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3 Transcendence
Since 1937 it is known that the constant of Champernowne is, in addition to normal, also transcendent. In that year, Kurt Mahler composed a proof that showed that a whole set of constants, including the constant of Champernowne, are transcendental [3]. Let us first state what it means for a constant to be transcendent.
Definition 3.1. A constant $x$ is said to be transcendental when it cannot be a zero of some polynomial with integer coefficients.
While the notion of transcendence is easy to state, the proofs that show that a specific constant is transcendental are quite involved. In fact, to give an idea of the complexity of the notion, it created an entire branch of number theory on its own: Transcendental number theory.
To get more feeling for what it means for a number to be transcendental, we can see that it is a subset of the irrational numbers. This can be illustrated by taking any rational constant $z=a / b$. That constant is namely the solution of the equation $b x=a$. The subset is even a strict subset since every $n$-root of rational numbers also cannot be transcendental since it would be the solution of the polynomial $b x^n=a$.
The study of transcendental numbers practically started with Liouville, who has stated the first criterion whereby transcendental numbers can be constructed [5, p.1]. The transcendental numbers that can be constructed in this way are called Liouville numbers. Examples of transcendental numbers which are not Liouville numbers are $e$ and $\pi$.
3.1 Transcendence proof by Mahler
As previously stated, Mahler proved in his paper the transcendence of a certain set of constants, more specifically, special decimal numbers which are not Liouville numbers. We will investigate his proof in the case of the constant of Champernowne. The general proof for the set for which Mahler proved transcendence can be found in his paper [3].
The set of constants for which Mahler has proved transcendence can be described as follows. Let $f(k)$ be a strictly increasing polynomial with natural numbers as outcome for all natural numbers $k$. Now consider the sequence $f(1), f(2), f(3), \ldots$ and let us call each $f(k)$ a member of that sequence. We define the constant $x$ to be the decimal number we get if we concatenate the sequence of members $f(k)$ after the period. Thus the constant is defined as $x=0 . f(1), f(2), f(3), \ldots$ for some given polynomial $f(k)$.
If we, for example, take the function $f(k)=\left(k^2+k\right) / 2$, the resulting constant will be given by the decimal number $x=0.1,3,6,10,15,21,28,36, \ldots$. It is obvious that constant of Champernowne belongs to this set of constants since that constant is created by simply taking the polynomial $f(k)=k$.
The theorem Mahler proved in his paper is given by the following theorem.
Theorem 3.1. Let $x$ be a decimal number given by
$$
x=0 . f(1), f(2), f(3), \ldots
$$
where $f(\cdot)$ is a strictly increasing polynomial with $f(k) \in \mathbb{N}$ for all $k \in \mathbb{N}$. Then $x$ is transcendental, without being a Liouville number.
The theorem we are going to prove will only consider the case of $f(k)=k$ since we are only interested in the transcendence of the constant of Champernowne.
Theorem 3.2. The constant of Champernowne in base ten is transcendental.
With the statement that the set of constants are not Liouville numbers, the first theorem implies that transcendence of those constants can not be shown to be transcendental in the same way as it is proven for Liouville numbers. Mahler shows that the constants which are described in Theorem 3.1 cannot be a Liouville number in the last part of his paper [3]. We also will proof that the constant of Champernowne is not a Liouville number in Section 3.1.2.
The results of Theorem 3.1 are similar for decimal numbers in every base $q$ greater or equal to 2 , hence Mahler left the base as a variable. We, on the other hand, are only interested in base ten and thus, in our case, $q$ will be the fixed number 10 .
The theorem will be proven with the use of a result by Schneider, like Mahler did in his paper. Details on the theorem of Schneider can be found by another paper by Mahler, which he published in 1936 [4]. The intuition behind the theorem is that a transcendental constant can be approximated very accurately by a sequence of fractions. By constructing that specific sequence of fractions and subtracting the constant by those fractions, we can prove that the remainder term can always be made so small such that there can not exist a polynomial with integer coefficients having that constant as an solution.
Following this method, the proof will concentrate on creating this fraction and prove the transcendence of the constant of Champernowne by using the theorem of Schneider. Without further ado, we will start with proving the theorem.
3.1.1 The argument of Mahler in the case of the constant of Champernowne
We will be proving the transcendence of the constant of Champernowne with the use of the following steps.
1. First we create a workable representation of the constant.
2. Next, we consider the construction of the constant in terms of its members;
3. Thereafter, we will construct a sequence of fractions that approximate the constant;
4. After subtracting the fractions from the constant, we will investigate the remainder;
5. Finally, we complete the proof with the use of the theorem of Schneider;
Representation of the constant of Champernowne in digits $Z_{k, l}$
Let the constant of Champernowne be given by $c$ and let $f$ be the first order polynomial $f(x)=x$. Let us again call each natural number $f(k)=k$ a member of the constant $c$ and let each member $k$ be a decimal number which has $v$ digits. Furthermore, let us denote the digits as $Z_{k, l}$ where the second subscript stands for the position of the digit inside the member $k$. Hence we can represent each member $k$ as
$$
k=Z_{k, 0} Z_{k, 1} \ldots Z_{k, v-2} Z_{k, v-1}=10^{v-1} Z_{k, 0}+10^{v-2} Z_{k, 1}+\ldots+10 Z_{k, v-2}+Z_{k, v-1}=\sum_{l=0}^{v-1} Z_{k, l} 10^{v-l} .
$$
With this expression for $k$ we write $c$ as
$$
c=0.1,2,3, \ldots, 25, \ldots, k, \ldots=0 . Z_{1,0}, Z_{2,0}, Z_{3,0}, \ldots, Z_{25,0} Z_{25,1}, \ldots, Z_{k, 0} Z_{k, 1} \ldots Z_{k, v-1}, \ldots
$$
Because it is difficult to investigate the properties of this expression, we will simplify representation of $c$.
Construction of the constant $c$ in terms of its members $f(k)$
Assume again that member $k$ has $v$ digits, from this we know that $10^{v-1} \leq k \leq 10^v-1$. Let us now create the sets $K_v$ of all members $k$ which have exactly $v$ digits. Observe the number of members $k$ contained in the set $K_v$, thus $\# K_v$, is equal to $\left(10^v-1\right)-\left(10^{v-1}-1\right)=9 \cdot 10^{v-1}$. Also note that the total number of digits of all members $k$ with $v$ digits together will be $v \cdot 9 \cdot 10^{v-1}$. If we count the total number of digits of all members with at most $v-1$ digits, we get
$$
\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}
$$
The total number of members with $v$ digits, up to the member $k$ is equal to $k-\left(10^{v-1}-1\right)$. Therefore the total number of digits of those members is equal to $v\left(k-10^{v-1}+1\right)$. If we add this number to the number of digits of all members up to $v-1$ digits, we get the total amount of digits of the 1st to the $k$ th member,
$$
\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(k-10^{v-1}+1\right)
$$
Note that this value is equal to the position of the last digit of the member $k$ in the constant $c$. Therefore, the contribution of this member to $c$ is equal to
$$
k \cdot 10^{-}\left(\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(k-10^{v-1}+1\right)\right) .
$$
For the contribution of all members that are contained in the constant $c$, which sums up to $c$ again, we sum the result of equation (32) for every natural number $k$.
$$
\begin{aligned}
c & =\sum_{k=1}^{\infty} k \cdot 10^{-}\left(\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(k-10^{v-1}+1\right)\right) \\
& =\sum_{k=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(10^{v-1}-1\right)} k \cdot 10^{-v k}
\end{aligned}
$$
Note that the summation over the members $k$ is equivalent to summing over the lengths of members $v$ and then summing over members $k$ with a certain length $v$. The latter will be a summation from the member $k=10^{v-1}$ to the member $k=10^v-1$ since those are the first and last member with $v$ digits. Hence, let us substitute the sum by the sums we just mentioned.
$$
\begin{aligned}
c & =\sum_{v=1}^{\infty} \sum_{k=10^{v-1}}^{10^v-1} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(10^{v-1}-1\right)} k \cdot 10^{-v k} \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(10^{v-1}-1\right)} \sum_{k=10^{v-1}}^{10^v-1} k \cdot 10^{-v k}
\end{aligned}
$$
Simplifying the summations
To proceed, we need the a formula from difference calculus that can be found in a book by Ernesto Cesaro and Gerhard Kowalewski [6], which is given as follows.
$$
\sum_{z=0}^{\infty} F(z) x^z=\sum_{h=0}^{\infty} \frac{x^h \partial^h F(0)}{(1-x)^{h+1}} .
$$
In this formula, the function $F(z)$ is defined for all non-negative integers $z$ and has the following as its $h$ th differential for a given number $z$.
$$
\partial^h F(z)=\sum_{i=0}^h\binom{h}{i}(-1)^i F(z+h-i)
$$
We can rewrite the sum at the end of equation (33) in such a way that we can use the formula of equation (34). Let us call this term $S_v$ and let us denote $10^{-v}$ as $x$.
$$
\begin{aligned}
S_v & =\sum_{k=10^{v-1}}^{10^v-1} k \cdot 10^{-v k}=\sum_{k=10^{v-1}}^{\infty} k \cdot 10^{-v k}-\sum_{l=10^v}^{\infty} l \cdot 10^{-v l}, \quad \text { take } z=k-10^{v-1}=l-10^v, \\
& =\sum_{z=0}^{\infty}\left(z+10^{v-1}\right) 10^{-v\left(z+10^{v-1}\right)}-\sum_{z=0}^{\infty}\left(z+10^v\right) 10^{-v\left(z+10^v\right)} \\
& =10^{-v 10^{v-1}} \sum_{z=0}^{\infty}\left(z+10^{v-1}\right) 10^{-v z}-10^{-v 10^v} \sum_{z=0}^{\infty}\left(z+10^v\right) 10^{-v z} \\
& =10^{-v 10^{v-1}} \sum_{z=0}^{\infty}\left(z+10^{v-1}\right) x^z-10^{-v 10^v} \sum_{z=0}^{\infty}\left(z+10^v\right) x^z .
\end{aligned}
$$
Note that the polynomials used in the sums of equation (35), $F_1(z)=z+10^{v-1}$ and $F_2(z)=z+10^v$, are first order and become equal to 1 after differentiating once and thus it is only necessary for $h$ to sum over 0 and 1 . By using equation (34) we can rewrite $S_v$ with the use of these polynomials.
$$
\begin{aligned}
S_v & =10^{-v\left(10^{v-1}\right)} \sum_{h=0}^1 \frac{x^h \partial^h F_1(0)}{(1-x)^{h+1}}-10^{-v 10^v} \sum_{h=0}^1 \frac{x^h \partial^h F_2(0)}{(1-x)^{h+1}} \\
& =10^{-v\left(10^{v-1}\right)} \sum_{h=0}^1 \frac{10^{-v h} \partial^h F_1(0)}{\left(1-10^{-v}\right)^{h+1}}-10^{-v 10^v} \sum_{h=0}^1 \frac{10^{-v h} \partial^h F_2(0)}{\left(1-10^{-v}\right)^{h+1}} \\
& =10^{-v\left(10^{v-1}-1\right)} \sum_{h=0}^1 \frac{\partial^h F_1(0)}{\left(10^v-1\right)^{h+1}}-10^{-v\left(10^v-1\right)} \sum_{h=0}^1 \frac{\partial^h F_2(0)}{\left(10^v-1\right)^{h+1}} \\
& =10^{-v\left(10^{v-1}-1\right)} A_v-10^{-v\left(0^v-1\right)} B_v
\end{aligned}
$$
where the terms $A_v$ and $B_v$ are given by
$$
\begin{aligned}
A_v & =\sum_{h=0}^1 \frac{\partial^h F_1(0)}{\left(10^v-1\right)^{h+1}} \\
& =\frac{10^{v-1}}{\left(10^v-1\right)^1}+\frac{1}{\left(10^v-1\right)^2} \\
& =\frac{10^{v-1}\left(10^v-1\right)+1}{\left(10^v-1\right)^2} \\
& =\frac{10^{2 v-1}-10^{v-1}+1}{10^{2 v}-2 \cdot 10^v+1}
\end{aligned}
$$
$$
\begin{aligned}
B_v & =\sum_{h=0}^1 \frac{\partial^h F_2(0)}{\left(10^v-1\right)^{h+1}} \\
& =\frac{10^v}{\left(10^v-1\right)^1}+\frac{1}{\left(10^v-1\right)^2} \\
& =\frac{10^v\left(10^v-1\right)+1}{\left(10^v-1\right)^2} \\
& =\frac{10^{2 v}-10^v+1}{10^{2 v}-2 \cdot 10^v+1}
\end{aligned}
$$
We can rewrite equation (33) using this new expression for $S_v$ in terms of $A_v$ and $B_v$.
$$
\begin{aligned}
c & =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}+v\left(10^{v-1}-1\right)}\left(10^{-v\left(10^{v-1}-1\right)} A_v-10^{-v\left(10^v-1\right)} B_v\right) \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}}\left(A_v-10^{-v\left(10^v-10^{v-1}\right)} B_v\right) \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}}\left(A_v-B_v^{\prime}\right)
\end{aligned}
$$
Using the expressions for $A_v$ and $B_v$ we can write out the expression for $c$.
$$
\begin{aligned}
c & =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}}\left(\frac{10^{2 v-1}-10^{v-1}+1}{10^{2 v}-2 \cdot 10^v+1}-10^{-v\left(10^v-10^{v-1}\right)} \frac{10^{2 v}-10^v+1}{10^{2 v}-2 \cdot 10^v+1}\right) \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}}\left(10^{-1} \frac{10^{2 v}-10^v+10}{10^{2 v}-2 \cdot 10^v+1}-10^{-v\left(10^v-10^{v-1}\right)}\left(1-\frac{-10^v}{\left(1-10^v\right)^2}\right)\right) \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}}\left(10^{-1}\left(1-\frac{-10^v+9}{\left(1-10^v\right)^2}\right)-10^{-v\left(10^v-10^{v-1}\right)}\left(1-\frac{-10^v}{\left(1-10^v\right)^2}\right)\right) \\
& =\sum_{v=1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}} E_v
\end{aligned}
$$
We leave the expression for $c$ as it is and we will denote the term between the brackets as $E_v$. Note that this term is approximately equal to one over ten since the terms between the smaller brackets will tend to one very fast and the second factor will already be very small when $v$ equals 1 .
The fraction that approximates the constant
Using the result from equation (37), we can create a sequence of fractions, which we will denote as $P_s / Q_s$, that will converge to the constant $c$ very quickly. This sequence will make it possible to eventually prove the transcendence of $c$. Let $D_s$ denote the least common multiple of the numbers $10-1,10^2-1, \ldots, 10^s-1$ for
$s \geq 1$. Let the denominator and the numerator of the fractions be
$$
Q_s=D_s^2 10^{\sum_{m=1}^{s-1} m \cdot 9 \cdot 10^{m-1}} \quad \text { and } \quad P_s=Q_s\left(\sum_{v=1}^s\left(10{ }^{10-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}} E_v\right)+B_s^{\prime}\right)
$$
and let us have a remainder term which is given as follows.
$$
R_s=\sum_{v=s+1}^{\infty}\left(10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}} E_v\right)-B_s^{\prime}
$$
Note that the terms $Q_s, P_s$ and $R_s$ are created in such a way that we have
$$
c-\frac{P_s}{Q_s}=R_s
$$
To give an idea of these sequences we will work out the first four term of every sequence.
Table 2: The first four terms of the sequences $D_s^2, Q_s, A_v, B_v^{\prime}, E_s, P_s, \frac{P_s}{Q_s}$ and $R_s$.
It is obvious that $Q_s$ is a natural number and by Table 2 we see that $P_s$ is also natural since $D_s^2$ is a multiple of the divisor of $E_s$ and the addition of $B_s^{\prime}$ removes the remaining negative term of 10 to a negative power of the last $E_v$ term of the sum inside $P_s$. Since both $Q_s$ and $P_s$ are natural numbers, we may use $Q_s$ as the denominator of the fraction. This was, for example, not the case if $P_s$ was rational instead of an integer.
The value of the remainder term $R_s$ in terms of $Q_s$
By increasing the number $s$, the fraction given by equation (40) will approximate the constant better and the remainder term $R_s$ will get smaller. However, for the constant to be transcendent, the remainder term needs to decrease faster than the extent to which the denominator increases. Thus let us consider the construction of the fraction and determine the remainder term in terms of the denominator.
To show this, we will derive asymptotic formulas for $\log Q_s$ and $\log R_s$. To obtain them, we need the following formula, denoting the number of digits of all member with at most $v-1$ digits.
$$
\begin{aligned}
i_v & =\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}=\sum_{m=1}^{v-1} m\left(10^m-10^{m-1}\right)=\sum_{m=1}^{v-1} m 10^m-\sum_{n=0}^{v-2}(n+1) 10^n, \quad \text { with } n=m-1, \\
& =\sum_{m=1}^{v-1} m 10^m-\sum_{n=1}^{v-1}(n+1) 10^n+v 10^{v-1}-1=v 10^{v-1}-\sum_{m=1}^{v-1} 10^m-1=(v-1) 10^{v-1}+O\left(10^{v-1}\right) .
\end{aligned}
$$
First note that $D_s$, the least common multiple of $10^t-1$ where $1 \leq t \leq s$, is surely smaller than just the product of these $10^t$. With that in mind, observe the following representations of $Q_s$ using the formula of $i_v$.
$$
\begin{aligned}
Q_s & =D_s^2 10^{\sum_{m=1}^{s-1} m \cdot 9 \cdot 10^{m-1}}<\left(\prod_{t=1}^s 10^t\right)^2 10^{i_s}=10^{s(s+1)} 10^{(s-1) 10^{s-1}+O\left(10^{s-1}\right)} \\
\log Q_s & \sim(s-1) 10^{s-1} \log 10 \\
\lim _{s \rightarrow \infty} \frac{\log Q_{s+1}}{\log Q_s} & =\lim _{s \rightarrow \infty} \frac{(s) 10^s \log 10}{(s-1) 10^{s-1} \log 10}=10 .
\end{aligned}
$$
Secondly, we will represent $R_s$ in the following way. First we will determine $R_s-B_s^{\prime}$. The second expression is justified since every following term in the sum will be 10 times smaller than the previous term and the denominator in the fraction between the brackets is similar to 1 since $10^{-v}$ is negligible compared to 1 .
$$
\begin{aligned}
R_s-B_s^{\prime} & =\sum_{v=s+1}^{\infty} 10^{-\sum_{m=1}^{v-1} m \cdot 9 \cdot 10^{m-1}} E_v=\sum_{v=s+1}^{\infty} 10^{-i_v} E_v & & \Rightarrow \\
R_s-B_s^{\prime} & \sim 10^{-i_{s+1}} & & \Rightarrow \\
& \sim 10^{-s 10^s} & & \Rightarrow
\end{aligned}
$$
To determine $R_s$, we will also need to evaluate $B_s^{\prime}$, which can be obtained from equation (36).
$$
\begin{aligned}
B_s^{\prime} & =10^{-s\left(10^s-10^{s-1}\right)} B_v \sim 10^{-s 10^s} \\
R_s & =\left(R_s-B_s^{\prime}\right)+B_s^{\prime} \sim 10^{-s 10^s} \\
\log \left|R_s\right| & \sim \log \left|10^{-s 10^s}\right|=-s 10^s \log 10 .
\end{aligned}
$$
We can observe from the previous equation that, if $s$ is large enough, the term $R_s$ will not vanish. Multiplying this term with the denominator will provide us with a notion of how well the fraction approximates the constant. If the absolute value of that term gets smaller if we increase $s$ linearly, we know that the
accuracy of the approximation increases more that linear. Using equation (41) and (43) we get
$$
\begin{aligned}
\log \left|Q_s R_s\right| & =\log Q_s+\log \left|R_s\right| \\
& \sim(s-1) 10^{s-1} \log 10-s 10^s \log 10 \\
& \sim s\left(10^{-1}-1\right) 10^s \log 10 \\
\left|Q_s R_s\right| & <1 \\
\left|Q_s R_s\right| & \sim 10^{-s\left(1-10^{-1}\right) 10^s} \\
& =10^{-1} 10^{-(s-1)\left(1-10^{-1}\right) 10^1 10^{s-1}} \\
& =10^{-1}\left(10^{-(s-1)\left(1-10^{-1}\right) 10^{s-1}}\right)^{10} \\
\left|Q_s R_s\right| & \sim 10^{-1}\left|Q_{s-1} R_{s-1}\right|^{10}
\end{aligned}
$$
From equations (44) and (46) we see that the term $\left|Q_s R_s\right|$ is similar to 10 times the previous step to the power 10 while $\left|Q_s R_s\right|$ is smaller than 1 for all sufficiently large $s$. Therefore we can conclude the following.
$$
\left|Q_s R_s\right|<\left|Q_{s-1} R_{s-1}\right|
$$
And thus we can conclude that the fraction from equation (40) indeed tends to the constant $c$ very quickly.
Completing the proof by using the proof of Schneider
To prove transcendence with the results we previously have obtained, we need the following theorem which is proven by Schneider.
Theorem 3.3. For the real number $x$, if there is a constant $k>1$ for a sequence of fractions
$$
\frac{p_1}{q_1}, \frac{p_2}{q_2}, \frac{p_3}{q_3}, \ldots, \frac{p_s}{q_s}, \ldots, \quad p_i \in \mathbb{Z}, q_i \in \mathbb{N}, \operatorname{gcd}\left(p_i, q_i\right)=1, \forall i \in \mathbb{N}
$$
which has for an increasing s that the denominator $q_s$ is a power of a natural number $q$ and increases strictly in such a way that
$$
0<\left|x-\frac{p_s}{q_s}\right| \leq q_s^{-k} \quad \text { and } \quad \lim _{s \rightarrow \infty} \sup _s \frac{\log q_{s+1}}{\log q_s}<\infty
$$
Then the constant $x$ is transcendent.
This theorem is proven in an paper [4] which Mahler also produced, a year before he published the paper that proves that the constant of Champernowne is transcendental.
To prove that the constant $c$ is transcendent, we actually need to modify the theorem a bit. If we look to the proof of the theorem, we see that it shows effortlessly that the constant $x$, which is specified in the theorem, is indeed transcendent. But the proof will also hold if we take a weaker requirement for $q_s$, namely, take $q_s=q_s^{\prime} q_s^{\prime \prime}$ with $q_s^{\prime}$ a natural number such that
$$
\lim _{s \rightarrow \infty} \frac{\log q_s^{\prime}}{\log q_s}=0
$$
and $q_s^{\prime \prime}$ a power of $q$. Therefore, the actual theorem which we are going to use to prove transcendence of the constant $c$ is given as follows.
Theorem 3.4. For the real number $x$, if there is a constant $k>1$ for a sequence of fractions
$$
\frac{p_1}{q_1}, \frac{p_2}{q_2}, \frac{p_3}{q_3}, \ldots, \frac{p_s}{q_s}, \ldots, \quad p_i \in \mathbb{Z}, q_i \in \mathbb{N}, \operatorname{gcd}\left(p_i, q_i\right)=1, \forall i \in \mathbb{N}
$$
which has for an increasing $s$ that the denominator $q_s=q_s^{\prime} q_s^{\prime \prime}$, where $q_s^{\prime}$ is a natural number and $q_s^{\prime \prime}$ is a power of a natural number $q$, strictly increases in such a way that the following three requirements hold:
$$
0<\left|x-\frac{p_s}{q_s}\right| \leq q_s^{-k}, \quad \lim _{s \rightarrow \infty} \sup _s \frac{\log q_{s+1}}{\log q_s}<\infty \quad \text { and } \quad \lim _{s \rightarrow \infty} \frac{\log q_s^{\prime}}{\log q_s}=0
$$
Then the constant $x$ is transcendent.
To prove that the constant $c$ is transcendental according to Theorem 3.4, we will be considering the chain of fractions $P_s / Q_s$, where the numerator and denominator are given by the equations (38) for all natural numbers $s$. Therefore, let us see if $Q_s$ satisfies the three requirements of the theorem.
Fulfilling the requirements of Theorem 3.4
First we need to prove that $Q_s$ fits the requirements. If we observe the way $Q_s$ is defined in equation (38), we see that it is indeed a product of a natural number and a power of some natural number. Thus let us have that
$$
q_s^{\prime}=D_s^2, \quad q=10 \quad \text { and } \quad q_s^{\prime \prime}=10^{\sum_{m=1}^{s-1} m \cdot 9 \cdot 10^{m-1}}
$$
Now consider the first requirement and let us fill in the constant $c$ and its corresponding variables.
$$
0<\left|c-\frac{P_s}{Q_s}\right|=\left|R_s\right| \leq Q_s^{-k}
$$
To fulfill the first requirement, $R_s$ may not be equal to zero and $|R(s)|$ needs to be less or equal to some negative power of $Q_s$. The first inequality already holds by the definition of $R_s$, give by equation (39), since it is not possible for $R_s$ to be zero. For the second inequality, we need equations (41) and (43).
$$
\begin{aligned}
\left|R_s\right| & =Q_s^{-a} \quad \Rightarrow \quad-a=\frac{\log \left|R_s\right|}{\log Q_s} \\
\lim _{s \rightarrow \infty} \frac{\log \left|R_s\right|}{\log Q_s} & =\lim _{s \rightarrow \infty} \frac{-s 10^s \log 10}{(s-1) 10^{s-1} \log 10}=-10
\end{aligned}
$$
From this, we may conclude that, for a sufficiently large $s$, the second inequality in equation (48) will hold for any number $k$ with $k<10$.
That the second requirement holds can easily be shown with equation (42) since it shows that the fraction will tend to 10 in stead of infinity when $s$ tends to infinity. For all other $s$, the fraction will just yield a rational number and thus the supremum will always be below infinity.
For the third requirement, we substitute the same $Q_s$ from equation (38). Recall that $D_s^2$ is smaller than $10^{s(s+1)}$.
$$
\begin{aligned}
\lim _{s \rightarrow \infty} \frac{\log D_s^2}{\log Q_s} & =\lim _{s \rightarrow \infty} \frac{s(s+1)}{s(s+1)+\sum_{m=1}^{s-1} m \cdot 9 \cdot 10^{m-1}} \\
& <\lim _{s \rightarrow \infty} \frac{s(s+1)}{s(s+1)+(s-1) 9 \cdot 10^{s-2}}
\end{aligned}
$$
$$
\leq \lim _{s \rightarrow \infty} \frac{s+4}{s+4+\frac{9}{10^2} \cdot 10^s} \rightarrow 0, \quad \text { since } s(s+1) \leq(s-1)(s+4) \text { for } s>1 .
$$
Now that we have shown that the sequence of fractions $P_s / Q_s$ that approximate the constant $c$ indeed meet the requirements of Theorem 3.4, we have proven that the constant of Champernowne $c$ is indeed transcendent. |
|
realnumber
发表于 2015-7-31 14:45
kuing
发表于 2015-7-31 16:29
