- #1
kostoglotov
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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
[tex]x_2 = x_1 - \frac{fg_y-f_yg}{f_xg_y-f_yg_x}[/tex]
and
[tex]y_2 = y_1 - \frac{f_xg-fg_x}{f_xg_y-f_y-g_x}[/tex]
Homework Equations
[itex]\vec{v} = \vec{n_f} \times \vec{n_g}[/itex]: direction vector of line of intersection of the two planes, basically the cross product of the normal vectors to the two planes
[itex]x_n = x_{n-1} - \frac{f(x_{n-1})}{f'(x_{n-1})}[/itex]: 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 [itex]\frac{\partial z}{\partial x} = \frac{f_xg_y-f_yg_x}{g_y-f_y}[/itex] and [itex]\frac{\partial z}{\partial y} = \frac{f_xg_y-f_yg_x}{f_x-g_x}[/itex] by taking the z component of [itex]\vec{v}[/itex] (the direction vector of the line formed from the cross product) and dividing it by the x and y components of [itex]\vec{v}[/itex] respectively.
I get to here with it:
[tex]x_2 = x_1 - z_1 \frac{g_y-f_y}{f_xg_y-f_yg_x}[/tex]
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
[tex]x_2 = x_1 - \frac{fg_y-f_yg}{f_xg_y-f_yg_x}[/tex]
The text shows the z = f when multiplied by the gy and z = g when multiplied by the fy.
Why is this?
A follow up question to this one is to find the points of intersection of [itex]x^x+y^y=1000[/itex] and [itex]x^y + y^x = 100[/itex] so I can see how it might be important to choose z = f for multiplication with gy and z = g for multiplication with fy, but I can't spy the reasoning behind it.