1400 Reputation

19 years, 106 days
University of Twente (retired)
Enschede, Netherlands

My "website" consists of a Maple Manual in Dutch

use double quotes...

(1) use " instead of ', e.g

RETURN ("Input is NOT a Lie algebra (',i,i,k,')=",A[i,i,k], "is not zero");

(2)

for l from 1 by 1 to n do

(you have the number "one" instead of letter l )

Perhaps you mean......

(edited)

flist := [0.1,0.2,2,3,3.5,6]:
xlist := [1,2,3,4,5,6]:
355.6

or

145.6

Another way...

a2:=(18*d^2*R^2*r-4*d1^3*r+5*R^6-z1*r^3*p-14*r1^3*r*rc+k1*d^2*R^2*r);
a4 := a2-a3;

Two arguments...

Do you mean that A contains the first en B the second argument of BesselY?
In that case (I changed B a little):

A := Matrix([[-13, -10], [21, 16]]): B := Matrix([[3, 1], [21, 16]]):
BesselY~(A,B);

Multiplication signs...

You forgot some multiplication signs inthe second equation:

restart;
eq1:=y=86:
eq2:=y=-0.0000054527*x^3+0.010903836*x^2+0.0714244709*x+74.18816:
sol:=solve({eq1,eq2},{x,y});
{x = 30.00404568, y = 86.}, {x = 2005.705515, y = 86.},  {x = -35.99639135, y = 86.}

evalf...

evalf( Int( 2.91*x*((1/(1+1.38*x^4)))^0.431 - 3.459*x^5/((1/(1+1.38*x^4))^0.569*(1+1.38*x^4)^2), x=0..1 ) );
1.001485791

A procedure...

ShowCols := proc(A)
print( seq(LinearAlgebra:-Column(A,i), i=1..LinearAlgebra:-ColumnDimension(A)) )
end proc:

with(LinearAlgebra):
A := RandomMatrix(5,7);
ShowCols(A);

Use tickmarks...

plots:-pointplot([seq([x,0],x=[-sqrt(16),-2,-3/4,0,exp(1),Pi])],view=[-4..4,0..0.01],
symbolsize=20,symbol=solidcircle,colour=blue,
tickmarks=[[-4.0=typeset(sqrt(`16`)), -2.=-2, -.75=typeset(-3/4),0.=0,
evalf(exp(1))=typeset(exp(1)),evalf(Pi)=typeset(Pi)],[]], scaling=constrained);

Number of terms...

You must ensure that the expression contains plus-signs, otherwise there is only one term. What about:

f := (b(t)*diff(a(t),t) + a(t)*b(t))*k ;
if type( expand(f),`+`) then nops(expand(f)) else 1 end if; #corrected

Use avoid option...

Solve Im(SS)=0, use the avoid option to get all solutions:

z1 := fsolve( Im(eval( SS, a=0.3 ))=0, {_Z2} );
{_Z2 = 0.}
z2 :=fsolve( Im(eval( SS, a=0.3 ))=0, {_Z2} , avoid=z1 );
{_Z2 = -1.}
z3 :=fsolve( Im(eval( SS, a=0.3 ))=0, {_Z2} , avoid=( z1 union z2) );
fsolve(-Im(LambertW(_Z2, -0.7 exp(-1))) = 0, {_Z2}, avoid = {_Z2 = -1., _Z2 = 0.})

No further solutions.

Use ListTools...

L := [1,2,3,7,6,5,4]:
m,pos := ListTools:-FindMaximalElement(L,position);

assumptions...

If you want to assume that a and b have the same sign, you can substiture b = k*a, and assume that k>0:

eq :=(4*a^3*b)^(1/2)/(-(a/(4*b))^(1/2))+(4*a^3*b*(4*b/a))^(1/2) = 0:
eq1 := eval(eq, b=k*a ):
simplify( eq1 ) assuming k>0 ;

MathType...

You could use MathType to convert LaTeX code to Word.

Use fsolve...

No chance to find exact solutions as af function if f1. So use fsolve for each desired value of f1:

ABC := t -> fsolve( eval({A,B,C},f1=t), {a,b,c} ):
ABC(0.1);

And to get a as a function of t:,

AA := t -> subs( ABC(t),a );

but the solutions seen rather chaotic (I did not check your formulas)

Indexing function...

The most elegant way to produce this kind of matrices is to use an indexing function:

A := M -> Matrix(2*M+1, (i,j) -> if (i=M+1 and j=M+1) then 1
elif j=M+1 then 1/(i-1-M)/Pi
elif i=M+1 then 1/Pi
elif (i+j)=2*M+2 then 1/abs(i-1-M)/Pi
else 0
end if):
A(3);

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