Proof that π is irrational

File:LambertContinuedFraction.JPG

In 1761, Johann Heinrich Lambert proved that $\backslash pi$ is irrational by first showing that this continued fraction expansion holds:

:$\backslash tan(x)\; =\; \backslash cfrac\{x\}\{1\; -\; \backslash cfrac\{x^2\}\{3\; -\; \backslash cfrac\{x^2\}\{5\; -\; \backslash cfrac\{x^2\}\{7\; -\; \{\}\backslash ddots\}\}\}\}.$

Then Lambert proved that if $x$ is non-zero and rational, then this expression must be irrational. Since $\backslash tan\backslash tfrac\backslash pi4\; =1$, it follows that $\backslash tfrac\backslash pi4$ is irrational, and thus $\backslash pi$ is also irrational.{{citation |editor1-last = Berggren |editor1-first = Lennart |editor2-last = Borwein |editor2-first = Jonathan M. |editor2-link = Jonathan M. Borwein |editor3-last = Borwein |editor3-first = Peter B. |editor3-link = Peter B. Borwein |last = Lambert |first = Johann Heinrich |orig-year = 1768 |chapter = Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques |title = Pi, a source book |place = New York |publisher = Springer-Verlag |year = 2004 |edition = 3rd |pages = 129–140 |isbn = 0-387-20571-3}}. A simplification of Lambert's proof is given below.

Written in 1873, this proof uses the characterization of $\backslash pi$ as the smallest positive number whose half is a zero of the cosine function and it actually proves that $\backslash pi^2$ is irrational.{{cite journal |last=Hermite |first=Charles |author-link=Charles Hermite |year=1873 |title=Extrait d'une lettre de Monsieur Ch. Hermite à Monsieur Paul Gordan |language=fr |journal=Journal für die reine und angewandte Mathematik |volume=76 |pages=303–311 |url=http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN002155435}}{{cite journal |last=Hermite |first=Charles |author-link=Charles Hermite |year=1873 |title=Extrait d'une lettre de Mr. Ch. Hermite à Mr. Carl Borchardt |language=fr |journal=Journal für die reine und angewandte Mathematik |volume=76 |pages=342–344 |url=http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN00215546X}} As in many proofs of irrationality, it is a proof by contradiction.

Consider the sequences of real functions $A\_n$ and $U\_n$ for $n\; \backslash in\; \backslash N\_0$ defined by:

: $\backslash begin\{align\}$

A_0(x) &= \sin(x), && A_{n+1}(x) =\int_0^xyA_n(y)\,dy \\[4pt]

U_0(x) &= \frac{\sin(x)}x, && U_{n+1}(x) =-\frac{U_n'(x)}x

\end{align}

Using induction we can prove that

:$\backslash begin\{align\}$

A_n(x) &=\frac{x^{2n+1}}{(2n+1)!!}-\frac{x^{2n+3}}{2\times(2n+3)!!}+\frac{x^{2n+5}}{2\times4\times(2n+5)!!}\mp\cdots \\[4pt]

U_n(x) &=\frac1{(2n+1)!!}-\frac{x^2}{2\times(2n+3)!!}+\frac{x^4}{2\times4\times(2n+5)!!}\mp\cdots

\end{align}

and therefore we have:

:$U\_n(x)=\backslash frac\{A\_n(x)\}\{x^\{2n+1\}\}.\backslash ,$

So

:$$

\begin{align}

\frac{A_{n+1}(x)}{x^{2n+3}} & =U_{n+1}(x)=-\frac{U_n'(x)}x=-\frac1x\frac {\mathrm{d}}{\mathrm{d}x}\left(\frac{A_n(x)}{x^{2n+1}}\right) \\[6pt]

& = -\frac{1}{x} \left( \frac{A_n'(x) \cdot x^{2n+1} - (2n+1) x^{2n} A_n(x)}{x^{2(2n+1)}} \right ) \\[6pt]

& = \frac{(2n+1)A_n(x)-xA_n'(x)}{x^{2n+3}}

\end{align}

which is equivalent to

:$A\_\{n+1\}(x)=(2n+1)A\_n(x)-x^2A\_\{n-1\}(x).\backslash ,$

Using the definition of the sequence and employing induction we can show that

:$A\_n(x)\; =\; P\_n(x^2)\; \backslash sin(x)\; +\; x\; Q\_n(x^2)\; \backslash cos(x),\backslash ,$

where $P\_n$ and $Q\_n$ are polynomial functions with integer coefficients and the degree of $P\_n$ is smaller than or equal to $\backslash bigl\backslash lfloor\; \backslash tfrac12n\backslash bigr\backslash rfloor.$ In particular, $A\_n\backslash bigl(\backslash tfrac12\backslash pi\backslash bigr)\; =\; P\_n\backslash bigl(\backslash tfrac14\backslash pi^2\backslash bigr).$

Hermite also gave a closed expression for the function $A\_n,$ namely

:$A\_n(x)=\backslash frac\{x^\{2n+1\}\}\{2^n\; n!\}\backslash int\_0^1(1-z^2)^n\backslash cos(xz)\backslash ,\backslash mathrm\{d\}z.\backslash ,$

He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to

:$\backslash frac\{1\}\{2^n\; n!\}\backslash int\_0^1(1-z^2)^n\backslash cos(x\; z)\backslash ,\backslash mathrm\{d\}z=\backslash frac\{A\_n(x)\}\{x^\{2n+1\}\}=U\_n(x).$

Proceeding by induction, take $n\; =\; 0.$

:$\backslash int\_0^1\backslash cos(xz)\backslash ,\backslash mathrm\{d\}z=\backslash frac\{\backslash sin(x)\}x=U\_0(x)$

and, for the inductive step, consider any natural number $n.$ If

:$\backslash frac\{1\}\{2^nn!\}\backslash int\_0^1(1-z^2)^n\backslash cos(xz)\backslash ,\backslash mathrm\{d\}z=U\_n(x),$

then, using integration by parts and Leibniz's rule, one gets

:$\backslash begin\{align\}$

&\frac{1}{2^{n+1}(n+1)!} \int_0^1\left(1-z^2\right)^{n+1}\cos(xz)\,\mathrm{d}z \\

&\qquad=\frac{1}{2^{n+1}(n+1)!}\Biggl(\,\overbrace{\left.(1-z^2)^{n+1}\frac{\sin(xz)}x\right|_{z=0}^{z=1}}^{=\,0} \ +\, \int_0^12(n+1)\left(1-z^2\right)^nz \frac{\sin(xz)}x\,\mathrm{d}z\Biggr)\\[8pt]

&\qquad= \frac1x\cdot\frac1{2^n n!}\int_0^1\left(1-z^2\right)^nz\sin(xz)\,\mathrm{d}z\\[8pt]

&\qquad= -\frac1x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1{2^nn!}\int_0^1(1-z^2)^n\cos(xz)\,\mathrm{d}z\right) \\[8pt]

&\qquad= -\frac{U_n'(x)}x \\[4pt]

&\qquad= U_{n+1}(x).

\end{align}

If $\backslash tfrac14\backslash pi^2\; =\; p/q,$ with $p$ and $q$ in $\backslash N$, then, since the coefficients of $P\_n$ are integers and its degree is smaller than or equal to $\backslash bigl\backslash lfloor\; \backslash tfrac12n\backslash bigr\backslash rfloor,$ $q^\{\backslash lfloor\; n/2\; \backslash rfloor\}P\_n\backslash bigl(\backslash tfrac14\backslash pi^2\backslash bigr)$ is some integer $N.$ In other words,

:$N=q^\{\backslash lfloor\; n/2\backslash rfloor\}\{A\_n\}\backslash bigl(\backslash tfrac12\backslash pi\backslash bigr)\; =q^\{\backslash lfloor\; n/2\backslash rfloor\}\backslash frac\{1\}\{2^nn!\}\backslash left(\backslash dfrac\; pq\; \backslash right)^\{n+\backslash frac\; 12\}\backslash int\_0^1(1-z^2)^n\; \backslash cos\; \backslash left(\backslash tfrac12\backslash pi\; z\; \backslash right)\backslash ,\backslash mathrm\{d\}z.$

But this number is clearly greater than $0.$ On the other hand, the limit of this quantity as $n$ goes to infinity is zero, and so, if $n$ is large enough, Thereby, a contradiction is reached.

Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the transcendence of $\backslash pi.$ He discussed the recurrence relations to motivate and to obtain a convenient integral representation. Once this integral representation is obtained, there are various ways to present a succinct and self-contained proof starting from the integral (as in Cartwright's, Bourbaki's or Niven's presentations), which Hermite could easily see (as he did in his proof of the transcendence of $e${{cite book |last=Hermite |first=Charles |author-link=Charles Hermite |editor-last=Picard |editor-first=Émile |editor-link=Charles Émile Picard |title=Œuvres de Charles Hermite |volume=III |publisher=Gauthier-Villars |year=1912 |language=fr |pages=150–181 |contribution=Sur la fonction exponentielle |url=http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=AAS7821.0003.001;didno=AAS7821.0003.001;view=pdf;seq=00000161 |orig-year=1873}}).

Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, $A\_n(x)$ is the "residue" (or "remainder") of Lambert's continued fraction for $\backslash tan\; x.${{cite journal |last=Zhou |first=Li |title=Irrationality proofs à la Hermite |journal=The Mathematical Gazette |year=2011 |volume=95 |issue=534|pages=407–413 |arxiv=0911.1929 |doi=10.1017/S0025557200003491|s2cid=115175505 }}

Harold Jeffreys wrote that this proof was set as an example in an exam at Cambridge University in 1945 by Mary Cartwright, but that she had not traced its origin.{{citation|last = Jeffreys|first = Harold|author-link = Harold Jeffreys|title = Scientific Inference|publisher = Cambridge University Press|year = 1973|page = [https://archive.org/details/scientificinfere0000jeff/page/268 268]|edition = 3rd|isbn = 0-521-08446-6|url = https://archive.org/details/scientificinfere0000jeff/page/268}} It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University.{{Cite web |title=Department of Pure Mathematics and Mathematical Statistics |url=http://www.dpmms.cam.ac.uk/study/IA/AnalysisI/ |access-date=2022-04-19 |website=www.dpmms.cam.ac.uk}}

Consider the integrals

:$I\_n(x)=\backslash int\_\{-1\}^1(1\; -\; z^2)^n\backslash cos(xz)\backslash ,dz,$

where $n$ is a non-negative integer.

Two integrations by parts give the recurrence relation

:$x^2I\_n(x)=2n(2n-1)I\_\{n-1\}(x)-4n(n-1)I\_\{n-2\}(x).\; \backslash qquad\; (n\; \backslash geq\; 2)$

If

:$J\_n(x)=x^\{2n+1\}I\_n(x),$

then this becomes

:$J\_n(x)=2n(2n-1)J\_\{n-1\}(x)-4n(n-1)x^2J\_\{n-2\}(x).$

Furthermore, $J\_0(x)\; =\; 2\; \backslash sin\; x$ and $J\_1(x)\; =\; -4x\backslash cos\; x\; +\; 4\backslash sin\; x.$ Hence for all $n\; \backslash in\; \backslash Z\_+,$

:$J\_n(x)=x^\{2n+1\}I\_n(x)=n!\backslash bigl(P\_n(x)\backslash sin(x)+Q\_n(x)\backslash cos(x)\backslash bigr),$

where $P\_n(x)$ and $Q\_n(x)$ are polynomials of degree $\backslash leq\; n,$ and with integer coefficients (depending on {{nobr|$n$).}}

Take $x\; =\; \backslash tfrac12\backslash pi,$ and suppose if possible that $\backslash tfrac12\backslash pi\; =\; a/b$ where $a$ and $b$ are natural numbers (i.e., assume that $\backslash pi$ is rational). Then

: $\backslash frac\{a^\{2n+1\}\}\{n!\}I\_n\backslash bigl(\backslash tfrac12\backslash pi\backslash bigr)\; =\; P\_n\backslash bigl(\backslash tfrac12\backslash pi\backslash bigr)b^\{2n+1\}.$

The right side is an integer. But since the interval $[-1,1]$ has length $2$ and the function being integrated takes only values between $0$ and $1.$ On the other hand,

: $\backslash frac\{a^\{2n+1\}\}\{n!\}\; \backslash to\; 0\; \backslash quad\; \backslash text\{\; as\; \}n\; \backslash to\; \backslash infty.$

Hence, for sufficiently large $n$

:

that is, we could find an integer between $0$ and $1.$ That is the contradiction that follows from the assumption that $\backslash pi$ is rational.

This proof is similar to Hermite's proof. Indeed,

:$\backslash begin\{align\}$

J_n(x)&=x^{2n+1}\int_{-1}^1 (1 - z^2)^n \cos(xz)\,dz\\[5pt]

&=2x^{2n+1}\int_0^1 (1 - z^2)^n \cos(xz)\,dz\\[5pt]

&=2^{n+1}n!A_n(x).

\end{align}

However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions $A\_n$ and taking as a starting point their expression as an integral.

This proof uses the characterization of $\backslash pi$ as the smallest positive zero of the sine function.{{citation|last = Niven|first = Ivan|author-link = Ivan Niven|year = 1947|title = A simple proof that {{pi}} is irrational|periodical = Bulletin of the American Mathematical Society|volume = 53|issue = 6|page = 509|url = https://www.ams.org/bull/1947-53-06/S0002-9904-1947-08821-2/S0002-9904-1947-08821-2.pdf|doi=10.1090/s0002-9904-1947-08821-2|doi-access = free}}

Suppose that $\backslash pi$ is rational, i.e. $\backslash pi\; =\; a/b$ for some integers $a$ and $b$ which may be taken without loss of generality to both be positive. Given any positive integer $n,$ we define the polynomial function:

:$f(x)\; =\; \backslash frac\{x^n(a\; -\; bx)^n\}\{n!\}$

and, for each $x\; \backslash in\; \backslash R$ let

:$F(x)\; =\; f(x)-f\text{'}\text{'}(x)+f^\{(4)\}(x)+\backslash cdots+(-1)^n\; f^\{(2n)\}(x).$

Claim 1: $F(0)\; +\; F(\backslash pi)$ is an integer.

Proof:

Expanding $f$ as a sum of monomials, the coefficient of $x^k$ is a number of the form $c\_k\; /n!$ where $c\_k$ is an integer, which is $0$ if Therefore, $f^\{(k)\}(0)$ is $0$ when and it is equal to $(k!\; /\; n!)\; c\_k$ if {{nobr|$n\; \backslash leq\; k\; \backslash leq\; 2n$;}} in each case, $f^\{(k)\}(0)$ is an integer and therefore $F(0)$ is an integer.

On the other hand, $f(\backslash pi-x)\; =\; f(x)$ and so $(-1)^kf^\{(k)\}(\backslash pi-x)\; =\; f^\{(k)\}(x)$ for each non-negative integer $k.$ In particular, $(-1)^kf^\{(k)\}(\backslash pi)\; =\; f^\{(k)\}(0).$ Therefore, $f^\{(k)\}(\backslash pi)$ is also an integer and so $F(\backslash pi)$ is an integer (in fact, it is easy to see that {{nobr|$F(\backslash pi)\; =\; F(0)$).}} Since $F(0)$ and $F(\backslash pi)$ are integers, so is their sum.

Claim 2:

:$\backslash int\_0^\backslash pi\; f(x)\backslash sin(x)\backslash ,dx=F(0)+F(\backslash pi)$

Proof: Since $f^\{(2n\; +\; 2)\}$ is the zero polynomial, we have

:$F\text{'}\text{'}\; +\; F\; =\; f.$

The derivatives of the sine and cosine function are given by sin' = cos and cos' = −sin. Hence the product rule implies

:$(F\text{'}\backslash cdot\backslash sin\{\}\; -\; F\backslash cdot\backslash cos\{\})\text{'}\; =\; f\backslash cdot\backslash sin$

By the fundamental theorem of calculus

: $\backslash left.\; \backslash int\_0^\backslash pi\; f(x)\backslash sin(x)\backslash ,dx=\; \backslash bigl(F\text{'}(x)\backslash sin\; x\; -\; F(x)\backslash cos\; x\backslash bigr)\; \backslash right|\_0^\backslash pi.$

Since $\backslash sin\; 0\; =\; \backslash sin\; \backslash pi\; =\; 0$ and $\backslash cos\; 0\; =\; -\; \backslash cos\; \backslash pi\; =\; 1$ (here we use the above-mentioned characterization of $\backslash pi$ as a zero of the sine function), Claim 2 follows.

Conclusion: Since $f(x)\; >\; 0$ and $\backslash sin\; x\; >\; 0$ for (because $\backslash pi$ is the smallest positive zero of the sine function), Claims 1 and 2 show that $F(0)\; +\; F(\backslash pi)$ is a positive integer. Since $0\; \backslash leq\; x(a\; -\; bx)\; \backslash leq\; \backslash pi\; a$ and $0\; \backslash leq\; \backslash sin\; x\; \backslash leq\; 1$ for $0\; \backslash leq\; x\; \backslash leq\; \backslash pi,$ we have, by the original definition of $f,$

:$\backslash int\_0^\backslash pi\; f(x)\backslash sin(x)\backslash ,dx\backslash le\backslash pi\backslash frac\{(\backslash pi\; a)^n\}\{n!\}$

which is smaller than $1$ for large $n,$ hence for these $n,$ by Claim 2. This is impossible for the positive integer $F(0)\; +\; F(\backslash pi).$ This shows that the original assumption that $\backslash pi$ is rational leads to a contradiction, which concludes the proof.

The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula

:$\backslash int\_0^\backslash pi\; f(x)\backslash sin(x)\backslash ,dx\; =\; \backslash sum\_\{j=0\}^n\; (-1)^j\; \backslash left\; (f^\{(2j)\}(\backslash pi)+f^\{(2j)\}(0)\backslash right\; )+(-1)^\{n+1\}\backslash int\_0^\backslash pi\; f^\{(2n+2)\}(x)\backslash sin(x)\backslash ,dx,$

which is obtained by $2n\; +\; 2$ integrations by parts. Claim 2 essentially establishes this formula, where the use of $F$ hides the iterated integration by parts. The last integral vanishes because $f^\{(2n+2)\}$ is the zero polynomial. Claim 1 shows that the remaining sum is an integer.

Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight. In fact,

:$\backslash begin\{align\}$

J_n(x)&=x^{2n+1}\int_{-1}^1(1-z^2)^n\cos(xz)\,dz\\

&=\int_{-1}^1\left (x^2-(xz)^2\right )^nx\cos(xz)\,dz.

\end{align}

Therefore, the substitution $xz\; =\; y$ turns this integral into

:$\backslash int\_\{-x\}^x(x^2-y^2)^n\backslash cos(y)\backslash ,dy.$

In particular,

:$\backslash begin\{align\}$

J_n\left(\frac\pi2\right)&=\int_{-\pi/2}^{\pi/2}\left(\frac{\pi^2}4-y^2\right)^n\cos(y)\,dy\\[5pt]

&=\int_0^\pi\left(\frac{\pi^2}4-\left(y-\frac\pi2\right)^2\right)^n\cos\left(y-\frac\pi2\right)\,dy\\[5pt]

&=\int_0^\pi y^n(\pi-y)^n\sin(y)\,dy\\[5pt]

&=\frac{n!}{b^n}\int_0^\pi f(x)\sin(x)\,dx.

\end{align}

Another connection between the proofs lies in the fact that Hermite already mentions that if $f$ is a polynomial function and

:$F=f-f^\{(2)\}+f^\{(4)\}\backslash mp\backslash cdots,$

then

:$\backslash int\; f(x)\backslash sin(x)\backslash ,dx=F\text{'}(x)\backslash sin(x)-F(x)\backslash cos(x)+C,$

from which it follows that

:$\backslash int\_0^\backslash pi\; f(x)\backslash sin(x)\backslash ,dx=F(\backslash pi)+F(0).$

Bourbaki's proof is outlined as an exercise in his calculus treatise.{{citation|last = Bourbaki|first = Nicolas|author-link = Nicolas Bourbaki|year=1949|title=Fonctions d'une variable réelle, chap. I–II–III|series = Actualités Scientifiques et Industrielles|volume = 1074|publisher = Hermann|language = fr|pages=137–138}} For each natural number b and each non-negative integer $n,$ define

:$A\_n(b)=b^n\backslash int\_0^\backslash pi\backslash frac\{x^n(\backslash pi-x)^n\}\{n!\}\backslash sin(x)\backslash ,dx.$

Since $A\_n(b)$ is the integral of a function defined on $[0,\backslash pi]$ that takes the value $0$ at $0$ and $\backslash pi$ and which is greater than $0$ otherwise, $A\_n(b)\; >\; 0.$ Besides, for each natural number $b,$ if $n$ is large enough, because

: $x(\backslash pi-x)\; \backslash le\; \backslash left(\backslash frac\backslash pi2\backslash right)^2$

and therefore

:$A\_n(b)\backslash le\backslash pi\; b^n\; \backslash frac\{1\}\{n!\}\; \backslash left(\backslash frac\backslash pi2\backslash right)^\{2n\}\; =\; \backslash pi\; \backslash frac\{(b\backslash pi^2/4)^n\}\{n!\}.$

On the other hand, repeated integration by parts allows us to deduce that, if $a$ and $b$ are natural numbers such that $\backslash pi\; =\; a/b$ and $f$ is the polynomial function from $[0,\; \backslash pi]$ into $\backslash R$ defined by

: $f(x)=\backslash frac\{x^n(a-bx)^n\}\{n!\},$

then:

:$\backslash begin\{align\}$

A_n(b) &= \int_0^\pi f(x)\sin(x)\,dx \\[5pt]

&= \Big[{-f(x)\cos(x)}\Big]_{x=0}^{x=\pi} \,- \Big[{-f'(x) \sin(x)} \Big]_{x=0}^{x=\pi} + \cdots \\[5pt]

&\ \qquad \pm \Big[ f^{(2n)}(x) \cos(x) \Big]_{x=0}^{x=\pi} \,\pm \int_0^\pi f^{(2n+1)}(x)\cos(x)\,dx.

\end{align}

This last integral is $0,$ since $f^\{(2n+1)\}$ is the null function (because $f$ is a polynomial function of degree {{nobr|$2n$).}} Since each function $f^\{(k)\}$ (with {{nobr|$0\; \backslash leq\; k\; \backslash leq\; 2n$)}} takes integer values at $0$ and $\backslash pi$ and since the same thing happens with the sine and the cosine functions, this proves that $A\_n(b)$ is an integer. Since it is also greater than $0,$ it must be a natural number. But it was also proved that if $n$ is large enough, thereby reaching a contradiction.

This proof is quite close to Niven's proof, the main difference between them being the way of proving that the numbers $A\_n(b)$ are integers.

Miklós Laczkovich's proof is a simplification of Lambert's original proof.{{citation|last = Laczkovich|first = Miklós| author-link = Miklós Laczkovich|year = 1997|title = On Lambert's proof of the irrationality of {{pi}}|periodical = American Mathematical Monthly|volume = 104|issue = 5|pages = 439–443|jstor = 2974737|doi=10.2307/2974737}} He considers the functions

:$f\_k(x)\; =\; 1\; -\; \backslash frac\{x^2\}k+\backslash frac\{x^4\}\{2!\; k(k+1)\}-\backslash frac\{x^6\}\{3!\; k(k+1)(k+2)\}\; +\; \backslash cdots\; \backslash quad\; (k\backslash notin\backslash \{0,-1,-2,\backslash ldots\backslash \}).$

These functions are clearly defined for any real number $x.$ Besides

:$f\_\{1/2\}(x)\; =\; \backslash cos(2x),$

:$f\_\{3/2\}(x)\; =\; \backslash frac\{\backslash sin(2x)\}\{2x\}.$

Claim 1: The following recurrence relation holds for any real number {{nobr|$x$:}}

:$\backslash frac\{x^2\}\{k(k+1)\}f\_\{k+2\}(x)=f\_\{k+1\}(x)-f\_k(x).$

Proof: This can be proved by comparing the coefficients of the powers of $x.$

Claim 2: For each real number $x,$

:$\backslash lim\_\{k\backslash to+\backslash infty\}f\_k(x)=1.$

Proof: In fact, the sequence $x^\{2n\}/n!$ is bounded (since it converges to {{nobr|$0$)}} and if $C$ is an upper bound and if $k\; >\; 1,$ then

:$\backslash left|f\_k(x)-1\backslash right|\backslash leqslant\backslash sum\_\{n=1\}^\backslash infty\backslash frac\; C\{k^n\}=C\backslash frac\{1/k\}\{1-1/k\}=\backslash frac\; C\{k-1\}.$

Claim 3: If $x\; \backslash neq\; 0,$ $x^2$ is rational, and $k\backslash in\backslash Q\backslash smallsetminus\backslash \{0,-1,-2,\backslash ldots\backslash \}$ then

:$f\_k(x)\backslash neq0\; \backslash quad\; \backslash text\{\; and\; \}\; \backslash quad\; \backslash frac\{f\_\{k+1\}(x)\}\{f\_k(x)\}\backslash notin\backslash Q.$

Proof: Otherwise, there would be a number $y\; \backslash neq\; 0$ and integers $a$ and $b$ such that $f\_k(x)\; =\; ay$ and $f\_\{k+1\}(x)\; =\; by.$ To see why, take $y\; =\; f\_\{k+1\}(x),$ $a\; =\; 0,$ and $b\; =\; 1$ if {{nobr|$f\_k(x)\; =\; 0$;}} otherwise, choose integers $a$ and $b$ such that $f\_\{k+1\}(x)\; /\; f\_k(x)\; =\; b/a$ and define $y\; =\; f\_k(x)/a\; =\; f\_\{k+1\}(x)/b.$ In each case, $y$ cannot be $0,$ because otherwise it would follow from claim 1 that each $f\_\{k+n\}(x)$ ($n\; \backslash in\; \backslash N$) would be $0,$ which would contradict claim 2. Now, take a natural number $c$ such that all three numbers $bc/k,$ $ck/x^2,$ and $c/x^2$ are integers and consider the sequence

:

Then

:$g\_0=f\_k(x)=ay\backslash in\backslash Z\; y\; \backslash quad\; \backslash text\{\; and\; \}\; \backslash quad\; g\_1=\backslash frac\; ckf\_\{k+1\}(x)=\backslash frac\{bc\}ky\backslash in\backslash Z\; y.$

On the other hand, it follows from claim 1 that

:$\backslash begin\{align\}$

g_{n+2}&=\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}\cdot\frac{x^2}{(k+n)(k+n+1)}f_{k+n+2}(x)\\[5pt]

& =\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n+1}(x)-\frac{c^{n+2}}{x^2k(k+1)\cdots(k+n-1)}f_{k+n}(x)\\[5pt]

&=\frac{c(k+n)}{x^2}g_{n+1}-\frac{c^2}{x^2}g_n\\[5pt]

&=\left(\frac{ck}{x^2}+\frac c{x^2}n\right)g_{n+1}-\frac{c^2}{x^2}g_n,

\end{align}

which is a linear combination of $g\_\{n+1\}$ and $g\_n$ with integer coefficients. Therefore, each $g\_n$ is an integer multiple of $y.$ Besides, it follows from claim 2 that each $g\_n$ is greater than $0$ (and therefore that {{nobr|$g\_n\; \backslash geq\; |y|$)}} if $n$ is large enough and that the sequence of all $g\_n$ converges to $0.$ But a sequence of numbers greater than or equal to $|y|$ cannot converge to $0.$

Since $f\_\{1/2\}(\backslash tfrac14\backslash pi)\; =\; \backslash cos\; \backslash tfrac12\backslash pi\; =\; 0,$ it follows from claim 3 that $\backslash tfrac1\{16\}\backslash pi^2$ is irrational and therefore that $\backslash pi$ is irrational.

On the other hand, since

:$\backslash tan\; x=\backslash frac\{\backslash sin\; x\}\{\backslash cos\; x\}=x\backslash frac\{f\_\{3/2\}(x/2)\}\{f\_\{1/2\}(x/2)\},$

another consequence of Claim 3 is that, if $x\; \backslash in\; \backslash Q\; \backslash smallsetminus\; \backslash \{0\backslash \},$ then $\backslash tan\; x$ is irrational.

Laczkovich's proof is really about the hypergeometric function. In fact, $f\_k(x)\; =\; \{\}\_0F\_1\; (k\; -\; x^2)$ and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation.{{citation | last = Gauss | first = Carl Friedrich | author-link = Carl Friedrich Gauss | title = Disquisitiones generales circa seriem infinitam | journal = Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores | volume = 2 | year = 1811–1813 | language = la}} This allowed Laczkovich to find a new and simpler proof of the fact that the tangent function has the continued fraction expansion that Lambert had discovered.

Laczkovich's result can also be expressed in Bessel functions of the first kind $J\_\backslash nu(x)$. In fact, $\backslash Gamma(k)J\_\{k-1\}(2x)\; =\; x^\{k-1\}f\_k(x)$ (where $\backslash Gamma$ is the gamma function). So Laczkovich's result is equivalent to: If $x\; \backslash neq\; 0,$ $x^2$ is rational, and $k\backslash in\backslash Q\backslash smallsetminus\backslash \{0,-1,-2,\backslash ldots\backslash \}$ then

:$\backslash frac\{x\; J\_k(x)\}\{J\_\{k-1\}(x)\}\backslash notin\backslash Q.$

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• Proof that e is irrational
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