## Archive for the ‘Rational and algebraic numbers’ Category

### Roger Apéry’s proof that zeta(3) is irrational

October 14, 2011

At a conference in 1978 Roger Apéry outlined a proof that

$\displaystyle \zeta(3) =\sum_{n=1}^\infty\frac{1}{n^3}$

is irrational.  He published a brief note on his results in the journal ‘Astérisque’ (Apery 1979) and a fuller, but still cryptic, account two years later (Apery (1981)).  Olivier and Batut (1980) provided a detailed explanation of Apéry’s method of generating rapidly converging continued fractions.  Because none of these papers is readily accessible I have provided links to pdf copies.   van der Poorten’s Mathematical Intelligencer article is widely cited and indeed sometimes given as the primary reference for Apéry’s proof.  Certainly, being in English, it served to publicise Apéry’s result and to highlight connections with other areas of mathematics.  But it provides virtually no insight into the ideas underlying Apéry’s approach.

In the pdf here, I attempt to provide a full explanation of Apéry’s method of proof.  This is outlined below.

The proof relies on the criterion that α is irrational if there exist integers pn and qsuch that 0<|qnα – pn|=Tn<ε  for any given 1>ε>0

For example it can be easily shown that e is irrational by taking

$\displaystyle q_n=n!$  and  $\displaystyle p_n=n!\sum_{k=0}^{n}\frac{1}{k!}$

as Tn is positive and decreases with 1/n.

In 1761 Lambert showed that π is irrational by finding a continued fraction whose partial numerators and partial denominators, pn and qn respectively, have the desired property (see earlier post).  Apéry’s method is also based on  use of continued fractions.  These are essentially recurrences of the form

$p_1=a_1$  $p_2=b_2p_1$  $p_{n+1}=a_{n+1}p_{n-1}+b_{n+1}p_n$ (n ≥2)

$q_1=b_1$  $q_2=a_2+b_2q_1$  $q_{n+1}=a_{n+1}q_{n-1}+b_{n+1}q_n$ (n ≥2)

where the an and bn are rational functions of n.  pn and qn are referred to as partial numerators and partial denominators respectively and their ratio as partial convergents.  If α can be represented as a series whose terms are rational functions of n, then it can also, quite easily, be represented as a continued fraction.   The connection with the irrationality criterion arises because

$\displaystyle\left \vert q_{n}\alpha-p_{n}\right\vert=\left \vert q_{n}\sum_{j=n+1}^{\infty}\frac{(-1)^{j}a_j \dots a_1}{q_{j}q_{j-1}}\right\vert =T_{n}$

Apéry developed a method by means of which, starting with partial numerators and denominators pn,0 and qn,0 derived from a series representation of a number, a hierarchy of continued fractions pn,1, qn,1; pn,2, qn,2 and so on can be derived and in a variety of cases, including that of ζ(3), these converge to the same value as the original continued fraction and the Tn,k increase less rapidly with increasing n as k increases. Moreover the diagonal elements pn,n and qn,n are also partial numerators and denominators of a continued fraction having the same value as the original and the corresponding Tn is even more strongly constrained.

For  ζ(3) this ‘diagonal’ continued fraction has the form

$a_{1}=6$   $b_1={5}$

$\displaystyle a_n=- \frac{(n-1)^3}{n^3}$   $\displaystyle b_n=34-\frac {51}{n}+\frac{27}{n^2}-\frac{5}{n^3}$  (n > 1)

It is not too difficult to show from the recurrence that qn increases with e3.4n and Tn is non-zero and decreases with e-3.4n. pn and qn are rational so to prove that ζ(3) is irrational it remains to show that they can be made integers by multiplying by some integer function of n whose product with Tn is still a decreasing function of n.

Apéry astounded his audience by claiming that all qn are integers and all pn become integers if multiplied by 2(Ln)3 where Ln is the smallest positive integer divisible by each of the integers between 1 and n. Ln is the exponent of the Ψ function used by Chebyshev in his investigations on the Prime Number Theorem and it is well known as a consequence that Ln increases with en.  Therefore Apéry’s claim, if true, immediately proves the irrationality of ζ(3).

Apéry used a generating function method to solve the recurrence and obtain ‘closed form’ representations of qn and pn.  However it is more straightforward to work with the algorith Apéry used to generate his hierarchy of continued fractions.  This is a form of partial difference equation, and closed form solutions can be found in terms of shifted factorials.  A shifted factorial (r)s is the product of s successive positive integers, the largest of which is r. The rationale for searching for solutions in this form is that (r)s in discrete calculus is the analogue of rs in the calculus of the real variable r.

Using this approach we find that

$\displaystyle q_n=\sum_{j=0}^n \left [ \frac {(n+j)_{2j}}{(j!)^2} \right ]^2$

$\displaystyle p_n=q_{n}\Sigma_n+\sum_{j=0}^{n-1} \left [\frac{(n+j)_{2j}}{(j!)^2} \right ] ^2\left [\Sigma_n-\Sigma_j\right ]$

where Σj is the sum of the first j inverse cubes and Σ0 and (r)0 are defined to be zero and one respectively.

Since the product of any n consecutive positive integers is divisible by n! the truth of Apéry’s claim is immediately evident.

### Lambert’s continued fractions for tanh(x) and tan(x) revisited

April 20, 2010

In earlier posts (since deleted) I discussed some of the properties of continued fractions and showed how Lambert’s continued fractions for tanh x and tan x could be derived and used to prove that these functions are always irrational numbers for non-zero, rational x and consequently that the exponential number, e, and pi are irrational.

In what follows, I simplify and generalise those discussions. The following is an outline, the detail can be found in the pdf here – revised in May 2016.

Irrationality criteria and continued fractions

Suppose f(x) is some function that we can represent in the following manner:

$f(x) = s_1(x)=\displaystyle\frac{a_1(x)}{b_1(x)+s_2(x)}$

$s_k(x)=\displaystyle\frac{a_k(x)}{b_k(x)+s_{k+1}(x)} (k\geq2)$

where the $a_k, b_k$ are polynomials in x with integer coefficients and $s_k$ has the property that it becomes rapidly smaller in absolute value for a given x as k becomes larger. We call this a continued fraction representation of f(x) and it is not too hard to derive from it the following expression

$q_n(x)f(x)-p_n(x)=\displaystyle\frac{(-1)^{n-1}(a_na_{n-1}...a_1)s_{n+1}(x)}{q_n(x)+q_{n-1}(x)s_{n+1}(x)}=R_n(x)$

where p, q are defined recursively by

$p_{-1}=1$ $p_0=0$

$q_{-1}=0$ $q_0=1$

and for $k\geq0$

$p_{k+1}(x)=a_{k+1}p_{k-1}(x)+b_{k+1}p_k(x)$

$q_{k+1}(x)=a_{k+1}q_{k-1}(x)+b_{k+1}q_k(x)$

Clearly $p_n$ and $q_n$ are integral polynomials in x. If d(n) is the degree of whichever of $p_n$ and $q_n$ has the higher degree, and if x = u/v is rational with u, v being integers having greatest common factor 1, then if we multiply the above expression by $v^{d(n)}$ then it takes the form

$kf(x)-h = v^{d(n)}R_n(x)$

where k, h are integers and we can make use of the following theorem:

f(x) is irrational if there exist integers h and k such that 0 < |kf(x) – h| < ε for any given 1 > ε > 0

Lambert’s continued fractions

Suppose $f_1(x)$ and $f_2(x)$ are functions that are can be expressed as even, absolutely convergent Maclaurin series. Then we can obtain a sequence of equations

$f_{n+1}(x)\displaystyle\frac{f_{n+1}(0)}{f_n(0)}f_n(x)=x^2f_{n+2}(x)$

If we take $f_1(x)=\cosh x$ and $f_2(x)=\sinh x/x$
then it is not too hard to show that $\tanh x/x$ can be expressed as a continued fraction of the form

$\tanh x/x=\displaystyle\frac{1}{1+s_2(x)}$

$s_k(x)=\displaystyle\frac{x^2}{(2k-1)+s_{k+1}(x)}$

Analogously

$\tan x/x=\displaystyle\frac{1}{1+s_2(x)}$

$s_k(x)=\displaystyle\frac{-x^2}{(2k-1)+s_{k+1}(x)}$

These are Lambert’s continued fractions. The process described above can be carried through for them to show that tanh x/x and tan x/x, and hence tanh x and tan x, are irrational for non-zero rational x.

### Weierstrass’ proof of the Lindemann-Weierstrass Theorem (Part 3 of 3)

February 27, 2010

In part 3 of his 1885 paper, Weierstrass proved the theorem, which in the form stated by him is: if each of z1, … , zn is algebraic and distinct, and A1, … , An are algebraic then

$A_1e^{z_1}+ ... + A_ne^{z_n}$

cannot be zero unless all of A1, … , An are zero.

In the attached pdf (revised on 6 Aug 2013 – I am grateful to Fang Sun for pointing out a problem with the original which is addressed in this revision), I reproduce part 3 of Weierstrass’ paper, with some minor modifications and some added explanation.

### Weierstrass’ proof of the Lindemann-Weierstrass Theorem (Part 2 of 3)

February 16, 2010

In part 2 of his 1885 paper, Weierstrass reproduced Lindemann’s proof that if z is algebraic,

$e^z+1$

is not zero.

Weierstrass’ proof of Lindemann’s result is concise and elegant, while still displaying its origins in Hermite’s ground-breaking proof that e is transcendental.  This contrasts with some of the ‘simplifications’ that have followed, for example the surprisingly cumbersome treatment given by Hardy and Wright.

In the attached pdf (updated 25 July 2013), I reproduce part 2 of Weierstrass’ paper, with some minor modifications and some added material on symmetric polynomials.  The latter draws on the description of Waring’s method in Tignol’s book ‘Galois’ Theory of Algebraic Equations’.

### Weierstrass’ proof of the Lindemann-Weierstrass Theorem (Part 1 of 3)

February 13, 2010

In part 1 of his 1885 paper, Weierstrass demonstrated the following lemma, which is both a simplification and a generalisation of the method developed by Hermite to prove the transcendence of e:

Lemma:  Let f(z) be a polynomial of degree n+1 with integer coefficients and whose roots z0, … , zn are all distinct.  Then there exists a system of polynomials g0(z), … , gn(z) of degree not greater than n in z, and with integer coefficients, such that (i) each of the differences

$g_{\nu}(z_0)e^{z_\lambda}-g_{\nu}(z_{\lambda})e^{z_0}$

(where ν,λ can take any of the values 0, 1, … , n) can be made arbitrarily small in absolute value, and (ii) the determinant whose elements are gν(zλ) is non-zero.

In the attached pdf (updated on 1 July 2013), I reproduce Weierstrass’ proof of this lemma, with some minor modifications.

### Lindemann-Weierstrass Theorem

February 6, 2010

Lindemann showed that pi is transcendental by showing that if the complex number z is algebraic (the root of an integral polynomial) then ez +1 cannot be zero.

Lindemann also claimed, and Weierstrass proved rigorously, that this result can be generalised to the following: if each of z1, … , zn is algebraic, and N1, … , Nn are algebraic and non-zero, then

$N_1e^{z_1}+ ... +N_ne^{z_n}$

cannot be zero.

It follows immediately that if z is algebraic and non-zero, ez, sin z and cos z are transcendental and corresponding results can be obtained for the inverse functions.

Although proofs of the transcendence of pi are found in many texts, for example the proof in Alan Baker’s book ‘Transcendental Number Theory’  discussed in the previous post, full proofs of the Lindemann-Weierstrass Theorem are harder to find.  Baker provides a two-page proof, which is widely cited, but it is extremely condensed and key steps are omitted.  I have been forced to go back to Weierstrass’ original 1885 paper to find a full proof.

Weierstrass’ paper was published in the rather fulsomely titled ‘Sitzungsberichte der Königlich Preußischen Akademie der Wissenschaften zu Berlin’, pages 1067 to 1085.  A link to the journal is provided on the ‘Journals’ page.

Weierstrass’ paper is in three parts, and in the next three posts I will discuss each of these parts in turn.

### Pi is transcendental

November 21, 2009

Lindemann’s proof that pi is transcendental was published in 1882, nine years after Hermite’s proof that e is transcendental.  Lindemann used Hermite’s methods, but generalised them considerably.

Lindemann showed that if z is complex and algebraic (the root of a polynomial with integer coefficients) then ez+1 cannot be zero. Since by Euler’s identity e+1 is zero, iπ is not algebraic. From this it is easy to show that pi is not algebraic.

The University of St Andrews mathematical biography series says of this work of Lindemann:

‘Many historians of science regret that Hermite, despite doing most of the hard work, failed to make the final step to prove the result concerning which would have brought him fame outside the world of mathematics. This fame was instead heaped on Lindemann but many feel that he was a mathematician clearly inferior to Hermite who, by good luck, stumbled on a famous result. Although there is some truth in this, it is still true that many people make their own luck and in Lindemann’s case one has to give him much credit for spotting the trick which Hermite had failed to see.’

I feel this opinion is a little unjust. To achieve his result, Lindemann exploited the symmetry properties of the roots of integral algebraic equations in an extremely subtle way, and his results were far more general.

Rather than provide a proof within this blog, I have attached a pdf:

pdf – pi is transcendental

This proof follows the concise and elegant proof given by Baker in his book ‘Transcendental Number Theory’ but enlarges on some aspects, in particular those to do with the properties of symmetric functions. The proof of the ‘Fundamental Theorem of Symmetric Functions’, given as Lemma 3 in the pdf, closely follows that in Tignol’s book ‘Galois’ Theory of Algebraic Equations’. The reference to H&W is of course to Hardy and Wright’s book on Number Theory.

### e is transcendental

August 19, 2009

In 1873 Charles Hermite showed that e is transcendental.  Liouville had earlier shown how to construct such numbers, but Hermite’s was the first proof that an ‘interesting’ mathematical constant was transcendental.

Hermite’s proof is long and complex.  It has since been simplified considerably and in particular a short and elegant proof appears in Baker’s book ‘Transcendental Number Theory’.  However Baker himself says that the motivation of the proof has been obscured by the process of simplification, and it is necessary to go back to Hermite’s original paper in order to understand it.

I think this is not entirely so but nevertheless in this post  I give in outline a version of Baker’s proof that I preface with some references to ideas in Hermite’s paper and I hope in so doing I make the motivation clearer.  A more detailed version of this outline is in a pdf here.

As discussed in an earlier post, by repeated integration by parts Hermite obtained an expression of the form

$i^{2m+1}\int_0^1e^{it}F_m(t)dt=k_i(m)e^i-h_i(m)$

where Fm(t)=tm(1-t)m , m and i are positive integers and ki, hi are polynomials with integer coefficients.  The expression on the left of the equation is not zero but can be made arbitrarily small in absolute value by taking m sufficiently large, thereby proving that ei is irrational for any integer i.

This suggests that one could obtain in a similar manner a set of equations of the form

$k(m)e^i-h_i(m)=I_i(m)$   ………………………………………….   (1)

where k(m) is a non-zero integer not dependent on i and the Ii(m) either decrease in absolute value or grow only slowly with increasing m.

Now let q0, q1, … , qn be any set of integers (provided that n > 0 and q0, qn ≠0).  If we multiply (1) by qi and sum over i then we have

$\sum_{i=0}^nq_iI_i(m)=-\sum_{i-0}^nq_ih_i(m)+k(m)(q_0+q_1e+ ... +q_ne^n)$

Then if we can find Ii(m) and hi(m) and non-zero k(m) such that for sufficiently large m,

$|\sum_{i=0}^nq_iI_i|<|\sum_{i=0}^nq_ih_i|$   ……………………………………   (2)

we have proved e is transcendental because then it is not possible to have

$q_0+q_1e+ ... +q_ne^n=0$

If we use in the integral ei-t rather than eit as used by Hermite we  immediately obtain an expression in the desired form (1).  Let

$I_i(m)=\int_0^ie^{i-t}f(t)dt$

where f(t) is a polynomial of degree m with integer coefficients.  For i > 0 repeated integration by parts gives

$I_i(m)=e^i\sum_{j=0}^mf^{(j)}(0)-\sum_{j=0}^mf^{(j)}(i)$

and the equation, which is identical in form to (1), is also correct in a formal sense for I0=0.  Now let

$f(t)=t^{p-1}(t-1)^p(t-2)^p ... (t-n)^p$

Then

$|I_i|\leq\int_0^i|e^{i-t}f(t)|dt<(2i)^mie^i$

and it is not hard to find that for sufficiently large p

$|\sum_0^nq_iI_i| < A^p$

where A is positive and independent of p.  This puts an upper bound on the left hand side of (2).  It is also not too hard to show (pdf) that

$k(m)=\sum_{j=0}^mf^{(j)}(0)$

cannot be zero if we make p a prime and greater than n (which we are quite at liberty to do).  Finally we can show that as long as p>|q0| then

$|\sum_{i=0}^nq_ih_i(m)| > (p-1)!$

giving a lower bound to the right hand side of (2).  But since (p-1)!>Ap for some large enough p, then (2) holds and we have proved e is transcendental.

### The irrational nature of zeta(3)

August 2, 2009

If n is an integer greater than one, one defines

$\displaystyle\zeta(n)=\sum_{n=1}^\infty\frac{1}{k^n}$

When n  is even, $\zeta(n)$ is irrational iff $\pi^n$ is irrational since

$\displaystyle\zeta(2n)=\frac{2^{2n-1}|B_{2n}|\pi^{2n}}{(2n)!}$

and $B_n$, the n th Bernoulli number, is rational.

The nature of $\zeta(n)$ for odd n remained a mystery until 1978, when it was shown by Roger Apéry that $\zeta(3)$ is irrational.  (The mystery remains for odd n>3).

As before we use the fact that

x is irrational if there exist integers h and k such that $0 < |kx - h| < \epsilon$ for any given real number $1 > \epsilon> 0$

Apéry [1] described sequences $h_n, k_n$ that satisfy this condition for sufficiently large  n when $x = \zeta(3)$.  In 1979 Beukers described [2] a simpler method of arriving at the same result.  This works as follows (to keep the post, and my editing time, short, I haven’t included proofs of the lemmas.  Instead, the full article with lemmas is available as a pdf here)

Firstly we note the following formulas (see Lemma 1)

$\displaystyle\int^1_0\int^1_0\frac{x^iy^i\ln xy}{1-xy}dydx=-2\sum^\infty_{k=i+1}\frac{1}{k^3}$

$\displaystyle =\begin{cases}-2\zeta(3) \text{ if i=0}\\-2\zeta(3)+2\sum^i_{k=1}\frac{1}{k^3} \text{ if }i>0\end{cases}$

When i>j

$\displaystyle\int^1_0\int^1_0\frac{x^iy^j\ln xy}{1-xy}dydx=\frac{-1}{i-j}\sum^i_{k=j+1}\frac{1}{k^2}$

Suppose $F_n(x)$ is some polynomial in x of degree n with integer coefficients.  Then we can represent $F_n(x)F_n(y)$ as

$\displaystyle F_n(x)F_n(y)=\sum^n_{i=0}a_{ii}x^iy^i+ \sum^n_{i=1}\sum^{i-1}_{j=0}a_{ij}x^iy^j$

with the aij integers.

Combining the three expressions above we can obtain

$\displaystyle\int^1_0\int^1_0\frac{F_n(x)F_n(y)\ln xy}{1-xy}dydx=-2a_{00}(n)\zeta(3)+$

$\displaystyle\sum^n_{i=1}a_{ii}(-2\zeta(3)+2\sum^i_{k=1}\frac{1}{k^3})+2\sum^n_{i=1}\sum^{i-1}_{j=0}a_{ij}\frac{-1}{i-j}\sum^i_{k=j+1}\frac{1}{k^2}$

Let dn be the least integer that is divisible by each of the integers 1,2, .. ,n.  Then it is not hard to see that the denominators of all the fractions on the right hand side of the above expression divide (dn)3.  Therefore we have

$\displaystyle(d_n)^3\int^1_0\int^1_0\frac{F_n(x)F_n(y)\ln xy}{1-xy}dydx=k_n\zeta(3)-h_n$

with kn, hn integers.  To complete the proof we need to find an $F_n(x)$ such that for large enough n, we can make the absolute value of the left hand side as close as we like to zero, without actually being zero.

It was established in connection with the proof of the Prime Number Theorem that [3] dn<e(1+δ)n for sufficiently large n and any given positive real number δ.  Consequently (dn)3<e(3+δ)n for sufficiently large n.

In an earlier post we made use of the polynomial

$(d^n/dx^n)(x^n(1-x)^n)$

This is a polynomial of degree n with integer coefficients, and each of its coefficients is divisible by n! (Lemma 2).  We take

$\displaystyle F_n(x)=\frac{1}{n!}(d^n/dx^n)(x^n(1-x)^n)$

It is not too difficult to establish (Lemma 3) that

$\displaystyle\int^1_0\int^1_0\frac{F_n(x)F_n(y)\ln xy}{1-xy}dydx<\exp^{-4(\ln(\sqrt{2}+1))n}

Consequently

$\displaystyle0<(d_n)^3\int^1_0\int^1_0\frac{F_n(x)F_n(y)\ln xy}{1-xy}dydx

And the result is proved.

[1] Roger Apéry, ‘Irrationalité de ζ(2)et ζ(3)’, Astérisque 61, 1979, pp 11-13; and ‘Interpolation de fractions continues et irrationalite de certaines constantes’, Bulletin de la Section des Sciences du C.T.H.S. III, 1981, pp 37-53.

[2] Frits Beukers, ‘A Note on the Irrationality of ζ(2)and ζ(3)’, Bulletin of the London Mathematical. Society, 11, 1979, pp 268-272.

[3]  In analytical number theory, the function ln dn is usually called ψ(n).  This function was introduced around 1850 by Chebyshev who showed that the conjecture limn→∞ψ(n)/n=1 (if such a limit exists) implies the prime number theorem.  (Chebyschev also obtained upper and lower bounds on ψ(n)/n and showed that if a limit exists, it must be 1).  Most proofs of the Prime Number Theorem whether by way of Riemann’s zeta function or by way of the ‘elementary proof’ proceed by proving the limit exists.  Consequently ln dn/n<1+δ for sufficiently large n and any given real positive δ.

### Niven’s proof that the trigonometric and inverse trigonometric functions are irrational for rational non-zero arguments

July 17, 2009

In the late 1940s, Niven discovered a way to apply Hermite’s method to the cos function.  The following proof that cos x is irrational for non-zero rational x is adapted from that given in Niven’s book on irrational numbers [1].  I have tried to simplify the proof and make the motivation a little clearer.

Once again, we begin with the fact that if x is a real number

x is irrational if there exist integers h and k such that 0<|kx – h|< ε for any given real number 1>ε>0

Now consider the integral ∫sin(ht)Fn(t)dt with h a positive integer and F(t) a polynomial in t with integer coefficients, and whose degree is 4n.  If we integrate repeatedly by parts we obtain

$h^{4n+1}\int \sin ht F_n(t)dt=\cos ht\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)+\\\indent\sin ht\sum_{(k+1)/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)$

where in the first and second sums k is permitted to take only even and odd values respectively.  Now suppose we take

$F_n(t)=(1-t)^{2n}(1-(1-t)^2)^n$

Since this is an even polynomial in (1-t), its odd derivatives are zero at t=1.  Hence the second sum vanishes when t=1, and since sin ht=0 when t=0, if we integrate from t=0 to t=1 we obtain

$h^{4n+1}\int_0^1 \sin ht F_n(t)dt=\cos h\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(1)-\\\indent\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(0)=(\cos h)k_n-h_n$

with hn, kn integers.  $F_n(t)$ has the properties that 1≥ $F_n(t)$≥0 if 1≥t≥ 0, and (see lemma below) $F^{(k)}_n(1)$ and $F^{(k)}_n(0)$ are divisible by n! so we have

$h\frac{(h^4)^n}{n!}\int_0^1\sin htF_n(t)dt=(\cos h)k_n^\prime+h_n^\prime$

with k’n(h) = kn(h)/n! and h’n(h) = hn(h)/n! integers.

The absolute value of the integral is less than 1.   (h4)n/n! can be made arbitrarily small by taking n large enough, so for all n greater than some sufficiently large value, nε say, the LHS of the equation can be made less in absolute value than any given positive real number ε.

To finish we need only show that for any given real number r and some n≥nε , rk’n+h’n≠0.  This must be the case if r is irrational so we suppose r = p/q with p, q>0 and g.c.d.(p,q) = 1.  rk’n+h’n cannot be zero if pk’n+qh’n is not zero.

By the Lemma, $F^{(k)}_n(1)/n!$ and hence k’n is divisible by n+1.  We take n even.  By the Lemma, all the terms $F^{(k)}_n(0)/n!$ in the sum

$h^\prime_n=\sum_{k/2=0}^{2n}(-1)^kh^{4n-k}F^{(k)}_n(0)$

are divisible by n + 1 with the exception of $F^{(n)}_n(0)$.  We take n + 1 greater than nε , and such that it contains a prime factor that is not a prime factor of either h or q.  Then qh’n is not divisible by n + 1.  If n + 1 divides pk’n but not qh’n , their sum cannot be zero.

The fact that cos h is irrational for a positive integer h implies that cos x is irrational for all non-zero rational numbers x.  Let x be positive and rational and equal to a/b, with a, b positive integers.  Then as a consequence of the recursive trigonometric identity

$\cos nt = 2\cos t \cos(n-1)t-\cos (n-2)t$

cos nt is a polynomial in cos t, and so cos bx = cos a is a polynomial in cos x.  If cos x is rational, then cos a is rational, so by the counterpositive, since cos a is irrational, cos x is irrational.  Since cos (-x)=cos x, the result is true also for negative rational x.

The following is also due to Niven.

If sin x were rational for rational x ≠ 0, so would be $1- 2\sin^2x=\cos 2x$.  Thus sin x is irrational for rational x ≠ 0.

If tan x were rational for rational x ≠ 0, so would be

$(1-\tan^2x)/(1+\tan^2x)=\cos 2x$

Thus tan x is irrational for rational x≠0.

As a consequence of the following

Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f-1(y) where y is the set of numbers f(x).  Let f be such that if x is rational then f(x) is irrational.  Then if y is rational f-1(y) is irrational.

the inverse trigonometric functions can be shown to be irrational for non-zero rational x.  Since $\arctan 1 = \pi/4$, $\pi$ is necessarily irrational.

Niven provides a proof along similar lines that $\pi^2$ is irrational.  However Lambert’s continued fraction method can also be used to prove the same result, and I will provide this proof in a later post.

Lemma:  Let

$F_n(t)=(1-t)^{2n}(1-(1-t)^2)n=(1-t)^{2n}t^n(2-t)^n$

Then $F^{(k)}_n(1)$ is divisible by n+1! and $F^{(k)}_n(0)$ is divisible by n!  $F^{(k)}_n(0)$ is divisible by n+1! when k ≠ n, but not when k = n if n+1 is not a power of two.

Proof$F^{(k)}_n(t)=(-1)^k\frac{d^kF_n(t)}{d(1-t)^k}$.  Since $F_n(t)$ is a polynomial in $1-t$ whose term of least degree is $(1-t)^{2n}$, $F^{(k)}_n(1)=0$ when k<2n.  When 2n≤ k≤ 4n, $F^{(k)}_n(t)=b_kk!+f(1-t)$.  n+1! divides k!, so $F^{(k)}_n(1)/(n+1!)$ is an integer.

$F_n(t)$ is a polynomial in t whose term of least degree is tn, so $F^{(k)}_n(0)=0$ when k < n.  $F^n_n(0) = 2^nn!$ so if n+1 is not a power of two, it is not divisible by n+1!  When k > n, $F^{(k)}_n(t)$ = akk! + f(t), where f(t) is either zero or has a factor t.  Hence $F^{(k)}_n(t)$ is divisible by k! and hence by n+1!

[1] I Niven, ‘Irrational Numbers’, Wiley, 1956 pp 16-19