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.

I came across a statement which says that $$F(x+e t) = F(x) + ct $$, for all $x \in \mathbb R^m$ where $c$ is constant , and $e = (1,1,....,1)$ is equivalent to saying that $$\sum_{j=1}^m \frac {\partial^2 F}{\partial x_i \partial x_j}(x) =0 ..\forall i \in 1,...., m , x \in \mathbb R^m$$

Here $F$ is twice differentiable function .

Can someone help me to see that both of them are equivalent statements . Thanks

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Using the chain-rule, we get $$ \sum_{j=1}^m\frac{\partial}{\partial x_j}F(x)=e\cdot\nabla F(x)=\frac{\partial}{\partial t}F(x+et)=c\tag{1} $$ Taking $\dfrac{\partial}{\partial x_i}$ of $(1)$ gives your result.

share|improve this answer

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.