273 Reputation

19 years, 280 days

It does work without the single quotes....

You are correct.  One must be careful to use a definite integral if the function one obtains from the integration is not known to Maple.  If one uses your technique with plot(int(exp(-x^4)/(1+x^2),x),z=-4..4), Maple gives an error.  If ones uses plot(int(exp(-x^4)/(1+x^2),x=0..z),z=-4..4), one does not get an error.

It does work without the single quotes....

You are correct.  One must be careful to use a definite integral if the function one obtains from the integration is not known to Maple.  If one uses your technique with plot(int(exp(-x^4)/(1+x^2),x),z=-4..4), Maple gives an error.  If ones uses plot(int(exp(-x^4)/(1+x^2),x=0..z),z=-4..4), one does not get an error.

That does not look right...

@Husker Since I am not working with MapleTA 9, I cannot help with that.  Try contacting Maple support.

That does not look right...

@Husker Since I am not working with MapleTA 9, I cannot help with that.  Try contacting Maple support.

Questions not in groups...

@Husker  In MapleTA 8 there is a "questions not in groups" check box at the top of the questions repository left panel.  Try checking that box and hitting enter.  That may work.

Questions not in groups...

@Husker  In MapleTA 8 there is a "questions not in groups" check box at the top of the questions repository left panel.  Try checking that box and hitting enter.  That may work.

Look at the responses to your last quest...

The hint on how to do this is in Kitonum's reply to your last post.  That will give you the substitution.

(Oops.  I was trying not to give out the answer.  )

From the substitution...

@coach299   The idea is to get the form 1/(1+u^2) . To get the 1 in the denominator you need to divide the whole denominator by 2.  This gives (1/2)*(1/(1+(x^2/2)).  One then uses u^2=(x/sqrt(2))^2 or u=x/sqrt(2).  That is where one gets the square root of 2.

From the substitution...

@coach299   The idea is to get the form 1/(1+u^2) . To get the 1 in the denominator you need to divide the whole denominator by 2.  This gives (1/2)*(1/(1+(x^2/2)).  One then uses u^2=(x/sqrt(2))^2 or u=x/sqrt(2).  That is where one gets the square root of 2.

Open linux releases...

Why not add the open releases of SUSE and Red Hat?  They are Fedora Core and OpenSUSE.  Many students and professors use the open releases.

Try using the inert command Sum...

@Aritra It could be that Maple is doing too many symbolic computations early.  You might try using the inert command "Sum" instead of "add" and then evaluating the result later.

Try using the inert command Sum...

@Aritra It could be that Maple is doing too many symbolic computations early.  You might try using the inert command "Sum" instead of "add" and then evaluating the result later.

One method...

@Aritra   Here is a simple way to do this.  It is based on the fact that sequences are technically function on sets of integers or on sets of ordered tuples of integers.  Two methods of evaluation are given.

 > restart;
 > a := (i,j) -> (k[i]-k[j])^2;
 > K1 := [seq(i,i=1..10)]; K := {seq(k[i]=K1[i],i=1..10)}; subs(K,Sum1);
 > k := K1; Sum1;
 >

Subscript3.mw

One method...

@Aritra   Here is a simple way to do this.  It is based on the fact that sequences are technically function on sets of integers or on sets of ordered tuples of integers.  Two methods of evaluation are given.

 > restart;
 > a := (i,j) -> (k[i]-k[j])^2;
 > K1 := [seq(i,i=1..10)]; K := {seq(k[i]=K1[i],i=1..10)}; subs(K,Sum1);
 > k := K1; Sum1;
 >

A little closer to your prototype than M...

With a couple changes one can simplify Markiyan Hirnyk's code.  This may be more like what you indicated you want.  The graph will look much better when you run the code.

 > with(plots): with(plottools): f := (psi-(1/2)*Pi)^2+(theta-.5)^2; g := transform((psi, theta) -> [cos(psi)*sin(theta), sin(psi)*sin(theta), cos(theta)]): c:=contourplot(f, psi = 0 .. 2*Pi, theta = 0 .. Pi, contours = [seq(i,i=-1..10)],coloring=["LightGreen",blue],filledregions = true,color=black,numpoints=10000): display(g(c));
 >