Difference between revisions of "Pi is transcendental"

From arguably.io
Jump to navigation Jump to search
m
(completed the proof)
 
(4 intermediate revisions by the same user not shown)
Line 4: Line 4:


==Proof==
==Proof==
The proof sketched out here is not the original proof by Lindemann but a later proof by David Hilbert which is more accessible.
The proof shown here is not the original proof by Lindemann but a later proof by David Hilbert which is a bit more accessible.


====Preparation====
Suppose <math>\pi</math> is algebraic.  
Suppose <math>\pi</math> is algebraic.  
Set <math>\alpha_{1}=i\pi</math>  
Set <math>\alpha_{1}=i\pi</math>  
where <math>i</math> is the imaginary unit <math>\left(i^2=-1\right)</math>.
where <math>i</math> is the imaginary unit <ref><math>i^2=-1</math></ref>.
Then <math>\alpha_{1}</math> is also algebraic, so it is the root of an  
Then <math>\alpha_{1}</math> is also algebraic, so it is the root of an  
<math>n</math>-th degree polynomial with integer coefficients.  
<math>n</math>-th degree polynomial with integer coefficients.  
Let <math>\alpha_{2}\text{, . . . , }\alpha_{n}</math> be the other roots of this polynomial.  
Let <math>\alpha_{2}\text{, . . . , }\alpha_{n}</math> be the other roots of this polynomial.  


Since <math>1+e^{i\pi}=0</math> (Euler's identity) we have  
Since <math>1+e^{i\pi}=0</math> <ref>Euler's identity</ref> we have  
<math>(1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0</math>   
<math>(1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0</math>   
for some <math>\beta_{1}\text{, . . . , }\beta_{N}</math>.
for some <math>\beta_{1}\text{, . . . , }\beta_{N}</math>.
Line 25: Line 26:
and because no <math>\beta_i</math> is equal to zero <math>b_M\neq 0</math> also.
and because no <math>\beta_i</math> is equal to zero <math>b_M\neq 0</math> also.


===First Part===
====Split====
Now we will multiply our equation with the integral <math>\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz</math>,  
First we will multiply our equation with the integral
where <math>\rho</math> is some positive integer.
<math>
\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz
</math>
<ref>
This is a modified Version of an alternative formula for the factorial:
<math>n!=\int_0^{\infty}z^{n}e^{-z}dz</math>.
We can use this formula as a "bridge" between the exponential function and the integers.
</ref>,  
where <math>\rho</math> is some positive integer and <math>g(z)=f(z)\cdot b^M </math>.
We will write <math>\int_0^{\infty}</math> as a shorthand.
We will write <math>\int_0^{\infty}</math> as a shorthand.
This allows us to split the sum into two Parts:
This allows us to split the sum into two Parts:
<ref>
Where <math>\int_0^{\beta_i}</math> is a line integral along the line segment from <math>0</math> to <math>\beta_i</math> in the complex plane,
and <math>\int_{\beta_i}^{\infty}</math> is also a line integral
obtained by integrating over the line from <math>\beta_i</math> to <math>\infty</math> parallel to the
real axis in the complex plane.
If <math>\beta_i</math> is a real number this just boils down to splitting the interval of integration.
But if <math>\beta_i</math> is a non-real complex number Cauchy's integral theorem has to be utilised
to show that integrating along these different paths gives the same result.
</ref>
:<math>P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty}</math>
:<math>P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty}</math>


:<math>P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M}</math>
:<math>P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M}</math>


where <math>\int_0^{\beta_i}</math> is a line integral along the line segment from <math>0</math> to <math>\beta_i</math> in the complex plane,  
====First Sum, First Term====
and <math>\int_{\beta_i}^{\infty}</math> is also a line integral obtained by integrating over the line from <math>\beta_i</math> to <math>\infty</math> parallel to the
Then, one can show that
real axis in the complex plane.
<math>
\int_0^{\infty}
=\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz
</math>
is an integer that is divisible by <math>\rho!</math> by expanding
<math>
\left[g(z)\right]^{\rho+1}=\left[f(z)b^M\right]^{\rho + 1}
</math>
to an (integer) polynomial and then using the fact that
<math>
n!=\int_0^{\infty}z^{n}e^{-z}dz
</math> 
for positive integers <math>n</math> to get rid of the <math>e</math>-terms and integrals.
<ref>
<math>
\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz
=b^{M\left(\rho+1\right)}\int_0^{\infty}\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)z^{\rho}e^{-z}dz
= b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right)
=b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right)
</math>
where the <math>c_i</math> are integers; if <math>f</math> is a polynomial with integer coefficients so is <math>f^{\rho+1}</math>.
Note that the fractions
<math>
\frac{\left(\rho+k\right)!}{\rho!}
</math>
are also integers.
</ref>
 
More specifically, if divided by <math>(\rho+1)</math>, one gets the remainder
<math>  
b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho!
</math>.
<ref>
Note that in the parentheses above every term is divisible by <math>(\rho+1)</math> except the last one <math>c_{M\rho+M}</math>.
Hence, modulo <math>(\rho+1)</math> we have the congruence
<math>
\int_0^{\infty}\equiv b^{M\rho+M}\rho! c_{M\rho+M}\equiv b^{M\rho+M}\rho! b_M^{\rho+1}
</math>
</ref>
 
====First Sum, Other Terms====
Something slightly weaker can be shown for the terms
<math>
e^{\beta_i}\int_{\beta_i}^{\infty}
=e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz
</math>.
Using the substitution
<math>
\omega = z+\beta_i
</math>
we can set the bounds of integration to <math>0</math> and <math>\infty</math>,
<ref>
<math>
e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz
= e^{\beta_i}\int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega-\beta_i}d\omega
= \int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega
</math>
</ref>
and then perform some similar algebraic manipulations so that we can use
<math>
n!=\int_0^{\infty}z^{n}e^{-z}dz
</math>
again,
<ref>
We can expand the first factor using the binomial theorem
<math>
\int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega
= b^{M\rho+M}\int_0^{\infty}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\omega^k\beta_i^{\rho-k}\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega
= b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega
</math>
 
: The expression <math>f(\omega+\beta_i)</math> will again be some kind of <math>M</math>-th degree polynomial. But because the polynomial is zero for <math>\omega = 0</math> we know that the last coefficient is also equal to zero.
 
<math>
b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega
= b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[b(\omega+\beta_i)^M + \text{ . . . }+b_M\right]^{\rho + 1}\omega^k e^{-\omega}d\omega
= b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega
</math>
 
: So, raised to the <math>\rho+1</math>-th power we get a polynomial again, but with the last <math>\rho</math> coefficients equal to zero.
 
<math>
b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega
= b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega
</math>
 
: We can "distribute" the integral into the sum
 
<math>
b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega 
= b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\int_0^{\infty}\omega^{M\rho+M}\omega^k e^{-\omega}d\omega + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\int_0^{\infty}\omega^{\rho+1}\omega^k e^{-\omega}d\omega\right)
</math>
 
: As above we can now use the fact that <math>\int_0^{\infty}\omega^{\rho}e^{-\omega}d\omega = \rho!</math> to get rid of the integrals; the expression becomes
 
<math>
b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\left(M\rho+M+k\right)! + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\left(\rho+1+k\right)!\right)
= b^{M\rho+M}\left(\rho+1\right)!\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\frac{\left(M\rho+M+k\right)!}{\left(\rho+1\right)!} + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\frac{\left(\rho+1+k\right)!}{\left(\rho+1\right)!}\right)
= \left(\rho+1\right)!\cdot G\left(\beta_i\right)
</math>
 
: where <math>G(\beta_i)</math> is a polynomial in <math>\beta_i</math> with integer coefficients.
</ref>
which tells us that the integral is equal to
<math>
\left(\rho+1\right)! G\left(\beta_i\right)
</math>, 
where <math>G(\beta_i)</math> is a polynomial in <math>\beta_i</math> with integer coefficients.


If <math>\beta_i</math> is a real number these integrals are just normal Riemann integrals over the real numbers, and it's easy to see that it's valid to split them up in this way.
Even though the <math>\beta_i</math> are themselves no integers,  
So let <math>\beta_i</math> be a complex number with imaginary part <math>\text{Im}\left(\beta_i\right)\neq 0</math>.  
since they are the roots of the integer polynomial <math>f</math>
As above, <math>\int_0^{\beta_i}</math> is the integral along the line segment from <math>0</math> to <math>\beta_i</math>,  
we get an integer again if we sum them all up.
<math>\int_{\beta_i}^{R'}</math> along the line segment parallel to the real axis from <math>\beta_i</math> to a number <math>R'</math> with real part <math>\text{Re}\left(R'\right)= R</math>,
Likewise, the sums of the squares, third powers etc. are all integers aswell.
<math>\int_{R'}^{R}</math> along the line segment parallel to the imaginary axis from <math>R'</math> to the real number <math>R</math>,
<ref>
and <math>\int_{R}^0</math> along the real axis from <math>R</math> to <math>0</math>.
See Newton-Girard Formulas
</ref>
So, even though the <math>G(\beta_i)</math> are no integers, the sum
<math>
G(\beta_1)+G(\beta_2)+\text{ . . . }+G(\beta_M)
</math>
is.  


So <math>\int_0^{\beta_i} + \int_{\beta_i}^{R'} + \int_{R'}^{R} + \int_{R}^0</math> is an integral along a closed path in the complex plane,  
So we can bundle all terms of the first sum, except the first, together to get an integer, that is furthermore divisible by <math>\left(\rho +1\right)!</math>.
which is not self-intersecting if we choose <math>R</math> large enough (<math>R>\text{Re}\left(\beta_i\right)</math>).
Since the function which is being integrated is analytical on the whole complex plane this integral evaluates to zero according to the Cauchy integral theorem.


Note that if we let <math>R\longrightarrow\infty</math> the lenght of the line segment from  <math>R'</math> to  <math>R</math> stays the same
On the whole we have thus shown that <math>P_1</math> is an integer that is divisible by <math>\rho !</math>.  
while the function <math>z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}</math> tends towards zero on that line segment.
Using the estimation lemma for contour integrals it follows that  
<math>\left|\int_{R'}^{R}\right|\leq\left|R-R'\right|\cdot\max_{\left[R';R\right]}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}\longrightarrow 0</math>.


:<math>\lim_{R\rightarrow\infty}\left(\int_0^{\beta_i} + \int_{\beta_i}^{R'} + \int_{R'}^{R} + \int_{R}^0\right) = 0 \Rightarrow\int_0^{\beta_i} + \int_{\beta_i}^{\infty} +\text{ } 0 + \int_{\infty}^0 = 0\Rightarrow \int_0^{\beta_i} + \int_{\beta_i}^{\infty} = \int_{0}^{\infty}</math>
That means <math>\frac{P_1}{\rho !}</math> is also an integer, and dividing it by <math>\rho + 1</math> would give one the remainder
<math>
b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}  
</math>.
(This is just important to make sure that <math>P_1</math> is not always equal to zero.)


===Second Part===
====Second Sum====
Now we will show that <math>\int_0^{\infty}=\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz</math> is an integer
that is divisible by <math>\rho!</math> by using the fact that <math>\int_0^{\infty}z^{\rho}e^{-z}dz = \rho!</math>.


:<math>\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz = b^{M\left(\rho+1\right)}\int_0^{\infty}\left(bz^M+\text{ . . . }+b_M\right)^{\rho+1}z^{\rho}e^{-z}dz = b^{M\left(\rho+1\right)}\int_0^{\infty}\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)z^{\rho}e^{-z}dz </math>
For the terms in the second sum we just calculate upper bounds.
:::<math>= b^{M\left(\rho+1\right)}\left(c\int_0^{\infty}z^{M\rho+M+\rho}e^{-z}dz+\text{ . . . }+c_{M\rho+M}\int_0^{\infty}z^{\rho}e^{-z}dz\right) = b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right)</math>
Let <math>K</math> be the maximum (absolute) value of <math>z\cdot g(z)</math> on all the line segments <math>\left[0;\beta_i\right]</math>
:::<math>=b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right)</math>
and <math>k</math> the maximum (absolute) value of <math>g(z)\cdot e^{-z}</math> also on these line segments. Then:
where the <math>c_i</math> are integers; if <math>f</math> is a polynomial with integer coefficients so is <math>f^{\rho+1}</math>.
:<math>\left|\int_0^{\beta_i}\right|\leq \left|\beta_i\right|\cdot k \cdot K^{\rho}</math>
The fractions are also integers because <math>\frac{\left(\rho+k\right)!}{\rho!}=\frac{(\rho+k)(\rho+k-1)\text{ . . . }\cdot2\cdot 1}{\rho(\rho-1)\text{ . . . }\cdot2\cdot 1}=(\rho+k)\text{ . . . }(\rho+2)(\rho+1)</math>.
So, if we abbreviate
<math>
\kappa =\left(\left|\beta_1\cdot e^{\beta_1}\right|
+\left|\beta_2\cdot e^{\beta_2}\right|
+\text{ . . . }
+\left|\beta_M\cdot e^{\beta_M}\right|
\right)\cdot k
</math>
we simply get
:<math>\left|P_2\right|\leq\kappa\cdot K^{\rho}</math>.
Note that neither <math>\kappa</math> nor <math>K</math> depend on <math>\rho</math>.


So the whole expression is an integer that is divisible by <math>\rho!</math>.
====Conclusion====
Note that in the parentheses every term is divisible by <math>(\rho+1)</math> except the last one <math>c_{M\rho+M}</math>.
Now we come back to our original equation
Hence, modulo <math>(\rho+1)</math> we have the congruence <math>\int_0^{\infty}\equiv b^{M\rho+M}\rho! c_{M\rho+M}</math>,
<math>
and since
P_1+P_2=0
<math>\left(bz^M+\text{ . . . }+b_M\right)^{\rho+1}=\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)\Rightarrow c_{M\rho+M}=b_M^{\rho+1}\Rightarrow\int_0^{\infty}\equiv b^{M\rho+M}\rho! b_M^{\rho+1}\text{    (mod } \rho+1)</math>
</math>.  
By dividing with <math>\rho !</math> on both sides we get
:<math>\frac{P_1}{\rho !}+\frac{P_2}{\rho !}=0</math>
Remember that <math>\rho </math> was an arbitrary positive integer.
According to the inequality we proved for <math>P_2</math> we can get <math>\frac{P_2}{\rho !}</math>  
as close to zero as we want if we choose <math>\rho </math> large enough.
But we also proved that <math>\frac{P_1}{\rho !}</math> is always an integer, and there are infinitely many choices
<ref>
multiples of <math>a\cdot b\cdot b_M</math>
</ref>
for <math>\rho</math> such that
<math>\frac{P_1}{\rho !}</math> is not divisible by <math>\rho + 1</math> so it can't be zero.
One can easily see that this equation can't hold.
Therefore our initial premise was faulty and <math>\pi</math> is transcendent.


===Third Part===


{{Claim
{{Claim

Latest revision as of 22:09, 8 February 2022

Pi is transcendental is a famous claim regarding the nature of the number [math]\displaystyle{ \pi }[/math]. It was first proven by Ferdinand von Lindemann in 1882, buildung on methods developed by Charles Hermite, who was able to prove that the number [math]\displaystyle{ e }[/math] is transcendental roughly 10 years earlier in 1873. With the help of Karl Weierstrass their findings could be generalized to the Lindemann-Weierstrass theorem in 1885.

Proof

The proof shown here is not the original proof by Lindemann but a later proof by David Hilbert which is a bit more accessible.

Preparation

Suppose [math]\displaystyle{ \pi }[/math] is algebraic. Set [math]\displaystyle{ \alpha_{1}=i\pi }[/math] where [math]\displaystyle{ i }[/math] is the imaginary unit [1]. Then [math]\displaystyle{ \alpha_{1} }[/math] is also algebraic, so it is the root of an [math]\displaystyle{ n }[/math]-th degree polynomial with integer coefficients. Let [math]\displaystyle{ \alpha_{2}\text{, . . . , }\alpha_{n} }[/math] be the other roots of this polynomial.

Since [math]\displaystyle{ 1+e^{i\pi}=0 }[/math] [2] we have [math]\displaystyle{ (1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0 }[/math] for some [math]\displaystyle{ \beta_{1}\text{, . . . , }\beta_{N} }[/math]. These [math]\displaystyle{ \beta_i }[/math] are also algebraic, since they are the sums of other algebraic numbers (the [math]\displaystyle{ \alpha_i }[/math]). Disregarding the [math]\displaystyle{ \beta_{i} }[/math] that are equal to [math]\displaystyle{ 0 }[/math], we end up with [math]\displaystyle{ a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0 }[/math] for some positive integer [math]\displaystyle{ a }[/math] (since [math]\displaystyle{ e^0=1 }[/math]). Because the [math]\displaystyle{ \beta_1\text{, . . . , }\beta_M }[/math] are algebraic, they are the roots of a polynomial [math]\displaystyle{ f(z)=b z^M+ b_1 z^{M-1}+\text{ . . . }+b_{M-1} z + b_M }[/math] with integer coefficients [math]\displaystyle{ b\text{, }b_1\text{, . . . , }b_M }[/math], and because no [math]\displaystyle{ \beta_i }[/math] is equal to zero [math]\displaystyle{ b_M\neq 0 }[/math] also.

Split

First we will multiply our equation with the integral [math]\displaystyle{ \int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }[/math] [3], where [math]\displaystyle{ \rho }[/math] is some positive integer and [math]\displaystyle{ g(z)=f(z)\cdot b^M }[/math]. We will write [math]\displaystyle{ \int_0^{\infty} }[/math] as a shorthand. This allows us to split the sum into two Parts: [4]

[math]\displaystyle{ P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty} }[/math]
[math]\displaystyle{ P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M} }[/math]

First Sum, First Term

Then, one can show that [math]\displaystyle{ \int_0^{\infty} =\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }[/math] is an integer that is divisible by [math]\displaystyle{ \rho! }[/math] by expanding [math]\displaystyle{ \left[g(z)\right]^{\rho+1}=\left[f(z)b^M\right]^{\rho + 1} }[/math] to an (integer) polynomial and then using the fact that [math]\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }[/math] for positive integers [math]\displaystyle{ n }[/math] to get rid of the [math]\displaystyle{ e }[/math]-terms and integrals. [5]

More specifically, if divided by [math]\displaystyle{ (\rho+1) }[/math], one gets the remainder [math]\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho! }[/math]. [6]

First Sum, Other Terms

Something slightly weaker can be shown for the terms [math]\displaystyle{ e^{\beta_i}\int_{\beta_i}^{\infty} =e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }[/math]. Using the substitution [math]\displaystyle{ \omega = z+\beta_i }[/math] we can set the bounds of integration to [math]\displaystyle{ 0 }[/math] and [math]\displaystyle{ \infty }[/math], [7] and then perform some similar algebraic manipulations so that we can use [math]\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }[/math] again, [8] which tells us that the integral is equal to [math]\displaystyle{ \left(\rho+1\right)! G\left(\beta_i\right) }[/math], where [math]\displaystyle{ G(\beta_i) }[/math] is a polynomial in [math]\displaystyle{ \beta_i }[/math] with integer coefficients.

Even though the [math]\displaystyle{ \beta_i }[/math] are themselves no integers, since they are the roots of the integer polynomial [math]\displaystyle{ f }[/math] we get an integer again if we sum them all up. Likewise, the sums of the squares, third powers etc. are all integers aswell. [9] So, even though the [math]\displaystyle{ G(\beta_i) }[/math] are no integers, the sum [math]\displaystyle{ G(\beta_1)+G(\beta_2)+\text{ . . . }+G(\beta_M) }[/math] is.

So we can bundle all terms of the first sum, except the first, together to get an integer, that is furthermore divisible by [math]\displaystyle{ \left(\rho +1\right)! }[/math].

On the whole we have thus shown that [math]\displaystyle{ P_1 }[/math] is an integer that is divisible by [math]\displaystyle{ \rho ! }[/math].

That means [math]\displaystyle{ \frac{P_1}{\rho !} }[/math] is also an integer, and dividing it by [math]\displaystyle{ \rho + 1 }[/math] would give one the remainder [math]\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)} }[/math]. (This is just important to make sure that [math]\displaystyle{ P_1 }[/math] is not always equal to zero.)

Second Sum

For the terms in the second sum we just calculate upper bounds. Let [math]\displaystyle{ K }[/math] be the maximum (absolute) value of [math]\displaystyle{ z\cdot g(z) }[/math] on all the line segments [math]\displaystyle{ \left[0;\beta_i\right] }[/math] and [math]\displaystyle{ k }[/math] the maximum (absolute) value of [math]\displaystyle{ g(z)\cdot e^{-z} }[/math] also on these line segments. Then:

[math]\displaystyle{ \left|\int_0^{\beta_i}\right|\leq \left|\beta_i\right|\cdot k \cdot K^{\rho} }[/math]

So, if we abbreviate [math]\displaystyle{ \kappa =\left(\left|\beta_1\cdot e^{\beta_1}\right| +\left|\beta_2\cdot e^{\beta_2}\right| +\text{ . . . } +\left|\beta_M\cdot e^{\beta_M}\right| \right)\cdot k }[/math] we simply get

[math]\displaystyle{ \left|P_2\right|\leq\kappa\cdot K^{\rho} }[/math].

Note that neither [math]\displaystyle{ \kappa }[/math] nor [math]\displaystyle{ K }[/math] depend on [math]\displaystyle{ \rho }[/math].

Conclusion

Now we come back to our original equation [math]\displaystyle{ P_1+P_2=0 }[/math]. By dividing with [math]\displaystyle{ \rho ! }[/math] on both sides we get

[math]\displaystyle{ \frac{P_1}{\rho !}+\frac{P_2}{\rho !}=0 }[/math]

Remember that [math]\displaystyle{ \rho }[/math] was an arbitrary positive integer. According to the inequality we proved for [math]\displaystyle{ P_2 }[/math] we can get [math]\displaystyle{ \frac{P_2}{\rho !} }[/math] as close to zero as we want if we choose [math]\displaystyle{ \rho }[/math] large enough. But we also proved that [math]\displaystyle{ \frac{P_1}{\rho !} }[/math] is always an integer, and there are infinitely many choices [10] for [math]\displaystyle{ \rho }[/math] such that [math]\displaystyle{ \frac{P_1}{\rho !} }[/math] is not divisible by [math]\displaystyle{ \rho + 1 }[/math] so it can't be zero. One can easily see that this equation can't hold. Therefore our initial premise was faulty and [math]\displaystyle{ \pi }[/math] is transcendent.


Claim
Statement of the claim Pi is transcendental
Level of certainty Proven
Nature Theoretical
Counterclaim Pi is algebraic
Dependent on


Dependency of


References

  1. [math]\displaystyle{ i^2=-1 }[/math]
  2. Euler's identity
  3. This is a modified Version of an alternative formula for the factorial: [math]\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }[/math]. We can use this formula as a "bridge" between the exponential function and the integers.
  4. Where [math]\displaystyle{ \int_0^{\beta_i} }[/math] is a line integral along the line segment from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \beta_i }[/math] in the complex plane, and [math]\displaystyle{ \int_{\beta_i}^{\infty} }[/math] is also a line integral obtained by integrating over the line from [math]\displaystyle{ \beta_i }[/math] to [math]\displaystyle{ \infty }[/math] parallel to the real axis in the complex plane. If [math]\displaystyle{ \beta_i }[/math] is a real number this just boils down to splitting the interval of integration. But if [math]\displaystyle{ \beta_i }[/math] is a non-real complex number Cauchy's integral theorem has to be utilised to show that integrating along these different paths gives the same result.
  5. [math]\displaystyle{ \int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz =b^{M\left(\rho+1\right)}\int_0^{\infty}\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)z^{\rho}e^{-z}dz = b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right) =b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right) }[/math] where the [math]\displaystyle{ c_i }[/math] are integers; if [math]\displaystyle{ f }[/math] is a polynomial with integer coefficients so is [math]\displaystyle{ f^{\rho+1} }[/math]. Note that the fractions [math]\displaystyle{ \frac{\left(\rho+k\right)!}{\rho!} }[/math] are also integers.
  6. Note that in the parentheses above every term is divisible by [math]\displaystyle{ (\rho+1) }[/math] except the last one [math]\displaystyle{ c_{M\rho+M} }[/math]. Hence, modulo [math]\displaystyle{ (\rho+1) }[/math] we have the congruence [math]\displaystyle{ \int_0^{\infty}\equiv b^{M\rho+M}\rho! c_{M\rho+M}\equiv b^{M\rho+M}\rho! b_M^{\rho+1} }[/math]
  7. [math]\displaystyle{ e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz = e^{\beta_i}\int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega-\beta_i}d\omega = \int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega }[/math]
  8. We can expand the first factor using the binomial theorem [math]\displaystyle{ \int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\int_0^{\infty}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\omega^k\beta_i^{\rho-k}\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega }[/math]
    The expression [math]\displaystyle{ f(\omega+\beta_i) }[/math] will again be some kind of [math]\displaystyle{ M }[/math]-th degree polynomial. But because the polynomial is zero for [math]\displaystyle{ \omega = 0 }[/math] we know that the last coefficient is also equal to zero.
    [math]\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[b(\omega+\beta_i)^M + \text{ . . . }+b_M\right]^{\rho + 1}\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega }[/math]
    So, raised to the [math]\displaystyle{ \rho+1 }[/math]-th power we get a polynomial again, but with the last [math]\displaystyle{ \rho }[/math] coefficients equal to zero.
    [math]\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega }[/math]
    We can "distribute" the integral into the sum
    [math]\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\int_0^{\infty}\omega^{M\rho+M}\omega^k e^{-\omega}d\omega + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\int_0^{\infty}\omega^{\rho+1}\omega^k e^{-\omega}d\omega\right) }[/math]
    As above we can now use the fact that [math]\displaystyle{ \int_0^{\infty}\omega^{\rho}e^{-\omega}d\omega = \rho! }[/math] to get rid of the integrals; the expression becomes
    [math]\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\left(M\rho+M+k\right)! + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\left(\rho+1+k\right)!\right) = b^{M\rho+M}\left(\rho+1\right)!\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\frac{\left(M\rho+M+k\right)!}{\left(\rho+1\right)!} + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\frac{\left(\rho+1+k\right)!}{\left(\rho+1\right)!}\right) = \left(\rho+1\right)!\cdot G\left(\beta_i\right) }[/math]
    where [math]\displaystyle{ G(\beta_i) }[/math] is a polynomial in [math]\displaystyle{ \beta_i }[/math] with integer coefficients.
  9. See Newton-Girard Formulas
  10. multiples of [math]\displaystyle{ a\cdot b\cdot b_M }[/math]