Difference between revisions of "Pi is transcendental"

Latest revision as of 22:09, 8 February 2022

Pi is transcendental is a famous claim regarding the nature of the number $π$ . 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 $e$ 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 $π$ is algebraic. Set $α_{1} = i π$ where $i$ is the imaginary unit ^[1]. Then $α_{1}$ is also algebraic, so it is the root of an $n$ -th degree polynomial with integer coefficients. Let $α_{2}, . . . , α_{n}$ be the other roots of this polynomial.

Since $1 + e^{i π} = 0$ ^[2] we have $(1 + e^{α_{1}}) (1 + e^{α_{2}}) . . . (1 + e^{α_{n}}) = 1 + e^{β_{1}} + e^{β_{2}} + . . . + e^{β_{N}} = 0$ for some $β_{1}, . . . , β_{N}$ . These $β_{i}$ are also algebraic, since they are the sums of other algebraic numbers (the $α_{i}$ ). Disregarding the $β_{i}$ that are equal to $0$ , we end up with $a + e^{β_{1}} + e^{β_{2}} + . . . + e^{β_{M}} = 0$ for some positive integer $a$ (since $e^{0} = 1$ ). Because the $β_{1}, . . . , β_{M}$ are algebraic, they are the roots of a polynomial $f (z) = b z^{M} + b_{1} z^{M - 1} + . . . + b_{M - 1} z + b_{M}$ with integer coefficients $b, b_{1}, . . . , b_{M}$ , and because no $β_{i}$ is equal to zero $b_{M} \neq 0$ also.

Split

First we will multiply our equation with the integral $\int_{0}^{\infty} z^{ρ} {[g (z)]}^{ρ + 1} e^{- z} d z$ ^[3], where $ρ$ is some positive integer and $g (z) = f (z) \cdot b^{M}$ . We will write $\int_{0}^{\infty}$ as a shorthand. This allows us to split the sum into two Parts: ^[4]

P_{1} = a \int_{0}^{\infty} + e^{β_{1}} \int_{β_{1}}^{\infty} + . . . + e^{β_{M}} \int_{β_{M}}^{\infty}

P_{2} = 0 + e^{β_{1}} \int_{0}^{β_{1}} + . . . + e^{β_{M}} \int_{0}^{β_{M}}

First Sum, First Term

Then, one can show that $\int_{0}^{\infty} = \int_{0}^{\infty} z^{ρ} {[g (z)]}^{ρ + 1} e^{- z} d z$ is an integer that is divisible by $ρ!$ by expanding ${[g (z)]}^{ρ + 1} = {[f (z) b^{M}]}^{ρ + 1}$ to an (integer) polynomial and then using the fact that $n! = \int_{0}^{\infty} z^{n} e^{- z} d z$ for positive integers $n$ to get rid of the $e$ -terms and integrals. ^[5]

More specifically, if divided by $(ρ + 1)$ , one gets the remainder $b^{(M ρ + M)} \cdot b_{M}^{(ρ + 1)} \cdot ρ!$ . ^[6]

First Sum, Other Terms

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

Even though the $β_{i}$ are themselves no integers, since they are the roots of the integer polynomial $f$ 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 $G (β_{i})$ are no integers, the sum $G (β_{1}) + G (β_{2}) + . . . + G (β_{M})$ is.

So we can bundle all terms of the first sum, except the first, together to get an integer, that is furthermore divisible by $(ρ + 1)!$ .

On the whole we have thus shown that $P_{1}$ is an integer that is divisible by $ρ!$ .

That means $\frac{P_{1}}{ρ!}$ is also an integer, and dividing it by $ρ + 1$ would give one the remainder $b^{(M ρ + M)} \cdot b_{M}^{(ρ + 1)}$ . (This is just important to make sure that $P_{1}$ is not always equal to zero.)

Second Sum

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

| \int_{0}^{β_{i}} | \leq | β_{i} | \cdot k \cdot K^{ρ}

So, if we abbreviate $κ = (| β_{1} \cdot e^{β_{1}} | + | β_{2} \cdot e^{β_{2}} | + . . . + | β_{M} \cdot e^{β_{M}} |) \cdot k$ we simply get

| P_{2} | \leq κ \cdot K^{ρ}

.

Note that neither $κ$ nor $K$ depend on $ρ$ .

Conclusion

Now we come back to our original equation $P_{1} + P_{2} = 0$ . By dividing with $ρ!$ on both sides we get

\frac{P_{1}}{ρ!} + \frac{P_{2}}{ρ!} = 0

Remember that $ρ$ was an arbitrary positive integer. According to the inequality we proved for $P_{2}$ we can get $\frac{P_{2}}{ρ!}$ as close to zero as we want if we choose $ρ$ large enough. But we also proved that $\frac{P_{1}}{ρ!}$ is always an integer, and there are infinitely many choices ^[10] for $ρ$ such that $\frac{P_{1}}{ρ!}$ is not divisible by $ρ + 1$ so it can't be zero. One can easily see that this equation can't hold. Therefore our initial premise was faulty and $π$ 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

↑ $i^{2} = - 1$
↑ Euler's identity
↑ This is a modified Version of an alternative formula for the factorial: $n! = \int_{0}^{\infty} z^{n} e^{- z} d z$ . We can use this formula as a "bridge" between the exponential function and the integers.
↑ Where $\int_{0}^{β_{i}}$ is a line integral along the line segment from $0$ to $β_{i}$ in the complex plane, and $\int_{β_{i}}^{\infty}$ is also a line integral obtained by integrating over the line from $β_{i}$ to $\infty$ parallel to the real axis in the complex plane. If $β_{i}$ is a real number this just boils down to splitting the interval of integration. But if $β_{i}$ 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.
↑ $\int_{0}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} (c z^{M ρ + M} + . . . + c_{M ρ + M}) z^{ρ} e^{- z} d z = b^{M (ρ + 1)} (c (M ρ + M + ρ)! + . . . + c_{M ρ + M} ρ!) = b^{M ρ + M} ρ! (c \frac{(M ρ + M + ρ)!}{ρ!} + . . . + c_{M ρ + M})$ where the $c_{i}$ are integers; if $f$ is a polynomial with integer coefficients so is $f^{ρ + 1}$ . Note that the fractions $\frac{(ρ + k)!}{ρ!}$ are also integers.
↑ Note that in the parentheses above every term is divisible by $(ρ + 1)$ except the last one $c_{M ρ + M}$ . Hence, modulo $(ρ + 1)$ we have the congruence $\int_{0}^{\infty} \equiv b^{M ρ + M} ρ! c_{M ρ + M} \equiv b^{M ρ + M} ρ! b_{M}^{ρ + 1}$
↑ $e^{β_{i}} \int_{β_{i}}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = e^{β_{i}} \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω - β_{i}} d ω = \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω$
↑ We can expand the first factor using the binomial theorem $\int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \int_{0}^{\infty} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) ω^{k} β_{i}^{ρ - k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} ω^{k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω$
The expression $f (ω + β_{i})$ will again be some kind of $M$ -th degree polynomial. But because the polynomial is zero for $ω = 0$ we know that the last coefficient is also equal to zero.
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} ω^{k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[b (ω + β_{i})^{M} + . . . + b_{M}]}^{ρ + 1} ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[l ω^{M} + . . . + ω l_{M - 1}]}^{ρ + 1} ω^{k} e^{- ω} d ω$
So, raised to the $ρ + 1$ -th power we get a polynomial again, but with the last $ρ$ coefficients equal to zero.
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[l ω^{M} + . . . + ω l_{M - 1}]}^{ρ + 1} ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} (l^{'} ω^{M ρ + M} + . . . + ω^{ρ + 1} l_{M (ρ + 1) - ρ - 1}^{'}) ω^{k} e^{- ω} d ω$
We can "distribute" the integral into the sum
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} (l^{'} ω^{M ρ + M} + . . . + ω^{ρ + 1} l_{M (ρ + 1) - ρ - 1}^{'}) ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} \int_{0}^{\infty} ω^{M ρ + M} ω^{k} e^{- ω} d ω + . . . + l_{M (ρ + 1) - ρ - 1}^{'} \int_{0}^{\infty} ω^{ρ + 1} ω^{k} e^{- ω} d ω)$
As above we can now use the fact that $\int_{0}^{\infty} ω^{ρ} e^{- ω} d ω = ρ!$ to get rid of the integrals; the expression becomes
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} (M ρ + M + k)! + . . . + l_{M (ρ + 1) - ρ - 1}^{'} (ρ + 1 + k)!) = b^{M ρ + M} (ρ + 1)! \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} \frac{(M ρ + M + k)!}{(ρ + 1)!} + . . . + l_{M (ρ + 1) - ρ - 1}^{'} \frac{(ρ + 1 + k)!}{(ρ + 1)!}) = (ρ + 1)! \cdot G (β_{i})$
where $G (β_{i})$ is a polynomial in $β_{i}$ with integer coefficients.
↑ See Newton-Girard Formulas
↑ multiples of $a \cdot b \cdot b_{M}$

[1] $i^{2} = - 1$

[2] Euler's identity

[3] This is a modified Version of an alternative formula for the factorial: $n! = \int_{0}^{\infty} z^{n} e^{- z} d z$ . We can use this formula as a "bridge" between the exponential function and the integers.

[4] Where $\int_{0}^{β_{i}}$ is a line integral along the line segment from $0$ to $β_{i}$ in the complex plane, and $\int_{β_{i}}^{\infty}$ is also a line integral obtained by integrating over the line from $β_{i}$ to $\infty$ parallel to the real axis in the complex plane. If $β_{i}$ is a real number this just boils down to splitting the interval of integration. But if $β_{i}$ 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] $\int_{0}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} (c z^{M ρ + M} + . . . + c_{M ρ + M}) z^{ρ} e^{- z} d z = b^{M (ρ + 1)} (c (M ρ + M + ρ)! + . . . + c_{M ρ + M} ρ!) = b^{M ρ + M} ρ! (c \frac{(M ρ + M + ρ)!}{ρ!} + . . . + c_{M ρ + M})$ where the $c_{i}$ are integers; if $f$ is a polynomial with integer coefficients so is $f^{ρ + 1}$ . Note that the fractions $\frac{(ρ + k)!}{ρ!}$ are also integers.

[6] Note that in the parentheses above every term is divisible by $(ρ + 1)$ except the last one $c_{M ρ + M}$ . Hence, modulo $(ρ + 1)$ we have the congruence $\int_{0}^{\infty} \equiv b^{M ρ + M} ρ! c_{M ρ + M} \equiv b^{M ρ + M} ρ! b_{M}^{ρ + 1}$

[7] $e^{β_{i}} \int_{β_{i}}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = e^{β_{i}} \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω - β_{i}} d ω = \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω$

[8] We can expand the first factor using the binomial theorem $\int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \int_{0}^{\infty} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) ω^{k} β_{i}^{ρ - k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} ω^{k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω$
The expression $f (ω + β_{i})$ will again be some kind of $M$ -th degree polynomial. But because the polynomial is zero for $ω = 0$ we know that the last coefficient is also equal to zero.
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} ω^{k} {[f (ω + β_{i})]}^{ρ + 1} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[b (ω + β_{i})^{M} + . . . + b_{M}]}^{ρ + 1} ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[l ω^{M} + . . . + ω l_{M - 1}]}^{ρ + 1} ω^{k} e^{- ω} d ω$
So, raised to the $ρ + 1$ -th power we get a polynomial again, but with the last $ρ$ coefficients equal to zero.
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} {[l ω^{M} + . . . + ω l_{M - 1}]}^{ρ + 1} ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} (l^{'} ω^{M ρ + M} + . . . + ω^{ρ + 1} l_{M (ρ + 1) - ρ - 1}^{'}) ω^{k} e^{- ω} d ω$
We can "distribute" the integral into the sum
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} \int_{0}^{\infty} (l^{'} ω^{M ρ + M} + . . . + ω^{ρ + 1} l_{M (ρ + 1) - ρ - 1}^{'}) ω^{k} e^{- ω} d ω = b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} \int_{0}^{\infty} ω^{M ρ + M} ω^{k} e^{- ω} d ω + . . . + l_{M (ρ + 1) - ρ - 1}^{'} \int_{0}^{\infty} ω^{ρ + 1} ω^{k} e^{- ω} d ω)$
As above we can now use the fact that $\int_{0}^{\infty} ω^{ρ} e^{- ω} d ω = ρ!$ to get rid of the integrals; the expression becomes
$b^{M ρ + M} \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} (M ρ + M + k)! + . . . + l_{M (ρ + 1) - ρ - 1}^{'} (ρ + 1 + k)!) = b^{M ρ + M} (ρ + 1)! \sum_{k = 0}^{ρ} (\begin{matrix} ρ \\ k \end{matrix}) β_{i}^{ρ - k} (l^{'} \frac{(M ρ + M + k)!}{(ρ + 1)!} + . . . + l_{M (ρ + 1) - ρ - 1}^{'} \frac{(ρ + 1 + k)!}{(ρ + 1)!}) = (ρ + 1)! \cdot G (β_{i})$
where $G (β_{i})$ is a polynomial in $β_{i}$ with integer coefficients.

[9] See Newton-Girard Formulas

[10] ultiples of $a \cdot b \cdot b_{M}$

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

[9]

[10]

Difference between revisions of "Pi is transcendental"

Latest revision as of 22:09, 8 February 2022

Contents

Proof

Preparation

Split

First Sum, First Term

First Sum, Other Terms

Second Sum

Conclusion

References

Navigation menu

@@ Line 1: / Line 1: @@
-Just creating some pages I'd like to see filled, if any of you want to try their hand!
+'''Pi is transcendental''' is a famous claim regarding the nature of the number <math>\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>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>\pi</math> is algebraic.
+Set <math>\alpha_{1}=i\pi</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
+<math>n</math>-th degree polynomial with integer coefficients.
+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
+<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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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( $\begin{matrix} ρ \\ k \end{matrix}$ \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.
+{{Claim
+|Claim=Pi is transcendental
+|Level=Proven
+|Nature=Theoretical
+|Counterclaim=Pi is algebraic
+|DependentOn1=
+|DependencyOf1=
+}}
+==References==
+<references/>
+[[Category:Mathematics]]

Difference between revisions of "Pi is transcendental"

Latest revision as of 22:09, 8 February 2022

Proof

Preparation

Split

First Sum, First Term

First Sum, Other Terms

Second Sum

Conclusion

References

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