i get error can any one help

erro in my code pleas help me

my code:

f= @(x,y,z) [((x-4)^2 - z) ;( 2*y-4*z); z+16*(x-8)+1/y -21]
jacobian_matrix = @(x,y,z) [2*(x-10), 0 ,-1;0,2,-4;16,(-1/y^2),1]
x=x0(1,1)
y=x0(2,1)
z=x0(3,1)
itar=0
x1=zeros(3)
while ((det(jacobian_matrix(x,y,z))~=0 && norm(f(x,y,z))>epsilon))
    itar=itar+1
    hfresh=((jacobian_matrix(x,y,z))\(-1*(f(x,y,z))))
    x1=x0+hfresh
    x=x1(1,1)
    y=x1(2,1)
    z=x1(3,1)
    
    x0=x1
end

my input

newton_method([3;4;1],10^-6)

Error using vertcat

Dimensions of arrays being

concatenated are not consistent.

Error in

newton_method>@(x,y,z)[((x-4)^2-z);(2*y-4*z);z+16*(x-8)+1/y,-21]

(line 2)

f= @(x,y,z) [((x-4)^2 - z) ;(

2*y-4*z); z+16*(x-8)+1/y -21]

[Walter Roberson]

How does

newton_method([3;4;1],10^-6)

map into a call to f (which expects three separate parameters) ?

[mohammad massarwe]

what i should do i want to iput a victor and an limit and the output is the itar and xpitaron and err

Answer 1

[Walter Roberson]

f= @(x,y,z) [((x-4)^2 - z) ;( 2*y-4*z); z+16*(x-8)+1/y -21]

Inside [] if you have an expression of the form A -B then that does not mean to subtract B from A. Instead it is treated as [A, -1*B] .

Spacing is important. [A-B] and [A- B] and [A - B] are all subtraction, but [A -B] treats the - as unary minus. Consider the expression

[0 -1 1 2]

then it becomes obvious in context that you want a list of four elements, equivalent to [0, -1, 1, 2]; and if you had [0 -1 1] it would still be obvious you were creating a list with negative values... and the same parsing rule holds for [0 -1] which does not suddenly become interpreted as a subtraction 0 minus 1.