Ronan

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

Answered: Ronan 1471

December 23 2023

3 0

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;

LinePts := proc (p1, p2) description " Calculates a line through two points"; (p2[2]-p1[2])*x+(p1[1]-p2[1])*y-p1[1]*p2[2]+p1[2]*p2[1] end proc

proc (p1, p2) description " Calculates a line through two points"; (p2[2]-p1[2])*x+(p1[1]-p2[1])*y-p1[1]*p2[2]+p1[2]*p2[1] end proc

(1)

(2)

Firstly Paramaterise a conic centred on the Origin using projective points

(3)

(4)

(5)

(6)

[[x = -(-d/a)^(1/2), y = 0], [x = -(-d/a)^(1/2)*(b*t^2+d)/(c*(-d/a)^(1/2)*t+b*t^2-d), y = t*(c*(-d/a)^(1/2)*t-2*d)/(c*(-d/a)^(1/2)*t+b*t^2-d)]]

(7)

Vector[column](%id = 36893490806110780580)

(8)

$proc (t) options operator, arrow; rtable(1 .. 3, {1 = -(-d/a)^(1/2)*(b*t^2+d)/(c*(-d/a)^(1/2)*t+b*t^2-d), 2 = t*(c*(-d/a)^(1/2)*t-2*d)/(c*(-d/a)^(1/2)*t+b*t^2-d), 3 = 1}, datatype = anything, subtype = Vector[column], storage = rectangular, order = Fortran_order) end proc$

(9)

Vector[column](%id = 36893490806134464372)

(10)

Create 6 Projective Ponits on the conic

Vector[column](%id = 36893490806134620156)

(11)

Vector[column](%id = 36893490806134630388)

(12)

Vector[column](%id = 36893490806134722556)

(13)

Vector[column](%id = 36893490806134740980)

(14)

Vector[column](%id = 36893490806134845436)

(15)

Vector[column](%id = 36893490806127228916)

(16)

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

Vector[column](%id = 36893490806134060444)

(17)

Vector[column](%id = 36893490806134290060)

(18)

Vector[column](%id = 36893490806188406652)

(19)

Vector[column](%id = 36893490806188440500)

(20)

Vector[column](%id = 36893490806188465316)

(21)

Vector[column](%id = 36893490806130202612)

(22)

Get 3 intersection points in projective form

Vector[column](%id = 36893490806134736044)

(23)

Vector[column](%id = 36893490806192605772)

(24)

Vector[column](%id = 36893490806145831988)

(25)

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

Matrix(%id = 36893490806102267172)

(26)

(27)

Plotting

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

(28)

for i from 0 to 5 do p || i := convert(P || i[1 .. 2], list) end do

[-(-d/a)^(1/2)*(b*t[5]^2+d)/(c*(-d/a)^(1/2)*t[5]+b*t[5]^2-d), t[5]*(c*(-d/a)^(1/2)*t[5]-2*d)/(c*(-d/a)^(1/2)*t[5]+b*t[5]^2-d)]

(29)

(30)

x*((-d/a)^(1/2)*t[0]*a*c+(-d/a)^(1/2)*t[4]*a*c-2*t[0]*t[4]*a*b+t[0]*t[4]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[0]*a*b+2*(-d/a)^(1/2)*t[4]*a*b+t[0]*t[4]*b*c-d*c)-2*(-d/a)^(1/2)*t[0]*t[4]*a*b+2*(-d/a)^(1/2)*a*d+t[0]*c*d+t[4]*c*d

(31)

x*((-d/a)^(1/2)*t[0]*a*c+(-d/a)^(1/2)*t[5]*a*c-2*t[0]*t[5]*a*b+t[0]*t[5]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[0]*a*b+2*(-d/a)^(1/2)*t[5]*a*b+t[0]*t[5]*b*c-d*c)-2*(-d/a)^(1/2)*t[0]*t[5]*a*b+2*(-d/a)^(1/2)*a*d+t[0]*c*d+t[5]*c*d

(32)

x*((-d/a)^(1/2)*t[1]*a*c+(-d/a)^(1/2)*t[3]*a*c-2*t[1]*t[3]*a*b+t[1]*t[3]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[1]*a*b+2*(-d/a)^(1/2)*t[3]*a*b+t[1]*t[3]*b*c-d*c)-2*(-d/a)^(1/2)*t[1]*t[3]*a*b+2*(-d/a)^(1/2)*a*d+t[1]*c*d+t[3]*c*d

(33)

x*((-d/a)^(1/2)*t[1]*a*c+(-d/a)^(1/2)*t[5]*a*c-2*t[1]*t[5]*a*b+t[1]*t[5]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[1]*a*b+2*(-d/a)^(1/2)*t[5]*a*b+t[1]*t[5]*b*c-d*c)-2*(-d/a)^(1/2)*t[1]*t[5]*a*b+2*(-d/a)^(1/2)*a*d+t[1]*c*d+t[5]*c*d

(34)

x*((-d/a)^(1/2)*t[2]*a*c+(-d/a)^(1/2)*t[3]*a*c-2*t[2]*t[3]*a*b+t[2]*t[3]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[2]*a*b+2*(-d/a)^(1/2)*t[3]*a*b+t[2]*t[3]*b*c-d*c)-2*(-d/a)^(1/2)*t[2]*t[3]*a*b+2*(-d/a)^(1/2)*a*d+t[2]*c*d+t[3]*c*d

(35)

x*((-d/a)^(1/2)*t[2]*a*c+(-d/a)^(1/2)*t[4]*a*c-2*t[2]*t[4]*a*b+t[2]*t[4]*c^2-2*a*d)+y*(2*(-d/a)^(1/2)*t[2]*a*b+2*(-d/a)^(1/2)*t[4]*a*b+t[2]*t[4]*b*c-d*c)-2*(-d/a)^(1/2)*t[2]*t[4]*a*b+2*(-d/a)^(1/2)*a*d+t[2]*c*d+t[4]*c*d

(36)

The 3 collinear points

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

(37)

Download 2023-12-23_A_Pascal_Conic_parameterised.mw

5th Order Polynomial in h...

Answered: Ronan 1471

October 27 2023

0 1

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...

Answered: Ronan 1471

April 13 2023

0 0

I asked the same question a couple on months age

@C_R answered it as follows

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?...

Answered: Ronan 1471

April 12 2023

2 0

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

>

(1)

>

for i in l do A[i] := (x/a)^(i+1)*(1-x/a)^2 end do

x^2*(1-x/a)^2/a^2

x^4*(1-x/a)^2/a^4

x^6*(1-x/a)^2/a^6

x^8*(1-x/a)^2/a^8

x^10*(1-x/a)^2/a^10

x^12*(1-x/a)^2/a^12

x^14*(1-x/a)^2/a^14

x^16*(1-x/a)^2/a^16

x^18*(1-x/a)^2/a^18

x^20*(1-x/a)^2/a^20

(2)

>	~# Or use seq

>

[x^2*(1-x/a)^2/a^2, x^4*(1-x/a)^2/a^4, x^6*(1-x/a)^2/a^6, x^8*(1-x/a)^2/a^8, x^10*(1-x/a)^2/a^10, x^12*(1-x/a)^2/a^12, x^14*(1-x/a)^2/a^14, x^16*(1-x/a)^2/a^16, x^18*(1-x/a)^2/a^18, x^20*(1-x/a)^2/a^20]

(3)

>

A[4]

x^8*(1-x/a)^2/a^8

(4)

>

Download 3a.mw

Incorrect square brackets []use ()...

Answered: Ronan 1471

April 12 2023

2 0

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...

Answered: Ronan 1471

February 23 2023

1 0

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

Download A_23-2-23-how_maple_gets_answer.mw

B:=...

Answered: Ronan 1471

September 28 2022

0 0

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

Download assumetest-1.mw

Try.....

Answered: Ronan 1471

September 27 2022

0 2

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

Two ways around it....

Answered: Ronan 1471

July 22 2022

0 2

Here are two potential ways around you issue.

Download 22-7-22_A_List.mw

expand may work for you...

Answered: Ronan 1471

July 11 2022

1 2

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))

answers from Maple2022

eq := 1.411454823, 0.

answers from Maple 18

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...

Answered: Ronan 1471

May 28 2022

1 2

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])]]);

Matrix(4, 4, {(1, 1) = lambda[1], (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = lambda[1], (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = conjugate(lambda[1]), (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 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])]]);

Matrix(4, 4, {(1, 1) = lambda[2], (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = lambda[2], (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = conjugate(lambda[2]), (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = conjugate(lambda[2])})

(7)

>	H11:=H[1].Lambda[1];

(8)

>	H12:=H[2].Lambda[2];

(9)

>	H13:=H[1].Lambda[1].Lambda[1];

$Matrix(4, 4, {(1, 1) = `ε`[1]*phi(x, t)[1]*lambda[1]^2, (1, 2) = conjugate(`ε`[2])*phi(x, t)[1]*lambda[1]^2, (1, 3) = conjugate(chi(x, t)[1])*conjugate(lambda[1])^2, (1, 4) = 0, (2, 1) = `ε`[2]*psi(x, t)[1]*lambda[1]^2, (2, 2) = -conjugate(`ε`[1])*psi(x, t)[1]*lambda[1]^2, (2, 3) = 0, (2, 4) = conjugate(rho(x, t)[1])*conjugate(lambda[1])^2, (3, 1) = -chi(x, t)[1]*lambda[1]^2, (3, 2) = 0, (3, 3) = conjugate(`ε`[1])*conjugate(phi(x, t)[1])*conjugate(lambda[1])^2, (3, 4) = conjugate(`ε`[2])*conjugate(phi(x, t)[1])*conjugate(lambda[1])^2, (4, 1) = 0, (4, 2) = rho(x, t)[1]*lambda[1]^2, (4, 3) = -`ε`[2]*conjugate(psi(x, t)[1])*conjugate(lambda[1])^2, (4, 4) = `ε`[1]*conjugate(psi(x, t)[1])*conjugate(lambda[1])^2})$

(10)

>	H14:=H[2].Lambda[2].Lambda[2];

$Matrix(4, 4, {(1, 1) = `ε`[1]*phi(x, t)[2]*lambda[2]^2, (1, 2) = conjugate(`ε`[2])*phi(x, t)[2]*lambda[2]^2, (1, 3) = conjugate(chi(x, t)[2])*conjugate(lambda[2])^2, (1, 4) = 0, (2, 1) = `ε`[2]*psi(x, t)[2]*lambda[2]^2, (2, 2) = -conjugate(`ε`[1])*psi(x, t)[2]*lambda[2]^2, (2, 3) = 0, (2, 4) = conjugate(rho(x, t)[2])*conjugate(lambda[2])^2, (3, 1) = -chi(x, t)[2]*lambda[2]^2, (3, 2) = 0, (3, 3) = conjugate(`ε`[1])*conjugate(phi(x, t)[2])*conjugate(lambda[2])^2, (3, 4) = conjugate(`ε`[2])*conjugate(phi(x, t)[2])*conjugate(lambda[2])^2, (4, 1) = 0, (4, 2) = rho(x, t)[2]*lambda[2]^2, (4, 3) = -`ε`[2]*conjugate(psi(x, t)[2])*conjugate(lambda[2])^2, (4, 4) = `ε`[1]*conjugate(psi(x, t)[2])*conjugate(lambda[2])^2})$

(11)

>	H115:=Matrix([H[1],H[2]]);

(12)

>	H15:=Matrix([H11,H12]);

(13)

>	H116:=Matrix([[H115],[H15]]);

(14)

>	H16:=Matrix([H13,H14]);

Matrix(%id = 36893490284340821276)

(15)

>

Matrix(%id = 36893490284313909236)

(16)

>

Download matrixop-1.mw

Try this...

Answered: Ronan 1471

March 31 2022

1 1

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...

Answered: Ronan 1471

March 26 2022

0 2

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.

Download 27-3-22_A_add_to_Array_or_List.mw

Is this what you mean....

Answered: Ronan 1471

March 23 2022

2 2

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)

proc (x) options operator, arrow; -(1/2)*(2*x+8)/(x^2+8*x+9)^(1/2) end proc

(3)

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-18-(7/2)*ln(7)+3*43^(1/2)+(7/2)*ln(43^(1/2)+6)-(int(6-(x^2+8*x+9)^(1/2), x = -4+43^(1/2) .. k)) = 10

(8)

24-3*43^(1/2)-(7/2)*ln(43^(1/2)+6)+6*k-(1/2)*(k^2+8*k+9)^(1/2)*k-2*(k^2+8*k+9)^(1/2)+(7/2)*ln(k+4+(k^2+8*k+9)^(1/2))

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Download 23-3-22_A_integration.mw

How did you install...

Answered: Ronan 1471

March 16 2022

2 2

If you are using windows

Always with this programs like Maple I install them using the hidden administrator account in windows. i learn't this the hard way using the Solidworks CAD systtem.

When I first used Maple in 2012 I forgot to do the above and had all sorts of problems until I reinstalled as above.

Might help.

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