Pi is transcendental

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 . 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] 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] , 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:

[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.

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].

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], 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, 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. 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 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