jaytreiman

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19 years, 229 days

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These are replies submitted by jaytreiman

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.

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.

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

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

@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.

@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.

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.  )

@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.

@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.

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.

@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.

@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.

@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;

Sum1 := add(add(a(i,j),i=1..10),j=1..10);

K1 := [seq(i,i=1..10)];
K := {seq(k[i]=K1[i],i=1..10)};
subs(K,Sum1);

k := K1;
Sum1;

 

Subscript3.mw

@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;

Sum1 := add(add(a(i,j),i=1..10),j=1..10);

K1 := [seq(i,i=1..10)];
K := {seq(k[i]=K1[i],i=1..10)};
subs(K,Sum1);

k := K1;
Sum1;

 


Download Subscript3.mw

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));

(psi-(1/2)*Pi)^2+(theta-.5)^2

 

 

 


Download a3DContours.mw

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