I need a little bit of help with a question.
Whats the second degree taylor polynomial (taylor vector quadratic) for the function
f(x) = x1^3 x2^2 about the point [2, 3] <------- Vector
Thanks if anyone can help
It is rather difficult to understand what function you are referring to. Perhaps you mean:
$\displaystyle f(x_1,x_2)=x_1^3x_2^2$
At any rate:
$\displaystyle p(x_1,x_2)=f(2,3)+\dfrac{{\partial f}}{{\partial x_1}}(2,3)(x_1-2)+\dfrac{{\partial f}}{{\partial x_2}}(2,3)(x_2-3)+$
$\displaystyle \dfrac{{\partial^2 f}}{{\partial x_1}^2}(2,3)(x_1-2)^2+2\dfrac{{\partial^2 f}}{{\partial x_1\partial x_2}}(2,3)(x_1-2)(x_2-3)+\dfrac{{\partial^2 f}}{{\partial x_2}^2}(2,3)(x_2-3)^2
=\ldots$
Regards.
Fernando Revilla
Another way to do it:
A polynomial already is a Taylor's polynomial. To get it around (2, 3) (which is NOT a vector), let $\displaystyle u= x_1- 2$ and $\displaystyle v= x_2- 3$, so that $\displaystyle x_1= u+ 2$ and $\displaystyle x_2= v+ 3$. Then $\displaystyle x_1^3x_2^2= (u+ 2)^3(v+ 3)^2= (u^3+ 6u^2+ 12u+ 8)(v^2+ 6v+ 9)$$\displaystyle = u^3(v^2+ 6v+ 9)+ 6u^2(v^2+ 6v+ 9)+ 12u(v^2+ 6v+ 9)+ 8(v^2+ 6v+ 9)$$\displaystyle = u^3v^2+ 26u^2+v+ 6u^3v+ 9u^3+ 6u^2v^2+ 36u^2v+ 54u^2v+ 12uv^2+ 72uv+ 108u+ 8v^2+ 48v+ 72$.
To get the second degree polynomial, just drop all terms where the exponents sum to 3 or larger, then replace u and v with $\displaystyle x_1- 2$ and $\displaystyle x_2- 3$ again.