I've been able to make following sum: "for i from 1 to 4 do total:=total+i end do;" The result is logic: 1,3,6,10. Now I want to make the sum of some terms I've already calculated and named by opl[i]. With I a posint. Now I write the command: "for i from 1 to 4 do total:=total+opl[i] end do;" This gives me the error of to many levels of recursion. I am not able to see the problem in what I've programmed. Thank you in advance,

Thank you. Problem is clear now.

Thank you. Problem is clear now.

Thank you! Where logic thinking could bring you to... Now I have an extra method to check myself on my exam mathematics next week.

Thank you! Where logic thinking could bring you to... Now I have an extra method to check myself on my exam mathematics next week.

It helped me a lot. Now I just have to insert the values manually and calculate the solutions. It's nice I have came so far but maybe there is a way to substitute these paramters also with Maple. I tried this: >F1 := Pi*(0.5)^2*x(t)+Pi/3*(y(t)/2)^2*y(t)=Pi/3; >dF1 := diff( F1, t ); <\prev> As solution I've got: dF1 := .25*Pi*diff(x(t),t)+1/4*Pi*y(t)^2*diff(y(t),t) = 0 <\prev> Now I try to substitute some values: subs(x(t)=-0.5,y(t)=(5/2)^(1/3),diff(x(t),t)=-0.01,dF1); <\prev> As solution I get: .25*Pi*diff(-.5,t)+1/8*Pi*5^(2/3)*2^(1/3)*diff(1/2*5^(1/3)*2^(2/3),t) = 0 <\prev> As you see Maple substitutes the values for x(t) and y(t) perfectly but with substituting dx/dt he has a little problem. Is there a way to get around this? Edit: Have post a little picture to make my problem clearer. http://img120.imageshack.us/img120/1106/probsxs8.jpg

It helped me a lot. Now I just have to insert the values manually and calculate the solutions. It's nice I have came so far but maybe there is a way to substitute these paramters also with Maple. I tried this: >F1 := Pi*(0.5)^2*x(t)+Pi/3*(y(t)/2)^2*y(t)=Pi/3; >dF1 := diff( F1, t ); <\prev> As solution I've got: dF1 := .25*Pi*diff(x(t),t)+1/4*Pi*y(t)^2*diff(y(t),t) = 0 <\prev> Now I try to substitute some values: subs(x(t)=-0.5,y(t)=(5/2)^(1/3),diff(x(t),t)=-0.01,dF1); <\prev> As solution I get: .25*Pi*diff(-.5,t)+1/8*Pi*5^(2/3)*2^(1/3)*diff(1/2*5^(1/3)*2^(2/3),t) = 0 <\prev> As you see Maple substitutes the values for x(t) and y(t) perfectly but with substituting dx/dt he has a little problem. Is there a way to get around this? Edit: Have post a little picture to make my problem clearer. http://img120.imageshack.us/img120/1106/probsxs8.jpg

They are in function of t. I have the numeric value of dx/dt. I want Maple do write something like dy/dt + dx/dt and not really solve dy/dt so I could substitute in that equation that dx/dt is for example 5 and that I could calculate dy/dt. I don't now really if its possible. Can Maple not solve something like that symbolic?

They are in function of t. I have the numeric value of dx/dt. I want Maple do write something like dy/dt + dx/dt and not really solve dy/dt so I could substitute in that equation that dx/dt is for example 5 and that I could calculate dy/dt. I don't now really if its possible. Can Maple not solve something like that symbolic?

Great. That's all I needed to know. Thank you for the help and patience!

Great. That's all I needed to know. Thank you for the help and patience!

I'm starting all over. Problems in my worksheet: 1) How do I convert an integer to a sqrt-form? Things I want to do better: 1) When I need to calculate for which t value a function is zero and for which t value a function is one then I use two solve commando's. Now I want combine this in one commando (If possible). Something like solve(y=0 and y=1,t). So that I get something of the form t=.. for y(0) and t=.. for y(1) on one line. Is this possible? 2) Same as above but with the commando subs(...). Can I calculate the function value for two different values of the variable. Something like subs(t=2 and t=3,y). So that I get something of the form y=.. for t=2 and y=.. for t=3. I tought that in both examples you could make use of AND but it would not work. In these functions I have took as example something of the form y(t). I hope it is explained well :)

I'm starting all over. Problems in my worksheet: 1) How do I convert an integer to a sqrt-form? Things I want to do better: 1) When I need to calculate for which t value a function is zero and for which t value a function is one then I use two solve commando's. Now I want combine this in one commando (If possible). Something like solve(y=0 and y=1,t). So that I get something of the form t=.. for y(0) and t=.. for y(1) on one line. Is this possible? 2) Same as above but with the commando subs(...). Can I calculate the function value for two different values of the variable. Something like subs(t=2 and t=3,y). So that I get something of the form y=.. for t=2 and y=.. for t=3. I tought that in both examples you could make use of AND but it would not work. In these functions I have took as example something of the form y(t). I hope it is explained well :)

In my worksheet I had calculated the two values of y (6 and -6). Normally I would now calculate subs(y=6,K) and subs(y=-6,K) to find my two x values. This does not result in the correct values because I had already substituted y=2y in the equation. Now was my question if there is some way to reset the function or something like that?

In my worksheet I had calculated the two values of y (6 and -6). Normally I would now calculate subs(y=6,K) and subs(y=-6,K) to find my two x values. This does not result in the correct values because I had already substituted y=2y in the equation. Now was my question if there is some way to reset the function or something like that?