## 1350 Reputation

15 years, 9 days
University of Twente (retired)
Enschede, Netherlands

My "website" consists of a Maple Manual in Dutch

## Something is missing...

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!

## Something is missing...

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!

## No need to evaluate the integral symboli...

@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.

## No need to evaluate the integral symboli...

@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.

## ranges...

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

## ranges...

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

## combination of typeset and comma...

A little improvement...

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

## combination of typeset and comma...

A little improvement...

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

## Oops...

@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 vectoru := VectorField(<2,3>,coords=cartesian[x,y]);   # shows a vectorfield (barred unit vectors) u-w;  # gives a zero vectorfield#butis(u=w);                             false# andAbout(u),About(w);# shows that w is in cartesian coordinates, with no coordinate names. `

## Oops...

@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 vectoru := VectorField(<2,3>,coords=cartesian[x,y]);   # shows a vectorfield (barred unit vectors) u-w;  # gives a zero vectorfield#butis(u=w);                             false# andAbout(u),About(w);# shows that w is in cartesian coordinates, with no coordinate names. `

## Edited......

...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 );`

## Edited......

...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 );`

## Meaning of x -> ......

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.

## Meaning of x -> ......

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.

## Not fsolve, of course...

@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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