Questions - MaplePrimes

Equivalent of lookup function?...

Asked by: brian bovril 914

October 28 2014

1 3

Hello. As you can see the minimum of set X is 2.

the time pair in set W whose difference is 2 corresponds to elements 54 and 55. How do i locate and print these automatically?

Digits:=4:

U := [1.010, 1.110, 1.210, 1.310, 1.410, 1.510, 1.610, 1.710, 1.810, 1.910, 2.020, 2.120, 2.220, 2.320, 2.420, 2.520, 2.620, 2.720, 2.820, 2.920, 3.030, 3.130, 3.230, 3.330, 3.430, 3.530, 3.630, 3.730, 3.830, 3.930, 4.040, 4.140, 4.240, 4.340, 4.440, 4.540, 4.640, 4.740, 4.840, 4.940, 5.050, 5.150, 5.250, 5.350, 5.450, 5.550, 5.650, 5.750, 5.850, 5.950, 6.060, 6.160, 6.260, 6.360, 6.460, 6.560, 6.660, 6.760, 6.860, 6.960, 7.070, 7.170, 7.270, 7.370, 7.470, 7.570, 7.670, 7.770, 7.870, 7.970, 8.080, 8.180, 8.280, 8.380, 8.480, 8.580, 8.680, 8.780, 8.880, 8.980, 9.090, 9.190, 9.290, 9.390, 9.490, 9.590, 9.690, 9.790, 9.890, 9.990, 10.01, 11.11, 12.21]:

W := select(proc (t) options operator, arrow; .59 >= frac(t) end proc, U);

[1.010, 1.110, 1.210, 1.310, 1.410, 1.510, 2.020, 2.120, 2.220, 2.320, 2.420, 2.520, 3.030, 3.130, 3.230, 3.330, 3.430, 3.530, 4.040, 4.140, 4.240, 4.340, 4.440, 4.540, 5.050, 5.150, 5.250, 5.350, 5.450, 5.550, 6.060, 6.160, 6.260, 6.360, 6.460, 6.560, 7.070, 7.170, 7.270, 7.370, 7.470, 7.570, 8.080, 8.180, 8.280, 8.380, 8.480, 8.580, 9.090, 9.190, 9.290, 9.390, 9.490, 9.590, 10.01, 11.11, 12.21]

(1)

X:=[seq(trunc(W[i+1])*60+frac(W[i+1])*100-(trunc(W[i])*60+frac(W[i])*100),i=1..nops(W)-1)];

[10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 11.00, 10.00, 10.00, 10.00, 10.00, 10.00, 2.00, 70.00, 70.00]

(2)

min(X),{W[54],W[55]};

(3)

Download test.mw

How to plot two or more plane equations in maple?...

Asked by: ANANDMUNAGALA 175

October 28 2014

0 2

Good afternoon sir.

I request your kind suggestion to my cited query.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

finite summation evaluation...

Asked by: Aakanksha 30

October 28 2014

0 6

Need to evaluate above summation in maple. For simplicity infinity can be replaced by some constant say 1000.

The value of parameters used are:

alpha=73.11

beta=76.11

E=4.75

help....

2D numerical Integration Error...

Asked by: Hb 10

October 27 2014

0 0

Hi,

I have been trying to solve 2D Diffusion Equation with zero Neumann BC over the unit disk. If I use Gaussian type function with a sharp peak as initial condition, I get huge errors between initial values. Let's say u(r,phi,t) is the solution of the PDE and f(r,phi) is initial value function. The expectation is for the point (r*,phi*) , u(r*,phi*,0)=f(r*,phi*), but it is not.

Is Numerical integration in Maple not able to handle such sharp peak? I tried some of the built-in methods such as MonteCarlo,CubaVegas but no difference.

It might be a good idea to specify some nodes arround the peak. There is a command called "peaks", but I could not use it, error message says "invalid arguments".

Thanks in advance.

Soln_2D_Gaussian.mw

Could possibly (still) be a bug? Rank of a symboli...

Asked by: casperyc 962

October 27 2014

1 4

I am a research student and I am writing up my thesis. I was reading one of the paper written earlier by authors in 2003.

The authors calculated a symbolic rank of a matrix A, and got 9, using Maple 6.

Then new methods developed in 2012 and proved that the "true" rank should be 8. The later paper that "Most recent versions of Maple have this simplification built in and are able to return a rank of 8.".

I just noticed that in Maple 18.01, this matrix was still evaluted for a symbolic rank 9, if no simplifcaiton was done before using Rank(), see attached.

I didnt explore a lot, but just as a notice. I am a bit concerned as most of my research was trying to deal with exponentials.

Is that something to be fixed in future versions?

Thanks!

Bug.mw

piecewise function simplify / cancel out a variabl...

Asked by: Stephan 45

October 27 2014

1 3

Hello,
Maple does not cancel out a variable.

Why is that?

Is there a way to solve this?

(I pasted my code on the bottom of this message)

Thanks for your help/advice,

Stephan

restart:
M(x):=piecewise(x<=l,1/2*(q*x^2)/(EI)-3/8*(q*l*x)/(EI),l<x,1/2*(q*x^2)/(EI)-13/8*(q*l*x)/(EI)+5/4*(q*l^2)/(EI)):
M(x):=M(x)*(-EI);
# simplify() does not work.....?
M(x):=simplify(%) assuming EI>0;
# Wiht EI cancelled out by hand it schould look like:
M(x):=piecewise(x<=l,1/2*(q*x^2)-3/8*(q*l*x),l<x,1/2*(q*x^2)-13/8*(q*l*x)+5/4*(q*l^2));

Dos and don'ts of Maple programming w.r.t. the phy...

Asked by: NOh 55

October 27 2014

2 5

EDIT 2: Ok, here is what I am trying to do with Maple.

I have the following expression:

with

and

(j times)

Now the above expression should be zero. I get a set of equations in orders of lambda. The H matrices are known. I only need to find the S matrices.

Anyways, the HOffDiag is the sum above and the GetAdvCommutator0 is this nested commutator. And I need this sum evaluated up to order n (n is a given input-number) in lambda. The rest is really not that performance hungry.

EDIT: Ok, it realy seems that this sum of commutators is something that takes Maple forever. I have written a new MWE. Is it possible to speed things up? Maybe using assumptions or so?

>	restart;with(Physics):with(LinearAlgebra): Setup(mathematicalnotation=true):Setup(noncommutativeprefix={M,H,S});

(1)

>

GetAdvCommutator0:=proc(power::integer,order::integer,M)
if power <= 0 then return M fi:
if power = 1 then

return mtaylor(Commutator(M,add(lambda^l*S[l],l=1..order)),lambda, order+1) fi:

return mtaylor(Commutator(GetAdvCommutator0(power-1,order,M),add(lambda^l*S[l],l=1..order)),lambda,order+1):
end proc:

>	HOffDiag:=n->eval(add(1/((2j+1)!)'GetAdvCommutator0'(2j+1,n,H[0]),j=0..iquo(n-1,2)+1)+add(1/((2j+1)!)'GetAdvCommutator0'(2j+1,n,lambdaH[1]),j=0..iquo(n-2,2)+1)+add(1/((2j)!)'GetAdvCommutator0'(2j,n,lambda*H[2]),j=0..iquo(n-1,2)+1)):

>	HOffDiag(3); # still fast

lambda^3*Physics:-Commutator(H[0], S[3])+lambda^2*Physics:-Commutator(H[0], S[2])+lambda*Physics:-Commutator(H[0], S[1])+(1/6)*lambda^3*Physics:-Commutator(Physics:-Commutator(Physics:-Commutator(H[0], S[1]), S[1]), S[1])+lambda^2*Physics:-Commutator(H[1], S[1])+lambda^3*Physics:-Commutator(H[1], S[2])+lambda*H[2]+(1/2)*lambda^3*Physics:-Commutator(Physics:-Commutator(H[2], S[1]), S[1])

(2)

>	HOffDiag(6); # takes a very long time, I didn't even try entering higher numbers

Warning, computation interrupted

>

SumOfCommutators.mw

Hi,

I know this question is really unspecific. However, since I am still new to Maple, and since you don't program with Maple as you would do with c++, I would appreciate some advise. I have written a program in Maple and I don't know if it works correctly. All I know is, it is pretty slow :(.

So if anyone of you could just have a short look at my attached worksheet and could tell me, if I at some point used an obvious "don't", I would greatly appreciate it.

Cheers

NOh

A Maple worksheet

Setup Maple

>	restart;with(Physics):with(LinearAlgebra):Setup(mathematicalnotation=true):Setup(noncommutativeprefix={MA,MB,H,S});

(1.1)

Input parameter

>	H0:=Matrix([[ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , U , 0 ], [ 0 , 0 , 0 , U ]]);

H0 := Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = U, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = U})

(2.1)

>	H1:=Matrix([[ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ]]);

H1 := Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0})

(2.2)

>	H2:=Matrix([[ 0 , 0 , -t , -t ], [ 0 , 0 , t , t ], [ -t , t , 0 , 0 ], [ -t , t , 0 , 0 ]]);

H2 := Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = -t, (1, 4) = -t, (2, 1) = 0, (2, 2) = 0, (2, 3) = t, (2, 4) = t, (3, 1) = -t, (3, 2) = t, (3, 3) = 0, (3, 4) = 0, (4, 1) = -t, (4, 2) = t, (4, 3) = 0, (4, 4) = 0})

(2.3)

>	orderNumber:= 4 ;

(2.4)

Define all necessary functions

>

GetAdvCommutator0:=proc(power::integer,order::integer)
    local Comm:
    if power <= 0 then return H[0] fi:
    if power = 1 then return add(coeff(Commutator(H[0],add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order) fi:

    return add(coeff(Commutator(GetAdvCommutator0(power-1,order),add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order):
end proc:

>

GetAdvCommutator1:=proc(power::integer,order::integer)
    local Comm:
    if power <= 0 then return lambda*H[1] fi:
    if power = 1 then return add(coeff(Commutator(lambda*H[1],add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order) fi:

    return add(coeff(Commutator(GetAdvCommutator1(power-1,order),add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order):
end proc:

>

GetAdvCommutator2:=proc(power::integer,order::integer)
    local Comm:
    if power <= 0 then return lambda*H[2] fi:
    if power = 1 then return add(coeff(Commutator(lambda*H[2],add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order) fi:

    return add(coeff(Commutator(GetAdvCommutator2(power-1,order),add(lambda^l*S[l],l=1..order)),lambda,n)*lambda^n,n=0..order):
end proc:

>	HOffDiag:=n->eval(add(1/((2j+1)!)'GetAdvCommutator0'(2j+1,n),j=0..iquo(n-1,2)+1)+add(1/((2j+1)!)'GetAdvCommutator1'(2j+1,n),j=0..iquo(n-2,2)+1)+add(1/((2j)!)'GetAdvCommutator2'(2*j,n),j=0..iquo(n-1,2)+1)):

>	HOnDiag:=n->eval(add(1/((2j)!)'GetAdvCommutator0'(2j,n),j=0..iquo(n,2)+1)+add(1/((2j)!)'GetAdvCommutator1'(2j,n),j=0..iquo(n-1,2)+1)+add(1/((2j+1)!)'GetAdvCommutator2'(2*j+1,n),j=0..iquo(n-2,2)+1)):

>

computeHBlockDiag:=proc( n::posint,firstBlockWidth::posint, H0::Matrix, H1::Matrix, H2::Matrix )
    local expr, Orders, i, eq, j, eq2, orders,smatrices,rows::posint,evallist,m,l,cols,Hdiag,ff;


    expr:=HOffDiag(n);       # !!!!! this one is one of the slowest parts I think
    Orders:=[$1..n];
    for i from 1 to n do
        eq:=add(coeff(expr,lambda,l)*lambda^l,l=0..i)=0;
        for j from 1 to i-1 do
            eq:=eval(eq,Commutator(H[0],S[j])=Orders[j]);
        od;
        eq:=subs(Commutator(H[0],S[i])=MX,eq);
        eq2:=solve(eq,MX);
        Orders[i]:=eq2;
    od;


    rows,cols:=Dimension(H0);
    smatrices:=[seq(H0,i=1..n)];
    for i from 1 to n do
        evallist:=[`*`=`.`,H[1]=H1,H[2]=H2];
        for j from 1 to i-1 do
            evallist:=[op(evallist),S[j]=smatrices[j]];
        od;
        smatrices[i]:=eval(Orders[i],evallist);
        for m from 1 to firstBlockWidth do
            for l from firstBlockWidth+1 to rows do
                smatrices[i][m,l]:=smatrices[i][m,l]*1/(H0[m][m]-H0[l][l]);
                smatrices[i][l,m]:=smatrices[i][l,m]*1/(H0[m][m]-H0[l][l]);
            od;
        od;
    od;

    expr:=HOnDiag(n);            # !!!!! and this one of course (almost identical to the one above)
    Hdiag:=eval(add(coeff(expr,lambda,l)*lambda^l,l=0..n),lambda=1);
    evallist:= [op(evallist), H[0]=H0,S[n]=smatrices[n]];
    Hdiag:=eval(Hdiag,evallist);


    return Hdiag;
end proc:

>

>	computeHBlockDiag(2,2,H0,H1,H2);

$Matrix(4, 4, {(1, 1) = -2*t^2/U, (1, 2) = 2*t^2/U, (1, 3) = 0, (1, 4) = 0, (2, 1) = 2*t^2/U, (2, 2) = -2*t^2/U, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = U+2*t^2/U, (3, 4) = 2*t^2/U, (4, 1) = 0, (4, 2) = 0, (4, 3) = 2*t^2/U, (4, 4) = U+2*t^2/U})$

(1)

>	Eigenvalues(%);

$Vector(4, {(1) = 0, (2) = U, (3) = -4*t^2/U, (4) = (U^2+4*t^2)/U})$

(2)

>

Testing

#Fast

>	computeHBlockDiag(2,2,H0,H1,H2);

$Matrix(4, 4, {(1, 1) = -2*t^2/U, (1, 2) = 2*t^2/U, (1, 3) = 0, (1, 4) = 0, (2, 1) = 2*t^2/U, (2, 2) = -2*t^2/U, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = U+2*t^2/U, (3, 4) = 2*t^2/U, (4, 1) = 0, (4, 2) = 0, (4, 3) = 2*t^2/U, (4, 4) = U+2*t^2/U})$

(4.1)

#Not so fast

>	computeHBlockDiag(4,2,H0,H1,H2);

$Matrix(4, 4, {(1, 1) = -2*t^2/U-(8/3)*t^4/U^3, (1, 2) = 2*t^2/U+(8/3)*t^4/U^3, (1, 3) = 0, (1, 4) = 0, (2, 1) = 2*t^2/U+(8/3)*t^4/U^3, (2, 2) = -2*t^2/U-(8/3)*t^4/U^3, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = U+2*t^2/U+(8/3)*t^4/U^3, (3, 4) = 2*t^2/U+(8/3)*t^4/U^3, (4, 1) = 0, (4, 2) = 0, (4, 3) = 2*t^2/U+(8/3)*t^4/U^3, (4, 4) = U+2*t^2/U+(8/3)*t^4/U^3})$

(4.2)

#Already pretty slow :(

>	computeHBlockDiag(6,2,H0,H1,H2);

Warning, computation interrupted

>

Download MapleWorksheet.mw

How to correct error of invalid range...

Asked by: emmantop 40

October 27 2014

0 1

How can error of invalid range be corrected FDM.mw

Memory allocation question...

Asked by: Mac Dude 1576

October 27 2014

1 2

I am running a Maple procedure in CLI Maple (mainly because it just provides colored random numbers for other work). When the procedure runs I am getting this kind of output scrolling by:

memory used=10732.9MB, alloc=344.0MB, time=895.85
memory used=10809.2MB, alloc=345.6MB, time=904.26
memory used=10885.5MB, alloc=347.3MB, time=912.67
memory used=10961.8MB, alloc=348.9MB, time=921.27
.
.
.

So I am wondering what "memory used" and "memory allocated" mean here. In "Activity Monitor" (this is Mac OS X) I see that the "memory allocated" values correspond rather closely to the "real memory" usage. No idea what the "memory used" column is. Note that this is a 32 bit process running on a PPS Mac, and it merrily "uses" unlimited amounts of memory that don't exist on the machine and that a 32 bit process should not be able to use (I think Mac OS X limits 32 bit processes to 4 GB). It is clearly not paging as the machine stays perfectly useable.

So, what am I to make of these numbers?

Mac Dude.

Is it possible to do a partial fraction decomposit...

Asked by: descartes0000 25

October 26 2014

0 3

I was just wondering if there is such thing as a technique for solving integrals using a PFD(partial fraction decomposition) if the denominator's rational roots are complex numbers???

Multiple pointplots...

Asked by: jonlg 55

October 26 2014

1 6

Hi there,

I am trying to display multiple pointplots in the same plot. I would like to either distinguish the plots by their color or by the symbol (square, traiangle, circle) at each data-point. May be both, although I ignore whether using symbols at data-points is possible in Maple.

The plot i would like to achieve is this:

Maple says that my points are not in the correct format: MapleSimulation_test6.mw

My data array N has two parameters, and the free variable. The two parameters are also varied over a certain range. For plotting purposes, I would like to have one of the variables fixed, select three parameters for the other variable, and plot all data-points for them.

The resulting data array N seems to be OK, but I am aware that my attempt for plotting multiple arrays may be wrong. May be pointplot is not even the best choice.

Any ideas?

Thank you,

jon