Increments

Definition

Basically, an increment is a slight change in a function. 

Let w = $f(x,y)$ and let $\Delta{}x$ and $\Delta{}y$ be increments of x and y, respectively. So, the increment, $\Delta{}w$, is $$\Delta{}w = f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$$

Alternate Definition

Basically, our goal is to rewrite the expression above, because that is how we will define differentiation. 

Let's start by thinking about this in 2D. Instead of $w$, let's let $u$ equal a differentiable function of one variable, x, and let $u = f(x)$. Let $x_0$ be an arbitrary value of x. By our last definition, the increment of $u$ is $$\Delta{}u = f(x_0 + \Delta{}x) - f(x_0)$$ Now, let's divide through by $\Delta{}x$ and see what happens. $$\dfrac{\Delta{}u}{\Delta{}x} = \dfrac{f(x_0 + \Delta{}x) - f(x_0)}{\Delta{}x}$$ The right side of the equation is the definition of a derivative, so if we take the limit of both sides as $\Delta{}x$ approaches 0, we get $$\lim_{\Delta{}x \to 0} \dfrac{\Delta{}u}{\Delta{}x} = f'(x_0)$$ We can move it all to one side and get $$lim_{\Delta{}x \to 0} \left( \dfrac{\Delta{}u}{\Delta{}x} - f'(x_0)\right) = 0$$ So, we can define $\varepsilon$ as $\dfrac{\Delta{}u}{\Delta{}x} - f'(x_0)$, which approaches 0 as $\Delta{}x$ approaches 0. We get $$\Delta{}u = f'(x_0)\Delta{}x + \varepsilon\Delta{}x$$ If we extend that to two variables, we get $$\Delta{}w = f_x(x_0, y_0)\Delta{}x + f_y(x_0, y_0)\Delta{}y + \varepsilon_1\Delta{}x + \varepsilon_2\Delta{}y$$ Both $\varepsilon$'s approach 0 as $(\Delta{}x, \Delta{}y) \to (0,0)$

You can prove that, but it's too long to type out. Basically, you just apply the Mean Value Theorem on both x and y, and you get that.

Example

Find values of $\varepsilon_1$ and $\varepsilon_2$ that satisfy Definition (16.12). $$f(x,y) = (2x-y)^2$$ Recall that 16.12, or our definition for $$\Delta{}w = f_x(x_0,y_0)\Delta{}x + f_y(x_0, y_0)\Delta{}y + \varepsilon_1\Delta{}x + \varepsilon_2\Delta{}y = f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$$ So, what we have to do is find $f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$, and make sure it equals our real definition. $$f(x + \Delta{}x, y + \Delta{}y) - f(x,y) = ((2x + 2\Delta{}x) - (y + \Delta{}y))^2 - (2x - y)^2$$ $$= (2x + 2\Delta{}x)^2 + (y + \Delta{}y)^2 - 2(2x + 2\Delta{}x)(y + \Delta{}y) - 4x^2 + 4xy - y^2$$ $$ = 4x^2 + 8x\Delta{}x + 4\Delta{}x^2 + y^2 + 2y\Delta{}y + \Delta{}y^2 - 2(2xy + 2x\Delta{}y + 2y\Delta{}x + 2\Delta{}x\Delta{}y) - 4x^2 + 4xy - y^2$$ $$=8x\Delta{}x + 4\Delta{}x^2 + 2y\Delta{}y + \Delta{}y^2 - 4x\Delta{}y - 4y\Delta{}x - 4\Delta{}x\Delta{}y$$ $$f_x(x,y) = 8x - 4y, f_y(x,y) = 2y - 4x$$ Therefore, we must have terms like $$(8x - 4y)\Delta{}x, (2y - 4x)\Delta{}y$$ So, we get $$(8x-4y)\Delta{}x + (2y-4x)\Delta{}y + 4\Delta{}x^2 + \Delta{}y^2 - 4\Delta{}x\Delta{}y$$ $$(8x-4y)\Delta{}x + (2y-4x)\Delta{}y + (4\Delta{}x - 4\Delta{}y)\Delta{}x + (\Delta{}y)\Delta{}y$$ $$\varepsilon_1 = 4\Delta{}x - 4\Delta{}y$$ $$\varepsilon_2 = \Delta{}y$$ Note that how you partition the epsilons doesn't matter, as long as they approach zero as $\Delta{}x,y \to 0$, then it's fine.