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Carl Love

Carl Love

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12 years, 300 days
Himself
Wayland, Massachusetts, United States
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My name was formerly Carl Devore.

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These are answers submitted by Carl Love

Is n defined?...

Carl Love 28020
August 29 2013
0 1

Does n have a definite integer value equal to the number of rows of the Matrix? So the Matrix should be n x J. It is probably i rather than j that is causing the problem.

Also, you should use add instead of sum, and then you don't need the quotes on A[i,j]. Use sum for symbolic summation.

If that doesn't fix the problem, then you'll need to post more of the code.

sum or product...

Carl Love 28020
August 29 2013
2 0

I think that in general what you are trying to do is impossible. The limit command, unlike int, does not have a procedure option. But if you could rethink your procedure and express it in terms of sum or product then you may be able to use limit.

x=3, y=13...

Carl Love 28020
August 29 2013
1 0



 

(**)

restart:
(**)

eqn:= arctan(x) + arccos(y/sqrt(y^2+1)) - arctan(4):
(**)

f:= numer(simplify(expand(tan(eqn)))) assuming x>=0, y>=0;

x*y+4*x-4*y+1

(**)

Sol:= [isolve](f);

[{x = -13, y = -3}, {x = 3, y = 13}, {x = 5, y = -21}, {x = 21, y = -5}]

(**)

remove(hastype, Sol, negative);

[{x = 3, y = 13}]

I am somewhat skeptical about isolve's ability to return all the solutions. So, investigating further,

(**)

solve(f,x);

(4*y-1)/(y+4)

(**)

convert(%, parfrac);

4-17/(y+4)

Now it is clear that y = 13 is the only positive integer solution.

 



Download tan_eqn.mw

 

964...

Carl Love 28020
August 29 2013
0 0

nops([StringTools:-SearchAll("0", sprintf("%d", 2014!))]);

964

(* ... *)...

Carl Love 28020
August 28 2013
0 3

Anything between (* and *) is commented out, even if it is multiple lines.

Adding text with fake multiplication...

Carl Love 28020
August 28 2013
2 0

Anything can be a ?symbol (a variable's name) in Maple if you enclose it in back quotes (``), aka accent grave. And symbols can be multiplied with `*`. The multiplication appears as simple juxtaposition when it is printed.

Your original expression is not mathematically correct. It should have a factor of "delta x" on the left side of the equation. Putting it all together, we get this:



(**)

f := x-> x^2; a := 0; b := 1;
for i to 5 do
     print(

          `The answer is `*'f'(x[i])*Delta*x

          = f(a+i*(b-a)/5)*(b-a)/5

     )
end do:

proc (x) options operator, arrow; x^2 end proc

0

1

`The answer is `*f(x[1])*Delta*x = 1/125

`The answer is `*f(x[2])*Delta*x = 4/125

`The answer is `*f(x[3])*Delta*x = 9/125

`The answer is `*f(x[4])*Delta*x = 16/125

`The answer is `*f(x[5])*Delta*x = 1/5

(**)

 



Download fake_mult.mw

Use unevaluated form, 'f'...

Carl Love 28020
August 28 2013
3 3

printf (and lprint) produce only 1d output. Perhaps you can acheive what you want by using the unevaluated form of f, i.e., by enclosing it in forward single quotes 'f'.



(**)

f := x-> x^2; a := 0; b := 1;
for i to 5 do  

     print('f'(x[i])=f(a+i*(b-a)/5)*(b-a)/5)

end do;

proc (x) options operator, arrow; x^2 end proc

0

1

f(x[1]) = 1/125

f(x[2]) = 4/125

f(x[3]) = 9/125

f(x[4]) = 16/125

f(x[5]) = 1/5

(**)

 



Download unevaluated.mw

Is that what you want?

 

a(n) = 3*2^(n-1)...

Carl Love 28020
August 28 2013
0 2

a(0) = 1

a(n) = 3*2^(n-1), n > 0

final evalhf module to do it...

Carl Love 28020
August 28 2013
1 4

Here is the final evalhf/module version of your program. It runs in about 7 seconds; so, about a factor-of-8 improvement over the regular version.

Combi:= module()
export
    probVector::Vector(realcons),  #probability vector
    lambdaVector::Vector(realcons)
;
local
    Combi:= proc(
        k::integer,
        boxContent::Vector,
        activeBox::integer,
        marblesRemaining::integer,
        Output::Matrix,
        Index::Vector,
        p::Vector
    )       
    local j::integer;
        
        if activeBox < k then
            for j from 0 to marblesRemaining do
                boxContent[activeBox]:= j;
                thisproc(k, boxContent, activeBox+1, marblesRemaining-j, Output, Index, p)
            end do
        else
            boxContent[activeBox]:= marblesRemaining;
            Index[1]:= Index[1] + 1;
            Output[1, Index[1]]:= n!*mul(p[j]^boxContent[j], j= 1..k)/mul(boxContent[j]!, j= 1..k);
            Output[2, Index[1]]:= `if`(
                mul(boxContent[j], j= 1..k) = 0,
                undefined,
                -2*add(boxContent[j]*ln(n*p[j]/boxContent[j]), j= 1..k)
            )    
        end if
    end proc,

    ModuleApply:= proc(n::posint, k::posint, p::~Vector(realcons))::identical();
    local
        Index::Vector(realcons), Output::Vector(realcons),
        boxContent::Vector(realcons)  #A vector whose i'th entry holds the number of marbles in the i'th box
    ;
        boxContent:= Vector(k, datatype= float[8]);
        Output:= Matrix(2, binomial(n+k-1, k-1), datatype= float[8]);
        Index:= Vector(1);
        evalhf(Combi(k, boxContent, 1, n, Output, Index, p));
        probVector:= Output[1, ..];
        lambdaVector:= Output[2, ..];
        [][]
    end proc
;
end module;


A recursive procedure

restart

with(Statistics):

n := 40:

k := 6:

p := Vector(k, proc (i) options operator, arrow; 1/k end proc):

(**)

 

Combi:= module()

module () local Combi, ModuleApply; export probVector::(Vector(realcons)), lambdaVector::(Vector(realcons)); end module

``

B := CodeTools[Usage](Combi(n, k, p));

``

Combi:-probVector[1221759];

HFloat(7.480833428389761e-32)

Combi:-lambdaVector[1221759];

HFloat(HFloat(undefined))

NULL


Download A_recursive_procedu.mw

194785...

Carl Love 28020
August 27 2013
1 0


(**)

restart:
(**)

d:= proc(s)  option remember;  parse(s)  end proc:
d("0"):= 1:
(**)

p:= n-> `*`(map(d, StringTools:-Explode(sprintf("%d", n)))[]):
(**)

add(p(n), n= 1..2014);

194785

(**)

 


Download product_digits.mw

Law of sines...

Carl Love 28020
August 27 2013
0 0


1.

(**)

restart:
(**)

B:= 2*A:  C:= 4*A:
(**)

solve(A+B+C = Pi, A);

(1/7)*Pi

(**)

A:= %:

By law of sines:

(**)

b:= sin(B)*a/sin(A):  c:= sin(C)*a/sin(A):
(**)

simplify(1/b + 1/c);

1/a

2.

(**)

simplify(cos(A)^2 + cos(B)^2 + cos(C)^2);

5/4

(**)

 


Download Geometry.mw

6...

Carl Love 28020
August 27 2013
0 2


(**)

x:= 2014!/5^20:
(**)

d:= 0:
(**)

while d=0 do x:= iquo(x,10,'d') end do:
(**)

d;

6

(**)

 


Download 2014.mw

1009...

Carl Love 28020
August 27 2013
0 1


(**)

restart:
(**)

solve(t^2-2016*t-1, t);

1008+1016065^(1/2), 1008-1016065^(1/2)

(**)

lambda:= %[1]:
(**)

x:= proc(n::posint)
option remember;
     floor(lambda*thisproc(n-1))
end proc;
x(0):= 1;

proc (n::posint) option remember; floor(lambda*thisproc(n-1)) end proc

1

(**)

x(2016) mod 2016;

1009

(**)

 


Download 2016.mw

125...

Carl Love 28020
August 27 2013
2 0


> 

x:= 4/mul(5^(1/2^k)+1, k= 1..4);

4/((5^(1/2)+1)*(5^(1/4)+1)*(5^(1/8)+1)*(5^(1/16)+1))

(**)

rationalize(%);

`[Length of output exceeds limit of 100000]`

(**)

expand(%);

-1+5^(1/16)

(**)

(%+1)^48;

125

(**)

 


Download 125.mw

Example of Multivariate Newton's method...

Carl Love 28020
August 26 2013
1 19

Here is an example of multivariate Newton's method. I didn't have access to your equations, so I made some random similar examples.


> 

F:= ['randpoly([y[k] $ k= 1..5], degree= 3) +
      randpoly([y[k] $ k= 1..5], degree= 1, dense)'
    $ 5];

[57*y[1]^2*y[3]-49*y[1]*y[2]*y[5]+31*y[2]^3+73*y[2]*y[5]^2-84*y[1]*y[3]+95*y[1]+148*y[2]-29*y[3]+5*y[4]-26*y[5]-51, 80*y[1]*y[2]*y[5]+90*y[1]*y[3]*y[5]-57*y[1]*y[3]+85*y[4]^2+74*y[1]+27*y[2]+9*y[3]-127*y[4]+81*y[5]+70, 36*y[1]^2*y[3]-8*y[2]^3+30*y[2]*y[3]^2-3*y[3]^2*y[5]-56*y[3]*y[4]^2-5*y[2]^2-91*y[1]-70*y[2]+42*y[3]+9*y[4]-21*y[5]-27, 49*y[1]*y[2]*y[4]-58*y[3]^2*y[5]+49*y[3]*y[5]^2-44*y[2]*y[4]-31*y[2]*y[5]+45*y[5]^2+y[1]-95*y[2]+86*y[3]-97*y[4]-14*y[5]+83, 96*y[1]^2*y[3]-51*y[1]*y[2]*y[3]+89*y[1]*y[3]*y[5]+14*y[2]*y[5]^2-79*y[3]*y[4]^2-58*y[1]-95*y[2]+123*y[3]-2*y[4]+86*y[5]+57]

(**)

fsolve(F);

{y[1] = -1.52634660833751, y[2] = 1.14614000845494, y[3] = -0.912884720559115e-1, y[4] = -.156870015505402, y[5] = 0.268564748179879e-1}

(**)

J:= VectorCalculus:-Jacobian(F, [y[k] $ k= 1..5]):
(**)

Y:= <[1$5]>:  #Dummy first iteration.
(**)

Y1:= <[0$5]>:  #Initial guess
(**)

for n to 100 while LinearAlgebra:-Norm(Y-Y1) > 1e-15 do
     Y:= evalf(Y1);
     Y1:= Y - LinearAlgebra:-LinearSolve(
          eval(J, [seq](y[k]= Y[k], k= 1..5)),
          <eval(F, [seq](y[k]= Y[k], k= 1..5))>
     )
end do:
(**)

n;

7

(**)

Y1;

Vector(5, {(1) = -.937327749126478, (2) = .819756118294004, (3) = -0.193900192863524e-1, (4) = .123223922590096, (5) = -.340268950372420})

Note that we obtained a different solution than fsolve. There are multiple solutions.

(**)

eval(F, [seq](y[k]= Y[k], k= 1..5));

[HFloat(1.4210854715202004e-14), HFloat(0.0), HFloat(0.0), HFloat(0.0), HFloat(7.105427357601002e-15)]

(**)

 


Download MVNewton.mw

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