Examples from Multivariable approximation All examples from Multivariable approximation.
[ EXAMPLE ] Let
f ( x , y ) = x 2 + y 2 − 2 x y = ( x − y ) 2 . f(x,y)=x^2+y^2-2xy=(x-y)^2. f ( x , y ) = x 2 + y 2 − 2 x y = ( x − y ) 2 . At ( 1 , 3 ) (1,3) ( 1 , 3 )
f ( 1 , 3 ) = 4 , f x = 2 x − 2 y , f y = 2 y − 2 x , f(1,3)=4,
\qquad
f_x=2x-2y,
\qquad
f_y=2y-2x, f ( 1 , 3 ) = 4 , f x = 2 x − 2 y , f y = 2 y − 2 x , so
f x ( 1 , 3 ) = − 4 , f y ( 1 , 3 ) = 4. f_x(1,3)=-4,
\qquad
f_y(1,3)=4. f x ( 1 , 3 ) = − 4 , f y ( 1 , 3 ) = 4. The tangent-plane approximation is
L ( x , y ) = 4 − 4 ( x − 1 ) + 4 ( y − 3 ) = − 4 x + 4 y − 4. \begin{aligned}
L(x,y)
&=4-4(x-1)+4(y-3)\\
&=-4x+4y-4.
\end{aligned} L ( x , y ) = 4 − 4 ( x − 1 ) + 4 ( y − 3 ) = − 4 x + 4 y − 4. Thus, close to ( 1 , 3 ) (1,3) ( 1 , 3 )
( x − y ) 2 ≈ − 4 x + 4 y − 4 . {(x-y)^2\approx -4x+4y-4}. ( x − y ) 2 ≈ − 4 x + 4 y − 4 . Writing x = 1 + h x=1+h x = 1 + h y = 3 + k y=3+k y = 3 + k
f ( 1 + h , 3 + k ) = 4 − 4 h + 4 k + ( h − k ) 2 . f(1+h,3+k)=4-4h+4k+(h-k)^2. f ( 1 + h , 3 + k ) = 4 − 4 h + 4 k + ( h − k ) 2 . The approximation keeps the constant and first-degree terms and discards the
quadratic error ( h − k ) 2 (h-k)^2 ( h − k ) 2 ( h , k ) → ( 0 , 0 ) (h,k)\to(0,0) ( h , k ) → ( 0 , 0 )
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
f ( x , y ) = e x − y . f(x,y)=e^{x-y}. f ( x , y ) = e x − y . Set u = x − y u=x-y u = x − y u u u
T 3 ( x , y ) = 1 + ( x − y ) + ( x − y ) 2 2 + ( x − y ) 3 6 . {
T_3(x,y)
=1+(x-y)+\frac{(x-y)^2}{2}
+\frac{(x-y)^3}{6}
}. T 3 ( x , y ) = 1 + ( x − y ) + 2 ( x − y ) 2 + 6 ( x − y ) 3 . Each power ( x − y ) k (x-y)^k ( x − y ) k k k k u u u u 3 u^3 u 3
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y ) = x 2 y , f(x,y)=x^2y, f ( x , y ) = x 2 y , the function has total degree three. Hence its Taylor polynomial through
degree three about ( 0 , 0 ) (0,0) ( 0 , 0 )
T 3 ( x , y ) = x 2 y . {T_3(x,y)=x^2y}. T 3 ( x , y ) = x 2 y . The derivative construction gives the same result: all lower-degree terms
vanish at the origin and the third-order term reproduces x 2 y x^2y x 2 y
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] For
f ( x , y , z ) = e x + y + z f(x,y,z)=e^{x+y+z} f ( x , y , z ) = e x + y + z near ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 ) s = x + y + z s=x+y+z s = x + y + z
T 2 ( x , y , z ) = 1 + ( x + y + z ) + 1 2 ( x + y + z ) 2 . {
T_2(x,y,z)
=1+(x+y+z)+\frac12(x+y+z)^2
}. T 2 ( x , y , z ) = 1 + ( x + y + z ) + 2 1 ( x + y + z ) 2 . The total degree counts the total number of factors drawn from all coordinate
increments, so the same construction works in any number of variables.
[RIGHT CLICK] [ASK AI] Multivariable approximation Builds on Derivatives and level-set geometry .