Timeline for Interesting applications (in pure mathematics) of first-year calculus
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 11, 2010 at 23:03 | comment | added | John Stillwell | Nice proof, Per. I also like to see applications of combinatorics to first-year calculus :) | |
| Aug 11, 2010 at 16:22 | comment | added | Per Vognsen | The combinatorial proof is more conceptual and almost as short. The left-hand side is the data type for $(A \cup B)$-multisets. The right-hand side is the data type for pairs of $A$-multisets and $B$-multisets. Obviously they are isomorphic. (You can now take images to analytic power series.) This is essentially the same as the combinatorial proof of the binomial formula but without the messy detour through binomial coefficients. | |
| Aug 11, 2010 at 3:50 | comment | added | John Stillwell | Ryan, thanks; I think that's the formal argument behind my rough idea. | |
| Aug 11, 2010 at 1:45 | comment | added | Ryan Reich | John, by "same computation" do you mean the following: we would like to interpret $e^A e^B = e^{A + B}$ as an identity in formal power series with commuting variables $A, B$. To do so, we observe the map, given by Taylor series, from two-variable analytic real functions to $R[[A,B]]$, contains both sides of the equation, and that the equation holds of the actual functions by means of some calculus argument. | |
| Aug 11, 2010 at 0:34 | history | edited | John Stillwell | CC BY-SA 2.5 |
Corrected "exponential"
|
| Aug 11, 2010 at 0:34 | comment | added | Victor Protsak | This proof is slick, but I am not convinced of its merits: are you proposing to prove an $\textit{algebraic}$ identity using a functional interpretation? In principle (although not in this particular instance), it could happen that different algebraic expressions evaluate to the same function (if this had happened for numbers, but not for matrices, it wouldn't have worked). Perhaps, I am misunderstanding what you mean when you say "This follows, without actually computing the two sides, by the same computation as for real numbers". | |
| Aug 10, 2010 at 23:28 | history | answered | John Stillwell | CC BY-SA 2.5 |