Taylor series third order approximation

There has been this question that had been bothering me for a while and I could not find a satisfying answer on the internet or any of the books even though it is not very complex.

i) Its because if I have to find a third order polynomium approximation using taylor series for a 2 variable function, then is it correct to write that the third term will look something like this,

$$... + \frac{1}{3!}[f_{xxx}(x_0,y_0)(x-a)^3 + 6f_{xxy}(x_0,y_0)(x-a)(y-b)+f_{yyy}(x_0,y_0)(y-b)^3] + ....$$

I was a bit unsure about the middle part.

ii) About the hessian matrix, how would I write a hessian matrix if I have to make one for a third order like the one above. I know that for second order it looks like,

$$H_f(x,y) = \left(\begin{array}{cccc} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array}\right)$$

Thank You :)

2 Answers

i) Looks like you want: $$\frac{1}{6} \left(3 (x-a)^2 (y-b) f_{\text{xxy}}\left(x_0,y_0\right)+3 (x-a) (y-b)^2 f_{\text{xyy}}\left(x_0,y_0\right)+(x-a)^3 f_{\text{xxx}}\left(x_0,y_0\right)+(y-b)^3 f_{\text{yyy}}\left(x_0,y_0\right)\right)$$

ii) There's no third order "hessian matrix"

1st order derivative -> 2D vector

2nd order derivative -> 2x2 symmetric matrix

3rd order derivative -> 2x2x2 symmetric tensor.

Check wiki link.

The third order term (it can be genralized to any order term) of the multivariable Taylor series of $$f(\mathbf x):\mathbf R^n\to\mathbf R$$ around $$\mathbf a=(a,b,...)$$ is \begin{aligned} \frac{1}{3!}((\mathbf x-\mathbf a)\cdot\boldsymbol\nabla|_{\mathbf a})^3f(\mathbf x)&\overset{\text{in }\mathbf R^2}{=}\frac{1}{3!}((x-a)\partial_x|_{\mathbf a}+(y-b)\partial_y|_{\mathbf a})^3f(x,y)\\ &\ \ =\frac{1}{3!}((x-a)^3\partial_{xxx}|_{\mathbf a}+3(x-a)^2(y-b)\partial_{xxy}|_{\mathbf a}\\ &\hspace{1.5cm}+3(x-a)(y-b)^2\partial_{xyy}|_{\mathbf a}+(y-b)^3\partial_{yyy}|_{\mathbf a})f(x,y)\\ &\ \ = \frac{1}{3!}[(x-a)^3f_{xxx}(a,b)+3(x-a)^2(y-b)f_{xxy}(a,b)\\ &\hspace{1.5cm}+3(x-a)(y-b)^2f_{xyy}(a,b)+(y-b)^3f_{yyy}(a,b)] \end{aligned} or in matrix language, though in your case it is a $$2\times 2\times 2$$ tensor, I found it could be expressed as (example in $$\mathbf R^2$$) $$\frac{1}{3!}(x-a,y-b)\begin{pmatrix} (\mathbf{x-a})\cdot\boldsymbol\nabla f_{xx}(\mathbf a) & (\mathbf{x-a})\cdot\boldsymbol\nabla f_{xy}(\mathbf a)\\ (\mathbf{x-a})\cdot\boldsymbol\nabla f_{yx}(\mathbf a) & (\mathbf{x-a})\cdot\boldsymbol\nabla f_{xx}(\mathbf a) \end{pmatrix} \begin{pmatrix} x-a\\ y-b \end{pmatrix}.$$ Using Schwarz theorem ($$f_{xy}=f_{yx}$$) you can get to your expression.