Advanced Search

Ronan

1406 Reputation

16 Badges

14 years, 53 days
East Grinstead, United Kingdom
Contact Ronan

MaplePrimes Activity


These are questions asked by Ronan

Convert Vector to diagonal Matrix...

Ronan 1406 Maple 2024
matrix vector
December 26 2024
1 0

I am looking for a more eligent way to convert a Vector to a Diagonal Matrix.

> 

restart
> 

 

> 

 

> 

with(LinearAlgebra):
> 

 

> 

V:=Vector[column](3, [0.5863730366, 0.1171249270, 0.2965020364])

Vector(3, {(1) = .5863730366, (2) = .1171249270, (3) = .2965020364})

 (1)
> 

Vm:=Matrix(3,[[V[1],0,0],[0,V[2],0],[0,0,V[3]]])

Matrix(3, 3, {(1, 1) = .5863730366, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = .1171249270, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = .2965020364})

 (2)
> 

Vm1:=Matrix(3,3):
> 

for i to 3 do
Vm1[i,i]:=V[i];
end do:
> 

Vm1

Matrix(3, 3, {(1, 1) = .5863730366, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = .1171249270, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = .2965020364})

 (3)
 

 

Download 2024-12-26_Q_Diagonal_Matrix_from_Vector.mw

Can gcd be mapped across a list?...

Ronan 1406 Maple
gcd map
December 24 2024
2 2

I knowI can use a loop but I would like something neater. igcd can be applied to a list. Is there a way to map or use @ to apply gcd to a list?

lst:= [4, 2, 8, 4];
lst1 := [a^3/10, -a^2/2, a, a^4/4]
                      lst := [4, 2, 8, 4]

                        [1   3    1  2     1  4]
                lst1 := [-- a , - - a , a, - a ]
                        [10       2        4   ]


gd:=igcd(lst);



                            gd := 2

gd1=(gcd@op)(lst1);

 

Determine if a point lies on a 3D line...

Ronan 1406 Maple 2024
geometry 3d
December 21 2024
2 3

It is not that this a terribly difficult to work out, but I feel I am probably missing something. I need to check if a 3D point lies on a 3D line. What is a good approach here. I started of with the idea all alpha's are equal. but there are exceptions. See P3 and P4

restart

NULL

l := `<,>`(3+2*alpha, 1+6*alpha, 4-5*alpha)

Vector[column](%id = 36893489809910741940)

 (1)

NULL

P := [9, 19, -11]

[9, 19, -11]

 (2)

seq(solve({l[i] = P[i]}, alpha), i = 1 .. 3)

{alpha = 3}, {alpha = 3}, {alpha = 3}

 (3)

l1 := `<,>`(3+2*alpha, 1+0*alpha, 4-5*alpha)

Vector[column](%id = 36893489809910721460)

 (4)

P1 := [9, 1, -11]

[9, 1, -11]

 (5)

seq(solve({l1[i] = P1[i]}, alpha), i = 1 .. 3)

{alpha = 3}, {alpha = alpha}, {alpha = 3}

 (6)

l2 := `<,>`(3+2*alpha, 1+0*alpha, 4-0*alpha)

Vector[column](%id = 36893489809910705556)

 (7)

P2 := [9, 1, 4]

[9, 1, 4]

 (8)

seq(solve({l2[i] = P2[i]}, alpha), i = 1 .. 3)

{alpha = 3}, {alpha = alpha}, {alpha = alpha}

 (9)

l3 := `<,>`(3+2*alpha, 0+0*alpha, 4-0*alpha)

Vector[column](%id = 36893489809963852012)

 (10)

P3 := [9, 0, 4]

[9, 0, 4]

 (11)

seq(solve({l3[i] = P3[i]}, alpha), i = 1 .. 3)

{alpha = 3}, {alpha = alpha}

 (12)

P4 := [9, 0, -2]

[9, 0, -2]

 (13)

seq(solve({l3[i] = P4[i]}, alpha), i = 1 .. 3)

{alpha = 3}, {alpha = alpha}

 (14)

 

Download 2024-12-21_Q_3D_point_lies_on_3D_line.mw

How is the simplification achieved...

Ronan 1406 Maple 2024
radical simplification
December 20 2024
0 2

2024-12-20_Q_simplification_Question.mw
Solve the general cubic. Apply values and simplify. 

Could someone show how Maple simplifies to the value of X=3? I tried doing it manually and I could not figure it out. 

Also is there a Help assistant to see the setps?

> 

restart
> 

 

> 

 

> 

X^3+a*X=b

X^3+X*a = b

 (1)
> 

 

> 

 

> 

sol:=solve(X^3+a*X=b,[X])

[[X = (1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)-2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)], [X = -(1/12)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+((1/2)*I)*3^(1/2)*((1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3))], [X = -(1/12)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)-((1/2)*I)*3^(1/2)*((1/6)*(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3)+2*a/(108*b+12*(12*a^3+81*b^2)^(1/2))^(1/3))]]

 (2)
> 

vals:=[a=6,b=45]

[a = 6, b = 45]

 (3)
> 

Nans:=(map(eval,sol,vals))

[[X = (1/6)*(4860+12*166617^(1/2))^(1/3)-12/(4860+12*166617^(1/2))^(1/3)], [X = -(1/12)*(4860+12*166617^(1/2))^(1/3)+6/(4860+12*166617^(1/2))^(1/3)+((1/2)*I)*3^(1/2)*((1/6)*(4860+12*166617^(1/2))^(1/3)+12/(4860+12*166617^(1/2))^(1/3))], [X = -(1/12)*(4860+12*166617^(1/2))^(1/3)+6/(4860+12*166617^(1/2))^(1/3)-((1/2)*I)*3^(1/2)*((1/6)*(4860+12*166617^(1/2))^(1/3)+12/(4860+12*166617^(1/2))^(1/3))]]

 (4)
> 

simplify(Nans)

[[X = 3], [X = (1/4)*(I*3^(1/2)*(180+44*17^(1/2))^(2/3)+(8*I)*3^(1/2)-(180+44*17^(1/2))^(2/3)+8)/(180+44*17^(1/2))^(1/3)], [X = -3/2-((1/2)*I)*51^(1/2)]]

 (5)
 

 

Download 2024-12-20_Q_simplification_Question.mw

DataFrame display alpha as a Greek lette...

Ronan 1406 Maple 2024
typesetting greek dataframe
December 01 2024
1 3

How do I get alpha to display as the Greek letter in DataFrame Row column?

> 

restart
> 

#varp:=alpha
> 

QQFProj := proc(q12::algebraic, q23::algebraic,
                q34::algebraic, q14::algebraic,{varp:=:-alpha},
                  prnt::boolean:=true) #{columns:=[QQFproj,Q13proj,Q24proj]}
  description "Projective quadruple quad formula and intermediate 13 and 24 quads. Useful for cyclic quadrilaterals";
  local qqf,q13,q24, sub1,sub2,sub3, R,values,DF,lens;
  uses   DocumentTools;
  sub1:= (q12 + q23 + q34 + q14);
  sub2:=-4*(q12*q23*q34);
  sub3:=64*q12*q23;
  qqf:=(sub1+sub2)^2-sub3;
  q13:=(q12-q23)^2;
  q24:=varp*(q23-q34)^2;
  if prnt then
  
   values:=<qqf,q13,q24>;
   DF:=DataFrame(<values>, columns=[`"Values Equations"`],rows=[`#1  QQF`,`#2  Q13`,cat(`#3  Q24 (`,varp,`)`)]);
   lens := [4 +8* max(op(length~(RowLabels(DF)))),4+ min(max( 10*(length~(values))),1000)];#op(length~(ColumnLabels(DF)0)
   Tabulate(DF,width=add(lens),widthmode = pixels,weights = lens);
  return qqf,q13,q24
  end if;
  return qqf,q13,q24
end proc:
> 

 q12:=1/2:q23:=9/10:q34:=25/26:q41:=9/130:#Cyclic quadrilateral
q12:=sqrt(17+a)/2:q23:=(r^2+t^2)^2/10:q34:=((a+b+c)^4/26):q41:=sqrt(17+b)/130:
> 

Q:=QQFProj(q12,q23,q34,q41,true):

Download 2024-12-01_Q_Data_Table_alpha_as_Greek_Letter.mw

1 2 3 4 5 6 7 Last Page 3 of 36