This one can also be done using int, without changing variables. (A matter of taste, whether figuring out the intersections is easier or not, for this example.)

> int(piecewise(x<1,2*x,2/x)-piecewise(x<1,1/x,x),x=1/sqrt(2)..sqrt(2));
                                   1      
                                   - ln(2)
                                   2      

acer

1) The command with should not be used inside a proc or a module. The help-page for with says that. I quote:

- The with command is effective only at the top level, and
  intended primarily for interactive use. Because with operates
  by using lexical scoping, it does not work inside the bodies
  of procedures, module definitions, or within statements.
  See the examples at the end of this help topic.

2) You have saved the module Phys_Quant_ex_0 with the global binding for diffindice. Rebinding diffindice interactively by issuing with(Math_Tools_ex) at the top-level in a new session will not affect the name diffindice that was saved in the module. (That is an intended part of the design of modules, and lots of stuff depends on that desirable behaviour.)

Either look at the help-page for use, or utilize the syntax Math_Tools_ex:-diffindice inside the body of module Phys_Quant_ex_0.

acer

> with(LinearAlgebra):
> M := Matrix(2, 2, [[r, r^2], [r^2, 1/r]]):
> vals,vecs := Eigenvectors(M):

> simplify(M.vecs-vecs.DiagonalMatrix(vals));
                                   [0    0]
                                   [      ]
                                   [0    0]

> simplify(vals[1]*vecs[1..-1,1] - M.vecs[1..-1,1]);
                                      [0]
                                      [ ]
                                      [0]

acer

It wasn't clear to me whether you wanted 1000 decimal digits or 1000 dozenal digits.

> F:=proc(x,b::posint,N::posint)
> local y,p:
> if b>16 then error; end if;
> y:=subs([10=A,11=B,12=C,13=D,14=E,15=F],
>         ListTools:-Reverse(
>   convert(floor(evalf[floor(N*ln(b)/ln(10))](x*b^N)),base,b))):
> p:=nops(y)-N;
> cat(op(y[1..p]),".",op(y[p+1..N]));
> end proc:

> # simple test case
> ans := F(sqrt(31793),12,60);
     ans := 12A.38075AB57660B4819B0B955BAA84365B892465A102072B7A53109A398

> Digits:=floor(60*ln(12)/ln(10)):

> convert(ans,decimal,12);
       178.3059168956543629146978561044118852999755964655120186918299562

> evalf(sqrt(31793)); # should be same as above
       178.3059168956543629146978561044118852999755964655120186918299562

> Digits:=10: # want F to work independent of top-level Digits setting

> # now, the posted question

> F(sqrt(2),12,1000);
1.4B79170A07B85737704B085486853504563650B559B8B79A401387B342380A998A173A951\
    303434821B55419A068816958B64282342A358A8947369B97237B9B04B656A072334932\
    8A219013A8B21AB42844A5758BA27B3A14317B17B28A4354B796260136269A55A79598A\
    4619BA2352A310A3373251B0598676B4537681A191A6901560B13362953A3B373054251\
    593693051410425656527080871A620766432B006383A272876409AB560250154713653\
    46AA731A9248B86B009972A5059115A10537765A3727300B71615798551101BB025B5A1\
    19781083699746484A9A0A5807960910B945AB250B74A6594723624594035156BB3A6A9\
    6559A453899500B6BB8811032B2332A74BB8070401B50A8A15BA2096184636714AB8894\
    749356151A36BA8AA424B6511A6AA35635A55848B5B4A9953B96478B317223B62700B28\
    4559B59A0AA34B6724497A247B53B8256881993B18A90A575862342586554334ABAB283\
    AA091186977782BB99734B16373A27B60A882935333325A1167A98A42B053831634948A\
    444A7572A993929440A412296997B297AA1810B79145B39974988B968B6731343532269\
    14236833678A02694B9A563B1017B953268692960AB384B15488A8B26808164413967B0\
    11056BA0A08BBAB3022935A1B6A096AA9A044836568294400477129BAA8048102482911\
    2B5

> length(%); # including the period
                                     1001

acer

Your data file is all ones and zeroes, so I presume you want that interpreted as a data set, say as discrete measurements of a square wave.

If that's right, then you could read it into Matrix using ImportMatrix, convert it to a Vector, and then run DiscreteTransforms:-FourierTransform on that. How you then wish to plot the frequency information is unclear, as you haven't mentioned anything about the sampling rate or time.

You can compare such DFT results with the continuous equivalent by searching  in google for  Fourier transform of a square  wave. (Eg, here, or the graphs here.)

I wonder, would it be useful for Maple to have something like Matlab's fftshift function?

acer

Could you use tensor[create] instead?

tensor[entermetric] eventually calls readline(), which calls iolib(2,..). And iolib is't checking IsWorksheetInterface() to acertain which interface is being used. So I don't see any way to trick it into behaving like it does in the commandline interface.

acer

Ok, so you can stuff all that data into a float[8] rtable (eg. Array), and plot that directly. See quickplot from R.Israel's Maple Advisor Database for a nice user-friendly routine for doing that. Here's an example of the kind of thing it does under the hood to make it more efficient for large numbers of points,

PLOT(POINTS(Array(1..3,1..2,
                  [[-0.5,1],[0.25,0],[0.5,-0.5]],
                  datatype=float[8])));

But a plot of a million points will likely still render too slowly and use a lot of memory. So instead of making a plot perhaps you should be thinking about making an image file directly, using the ImageTools package. In particular, ImageTools:-Create might help.

ps. There may also be an issue with your code. If you post it or upload it here, someone might be able to offer useful suggestions.

acer

Just use fclose on the file.

acer

> M:=ImportMatrix("foo.txt",source=Matlab):

> for i from 1 to 3 do
>   p[i],n[i] := M[i][1..3]^%T, M[i][4..-1]^%T;
> od:

> p[1],n[1],p[2],n[2],p[3],n[3];
                        [1]  [4]  [2]  [ 8]  [3]  [12]
                        [ ]  [ ]  [ ]  [  ]  [ ]  [  ]
                        [2], [5], [4], [10], [6], [15]
                        [ ]  [ ]  [ ]  [  ]  [ ]  [  ]
                        [3]  [6]  [6]  [12]  [9]  [18]

acer

> L1/2+L2;
                                                              -1
              [1, -1/2, 1/3, -1/4, 1/5, -1/6, 1/7, -1/8, 1/9, --]
                                                              10

> L1/2+L2/2;
                              -3                    -3        -3
             [3/4, -3/8, 1/4, --, 3/20, -1/8, 3/28, --, 1/12, --]
                              16                    32        40

...and so on.

If that's not what you want, then I cannot see what your goal is. What do you want to add together, to get the first element of the final result? Is it L1[1]/2 |+ L2[1]/2, or some mix of indices used for each list, or...?

acer

> subs((x-y)^(-1/2)=1/g,f2);
                                    a/g + b

acer

> xy:=Matrix([[10,10],[-14,-2],[-10,-10],[4,-24],[10,10]]):

> x:=xy[1..-1,1]:
> y:=xy[1..-1,2]:

> n := LinearAlgebra[RowDimension](xy):

> A := (1/2)*(add(x[i]*y[i+1]-y[i]*x[i+1], i = 1 .. n-1));
                                   A := 400
 
> 1/2*( x[1..-2].y[2..-1] - y[1..-2].x[2..-1] );
                                      400

Sometimes the inadvertant use of `sum` instead of `add` can work successfully for doing summation (adding) of finitely many terms. But not when there are unspecified Vector, Matrix or Array entries in the summed expression such as x[i]. The x[i] term is cannot be evaluated and is not allowed, prior to i being specified by an actual integer. And `sum` attempts that evaluation, while `add` delays it.

The evaluation can also be delayed for this example using so-called uneval-quotes. (That doesn't make it easier. I just mention it for completeness.)

> A := (1/2)*(sum('x[i]'*'y[i+1]'-'y[i]'*'x[i+1]', i = 1 .. n-1));
                                   A := 400

Who knows... maybe that behaviour of accessing unknown rtable entries like y[i] will change some day.

This topic may also be a common source of mistakes by new users, if the palette entries (Greek Sigma) supply `sum`.

acer

with(CurveFitting):
with(plots):
data := [[2005,2.85],[2006,5.70],[2007,10.0],[2008,14.8],[2011,25.0]]:
curve:=LeastSquares(data,v);
P1:=pointplot(data):
P2:=plot(curve,v=2005..2011):
display(P1,P2);

acer

Try it with a full colon after end do, but wrap the pointplot3d call in print.

 acer

Try here or here.

You also simply enter the word distance into Maple's help browser.

acer