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Mindscrape
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I am linearizing the function [tex]f(x,y,z) = tan^{-1}(xyz)[/tex] at the point (1,1,1).
Since [tex]f(x_0,y_0,z_0)= \frac{\pi}{4} + \pi*n[/tex] should I just take the first value or do I have to carry all the solutions through the linearization process?
Um, anybody remember this? I can put up some work if it helps.
[tex]L(x,y,z) = \Delta f = f_x \Delta x + f_y \Delta y + f_z \Delta z[/tex]
So
[tex]L(x,y,z) = f(x_0,y_0,z_0) + f_x(x-x_0) + f_y (y-y_0) + f_z (z-z_0)[/tex]
where [tex]f_x = \frac{\partial f}{\partial x}[/tex] (and f_y and f_z)
So does [tex] L(x,y,z) = \frac{\pi}{4} +\pi n + \frac{1}{2}[(x-1)+(y-1)+(z-1)] [/tex]
?
Since [tex]f(x_0,y_0,z_0)= \frac{\pi}{4} + \pi*n[/tex] should I just take the first value or do I have to carry all the solutions through the linearization process?
Um, anybody remember this? I can put up some work if it helps.
[tex]L(x,y,z) = \Delta f = f_x \Delta x + f_y \Delta y + f_z \Delta z[/tex]
So
[tex]L(x,y,z) = f(x_0,y_0,z_0) + f_x(x-x_0) + f_y (y-y_0) + f_z (z-z_0)[/tex]
where [tex]f_x = \frac{\partial f}{\partial x}[/tex] (and f_y and f_z)
So does [tex] L(x,y,z) = \frac{\pi}{4} +\pi n + \frac{1}{2}[(x-1)+(y-1)+(z-1)] [/tex]
?
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