3. The answer to the previous questions.

Now it is time for us to answer the preceding three questions.

We are going to show the answer of the three questions by Example 5, Lemma 2 and Theorem 1. Here you need the knowledge of Sterling's formula.

In fact it took us more than 2 years to get the answer. We had presented this beautiful sectors to many mathematicians and computer scientists three times in international symposiums and workshops, but no one came up with a good explanation for this. For example we presented this fact in the International Mathematica Syposium 2003. Please see Miyadera [3] in the reference.

One day one student proposed a very good idea. "We are paying too much attention to the beautiful figure made of a matrix of numbers. In reality it is only a sequence of digits and not a matrix. Let's count the numbers of 0 and 9 between other numbers carefully, and we may find the law that governs the structure of the sequence." This was indeed a very good idea. After this proposal it took only 3 days for us to find the answer. Once the answer was found, the structure looked very simple, but it was like Egg of Columbus.

The key element is the width of the matrix of numbers.

Example 5. We are going to study the structure of Figure (2) in Example 4.

$\includegraphics[height=6cm,clip]{graph9-15.eps}$

…Figure (2)

$1000000000000000000000000000000^{13}$

$-13 \times1^{1} \times1000000000000000000000000000000^{12}$

$78 \times1^{2} \times1000000000000000000000000000000^{11}$

$-286 \times1^{3} \times1000000000000000000000000000000^{10}$

$715 \times1^{4} \times1000000000000000000000000000000^{9}$

$-1287 \times1^{5} \times1000000000000000000000000000000^{8}$

$1716 \times1^{6}\times1000000000000000000000000000000^{7}$

$-1716\times1^{7}\times1000000000000000000000000000000^{6}$

$1287 \times1^{8}\times1000000000000000000000000000000^{5}$

$-715 \times1^{9}\times1000000000000000000000000000000^{4}$

$286 \times1^{10}\times1000000000000000000000000000000^{3}$

$-78\times1^{11}\times1000000000000000000000000000000^{2}$

$13 \times1^{12}\times1000000000000000000000000000000^{1}$

$-1^{13}$

…Figure (3)

Figure (3) is the table of terms we get when we apply the binomial theorem to the number $\displaystyle 999999999999999999999999999999^{13}$ = (10³⁰-1)¹³that we studied in Example 4. We are going to use Figure (3) here to analyze the structure of Figure(2). Approximately the first row of Figure (2) is made by summing the first and the second row of Figure (3). Similarly the kth row of Figure (2) is made by summing the kth and (k+1)th row of Figure (3).

For reader who are not familiar with Sterling's formula we are going to explain about it. You only have to understand how to use the formula to understand our article.

Lemma 1. $n!\sim \sqrt{2\pi}{n^{n+{\frac{1}{2}}}}e^{-n}$

We omit the proof, since this is the famous Sterling's formula. It asserts that the right part of the above formula is approximately equal to the left part when n is big enough.

Lemma 2. For sufficiently large n and t such that we have
$\frac{\log_{10}t!}{n}\sim \frac{t\log_{10}t-t\log_{10}e}{n}$

Proof. By using Lemma 1 for t we have
$\frac{\log_{10}t!}{n}\sim \frac{\log_{10}(\sqrt{2\pi}{t^{t+{\frac{1}{2}}}}e^{-t})}{n}$
. The conclusion of Lemma 2 is a direct result of this formula , since $\log_{10}\sqrt{2\pi}$ and $\frac{1}{2}\log_{10}t$ are relatively very small compared to n when n is big enough.

Lemma 3. If we make n and m big enough while keeping $\frac{m}{n}$ as an almost the same size, then we have
$\frac{\log_{10}{}_{n}C_{m}}{n}$
$= - (\frac{n - m}{n}\log_{10}(\frac{n-m}{n}) + \frac{m}{n}\log_{10}(\frac{m}{n}))$

Proof. In this proof we assume that n and m are big enough and we can use Lemma 2 for n ,m and n-m.
$\frac{\log_{10}{}_{n}C_{m}}{n} = \frac{1}{n}\log_{10}\frac{n!}{(n-m)!m!}$
$= \frac{\log_{10}n! - \log_{10}(n-m)! - \log_{10}m!}{n}$
$\sim \frac{n\log_{10}n - n\log_{10}e -((n-m) \log_{10}(n-m) - (n-m) \log_{10}e) - (m\log_{10}m - m\log_{10}e)}{n}$
$= \frac{n\log_{10}n -(n-m) \log_{10}(n-m) - m\log_{10}m }{n}$
$= \frac{(n-m)\log_{10}n -(n-m) \log_{10}(n-m)+m\log_{10}n - m\log_{10}m }{n}$
$= - (\frac{n - m}{n}\log_{10}(\frac{n-m}{n}) + \frac{m}{n}\log_{10}(\frac{m}{n}))$

Remark. This proof for Lemma 3 was already published in Miyadera and Kotera [1]. We used this for another theorem. We reproduced this again for reader who cannot read Italian Language.

Theorem 1. If we expand and express it as a matrix whose length of row is k, then we have the following (a), (b) and (c).
(a) The matrix has the structure which is described in the following table. See Figure (4). Here we denote by $L({}_{n}C_{m})$ the length of the digits of ${}_{n}C_{m}$ . As is well known $L({}_{n}C_{m}) \sim log_{10}({}_{n}C_{m})+1$ .
(b) The number of 0 and 9 is approximately
$n\times k - (log_{10}({}_{n}C_{1}) + \ldots + log_{10}({}_{n}C_{n}))$ ,
so it increases if we make $k \to \infty$ while we keep n constant or we decrease n.
(c) If we make $n \to \infty$ , then the figure made of numbers 1,2,3,4,5,6,7,8 is getting nearer and nearer to the graph of the function
$y = -(x\log_{10}x + (1-x)log_{10}(1-x))$
, where the x-coordinate is vertical.

the number of 0 and 9 the number of 1,2,3,4,5,6,7,8

$k-L({}_{n}C_{1})$ $L({}_{n}C_{1})$

$k-L({}_{n}C_{2})$ $L({}_{n}C_{2})$

$k-L({}_{n}C_{3})$ $L({}_{n}C_{3})$

$k-L({}_{n}C_{4})$ $L({}_{n}C_{4})$

$k-L({}_{n}C_{5})$ $L({}_{n}C_{5})$

$\vdots$ $\vdots$

$k-L({}_{n}C_{n})$ $L({}_{n}C_{n})$

…Figure (4)

Proof. (a) If we study the figure (2) and (3) carefully, then it is not difficult to generalize the example to the case of . Then we can easily get this table.
(b) As is easily seen the number of 0 and 9 is approximately
$n\times k - (log_{10}({}_{n}C_{1}) + \ldots + log_{10}({}_{n}C_{n}))$ .
Therefore it increases if we $k \to \infty$ while we keep the size of n almost constant or decrease it.
(c) If we make $n \to \infty$ , then the figure made of numbers 1,2,3,4,5,6,7,8 will become bigger. To make the figure almost the same size we divide it by n. By the above table we can approximate the figure by the list
$(\frac{0}{n},\frac{L({}_{n}C_{0})}{n}),(\frac{1}{n},\frac{L({}_{n}C_{1})}{n}),... ...c{m}{n},\frac{L({}_{n}C_{m})}{n}),\ldots,(\frac{n}{n},\frac{L({}_{n}C_{n})}{n})$ … (1)
using the x-y coordinates.

Here we use x-coordinate vertically.

Since $\frac{L({}_{n}C_{m})}{n} \sim \frac{\log_{10}({}_{n}C_{m})}{n}$
, we can use Lemma 2 and finish the proof of this theorem.

The answer to the three questions. Theorem 1 (b) gives the answer of Question 1 and 2, and Theorem 1 (c) gives the answer of Question 3. The function
$y = -(x\log_{10}x + (1-x)log_{10}(1-x))$
is the so-called entropy function and no doubt it has a beautiful curve.

Example 6. Once we know the structure of a sector, then we can make a beautiful sector easily. This time we use

$x = 99999\dotsb99$ that is bigger than the number we used in Example 4. It is easy to make a beautiful figure with a bigger numbers.

$\includegraphics[height=8cm,clip]{graph9-16.eps}$

Conclusion. Our investigation of the number $99999\dotsb99 ^{n}$ ends here. We really enjoyed it. Once we know the truth, the truth was very simple. We think that this is one of the most beautiful application of the binomial Theorem. Many high school students study the binomial Theorem, but they do not know that the theorem can make this kind of beautiful picture! We want many high school students and high school teachers know our findings and enjoy the beauty of the binomial Theorem!

the number of 0 and 9	the number of 1,2,3,4,5,6,7,8
$k-L({}_{n}C_{1})$	$L({}_{n}C_{1})$
$k-L({}_{n}C_{2})$	$L({}_{n}C_{2})$
$k-L({}_{n}C_{3})$	$L({}_{n}C_{3})$
$k-L({}_{n}C_{4})$	$L({}_{n}C_{4})$
$k-L({}_{n}C_{5})$	$L({}_{n}C_{5})$
$\vdots$	$\vdots$
$k-L({}_{n}C_{n})$	$L({}_{n}C_{n})$