Difference between revisions of "Pi is transcendental"

Revision as of 20:32, 24 January 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.

Suppose $π$ is algebraic. Set $α_{1} = i π$ where $i$ is the imaginary unit $(i^{2} = - 1)$ . 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 $1 + e^{i π} = 0$ (Euler's identity) 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.

First Part

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

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}}

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. ^[1]

Second Part

The reason to do this is that, surprisingly, this turns most of the equation into integers.

First, one can show that $\int_{0}^{\infty} = \int_{0}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z$ is an integer that is divisible by $ρ!$ by expanding ${[f (z) b^{M}]}^{ρ + 1}$ to an integer polynomial and then using the fact that $\int_{0}^{\infty} z^{σ} e^{- z} d z = σ!$ for positive integers $σ$ . ^[2]

More specifically, if divided by $(ρ + 1)$ , one get's the remainder $b^{(M ρ + M)} \cdot b_{M}^{(ρ + 1)} \cdot ρ!$ . ^[3]

Third Part

Something slightly weaker can be shown for the terms $e^{β_{i}} \int_{β_{i}}^{\infty} = e^{β_{i}} \int_{β_{i}}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z$ . Using the substitution $ω = z + β_{i}$ we can set the bounds of integration to $0$ and $\infty$ , ^[4] and then perform some similar algebraic manipulations so that we can use $\int_{0}^{\infty} z^{σ} e^{- z} d z = σ!$ again, ^[5] which tells us that the integral is equal to $(ρ + 1)! \cdot G (β_{i})$ , where $G (β_{i})$ is a polynomial in $β_{i}$ with integer coefficients.

Fourth Part

...

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

References

↑ If $β_{i}$ 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. So let $β_{i}$ be a complex number with imaginary part $Im (β_{i}) \neq 0$ . As above, $\int_{0}^{β_{i}}$ is the integral along the line segment from $0$ to $β_{i}$ , $\int_{β_{i}}^{R^{'}}$ along the line segment parallel to the real axis from $β_{i}$ to a number $R^{'}$ with real part $Re (R^{'}) = R$ , $\int_{R^{'}}^{R}$ along the line segment parallel to the imaginary axis from $R^{'}$ to the real number $R$ , and $\int_{R}^{0}$ along the real axis from $R$ to $0$ . So $\int_{0}^{β_{i}} + \int_{β_{i}}^{R^{'}} + \int_{R^{'}}^{R} + \int_{R}^{0}$ is an integral along a closed path in the complex plane, which is not self-intersecting if we choose $R$ large enough ( $R > Re (β_{i})$ ). 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 $R ⟶ \infty$ the lenght of the line segment from $R^{'}$ to $R$ stays the same while the function $z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z}$ tends towards zero on that line segment. Using the estimation lemma for contour integrals it follows that $| \int_{R^{'}}^{R} | \leq | R - R^{'} | \cdot max_{[R^{'}; R]} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} ⟶ 0$ .
$lim_{R \to \infty} (\int_{0}^{β_{i}} + \int_{β_{i}}^{R^{'}} + \int_{R^{'}}^{R} + \int_{R}^{0}) = 0 \Rightarrow \int_{0}^{β_{i}} + \int_{β_{i}}^{\infty} + 0 + \int_{\infty}^{0} = 0 \Rightarrow \int_{0}^{β_{i}} + \int_{β_{i}}^{\infty} = \int_{0}^{\infty}$
↑
$\int_{0}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} {(b z^{M} + . . . + b_{M})}^{ρ + 1} z^{ρ} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} (c z^{M ρ + M} + . . . + c_{M ρ + M}) z^{ρ} e^{- z} d z$
$= b^{M (ρ + 1)} (c \int_{0}^{\infty} z^{M ρ + M + ρ} e^{- z} d z + . . . + c_{M ρ + M} \int_{0}^{\infty} z^{ρ} e^{- z} d z) = b^{M (ρ + 1)} (c (M ρ + M + ρ)! + . . . + c_{M ρ + M} ρ!)$

$= b^{M ρ + M} ρ! (c \frac{(M ρ + M + ρ)!}{ρ!} + . . . + c_{M ρ + M})$
where the $c_{i}$ are integers; if $f$ is a polynomial with integer coefficients so is $f^{ρ + 1}$ . The fractions are also integers because $\frac{(ρ + k)!}{ρ!} = \frac{(ρ + k) (ρ + k - 1) . . . \cdot 2 \cdot 1}{ρ (ρ - 1) . . . \cdot 2 \cdot 1} = (ρ + k) . . . (ρ + 2) (ρ + 1)$ .
↑ 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 $\int_{0}^{\infty} \equiv b^{M ρ + M} ρ! c_{M ρ + M}$ , and since ${(b z^{M} + . . . + b_{M})}^{ρ + 1} = (c z^{M ρ + M} + . . . + c_{M ρ + M}) \Rightarrow c_{M ρ + M} = b_{M}^{ρ + 1} \Rightarrow \int_{0}^{\infty} \equiv b^{M ρ + M} ρ! b_{M}^{ρ + 1} (mod ρ + 1)$ .
↑
$e^{β_{i}} \int_{β_{i}}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = e^{β_{i}} \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω - β_{i}} d ω = \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω$
↑ 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.

[1] If $β_{i}$ 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. So let $β_{i}$ be a complex number with imaginary part $Im (β_{i}) \neq 0$ . As above, $\int_{0}^{β_{i}}$ is the integral along the line segment from $0$ to $β_{i}$ , $\int_{β_{i}}^{R^{'}}$ along the line segment parallel to the real axis from $β_{i}$ to a number $R^{'}$ with real part $Re (R^{'}) = R$ , $\int_{R^{'}}^{R}$ along the line segment parallel to the imaginary axis from $R^{'}$ to the real number $R$ , and $\int_{R}^{0}$ along the real axis from $R$ to $0$ . So $\int_{0}^{β_{i}} + \int_{β_{i}}^{R^{'}} + \int_{R^{'}}^{R} + \int_{R}^{0}$ is an integral along a closed path in the complex plane, which is not self-intersecting if we choose $R$ large enough ( $R > Re (β_{i})$ ). 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 $R ⟶ \infty$ the lenght of the line segment from $R^{'}$ to $R$ stays the same while the function $z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z}$ tends towards zero on that line segment. Using the estimation lemma for contour integrals it follows that $| \int_{R^{'}}^{R} | \leq | R - R^{'} | \cdot max_{[R^{'}; R]} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} ⟶ 0$ .
$lim_{R \to \infty} (\int_{0}^{β_{i}} + \int_{β_{i}}^{R^{'}} + \int_{R^{'}}^{R} + \int_{R}^{0}) = 0 \Rightarrow \int_{0}^{β_{i}} + \int_{β_{i}}^{\infty} + 0 + \int_{\infty}^{0} = 0 \Rightarrow \int_{0}^{β_{i}} + \int_{β_{i}}^{\infty} = \int_{0}^{\infty}$

[2] 
$\int_{0}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} {(b z^{M} + . . . + b_{M})}^{ρ + 1} z^{ρ} e^{- z} d z = b^{M (ρ + 1)} \int_{0}^{\infty} (c z^{M ρ + M} + . . . + c_{M ρ + M}) z^{ρ} e^{- z} d z$
$= b^{M (ρ + 1)} (c \int_{0}^{\infty} z^{M ρ + M + ρ} e^{- z} d z + . . . + c_{M ρ + M} \int_{0}^{\infty} z^{ρ} e^{- z} d z) = b^{M (ρ + 1)} (c (M ρ + M + ρ)! + . . . + c_{M ρ + M} ρ!)$

$= b^{M ρ + M} ρ! (c \frac{(M ρ + M + ρ)!}{ρ!} + . . . + c_{M ρ + M})$
where the $c_{i}$ are integers; if $f$ is a polynomial with integer coefficients so is $f^{ρ + 1}$ . The fractions are also integers because $\frac{(ρ + k)!}{ρ!} = \frac{(ρ + k) (ρ + k - 1) . . . \cdot 2 \cdot 1}{ρ (ρ - 1) . . . \cdot 2 \cdot 1} = (ρ + k) . . . (ρ + 2) (ρ + 1)$ .

[3] 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 $\int_{0}^{\infty} \equiv b^{M ρ + M} ρ! c_{M ρ + M}$ , and since ${(b z^{M} + . . . + b_{M})}^{ρ + 1} = (c z^{M ρ + M} + . . . + c_{M ρ + M}) \Rightarrow c_{M ρ + M} = b_{M}^{ρ + 1} \Rightarrow \int_{0}^{\infty} \equiv b^{M ρ + M} ρ! b_{M}^{ρ + 1} (mod ρ + 1)$ .

[4] 
$e^{β_{i}} \int_{β_{i}}^{\infty} z^{ρ} {[f (z) b^{M}]}^{ρ + 1} e^{- z} d z = e^{β_{i}} \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω - β_{i}} d ω = \int_{0}^{\infty} {(ω + β_{i})}^{ρ} {[f (ω + β_{i}) b^{M}]}^{ρ + 1} e^{- ω} d ω$

[5] 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.

[1]

[2]

[3]

[4]

[5]

@@ Line 91: / Line 91: @@
 <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.
+: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.
+: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
+: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
+: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.
+: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

Difference between revisions of "Pi is transcendental"

Revision as of 20:32, 24 January 2022

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Proof

First Part

Second Part

Third Part

Fourth Part

References

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Difference between revisions of "Pi is transcendental"

Revision as of 20:32, 24 January 2022

Proof

First Part

Second Part

Third Part

Fourth Part

References

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