1

Context:

As I understand this limit, it can be interpreted as a Riemann Sum with tag $c_i=\frac{(2i-1)}{n}$.

$$\lim_{n\to\infty}\sum_{i=1}^n \frac{2}{n}\left(1+\frac{2i-1}{n}\right)^\frac{1}{3}$$

Therefore it represents: $$\int_0^2(1+x)^\frac{1}{3}dx$$

If $x_i=\frac{2i}{n}$ then $c_i$ falls neatly between $x_{i-1}$ and $x_i$:

$$x_{i-1}=\frac{(2i-2)}{n}<c_i<\frac{2i}{n}=x_i$$

So in that sense it is a valid tag to be able to interpret it as a Riemann Sum.

Question:

How about this limit?

$$\lim_{n\to\infty}\sum_{i=1}^n \frac{2}{n}\left(1+\frac{2i-1000}{n}\right)^\frac{1}{3}$$

Clearly the tag doesn't fall between $x_{i-1}$ and $x_i$ anymore. Still I wonder: does this matter if $n\to\infty$? Can we still interpret this limit as the same definitive integral?

I noticed that when calculating this limit in Mathematica it gives exactly the same answer as with the limit above...

GambitSquared
  • 3,662
  • We can prove the limit is the same, but I know no general principle one could appeal to, to show this. – Did Sep 04 '17 at 08:11
  • You are likely to get a better response if you use standard mathematical terminology that everybody understands, and avoid slangy terms like "tag". – bof Sep 04 '17 at 08:14
  • @bof What term for "tag" would you suggest then? I took this term from "Calculus, a complete course" by Adams and Essex. – GambitSquared Sep 04 '17 at 08:16
  • @bof "tag" is standard, look for "tagged partition". – Did Sep 04 '17 at 08:26
  • If I needed a name for the points where the function is being evaluated, I'd probably just call them "evaluation points". Not very clever, maybe not as fashionable as "tag", but more self-explanatory. But I stand corrected, the term "tag" really exists, it's my fault if I don't know it. – bof Sep 04 '17 at 08:51
  • @Did OK, but it's a neologism, right? Probably only goes back to the 60s, when a lot of things started to go wrong. – bof Sep 04 '17 at 08:54
  • @bof A neologism dating back from the 15th century then, if I am to believe M-W (scroll down to "tag" the verb). So, maybe "a lot of things started to go wrong" back then, actually? – Did Sep 04 '17 at 08:59

1 Answers1

1

(2017.09.16) Is it me or is the plague of silent revenge downvotes spreading? Yet one more...

Let us prove a more general result, to make as salient as possible the relevant features of the situation.

Assume that some function $f$ is continuous on $[0,1]$ and bounded on $[-\epsilon,1+\epsilon]$, for some positive $\epsilon$, and consider, for every fixed $c$, $$S_n^c(f)=\frac1n\sum_{k=1}^nf\left(\frac{k+c}n\right)$$ If $c=0$, each $S_n^0(f)$ is simply a Riemann sum of $f$ on $[0,1]$ and $f$ is Riemann integrable on $[0,1]$ with integral $I(f)$, say, hence $S_n^0(f)\to I(f)$ when $n\to\infty$.

If $c\ne0$, $S_n^c(f)$ involves the value of $f$ at some points not in $[0,1]$, but, for every $n$ large enough (say, every $n\geqslant|c|/\epsilon$), these points are in $[-\epsilon,1+\epsilon]$ hence $S_n^c(f)$ is well defined. Furthermore, assuming that $c$ is a positive integer to simplify the proof (but the result is general), one sees that $$S_n^c(f)=S_n^0(f)+\frac1n\sum_{k=n+1}^{n+c}f\left(\frac{k}n\right)-\frac1n\sum_{k=1}^cf\left(\frac{k}n\right)$$ hence $$|S_n^c(f)-S_n^0(f)|\leqslant\frac1n\sum_{k=n+1}^{n+c}\left|f\left(\frac{k}n\right)\right|+\frac1n\sum_{k=1}^c\left|f\left(\frac{k}n\right)\right|\leqslant2c\sup|f|$$ which suffices to show that every such sequence of "shifted Riemann sums" $S_n^c(f)$ also converges to $I(f)$ when $n\to\infty$.

The sums in the question correspond to $2S_n^c(f)$, for $$f(x)=(1+2x)^{1/3}\qquad c=-500$$

Did
  • 279,727