Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From Wikipedia:

The natural analog of second, third, and higher-order total derivatives is not a linear transformation, is not a function on the tangent bundle, and is not built by repeatedly taking the total derivative.

I think the $n$-th order (total) derivative of a mapping is the derivative of its $n-1$-th order derivative. In other words, derivative can be defined recursively. So I was wondering how to understand that "the natural analog of second, third, and higher-order total derivatives ... is not built by repeatedly taking the total derivative"?

Thanks and regards!

share|improve this question
    
I just thought the same thing about that Wikipedia article. As it is it looks plain wrong and Didier's answer below corroborates this. –  Giuseppe Negro Mar 21 '12 at 0:51
add comment

1 Answer

As explained on the WP page, if $f:E\to F$ for finite dimensional vector spaces $E$ and $F$, then $f':E\to V$ where $V=L(E,F)$ is the space of linear maps from $E$ to $F$. Hence $f'':E\to W$ where $W=L(E,V)$ is the space of linear maps from $E$ to $V$, and so on.

In general, the vector spaces $V$, $W$ and their analogues for higher derivatives are all different, for example if $E$ and $F$ have dimensions $n$ and $m$, the dimension of $V$ is $nm$, the dimension of $W$ is $n^2m$, and so on, hence the objects $f'$, $f''$ and so on, are each in a different space. However, when $E=\mathbb R$, then for every finite dimensional vector space $U$ there is a natural identification of $L(E,U)=U^*$ with $U$ since every $\varphi$ in $L(\mathbb R,U)$ is $\varphi:x\mapsto xu$ for a given $u$ in $U$.

Using these identifications, when $E=\mathbb R$ and $F$ is finite dimensional, for each $n\geqslant1$, $f^{(n)}$ corresponds to a function from $\mathbb R$ to $L(\mathbb R,F)=F^*\simeq F$.

share|improve this answer
    
Thanks! (1) "f(n) corresponds to a function from R to F", probably you wanted to say "f(n) corresponds to a function from R to U"? (2) So the n-th order derivative is defined as the derivative of the (n-1)-th order derivative, isn't it? –  Tim Nov 16 '11 at 8:24
    
(3) Isn't derivative defined at least for mappings between Banach spaces, instead of between general vector spaces? –  Tim Nov 16 '11 at 8:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.