Pi is transcendental

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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.

Suppose [math]\displaystyle{ \pi }[/math] is algebraic. Set [math]\displaystyle{ \alpha_{1}=i\pi }[/math] where [math]\displaystyle{ i }[/math] is the imaginary unit [math]\displaystyle{ \left(i^2=-1\right) }[/math]. 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] (Euler's identity) 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.

First Part

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

[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]

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. [1]

Second Part

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

First, one can show that [math]\displaystyle{ \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]\displaystyle{ \rho! }[/math] by expanding [math]\displaystyle{ \left[f(z)b^M\right]^{\rho + 1} }[/math] to an integer polynomial and then using the fact that [math]\displaystyle{ \int_0^{\infty}z^{\sigma}e^{-z}dz = \sigma! }[/math] for positive integers [math]\displaystyle{ \sigma }[/math]. [2]

More specifically, if divided by [math]\displaystyle{ (\rho+1) }[/math], one get's the remainder [math]\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho! }[/math]. [3]

Third Part

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[f(z)b^M\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], [4] and then perform some similar algebraic manipulations so that we can use [math]\displaystyle{ \int_0^{\infty}z^{\sigma}e^{-z}dz = \sigma! }[/math] again, [5] which tells us that the integral is equal to [math]\displaystyle{ \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.

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

  1. If [math]\displaystyle{ \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. So let [math]\displaystyle{ \beta_i }[/math] be a complex number with imaginary part [math]\displaystyle{ \text{Im}\left(\beta_i\right)\neq 0 }[/math]. As above, [math]\displaystyle{ \int_0^{\beta_i} }[/math] is the integral along the line segment from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \beta_i }[/math], [math]\displaystyle{ \int_{\beta_i}^{R'} }[/math] along the line segment parallel to the real axis from [math]\displaystyle{ \beta_i }[/math] to a number [math]\displaystyle{ R' }[/math] with real part [math]\displaystyle{ \text{Re}\left(R'\right)= R }[/math], [math]\displaystyle{ \int_{R'}^{R} }[/math] along the line segment parallel to the imaginary axis from [math]\displaystyle{ R' }[/math] to the real number [math]\displaystyle{ R }[/math], and [math]\displaystyle{ \int_{R}^0 }[/math] along the real axis from [math]\displaystyle{ R }[/math] to [math]\displaystyle{ 0 }[/math]. So [math]\displaystyle{ \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, which is not self-intersecting if we choose [math]\displaystyle{ R }[/math] large enough ([math]\displaystyle{ R\gt \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]\displaystyle{ R\longrightarrow\infty }[/math] the lenght of the line segment from [math]\displaystyle{ R' }[/math] to [math]\displaystyle{ R }[/math] stays the same while the function [math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ \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]
  2. [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(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]
    [math]\displaystyle{ = 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]
    [math]\displaystyle{ =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]. The fractions are also integers because [math]\displaystyle{ \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].
  3. 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} }[/math], and since [math]\displaystyle{ \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].
  4. [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]
  5. 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.