Recall the definition of a Taylor polynomial:

We have a similar formula for functions $F:\R ^n\to \R$.

As you can see, it is more complex than the formula for a single variable. Good news everyone: In this class, we will only compute the degree two polynomial for functions of two variables. In this case $P_2$ is: $$\begin{align*} P_2(\vec{x})=F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c}) (x-c_1)^2 \\ &+ F^{(1,1)}(\vec{c})(x-c_1)(y-c_2) \\ &+ \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2. \end{align*}$$

Basically, given a function $F:\R ^2\to \R$, the second degree Taylor polynomial $P_2$ at a point $\vec {c}$ is a polynomial “cooked-up” so that:

• The values are equal: $P_2(\vec {c}) = F(\vec {c})$.
• The first partial derivatives are equal: $P_2^{(1,0)}(\vec {c}) = F^{(1,0)}(\vec {c})$ and $P_2^{(0,1)}(\vec {c}) = F^{(0,1)}(\vec {c})$.
• The second partial derivatives are equal: $P_2^{(2,0)}(\vec {c}) = F^{(2,0)}(\vec {c})$, $P_2^{(0,2)}(\vec {c}) = F^{(0,2)}(\vec {c})$, and $P_2^{(1,1)}(\vec {c}) = F^{(1,1)}(\vec {c})$.

Here’s the plan. Soon we will be trying to find maximums and minimums for functions of two variables. The second degree Taylor polynomial will be the key to developing a “second derivative test” for identifying these extrema.

Try it, you might like it

Computing the Taylor polynomial is not so bad, you just need to get the hang of it.

In other words

Now that we have a formula and we (hopefully!) can apply it. Let’s finish by talking about what is really going on. Given a function $f:\R \to \R$, the Taylor polynomial $$p_d(x) = \sum _{k=0}^d\frac {f^{(k)}(c)}{k!}(x-c)^k.$$ is a polynomial “cooked-up” to share the value of the function, meaning $$p_d(c)=f(c),$$ and share values of the first $d$ derivatives, meaning $$p_d^{(i)}(c) = f^{(i)}(c)$$ whenever $0\le i\le d$. The exact same idea is true for functions of several variables. Let’s explain the construction of the Taylor polynomial as an iterative process. Given $F:\R ^2\to \R$ (and similarly for functions $F:\R ^n\to \R$) the degree zero Taylor polynomial is just the value of the function $$P_0(x,y) = F(c_1,c_2)$$ where $\vec {c}=\vector {c_1,c_2}$ is the center of the Taylor polynomial. The degree one Taylor polynomial is just the degree zero polynomial plus the first partial derivatives with respect to $x_i$ multiplied by $(x_i-c_i)$ $$P_1(x,y) = P_0(x,y) + F^{(1,0)}(\vec {c})(x-c_1) + F^{(0,1)}(\vec {c})(y-c_2).$$ The degree two Taylor polynomial can be found by adding the degree one Taylor polynomial to one-half of all the second partial derivatives with respect to $x$ and $y$ multiplied by

• $(x-c_1)^2$ when taking the partial derivative with respect to $x$,
• $(y-c_2)^2$ when taking the partial derivative with respect to $y$, and
• $2(x-c_1)(y-c_2)$ when taking the partial derivative with respect to $x$ and $y$.

Putting this together, we see

$$\begin{align*} P_2(x,y) &= P_1(x,y) \\ &+\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c})(x-c_1)^2 \\ &+F^{(1,1)}(\vec{c})(x-c_1)(y-c_2)\\ &+\left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2. \end{align*}$$ The interested reader can (repeatedly) differentiate $P_2(x,y)$ to see that its value at $\vec {x}=\vec {c}$ and the values of the first two derivatives of $P_2(x,y)$ do indeed match those of $F(x,y)$.

Unpacking the general formula

This final section is for the interested student, and is not required for this course.

Recall that if $F:\R ^n\to \R$ is a function whose first $d$ derivatives exist at $\vec {x}=\vec {c}$. The Taylor polynomial of degree $d$ of $F$ centered at $\vec {x}=\vec {c}$ is $$P_d(\vec {x}) = \sum _{k=0}^d \eval {\eval {\frac {(\vec {a}\dotp \grad )^k F(\vec {x})}{k!}}_{\vec {x}=\vec {c}}}_{\vec {a}=\vec {x}-\vec {c}}$$

This formula is complex and will take some unpacking. We’ll walk you through this.

The degree zero Taylor polynomial

First note that

$$\begin{align*} P_0(\vec{x}) &= \sum_{k=0}^0 \eval{\eval{\frac{(\vec{a}\dotp \grad)^k F(\vec{x})}{k!}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=\eval{\eval{\frac{(\vec{a}\dotp \grad)^0 F(\vec{x})}{0!}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=\eval{\eval{F(\vec{x})}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=F(\vec{c}). \end{align*}$$ This means for any function $F:\R ^n\to \R$, the $0$th degree Taylor polynomial for $F$ at $\vec {x}=\vec {c}$ is just $$P_0(\vec {x})=F(\vec {c}).$$

The degree one Taylor polynomial

Now let’s look at the $1$st degree Taylor polynomial:

$$\begin{align*} P_1(\vec{x})&= \sum_{k=0}^1 \eval{\eval{\frac{(\vec{a}\dotp \grad)^k F(\vec{x})}{k!}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=P_0(\vec{x}) + \eval{\eval{\frac{(\vec{a}\dotp \grad)^1 F(\vec{x})}{1!}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=F(\vec{c}) + \eval{\eval{(\vec{a}\dotp \grad) F(\vec{x})}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}} \end{align*}$$ Now we ask, what is $(\vec {a}\dotp \grad )F$? Well, computing the dot product, $$(\vec {a}\dotp \grad ) = a_1 \pp {x} + a_2 \pp {y}$$ and to find $(\vec {a}\dotp \grad )F$, we distribute $F$ to obtain $$\begin{align*} (\vec{a}\dotp\grad)F &= a_1 \pp[F]{x} + a_2 \pp[F]{y}\\ &= \grad F(\vec{x})\dotp \vec{a} \end{align*}$$ So $$\begin{align*} P_1(\vec{x})&=F(\vec{c}) + \eval{\eval{\grad F(\vec{x})\dotp\vec{a}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}\\ &=F(\vec{c}) +\grad F(\vec{c})\dotp (\vec{x}-\vec{c}). \end{align*}$$ This means for any function $F:\R ^n\to \R$, the $1$st degree Taylor polynomial for $F$ at $\vec {x}=\vec {c}$ is just the tangent “plane” for $F$ at $\vec {x}= \vec {c}$.

The degree two Taylor polynomial

To get our hands on the $2$nd degree Taylor polynomial, we will specialize to functions $F:\R ^2\to \R$. Let $\vec {c}=\vector {c_1,c_2}$ and let $\vec {x} = \vector {x,y}$. Write with me: $$P_2(\vec {x}) = P_1(\vec {x}) + \eval {\eval {\frac {(\vec {a}\dotp \grad )^2 F(\vec {x})}{2!}}_{\vec {x}=\vec {c}}}_{\vec {a}=\vec {x}-\vec {c}}$$

$$\begin{align*} = F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+(1/2)\eval{\eval{(\vec{a}\dotp \grad)^2 F(\vec{x})}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}} \end{align*}$$ Now we ask ourselves, what is $(\vec {a}\dotp \grad )^2 F(\vec {x})$? Well, we know that $$(\vec {a}\dotp \grad )F = a_1 \pp [F]{x} + a_2 \pp [F]{y}$$ and $$\begin{align*} (\vec{a}\dotp\grad)^2F &=(\vec{a}\dotp\grad)(\vec{a}\dotp\grad)F\\ &= (\vec{a}\dotp\grad)\left(a_1 \pp[F]{x} + a_2 \pp[F]{y}\right)\\ &= \left(a_1 \pp[F]{x} + a_2 \pp[F]{y}\right)\left(a_1 \pp[F]{x} + a_2 \pp[F]{y}\right) \end{align*}$$ Now we use the distributively property and (since we are assuming that all derivatives of $F$ exist) we have that $$\frac {\partial ^2F}{\partial x\partial y} = \frac {\partial ^2F}{\partial y\partial x}$$ so $$(\vec {a}\dotp \grad )^2F = a_1^2\frac {\partial ^2F}{\partial x^2} + 2a_1 a_2\frac {\partial ^2F}{\partial x\partial y} + a_2^2\frac {\partial ^2F}{\partial y^2}.$$ We can now unpack our second degree Taylor polynomial: $$\begin{align*} P_2(\vec{x})= F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+(1/2)\eval{\eval{(\vec{a}\dotp\grad)^2F}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}, \end{align*}$$ as $$\begin{align*} P_2(\vec{x})=F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+(1/2)\eval{\eval{a_1^2\frac{\partial^2F}{\partial x^2} + 2a_1 a_2\frac{\partial^2F}{\partial x\partial y} + a_2^2\frac{\partial^2F}{\partial y^2}}_{\vec{x}=\vec{c}}}_{\vec{a}=\vec{x}-\vec{c}}. \end{align*}$$ and finally we see: $$\begin{align*} P_2(\vec{x})=F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c}) (x-c_1)^2 \\ &+ F^{(1,1)}(\vec{c})(x-c_1)(y-c_2) \\ &+ \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2. \end{align*}$$

Whew. That was a lot of work. However, as we have said before, all you need to know right now, is how to compute the degree two Taylor polynomial for functions of two variables.