Maple has a built in sum command. Take a look at the help files: >?sum Using a capital "S" for the sum command, as in "Sum", will give you the inert form of "sum". This is explained in the help files as well. You can also open up the help files and do a search for sum. There are plenty of examples there that you can use as a starting point. Examples from the help files: >sum(k,k=0..n-1); 1/2*n^2-1/2*n Inert form: >Sum(k,k=0..n-1); Sum(k,k = 0 .. n-1) Regards, Georgios Kokovidis Dräger Medical

As mentioned above, this is a precision issue. The attached worksheet correlates with the answer from your plot. View 221_ratio.mw on MapleNet or Download 221_ratio.mw
Regards, Georgios Kokovidis Dräger Medical

After your output, add the following line: >evalf(%); This is what I got: {x = 2.828427124, y = .656854248} You can do it directly like this: >fsolve({y=2*x-5,y=2*x^2+2*x-21},{x,y}); Regards, Georgios Kokovidis Dräger Medical

Could you post the code that generated your plot above? Regards, Georgios Kokovidis Dräger Medical

Yes, but you might not like what you see. An example below. This might not be what you are looking for exactly. Here the larger scale (0..10) is displayed while the plot from (0.01..0.1) shows up as well but it is not clear what range of values are used just by looking at the plot for the minor scale. >restart: >eq1:=x+20; >eq2:=100*x+50; >p1:=plot(eq1,x=1..10): >p2:=plot(eq2,x=0.01..0.1,color=blue): >with(plots):display(p1,p2); Regards, Georgios Kokovidis Dräger Medical

Maybe you are looking to use "add" instead of "sum". >add(f,n=0...0); 1.0 >?sum/details for more help on this. Regards, Georgios Kokovidis Dräger Medical

There are nice tutorials at the Maple Application Center. Two of them can be found below. Tangent and Normal Lines Tangent Lines and Differentiability Regards, Georgios Kokovidis Dräger Medical

There is some useful information here from a previous posting. The leastsquares command accepts functions that are not linear. They can be exponentials or polynomials of higher order. The older package stats might be of interest as well. >?stats/fit for more help on this Regards, Georgios Kokovidis Dräger Medical

Using the standard interface, you have to use "evalf" to get a numerical version of the expression. So for something that involves "Pi", you would go about it like this: >2*Pi; 2*Pi >evalf(%); 6.283185308 If you type the following you will get a symbol as an answer. >exp(1); exp(1) Now you can call evalf as with the example using "Pi" or you can use exp(1.0). Notice that exp(1.0) is different that exp(1) because you are forcing Maple to use floating point numbers. >exp(1.0); 2.718281828 I have not answered your direct question for the non-standard interface, but now you know why Maple in general behaves the way it does with symbols and floating point numbers. Regards, Georgios Kokovidis Dräger Medical

Display it twice, once in the up direction and once in the down direction. Play the animation below to see it. >restart:with(plots):with(plottools): >c:=seq(rectangle([i,i+1], [i+1,i]),i=0..10,.5): >d:=seq(rectangle([i,i+1],[i+1,i]),i=10..0,-.5): >display(c,d, scaling=constrained,insequence=true); Regards, Georgios Kokovidis Dräger Medical

>restart: >limit( (1+(a/x))^(b*x), x=infinity ); exp(a*b) Regards, Georgios Kokovidis Dräger Medical

Your files did not post correctly. Try again. Regards, Georgios Kokovidis Dräger Medical

Not an option, regardless of Maple version. Regards, Georgios Kokovidis Dräger Medical

You do not need to solve for conductivity, since you already have an equation that defines it. What you are looking to do, is evaluate p for a given set of numbers for the other variables. >restart: L:=20; R:=10; A:=5; >p:=L/(R*A); The above will evaluate p at the defined values of L,R and A. >restart: >solve(p=L/R*A,A); The above will solve for A in the equation in terms of p,R and L. Regards, Georgios Kokovidis Dräger Medical

The original code for displaying the sliding block was crude. The version below is better and it might make the rest of your problem easier to solve. >restart:with(plots):with(plottools): >c := seq(rectangle([i,i+1], [i+1,i]),i=0..10,.1): >display(c, scaling=constrained,insequence=true); The i=0..10,.1 is part of the sequence statement and gives you control of the starting point and ending point along with the number of steps taken to get there. You can change these values to see what happens. Regards, Georgios Kokovidis Dräger Medical