Quadratic approximations and concavity

Second order partial derivatives are connected with 'concavity', but the relationship is more subtle than in the one variable case. If the tangent plane at a point $P(a,\,b,\, f(a,\,b))$ on the graph of $z = f(x,\, y)$ gives the best Linear Approximation $$L(x,\,y) \ = \ f(a,\,b) + f_x\Bigl|_{(a,\,b)} (x-a) + f_y\Bigl|_{(a,\,b)}(y-b) $$ to $f$ near $P$, then we can expect that some Quadric surface will give the best Quadratic Approximation to $f$ near $P$. Just as the best Linear Approximation is the degree 1 Taylor polynomial centered at $(a,\, b)$ for $f$, so this best Quadratic Approximation is the degree 2 Taylor polynomial. For brevity we'll speak of these degree one and degree two Taylor polynomials as simply the respective Linear and Quadratic Approximations to $z = f(x,\,y)$ at $(a,\,b)$.

The Quadratic Approximation to a function $z = f(x,\,y)$ at $(a,\, b)$ is given by $$Q(x,\,y) \ = \ f(a,\,b) + f_x\Bigl|_{(a,\,b)} (x-a) + f_y\Bigl|_{(a,\,b)}(y-b) \hskip1.0in $$ $$ \hskip1.0in + \frac{1}{2} f_{xx}\Bigl|_{(a,\,b)} (x-a)^2+ f_{xy}\Bigl|_{(a,\,b)} (x-a)(y-b) + \frac{1}{2} f_{yy}\Bigl|_{(a,\,b)} (y-b)^2 \,.$$ At $(x,y)=(a,b)$, the value, first derivatives and second derivatives of $Q(x,y)$ equal the value, first derivatives and second derivatives of $f(x,y)$. In fact, $Q(x,y)$ is the unique quadratic polynomial with this property.

Example: Find the linear and quadratic approximations to $f(x,\,y) =\displaystyle{\frac{2}{xy}}$ near $(1,2)$. Solution: We first compute derivatives at $(1,2)$: $$f(1,\,2) = 1, \quad f_x(1,\,2) = -1, \quad f_y(1,\,2) = - \frac{1}{2},$$ $$f_{xx}(1,\,2) = 2, \quad f_{xy}(1,\,2) = \frac{1}{2}, \quad f_{yy}(1,\,2)= \frac{1}{2}\,,$$ so the Linear Approximation to $\displaystyle{ f (x,\,y) = \frac{2}{xy}}$ near $(1,\,2)$ is given by

\begin{eqnarray*}\frac{2}{xy} \ \approx \ L(x,\,y) & = & 1 - (x-1) - \frac{1}{2}(y-2) \cr &= & 3 - x -\frac{1}{2}y\,, \end{eqnarray*} while the Quadratic Approximation is \begin{eqnarray*}\frac{2}{xy} \ \approx \ Q(x,\,y) & = & 1 - (x-1) - \frac{1}{2}(y-2) \cr && +(x-1)^2 + \frac{1}{2}(x-1)(y-2) + \frac{1}{4}(y-2)^2\,,\end{eqnarray*} which after some algebra becomes $$ \frac{2}{xy}\ \approx \ 6 - 4x - 2y +x^2 + \frac{1}{2}xy + \frac{1}{4}y^2\,.$$

But what does all this mean graphically for a function $z = f(x,\,y)$? Well, the graph of a linear equation $Ax + by+Cz=D$ is a plane, while the graph of a quadratic equation is a quadric surface. So near a point $(a,\, b)$ the Linear Approximation $L(x,\, y)$ at $(a,\, b)$ approximates the graph of $z = f(x,\,y)$ by a plane - the Tangent Plane - while the Quadratic Approximation $Q(x,\, y)$ at $(a,\,b)$ approximates the graph of $z = f(x,\,y)$ by a quadric surface such as a paraboloid, hyperbolic paraboloid, or a hyperboloid.

   Thus: when $f(x,\, y) = \sin x \sin y$,

  near $(0,\,0): \quad \displaystyle{ f(x,\, y) \ \ \approx \ \ xy\,, \quad }$ a hyperbolic paraboloid;

  near $\displaystyle{\Big(\frac{\pi}{2},\ \frac{\pi}{2}\Big): \quad f(x,\, y) \ \ \approx \ \ 1 - \frac{1}{2} \Big(x - \frac{\pi}{2}\Big)^2 - \frac{1}{2} \Big(y - \frac{\pi}{2}\Big)^2\,,\quad }$ a paraboloid opening downwards;

  near $\displaystyle{\Big(\frac{\pi}{2},\ -\frac{\pi}{2}\Big): \quad f(x,\, y) \ \ \approx \ \ -1 + \frac{1}{2} \Big(x - \frac{\pi}{2}\Big)^2 + \frac{1}{2} \Big(y +\frac{\pi}{2}\Big)^2\,,\quad }$ a paraboloid opening upwards.

Do these seem reasonable given the graph of $f$ shown above? For a function $y = f(x)$ of one variable, the degree two approximating polynomial in $x$ would be a parabola, opening up or down, but in two variables things are more subtle because there are several possible approximating quadric surfaces.

Just as in one dimension, we can use higher derivatives to get an even more accurate approximation, and to express functions as power series.

Taylor Series in Two Variables: If $f(x,y)$ is an analytic function of two variables, then $$f(x,y) = \sum_{n,m=0}^\infty c_{n,m} (x-a)^n (y-b)^m,$$ where $$c_{n,m}= \frac{1}{n!m!}\frac{\partial^{n+m}f}{\partial x^n\partial y^m}(a,b).$$