## 1162 Reputation

12 years, 159 days

## My approach to Pascal's theorem (from a ...

I parameterised the points on the ellipse. I made them projective points [x,y,1] in the form of vectors. Reason the equation of the  line through two projective points is the Cross Product of the two points as another vector . P1 X P2  = L12 Now the intersect point of two lines as vectors is L1 X L2= Pintersect.

A line in vector form <a , b, c>   represents a*x + b*y + c=0

The use of factrix by @Robert B. Isreal  reduces the projective vectors, as in projective geometry vectors can be scaled without changing the answers.

Edit:- Correction.  Replece this into the document.  The intersection points were scaled incorrectly.

```The 3 collinear points

for i to 3 do
pPint || i := convert(Pint || i[1 .. 2]/Pint || i[3], list);
eval(pPint || i, pramsplot);
evalf(%);
end do;
```

 (1)

 (2)

Firstly Paramaterise a conic centred on the Origin using projective points

 (3)

 (4)

 (5)

 (6)

 (7)

 (8)

 (9)

 (10)

Create 6 Projective Ponits on the conic

 (11)

 (12)

 (13)

 (14)

 (15)

 (16)

Create 6 Projective Lines through pairs of Ponits P0P4, P0P5:  P1P3, P1P5:  P2P3, P2P4 uning crossproduct of vectors

 (17)

 (18)

 (19)

 (20)

 (21)

 (22)

Get 3 intersection points in projective form

 (23)

 (24)

 (25)

From a square matrix using the 3 points. If the determinant is 0 the points are colinear

 (26)

 (27)

Plotting

play with t_0 ..t_5 to move the points on the Conic

 (28)

 (29)

 (30)

 (31)

 (32)

 (33)

 (34)

 (35)

 (36)

The 3 collinear points

 (37)

## 5th Order Polynomial in h...

You basically have a 5th order polynomail in h. In general there is no explicit solution for a 5th order or higher polynomial.

## Use Ctrl-K...

I asked the same question a couple on months age

inserts a new execution group before the cursor.

For pasted text you could try F3 to split the text at the postition of the cursor.

from the menu should also work.

## Is this what you mean?...

Here are two possible ways assuming I have understood you question correctly. I renamed your list to l to avoid confusion.

 >
 >
 >
 (1)
 >
 (2)
 > ~# Or use seq
 >
 (3)
 > A[4]
 (4)
 >

## Incorrect square brackets []use ()...

Square brackets are incorrect use parenthesis ()

`1/2*int(D__11*diff(w, x, x)^2 + 2*D__12*diff(w, x, x)*diff(w, y, y) + 4*D[66]*diff(w, x, y)^2 + D[22]*diff(w, y, y)^2 - 2*q*w, [x = 0 .. b, y = 0 .. a])`

I got

`(((144*D__11*c[1]^2)/(5*a^8*b^3) - 72*D__11*c[1]^2/(a^7*b^4) + 64*D__11*c[1]^2/(a^6*b^5) - 24*D__11*c[1]^2/(a^5*b^6) + 4*D__11*c[1]^2/(a^4*b^7))*a^9)/18 + ((-(576*D__11*c[1]^2)/(5*a^8*b^2) + 288*D__11*c[1]^2/(a^7*b^3) - 256*D__11*c[1]^2/(a^6*b^4) + 96*D__11*c[1]^2/(a^5*b^5) - 16*D__11*c[1]^2/(a^4*b^6))*a^8)/16 + ((((288*D__12*c[1]^2/(a^8*b^8) + 1024*D[66]*c[1]^2/(a^8*b^8))*b^7)/7 + ((-864*D__12*c[1]^2/(a^7*b^8) - 3072*D[66]*c[1]^2/(a^7*b^8))*b^6)/6 + ((864*D__11*c[1]^2/(a^8*b^6) + 912*D__12*c[1]^2/(a^6*b^8) + 3328*D[66]*c[1]^2/(a^6*b^8))*b^5)/5 + ((-1728*D__11*c[1]^2/(a^7*b^6) - 384*D__12*c[1]^2/(a^5*b^8) - 1536*D[66]*c[1]^2/(a^5*b^8))*b^4)/4 + ((1152*D__11*c[1]^2/(a^6*b^6) + 48*D__12*c[1]^2/(a^4*b^8) + 256*D[66]*c[1]^2/(a^4*b^8))*b^3)/3 - 144*D__11*c[1]^2/(a^5*b^4) + 24*D__11*c[1]^2/(a^4*b^5))*a^7)/14 + ((((-864*D__12*c[1]^2/(a^8*b^7) - 3072*D[66]*c[1]^2/(a^8*b^7))*b^7)/7 + ((2592*D__12*c[1]^2/(a^7*b^7) + 9216*D[66]*c[1]^2/(a^7*b^7))*b^6)/6 + ((-576*D__11*c[1]^2/(a^8*b^5) - 2736*D__12*c[1]^2/(a^6*b^7) - 9984*D[66]*c[1]^2/(a^6*b^7))*b^5)/5 + ((1152*D__11*c[1]^2/(a^7*b^5) + 1152*D__12*c[1]^2/(a^5*b^7) + 4608*D[66]*c[1]^2/(a^5*b^7))*b^4)/4 + ((-768*D__11*c[1]^2/(a^6*b^5) - 144*D__12*c[1]^2/(a^4*b^7) - 768*D[66]*c[1]^2/(a^4*b^7))*b^3)/3 + 96*D__11*c[1]^2/(a^5*b^3) - 16*D__11*c[1]^2/(a^4*b^4))*a^6)/12 + ((16*D[22]*c[1]^2*b/a^8 - 72*D[22]*c[1]^2/a^7 + ((912*D__12*c[1]^2/(a^8*b^6) + 3328*D[66]*c[1]^2/(a^8*b^6) + 864*D[22]*c[1]^2/(a^6*b^8))*b^7)/7 + ((-2736*D__12*c[1]^2/(a^7*b^6) - 9984*D[66]*c[1]^2/(a^7*b^6) - 576*D[22]*c[1]^2/(a^5*b^8))*b^6)/6 + ((144*D__11*c[1]^2/(a^8*b^4) + 2888*D__12*c[1]^2/(a^6*b^6) + 10816*D[66]*c[1]^2/(a^6*b^6) + 144*D[22]*c[1]^2/(a^4*b^8) - 2*q*c[1]/(a^4*b^4))*b^5)/5 + ((-288*D__11*c[1]^2/(a^7*b^4) - 1216*D__12*c[1]^2/(a^5*b^6) - 4992*D[66]*c[1]^2/(a^5*b^6) + 4*q*c[1]/(a^3*b^4))*b^4)/4 + ((192*D__11*c[1]^2/(a^6*b^4) + 152*D__12*c[1]^2/(a^4*b^6) + 832*D[66]*c[1]^2/(a^4*b^6) - 2*q*c[1]/(a^2*b^4))*b^3)/3 - 24*D__11*c[1]^2/(a^5*b^2) + 4*D__11*c[1]^2/(a^4*b^3))*a^5)/10 + ((-32*D[22]*c[1]^2*b^2/a^8 + 144*D[22]*c[1]^2*b/a^7 + ((-384*D__12*c[1]^2/(a^8*b^5) - 1536*D[66]*c[1]^2/(a^8*b^5) - 1728*D[22]*c[1]^2/(a^6*b^7))*b^7)/7 + ((1152*D__12*c[1]^2/(a^7*b^5) + 4608*D[66]*c[1]^2/(a^7*b^5) + 1152*D[22]*c[1]^2/(a^5*b^7))*b^6)/6 + ((-1216*D__12*c[1]^2/(a^6*b^5) - 4992*D[66]*c[1]^2/(a^6*b^5) - 288*D[22]*c[1]^2/(a^4*b^7) + 4*q*c[1]/(a^4*b^3))*b^5)/5 + ((512*D__12*c[1]^2/(a^5*b^5) + 2304*D[66]*c[1]^2/(a^5*b^5) - 8*q*c[1]/(a^3*b^3))*b^4)/4 + ((-64*D__12*c[1]^2/(a^4*b^5) - 384*D[66]*c[1]^2/(a^4*b^5) + 4*q*c[1]/(a^2*b^3))*b^3)/3)*a^4)/8 + (((64*D[22]*c[1]^2*b^3)/(3*a^8) - 96*D[22]*c[1]^2*b^2/a^7 + ((48*D__12*c[1]^2/(a^8*b^4) + 256*D[66]*c[1]^2/(a^8*b^4) + 1152*D[22]*c[1]^2/(a^6*b^6))*b^7)/7 + ((-144*D__12*c[1]^2/(a^7*b^4) - 768*D[66]*c[1]^2/(a^7*b^4) - 768*D[22]*c[1]^2/(a^5*b^6))*b^6)/6 + ((152*D__12*c[1]^2/(a^6*b^4) + 832*D[66]*c[1]^2/(a^6*b^4) + 192*D[22]*c[1]^2/(a^4*b^6) - 2*q*c[1]/(a^4*b^2))*b^5)/5 + ((-64*D__12*c[1]^2/(a^5*b^4) - 384*D[66]*c[1]^2/(a^5*b^4) + 4*q*c[1]/(a^3*b^2))*b^4)/4 + ((8*D__12*c[1]^2/(a^4*b^4) + 64*D[66]*c[1]^2/(a^4*b^4) - 2*q*c[1]/(a^2*b^2))*b^3)/3)*a^3)/6 + ((-(16*D[22]*c[1]^2*b^4)/(3*a^8) + 24*D[22]*c[1]^2*b^3/a^7 - (288*D[22]*c[1]^2*b^2)/(7*a^6) + 32*D[22]*c[1]^2*b/a^5 - (48*D[22]*c[1]^2)/(5*a^4))*a^2)/4 + (2*D[22]*c[1]^2*b^5)/(9*a^7) - D[22]*c[1]^2*b^4/a^6 + (12*D[22]*c[1]^2*b^3)/(7*a^5) - (4*D[22]*c[1]^2*b^2)/(3*a^4) + (2*D[22]*c[1]^2*b)/(5*a^3)`

## Roots are real on x axis but give comple...

This shows basically what is happening and why Maple gives answers you have. The are correct.

## B:=...

You had set B= Omega. Needs to be B:=Omega

 >
 >

 >

 >

## Try.....

Try setting q:=p^2. That will get rid of the fractional exponents.

## Two ways around it....

Here are two potential ways around you issue.

 could use double delayed evaluation quotes   '' ...   '' but then can't use a^2*b^5 or  enclose computation an square brackets

## expand may work for you...

I checked this in Maple 18 (not Maple 2018) and Maple 2022.  I don't have Maple 2020 installed anymore.

`eq := solve(expand((20 + 20*T + 2*T*(T + 1))*exp(-T) - 10*exp(-2*T) - 2*T - 10.0))`

`eq := 1.411454823, 0.`

`1.411454823, -0., 0.`

If you use 10 instead of 10.0 the numerical answers need to be extracted using "allvalues(%)" In this case Maple 18 returns four root solutions but only evaluates two of the to numerical values. Maple 2022 returns two root solutions

Beyond that I have no explination.

## 1st use the LinearAlgebra package then d...

Firstly you are better off using the modern LinearAlgebra package. I converted some of you code to suit.

I created a matrix Hnew is what you are looking for

 > restart;
 > kernelopts(version);
 (1)
 > with(LinearAlgebra);
 (2)
 > alias(phi = phi(x, t), chi = chi(x, t), psi = psi(x, t), rho = rho(x, t));
 (3)
 >
 > H[1]:=Matrix([[epsilon[1]*phi[1],conjugate(epsilon[2])*(phi[1]),conjugate(chi[1]),0],
 > [epsilon[2]*psi[1],-conjugate(epsilon[1])*(psi[1]),0,conjugate(rho[1])],
 > [-chi[1],0,conjugate(epsilon[1])*conjugate(phi[1]),conjugate(epsilon[2])*conjugate(phi[1])],
 > [0,rho[1],-epsilon[2]*conjugate(psi[1]),epsilon[1]*conjugate(psi[1])]]);
 >
 >
 (4)
 > H[2]:=Matrix([[epsilon[1]*phi[2],conjugate(epsilon[2])*(phi[2]),conjugate(chi[2]),0],
 > [epsilon[2]*psi[2],-conjugate(epsilon[1])*(psi[2]),0,conjugate(rho[2])],
 > [-chi[2],0,conjugate(epsilon[1])*conjugate(phi[2]),conjugate(epsilon[2])*conjugate(phi[2])],
 > [0,rho[2],-epsilon[2]*conjugate(psi[2]),epsilon[1]*conjugate(psi[2])]]);
 (5)
 > Lambda[1]:=Matrix([[lambda[1],0,0,0],[0,lambda[1],0,0],[0,0,conjugate(lambda[1]),0],[0,0,0,conjugate(lambda[1])]]);
 (6)
 > Lambda[2]:=Matrix([[lambda[2],0,0,0],[0,lambda[2],0,0],[0,0,conjugate(lambda[2]),0],[0,0,0,conjugate(lambda[2])]]);
 (7)
 > H11:=H[1].Lambda[1];
 (8)
 > H12:=H[2].Lambda[2];
 (9)
 > H13:=H[1].Lambda[1].Lambda[1];
 (10)
 > H14:=H[2].Lambda[2].Lambda[2];
 (11)
 > H115:=Matrix([H[1],H[2]]);
 (12)
 > H15:=Matrix([H11,H12]);
 (13)
 > H116:=Matrix([[H115],[H15]]);
 (14)
 > H16:=Matrix([H13,H14]);
 (15)
 >
 (16)
 >

## Try this...

A is the name of the vector.

A[1]:=5

In you other equation. use A[1]....  C[1]... not a[1]... c[1]...

## Two Ways...

There are two basic ways to do this. 1st extend a list. This is highly ineffecient when the list grows long as Maple creats copies of the list.

The better way is to use a programmable Array( ) as opposed to Array[ ]

I have shown both is the attached.

This is an example as I do not know how you achieved you results. There are two basic ways.

 (1)

 (2)

 (3)

 (4)

 (5)

 (6)

 (7)

 (8)

## Is this what you mean....

Edit:- @max125 correctly pointed out I missed the minus sign in front of the x^2. I posted the revised answer in the reply to him.

Look at this and see if that is what you want.

 (1)

 (2)

 (3)

 (4)

 (5)

 (6)

 (7)

 (8)

 (9)

 (10)

 (11)

 (12)

 (13)