Approximation of a sum with an integral...................

Question

Let $G$ a continuous function in $C([0,1], \mathbb R)$. I think that $$\frac{1}{N}\sum_{x \;\text{odd}\in \{1,\ldots, N\}}G\Big (\frac{x}{N}\Big )\xrightarrow{N\to +\infty}\frac{1}{2}\int_0^1G(r)dr,$$ and I would like to prove that. I think to do a change of variables as follows: $$\frac{1}{N}\sum_{x \;\text{odd}\in \{1,\ldots, N\}}G\Big (\frac{x}{N}\Big )=\frac{1}{N}\sum_{k=0}^{\frac{N-1}{2}}G\Big(\frac{2k+1}{N}\Big)=\frac{1}{2}\frac{1}{N}\sum_{z=1}^NG\Big(z\Big),$$ where in the last step I defined $z=2k+1$ and I applied a kind of change of variables for the series that usually holds for integrals ($dk=\frac{1}{2}dz$). Is that correct?

Answer 1

It would perhaps be worth recalling that for a sequence of partitions $ \Big \{ 0=y_0^{(m)}< y_1^{(m)}<....< y_{n^{(m)}}^{(m)} =1 \Big\}_{m=1}^\infty $ of $[0,1]$ such that $\underset{k\in n^{(m)}}{\min} \vert y_k^{(m)}-y_{k-1}^{(m)} \vert \overset {m\rightarrow \infty}{\rightarrow}0$, that:

$\underset{k=1}{ \overset{n^{(m)} }{\sum} } G(\zeta_k^{(m)}) \cdot (y_k^{(m)}-y_{k+1}^{(m)}) \overset{m\rightarrow \infty}{\rightarrow} \int_0^1 G(t) dt $

where $\zeta_k^{(m)}\in [y_k^{(m)}, y_{k+1}^{(m)}]$ for all $k\in \{1,...,n^{(m)} \}$.

Answer 2

The points $x/N$ (add $0,1$ if needed) for odd $x$ form a partition of $[0,1]$ with norm $2/N$ and hence the Riemann sum $$\frac{2}{N}\sum_{x\text{ odd}} G(x/N) $$ tends to $\int_{0}^{1}G(r)\,dr$ as $N\to\infty $ and the proof of the result in question in complete. Note that the result holds for all Riemann integrable functions $G$ and not just for continuous functions.

Answer 3

As stated in the comments above, the second formula in the original post is not correct. However, the statement about the limit is still true.

In the second formula, the use of the change-of-variables formula for differential forms is incorrect. The correct discrete analog of such a formula is as follows. Suppose you have values $z_0,\ldots,z_n$ in the domain and you want to calculate the right-hand Riemann sum:

$$S_n = \sum_{i=1}^n G(z_i) \Delta z_i$$

Where $\Delta z_i := z_i - z_{i-1}$. Then you may make a change of variable $z_i = 2k_i + 1$. The discrete difference then satisfies $\Delta k_i = \frac12 \Delta z_i$ so that $S_n = 2\cdot\sum_{i=1}^n G(2k_i+1)\Delta k_i$.

As for the limit formula, you may think of it as a midpoint Riemann sum ($1/N$ is midpoint from $0$ to $2/N$, $3/N$ is the midpoint from $2/N$ to $4/N$, etc.) except $1/N$ has been used when the intervals really have size $2/N$. (Technically the midpoint sum doesn’t work out for $N$ even but this missing term dies in the limit.)