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What is a ceiling function in mathematics?
What is the difference between ceiling function and floor function?
What is the main difference between the ceiling function and the floor function?
Note that $\lceil x\rceil = -\lfloor -x\rfloor$.
I've tried my hand at writing one; I'd be interested to hear what you think. Assume $x\in\mathbb{R}$.
$$\lfloor x \rfloor:=\sup\{n\,\big\vert\,n\in\mathbb{Z}\quad n\leq x\}$$
$$\lceil x \rceil:=\inf\{n\,\big\vert\,n\in\mathbb{Z}\quad n\geq x\}$$
You can convert $L=\lceil 0.06 M \rceil$ into two inequalities,
$$0.06M\le L < 0.06M + 1\;,$$
together with the condition that $L$ is an integer. Then you can substitute $M$ from the first equation:
$$0.06(C+L-1)\le L < 0.06(C+L-1) + 1$$
and "solve" for $L$:
$$0.06(C-1)\le 0.94L < 0.06(C-1) + 1\;,$$
$$\frac{0.06(C-1)}{0.94}\le L < \frac{0.06(C-1) + 1}{0.94}\;.$$
Since the difference between the two bounds is $1/0.94\approx 1.06$, this can have either one or two integer solutions for $L$, depending on $C$, which you can substitute into the first equation to get the corresponding values for $M$.
You can rewrite your problem as the intersection of two objects. Let $x=M$ and $y=L$ then the first equation is the parametric line $y=x+1-c$ and the second equation $y = \lceil 0.06x \rceil$ can be thought as a collection of horizontal lines (I let you figure out precisely what is that collection). Now all you want is to find the intersection of these two things (which clearly will involve your parameter $c$.
You can replace $\lfloor x \rfloor$ with $x - \theta$, where $\theta \in [0,1)$ is some unknown quantity. Similarly, $\lceil x \rceil = x + \theta$ (a different $\theta$ within the same range).
Another helpful identity is $\lfloor x \rfloor + n = \lfloor x + n \rfloor$ for any integer $n$.
Your final expression gives you the number you want.
According to your blog post, you're looking for the smallest integer $n$ (i.e., the "first Fibonacci number with 1000 digits") that satisfies $$G(n) = \left\lfloor n \log \varphi - \frac{\log 5}{2} \right\rfloor + 1.$$ There may, of course, be more than one integer $n$ for which this is true.
By definition of the floor function, the values of $n$ that satisfy this are the values that satisfy $$G(n) - 1 \leq n \log \varphi - \frac{\log 5}{2} < G(n),$$ which, since $\log \phi > 0$, are the values that satisfy $$\frac{G(n) + \frac{\log 5}{2}}{\log \varphi} - \frac{1}{\log \varphi} \leq n < \frac{G(n) + \frac{\log 5}{2}}{\log \varphi}.$$
Since $\frac{1}{\log \varphi} \approx 4.78$, there are either four or five integers in this interval. But the smallest one is obtained by taking the ceiling of the lower endpoint of the interval; i.e., $$\left\lceil\frac{G(n) + \frac{\log 5}{2} - 1}{\log \varphi}\right\rceil.$$
Incidentally, this argument also apparently shows that there are either four or five Fibonacci numbers that have a given number of digits. (Except in the single-digit case, where there are six (not counting 0). But your formula for $G(n)$ doesn't hold when $n=1$, so we shouldn't expect this calculation to be true in the single-digit case anyway.)
$$ \lceil x \rceil = \min\{k \in \mathbb{Z} \mid k \geq x\}$$
So, basically:
$$\left\lceil {n\over 4} \right\rceil \ge 3 \iff {n\over 4} > 2 \iff n > 8$$
Yes, you should use the definition. However, the definition should be $\lceil x \rceil = n$ iff $n-1 < x \leq n$. (Otherwise the ceiling of an integer could take on two different values - e.g. $2-1\leq 2\leq 2$ and $3-1 \leq 2 \leq 3$ so $\lceil 2\rceil = 2$ and $\lceil 2 \rceil =3$, which is obviously not what we want).
Now you want to find which $n$ satisfy $\lceil n/4 \rceil \geq 3$. This happens if and only if $n/4>3-1$. To see why, suppose $\lceil n/4\rceil =k \geq 3$. Then $k-1 < n/4 \leq k$, so $n/4 >k-1 \geq 3-1$. Conversely, suppose $n/4>3-1=2$. Let $k=\lceil n/4 \rceil$. Then $k-1<n/4 \leq k$. Since we also have $n/4>2$, it follows that $k \geq n/4>2$, and thus $\lceil n/4 \rceil = k \geq 3$.
So the inequality becomes $2<n/4$, and thus $n>8$.