Sometimes I read that directional derivatives for function in one variables are right and left derivatives. But this doesn't make sense to me. The only unit vector in $\mathbb{R}$ are $\pm1 $ , so we are saying that: $$D_1f(x)=f'_+(x)=\lim_{t\to 0^+} \frac{f(x+t)-f(x)}{t}$$ $$D_{-1}f(x)=f'_-(x)=\lim_{t\to 0^-} \frac{f(x+t)-f(x)}{t}$$ But by definition: $$D_{\mathbf{v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$ By this logic: $$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}-t\mathbf{v})-f(\mathbf{x})}{t}$$ I can substitute $-t=u$ ($-t$ doesn't assume infinite times the value $0$ so I can apply composite function limit theorem), so: $$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{-u}=-\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{u}=-D_{\mathbf{v}} f(\mathbf{x})$$ This should mean that: $$f'_+(x)=-f'_-(x)$$ That in general is false! Wouldn't be more correct to say that by convention in one variable we use derivative along the unit vector $1$ and that right and left derivatives are simply the right and left directional derivatives along the unit vector $1$. Thanks in advance.
- $\begingroup$ If $f'_+(x)=-f'_-(x)$ is false, the function $f$ is not differentiable at $x$. $\endgroup$K.defaoite– K.defaoite2020-11-29 17:25:27 +00:00Commented Nov 29, 2020 at 17:25
- 2$\begingroup$ Left and right derivative and directional derivative have somewhat different purposes. You can define the left and right directional derivative along the vector $\pmb{v}$ as $\lim_{t\to0\pm}\frac{f(\pmb{x}+t\pmb{v})-f(\pmb{x})}{t}$. The two limits may be different, indicating a discontinuity in that particular direction. This definition generalizes (and is very useful) on differential manifolds. We have of course that the left derivative w.r.t. $\pmb{v}$ is the same as the right derivative w.r.t. $-\pmb{v}$. $\endgroup$pglpm– pglpm2020-11-29 19:39:27 +00:00Commented Nov 29, 2020 at 19:39
- 1$\begingroup$ Sorry I meant "the left derivative w.r.t. $\pmb{v}$ is the same as minus the right derivative w.r.t. $-\pmb{v}$". So, as I see it, in 1D the directional derivative is a somewhat trivial notion, yet it isn't the same as the left or right derivative. But in 1D too we have $\partial_{\pmb{v}^+}f = -\partial_{(-\pmb{v})^-}f$. $\endgroup$pglpm– pglpm2020-11-29 19:47:54 +00:00Commented Nov 29, 2020 at 19:47
- 1$\begingroup$ On top of that, the directional derivative is not limited to unit vectors – this is its main point, since unit vectors are undefined in a generic vector space or on a differential manifold. This notion is somewhat related to that of velocity. So in a generic 1D vector space the left and right derivative are always specific definitions of a directional derivative – we must specify the 1D vector along which the derivation takes place, since there's no unit vector. $\endgroup$pglpm– pglpm2020-11-29 19:57:21 +00:00Commented Nov 29, 2020 at 19:57
- 1$\begingroup$ I agree, but I think that we cannot write this using "$\ne$": this expresses that they're the same kind of object but have different values, which is not what we're saying here. $\endgroup$pglpm– pglpm2020-11-29 20:08:45 +00:00Commented Nov 29, 2020 at 20:08
2 Answers
You are right. Left & right derivatives and directional derivative are different notions – they coexist in fact, also in one-dimensional spaces such as $\mathbf{R}$.
The directional derivative can be defined on a generic vector- or affine space or on a differential manifold, without the existence of a norm or scalar product. It's therefore useful not only in many dimensions, but also in one dimension when the notion of "unit vector" is undefined.
More precisely we define, at a point $\pmb{x}$, the right directional derivative $\partial_{\pmb{v}^+}f$ of a function along the vector $\pmb{v}$ as $$\lim_{t\to 0+} \frac{f(\pmb{x}+t\pmb{v}) - f(\pmb{x})}{t}$$ or equivalently $$\lim_{n\to +\infty} \frac{f(\pmb{x}+\pmb{v}/n) - f(\pmb{x})}{1/n}.$$ The left directional derivative is defined identically except for $t \to 0-$ or $n\to -\infty$. From these definitions we have $\partial_{\pmb{v}^+}f \equiv -\partial_{-\pmb{v}^-}f$ (so I'm not sure about your last statement "this in general is false" – it depends on how you're defining $f_{\pm}'(x)$). Note how this limit doesn't involve any scalar product or norm in the space of $\pmb{v}$.
If both left and right directional derivatives along $\pmb{v}$ exist and are the same, then we just speak of the directional derivative along $\pmb{v}$. In differential geometry it is also denoted simply as "$\pmb{v}(f)$". We can have an interesting interplay of these two notions on the boundary of a manifold with boundary, even a 1D one, where it can happen that both $\pmb{v}$ and $-\pmb{v}$ can be defined at the boundary and yet we can only speak of the left or right derivative with respect to one or the other.
Intuitively and informally speaking, the directional derivative tells us something about the rate of change of a function as we move on the vector/affine space or manifold along some direction with some velocity, represented together by the vector $\pmb{v}$. The left and right derivatives tell us something about possible discontinuities in such rate of change.
So no, it isn't true that "directional derivatives for function in one variables are right and left derivatives", but we can say that the left derivative is minus the right derivative with respect to $-\pmb{e}$, where $\pmb{e}\equiv 1$ is the canonical unit vector, and vice versa.
For derivatives on vector spaces and differential manifolds, also with boundary, see eg Choquet-Bruhat, DeWitt-Morette, Dillard-Bleick's Analysis, Manifolds and Physics. Part I: Basics (rev. ed., Elsevier 1996), and Curtis, Miller's Differential Manifolds and Theoretical Physics (Academic Press 1985). There are of course many other good books out there on these topics.
The directional derivative is not the same as the limit approaching from the left or the right. If you use the negative unit vector, you're asking what the slope is if you're traveling to the right vs. to the left. You get the negative of the slope if you use the negative unit vector. But that is correct. The divide by the magnitude of v ($|v|$) is superfluous if you restrict v to be a unit vector.
$$\nabla_v f(x) = \nabla f(x) . \frac{v}{|v|}$$
- $\begingroup$ Thank you for the contribution, but what should it mean dividing by the vector tv ? I simply used the definition of directional derivative in that step. $\endgroup$Kandinskij– Kandinskij2020-11-29 17:40:00 +00:00Commented Nov 29, 2020 at 17:40
- $\begingroup$ Should have used the $|v|$ below but that = 1 for the unit vector. $\endgroup$IanJ– IanJ2020-11-29 17:58:47 +00:00Commented Nov 29, 2020 at 17:58
- $\begingroup$ No offense, but your answer is a bit nonsensical to me. $\endgroup$Kandinskij– Kandinskij2020-11-29 18:09:54 +00:00Commented Nov 29, 2020 at 18:09
- 1$\begingroup$ This is incorrect. Both the gradient $\nabla f$ and $v$ are vectors. You cannot simply multiply them. Perhaps you meant to take the dot product? $\endgroup$K.defaoite– K.defaoite2020-11-29 20:49:51 +00:00Commented Nov 29, 2020 at 20:49
- 1$\begingroup$ @K.defaoite the lack of multiplication sign can be seen as correct if $\nabla$ is intended as the covariant derivative, because in this case $\nabla f$ is a covector, which operates on a vector, the operation denoted by juxtaposition. But the division by $\lvert v\rvert$ is incorrect. $\endgroup$pglpm– pglpm2020-11-30 14:26:41 +00:00Commented Nov 30, 2020 at 14:26