Timeline for Is there a reason why integrals are so much easier to evaluate than sums?
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| Feb 19, 2011 at 21:21 | comment | added | Richard Stanley | Kevin, as an example of what I mean consider $$ \int_0^1 x^2dx =\lim_{n\to\infty}\frac 1n\sum_{k=1}^n\left( \frac kn \right)^2 = \lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}. $$ It is trivial to compute that the limit on the right is 1/3. Thus the sum is harder to compute than the integral; once we know the sum, the integral is easy (but not conversely). | |
| Feb 19, 2011 at 5:13 | comment | added | Kevin O'Bryant | "integrals are easier since if we can evaluate the sums explicitly then the limit is straightforward" --- that's a nonsequitor. The conclusion should be: "integrals are often not much harder". Or perhaps I misunderstand your whole point... | |
| Feb 18, 2011 at 17:39 | history | answered | Richard Stanley | CC BY-SA 2.5 |