If d/dx is an operator, on what does it operate? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:38:18Z http://mathoverflow.net/feeds/question/115416 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate If d/dx is an operator, on what does it operate? Jason Howald 2012-12-04T16:21:35Z 2012-12-07T07:41:42Z <p>If $\frac{d}{dx}$ is a differential operator, what are its inputs? If the answer is "(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs), we have difficulty distinguishing between $\frac{d}{dx}$ and $\frac{d}{dt}$, which in practice have different meanings. If the answer is "(differentiable) functions of $x$", what does that mean? It sounds like a peculiar hybrid of mathematical object (function) with mathematical notation (variable $x$). </p> <p>Does $\frac{d}{dx}$ have an interpretation as an operator, distinct from $\frac{d}{dt}$, and consistent with its use in first-year Calculus?</p> http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115473#115473 Answer by Bazin for If d/dx is an operator, on what does it operate? Bazin 2012-12-05T07:14:31Z 2012-12-05T07:14:31Z <p>Two answers: (1) Distribution theory. On the space $\mathcal D'(\mathbb R)$ of continuous linear forms on $\mathcal D(\mathbb R)=C_c^\infty(\mathbb R)$ it is easy to define the first derivative: $$\langle\frac{du}{dx},\phi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}= -\langle u,\frac{d\phi}{dx}\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}.$$ You get the ordinary derivative of a differentiable function, also $H'=\delta$ ($H$ is the Heaviside function, characteristic function of $\mathbb R_+$, $\delta$ the Dirac mass), $$\frac{d}{dx}(\ln \vert x\vert)=\text{pv}\frac{1}{x}$$ and many other classical formulas. In particular, you can define the derivative of any $L^1_{loc}$ function, of course not pointwise but as above.</p> <p>(2) Operator theory. In $L^2(\mathbb R)$, you consider the subspace $H^1(\mathbb R)=${$u\in L^2(\mathbb R), u'\in L^2(\mathbb R)$}, where the derivative is taken in the distribution sense. Then the operator $d/dx$ is an unbounded operator with domain $H^1(\mathbb R)$. It is even possible to prove that the operator $\frac{d}{idx}$ is selfadjoint.</p> http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115485#115485 Answer by David Roberts for If d/dx is an operator, on what does it operate? David Roberts 2012-12-05T09:18:10Z 2012-12-05T23:25:28Z <p>Edit: obviously some people didn't realise this answer was tongue-in-cheek. Also, I read the question differently to others, given its ambiguity, and didn't bother with the last (possibly most crucial) part of the question.</p> <p>The gist of my answer was an expansion of Sam Gunningham's comment, namely that the operator "$\frac{\mathrm{d}}{\mathrm{d}x}$" is actually the restriction of a functor on the category of pointed smooth manifolds and pointed maps, to the subcategory consisting of 1-dimensional vector spaces over $\mathbb{R}$. The idea of coordinate-independence is captured in the <a href="http://ncatlab.org/nlab/show/principle+of+equivalence" rel="nofollow">principle of equivalence</a> (the violation of which used to jokingly be called "evil" by some people), in that mathematics can't tell the difference between diffeomorphic manifolds, and so whatever we call this operator, $\frac{\mathrm{d}}{\mathrm{d}x}$ or $\frac{\mathrm{d}}{\mathrm{d}t}$ or what-have-you, they are all naturally isomorphic, and so indistinguishable in the 1-variable case. I do take the point, hashed out in the comments below, that in the multi-variable case things are more subtle, and I cede to <a href="http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115494#115494" rel="nofollow">Golderstern's answer</a>.</p> <p>But the punchline is that the category of manifolds can be defined in many different ways: from material sets, from structural sets, via synthetic differential geometry or via Fermat theories, so I contend there is not a single answer to (my reading of the) question. </p> <hr> <p>I find the statement </p> <blockquote> <p>"(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs)</p> </blockquote> <p>exceeding peculiar. A differentiable function is a certain arrow in the category of smooth manifolds, and even better, it's a arrow in the category where objects are finite-dimensional $\mathbb{R}$-vector spaces $E^n$ (for all $n$) with the usual topology. The tangent bundle functor takes a smooth function $f\colon E^n \to E^m$ and returns a smooth function $df\colon TE^n \to TE^m$ (the tangent bundle of $E^n$ is diffeomorphic to $E^{2n}$, hence again a vector space). Let us say we are in the case $n=1$. We can restrict this function to the tangent space of $E^1$ at $0$ and get a smooth function $E \to E^m$. No coordinates were chosen here.</p> <p>But how did you get this category of manifolds? I hear you ask. Well, I started with the category of sets and did the usual thing. But how did this category of sets turn up? Well, to give the short answer, <a href="http://ncatlab.org/nlab/show/ETCS" rel="nofollow">ETCS</a>. The longer answer is that the category of sets (or rather, a category of sets strong enough to formalise all of undergraduate calculus and in fact most of mathematics) can be defined in terms of a first order theory. (Aside, if it irks you to miss out of the more hard-core parts of ZFC, use the foundational theory <a href="http://ncatlab.org/nlab/show/SEAR" rel="nofollow">SEAR-C</a> instead - it likewise doesn't define functions as sets of ordred pairs.)</p> <p>At no point did I define a function to be a set of ordered pairs, and everything is independent of choices of coordinates.</p> <p>Alternatively, we just say that $d/dx$ is an operation in the <a href="http://ncatlab.org/nlab/show/Fermat+theory" rel="nofollow">Fermat theory</a> of $C^\infty$-rings. In this sense, smooth functions can be seen as models for a theory which is far more focussed than set theory, and there is no flab, in that this theory <em>only</em> talks about smooth functions.</p> <p>[If you are asking questions that assume $df(t)/dt$ and $df(x)/dx$ are somehow distinguishable, and bringing foundational definitions into basic calculus, then expect answers that answer with a similar level of chutzpah]</p> http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115494#115494 Answer by Goldstern for If d/dx is an operator, on what does it operate? Goldstern 2012-12-05T11:54:22Z 2012-12-05T11:54:22Z <p>I repeat (a variant of) my comment, even though I agree that it is shallow and has low entertainment value. </p> <p>As long as we are only looking at functions in one variable, there is only one differential operator $D$, which may be called $\frac d{dx}$ or $\frac{d}{dt}$ depending on the context. </p> <p>If you look at a composite function $f \circ g$, you may introduce the notation/abbreviation $x=g(t)$, $y=f(x)$, then</p> <ul> <li>$\frac {d}{dx} f$ or $\frac d {dx} y$ is just $D(f)$, </li> <li>and by $\frac{d}{dt} f$ or $\frac d{dt} y$ you mean $D(f\circ g)$. </li> </ul> <p>So here both $\frac {d}{dx}$ and $\frac {d}{dt}$ have a meaning, and the meaning is different. </p> <p>When we look at functions in, say, two variables (do they appear in first year calculus?), we implicitly introduce an (arbitrary) order of variables, say x is the first and t the second, and $\frac{\partial}{\partial x}$ is the partial derivative with respect to the first variable. This makes sense even if you treat functions as "variable-agnostic" sets of ordered pairs. (Which I do all the time, and do not find peculiar at all. Tastes differ.) </p> <p>Of course, the intended meaning always depends on the context. If $f$ is a binary function, $\frac d {dt} f$ may be a variant notation for $\frac{\partial}{\partial t}f$, or it may be understood that we are really looking at a unary function $\hat f$ obtained by composing $f$ with some function $t \mapsto (x(t), y(t))$.</p> http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115581#115581 Answer by Joel David Hamkins for If d/dx is an operator, on what does it operate? Joel David Hamkins 2012-12-06T06:58:41Z 2012-12-06T06:58:41Z <p>(From the post <a href="http://jdh.hamkins.org/the-differential-operator-ddx-binds-variables/" rel="nofollow">on my blog</a>:)</p> <p>To my way of thinking, this is a serious question, and I am not really satisfied by the other answers and comments, which seem to answer a different question than the one that I find interesting here.</p> <p>The problem is this. We want to regard $\frac{d}{dx}$ as an operator in the abstract senses mentioned by several of the other comments and answers. In the most elementary situation, it operates on a functions of a single real variable, returning another such function, the derivative. And the same for $\frac{d}{dt}$.</p> <p>The problem is that, described this way, the operators $\frac{d}{dx}$ and $\frac{d}{dt}$ seem to be the <em>same</em> operator, namely, the operator that takes a function to its derivative, but nevertheless we cannot seem freely to substitute these symbols for one another in formal expressions. For example, if an instructor were to write $\frac{d}{dt}x^3=3x^2$, a student might object, "don't you mean $\frac{d}{dx}$?" and the instructor would likely reply, "Oh, yes, excuse me, I meant $\frac{d}{dx}x^3=3x^2$. The other expression would have a different meaning."</p> <p>But if they are the same operator, why don't the two expressions have the same meaning? Why can't we freely substitute different names for this operator and get the same result? What is going on with the logic of reference here?</p> <p>The situation is that the operator $\frac{d}{dx}$ seems to make sense only when applied to functions whose independent variable is described by the symbol "x". But this collides with the idea that what the function is at bottom has nothing to do with the way we represent it, with the particular symbols that we might use to express which function is meant. That is, the function is the abstract object (whether interpreted in set theory or category theory or whatever foundational theory), and is not connected in any intimate way with the symbol "$x$". Surely the functions $x\mapsto x^3$ and $t\mapsto t^3$, with the same domain and codomain, are simply different ways of describing exactly the same function. So why can't we seem to substitute them for one another in the formal expressions?</p> <p>The answer is that the syntactic use of $\frac{d}{dx}$ in a formal expression involves a kind of binding of the variable $x$.</p> <p>Consider the issue of <em>collision</em> of bound variables in first order logic: if $\varphi(x)$ is the assertion that $x$ is not maximal with respect to $\lt$, expressed by $\exists y\ x\lt y$, then $\varphi(y)$, the assertion that $y$ is not maximal, is not correctly described as the assertion $\exists y\ y\lt y$, which is what would be obtained by simply replacing the occurrence of $x$ in $\varphi(x)$ with the symbol $y$. For the intended meaning, we cannot simply syntactically replace the occurrence of $x$ with the symbol $y$, if that occurrence of $x$ falls under the scope of a quantifier.</p> <p>Similarly, although the functions $x\mapsto x^3$ and $t\mapsto t^3$ are equal as functions of a real variable, we cannot simply syntactically substitute the expression $x^3$ for $t^3$ in $\frac{d}{dt}t^3$ to get $\frac{d}{dt}x^3$. One might even take the latter as a kind of ill-formed expression, without further explanation of how $x^3$ is to be taken as a function of $t$.</p> <p>So the expression $\frac{d}{dx}$ causes a binding of the variable $x$, much like a quantifier might, and this prevents free substitution in just the way that collision does. But the case here is not quite the same as the way $x$ is a bound variable in $\int_0^1 x^3\ dx$, since $x$ remains free in $\frac{d}{dx}x^3$, but we would say that $\int_0^1 x^3\ dx$ has the same meaning as $\int_0^1 y^3\ dy$.</p> <p>Of course, the issue evaporates if one uses a notation, such as the $\lambda$-calculus, which insists that one be completely explicit about which syntactic variables are to be regarded as the independent variables of a functional term, as in $\lambda x.x^3$, which means the function of the variable $x$ with value $x^3$. And this is how I take several of the other answers to the question, namely, that the use of the operator $\frac{d}{dx}$ indicates that one has previously indicated which of the arguments of the given function is to be regarded as $x$, and it is with respect to this argument that one is differentiating. In practice, this is almost always clear without much remark. For example, our use of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ seems to manage very well in complex situations, sometimes with dozens of variables running around, without adopting the onerous formalism of the $\lambda$-calculus, even if that formalism is what these solutions are essentially really about.</p> <p>Meanwhile, it is easy to make examples where one must be very specific about which variables are the independent variable and which are not, as Todd mentions in his comment to David's answer. For example, cases like</p> <p>$$\frac{d}{dx}\int_0^x(t^2+x^3)dt\qquad \frac{d}{dt}\int_t^x(t^2+x^3)dt$$</p> <p>are surely clarified for students by a discussion of the usage of variables in formal expressions and more specifically the issue of bound and free variables.</p> http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate/115684#115684 Answer by Dirk for If d/dx is an operator, on what does it operate? Dirk 2012-12-07T07:41:42Z 2012-12-07T07:41:42Z <p>I am a little late to the question but wanted to add a low-tech answer which somehow complements JDH's answer:</p> <blockquote> <p>The operators $\frac{d}{dx}$ and $\frac{d}{dt}$ are as distinguishable as $f(x)$ and $f(t)$.</p> </blockquote> <p>Probably this formulation is a little too vague but it should just reflect that writing $d/dx$ says how one has named the free variables. As already illustrated, one gets into notational ambiguities in cases as $\frac{d}{dt} f(t,x(t))$...</p>