Multivariable calculus and vector analysis

Talyor polynomial example

Calculate the second-degree Taylor polynomial of

$$\begin{array}{lll}\hfill f\left(x,y\right)=e^{-\left(x^2+y^2\right)}& & \hfill \end{array}$$

at the point $\left(0,0\right)$ and at the point $\left(1,2\right)$

Solution: The second-degree Taylor polynomial at the point $\left(a,b\right)$ is

$$\begin{array}{llll}\hfill p_2\left(x,y\right)& =f\left(a,b\right)+Df\left(a,b\right)\cdot \left(\left(x,y\right)-\left(a,b\right)\right)& \hfill & \\ \hfill & +\frac{1}{2}\left[\begin{array}{cc}\hfill x-a\hfill & \hfill y-b\hfill \end{array}\right]Hf\left(a,b\right)\left[\begin{array}{c}\hfill x-a\hfill \\ \hfill y-b\hfill \end{array}\right]& \hfill & \end{array}$$

First compute all the derivatives:

$$\begin{array}{llll}\hfill \frac{\partial f}{\partial x}\left(x,y\right)& =-2x{e}^{-\left(x^2+y^2\right)}& \hfill & \\ \hfill \frac{\partial f}{\partial y}\left(x,y\right)& =-2y{e}^{-\left(x^2+y^2\right)},& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x^2}\left(x,y\right)& =\left(-2+4x^2\right)e^{-\left(x^2+y^2\right)}& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial y^2}\left(x,y\right)& =\left(-2+4y^2\right)e^{-\left(x^2+y^2\right)}& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x\partial y}\left(x,y\right)=\frac{{\partial }^{2}f}{\partial y\partial x}\left(x,y\right)& =4xy{e}^{-\left(x^2+y^2\right)}& \hfill & \end{array}$$

At the point $\left(a,b\right)=\left(0,0\right)$

$$\begin{array}{llll}\hfill f\left(0,0\right)& =e^0=1& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x^2}\left(0,0\right)& =-2,& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial y^2}\left(0,0\right)& =-2& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x\partial y}\left(0,0\right)& =0& \hfill & \end{array}$$

The second-degree Taylor polynomial at the point (0,0) is

$$\begin{array}{llll}\hfill p_2\left(x,y\right)& =f\left(0,0\right)+Df\left(0,0\right)\cdot \left(\left(x,y\right)-\left(0,0\right)\right)& \hfill & \\ \hfill & +\frac{1}{2}\left[\begin{array}{cc}\hfill x-0\hfill & \hfill y-0\hfill \end{array}\right]Hf\left(0,0\right)\left[\begin{array}{c}\hfill x-0\hfill \\ \hfill y-0\hfill \end{array}\right]& \hfill & \\ \hfill & =1+\left(0,0\right)\cdot \left(x,y\right)+\frac{1}{2}\left[\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill -2\end{array}\right]\left[\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right]& \hfill & \\ \hfill & =1-x^2-y^2& \hfill & \end{array}$$

At the point $\left(a,b\right)=\left(1,2\right)$, things get uglier. But it is good practice.

$$\begin{array}{llll}\hfill f\left(1,2\right)& =e^{-5}& \hfill & \\ \hfill \frac{\partial f}{\partial x}\left(1,2\right)& =-2e^{-5}& \hfill & \\ \hfill \frac{\partial f}{\partial y}\left(1,2\right)& =-4e^{-5},& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x^2}\left(1,2\right)& =2e^{-5}& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial y^2}\left(1,2\right)& =14e^{-5}& \hfill & \\ \hfill \frac{{\partial }^{2}f}{\partial x\partial y}\left(1,2\right)=\frac{{\partial }^{2}f}{\partial y\partial x}\left(1,2\right)& =16e^{-5}& \hfill & \end{array}$$

The second-degree Taylor polynomial at the point (1,2) is

$$\begin{array}{llll}\hfill & p_2\left(x,y\right)=f\left(1,2\right)+Df\left(1,2\right)\cdot \left(\left(x,y\right)-\left(1,2\right)\right)+\frac{1}{2}\left[\begin{array}{cc}\hfill x-1\hfill & \hfill y-2\hfill \end{array}\right]Hf\left(1,2\right)\left[\begin{array}{c}\hfill x-1\hfill \\ \hfill y-2\hfill \end{array}\right]& \hfill & \\ \hfill & =e^{-5}+e^{-5}\left(-2,-4\right)\cdot \left(x-1,y-2\right)+\frac{1}{2}\left[\begin{array}{cc}\hfill x-1\hfill & \hfill y-2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2e^{-5}& \hfill 16e^{-5}\\ \hfill 16e^{-5}& \hfill 14e^{-5}\end{array}\right]\left[\begin{array}{c}\hfill x-1\hfill \\ \hfill y-2\hfill \end{array}\right]& \hfill & \\ \hfill & =e^{-5}-e^{-5}\left(2\left(x-1\right)+4\left(y-2\right)\right)+e^{-5}{\left(x-1\right)}^2+e^{-5}16\left(x-1\right)\left(y-2\right)+e^{-5}7{\left(y-2\right)}^2& \hfill & \\ \hfill & =e^{-5}\left(1-2\left(x-1\right)-4\left(y-2\right)+{\left(x-1\right)}^2+16\left(x-1\right)\left(y-2\right)+7{\left(y-2\right)}^2\right)& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

2d. Applications of differential calculus

• Taylor's theorem
• Taylor polynomial example
• Local extrema
• Extrema examples

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material