Use the previous exercise (1.11, #4 parts (d) and (e)) to deduce a formula for and prove this formula is correct.
Claim:
Proof. Let , then we have
Thus, there exists some with
such that
Hence, . Thus,
Then, for each with
we have
Then, for , we have
So,
Hmmm… the latex formatting seems to have gone badly awry on my last comment. You can get it as a PDF at https://hursts.org.uk/proof-%5Bnx%5D-2.pdf
Claim:
Proof:
(1)![Rendered by QuickLaTeX.com \begin{equation*} \left[x + \frac{k}{n}\right]= \begin{cases} [x] &\quad \mathrm{if}\> x x\geq [x] + \frac{n-k}{n}\\ \end{cases} \quad ;k\in\mathbb{Z},0\leq k\leq\> x \> [x]+\frac{j+1}{n}\quad ;j\in\mathbb{Z}, 0\leq j<n \end{equation*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d3b97e1e4614e8442166f0a309be4c75_l3.png)
This allows us to derive a closed form of the summation:
(2)![Rendered by QuickLaTeX.com \begin{equation*} \sum_{k=0}^{n-1}\left[ x+\frac{k}{n}\right] = \sum_{k=0}^{n-j-1}\left[ x+\frac{k}{n}\right] + \sum_{k=n-j}^{n-1}\left[ x+\frac{k}{n}\right] \end{equation*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8217528f06ca1c2f9f1bef20c32ac171_l3.png)
For the first term on the RHS of \eqref{e:3}
Applying \eqref{e:2} we get
(3)![Rendered by QuickLaTeX.com \begin{equation*} x\leq\> nx \> n[x]+j+1\quad \Rightarrow\quad [nx]=n[x] + j \end{equation*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-865b629386141595993e885572c8e519_l3.png)
Thus, for condition \eqref{e:2}
and since
is not involved in the formula, it is true
.
Thanks for the work your doing here. But I’ve not been able to figure out the introduction of j into into the proof, even after reading the explanation you gave in the comment section.Please can you like explain again how your going about your proof, like how you know where your headed….
Sorry to be commenting so much, but how do you know that you can choose j, such that 0 <= j < n in equation 3?
Hi, no problem, comment away! (Though if I get busy, responses could start to lag.)
We have
But we know for any number
that it is between
and
for exactly one integer
. But since
we know that integer
is somewhere between
and
, which means it is
for some integer
. So, then we have that
is between
and
. Does that make sense? Basically, we are just saying that we know
for some integer
, but then it is more convenient to write it as
since that will help us out later.
Not sure if this actually clears up your question. Let me know if not.