7)-(3+2*sqrt(2))^(1/7)-(3-2*sqrt(2))^(1/7).

Here is its verification:

Is it possible to find that in Maple? I unsuccessfully tried the solve command with the explicit option.

I was required to purchase Maple 17 for my upcoming Calculus III course, and until now, I have been using my TI-Nspire CAS CX for all of my CAS needs.  I am going through various tutorials/labs in an effort to learn how to use the Maple 17 Software. As a part of this process, I am attempting to solve a system of equations and was told to use the following command:

>solve({2*x+3*y=7,5*x+8*y=9},{x,y}); 

in order to receive the answer 

{y=-17,x=29}.

Instead, I have received the following error message, which has no help attached to it through the help page.

solve({2*x+3*y = 7, 5*x+8*y = 9}, {x, y});
Warning, solving for expressions other than names or functions is not recommended.

I am hoping this has something to do with Mac vs. Windows software, and that it has a simple solution.  I would greatly appreciate any guidance!