Differentiability, linearization.

Theorem 5.5   Let f be a function defined on an open domain $\mathcal{D}$.Suppose that f has first order partial derivatives at every point of $\mathcal{D}$and that these partial derivatives are continuous at the point (x₀,y₀). Denote $\Delta f = f(x_0 + \Delta x ,y_0 + \Delta y) = f(x_0,y_0)$.

Then:

$$\Delta f = f_x(x_0,y_0) \Delta x + f_y(x_0,y_0) \Delta y + \... ...lta y) \Delta x + \varepsilon_2 (\Delta x, \Delta y) \Delta y$$

where $\underset{(\Delta x, \Delta y) \rightarrow (0,0)}{\lim}\varepsilon_1 (\Delta x, \Delta y) =0$and $\underset{(\Delta x, \Delta y) \rightarrow (0,0)}{\lim}\varepsilon_2 (\Delta x, \Delta y) =0$.

Definition 5.6   A function f is differentiable at the point (x₀,y₀) if the partial derivatives f_x(x₀,y₀) and f_y(x₀,y₀) exist and the equation  2.1 holds for f at (x₀,y₀).

The function f is differentiable on the open domain $\mathcal{D}$if it is differentiable at every point of $\mathcal{D}$.

Corollary 5.7   If the first order partial derivatives of f are continuous on the open domain $\mathcal{D}$, then f is differentiable on $\mathcal{D}$.

Theorem 5.8   If f is differentiable at the point (x₀,y₀), then f is continuous at the point (x₀,y₀).

Definition 5.9   The linearization of a function f at a point (x₀,y₀) is the function

Lf_(x0,y0) (x,y)=f(x₀,y₀)+ f_x(x₀,y₀) (x-x₀)+f_y(x₀,y₀)(y-y₀)

Example 5.10   Let $f(x,y)=x^2 \ln (y+1)$and (x₀,y₀)=(1,0). Then $f_x=2x \ln (y+1)$and $f_y=\frac {x^2}{y+1}$. Thus, the linearization of f at (1,0) is:

$$Lf_{(1,0)} (x,y) = 2 \cdot 0 (x-1) + 1 (y-0) = y.$$

We can use this linearization to compute an approximation of f(1.02, 0.15):

$$f (1.02, 0.15) \approx Lf_{(1,0)}(1.02, 0.15) = 0.15.$$

The ``true'' value is $1.02^2 \ln (0.15+1)= 0.1454$.

Question: Can we have an estimation of the error when using the linearization instead of the actual function?

The positive answer is described in the following result:

Proposition 5.11   Suppose that f has continuous partial derivatives of order 1 and 2 in an open domain $\mathcal{D}$. Let $\mathcal{R}$be a rectangle centered at the point (x₀,y₀) and contained in $\mathcal{D}$. We denote by M an upper bound for |f_xx|, |f_xy| and |f_yy|. Then the error made when replacing f(x,y) by its linearization for $(x,y) \in \mathcal{R}$satisfies the following inequality:

$$\vert E(x,y)\vert \leq \frac M2 ( \vert x-x_0\vert+\vert y-y_0\vert)^2.$$

Example 5.12   Take f(x,y)=x²y+xy³-1, (x₀,y₀)=(1,2) and $\mathcal{R}= \{ (x,y) ; \vert x-1\vert<0.1 and \vert y-2\vert<0.1 \}$.

We have: f_x=2xy+y³, f_y=x²+3xy², f_xx= 2y, f_xy= 2x+y² and f_yy=6xy.

On the rectangle $\mathcal{R}$, we can take: M=. Then we have the inequality:

$$\vert E(x,y)\vert \leq \frac {13.86}{2} ( 0.1 + 0.1 )^2 = 0.5544.$$

As we have f(1,2)=9, the error percentage is less than $\frac {0.5544}{9}=0.0616$, i.e. we made an error of at most $6.1 \%$.