I am thinking about the directional derivative. I think that the easiest way how to express it is

\begin{equation} \frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{||v|| \rightarrow 0} \frac{f(\mathbf{x} + \mathbf{v})-f(\mathbf{x})}{||\mathbf{v}||}, \end{equation}

but the directional derivative is usually defined as

\begin{equation} \frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{h \rightarrow 0} \frac{f(\mathbf{x} + h \mathbf{v})-f(\mathbf{x})}{h}. \end{equation}

Can you rigorously explain the transition from first and second definition?



Just to make it clear.

The first definition is primarily wrong because the orientation of the directional vector is not fixed. So if I correct it like this (switching to conventional notation)

\begin{equation} \nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{||\mathbf{v}|| \to 0} \frac{f(\mathbf{x} + ||\mathbf{v}||\mathbf{\hat{v}})-f(\mathbf{x})}{||\mathbf{v}||}, \end{equation}

it makes a little bit more sense ($ \mathbf{\hat{v}}$ denotes unit vector). BUT the norm allows to get close to zero just from right side ($ ||\mathbf{v}|| \to 0+ $) and the limit makes sense even for vector reversal (~ negative norm). So we can actually use any scalar $h$ scaling the vector and write the derivative as

\begin{equation} \nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{\hat{v}})-f(\mathbf{x})}{h} \end{equation}

or you can find it equivalently written as

\begin{equation} \nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{v})-f(\mathbf{x})}{h||\mathbf{v}||}. \end{equation}

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    $\begingroup$ If it is a "directional" derivative, in what direction is the derivative. If you wanted the derivative in the "direction" of $(1,1,2)$, how would your definition do that? Your definition doesn't restrict $v$ to be "in that direction." $\endgroup$ – Thomas Andrews Feb 28 '13 at 16:13
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    $\begingroup$ As an aside, if you want to allow non-unit $\mathbf{v}$, you get a better operator if you don't normalize: for example, you want to have $$\nabla_{2 \mathbf{v}} f = 2 \nabla_{\mathbf{v}} f$$ An example of the benefit is that this version still satisfies the formula $$ \nabla_{\mathbf{v}} f = \mathbf{v} \cdot \nabla f$$ $\endgroup$ – user14972 Mar 1 '13 at 9:03

Your definition is not "directional," since $v$ in your definition can be in any direction. Note, if you were just in $1$ dimension, the limit $$\lim_{\delta\to 0} \frac{f(x+\delta)-f(x)}{|\delta|}$$ isn't even defined when $f$ is normally differentiable, because when $\delta\to 0-$ the limit is $-f'(x)$and when $\delta\to 0+$ the limit is $f'(x)$.

Essentially, the "directional derivative" can be seen as taking $\mathbf x$ and $\mathbf v$ and defining a new function on the real numbers: $g(h) = f(\mathbf x + h\mathbf v)$. This $g$ computes the value of $f$ on the line through $\mathbf x$ in the direction of $\mathbf v$, and the directional derivative of $f$ at $\mathbf x$ in the direction $\mathbf v$ is defined as $g'(0)$.

  • $\begingroup$ It's worse than that; his definition doesn't even make sense from a notational viewpoint! $\endgroup$ – Noldorin Feb 28 '13 at 16:14
  • $\begingroup$ thanks, now I understand ... :-) $\endgroup$ – OukiDouki Feb 28 '13 at 16:30
  • $\begingroup$ Please, look at the edit above. I want to know if I get it right. (Sorry for asking to obvious/stupid question.) $\endgroup$ – OukiDouki Mar 1 '13 at 9:01
  • $\begingroup$ That still doesn't deal with the case of $h<0$. You are allowing, which is not the same as my definition. For example, in your definition, $f(\mathbf v)=|\mathbf v|$ has direction derivative at $\mathbf v=0$, but in my definition, it does not. That's because you have $|v|\to 0$. Also, your definition is confusing since nothing make $\mathbf v$ be in the same vector as $\hat{\mathbf{v}}$ - you are essentially using $\hat{\mathbf{v}}$ in your limit as a stand-in for a positive real - you might as well say $$\lim_{h\to 0+}$$ $\endgroup$ – Thomas Andrews Mar 1 '13 at 12:14

Your first definition makes no sense really, since you're using $\Delta \mathbf{x}$ simultaneously as a parameter and variable (which you're taking a limit over). The second (correct) definition thus uses a separate variable over which the limit is taken, used to scale the positional vector.

  • $\begingroup$ sorry, corrected $\endgroup$ – OukiDouki Feb 28 '13 at 15:57
  • $\begingroup$ It still doesn't make sense I'm afraid; the RHS is what doesn't work, as explained above. As for the LHS, you're using unconventional notation (the nabla symbol is almost always used), but I get what you mean. $\endgroup$ – Noldorin Feb 28 '13 at 16:02

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