Carl Love

Carl Love

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12 years, 316 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are answers submitted by Carl Love

The defined coordinate systems are stored in an unexported table of an  undocumented package, as given in the title. Here's a procedure to access it:

Coords:= proc(C::{name,function}, V::list(name))
uses PC= Plot:-CoordinateSystems;  #undocumented package
local
    opq:= kernelopts(opaquemodules= false),    
    r:= PC:-CoordinateSystemsTable[String(`if`(C::name, C, op(0,C)))](
        V[],`if`(C::name, [][], op(C))
    )
;
    kernelopts(opaquemodules= opq);
    r
end proc
:
Coords(spherical_physics, [r, theta, phi]);
      [r*cos(phi)*sin(theta), r*sin(theta)*sin(phi), r*cos(theta)]

Since the package is undocumented, don't rely too much on the syntax remaining exactly the same as the way I just accessed it in the procedure.

By using a Newton's method iteration with pure-integer builtin (i.e., GMP) arithmetic, I can find roots of integers significantly faster than with any method that uses evalf. While most of Maple's integer-arithmetic commands whose names begin with i are builtin, that isn't true of iroot: It's written in Maple, and it uses evalf.

My Newton procedure is the second procedure listed in the table methods in the worksheet below. The Newton iteration for finding the cube root r of is simply

r-> (2*r^3 + n)/(3*r^2)

restart
:

#Make garbage collection single threaded, because multi-threaded makes
#"real-time" comparisons difficult:
kernelopts(gcmaxthreads= 1)
:

#
# The procedures to be tested and compared
# ----------------------------------------
methods:= table([
    #iroot with floor correction:
    Iroot= (n-> local r:= iroot(n,3); `if`(r^3 > n, r-1, r)),

    #Integerized Newton's method:
    Newton= proc(n)
    local r1:= integermul2exp(1, trunc(ilog2(1+n)/3 + 1/2)), r, r2;
        do r1:= iquo(2*(r2:= (r:= r1)^2)*r + n, 3*r2) until r-r1 in {-1,0,1};
        min(r,r1)
    end proc,

    #Two of @dharr's methods:
    root@evalf= (n-> floor(root(evalf(n),3))),
    evalf@root= (n-> floor(evalf(root(n,3)))),

    #iroot's own method, minimalized:
    evalf@`^`= (n-> trunc(evalf(n^(1/3))))
]):
meth_names:= [indices](methods, nolist);

[`@`(evalf, `^`), `@`(root, evalf), Iroot, Newton, `@`(evalf, root)]

# Generate random test cases:
#
nWords:= 3:  iterations:= 2^16:
R:= rand(2^((nWords-1)*64)..2^(nWords*64)-1):
#Arbitrary key for consistent test data on different runs and platforms.
#Change the key to generate different random test data.
randomize(23):
testdata:= table(
    [meth_names[], accuracy]=~ ['['R'()$iterations]' $ 1+numelems(methods)]
):

#Set sufficient floating-point precision (see showstat(iroot), line 39)
#(only needed for methods that directly use evalf):
d:= length(2^(nWords*64)-1):  Digits:= iquo(d,3)+5+length(d);

26

# The tests:
#
#Accuracy test: If all methods agree, then all will be listed in a single set:
entries(ListTools:-Classify(j-> methods[j]~(testdata[accuracy]), meth_names), nolist);

{Iroot, Newton, `@`(evalf, `^`), `@`(evalf, root), `@`(root, evalf)}

#Efficiency test:
for j in meth_names do
    f:= methods[j];
    printf("%a:\n", j);
    CodeTools:-Usage(seq[reduce= ()](f(n), n= testdata[j]));
    printf("\n")
od:

evalf@`^`:
memory used=1.59GiB, alloc change=-16.00MiB, cpu time=4.27s, real time=7.89s, gc time=750.00ms

root@evalf:
memory used=1.79GiB, alloc change=0 bytes, cpu time=3.69s, real time=9.42s, gc time=578.12ms

Iroot:
memory used=485.51MiB, alloc change=0 bytes, cpu time=953.00ms, real time=1.96s, gc time=218.75ms

Newton:
memory used=211.56MiB, alloc change=0 bytes, cpu time=750.00ms, real time=1.42s, gc time=203.12ms

evalf@root:
memory used=3.62GiB, alloc change=0 bytes, cpu time=13.97s, real time=25.77s, gc time=1.73s
 

Newton's method is the winner.

 

The Newton-method procedure for any particular root be generated by this procedure:

NewtonIroot:= (k::And(posint, Not(1)))-> subs(
    _k= k,
    proc(n::nonnegint)
    local r1:= integermul2exp(1, trunc(ilog2(1+n)/_k + 1/2)), r, r2;
        do r1:= iquo((_k-1)*(r2:= (r:= r1)^(_k-1))*r + n, _k*r2)
        until r-r1 in {-1,0,1};
        min(r,r1)
    end proc
):

showstat((Iroot5:= NewtonIroot(5)));


proc(n::nonnegint)
local r1, r, r2;
   1   r1 := integermul2exp(1,trunc(1/5*ilog2(1+n)+1/2));
   2   do
   3       r1 := iquo(4*(r2 := (r := r1)^4)*r+n,5*r2)
       until r-r1 in {-1, 0, 1};
   4   min(r,r1)
end proc
 

r:= rand();

441235504072

n:= (r+1)^5-1:  Iroot5(n);

441235504072

 

Download Iroot.mw

The votes awarded after deletion do apply to the author's reputation. It has always been true (at least for the 10 years or so that I've been using MaplePrimes) that votes awarded to an item which is subsequently deleted continue to apply to the author's reputation.

I totally agree with acer that items Deleted As Spam should not be retained. There are now several Posts where the majority of Replies have been deleted as spam yet remain as clutter.

The syntax of Maple's hypergeom always has exactly three arguments. The first two arguments are arbitrary-length lists of algebraic expressions, and the third is a single algebraic expression rather than a list. If it's a 2F1 function, then the first list has 2 elements, and the second has 1.

The biggest and ugliest (IMO) differences between Maple and Mathematica syntax are that Maple's lists are enclosed in square brackets ([...]) rather than curly braces ({...}) and that Maple's function arguments are always in parentheses ((...)) rather than square brackets. (I consider the Mathematica to be uglier.)

In another Answer, @janhardo shows the correct syntax of a simple Maple hypergeom function (of class 2F1): 

hypergeom([1,1], [2], -z)

Your example could be entered as

-r^2/3*((1-beta)/`γ`)^n*
    hypergeom(
        [n, n/(1-beta)], [(beta-1-n)/(beta-1)], -r^(3/n*(1-beta))*chi^2/`γ`
    )

I used `γ` instead of gamma because gamma is by default a pre-defined constant (similar to Pi). If you mean the constant, then just use gamma. If it's intended to be a variable or parameter, there are several workarounds, one of which is `γ`.

My original Answer was wrong because I made M^*.M a symmetric matrix when actually it's hermitian. I thank dharr for pointing out my error. Here's a corrected Answer. The incorrect one is at the bottom of this Answer for the sake of continuity of reading.

I will use for your because the default sans-serif font on MaplePrimes essentially makes l unreadable. The trick to doing your problem is expressing L as Lx + I*Ly where Lx and Ly are assumed real, and simplifying conjugated expressions with simplify(..., conjugate) and evalc.

In Maple, M^%H (or M^*) is the HermitianTranspose of M, and A.B is the matrix product of matrices A and B. Spelling out MatrixMatrixMultiply or HermitianTranspose is unnecessary.

Because I use evalc (which assumes variables are real), the assume(a::complex) is needed.

The final eigenvectors below can be further simplified, but it requires subtlety.

restart:

interface(showassumed= 0):

assume(Lx::real, Ly::real, a::complex, k>0, Lx^2+Ly^2 < 1);

L:= Lx+I*Ly:

B:= simplify(k*(1-L*conjugate(L)));

-k*(Lx^2+Ly^2-1)

M:= <L-B, -a*B; 0, L+B>;

Matrix(2, 2, {(1, 1) = Lx+I*Ly+k*(Lx^2+Ly^2-1), (1, 2) = a*k*(Lx^2+Ly^2-1), (2, 1) = 0, (2, 2) = Lx+I*Ly-k*(Lx^2+Ly^2-1)})

MS:= simplify(evalc~(M^%H.M), conjugate);

Matrix(2, 2, {(1, 1) = (Lx^2+Ly^2-1)^2*k^2+2*Lx*(Lx^2+Ly^2-1)*k+Lx^2+Ly^2, (1, 2) = a*k*(Lx^2+Ly^2-1)*(Lx-I*Ly+Lx^2*k+Ly^2*k-k), (2, 1) = k*(Lx^2+Ly^2-1)*conjugate(a)*(Lx+I*Ly+k*(Lx^2+Ly^2-1)), (2, 2) = (Lx^2+Ly^2-1)^2*k^2*abs(a)^2+Ly^2+(Lx-k*(Lx^2+Ly^2-1))^2})

EV:= LinearAlgebra:-Eigenvectors(MS, output= vectors);

Matrix(2, 2, {(1, 1) = -2*a*((-Lx^2-Ly^2+1)*k+I*Ly-Lx)/(-sqrt((Lx^2+Ly^2-1)^2*k^2*abs(a)^4+(4*(Lx^2+Ly^2-1)^2*k^2+4*Lx^2+4*Ly^2)*abs(a)^2+16*Lx^2)+k*(Lx^2+Ly^2-1)*abs(a)^2-4*Lx), (1, 2) = -2*a*((-Lx^2-Ly^2+1)*k+I*Ly-Lx)/(sqrt((Lx^2+Ly^2-1)^2*k^2*abs(a)^4+(4*(Lx^2+Ly^2-1)^2*k^2+4*Lx^2+4*Ly^2)*abs(a)^2+16*Lx^2)+k*(Lx^2+Ly^2-1)*abs(a)^2-4*Lx), (2, 1) = 1, (2, 2) = 1})

eval(map(simplify, EV, {L=`&ell;`, Lx-I*Ly= CL}), CL= conjugate(`&ell;`));

Matrix(2, 2, {(1, 1) = -2*a*(`&ell;`*conjugate(`&ell;`)*k+conjugate(`&ell;`)-k)/(k*(-`&ell;`*conjugate(`&ell;`)+1)*abs(a)^2+sqrt(k^2*(`&ell;`*conjugate(`&ell;`)-1)^2*abs(a)^4+(4*k^2*(`&ell;`*conjugate(`&ell;`)-1)^2+4*`&ell;`*conjugate(`&ell;`))*abs(a)^2+4*(`&ell;`+conjugate(`&ell;`))^2)+2*conjugate(`&ell;`)+2*`&ell;`), (1, 2) = 2*a*(`&ell;`*conjugate(`&ell;`)*k+conjugate(`&ell;`)-k)/(k*(`&ell;`*conjugate(`&ell;`)-1)*abs(a)^2+sqrt(k^2*(`&ell;`*conjugate(`&ell;`)-1)^2*abs(a)^4+(4*k^2*(`&ell;`*conjugate(`&ell;`)-1)^2+4*`&ell;`*conjugate(`&ell;`))*abs(a)^2+4*(`&ell;`+conjugate(`&ell;`))^2)-2*conjugate(`&ell;`)-2*`&ell;`), (2, 1) = 1, (2, 2) = 1})

 

Download SymbEigenvectors.mw

 

Original incorrect worksheet:

restart:

L:= Lx+I*Ly:

B:= simplify(k*(1-L*conjugate(L))) assuming real;

-k*(Lx^2+Ly^2-1)

M:= <L-B, -a*B; 0, L+B>;

Matrix(2, 2, {(1, 1) = Lx+I*Ly+k*(Lx^2+Ly^2-1), (1, 2) = a*k*(Lx^2+Ly^2-1), (2, 1) = 0, (2, 2) = Lx+I*Ly-k*(Lx^2+Ly^2-1)})

MS:= Matrix(evalc~(M^%H.M), shape= symmetric) assuming a::complex;

Matrix(2, 2, {(1, 1) = (Lx+k*(Lx^2+Ly^2-1))^2+Ly^2, (1, 2) = a*((Lx+k*(Lx^2+Ly^2-1))*k*(Lx^2+Ly^2-1)-I*Ly*k*(Lx^2+Ly^2-1)), (2, 1) = a*((Lx+k*(Lx^2+Ly^2-1))*k*(Lx^2+Ly^2-1)-I*Ly*k*(Lx^2+Ly^2-1)), (2, 2) = conjugate(a*k*(Lx^2+Ly^2-1))*a*k*(Lx^2+Ly^2-1)+(Lx-k*(Lx^2+Ly^2-1))^2+Ly^2})

EV:= LinearAlgebra:-Eigenvectors(MS, output= vectors) assuming (Lx,Ly)::~real, abs(L)^2 < 1, k>0;

Matrix(2, 2, {(1, 1) = -2*a*((-Lx^2-Ly^2+1)*k+I*Ly-Lx)/(-sqrt(k^2*(Lx^2+Ly^2-1)^2*abs(a)^4-8*k*Lx*(Lx^2+Ly^2-1)*abs(a)^2+4*a^2*k^2*(Lx^2+Ly^2-1)^2-8*a^2*(I*Ly-Lx)*(Lx^2+Ly^2-1)*k+(4*a^2+16)*Lx^2-(8*I)*a^2*Lx*Ly-4*Ly^2*a^2)+k*(Lx^2+Ly^2-1)*abs(a)^2-4*Lx), (1, 2) = -2*a*((-Lx^2-Ly^2+1)*k+I*Ly-Lx)/(sqrt(k^2*(Lx^2+Ly^2-1)^2*abs(a)^4-8*k*Lx*(Lx^2+Ly^2-1)*abs(a)^2+4*a^2*k^2*(Lx^2+Ly^2-1)^2-8*a^2*(I*Ly-Lx)*(Lx^2+Ly^2-1)*k+(4*a^2+16)*Lx^2-(8*I)*a^2*Lx*Ly-4*Ly^2*a^2)+k*(Lx^2+Ly^2-1)*abs(a)^2-4*Lx), (2, 1) = 1, (2, 2) = 1})

eval(map(simplify, EV, {L=`&ell;`, Lx-I*Ly= CL}), CL= conjugate(`&ell;`));

Matrix(2, 2, {(1, 1) = -2*a*(k*`&ell;`*conjugate(`&ell;`)+conjugate(`&ell;`)-k)/(k*(-`&ell;`*conjugate(`&ell;`)+1)*abs(a)^2+sqrt(k^2*(`&ell;`*conjugate(`&ell;`)-1)^2*abs(a)^4-4*k*(`&ell;`+conjugate(`&ell;`))*(`&ell;`*conjugate(`&ell;`)-1)*abs(a)^2+4*a^2*(`&ell;`*conjugate(`&ell;`)-1)^2*k^2+8*a^2*conjugate(`&ell;`)*(`&ell;`*conjugate(`&ell;`)-1)*k+4*`&ell;`^2+8*`&ell;`*conjugate(`&ell;`)+4*conjugate(`&ell;`)^2*(a^2+1))+2*conjugate(`&ell;`)+2*`&ell;`), (1, 2) = 2*a*(k*`&ell;`*conjugate(`&ell;`)+conjugate(`&ell;`)-k)/(k*(`&ell;`*conjugate(`&ell;`)-1)*abs(a)^2+sqrt(k^2*(`&ell;`*conjugate(`&ell;`)-1)^2*abs(a)^4-4*k*(`&ell;`+conjugate(`&ell;`))*(`&ell;`*conjugate(`&ell;`)-1)*abs(a)^2+4*a^2*(`&ell;`*conjugate(`&ell;`)-1)^2*k^2+8*a^2*conjugate(`&ell;`)*(`&ell;`*conjugate(`&ell;`)-1)*k+4*`&ell;`^2+8*`&ell;`*conjugate(`&ell;`)+4*conjugate(`&ell;`)^2*(a^2+1))-2*conjugate(`&ell;`)-2*`&ell;`), (2, 1) = 1, (2, 2) = 1})

``

Below, I have put your data in exactly the same order as you gave it; however, I've grouped it with square brackets and commas. Please note where the brackets are; the pattern should be obvious. Line breaks, spacing, and indentation are irrelevant; they're only used to enhance readability.

Pts:= [
    [
        [80, 1, 1.80074340195803263], [80, 2, 1.79160585795803287], [80, 3, 1.7805446209580329], 
        [80, 4, 1.76755968995803269], [80, 5, 1.75265106495803280], [80, 6, 1.73581874695803272],
        [80, 7, 1.71706273595803272]
    ],
    [                        
        [85, 1, 1.71097991748770582], [85, 2, 1.70184237348770577], [85, 3, 1.69078113648770581], 
        [85, 4, 1.67779620548770559], [85, 5, 1.66288758048770571], [85, 6, 1.64605526248770562],
        [85, 7, 1.62729925148770591]
    ],                          
    [
       [95, 1, 1.53550477309879818], [95, 2, 1.52636722910871089], [95, 3, 1.51530599205435690], 
       [95, 4, 1.50232106105435669], [95, 5, 1.48741243605435680], [95, 6, 1.47058011805435672], 
       [95, 7, 1.45182410705435672]
    ],
    [                    
       [100, 1, 1.44826461660029512], [100, 2, 1.43912707260029623], [100, 3, 1.42806583560029569],               [100, 4, 1.41508090460029576], [100, 5, 1.40017227960029588], [100, 6, 1.38333996160029579],       
       [100, 7, 1.36458395060029579]
    ]
]: 
plots:-display(PLOT3D(MESH(Pts)), labels= [x,y,z]);

Note that PLOT3D and MESH are all uppercase letters. Some help for them can be found at ?plot,structure; however, the best way to learn about them is to deconstruct the results of plot3d (lowercase) commands with op.

An equivalent command, producing exactly the same plot, is

plots:-surfdata(Pts, labels= [x,y,z])

Either way, the trickier part is gettiing the data arranged correctly in a list of lists of lists (as shown above) or a 3D-Array (as shown by acer). Once that's done, displaying it is easy.

 

The trick is to insert a fake symbolic exponent and substitute 1 for it after the extraction, like this:

Powers:= (e, T::type)-> local One;
    eval(
        indets(combine(evalindets(e, T, `^`, One), power), T^anything),
        One= 1
    )
:
e:=_C1^2+_C1+_C1*_C2^3+a+b:
Powers(e, suffixed(_C, nonnegint));
                       /        2     3\ 
                      { _C1, _C1 , _C2  }
                       \               / 

By the way, suffixed(_C, ...is a subtype of symbol, so your And(symbol, ...) is redundant.

It's not a caching issue. What's happening is that the elapsed time between your calls to randomize is much shorter than the "ticks" of the clock. There's never a need to call randomize more than once. Indeed, the effect of calling randomize twice in a short time period is to reset the random-number generator to a specific entry in its sequence, so that's about as unrandomized as you can get.

My feeling about randomize is that it should only be called at the top level, never in a procedure.

Also, there's never a need to pass randomize a custom clock value. The simple call randomize() (with no arguments) already uses a clock-based value that's sufficient for any purpose.

Like this:

collect(T, indets(T, function));
                   
f(x) + 3 + (b + 1) g(x)

@Kitonum is correct that shading regions of polar plots is difficult in Maple, and a "polygon" approach is needed. Here's a way to do it that's perhaps more intuitive than his: I extract the polygon vertices directly from the output of polarplot commands. There's no need to look at the numerical vertices; it's all done rather automatically. Some additional benefits of my approach are that I put the plots on Maple's polar axes rather than Cartesian axes, and no explicit coordinate conversions are required.

I also show three distinct ways to get the area.

restart:

interface(prompt= "")
:

(r1, r2):= (-6*cos(theta),  2-2*cos(theta))
:

#Find theta values at the intersections:
theta__1:= rhs(solve({r1=r2, theta >= 0, theta <= Pi})[]);

(2/3)*Pi

I will extract the point-data matrices (rtables) for two of the relevant arcs for two purposes:
    1. To make filled polygon plots;
    2. To get the area of those polygons via the Shoelace formula (simply to verify integration accuracy). The formula is given below, but you don't
        need to understand it to understand the rest of this worksheet.

 

The point data is stored in Cartesian (rather than polar) coordinates; however, you don't need to know that to understand this worksheet.

(Circle__arc, Cardioid__arc):= (op@indets)~(
    plots:-polarplot~([r1, r2], theta=~ [Pi/2..theta__1, theta__1..Pi]),
    rtable
)[]:
#Reconnect stray end to origin to close the polygon:
Cardioid__arc:= <Cardioid__arc, <0|0>>:

The whole plot:

plots:-display(
    plots:-polarplot~(
        [r1, r2], theta=~ [Pi/2..Pi, 0..Pi], color=~ [red, blue],
        thickness= 2,
        legend=~ [r1, r2], coordinateview= [default, 0..Pi]
    ),
    plots:-polygonplot~(
        [Circle__arc, Cardioid__arc], transparency= .5,
        color=~ [pink, blue], legend=~ ["Area 1", "Area 2"]
    )
);

We want the area of the pink region + the area of the blue region, times 2. We can get that directly like this:

2*(Int(r1^2/2, theta= Pi/2..theta__1) + Int(r2^2/2, theta= theta__1..Pi)):
% = value(%);

2*(Int(18*cos(theta)^2, theta = (1/2)*Pi .. (2/3)*Pi))+2*(Int((1/2)*(2-2*cos(theta))^2, theta = (2/3)*Pi .. Pi)) = 5*Pi

An indirect way to get the area, just as easy, is the area of the circle minus the area the part of the circle outside the cardioid:

2*(Int(r1^2/2, theta= Pi/2..Pi) - Int(r1^2/2-r2^2/2, theta= theta__1..Pi)):
% = value(%);

2*(Int(18*cos(theta)^2, theta = (1/2)*Pi .. Pi))-2*(Int(18*cos(theta)^2-(1/2)*(2-2*cos(theta))^2, theta = (2/3)*Pi .. Pi)) = 5*Pi

And a way to approximate the area is the "Shoelace formula", which gives the area of any polygon simply from the Cartesian coordinates of its vertices. I do this because its easy since we already have the polygons.

Shoelace:= (XY::Matrix)->
    add(XY[..,1]*~XY[[2..,1],2]-XY[..,2]*~XY[[2.., 1],1])/2
:
2*Shoelace(<Circle__arc, Cardioid__arc>) = evalf(5*Pi);

HFloat(15.707812810646793) = 15.70796327

 

Download PolarArea.mw

Use sort with this key:

sort(F, key= [f-> subs(0= infinity, degree~(f, [x4,x3,x2,x1])), nops])

I assume that by "degree of nonlinearity in x4" you mean nothing more than simply "degree in x4 (regardless of the presence of other variables)". If that's not what you mean, let me know. Also, this solution assumes that the polynomials are expanded.

Change
yB_:= y -> sol4[1]:
to
yB_:= MakeFunction(sol4[1], y):

You are correct that the y in the solution is not being seen. This is because there is no y explicitly on the right side of the arrow. In order to make a non-explicit variable into a procedure parameter, you need to use MakeFunction instead of ->.

Creating the set of equations can be done as simply as this:

{seq}(Matrix(2, symbol= a) - Matrix(2, symbol= b) =~ Matrix(2, shape= identity));
       
{a[1, 1] - b[1, 1] = 1, a[1, 2] - b[1, 2] = 0, a[2, 1] - b[2, 1] = 0, a[2, 2] - b[2, 2] = 1}

@michaelvio 

The vast majority of the slowness that you were experiencing was due to your use of int (symbolic integration) versus Int (numeric integration, when it's used in the context that you were using).

It's true that using a larger epsilon and a smaller numpoints will save some time, but that savings is infinitesimal in this case. I recommend that you remove both options so that those things will be done at their default values (although epsilon= 1e-6 is sufficient for plotting purposes).

You should also set Digits:= 15 right after the restart at the top of the worksheet. The extra precision makes a huge difference in the final plot, and the extra time needed isn't even noticeable.

To construct the function J1 as you just described, all that you need to do is

J1:= unapply(subs(x0= 1/t0, J), t0);

This is your last plot at the default Digits:= 10:

And this is using Digits:= 15:

It's a similar issue as with your recent Question regarding applying eval to Int or Intateval uses rules, usually to ensure the mathematical validity of the proposed substitution. The rule in this case is the procedure `eval/limit`. To create a more-sophisticated rule, read the procedure (showstat(`eval/limit`), it's a fairly easy read with only a moderate amount of Maple knowledge, and it's only 42 lines (Maple 2023)) and either write an overload or, perhaps, modify the procedure directly (ToInertFromInert).

If you'd like it to follow rules additional to what's already in the procedure, and you can state them for me with a moderate amount of mathematical precision and generality, I may be able to suggest a way to code it. 

Let's consider your examples individually:

1. e:= A+B*limit(f(x), x= infinity);  eval(e, limit(f(x), x= infinity)= 1);
This case doesn't involve changing anything inside any limit. Thus, the rule isn't used (verified by using trace(`eval/limit`)), and you got your expected result.

2. e:= A+B*limit(2*f(x), x= infinity);  eval(e, limit(f(x), x= infinity)= 1);
In this case you're asking it to evaluate an expression containing a limit by using information about another limit (albeit, a very closely related one); the eval rule will be used. I think that the mathematical rule (i.e., abstract, not Maple) that you're implicitly invoking here is always valid, although I'm not 100% sure about all weird "corner cases", domains other than 2-sided real, and multivariate (more than one bound variable (e.g., limit(..., (x,y)= ...)) cases. That rule, stated generally, is that factors that don't depend on the bound variable(s) can be factored out of the limit. Command expand will do this, so a quick solution is 
eval(expand(e), limit(f(x), x= infinity)= 1)

[to be continued]

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