Editing Transcendental number (section)

==Proofs for specific numbers==
===A proof that {{mvar|e}} is transcendental===
The first proof that [[E (mathematical constant)|the base of the natural logarithms, {{mvar|e}}]], is transcendental dates from 1873. We will now follow the strategy of [[David Hilbert]] (1862–1943) who gave a simplification of the original proof of [[Charles Hermite]]. The idea is the following:

Assume, for purpose of [[Proof by contradiction|finding a contradiction]], that {{mvar|e}} is algebraic. Then there exists a finite set of integer coefficients {{math|''c''<sub>0</sub>, ''c''<sub>1</sub>, ..., ''c<sub>n</sub>''}} satisfying the equation:
<math display=block>
  c_{0} + c_{1}e + c_{2} e^{2} + \cdots + c_{n} e^{n} = 0, \qquad c_0, c_n \neq 0 ~.
</math>
It is difficult to make use of the integer status of these coefficients when multiplied by a power of the irrational {{mvar|e}}, but we can absorb those powers into an integral which “mostly” will assume integer values. For a positive integer {{mvar|k}}, define the polynomial
<math display=block>
  f_k(x) = x^{k} \left [(x-1)\cdots(x-n) \right ]^{k+1},
</math>
and multiply both sides of the above equation by
<math display=block>
  \int^{\infty}_{0} f_k(x) \, e^{-x}\, \mathrm{d}x\ ,
</math>
to arrive at the equation:
<math display=block>
  c_0 \left (\int^{\infty}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right) + 
  c_1 e \left( \int^{\infty}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right ) 
  + \cdots + 
  c_{n}e^{n} \left( \int^{\infty}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right) = 0 ~.
</math>

By splitting respective domains of integration, this equation can be written in the form
<math display=block>
  P + Q = 0
</math>
where
<math display=block>
  \begin{align}
    P &= c_{0} \left( \int^{\infty}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right)
    + c_{1} e \left( \int^{\infty}_{1} f_k(x) e^{-x} \,\mathrm{d}x \right)
    + c_{2} e^{2} \left( \int^{\infty}_{2} f_k(x) e^{-x} \,\mathrm{d}x \right) 
    + \cdots 
    + c_{n} e^{n} \left( \int^{\infty}_{n} f_k(x) e^{-x} \,\mathrm{d}x \right) 
  \\
    Q &= c_{1} e \left(\int^{1}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right)
    + c_{2}e^{2} \left( \int^{2}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right)
    + \cdots+c_{n} e^{n} \left( \int^{n}_{0} f_k(x) e^{-x} \,\mathrm{d}x \right) 
  \end{align}
</math>
Here {{mvar|P}} will turn out to be an integer, but more importantly it grows quickly with {{mvar|k}}.

====Lemma 1====
''There are arbitrarily large {{mvar|k}} such that <math>\ \tfrac{P}{k!}\ </math> is a non-zero integer.''

'''Proof.''' Recall the standard integral (case of the [[Gamma function]])
<math display=block>
  \int^{\infty}_{0} t^{j} e^{-t} \,\mathrm{d}t = j!
</math>
valid for any [[natural number]] <math>j</math>. More generally,

: if <math> g(t) = \sum_{j=0}^m b_j t^j </math> then <math> \int^{\infty}_{0} g(t) e^{-t} \,\mathrm{d}t = \sum_{j=0}^m b_j j! </math>.

This would allow us to compute <math>P</math> exactly, because any term of <math>P</math> can be rewritten as
<math display=block>
  c_{a} e^{a} \int^{\infty}_{a} f_k(x) e^{-x} \,\mathrm{d}x =
  c_{a} \int^{\infty}_{a} f_k(x) e^{-(x-a)} \,\mathrm{d}x =
  \left\{ \begin{aligned} t &= x-a \\ x &= t+a \\ \mathrm{d}x &= \mathrm{d}t \end{aligned} \right\} =
  c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t
</math>
through a [[Integration by substitution|change of variables]]. Hence
<math display="block">
  P = \sum_{a=0}^n c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t
  = \int_0^\infty \biggl( \sum_{a=0}^n c_a f_k(t+a) \biggr) e^{-t} \,\mathrm{d}t
</math>
That latter sum is a polynomial in <math>t</math> with integer coefficients, i.e., it is a linear combination of powers <math>t^j</math> with integer coefficients. Hence the number <math>P</math> is a linear combination (with those same integer coefficients) of factorials <math>j!</math>; in particular <math>P</math> is an integer.

Smaller factorials divide larger factorials, so the smallest <math>j!</math> occurring in that linear combination will also divide the whole of <math>P</math>. We get that <math>j!</math> from the lowest power <math>t^j</math> term appearing with a nonzero coefficient in <math>\textstyle \sum_{a=0}^n c_a f_k(t+a) </math>, but this smallest exponent <math>j</math> is also the [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiplicity]] of <math>t=0</math> as a root of this polynomial. <math>f_k(x)</math> is chosen to have multiplicity <math>k</math> of the root <math>x=0</math> and multiplicity <math>k+1</math> of the roots <math>x=a</math> for <math>a=1,\dots,n</math>, so that smallest exponent is <math>t^k</math> for <math>f_k(t)</math> and <math>t^{k+1}</math> for <math>f_k(t+a)</math> with <math> a>0 </math>. Therefore <math>k!</math> divides <math>P</math>.

To establish the last claim in the lemma, that <math>P</math> is nonzero, it is sufficient to prove that <math>k+1</math> does not divide <math>P</math>. To that end, let <math>k+1</math> be any [[prime number|prime]] larger than <math>n</math> and <math>|c_0|</math>. We know from the above that <math>(k+1)!</math> divides each of <math> \textstyle c_a \int_0^\infty f_k(t+a) e^{-t} \,\mathrm{d}t </math> for <math> 1 \leqslant a \leqslant n </math>, so in particular all of those ''are'' divisible by <math>k+1</math>. It comes down to the first term <math> \textstyle c_0 \int_0^\infty f_k(t) e^{-t} \,\mathrm{d}t </math>. We have (see [[falling and rising factorials]])
<math display=block>
  f_k(t) = t^k \bigl[ (t-1) \cdots (t-n) \bigr]^{k+1} =
  \bigl[ (-1)^{n}(n!) \bigr]^{k+1} t^k + \text{higher degree terms}
</math>
and those higher degree terms all give rise to factorials <math>(k+1)!</math> or larger. Hence
<math display=block>
  P \equiv 
  c_0 \int_0^\infty f_k(t) e^{-t} \,\mathrm{d}t \equiv
  c_0 \bigl[ (-1)^{n}(n!) \bigr]^{k+1} k! \pmod{(k+1)}
</math>
That right hand side is a product of nonzero integer factors less than the prime <math>k+1</math>, therefore that product is not divisible by <math>k+1</math>, and the same holds for <math>P</math>; in particular <math>P</math> cannot be zero.

====Lemma 2====
''For sufficiently large {{mvar|k}}, <math>\left| \tfrac{Q}{k!} \right| <1</math>.''

'''Proof.''' Note that

<math display=block>\begin{align} 
  f_k e^{-x} &= x^{k} \left[ (x-1)(x-2) \cdots (x-n) \right]^{k+1} e^{-x}\\
  &= \left (x(x-1)\cdots(x-n) \right)^k \cdot \left( (x-1) \cdots (x-n) e^{-x} \right) \\
  &= u(x)^k \cdot v(x)
\end{align}</math>

where {{math|''u''(''x''), ''v''(''x'')}} are [[Continuous function|continuous functions]] of {{mvar|x}} for all {{mvar|x}}, so are bounded on the interval {{math|[0, ''n'']}}. That is, there are constants {{math|''G'', ''H'' > 0}} such that

<math display=block>\ \left| f_k e^{-x} \right| \leq |u(x)|^k \cdot |v(x)| < G^k H \quad \text{ for } 0 \leq x \leq n ~.</math>

So each of those integrals composing {{mvar|Q}} is bounded, the worst case being

<math display=block>\left| \int_{0}^{n} f_{k} e^{-x}\ \mathrm{d}\ x \right| \leq \int_{0}^{n} \left| f_{k} e^{-x} \right| \ \mathrm{d}\ x \leq \int_{0}^{n}G^k H\ \mathrm{d}\ x = n G^k H ~.</math>

It is now possible to bound the sum {{mvar|Q}} as well:

<math display=block> |Q| < G^{k} \cdot n H \left( |c_1|e+|c_2|e^2 + \cdots+|c_n|e^{n} \right) = G^k \cdot M\ ,</math>

where {{mvar|M}} is a constant not depending on {{mvar|k}}. It follows that

<math display=block>\ \left| \frac{Q}{k!} \right| < M \cdot \frac{G^k}{k!} \to 0 \quad \text{ as } k \to \infty\ ,</math>

finishing the proof of this lemma.

====Conclusion====

Choosing a value of {{mvar|k}} that satisfies both lemmas leads to a non-zero integer <math>\left(\tfrac{P}{k!}\right)</math> added to a vanishingly small quantity <math>\left(\tfrac{Q}{k!}\right)</math> being equal to zero: an impossibility. It follows that the original assumption, that {{mvar|e}} can satisfy a polynomial equation with integer coefficients, is also impossible; that is, {{mvar|e}} is transcendental.

===The transcendence of {{mvar|π}}===
A similar strategy, different from [[Ferdinand von Lindemann|Lindemann]]'s original approach, can be used to show that the [[Pi|number {{mvar|π}}]] is transcendental. Besides the [[gamma-function]] and some estimates as in the proof for {{mvar|e}}, facts about [[symmetric polynomial]]s play a vital role in the proof.

For detailed information concerning the proofs of the transcendence of {{mvar|π}} and {{mvar|e}}, see the references and external links.