Newton's Method generalized to 3 dimensions

[kostoglotov]

Homework Statement

Prelim: my question is about a very specific part of a question whereby the student is asked to derive the final formula for the general solution in two vars, but I will post the entire question for clarify.

Newton's Method for approximating the roots of an equation f(x)=0 can be adapted to approximating solutions of a system of equations f(x,y)=0 and g(x,y) = 0. The surfaces z = f(x,y) and z = g(x,y) intersect in a curve that intersects the xy-plane at the point (r,s) which is the solution of the system.

If an initial approximation (x₁,y₁) is close to the point (r,s), then the tangent planes to the surface at (x₁,y₁) intersect in a straight line that should intersect the xy-plane closer to (r,s).

Show that

$$x_2 = x_1 - \frac{fg_y-f_yg}{f_xg_y-f_yg_x}$$

and

$$y_2 = y_1 - \frac{f_xg-fg_x}{f_xg_y-f_y-g_x}$$

Homework Equations

$\vec{v} = \vec{n_f} \times \vec{n_g}$: direction vector of line of intersection of the two planes, basically the cross product of the normal vectors to the two planes

$x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}$: single var form of Newton's Method to be adapted.

The Attempt at a Solution

So I consider this line of intersection of two planes from two perspectives. One looking at it "down" the y-axis to consider just the relationship between z and x, and then by looking "down" the x-axis to consider just the relationship between z and y.

I can see that $\frac{\partial z}{\partial x} = \frac{f_xg_y-f_yg_x}{g_y-f_y}$ and $\frac{\partial z}{\partial y} = \frac{f_xg_y-f_yg_x}{f_x-g_x}$ by taking the z component of $\vec{v}$ (the direction vector of the line formed from the cross product) and dividing it by the x and y components of $\vec{v}$ respectively.

I get to here with it:

$$x_2 = x_1 - z_1 \frac{g_y-f_y}{f_xg_y-f_yg_x}$$

Now, I can see that since all the points we are working with here are points of intersection, that the x,y and z's are all common for all the surfaces, planes and lines concerned. So z₁ = f(x₁,y₁) = g(x₁,y₁)...so I can accept that we could pick either f or g to be z in the equation...but the text specifically states

$$x_2 = x_1 - \frac{fg_y-f_yg}{f_xg_y-f_yg_x}$$

The text shows the z = f when multiplied by the g_y and z = g when multiplied by the f_y.

Why is this?

A follow up question to this one is to find the points of intersection of $x^x+y^y=1000$ and $x^y + y^x = 100$ so I can see how it might be important to choose z = f for multiplication with g_y and z = g for multiplication with f_y, but I can't spy the reasoning behind it.

 

[wabbit]

I don't quite get the logic of your solution - but going back to the "relevant equations" part, if you calculate the vector $\vec v$ then you are searching for the intersection of the line $\{(x,y,z)+t\vec v,t\in\mathbb{R}\}$ with the plane z=0. Would that work ?

 

[kostoglotov]

I don't quite get the logic of your solution - but going back to the "relevant equations" part, if you calculate the vector $\vec v$ then you are searching for the intersection of the line $\{(x,y,z)+t\vec v,t\in\mathbb{R}\}$ with the plane z=0. Would that work ?

Actually, I realized that I was just confusing myself. Provided f(x,y) = 0 and g(x,y) = 0, z = f = g when evaluated at the same x,y, so it doesn't matter which partial deriv you pair the z,f, or g with, they'll always be the same value anyway, so I think the text was just pairing f with gy and g with fy for aesthetic purposes, because regardless of which you choose that second problem I mentioned $x^x+y^y=1000$ and $x^y + y^x = 100$ will have the same solutions, which look just fine when you graph them.

 

[wabbit]

Well there is only one Newton formula you can arrive at, no aestethic choice involved - so I am not sure what you mean here.

Also my question in the previous post was rhetorical. It will work.