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

Have a look here.

This brings up an interesting point. The Online Help is for the current release (which is Maple 13 at the time I write this). It would be useful if each help-page included a note as to the release in which the page was introduced or became relevant. Even if that could not be done comprehensively for all existing pages, it might still be useful were it done for all pages added since, say, Maple 12.

acer

Replace printf below with a call to fprintf (with a file name as the new first argument). That will make it print to a file rather than to the Maple interface.

> listA:=[40,50,60]:
> listB:=[80,100]:
> values:=[[1,2],[4,5],[6,7]]:

> for i from 1 to nops(listA) do
>   for j from 1 to nops(listB) do
>      printf("name_%d_%d = %d\n", listA[i],listB[j],values[i,j]);
>   end do:
> end do:

name_40_80 = 1
name_40_100 = 2
name_50_80 = 4
name_50_100 = 5
name_60_80 = 6
name_60_100 = 7

If values[i,j] are floating-point then you might use %g or %e or %f rather than %d. See the fprintf help-page for descriptions of the various formats.

acer

Presumably you are supposed to implement the elementwise parts of that task, otherwise you could write the absurdly trivial procedure,

inv2x2 := (A::Matrix(2,2)) -> A^(-1):

One way to get the formula for this is to find the inverse of a "general" 2x2 Matrix. By "general" I mean that every element has its own unique and mathematically unrelated name.

> M := Matrix(2,2,symbol=m);
                                [m[1, 1]    m[1, 2]]
                           M := [                  ]
                                [m[2, 1]    m[2, 2]]
 

So just issue the command M^(-1) and implement inside your precdure the formulae you see for each entry of the result .

Look for a common expression as the denominator of each entry, and have your procedure compute that first. Ask youself, what should your procedure do if that denominator is zero?

This may not always be the best way to go about such problems, but in this case it works easily.

acer