acer

N:=48:
a:=4:
b:=5:
Boundaries:=y(a)=0,y(b)=0;
P(x):=1;
Q(x):=1;
R(x):=1;
F(x):=cos(Pi*x^2)*x;
NLPart:=-exp(-sin(y(x)))*y(x)^2/(1+y(x));
Equation:=P(x)*diff(y(x),x$2)+Q(x)*diff(y(x),x$1)+R(x)*NLPart=F(x);
NewtonSolution:=dsolve({Equation,Boundaries},y(x),type=numeric,method=bvp);
PlotNewtonSolution:=odeplot(NewtonSolution,a....b);

diff(diff(y(x), x), x)+diff(y(x), x)-exp(-sin(y(x)))*y(x)^2/(1+y(x)) = cos(Pi*x^2)*x

>

delta[0]:=unapply(piecewise(j=k,1,j<>k,0),j,k):
delta[1]:=unapply(piecewise(j=k,0,j<>k,((-1)^(k-j))/(k-j)),j,k):
delta[2]:=unapply(piecewise(j=k,(-Pi^2)/3,j<>k,-2*(-1)^(k-j)/(k-j)^2),j,k):
d:=Pi/2:
h:=2/sqrt(N):
xk:=unapply((a+b*exp(k*h))/(1+exp(k*h)),k):
phi:=unapply(log((x-a)/(b-x)),x);
Dphi:=unapply(simplify(diff(phi(x),x)),x):
D2phi:=unapply(simplify(diff(phi(x),x$2)),x):
w:=unapply(1/Dphi(x),x):
Dw:=unapply(simplify(diff(w(x),x$1)),x):
D2w:=unapply(simplify(diff(w(x),x$2)),x):
MatrixSystem:=[]:
for p from -N to N
do
MatrixSystem:=[op(MatrixSystem),
h*(sum(c[j]*( (1/h^2)*delta[2](p,j)*(Dphi(xk(j))*subs(x=xk(j),P(x)*w(x)))+ (1/h^1)*delta[1](p,j)*(
(D2phi(xk(j))/Dphi(xk(j)))*subs(x=xk(j),P(x)*w(x))+2*subs(x=xk(j),
diff(P(x)*w(x),x))) + (1/h^0)*delta[0](p,j)*((subs(x=xk(j),
diff(P(x)*w(x),x$2))/Dphi(xk(j))))),j=-N..N)
-sum(c[j]*( (1/h^1)*delta[1](p,j)*(subs(x=xk(j),Q(x)*w(x)))+
(1/h^0)*delta[0](p,j)*(subs(x=xk(j),diff(Q(x)*w(x),x))) /Dphi(xk(j))
),j=-N..N)
+subs(y(x)=c[p],NLPart)*subs(x=xk(p),w(x)*R(x))/Dphi(xk(p))
-subs(x=xk(p),w(x)*F(x))/Dphi(xk(p)))=0]:
od:

>	CC:=fsolve(evalf(MatrixSystem)):

>	C:=Vector(2*N+1,(j)->eval(c[j-N-1],CC),datatype=float[8]);

$C := Vector(4, {(1) = ` 1 .. 97 `*Vector[column], (2) = `Data Type: `*float[8], (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})$

>	ApproximateSol:=unapply('add(evalf(C[j+N+1]sin(Pi(phi(x)-jh)/h)/(Pi(phi(x)- j*h)/h)),j=-N..N)',x,numeric): ApproximateSol(4.1);

>	a,b; plot({ApproximateSol(x)},x=a..b,color=green,style=point,symbol=solidcircle); `Sinc-GalerkinPlot`:=%:

>	display({`Sinc-GalerkinPlot`, PlotNewtonSolution }, title = "Sinc-GalerkinApproximation", labels=["x","y"]);

>

Download sinc_method_example_ac.mw

solve...

Answered: acer 32343

November 20 2018

4 1

restart;

C := [0,1/11],[1/11,1/9],[1/9, 1/7],[1/3,1/2],[1/2,1]:

op(map(`[]`@op,[solve(Or(op(map(r->__dum>=r[1] and __dum<=r[2],[C]))))]));

                         [   1]  [1   ]
                         [0, -], [-, 1]
                         [   7]  [3   ]

lprint(%);
  [0, 1/7], [1/3, 1]

select, isolve...

Answered: acer 32343

November 20 2018

2 3

Download solve_integer.mw

We can generate results for each a,b pair via a pair of nested seq calls. And we can accept or reject such candidates according to whether they each pass a conditonal test within the procedure which generates them.

Naturally, you could add whatever else you want to the conditional test (if...then) within procedure G. (note: the restrictions you were originally placing on a,b,x,y don't seem to follow from the problem description you gave. I mean, where you had nops({a, b, x, y}) = 4 and x*y*a*b <> 0)

>

restart;

>	G:=proc(a,b) local temp; temp := {isolve({(x-a)^2+(y-b)^2 = 225, (x+1)^2+(y+3)^2 = 125})}; if nops(temp)=2 then return {[a,b],temp}; else return NULL; end if; end proc:

>	ans := seq(seq(G(a,b),b=-20..20),a=-20..20):

>	seq(print(u), u=ans):

>

Download solve_integer_2.mw

one way...

Answered: acer 32343

November 19 2018

1 2

Here is one way.

student_md_question_ac.mw

You can find several older posts on this topic by searching this site for the word colorbar. Eg, [1], [2], [3] .

unapply...

Answered: acer 32343

November 19 2018

1 0

The 2+3 becomes 5 due to automatic simplification. That's not something that's going to be able to help you with your second example.

One way to construct a procedure (operator) from a given expression is to use the unapply command.

example2 := unapply( int(1,y), x );

                       example2 := x -> y

example3 := unapply( int(sin(x*y),y), x );

                                     cos(x y)
                  example3 := x -> - --------
                                        x    
example3( 17 );

                           1           
                         - -- cos(17 y)
                           17

pulling out the units...

Answered: acer 32343

November 19 2018

1 0

One way is to pull the units out of the the two columns, separately.

You haven't told us what units you want have for the results. Is it foot and day?

>

restart;

>	kernelopts(version);

>	M:=Matrix([[0., 0.], [0.2697Unit(ft), 0.3613Unit(ft^2d/inch^2)], [0.3821Unit(ft), 0.7225Unit(ft^2d/inch^2)], [0.5352Unit(ft), 1.4123Unit(ft^2*d/inch^2)]]);

$Matrix(4, 2, {(1, 1) = 0., (1, 2) = 0., (2, 1) = .2697*Units:-Unit(ft), (2, 2) = .3613*Units:-Unit(ft^2*d/`in`^2), (3, 1) = .3821*Units:-Unit(ft), (3, 2) = .7225*Units:-Unit(ft^2*d/`in`^2), (4, 1) = .5352*Units:-Unit(ft), (4, 2) = 1.4123*Units:-Unit(ft^2*d/`in`^2)})$

>	Units:-UseUnit(d): Units:-UseUnit(ft):

>	T1 := combine~(M[-1,1],units); #T1 := M[-1,1]; convert~(T1,unit_free,'u1'): u1; M1 := simplify(M[..,1]/u1);

Vector[column](%id = 18446884055612239614)

>	T2 := combine~(M[-1,2],units); #T2 := M[-1,2]; convert~(T2,unit_free,'u2'): u2; M2 := simplify(M[..,2]/u2);

Vector[column](%id = 18446884055612215654)

>	plot(M1, M2, labels=[u1,u2]);

>

Download plot_column_units.mw

prettyprint?...

Answered: acer 32343

November 15 2018

0 5

Which level did you use for the interface setting prettyprint? Was it 0, or 1?

I have been seeing weird behavior with interface(prettyprint=0) for a while now, when using the Standard GUI.

So it you are trying to use writeto from within the Standard GUI (and presuming you are not just using printf calls) then you might want to see if you can get by with the setting interface(prettyprint=1) .

kludge...

Answered: acer 32343

November 14 2018

1 2

This looks kludgy at best, and perhaps suspect at worst, but for this example...

Download kludge.mw

[edit] I think that I prefer the following form, with radnormal as an intermediate step, since the following form behaves nicer under simplify.

print...

Answered: acer 32343

November 14 2018

1 0

You can use the print command.

rand...

Answered: acer 32343

November 14 2018

1 0

You could simply use rand here.

restart;

LetList := [C, E, F, H, K, P, T, W, X, Y]:

gen:=rand(1..nops(LetList)):

gen();
                               7

LetList[gen()];
                               Y

LetList[gen()];
                               P

LetList[['gen()'$5]];
                        [E, H, P, K, C]

block matrix...

Answered: acer 32343

November 14 2018

3 1

It is, of course, trivial to scale the results below by the appropriate 1/2^k , simply by multiplying the generated Matrix by that factor. Or you can scale the two Matrix arguments passed into the procedure Gen.

>

restart;

>	interface(rtablesize=100):

>	with(LinearAlgebra):

>

M:=3:

>	F:=Matrix(M,M): F[1,1] := 2: F;

Matrix(3, 3, {(1, 1) = 2, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

>	L:=Matrix(M,M): L:=LinearAlgebra[BandMatrix]([ [seq(-sqrt(2i-1)/(2i-1)sqrt(2i-3),i=2..M)], [1,seq(0i,i=1..M-1)], [seq((sqrt(2i-3))/((2i-3)sqrt(2*i-1)),i=2..M)]]);

Matrix(3, 3, {(1, 1) = 1, (1, 2) = (1/3)*sqrt(3), (1, 3) = 0, (2, 1) = -(1/3)*sqrt(3), (2, 2) = 0, (2, 3) = (1/15)*sqrt(5)*sqrt(3), (3, 1) = 0, (3, 2) = -(1/5)*sqrt(5)*sqrt(3), (3, 3) = 0})

>	Gen:=proc(K::posint,A,B) Matrix(scan=triangular[upper],[seq([A,seq(B,i=j..2^(K-1)-1)],j=1..2^(K-1))]); end proc:

>

>	Gen(2, LL, FF);

Matrix(2, 2, {(1, 1) = LL, (1, 2) = FF, (2, 1) = 0, (2, 2) = LL})

>	Gen(3, LL, FF);

Matrix(4, 4, {(1, 1) = LL, (1, 2) = FF, (1, 3) = FF, (1, 4) = FF, (2, 1) = 0, (2, 2) = LL, (2, 3) = FF, (2, 4) = FF, (3, 1) = 0, (3, 2) = 0, (3, 3) = LL, (3, 4) = FF, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = LL})

>

>	Gen(2, L, F);

>

>	Gen(3, L, F);

>

Download block_band_matrix.mw

convert...

Answered: acer 32343

November 12 2018

1 2

Using Maple 2015.2,

convert(hypergeom([a, b], [c], 1), StandardFunctions);

                   GAMMA(c) GAMMA(c - a - b)
                   -------------------------
                   GAMMA(c - a) GAMMA(c - b)

convert( hypergeom([a, b], [c], 1), GAMMA_related );

                   GAMMA(c) GAMMA(c - a - b)
                   -------------------------
                   GAMMA(c - a) GAMMA(c - b)

Also,

restart;

kernelopts(version);

    Maple 2015.2, X86 64 LINUX, Dec 20 2015, Build ID 1097895

 simplify( hypergeom([a, b], [c], 1) );

              GAMMA(c) GAMMA(c - a - b)
              -------------------------
              GAMMA(c - a) GAMMA(c - b)

restart;

kernelopts(version);

    Maple 18.02, X86 64 LINUX, Oct 20 2014, Build ID 991181

simplify( hypergeom([a, b], [c], 1) );

              GAMMA(c) GAMMA(c - a - b)
              -------------------------
              GAMMA(c - a) GAMMA(c - b)

I don't understand why some of your Questions are marked as Maple version 2015, and some with version 18, and some with 2018.

like this?...

Answered: acer 32343

November 12 2018

1 0

If you expression attains it maximum at the end-point x=100, for each R, then could you not just divide by the value there? I mean, divide the original expression by the expression evaluated at x=100.

Let's pretend that your example would not have that behaviour, and that the maximum value is attained at a different numeric x for each different numeric R.

More generally, you can construct a procedure that admits the value val for the parameter name nm, and then optimizes w.r.t. x=1..100. And this procedure could be made to return unevaluated if its first argument val is not a numeric value. And then you could plot the original expression divided by a call to that maximizing procedure.

The plot3d command would pass in different values for val.

The procedure could attempt to remember the return value for given values of val, since plot3d will pump in the same val several times (as it computes for both different x and different R, say).

>

restart;

>

Xmax:=proc(val,expr,nm::name)
  option remember, system;
  local res, xvar;
  if not val::numeric then return 'procname'(args); end if;
  xvar:=(indets(expr,name) minus {nm})[1];
  try
    res:=Optimization:-Maximize(eval(expr,nm=val),
                                xvar=1..100,method='branchandbound');
  catch:
  end try;
  if not res[1]::numeric then
    return Float(undefined);
  else
    return res[1];
  end if;
end proc:

>

>	CARA2 := x^(1-R)/(1-R); #CARA2 := sqrt(abs(x-90))*x^(1-R)/(1-R);

>	Xmax(0.9,eval(CARA2,[R=Rdummy,x=xdummy]),Rdummy);

>	B := plot3d(CARA2, R = 0 .. .95, x = 1 .. 100): B;

>	# If CARA2 always attains its maximum (for each R) at x=100 then # this is simpler. plot3d(CARA2/eval(CARA2, x=100), R = 0 .. .95, x = 1 .. 100);

>	Bnew:=plot3d(CARA2/Xmax(R,eval(CARA2,[R=Rdummy,x=xdummy]),Rdummy), R = 0 .. .95, x = 1 .. 100): Bnew;

>	lnB := plot3d(ln(CARA2), R = 0 .. .95, x = 1 .. 100): lnB;

>	lnBnew:=plot3d(ln(CARA2/Xmax(R,eval(CARA2,[R=Rdummy,x=xdummy]),Rdummy)), R = 0 .. .95, x = 1 .. 100): lnBnew;

>	CARA1 := 1 - exp(-g*x);

>	A := plot3d(CARA1, g = 0 .. 1, x = 1 .. 100): A;

>	# Let's do this one just for fun, to illustrate how the use # of the Xmax command differs from the above example. Anew:=plot3d(CARA1/Xmax(g,eval(CARA1,[g=gdummy,x=xdummy]),gdummy), g = 0 .. 1, x = 1 .. 100): Anew;

>	plots:-display(A, Anew, color=[red,blue]);

>	plots:-display(Anew, Bnew, color=[green, blue]);

>

Download Cara_thing.mw

like this?...

Answered: acer 32343

November 12 2018

0 5

I'm not really sure what your goal is.

Is it something like this?

Download collect_numer.mw

If that's right, then it would be slightly more efficient to do it as follows, and thereby call `denom` only once.

temp := denom(f):
collect(numer(f),[k,GAMMA],u->u/temp);

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2D plot...

code...

solve...

select, isolve...

one way...

unapply...

pulling out the units...

prettyprint?...

kludge...

print...

rand...

block matrix...

convert...

like this?...

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