most recent 30 from http://mathoverflow.net 2013-06-19T12:21:02Z http://mathoverflow.net/feeds/user/10243 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84221/variable-centric-logical-foundation-of-calculus Jason Howald 2011-12-24T15:23:22Z 2012-12-07T11:34:20Z

<p>Since calculus originated long before our modern function concept, much of our language of calculus still focuses on variables and their interrelationships rather than explicitly on functions. For example, in the assertion "If $y=x^2$ then $\frac{dy}{dx}=2x$," the functions $f$ and $f'$ remain unnamed while the variables $x$ and $y$ take center stage. We interpret this as notational finesse, but there seems to be an important philosophical difference between what we say and what we mean.</p> <p>I have sometimes wondered: Is there an alternate logical foundation of Calculus in which variables, expressions, and equations are the central ideas, and functions per se are implicit?</p>

http://mathoverflow.net/questions/115416/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/115581#115581 Jason Howald 2012-12-06T19:35:47Z 2012-12-06T19:35:47Z

I am grateful to Joel for his support of the question, including this interesting answer. Certainly $\frac{d}{dx}$ is similar to a quantifier: It &quot;shields&quot; occurrences of the variable $x$ in its scope from direct substitution. It is defined in terms of the limit, which also binds a variable, as a quantifier could. It is a very strange quantifier, though, as $x$ once again occurs free in the (&quot;bound&quot;?) expression $\frac{d}{dx} x^3$ since $\frac{d}{dx} x^3 = 3x^2$.

http://mathoverflow.net/questions/115416/if-d-dx-is-an-operator-on-what-does-it-operate Jason Howald 2012-12-04T16:42:25Z 2012-12-04T16:42:25Z

Thank you for the criticisms. I have rephrased to clarify my intended meaning.