0

I am having a lot of difficulty understanding the given notations for Taylor Expansion for two variables, on a website they gave the expansion up to the second order:

$f(x,y)+[f_x+f_y]+\frac{1}{2!}[f_{xx}^2+2f_{xy}+f_{yy}^2]$

To continue the expansion, would it be correct to say:

$f(x,y)+[f_x+f_y]+\frac{1}{2!}[f_{xx}^2+2f_{xy}+f_{yy}^2]+\frac{1}{3!}[f_{xxx}^3+3f_{xy}+f_{yyy}^3]+...+\frac{1}{n!}[f_{x^n}^n+nf_{xy}+f_{y^n}^n]\space$ ?

EDIT: @user21820 thank you for providing a link, one of the answers for that question was

"For 3 variables: $$f(x,y,z)=f(x_0,y_0,z_0)$$ $$+\frac{\partial f_0}{\partial x}(x-x_0)+\frac{\partial f_0}{\partial y}(y-y_0)+\frac{\partial f_0}{\partial z}(z-z_0)\quad \Rightarrow Order 1$$ $$+\frac{1}{2} \bigg(\frac{\partial^2 f_0}{\partial x^2}(x-x_0)^2+\frac{\partial^2 f_0}{\partial y^2}(y-y_0)^2+\frac{\partial^2 f_0}{\partial z^2}(z-z_0)^2+2\frac{\partial^2 f_0}{\partial x\partial y}(x-x_0)(y-y_0) $$ $$+2\frac{\partial^2 f_0}{\partial x\partial z}(x-x_0)(z-z_0)+2\frac{\partial^2 f_0}{\partial z\partial y}(z-z_0)(y-y_0)\bigg)\quad \Rightarrow Order 2$$ And it goes like this to higher orders"

Given this follows the rules of a polynomial expansion like you stated above (i.e.. (1,1) then (1,2,1), etc) what would be the corresponding expansion for order 3?

mnmakrets
  • 251
  • 2
  • 6
  • 16
  • No. $(1,1)$ then $(1,2,1)$ then $(1,3,3,1)$ then $(1,4,6,4,1)$... – user21820 Aug 12 '15 at 15:28
  • And I'm not sure what your square brackets mean. It should be $f(a+x,b+y) = ( f + x f_x + y f_y + \frac{1}{2} ( x^2 f_{xx} + 2xy f_{xy} + y^2 f_{yy} ) + o(x^2) + o(y^2) )(a,b)$. – user21820 Aug 12 '15 at 15:32
  • So would it be $f(x,y)+[f_x+f_y]+\frac{1}{2!}[f_{xx}^2+2f_{xy}+f_{yy}^2]+\frac{1}{3!}[f_{xxx}^3+3f_{xy}+3f_{xy}+f_{yyy}^3]+\frac{1}{4!}[f_{xxxx}^4+4f_{xy}+6_f{xy}+4f_{xy}+f_{yyyy}^4+... $ and so on? Are the changes of $f_{x^n} $ correct for each term and order of expansion? – mnmakrets Aug 12 '15 at 15:36
  • See my second comment. Your terms do not make any sense to me. – user21820 Aug 12 '15 at 15:37
  • could you tell us the website on which you found the formula? – user251257 Aug 12 '15 at 15:45
  • You may want to try looking at http://math.stackexchange.com/q/331337/21820. – user21820 Aug 12 '15 at 16:01
  • Here is the website, http://people.rit.edu/pnveme/pigf/Derivatives/derv_pd_taylor.html – mnmakrets Aug 12 '15 at 16:10
  • @ user251257 Please see my comment above, I had to put it in an edit to my question above because it was exceeding the character limit here. – mnmakrets Aug 12 '15 at 16:21

1 Answers1

4

$$f({\bf x})=\left.\sum_{n=0}^{\infty} \frac 1 {n!}\left(\sum_{i=1}^{m}(x_i-x_{i_0})\frac {\partial}{\partial x_i} \right)^n f({\bf x})\right|_{{\bf x}={\bf x}_{0}} $$

where $m$ is the number of independent variables. In your case (around $(x,y)=(0,0)$), it simplifies to: $$f(x,y)=\sum_{n=0}^{\infty} \frac 1 {n!}\left(x\frac {\partial}{\partial x}+y\frac {\partial}{\partial y} \right)^n f(x,y)|_{(x,y)=(0,0)} $$ $$=f(0,0)+(xf_x(0,0)+yf_y(0,0))+\frac 1 2(x^2f_{xx}(0,0)+2xyf_{xy}(0,0)+y^2f_{yy}(0,0)) \dots$$

Ura Nogirag
  • 3
  • 2
GeorgSaliba
  • 4,611
  • 1
  • 18
  • 29
  • Im uncertain with a part of the sum notation you stated, what would be the expansion up to the 3rd or 4th order? – mnmakrets Aug 12 '15 at 16:25
  • @mnmakrets you can imagine it like $(a+b)^n$ and you're using Pascal's triangle to get the coefficients and powers (here partial derivatives) for $a$ and $b$. – GeorgSaliba Aug 12 '15 at 16:28
  • Your final expression makes no sense: you have $f(x,y)$ on the let and on the right, you don't have have any $x^iy^j$ monomials, and you're not evaluating the partial derivatives at $(0,0)$. Also, your coefficients of the second order terms are off. – Santiago Canez Aug 12 '15 at 16:28
  • @SantiagoCanez remember that $\partial / \partial x_i$ is an operator, and when raised to a certain power it means you're applying it several times. Plus I do have $x^i y^j$, the term in parentheses is raised to the $n$, so you have to apply the binomial formula to see it. – GeorgSaliba Aug 12 '15 at 16:31
  • @GeorgSaliba so for a taylor expansion for two variables up to a 3rd order be: $...+...+( f_{xx}+3f_{xy}+3f_{xy}+f_{yy} )$ ? – mnmakrets Aug 12 '15 at 16:32
  • Your final line has no monomial terms and is the expression I am referring to. – Santiago Canez Aug 12 '15 at 16:32
  • @SantiagoCanez oh sorry, I miscopied, I'll fix it right away. – GeorgSaliba Aug 12 '15 at 16:33
  • @mnmakrets no it should be $\frac 1 {3!}(x^3f_{xxx}+3x^2yf_{xxy}+3y^2xf_{yyx}+y^3f_{yyy})$ – GeorgSaliba Aug 12 '15 at 16:35
  • Still not correct. The $f(x,y)$ should be $f(0,0)$, each of the partial derivatives should also be evaluated at $(0,0)$, and each of the second-order terms should be multiplied by $\frac12$. – Santiago Canez Aug 12 '15 at 16:36
  • @SantiagoCanez now? – GeorgSaliba Aug 12 '15 at 16:39
  • Yes, this looks good. – Santiago Canez Aug 12 '15 at 16:39
  • Thank you, Im sorry for being so slow on this topic, but then for the expansion up to a 4th order would it be: $...+...+ \frac{1}{4!} (x^4f_{xxxx}+4x^3y f_{xxxy}+6xy f_{xxyy}+4y^3x f_{yyyx}+ y^4 f_{yyyy} )$ ? – mnmakrets Aug 12 '15 at 17:47
  • @mnmakrets it's okay, I should be sorry for being so hasty as to commit those mistakes. The middle term should be $6x^2y^2f_{xxyy}$ – GeorgSaliba Aug 12 '15 at 17:50