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I am readig Pugh's Analysis book:

Definition

Let $f:U \to \mathbb{R}^m$ be given where $U$ is an open subset of $\mathbb{R}^n$. The function $f$ is differentiable a $p \in U$ with derivative $(Df)_p = T$ if $T:\mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) \implies \lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$.

Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$\implies$" there rather than a "$\wedge$". Isn't it more natural to say

"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $\lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$"?

I'm having trouble seeing what the impact of changing these would be.

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    $\begingroup$ It would probably be clearer to write "...is a linear transformation and $\lim_{|v| \to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$. $\endgroup$ – Bungo Dec 18 '18 at 0:29
  • $\begingroup$ @Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $\implies$? Or to the one with $\wedge$? Or is the one with $\wedge$ nonsense? I have a vague feeling that $\implies$ and $\wedge$ won't matter because the derivative is unique, but I don't really understand it. $\endgroup$ – Ovi Dec 18 '18 at 0:35
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The author means

$$\forall R \left((\forall v ~ f(p+v) = f(p) + T(v) + R(v)) \implies \lim_{|v| \to 0} \frac{R(v)}{|v|} = 0\right)$$

which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like

$$ \bigg(R = v \mapsto f(p + v) - f(p) - T(v)\bigg) \land \bigg(\lim_{|v| \to 0} \frac{R(v)}{|v|} = 0\bigg)$$

The problem with the second equation is that it isn't defining $R$ as $v \mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,

$$\begin{cases} \text{define } R \text{ as } v \mapsto f(p + v) - f(p) - T(v) \\ \lim_{|v| \to 0} \frac{R(v)}{|v|} = 0 \end{cases}$$

is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.

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  • $\begingroup$ Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more. $\endgroup$ – Ovi Dec 18 '18 at 1:12
  • $\begingroup$ It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability? $\endgroup$ – Ovi Dec 18 '18 at 1:14
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    $\begingroup$ Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$\lim_{|v| \to 0} \frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$ $\endgroup$ – DanielV Dec 18 '18 at 1:36
  • $\begingroup$ If I could ask you another thing; can I formulate it like this? $f: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $p$ if $\exists T \exists R \forall v : \left( \dfrac{}{} f(p+v) = f(p)+T(v)+R(v) \right) \wedge \left( \lim_{v \to 0} \dfrac {|R(v)|}{|v|} = 0 \right) $? $\endgroup$ – Ovi Dec 20 '18 at 16:55
  • $\begingroup$ @Ovi You should take the $\exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $\exists y~(a = f(y) \land y = g(b))$ $\endgroup$ – DanielV Dec 20 '18 at 19:43
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I can't speak for Pugh, but your construction

T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $\lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$

is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.

Look at it this way, would you write the following?

T is the derivative if $\lim_{|v| \to 0} \dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$

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  • $\begingroup$ Thank you for the response! $\endgroup$ – Ovi Dec 18 '18 at 1:17

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