Difference between revisions of "Pi is transcendental"

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


==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.  
Then <math>\alpha_{1}:=i\pi</math> is also algebraic, so it is the root of an <math>n</math>-th degree polynomial with integer coefficients.  
Set <math>\alpha_{1}=i\pi</math>
Let <math>\alpha_{2}\text{, . . . , }\alpha_{n}</math> be the other roots of this polynomial.
where <math>i</math> is the imaginary unit <ref><math>i^2=-1</math></ref>.
Since <math>1+e^{i\pi}=0</math> we have  
Then <math>\alpha_{1}</math> is also algebraic, so it is the root of an  
<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>n</math>-th degree polynomial with integer coefficients.  
with some algebraic <math>\beta_{1}\text{, . . . , }\beta_{N}</math>.  
Let <math>\alpha_{2}\text{, . . . , }\alpha_{n}</math> be the other roots of this polynomial.  
 
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>
for some <math>\beta_{1}\text{, . . . , }\beta_{N}</math>.
These <math>\beta_i</math> are also algebraic, since they are the sums of other algebraic numbers (the <math>\alpha_i</math>).
Disregarding the <math>\beta_{i}</math> that are equal to <math>0</math>, we end up with  
Disregarding the <math>\beta_{i}</math> that are equal to <math>0</math>, we end up with  
<math>a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0</math> for some positive integer <math>a</math>.
<math>a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0</math> for some positive integer <math>a</math> (since <math>e^0=1</math>).
Because the <math>\beta_1\text{, . . . , }\beta_M</math> are algebraic,
they are the roots of a polynomial
<math>f(z)=b z^M+ b_1 z^{M-1}+\text{ . . . }+b_{M-1} z + b_M</math>
with integer coefficients <math>b\text{, }b_1\text{, . . . , }b_M</math>,
and because no <math>\beta_i</math> is equal to zero <math>b_M\neq 0</math> also.
 
====Split====
First we will multiply our equation with the integral
<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.
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_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>
\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.
 
Even though the <math>\beta_i</math> are themselves no integers,
since they are the roots of the integer polynomial <math>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.
<ref>
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 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>.
 
On the whole we have thus shown that <math>P_1</math> is an integer that is divisible by <math>\rho !</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 Sum====
 
For the terms in the second sum we just calculate upper bounds.
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>
and <math>k</math> the maximum (absolute) value of <math>g(z)\cdot e^{-z}</math> also on these line segments. Then:
:<math>\left|\int_0^{\beta_i}\right|\leq \left|\beta_i\right|\cdot k \cdot K^{\rho}</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>.
 
====Conclusion====
Now we come back to our original equation
<math>
P_1+P_2=0
</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.


We now have to multiply both sides with an improper integral which I will just call <math>I_{\rho}</math>, with some positive integer <math>\rho</math> as parameter.
This will allow us to split the sum into two parts <math>P_1+P_2=0</math>, where <math>P_1</math> and <math>P_2</math> have the following properties:
*<math>P_1</math> is an integer that can be divided by <math>\rho!</math> and <math>\frac{P_1}{\rho!}\equiv a b^{\rho M+M}b_{M}^{\rho +1}\text{ mod }(\rho+1)</math>
:where <math>b</math>, <math>b_M</math> are the first and last coefficient of the polynomial with integer coefficients that has the <math>\beta_{1}\text{, . . . , }\beta_{M}</math> as it's roots (so neither <math>a</math>, <math>b</math> nor <math>b_M</math> is zero).
*<math>|P_2|\lt\chi K^{\rho}</math> for some positive constants <math>\chi\text{ and }K</math>.
So, if we choose <math>\rho</math> big enough <math>\frac{P_2}{\rho!}</math> tends towards zero while <math>\frac{P_1}{\rho!}</math> always stays an integer.
Furthermore the complicated congruence guarantees that <math>\frac{P_1}{\rho!}</math> is not zero if <math>\rho</math> is a multiple of <math>a b b_M</math>.
Since <math>P_1+P_2=0</math>  it should also be the case that  <math>\frac{P_1}{\rho!}+\frac{P_2}{\rho!}=0</math>.
But that can't be when the first part get's arbitrarily small while the second part stays a non-zero integer.


{{Claim
{{Claim
Line 36: Line 236:
==References==
==References==
<references/>
<references/>
[[Category:Mathematics]]

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]