abbeykabir

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@Carl Love I hope i'll get this right. We stop iterating when the error difference between two consecutive iteration is relatively small. Maybe i should use the word converge. say abs(x[i+1]-x[i]) > 1e-10.

@Carl Love I know that Addition, Sub and mutiplication of matrix is possible. never heard of Division except scalar division.

@Carl Love I know that Addition, Sub and mutiplication of matrix is possible. never heard of Division except scalar division.

@Carl Love I cannot think of anything else other than the Jacobian of f : R^m -> R^m. i don't know how to type Matrix here

hope i'm wright

@Carl Love I cannot think of anything else other than the Jacobian of f : R^m -> R^m. i don't know how to type Matrix here

hope i'm wright

@Carl Love The iterative formula for Newton in single variable: x[new] = x[old] - f(x[old])/f'(x[old])

@Carl Love The iterative formula for Newton in single variable: x[new] = x[old] - f(x[old])/f'(x[old])

@Carl Love  Newton's method in a single variable x[i+1]=x[i] - f(x[i])/f'(x[i]) where x[i] is the initial guess. i=0..N

@Carl Love  Newton's method in a single variable x[i+1]=x[i] - f(x[i])/f'(x[i]) where x[i] is the initial guess. i=0..N

Amazing. To be honest with you, I KNOW NOTHING. I don't know where to even start from. Foremost, will download the worksheet for study and I pray I understand every bit of it bcos i am not conversant with some of the syntax therein. Pls i will contact you if I have any problem with any part of the code. Pls. here are my equations.

My y[0] is 1/51. and I use the following code to detemine x[i], i=1..5

x[0]:=1, h:=0.1 AND x[i]:=x[i-1] + i*h     i=1..5

after substitution I did the  fsolve({}) as you instructed. Separating the equations with a comma in the curly braces and I got the following results

I think the equivalent place in your code, you got(with ur randomly generated quest.)

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

I don't really understand this part of your code

(**)

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:

and also

(**)

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

Pls. don't mind my questions. I'm still a newbie

Amazing. To be honest with you, I KNOW NOTHING. I don't know where to even start from. Foremost, will download the worksheet for study and I pray I understand every bit of it bcos i am not conversant with some of the syntax therein. Pls i will contact you if I have any problem with any part of the code. Pls. here are my equations.

My y[0] is 1/51. and I use the following code to detemine x[i], i=1..5

x[0]:=1, h:=0.1 AND x[i]:=x[i-1] + i*h     i=1..5

after substitution I did the  fsolve({}) as you instructed. Separating the equations with a comma in the curly braces and I got the following results

I think the equivalent place in your code, you got(with ur randomly generated quest.)

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

I don't really understand this part of your code

(**)

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:

and also

(**)

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

Pls. don't mind my questions. I'm still a newbie

Yes I did but when I showed it to my lecturer, he said my answers were wrong and that I should try another method.

Yes I did but when I showed it to my lecturer, he said my answers were wrong and that I should try another method.

@abbeykabir I've said it befor and i'll say it again. Guy you are always on point!!! It works like magic

@abbeykabir I've said it befor and i'll say it again. Guy you are always on point!!! It works like magic

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