Taylor polynomials

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We introduce Taylor polynomials for functions of several variables.

Recall the definition of a Taylor polynomial:

Let $f:\R \to \R$ be a function whose first $d$ derivatives exist at $x=c$ . The Taylor polynomial of degree $d$ of $f$ centered at $x=c$ is $p_d(x) = \sum _{k=0}^d\frac {f^{(k)}(c)}{k!}(x-c)^k.$

Let $f(x) = \sin (x)$ . Compute the degree $7$ Taylor polynomial centered at $x=0$ . $p_7(x)\begin {prompt} = \answer {x-x^3/3!+x^5/5!-x^7/7!} \end {prompt}$

Let $f(x) = \cos (x)$ . Compute the degree $7$ Taylor polynomial centered at $x=0$ . $p_7(x)\begin {prompt} = \answer {1-x^2/2+x^4/4!-x^6/6!} \end {prompt}$

Let $f(x) = e^x$ . Compute the degree $7$ Taylor polynomial centered at $x=0$ . $p_7(x)\begin {prompt} = \answer {1 + x + x^2/2+ x^3/3! + x^4/4!+ x^5/5! + x^6/6!+x^7/7!} \end {prompt}$

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

Let $F:\R ^n\to \R$ be 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 $P_d(\vec {x}) = \sum _{k=0}^d \eval {\eval {\frac {(\vec {a}\dotp \grad )^k F(\vec {x})}{k!}}_{\vec {x}=\vec {c}}}_{\vec {a}=\vec {x}-\vec {c}}$

Don’t panic! 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 $P_0(\vec {x})=F(\vec {c}).$

Consider $F(x,y)= \sin (x+y)$ . Compute the $0$ th degree Taylor polynomial for $F$ at $\vector {x,y} =\vector {\pi /4,\pi /4}$ . $P_0(x,y) = \answer {1}$

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 $\vec {x}= \vec {c}$ .

Consider $F(x,y)= 3+4x-5y$ . Compute the $1$ st degree Taylor polynomial for $F$ at $\vector {x,y} =\vector {0,0}$ . $P_1(x,y) = \answer {3+4x-5y}$

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: $\begin{align*} P_2(\vec{x})=F(\vec{c}) &+ \grad F(\vec{c})\dotp (\vec{x}-\vec{c})\\ &+\left(\frac{1}{2}\right)F^{(2,0)}(\vec{c}) (x-c_1)^2 \\ &+ F^{(1,1)}(\vec{c})(x-c_1)(y-c_2) \\ &+ \left(\frac{1}{2}\right)F^{(0,2)}(\vec{c})(y-c_2)^2. \end{align*}$

Try it, you might like it

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

Compute the degree $2$ Taylor polynomial for: $F(x,y)=\sin (xy)$ centered at $(0,0)$ .

We’ll start by making a table of partial derivatives along with their value when evaluated at $(0,0)$ : $\begin {array}{lcl} F(x,y) = \sin (xy) & \Rightarrow &F(0,0) = 0\\ F^{(1,0)}(x,y) = \answer [given]{y\cos (xy)} & \Rightarrow & F^{(1,0)}(0,0) = \answer [given]{0}\\ F^{(0,1)}(x,y) = \answer [given]{x\cos (xy)} &\Rightarrow & F^{(0,1)}(0,0) = \answer [given]{0}\\ F^{(2,0)}(x,y) = \answer [given]{-y^2\sin (xy)} &\Rightarrow & F^{(2,0)}(0,0) = \answer [given]{0}\\ F^{(0,2)}(x,y) = \answer [given]{-x^2\sin (xy)} &\Rightarrow & F^{(0,2)}(0,0) = \answer [given]{0}\\ F^{(1,1)}(x,y) = \answer [given]{\cos (xy)-xy\sin (xy)} &\Rightarrow & F^{(1,1)}(0,0) = \answer [given]{1} \end {array}$ We see our polynomial is just $\begin{align*} P_2(x,y) &= \answer[given]{1}\cdot (x-0)\cdot (y-0)\\ &= \answer[given]{xy}. \end{align*}$

Compute the degree $2$ Taylor polynomial for: $F(x,y)= x^2 y + y^2 + x y$ centered at $(-1,0)$ .

We’ll start by making a table of partial derivatives along with their value when evaluated at $(-1,0)$ : $\begin {array}{lcl} F(x,y) = x^2y+y^2+xy & \Rightarrow &F(-1,0) = 0\\ F^{(1,0)}(x,y) = \answer [given]{2xy+y} & \Rightarrow & F^{(1,0)}(-1,0) = \answer [given]{0}\\ F^{(0,1)}(x,y) = \answer [given]{x^2+2y+x} &\Rightarrow & F^{(0,1)}(-1,0) = \answer [given]{0}\\ F^{(2,0)}(x,y) = \answer [given]{2y} &\Rightarrow & F^{(2,0)}(-1,0) = \answer [given]{0}\\ F^{(0,2)}(x,y) = \answer [given]{2} &\Rightarrow & F^{(0,2)}(-1,0) = \answer [given]{2}\\ F^{(1,1)}(x,y) = \answer [given]{1+2x} &\Rightarrow & F^{(1,1)}(-1,0) = \answer [given]{-1} \end {array}$ We see our polynomial is $\begin{align*} P_2(x,y) &= (1/2)\left(\answer[given]{2}\cdot (y-0)^2 + 2 (\answer[given]{-1})(x-(-1))(y-0) \right)\\ &= \answer[given]{y^2-(x+1)y}. \end{align*}$

Compute the degree $2$ Taylor polynomial for: $F(x,y)= x^3-3x-y^2+4y$ centered at $(-1,2)$ .

We’ll start by making a table of partial derivatives along with their value when evaluated at $(-1,2)$ : $\begin {array}{lcl} F(x,y) = x^3-3x-y^2+4y & \Rightarrow &F(-1,2) = 6\\ F^{(1,0)}(x,y) = \answer [given]{3x^2-3} & \Rightarrow & F^{(1,0)}(-1,2) = \answer [given]{0}\\ F^{(0,1)}(x,y) = \answer [given]{-2y+4} &\Rightarrow & F^{(0,1)}(-1,2) = \answer [given]{0}\\ F^{(2,0)}(x,y) = \answer [given]{6x} &\Rightarrow & F^{(2,0)}(-1,2) = \answer [given]{-6}\\ F^{(0,2)}(x,y) = \answer [given]{-2} &\Rightarrow & F^{(0,2)}(-1,2) = \answer [given]{-2}\\ F^{(1,1)}(x,y) = \answer [given]{0} &\Rightarrow & F^{(1,1)}(-1,2) = \answer [given]{0} \end {array}$ We see our polynomial is $\begin{align*} P_2(x,y) &= 6 + (1/2)\left(\answer[given]{-6}\cdot (x-(-1))^2 + (\answer[given]{-2})(y-2)^2 \right)\\ &= \answer[given]{6-3(x+1)^2-(y-2)^2}. \end{align*}$

Compute the degree $2$ Taylor polynomial for: $F(x,y)=\cos (xy)$ centered at $(0,0)$ .

Start by making a table of partial derivatives along with their value when evaluated at $(0,0)$ .

$P_2(x,y) = \answer {1}$

Compute the degree $2$ Taylor polynomial for: $F(x,y)= x^2 y + y^2 + x y$ centered at $(-1/2,1/8)$ .

Start by making a table of partial derivatives along with their value when evaluated at $(-1/2,1/8)$ .

$P_2(x,y) = \answer {-1/64 + (1/8)(x+1/2)^2+(y-1/8)^2}$

Compute the degree $2$ Taylor polynomial for: $F(x,y)= x^3-3x-y^2+4y$ centered at $(1,2)$ .

Start by making a table of partial derivatives along with their value when evaluated at $(1,2)$ .

$P_2(x,y) = \answer {2 + 3(x-1)^2 -(y-2)^2}$

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 $F(x,y)$ .