The Question  Solve for $(x,y,z)$ $$\left\{\begin{matrix} x+\left \lfloor y \right \rfloor + \left \{ z &gt; \right \}=1,1\\  z+\left \lfloor x \right \rfloor + \left \{ y \right &gt; \}=2,2\\  y+\left \lfloor x \right \rfloor + \left \{ z \right \}=3,3  \end{matrix}\right.$$  My Understanding From the definition of floor function we know that $n = \left \lfloor n \right \rfloor + \left \{n  \right \}$ and if you play with equation a bit $(1,1+,2,2=3,3)$ and do some cancelling it's easy to see that $$x+z=0 \Rightarrow x = -z.$$ After that you set $z=-x$ and rewrite the equation. You'll see from $[2(1,1)=2,2]$ and a bit cancelling $\left \{ y \right \} - \left \{ x \right \}=2,2$. Also an important note that $\left \lfloor x \right \rfloor \neq \left \lfloor -x \right \rfloor , \left \{ x \right \} \neq \left \{ -x \right \}$ example: $\left \lfloor 5,35 \right \rfloor = 5$ but $\left \lfloor -5,35 \right \rfloor = -6$. I got this so far and need help.
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The equation system is —  $x + \lfloor y \rfloor + \{ z \} = 13.2$  $\{ x \} + y + \lfloor z \rfloor = 15.1$  $\lfloor x \rfloor + \{ y \} + z = 14.3$  Now I've tried substituting $n$ with $\lfloor n \rfloor + \{ n \}$ everywhere possible and then gone on with algebraic manipulations. But everything gets messy from there. I tried solving the problem more than thrice over the past few days, but always ended up with different answers.
73676
I'm trying to prove rigorously the following:  $\lfloor x/a/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$   for $a,b&gt;1$  So far I haven't gotten far. It's enough to prove this instead  $\lfloor z/c \rfloor$ = $\lfloor \lfloor z \rfloor /c \rfloor$   for $c&gt;1$  since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.
123477
Mathematically, why is this true?  $$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$  Assume $a$ and $b$ are positive integers.  Is this also true if $a$ and $b$ are real numbers?
83874
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.  How should I solve this?  I can't think of a way with casework and I can't really simplify it more.  Thanks in advance for posting a proof!
115704
Prove using combinatorics   $$\sum\limits_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor{\frac{n-k}{2}}\rfloor}=\binom{2n+1}{n}$$  The right side seems easy for the right side we have floor function it seems we have to count the number of subsets that $A_i \notin A_j$ but when I try to prove I can't show the right side.Any hints?
20302
Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$   $$$$ Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$ I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$ I would be grateful for any help. Many thanks!
46104
The question says to solve this equation: $(z+1)^5 = z^5$  I did. Just want to find out if I did it properly and if my run-around logic makes sense.  First I begin my writing the equations as:  $$ (z+1)^5 = z^5$$ $$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$ So $$  \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$   $$  \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$   Now, because the natural logarithm is one-to-one, I write:  $$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$   So assign $ z = a +bi$   So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $  So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$  Now to find $b$  $$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$ $$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$  I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?  Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong...
19423
I realize that there is probably not a closed form, but is there an efficient way to calculate the following expression?  $$\sum_{k=1}^n \left\lfloor \frac{n}{k}\right\rfloor$$  I've noticed $$\sum_{k=1}^n \left\lfloor \frac{n}{k}\right\rfloor = \left\lfloor\frac{1}{2}\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor} \left\lfloor\frac{n}{k}\right\rfloor\right\rfloor + \sum_{k=1,\ odd(k)}^n \left\lfloor \frac{n}{k}\right\rfloor$$  But I can't see an easy way to let the second term enter in a recursion.
104873
My problem  Alright, so this is my dilemma and it all boils down to the very troublesome function floor ($\lfloor x \rfloor$) and piecewise constant functions in general ($\lfloor f(x) \rfloor$). Everyone knows what the indefinite integral is. Just in case it is not well known this is what it is:  $$F(x) = \int_{c}^{x} f(t)dt$$  The indefinite integral is an accumulation function with some base value that it starts out with. It varies by an arbitrary constant.  Now let me define a second operator that we will refer to as an implied integral ($\int^*$). It is best defined by example:  $\int^* \lfloor x \rfloor dx = x \lfloor x \rfloor + C(x)$  $\int^* x \lfloor x \rfloor dx = \frac 12 x^2 \lfloor x \rfloor + C(x)$  $\int^* \lfloor f(x) \rfloor dx = x \lfloor f(x) \rfloor + C(x)$  Basically, it is an indefinite integral that varies be piecewise constant functions (C(x)) and treats peicewise constant functions as constants.  Now,there is ambiguousness with the term "antiderivative" that I recently stumbled upon. The antiderivative seems to refer to both of these operations. For instance, see my question about the bug in matlab. The answer there claims that the implied integral is a valid antiderivative. I remember from elementary calculus that antiderivatives vary by constants. However, I am willing to give the benefit of the doubt since that user is a professor and probably knows way more than me.  Why is it important?  Well, while the area function is a set-in-stone idea differential equations rely upon use of the antiderivative and not the indefinite integral. Therefore, depending on which one the antiderivative follows... the result will change the solutions to differential equations. Specifically: It will decide whether or not their solutions must be continuous or if they are allowed to be discontinuous. A more concrete example is that we would no longer be able to just solve homogenous linear equations with constant coefficients. We would be able solve one's with piecewise constant coefficients trivially (it is possible to solve them for continuous solutions by doing things).  To sum it all up:     Do antiderivatives vary by constant functions or piecewise constant functions?   It truly is an important question.
42887
$$ \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? $$   I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.  Thanks in advance!
53752
So I'm trying to do a proof for this problem:  If $\displaystyle{k \le \left\lfloor \frac{n}{2} \right\rfloor}$ then $$\displaystyle{\binom{n}{k-1} &lt; \binom{n}{k}}$$  I can do it algebraically but my professor asked for a combinatorial proof.  How would one does it?
50528
My question is very simple. How would one take the floor of a complex number and what is the floor of i?
23982
There exists a formula for the $n$th term of this sequence  from the OEIS ("$n$ appears $n$ times") $$1, 2, 2, 3, 3, 3, 4, 4 ,4, 4, 5...$$ which is $$\left \lfloor \frac {1+\sqrt{1+8n}}{2} \right \rfloor.$$  Is there a better formula for the $n$th term? A formula that isn't a 'magic' one that just happens to work but one that comes from a well-founded observation. Preferably one not using the floor function. I also have suspicions about the sustainability of the above formula as I feel with very $n$ it may not function properly.
88464
Does $\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?$   I'm going to say this statement is false, and try to use the properties of limits  $$\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=\lim_{x \to 0+} x.\lim_{x \to 0+}\lfloor \frac{a}{x}\rfloor$$ =$0.\infty$ which is undefined.  Or is it $\infty $ because x ends up being a little bigger than 0?
105391
What is the smallest possible value of  $$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor$$  where $a,b,c,d&gt;0$?
110518
I have some samples of data of the form $x,y$ and $z=f(x,y)$. I wish to fit a plane $$z = Ax + By + C$$ to the data with the smallest mean square errors.  I have found an "answer" in section 3 of  document, but it is left in the form of some equations left to be solved. I just about have the ability to solve these equations, but the process gets so messy that the likelihood of me making a mistake is quite high. Surely someone has written out the full solution longhand somewhere (it might be called "closed form") anyway, in the form $$A=\ldots,$$ $$B=\ldots,$$ $$C=\ldots.$$  EDIT: Maybe "closed form" is the wrong phrase. So let me be clear. I want an explicit solution for $A$, $B$, and $C$, and not a solution that ends with "So now if you can solve these equations then you can discover the values of $A$, $B$ and $C$".
72169
Is there any way the following can be simplified?  $$\lfloor f(x)\cdot g(x) \rfloor - \lfloor f(x) \rfloor  \cdot \lfloor g(x) \rfloor$$