All Questions
13
questions
4
votes
1
answer
169
views
Prove that sum of integrals $= n$ for argument $n \in \mathbb{N}_{>1}$
ORIGINAL QUESTION (UPDATED):
I have a function $f:\mathbb{R} \rightarrow \mathbb{R}$ containing an integral that involves the floor function:
$$f(x):= - \lfloor x \rfloor \int_1^x \lfloor t \rfloor x \...
3
votes
1
answer
119
views
Does $\prod\limits_{n=1}^{\infty}\int\limits_{-1}^{1}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt$ converge to $1$?
This is the first $40$ partial products on Desmos (Desmos gave me "undefined" for everything higher):
Looking at this, it doesn't appear to converge but then, while attempting to get Desmos ...
- 1,688
5
votes
1
answer
179
views
What is $\int\limits_0^1\left\{\frac{1}{x\left\{\frac{1}{x}\right\}}\right\}dx$?
$\{x\}$ is the fractional part of $x$.
$\{x\}=x-\lfloor x\rfloor$
I ended up with this double summation:
$$\lim\limits_{\substack{a\to\infty\\b\to\infty}}\sum_{m=1}^{a}\sum_{n=0}^{b}\left(\frac{1}{m}\...
- 1,688
0
votes
1
answer
81
views
Prove that $\int_0^1\lfloor nx\rfloor^2 dx = \frac{1}{n}\sum_{k=1}^{n-1} k^2$
First of all apologies for the typo I made in an earlier question, I decided to delete that post and reformulate it
I am asked to prove that
$$\int_{(0,1)} \lfloor nx\rfloor^2\,\mathrm{d}x =\frac{1}{n}...
- 131
4
votes
1
answer
130
views
Solving the integral $\int\limits^{\infty}_{1}\frac{x-\sqrt{\lfloor x^2\rfloor}}{x}dx$
This problem was just for fun
Here was what I managed to come up with:
The integral is approximately equal to $0.242070053984$.
The integral equals
$$\sum_{n=1}^{\infty}\int_{\sqrt{n}}^{\sqrt{n+1}}...
- 1,688
2
votes
0
answers
61
views
Help with finding the equation of a summation
I have the summation:
$$F(n) := \sum_{i=1}^n \lfloor i\sqrt2 \rfloor $$
(i.e. $F(5) = 19$)
I would like to simplify the expression of $F$ and get rid of the sum.
I saw the proof that $\sum_{i=1}^n i = ...
- 121
0
votes
0
answers
101
views
Calculate the sum with floor function.
Let $a$ be a positive number. Calculate the sum $$\sum_{1\le n\le x}\left\lfloor \sqrt{n^{2}+a} \right\rfloor$$
I tried to calculate first $\left\lfloor \sqrt{n^{2}+a} \right\rfloor-n$. But probably ...
- 1,382
1
vote
2
answers
890
views
How to calculate sum of floor functions.
Let $f(x)$ is real valued function.
How to calulate or at least find lower and upper bound of sum:
$$\sum_{1\le n \le x}\left\lfloor f(n) \right\rfloor$$? For example when $f(n)=\frac{x}{n}$
...
- 1,382
13
votes
5
answers
675
views
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $
I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{...
user617243
0
votes
0
answers
100
views
If $a_1\le a_2 \le \dots \le a_n$ and $a_1 + 2a_2 + \cdots + na_n = 0$, then $a_1[x] + a_2[2x] + \cdots + a_n[nx] \geq 0$
Consider a positive integer $n$ and the real numbers $a_1 \leqslant a_2 \leqslant \cdots \leqslant a_n$ such that
$\displaystyle a_1 + 2a_2 + \cdots + na_n = 0$
Prove that
$\displaystyle a_1[x] +...
- 332
-1
votes
4
answers
137
views
Prove for every $n,\;\;$ $\sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^{k}}+\frac{1}{2} \right\rfloor=n $ [closed]
Problem
Prove that for every nonnegative integer $n$ we have,
$$\sum_{k=1}^{\infty}\left\lfloor \frac{n}{2^{k}}+\frac{1}{2} \right\rfloor=n$$
where $\lfloor x\rfloor$ denotes the integer part of ...
- 375
0
votes
1
answer
282
views
Kolmogorov's Truncation Lemma (iii)
Probability with Martingales:
In the definition of $f$, is that really $z$ and not $\lceil |z| \rceil$, $\lfloor |z| \rfloor + 1$ or something?
How exactly do we have the part in the $\color{red}{\...
- 13.7k
1
vote
2
answers
141
views
Growth of $\sum_{x=1}^{n-1} \left\lceil n-\sqrt{n^{2}-x^{2} } \right\rceil$
I'm interested in the growth of $$f(n):=\sum_{x=1}^{n-1} \left\lceil n-\sqrt{n^{2}-x^{2} } \right\rceil \quad \text{for}\quad n\rightarrow\infty $$
Progress
(From comments) I've got
$$\frac{f(n)}{...
- 21