Adri van der Meer

Adri vanderMeer

1345 Reputation

16 Badges

14 years, 39 days
University of Twente (retired)
Enschede, Netherlands

My "website" consists of a Maple Manual in Dutch

MaplePrimes Activity


These are replies submitted by Adri van der Meer

What is expected before the comma in

BasisVector(, tdeg(x, y, h121, h122, h123, h124, h125), 2), 
tdeg(x, y, h121, h122, h123, h124, h125),
'Q',2);

Error, `,` unexpected

Please try lprint(Q); to show what Q really is!

What is expected before the comma in

BasisVector(, tdeg(x, y, h121, h122, h123, h124, h125), 2), 
tdeg(x, y, h121, h122, h123, h124, h125),
'Q',2);

Error, `,` unexpected

Please try lprint(Q); to show what Q really is!

@mutaz As I said in my answer, the integral cannot be simplified further. But of course it can be used
for numerical approximations.

DE := (gamma*rho+(gamma*sin(x1)^2-(sigma*cos(x1))^2)*diff(V(x1),x1)) = 
        -kappa*sin(x1)/exp(x1);
params := {gamma=0.0048,rho=0.0069,sigma=0.0046,kappa=0.2096}:
s := dsolve( {DE,V(0)=0}, V(x1) );
Vs := unapply( subs(params,rhs(s)), x1 );

Now Vs is the solution with the parameters substituted. You can calculate:

evalf(Vs(1));
                        Float(undefined)
evalf(Vs(.05));
                          17.77048892

Note that the denominator under the integral is zero if _z1 approximately 0.066, and the integral diverges for x greater than this value. You can also make a plot

plot(Vs,0..0.06);

Anyway, this integral is a genuine function like the more familiar functions like sin, exp or erf.

@mutaz As I said in my answer, the integral cannot be simplified further. But of course it can be used
for numerical approximations.

DE := (gamma*rho+(gamma*sin(x1)^2-(sigma*cos(x1))^2)*diff(V(x1),x1)) = 
        -kappa*sin(x1)/exp(x1);
params := {gamma=0.0048,rho=0.0069,sigma=0.0046,kappa=0.2096}:
s := dsolve( {DE,V(0)=0}, V(x1) );
Vs := unapply( subs(params,rhs(s)), x1 );

Now Vs is the solution with the parameters substituted. You can calculate:

evalf(Vs(1));
                        Float(undefined)
evalf(Vs(.05));
                          17.77048892

Note that the denominator under the integral is zero if _z1 approximately 0.066, and the integral diverges for x greater than this value. You can also make a plot

plot(Vs,0..0.06);

Anyway, this integral is a genuine function like the more familiar functions like sin, exp or erf.

If the region is the triangle through the points (0,0), (0,1), (1,1), then the ranges must be:

x1=x2..1, x2=0..1

If the region is the triangle through the points (0,0), (0,1), (1,1), then the ranges must be:

x1=x2..1, x2=0..1

A little improvement...

caption = typeset(eta[12] = 0, ", ", eta[21] = 0)

A little improvement...

caption = typeset(eta[12] = 0, ", ", eta[21] = 0)

@Alejandro Jakubi That was one (#3) of the top ten Maple errors: I had assigned a value for v.

Playing around a little further, I discovered that the discrepancy is almost only in the way the vectorfields are displayed on the screen:

restart;
with(VectorCalculus):
f := (x, y) -> x^2+y^3;
gradf := unapply(Gradient(f(x, y), [x, y]), [x, y]):
w := gradf(1,1); # shows a column vector
u := VectorField(<2,3>,coords=cartesian[x,y]);
  # shows a vectorfield (barred unit vectors)
u-w;  # gives a zero vectorfield
#but
is(u=w);
                             false
# and
About(u),About(w);
# shows that w is in cartesian coordinates, with no coordinate names.


@Alejandro Jakubi That was one (#3) of the top ten Maple errors: I had assigned a value for v.

Playing around a little further, I discovered that the discrepancy is almost only in the way the vectorfields are displayed on the screen:

restart;
with(VectorCalculus):
f := (x, y) -> x^2+y^3;
gradf := unapply(Gradient(f(x, y), [x, y]), [x, y]):
w := gradf(1,1); # shows a column vector
u := VectorField(<2,3>,coords=cartesian[x,y]);
  # shows a vectorfield (barred unit vectors)
u-w;  # gives a zero vectorfield
#but
is(u=w);
                             false
# and
About(u),About(w);
# shows that w is in cartesian coordinates, with no coordinate names.


...because the MaplePrimes editor eated up the characters after the < in v := evalVF(Df,<1,1>

Addendum: If you want v to be a "common" vector, you can do:

ConvertVector( v, free );

...because the MaplePrimes editor eated up the characters after the < in v := evalVF(Df,<1,1>

Addendum: If you want v to be a "common" vector, you can do:

ConvertVector( v, free );

See the help page: In the assignment f1 := x -> ..., the expression on the right of the arrow operator is not evaluated. When you omit x ->, Maple tries to evaluate the expression, but the variable x has no value, ant that results in the errormessage "... cannot determine if this expression is true or false...".
But when you define f1 as a procedure, you can evaluate f1(0.1), etc.: i.e. for numerical values for x.

As regards the "square wave": Use signum(cos(x/2)) for your function.

See the help page: In the assignment f1 := x -> ..., the expression on the right of the arrow operator is not evaluated. When you omit x ->, Maple tries to evaluate the expression, but the variable x has no value, ant that results in the errormessage "... cannot determine if this expression is true or false...".
But when you define f1 as a procedure, you can evaluate f1(0.1), etc.: i.e. for numerical values for x.

As regards the "square wave": Use signum(cos(x/2)) for your function.

@acer once proved that f is a contraction on a closed interval - eventually using the result of

f := x -> (x+2)/(x+3):
solve( abs(D(f)(x)) < 1 );
 RealRange(-infinity, Open(-4)), RealRange(Open(-2), infinity)

and if solve doesn't find an answer, the limit is of course approximated by

 # L := evalf( (f@@10)(2) ); better:
 L := (f@@10)(2.0); f(L)-L;
                          0.7320508076
                               0.

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