It needs a minor change to make it work, something like this: restart;with(VectorCalculus): f := (a,b,c)->DirectionalDiff(x*exp(y)/(3*z^2+1), ,[x, y, z] ): f(2,-1,0); f(1,-2,3); Hope that helps, J. Tarr

It needs a minor change to make it work, something like this: restart;with(VectorCalculus): f := (a,b,c)->DirectionalDiff(x*exp(y)/(3*z^2+1), ,[x, y, z] ): f(2,-1,0); f(1,-2,3); Hope that helps, J. Tarr

What makes you think that Maple has not solved the simultaneous equations correctly? J. Tarr

The convention adopted by Maple is: sqrt(x) represents the "principal square root", defined by the formula sqrt(x) = exp(1/2 * ln(x)) Please see ?sqrt and ?sqrt,Definition. You are both right :-). J. Tarr

The convention adopted by Maple is: sqrt(x) represents the "principal square root", defined by the formula sqrt(x) = exp(1/2 * ln(x)) Please see ?sqrt and ?sqrt,Definition. You are both right :-). J. Tarr

Please see ?assume and ?assuming. You need to use these commands when the result of the command or action that you are taking depends on the properties of the variables. For example, the solution of a second order ODE depends upon the properties of the parameters - see your text books. J. Tarr

I am not sure why this did not work when I tried it about an hour ago, but this should do what you want: restart; with(inttrans): X:=k*(m[1]+m[2])/(s*(k*(m[1]+m[2])+m[1]*s^2*m[2])); invlaplace(X,s,t) assuming m[1]>0,m[2]>0,k>0; Does this answer your question? J. Tarr

If you look at the result of the inverse laplace transformation in your first posting above, you will see it is a hyperbolic cosine. You wanted a cosine function and Maple correctly converted that by the well known relationship cosh(x) = cos(i*x). That can be converted to 1/2*(exp(x) + exp(-x)), if you prefer. Please see ?convert. Maple doesn't know the values of m[1], m[2] or k, so it can only reduce the inverse laplace transformation to a hyperbolic cosine. In your second posting, you used a^2 and so Maple was able to simplify the inverse laplace transformation to a cosine function. Good luck. J. Tarr

You could do this: restart; with(inttrans): X:=k*(m[1]+m[2])/(s*(k*(m[1]+m[2])+m[1]*s^2*m[2])); invlaplace(X,s,t); convert(%,cos); Is that what you wanted? J. Tarr

Axel Vogt, I should have mentioned that most people in the newsgroup accepted that P^(-1)*A*P – J = 0, in which A is a square matrix of floating point data, J is its Jordan Form and P is the transition matrix corresponding to J. However, some were not convinced that this was a satisfactory test of accuracy, and so I produced the worksheet that compares the eigenvalues of A and J. Many thanks for your comment and the reference. Dave Linder, It all depends on what one means by "zero" in floating point. Setting Digits to, say, 50 results in all the terms on the leading diagonal of the Jordan form of your Matrix A being less than 10^(-8) and similarly for the eigenvalues of A. It might be worth including something about the instability of the algorithm in the help pages for linalg[jordan] and LinearAlgebra[JordanForm] and suggesting that the user consider the Schur Form instead. Many thanks for the references. All the best, J. Tarr

comp.soft-sys.math.maple often carries many postings each day and is not moderated. Some posts there are unconnected with Maple and some is near-spam, so the workload on MaplePrimes' moderator would increase sharply. Great if Maplesoft is happy to do this, but we should remember that Maplesoft is a commercial enterprise and we all want it to stay in business. J. Tarr

Developments in electronic chips will probably mean that within a decade, we will be carrying extremely powerful computers around in our pockets, unfolding them when required for use. My guess is that something like Maple will be available to run on these pocket computers of the future, together with custom calculators, spreadsheets and word processors. Just now I am wondering what the graphics will look like on an unrolled sheet of plastic... J. Tarr

The pdf mentioned above uses Maple V R3, or perhaps earlier, and so is a bit dated. There's nothing on transmission lines shown in the index. Out of curiosity I checked a simple example about radioactive decay: the answer shown was wrong! J. Tarr

Joe Riel, My post above was made in standard worksheet mode. I discovered a while back that spelling out a greek letter followed by an underscore and another character mysteriously produced the greek character joined to the succeeding character. It works for upper and lower case greek characters, and is very useful. I don't think it is documented. J. Tarr

Joe Riel, My post above was made in standard worksheet mode. I discovered a while back that spelling out a greek letter followed by an underscore and another character mysteriously produced the greek character joined to the succeeding character. It works for upper and lower case greek characters, and is very useful. I don't think it is documented. J. Tarr