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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?

Thanks!


Edit:

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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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." –  Thomas Andrews Feb 28 '13 at 16:13
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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 $$ \nabla_{\mathbf{v}} f = \mathbf{v} \cdot \nabla f$$ –  Hurkyl Mar 1 '13 at 9:03
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2 Answers

up vote 1 down vote accepted

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 $g'(0)$.

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It's worse than that; his definition doesn't even make sense from a notational viewpoint! –  Noldorin Feb 28 '13 at 16:14
    
thanks, now I understand ... :-) –  OukiDouki Feb 28 '13 at 16:30
    
Please, look at the edit above. I want to know if I get it right. (Sorry for asking to obvious/stupid question.) –  OukiDouki Mar 1 '13 at 9:01
    
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 $$\lim_{h\to 0+}$$ –  Thomas Andrews Mar 1 '13 at 12:14
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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.

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sorry, corrected –  OukiDouki Feb 28 '13 at 15:57
    
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. –  Noldorin Feb 28 '13 at 16:02
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