Part 3: Quadratic Approximations

Because functions of 2 variables have four second partial derivatives, the quadratic approximation of a function f( x,y) is defined using matrices, where a matrix A is a 2 dimensional rectangular array of numbers, such as for example

A = é ê ë 1 2 5 1 0 3 ù ú û

If the length of the rows of A is the same as the length of the columns of B, then the product AB is defined to be the matrix of inner products of rows of A with columns of B. For example, if

A = é ê ë 2 1 3 5 ù ú û     and    B = é ê ë -4 3 0 2 ù ú û

then their product is the matrix of inner products of rows of A with columns of B:

AB = é ê ë 2 1 3 5 ù ú û é ê ë -4 3 0 2 ù ú û = é ê ë 2( -4) +1·0 2·3+1·2 3( -4) +5·0 3·3+5·2 ù ú û = é ê ë -8 8 -12 19 ù ú û

Specifically, we define the Hessian matrix of f( x,y) at a point ( p,q) to be

Hf( p,q) = é ê ë fxx( p,q) fxy( p,q) fxy( p,q) fyy( p,q) ù ú û

We then use matrix arithmetic to define the quadratic approximation Q( x,y) of f at ( p,q) to be

Q( x,y) = L( x,y) +  1 2 [ x-p,    y-q]Hf( p,q) é ê ë x-p y-q ù ú û

The matrix form of Q( x,y) can subsequently be simplified using matrix arithmetic.      

EXAMPLE 4    Find the quadratic approximation of f(x,y) = x³-3x+y² at ( 1,2) .       

Solution: First, f_x = 3x²-3 implies that f_x(1,2) = 0, and f_y = 2y implies that f_y( 1,2) = 4. Since f( 1,2) = \allowbreak 2, the linearization of f(x,y) at ( 1,2) is

L( x,y) = 2+0( x-1) +4( y-2) = 2+4(y-2)

Since f_xx = 6x, f_xy = 0, and f_yy = 2, the Hessian at (1,2) is

Hf( 1,2) = é ê ë 6 0 0 2 ù ú û

Thus, in matrix form the quadratic approximation is given by

Q( x,y) = L( x,y) +  1 2 [ x-1,    y-2] é ê ë 6 0 0 2 ù ú û é ê ë x-1 y-2 ù ú û

Matrix arithmetic then leads to

Q( x,y) = L( x,y) +  1 2 [ x-1,  y-2] é ê ë 6( x-1) 2( y-2) ù ú û = L( x,y) +  1 2 ( x-1) ·6( x-1)+  1 2 ( y-2) ·2( y-2) = L( x,y) +3( x-1) 2+( y-2) 2

Substitution of L( x,y) = 2+4( y-2) then leads to

Q( x,y) = 2+4( y-2) +3( x-1) 2+(y-2) 2

           

It can be show that for all e > 0, there is a neighborhood of p on which

| f( x) -Qp( x)| < e || x-p || 2

Thus, a quadratic approximation is an even better approximation than is a linearization.

EXAMPLE 5    Find the quadratic approximation of f(x,y) = sin( x²+xy+y²) at ( 0,0)       

Solution: First, f_x = cos( x²+xy+y²) (2x+y) and f_y = cos( x²+xy+y²) (x+2y) . Thus, f( 0,0) = f_x( 0,0) = f_y( 0,0) = 0 and similarly, L( x,y) = 0. However,

fxx =  ¶ ¶x ( cos(x2+xy+y2) ( 2x+y) ) = 2cos( x2+xy+y2) -( 2x+y) 2sin(x2+xy+y2)

from which it follows that f_xx( 0,0) = 2. Likewise,

fxy = cos( x2+xy+y2) -sin(x2+xy+y2) ( x+2y) ( 2x+y) fyy = 2cos( x2+xy+y2) -sin(x2+xy+y2) ( x+2y) 2

Thus, f_xy( 0,0) = 1, f_yy( 0,0) = 2 and the hessian is

Hf( 0,0) = é ê ë 2 1 1 2 ù ú û

Consequently, the quadratic approximation is

Q( x,y) = L( x,y) +  1 2 [ x-0,  y-0] é ê ë 2 1 1 2 ù ú û é ê ë x-0 y-0 ù ú û

Matrix arithmetic and L( x,y) = 0 then leads to

Q( x,y) =  1 2 [ x,  y] é ê ë 2x+y x+2y ù ú û =  1 2 x( 2x+y) +  1 2 y( x+2y)

However, this in turn simplifies to Q( x,y) = x²+xy+y².

       

Below we have plotted Q( x,y) = x²+xy+y² in red versus f( x,y) = sin( x²+xy+y²) .

They are practically the same at ( 0,0) , and indeed, notice that both surfaces have a minimum at ( 0,0) .