I read in a book that the second order Taylor expansion of a function (around $x^0$) can be written as:
$$f(x)=f(x^0)+\sum_{j=1}^n df(x^0)/dx_j*(x_j-x_j^0)+\sum_{j=1}^n\sum_{i=1}^nd^2f(x^1)/dx_idx_j*(x_j-x_j^0)(x_i-x_i^0)$$
With $x^1=ax^0+(1-a)x$, for some $a \in (0,1)$. (Note that $x^1$ only affects the second derivative)
This seems weird to me because any textbook I have read says that I should have $x_0$ instead of $x_1$ (and also a remainder term). Is this formula correct? How does it relate to the well-known formula?
The reference is Intriligator (2002, p. 466), "Mathematical Optimization and Economic Theory". Furthermore, in page 22 the author suggests that this taylor expansion includes the remainder term.
Best Answer
The formula as given is more or less correct (a factor of ${1\over2}$ missing), modulo the absolutely awful notation.
Note that for a function $f:\>x\mapsto f(x)$ of one variable and a given point $a$ we have $$f(x)=f(a)+f'(a)(x-a)+{1\over2}f''(\xi)(x-a)^2\tag{1}$$ with $\xi$ between $a$ and $x$, i.e., $\xi=a+\tau(x-a)$ for some $\tau\in\>]0,1[\>$. The $n$-dimensional version of this is $$f(x)=f(a)+\sum_{k=1}^n f_{.k}(a)(x_k-a_k)+{1\over2}\sum_{j,\>k}f_{.jk}(\xi)(x_j-a_j)(x_k-a_k)\ ,$$ with $\xi=a+\tau(x-a)$ for some $\tau\in\>]0,1[\>$. The proof is via applying $(1)$ with $a:=0$ and $x=1$ to the auxiliary function $$\phi(t):=f\bigl(a+t(x-a)\bigr)\qquad(0\leq t\leq1)\ ,$$ and using the chain rule to compute $\phi'(0)$, $\phi''(\tau)$.