3
$\begingroup$

I have two definitions of a differential of a map $f: \mathbb{R}^n \rightarrow \mathbb{R}$ at $x$ denoted as $df_x$.

  1. $df_x(h) = \sum_{i = 1}^n \frac{\partial f (x)}{\partial x_i}h_i$
  2. $df_x(h) = \lim_{t \rightarrow 0} \frac{f(x + ht) - f(x)}{t}$

How do I go about showing this equivalence? Intuitively the equivalence makes sense. If we take $n = 2$, the partials in each direction give a sense of how much $f$ changes in each respective dimension, so (1) makes sense. I can also see how (2) gives us the differential if we think of the change of a surface in the direction of $h$.

I started with (2) and began by saying $$df_x(h) + E(t) = \frac{f(x + ht) - f(x)}{t}$$ such that $\lim_{t \rightarrow 0} E(t) = 0$. This gives us $f(x + h) = f(x) + df_x(h)t + o(t)$. I then tried to show that $f(x + h) = f(x) + (\sum_{i = 1}^n \frac{\partial f(x)} + {\partial x_i}h_i) + o(t)$ but I'm not sure if this helps and I am stuck here.

$\endgroup$
3
  • 1
    $\begingroup$ There's a crucial piece of information missing: The differential is linear as a function of $h$. Does that help? $\endgroup$ May 14 at 13:19
  • $\begingroup$ @AndrewD.Hwang - Thank you for the response. How do you know that it is linear? It is clear from (1) but not from (2). Is there a way we can show linearity with (2)? $\endgroup$ 2 days ago
  • $\begingroup$ In this setting, $f$ mapping is differentiable at $x$ if there exists a linear transformation $df(x)$ such that $$f(x+h)=f(x)+df(x)(h)+o(|h|).$$This cannot be deduced this algebraically from your conditions, but must be assumed. $\endgroup$ 2 days ago

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.