tag:blogger.com,1999:blog-264226589944705290.post4736520502979747784..comments2021-12-14T05:53:12.175-08:00Comments on God Plays Dice: Product rules for more complicated productsMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-264226589944705290.post-77582249998464220092008-03-09T21:18:00.000-07:002008-03-09T21:18:00.000-07:00In response to blaisepascal, although I don't know...In response to blaisepascal, although I don't know if you will later see this:<BR/><BR/>$+$ is a kind of "or", where for natural number arithmetic we're counting ways of doing something. You should think of $(u+v)$ as "you can do $u$ or $v$". $\times$ corresponds to "do one thing and then the next" (think of these as functions or matrices; "or" is naturally commutative, but "and then" is not). In any case, $(u+v)^n$ is "do $u$ or $v$, and then do $u$ or $v$, and then..." But $uv = vu$, so the binomial coefficient counts exactly the number of ways of doing that many $u$s and $v$s.<BR/><BR/>Now if "do $u$" is "differentiate $u$", then we get the coefficients on $(uv)^{(n)}$.<BR/><BR/>Let me give a different proof. If $U = e^{ux}$ and $V = e^{vx}$ for constants $u$ and $v$, then $UV = e^{(u+v)x}$, and differentiation corresponds to multiplication: $(UV)^{(n)} = (u+v)^n UV$. If you believe in Fourier, then you think that every function can be written as a sum of exponentials, and since both sides of the (general) Leibniz formula are linear in $U$ and $V$, the result follows.Theohttps://www.blogger.com/profile/03344294173628793721noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-11993216619521363782008-01-27T22:39:00.000-08:002008-01-27T22:39:00.000-08:00The extension of this formula also works for the p...The extension of this formula also works for the partial redivatives, with multiindexes instead of indexes, Lars Hormander calls it the Leibniz rule (see any of his books on PDEs). As for the presence of the binomial coefficients, it's rather natural, and can be proven by induction on the order of the differentiation, the same way that the binomial formula is.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-63982708152686219052008-01-24T14:33:00.000-08:002008-01-24T14:33:00.000-08:00I had never known that extension of the product ru...I had never known that extension of the product rule, but I'm glad I know it now.<BR/><BR/>I can't help but notice the similarity between the expansion of $(uv)^{(n)}$ and the expansion of $(u+v)^n$. Is there an underlying connection between them?<BR/><BR/>(and how do I do math notations in comments,anyway?)Buddha Buckhttps://www.blogger.com/profile/17167036913705912859noreply@blogger.com