Difference between revisions of "Rational approximation of famous numbers"
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− | + | '''Rational approximation''' is the application of [[Rational approximation|Dirichlet's theorem]] which shows that, for each irrational number <math>x\in\mathbb R</math>, the inequality <math>\left|x-\frac pq\right|<\frac 1{q^2}</math> has infinitely many solutions. On the other hand, sometimes it is useful to know that <math>x</math> cannot be approximated by rationals too well, or, more precisely, that <math>x</math> is not a [[Liouvillian number]], i.e., that for some power <math>M<+\infty</math>, the [[inequality]] <math>\left|x-\frac pq\right|\ge \frac 1{q^M}</math> holds for all sufficiently large denominators <math>q</math>. | |
− | + | ==Theorem== | |
− | == | + | Suppose that there exist <math>0<\beta<\gamma<1</math>, <math>1<Q<+\infty</math> |
− | + | and a sequence of pairs of integers <math>(P_n,Q_n)</math> such that for all sufficiently large <math>n</math>, we have <math>|Q_n|\le Q^n</math> and | |
− | Suppose that there exist <math>\beta | + | <math>\beta^n< \left|P_n-Q_n x\right|<\gamma^n</math>. Then, for every <math>M>\frac{\log(Q/\beta)}{\log(1/\gamma)}</math>, the inequality <math>\left|x-\frac pq\right|<\frac 1{q^M}</math> has only finitely many solutions. |
− | and a sequence of | ||
− | <math> | ||
− | |||
---- | ---- | ||
− | The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math> but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of | + | The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math>, but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of small, but not too small, linear combinations of <math>x</math> and <math>1</math> with not too large integer coefficients. |
− | ==Proof | + | ===Proof=== |
− | Choose the least <math>n</math> such that <math> | + | Choose the least <math>n</math> such that <math>\gamma^n\le 2q</math>. Note that for such choice of <math>n</math>, we have <math>\gamma^n> \frac {\gamma}{2q}</math>. Also note that <math>Q_n\ne 0</math> (otherwise <math>|P_n|</math> would be an integer strictly between <math>0</math> and <math>1</math>). Now, there are two possible cases: |
− | '''Case 1:''' <math>\frac pq= | + | '''Case 1:''' <math>P_n-Q_n\frac pq=0</math>. |
+ | Then <math>\left|x-\frac pq\right|=\left|x-\frac {P_n}{Q_n}\right|>\frac{\beta^n}{|Q_n|}>(\beta/Q)^n | ||
+ | =(\gamma^n)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}> | ||
+ | \left(\frac\gamma{2q}\right)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}>\frac 1{q^M}</math> | ||
− | '''Case 2:''' <math>\frac pq\ne | + | if <math>q</math> is large enough. |
− | \left|\frac pq | + | |
− | + | '''Case 2:''' <math>P_n-Q_n\frac pq\ne 0</math>. Then | |
− | + | ||
− | == | + | <math>\frac 1q\le\left|P_n-Q_n\frac pq\right|\le \left|P_n-Q_n x\right|+|Q_n|\cdot\left|x-\frac pq\right|\le \frac 1{2q}+Q^n\left|x-\frac pq\right| |
− | + | </math> | |
− | * | + | |
− | + | Hence, in this case, | |
+ | |||
+ | <math>\left|x-\frac pq\right|\ge \frac 1{2q}Q^{-n}\ge \frac \gamma{2q}Q^{-n}=\frac \gamma{2q}(\gamma^n)^{\frac{\log Q}{\log(1/\gamma)}}\ge \left(\frac\gamma{2q}\right)^{1+\frac{\log(Q}{\log(1/\gamma)}}>\frac 1{q^M}</math> | ||
+ | |||
+ | if <math>q</math> is large enough. (recall that <math>\beta<\gamma</math>, so <math>1+\frac{\log Q}{\log(1/\gamma)} | ||
+ | =\frac{\log(Q/\gamma)}{\log(1/\gamma)}<\frac{\log(Q/\beta)}{\log(1/\gamma)} </math>). | ||
+ | |||
+ | ==Magic Polynomial== | ||
+ | Before proceeding to the applications of the main theorem, let us introduce one very useful polynomial that often appears in proofs of irrationality. It is the polynomial | ||
+ | |||
+ | <math>P(x)\frac 1{n!}\left(\frac d{dx}\right)^n [x^n(1-x)^n]</math> | ||
+ | |||
+ | Its coefficients can be easily computed using the [[Binomial Theorem|binomial theorem]]: | ||
+ | |||
+ | <math>P(x)=\sum_{k=0}^n (-1)^k{n+k\choose n}{n\choose k}x^k.</math> | ||
+ | |||
+ | The important points are that all the coefficients are integer and | ||
+ | the sum of their absolute values does not exceed | ||
+ | <math>\max_{0\le k\le n}{n+k\choose n}\sum_{0\le k\le n}{n\choose k}\le {2n\choose n}2^n\le 8^n</math>. | ||
+ | |||
+ | Another useful remark is that the first <math>n-1</math> derivatives of <math>x^n(1-x)^n</math> vanish at <math>0</math> and <math>1</math>, which makes the integration by parts extremely convenient: | ||
+ | |||
+ | <math>\int_0^1 F(x)P(x)\,dx=(-1)^n\frac 1{n!}\int_0^1 F^{(n)}(x) x^n(1-x)^n\,dx.</math> | ||
+ | |||
+ | ==Example== | ||
+ | This section will prove that the number <math>\ln 2</math> is not [[Liouvillian number|Liouvillian]]. It can be read right after its parent article [[rational approximation of famous numbers]]. The proof of the non-Liouvillian character of <math>\ln 2</math> is much easier than that for <math>\pi</math>, but somewhat more difficult than that for <math>e</math>. | ||
+ | |||
+ | ===A Useful Integral=== | ||
+ | Everyone knows an integral representation for <math>\ln 2</math>. It comes right from the definition of the natural [[logarithm]]: <math>\ln 2=\int_0^1\frac {1}{1+x}\,dx</math>. Let us look at what will happen if we replace <math>1</math> in the numerator by some polynomial <math>P(x)</math> of degree <math>n</math> with integer coefficients. Since <math>P(x)-P(-1)=(1+x)R(x)</math> where <math>R(x)=\sum_{k=0}^{n-1} c_k x^k</math> is some polynomial of degree <math>n-1</math> with integer coefficients, we see that | ||
+ | |||
+ | <math>\int_0^1\frac{P(x)}{1+x}\,dx=\sum_{k=0}^{n-1}\frac {c_k}{k+1}+P(-1)\ln 2.</math> | ||
+ | |||
+ | Let now <math>D_n</math> be the [[least common multiple]] of the numbers <math>1,2,\dots, n</math>. Then | ||
+ | |||
+ | <math>D_n\int_0^1\frac{P(x)}{1+x}\,dx=P_n-Q_n\ln 2</math> | ||
+ | |||
+ | with some integer <math>P_n</math> and <math>Q_n=-D_nP(-1)</math>. | ||
+ | It remains to choose a polynomial <math>P(x)</math> that makes the integral small. | ||
+ | |||
+ | ===Polynomial <math>P(x)</math>=== | ||
+ | We shall just take our "magic polynomial" <math>P(x):=\frac{1}{n!}\left(\frac{d}{dx}\right)^n[x^n(1-x)^n]</math> | ||
+ | (see the [[rational approximation of famous numbers|parent article]] for its properties). | ||
+ | |||
+ | ===Estimates of the Integral=== | ||
+ | Integrating by parts <math>n</math> times, we see that | ||
+ | <math>\int_0^1\frac{P(x)}{1+x}\,dx= | ||
+ | (-1)^n\int_0^1\left[\frac{x(1-x)}{1+x}\right]^n\,\frac{dx}{1+x}</math>. | ||
+ | Since <math>\max_{0<x<1}x(1-x)=\frac 14</math> and since it is attained only at the point <math>x=\frac 12</math> where <math>1+x>1</math>, we can conclude that <math>\Gamma=\max_{0<x<1}\frac{x(1-x)}{1+x}<\frac 14</math>. The absolute value of our integral does not exceed <math>\Gamma^n</math> for all <math>n</math>. To estimate it from below, just notice that the integrand is at least <math>\frac 12\left(\frac 2 {15}\right)^n</math> for <math>\frac 1 3 < x< \frac 2 3</math>, so the absolute value of our integral is at least <math>\frac 16\left(\frac 2 {15}\right)^n>\left(\frac 1 8\right)^n</math> for large <math>n</math>. | ||
+ | |||
+ | ===Estimate of <math>D_n</math>=== | ||
+ | Since the largest possible power of a given prime <math>p\le n</math> that can divide one of the numbers <math>1,2,\dots,n</math> is <math>\left\lfloor\frac{\log n}{\log p}\right\rfloor</math> where <math>\lfloor\cdot\rfloor</math> is the floor function, we have | ||
+ | |||
+ | <math>D_n=\prod_{p\mathrm{\ prime,}\,p\le n}p^{\left\lfloor\frac{\log n}{\log p}\right\rfloor}\le \left(\prod_{p\mathrm{\ prime,}\,p\le \sqrt n}n\right)\cdot\left( \prod_{p\mathrm{\ prime,}\,\sqrt n<p\le n}p\right)\le n^{\sqrt n}\cdot\prod_{p\mathrm{\ prime,}\,p\le n}p\le n^{\sqrt n}\cdot 4^n</math> | ||
+ | |||
+ | according to [[Prime Number Theorem|Chebyshev's estimate]]. Also, we clearly have <math>D_n\ge 1</math>. Thus the absolute value of the product <math>D_n\int_0^1\frac{P(x)}{1+x}\,dx</math> | ||
+ | is between <math>(1/8)^n</math> and <math>(4\Gamma)^n\cdot n^{\sqrt n}</math> for large <math>n</math>. Note that <math>n^{\sqrt n}</math> grows slower than any geometric progression, so the upper bound can be replaced by <math>\gamma^n</math> with any <math>\gamma\in(4\Gamma, 1)</math>. In order to apply the main theorem from the [[rational approximation of famous numbers|parent article]], it remains to show that <math>Q_n</math> do not grow too fast. | ||
+ | |||
+ | ===Estimate of <math>Q_n</math>=== | ||
+ | We already saw that <math>D_n</math> grows not much faster than <math>4^n</math>. As to <math>|P(-1)|</math>, it does not exceed the sum of the absolute values of the coefficients of <math>P(x)</math>, which is not greater than <math>8^n</math>. Thus <math>|Q_n|</math> grow not much faster than <math>32^n</math>, and we are done. | ||
+ | |||
+ | ==See Also== | ||
+ | *[[Euler's constant]] | ||
+ | *[[Pi]] | ||
+ | |||
+ | [[Category:Number theory]] |
Latest revision as of 13:48, 26 January 2008
Rational approximation is the application of Dirichlet's theorem which shows that, for each irrational number , the inequality
has infinitely many solutions. On the other hand, sometimes it is useful to know that
cannot be approximated by rationals too well, or, more precisely, that
is not a Liouvillian number, i.e., that for some power
, the inequality
holds for all sufficiently large denominators
.
Contents
Theorem
Suppose that there exist ,
and a sequence of pairs of integers
such that for all sufficiently large
, we have
and
. Then, for every
, the inequality
has only finitely many solutions.
The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of , but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that
cannot be approximated by rationals too well, one needs to find plenty of small, but not too small, linear combinations of
and
with not too large integer coefficients.
Proof
Choose the least such that
. Note that for such choice of
, we have
. Also note that
(otherwise
would be an integer strictly between
and
). Now, there are two possible cases:
Case 1: .
Then
if is large enough.
Case 2: . Then
Hence, in this case,
if is large enough. (recall that
, so
).
Magic Polynomial
Before proceeding to the applications of the main theorem, let us introduce one very useful polynomial that often appears in proofs of irrationality. It is the polynomial
Its coefficients can be easily computed using the binomial theorem:
The important points are that all the coefficients are integer and
the sum of their absolute values does not exceed
.
Another useful remark is that the first derivatives of
vanish at
and
, which makes the integration by parts extremely convenient:
Example
This section will prove that the number is not Liouvillian. It can be read right after its parent article rational approximation of famous numbers. The proof of the non-Liouvillian character of
is much easier than that for
, but somewhat more difficult than that for
.
A Useful Integral
Everyone knows an integral representation for . It comes right from the definition of the natural logarithm:
. Let us look at what will happen if we replace
in the numerator by some polynomial
of degree
with integer coefficients. Since
where
is some polynomial of degree
with integer coefficients, we see that
Let now be the least common multiple of the numbers
. Then
with some integer and
.
It remains to choose a polynomial
that makes the integral small.
Polynomial ![$P(x)$](//latex.artofproblemsolving.com/6/e/f/6efaaf74e576202fad965134bc49c6a34aac4c30.png)
We shall just take our "magic polynomial"
(see the parent article for its properties).
Estimates of the Integral
Integrating by parts times, we see that
.
Since
and since it is attained only at the point
where
, we can conclude that
. The absolute value of our integral does not exceed
for all
. To estimate it from below, just notice that the integrand is at least
for
, so the absolute value of our integral is at least
for large
.
Estimate of ![$D_n$](//latex.artofproblemsolving.com/2/e/a/2eacbcb991af87d495f8689ac0e952b51317dbaa.png)
Since the largest possible power of a given prime that can divide one of the numbers
is
where
is the floor function, we have
according to Chebyshev's estimate. Also, we clearly have . Thus the absolute value of the product
is between
and
for large
. Note that
grows slower than any geometric progression, so the upper bound can be replaced by
with any
. In order to apply the main theorem from the parent article, it remains to show that
do not grow too fast.
Estimate of ![$Q_n$](//latex.artofproblemsolving.com/d/2/9/d29e04bbd77de0b9795acf7763c8fa9a8182806c.png)
We already saw that grows not much faster than
. As to
, it does not exceed the sum of the absolute values of the coefficients of
, which is not greater than
. Thus
grow not much faster than
, and we are done.