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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.

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    $\begingroup$ It looks like the first order expansion. The term with $x_1$ is the remainder. $\endgroup$
    – A.Γ.
    Jul 1, 2016 at 4:12

1 Answer 1

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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 $\phi'(0)$, $\phi''(\tau)$.

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  • $\begingroup$ (just to be sure) Do I have to use the limit $\lim_{τ\to 0}\frac{(f(ξ)-f(a))(x−a)}{τ(x−a)}-\frac{1}{2} f''(ξ)(x-a)^2τ=f'(a)(x-a)$? I proceeded in a clumsy way until I obtained this equation, so I'm not sure if it is ok. $\endgroup$
    – Belisario
    Jul 1, 2016 at 23:10
  • $\begingroup$ What I wrote is the answer to your original question about a formula you found in a book. The formula in your comment does not make sense to me. The $\tau$ is not a variable we have control on. Taylor's theorem with the appropriate remainder just says that there exists such a value $\tau$. $\endgroup$ Jul 2, 2016 at 11:04

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