Guest wrote:Relevant Reference Link:
https://www.researchgate.net/post/Are_Diophantine_equations_in_four_variables_generally_solvable
Guest wrote:Guest wrote:Relevant Reference Link:
https://www.researchgate.net/post/Are_Diophantine_equations_in_four_variables_generally_solvable
Given the Diophantine equation,
1. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex],
of four nonzero integer variables, w, x, y, and z with known nonzero integer constants, [tex]a_{1 }[/tex], [tex]a_{2 }[/tex], [tex]a_{3}[/tex], and [tex]a_{4 }[/tex] with
known nonzero integer exponents, [tex]e_{1 }[/tex], [tex]e_{2 }[/tex], [tex]e_{3}[/tex], and [tex]e_{4 }[/tex] and with known integer constant, n,
can we develop an algorithm which generally solves equation one?
Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis, and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]w = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]w = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]w = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].
Step O: Guess n_{10}, n_{20}, n_{30 }, and n_{40 };
Step 1: Compute the corresponding w_{0 }, x_{0},
y_{0}, and z_{0};
...
Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis, and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex];
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
Guest wrote:Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis, and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex];
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Guest wrote:Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3}y^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Guest wrote:Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Guest wrote:"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Guest wrote:x_{1 }"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Remark: This is a challenging and beautiful problem.
We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.
Important Derived Equations/Functions:
5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];
6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];
7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];
8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];
Notes:
[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;
[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;
[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;
[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;
And of course,
[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.
Central Equation (CE) of our Algorithm:
We have,
(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }[/tex],
where the inverse of J (Jacobian Matrix), [tex]J^{-1}[/tex], and the function, [tex]n_{j }[/tex], are evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]).
And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.
We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.
Dave,
https://www.researchgate.net/profile/David_Cole29
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Here's our tentative definition of the Jacobian, J:
[tex]J = \frac{\partial ( n_1, n_2, n_3, n_4 ) }{\partial (w, x, y, z)} = \begin{bmatrix}0& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& 0& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].
Tentatively, we have calculated the determinant of J:
[tex]|J| = \frac{3e_{1 } * e_{2 } * e_{3 } * e_{4 } *(n - n_{1 })(n - n_{2 })(n - n_{3 })(n - n_{4 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}((n - n_{2 })/a_{2 })^{1/e_{2 }}((n - n_{3 })/a_{3 })^{1/e_{3 }}((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex].
...
Dave.
Guest wrote:x_{1 }"Zero hides data..."
Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence
E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].
1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];
3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];
4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];
5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];
7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];
8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].
Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],
[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.
Step 1: Compute the corresponding [tex]w_{0}[/tex],
[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];
...
Remark: This is a challenging and beautiful problem.
We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.
Important Derived Equations/Functions:
5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];
6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];
7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];
8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];
Notes:
[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;
[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;
[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;
[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;
And of course,
[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.
Central Equation (CE) of our Algorithm:
We have,
(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j(k) }[/tex],
where the inverse of J (Jacobian Matrix), [tex]J^{-1}[/tex], is evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]). We are dealing with a column matrix (4x1) and a 4x4 matrix in CE. Please forgive the current notation...
And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.
We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.
Dave,
https://www.researchgate.net/profile/David_Cole29
Relevant Reference Link (Newton's Method):
https://en.wikipedia.org/wiki/Newton%27s_method
Here's our tentative definition of the Jacobian, J:
[tex]J = \frac{\partial ( n_1, n_2, n_3, n_4 ) }{\partial (w, x, y, z)} = \begin{bmatrix}0& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& 0& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].
Tentatively, we have calculated the determinant of J:
[tex]|J| = \frac{3e_{1 } * e_{2 } * e_{3 } * e_{4 } *(n - n_{1 })(n - n_{2 })(n - n_{3 })(n - n_{4 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}((n - n_{2 })/a_{2 })^{1/e_{2 }}((n - n_{3 })/a_{3 })^{1/e_{3 }}((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex].
Thus, the inverse of J is,
[tex]J^{-1} = [m'_{kl }][/tex]
where
[tex]m'_{kl } = \frac{(-1)^{l + k} * M_{lk }}{|J|}[/tex]
and where [tex]M_{lk }[/tex] are the determinants of the appropriate minor matrices of J.
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Dave.
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