# Pi is transcendental

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

Since $\displaystyle{ 1+e^{i\pi}=0 }$ [2] we have $\displaystyle{ (1+e^{\alpha_{1}})(1+e^{\alpha_{2}})\text{ . . . }(1+e^{\alpha_{n}})=1+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{N}}=0 }$ for some $\displaystyle{ \beta_{1}\text{, . . . , }\beta_{N} }$. These $\displaystyle{ \beta_i }$ are also algebraic, since they are the sums of other algebraic numbers (the $\displaystyle{ \alpha_i }$). Disregarding the $\displaystyle{ \beta_{i} }$ that are equal to $\displaystyle{ 0 }$, we end up with $\displaystyle{ a+e^{\beta_{1}}+e^{\beta_{2}}+\text{ . . . }+e^{\beta_{M}}=0 }$ for some positive integer $\displaystyle{ a }$ (since $\displaystyle{ e^0=1 }$). Because the $\displaystyle{ \beta_1\text{, . . . , }\beta_M }$ are algebraic, they are the roots of a polynomial $\displaystyle{ f(z)=b z^M+ b_1 z^{M-1}+\text{ . . . }+b_{M-1} z + b_M }$ with integer coefficients $\displaystyle{ b\text{, }b_1\text{, . . . , }b_M }$, and because no $\displaystyle{ \beta_i }$ is equal to zero $\displaystyle{ b_M\neq 0 }$ also.

#### Split

First we will multiply our equation with the integral $\displaystyle{ \int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$ [3], where $\displaystyle{ \rho }$ is some positive integer and $\displaystyle{ g(z)=f(z)\cdot b^M }$. We will write $\displaystyle{ \int_0^{\infty} }$ as a shorthand. This allows us to split the sum into two Parts: [4]

$\displaystyle{ P_1 = a\int_0^{\infty} +\text{ } e^{\beta_1}\int_{\beta_1}^{\infty} + \text{ . . . } + e^{\beta_M}\int_{\beta_M}^{\infty} }$
$\displaystyle{ P_2 = 0 + e^{\beta_1}\int_0^{\beta_1} + \text{ . . . } + e^{\beta_M}\int_0^{\beta_M} }$

#### First Sum, First Term

Then, one can show that $\displaystyle{ \int_0^{\infty} =\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$ is an integer that is divisible by $\displaystyle{ \rho! }$ by expanding $\displaystyle{ \left[g(z)\right]^{\rho+1}=\left[f(z)b^M\right]^{\rho + 1} }$ to an (integer) polynomial and then using the fact that $\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }$ for positive integers $\displaystyle{ n }$ to get rid of the $\displaystyle{ e }$-terms and integrals. [5]

More specifically, if divided by $\displaystyle{ (\rho+1) }$, one gets the remainder $\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho! }$. [6]

#### First Sum, Other Terms

Something slightly weaker can be shown for the terms $\displaystyle{ e^{\beta_i}\int_{\beta_i}^{\infty} =e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz }$. Using the substitution $\displaystyle{ \omega = z+\beta_i }$ we can set the bounds of integration to $\displaystyle{ 0 }$ and $\displaystyle{ \infty }$, [7] and then perform some similar algebraic manipulations so that we can use $\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }$ again, [8] which tells us that the integral is equal to $\displaystyle{ \left(\rho+1\right)! G\left(\beta_i\right) }$, where $\displaystyle{ G(\beta_i) }$ is a polynomial in $\displaystyle{ \beta_i }$ with integer coefficients.

Even though the $\displaystyle{ \beta_i }$ are themselves no integers, since they are the roots of the integer polynomial $\displaystyle{ f }$ 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 $\displaystyle{ G(\beta_i) }$ are no integers, the sum $\displaystyle{ G(\beta_1)+G(\beta_2)+\text{ . . . }+G(\beta_M) }$ is.

So we can bundle all terms of the first sum, except the first, together to get an integer, that is furthermore divisible by $\displaystyle{ \left(\rho +1\right)! }$.

On the whole we have thus shown that $\displaystyle{ P_1 }$ is an integer that is divisible by $\displaystyle{ \rho ! }$.

That means $\displaystyle{ \frac{P_1}{\rho !} }$ is also an integer, and dividing it by $\displaystyle{ \rho + 1 }$ would give one the remainder $\displaystyle{ b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)} }$. (This is just important to make sure that $\displaystyle{ P_1 }$ is not always equal to zero.)

#### Second Sum

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

$\displaystyle{ \left|\int_0^{\beta_i}\right|\leq \left|\beta_i\right|\cdot k \cdot K^{\rho} }$

So, if we abbreviate $\displaystyle{ \kappa =\left(\left|\beta_1\cdot e^{\beta_1}\right| +\left|\beta_2\cdot e^{\beta_2}\right| +\text{ . . . } +\left|\beta_M\cdot e^{\beta_M}\right| \right)\cdot k }$ we simply get

$\displaystyle{ \left|P_2\right|\leq\kappa\cdot K^{\rho} }$.

Note that neither $\displaystyle{ \kappa }$ nor $\displaystyle{ K }$ depend on $\displaystyle{ \rho }$.

#### Conclusion

Now we come back to our original equation $\displaystyle{ P_1+P_2=0 }$. By dividing with $\displaystyle{ \rho ! }$ on both sides we get

$\displaystyle{ \frac{P_1}{\rho !}+\frac{P_2}{\rho !}=0 }$

Remember that $\displaystyle{ \rho }$ was an arbitrary positive integer. According to the inequality we proved for $\displaystyle{ P_2 }$ we can get $\displaystyle{ \frac{P_2}{\rho !} }$ as close to zero as we want if we choose $\displaystyle{ \rho }$ large enough. But we also proved that $\displaystyle{ \frac{P_1}{\rho !} }$ is always an integer, and there are infinitely many choices [10] for $\displaystyle{ \rho }$ such that $\displaystyle{ \frac{P_1}{\rho !} }$ is not divisible by $\displaystyle{ \rho + 1 }$ so it can't be zero. One can easily see that this equation can't hold. Therefore our initial premise was faulty and $\displaystyle{ \pi }$ is transcendent.

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

## References

1. $\displaystyle{ i^2=-1 }$
2. Euler's identity
3. This is a modified Version of an alternative formula for the factorial: $\displaystyle{ n!=\int_0^{\infty}z^{n}e^{-z}dz }$. We can use this formula as a "bridge" between the exponential function and the integers.
4. Where $\displaystyle{ \int_0^{\beta_i} }$ is a line integral along the line segment from $\displaystyle{ 0 }$ to $\displaystyle{ \beta_i }$ in the complex plane, and $\displaystyle{ \int_{\beta_i}^{\infty} }$ is also a line integral obtained by integrating over the line from $\displaystyle{ \beta_i }$ to $\displaystyle{ \infty }$ parallel to the real axis in the complex plane. If $\displaystyle{ \beta_i }$ is a real number this just boils down to splitting the interval of integration. But if $\displaystyle{ \beta_i }$ 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. $\displaystyle{ \int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz =b^{M\left(\rho+1\right)}\int_0^{\infty}\left(cz^{M\rho+M}+\text{ . . . }+c_{M\rho+M}\right)z^{\rho}e^{-z}dz = b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right) =b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right) }$ where the $\displaystyle{ c_i }$ are integers; if $\displaystyle{ f }$ is a polynomial with integer coefficients so is $\displaystyle{ f^{\rho+1} }$. Note that the fractions $\displaystyle{ \frac{\left(\rho+k\right)!}{\rho!} }$ are also integers.
6. Note that in the parentheses above every term is divisible by $\displaystyle{ (\rho+1) }$ except the last one $\displaystyle{ c_{M\rho+M} }$. Hence, modulo $\displaystyle{ (\rho+1) }$ we have the congruence $\displaystyle{ \int_0^{\infty}\equiv b^{M\rho+M}\rho! c_{M\rho+M}\equiv b^{M\rho+M}\rho! b_M^{\rho+1} }$
7. $\displaystyle{ e^{\beta_i}\int_{\beta_i}^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz = e^{\beta_i}\int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega-\beta_i}d\omega = \int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega }$
8. We can expand the first factor using the binomial theorem $\displaystyle{ \int_0^{\infty}\left(\omega+\beta_i\right)^{\rho}\left[f\left(\omega+\beta_i\right)b^M\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\int_0^{\infty}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\omega^k\beta_i^{\rho-k}\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega }$
The expression $\displaystyle{ f(\omega+\beta_i) }$ will again be some kind of $\displaystyle{ M }$-th degree polynomial. But because the polynomial is zero for $\displaystyle{ \omega = 0 }$ we know that the last coefficient is also equal to zero.
$\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\omega^k\left[f\left(\omega+\beta_i\right)\right]^{\rho + 1}e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[b(\omega+\beta_i)^M + \text{ . . . }+b_M\right]^{\rho + 1}\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega }$
So, raised to the $\displaystyle{ \rho+1 }$-th power we get a polynomial again, but with the last $\displaystyle{ \rho }$ coefficients equal to zero.
$\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left[l\omega^M + \text{ . . . }+\omega l_{M-1}\right]^{\rho + 1}\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega }$
We can "distribute" the integral into the sum
$\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\int_0^{\infty}\left(l'\omega^{M\rho+M} + \text{ . . . }+\omega^{\rho+1} l'_{M(\rho+1)-\rho -1}\right)\omega^k e^{-\omega}d\omega = b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\int_0^{\infty}\omega^{M\rho+M}\omega^k e^{-\omega}d\omega + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\int_0^{\infty}\omega^{\rho+1}\omega^k e^{-\omega}d\omega\right) }$
As above we can now use the fact that $\displaystyle{ \int_0^{\infty}\omega^{\rho}e^{-\omega}d\omega = \rho! }$ to get rid of the integrals; the expression becomes
$\displaystyle{ b^{M\rho+M}\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\left(M\rho+M+k\right)! + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\left(\rho+1+k\right)!\right) = b^{M\rho+M}\left(\rho+1\right)!\sum_{k=0}^{\rho}\left(\begin{array}{c} \rho \\ k\end{array}\right)\beta_i^{\rho-k}\left(l'\frac{\left(M\rho+M+k\right)!}{\left(\rho+1\right)!} + \text{ . . . }+l'_{M(\rho+1)-\rho -1}\frac{\left(\rho+1+k\right)!}{\left(\rho+1\right)!}\right) = \left(\rho+1\right)!\cdot G\left(\beta_i\right) }$
where $\displaystyle{ G(\beta_i) }$ is a polynomial in $\displaystyle{ \beta_i }$ with integer coefficients.
9. See Newton-Girard Formulas
10. multiples of $\displaystyle{ a\cdot b\cdot b_M }$