The linear approximation of a function

Undefdisfigure · #1 $Old$ October 21st, 2007, 12:09 AM

Find the linear approximation of the function f (x, y, z) = root(x^2 + y^2 + z^2) at (3, 2, 6) and use it to approximate the number

root((3.02)^2 + (1.97)^2 + (5.99)^2)

Thanks.

Jhevon · #2 $Old$ October 21st, 2007, 12:18 AM

Quote:

Originally Posted by Undefdisfigure $View Post$

Find the linear approximation of the function f (x, y, z) = root(x^2 + y^2 + z^2) at (3, 2, 6) and use it to approximate the number

root((3.02)^2 + (1.97)^2 + (5.99)^2)

Thanks.

use the formula, $f(x,y,z) \approx f(x,y,z) + f_x (x_0,y_0,z_0)(x - x_0) + f_y (x_0,y_0,z_0)(y - y_0) + f_z(x_0,y_0,z_0)(z - z_0)$

for $(x,y,z)$ close to $(x_0,y_0,z_0)$

here, use $(x,y,z) = (3.02,1.97,5.99)$ and $(x_0,y_0,z_0) = (3,2,6)$

note, $f_x (x_0,y_0,z_0), f_y (x_0,y_0,z_0), \mbox{ and } f_z (x_0,y_0,z_0)$ are the derivative with respect to x,y,and z respectively, evaluated at $(x_0,y_0,z_0)$

can you take it from here?

topsquark · #3 $Old$ October 21st, 2007, 07:28 AM

For the record the multidimensional Taylor series around the point $\bold{x} = \bold{a}$ is
$T( \bold{x} ) = \sum_{ | \alpha | \geq 0 } \frac{D^{ \alpha } f( \bold{a} ) }{ \alpha ! } ( \bold{x} - \bold{a} )^{ \alpha }$
where the expression employs multi-index notation. (I don't know the formula for the error term, sorry! But you can probably generalize it from the 1D Taylor series.)

-Dan

Undefdisfigure · #4 $Old$ October 21st, 2007, 01:03 PM

No, I got up to what you wrote in reply to my thread. That part I understand. I ran into problems when I was taking the partial derivative and was hoping you could show me how to take the partial derivatives. Thanks for your reply though and I'm hoping you can go a couple of steps further.

By the way the book gives the answer 3/7x + 2/7y + 6/7z. It appears the point (3, 2, 6) is used in the answer but I don't know how they got it.

Jhevon · #5 $Old$ October 21st, 2007, 01:26 PM

Quote:

Originally Posted by topsquark $View Post$

For the record the multidimensional Taylor series around the point $\bold{x} = \bold{a}$ is
$T( \bold{x} ) = \sum_{ | \alpha | \geq 0 } \frac{D^{ \alpha } f( \bold{a} ) }{ \alpha ! } ( \bold{x} - \bold{a} )^{ \alpha }$
where the expression employs multi-index notation. (I don't know the formula for the error term, sorry! But you can probably generalize it from the 1D Taylor series.)

-Dan

hehe, we did that in my class the other day. I don't remember the formula for the error term, and i can't bother reaching for my text right now. maybe i'll update this post later... or TPH may tell you, he remembers all this stuff off the top of his head after reading it once. Heck, he could probably tell you on what specific homework problems he used it in (the section in the text, the problem number, maybe the page number, and the exact question), again, of course, off the top of his head.

Quote:

Originally Posted by Undefdisfigure $View Post$

No, I got up to what you wrote in reply to my thread. That part I understand. I ran into problems when I was taking the partial derivative and was hoping you could show me how to take the partial derivatives. Thanks for your reply though and I'm hoping you can go a couple of steps further.

By the way the book gives the answer 3/7x + 2/7y + 6/7z. It appears the point (3, 2, 6) is used in the answer but I don't know how they got it.

$f(x,y,z) = \sqrt{x^2 + y^2 + z^2}$

i will find $f_x(x,y,z)$ , the partial derivative with respect to $x$ . we treat $y$ and $z$ as constants, and employ the Chain rue:

$\Rightarrow f_x = \frac 12 \left( x^2 + y^2 + z^2 \right)^{-1/2} \cdot 2x = \frac x{\sqrt{x^2 + y^2 + z^2}}$

this problem is "symmetric" in all variables, so for the other partial derivatives, just replace the x in the numerator with the variable you are differentiating with respect to. and continue