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 [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 ^[1]. 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] ^[2] 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] ^[3], 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: ^[4]

[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. ^[5]

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]. ^[6]

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], ^[7] 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, ^[8] 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. ^[9] 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 ^[10] 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

References

↑ [math]\displaystyle{ i^2=-1 }[/math]
↑ Euler's identity
↑ This is a modified Version of an alternative formula for the factorial: [math]\displaystyle{ 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.
↑ 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. If [math]\displaystyle{ \beta_i }[/math] is a real number this just boils down to splitting the interval of integration. But if [math]\displaystyle{ \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.
↑ [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(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]\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]. Note that the fractions [math]\displaystyle{ \frac{\left(\rho+k\right)!}{\rho!} }[/math] are also integers.
↑ 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}\equiv b^{M\rho+M}\rho! b_M^{\rho+1} }[/math]
↑ [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]
↑ 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.
↑ See Newton-Girard Formulas
↑ multiples of [math]\displaystyle{ a\cdot b\cdot b_M }[/math]

[1] [math]\displaystyle{ i^2=-1 }[/math]

[2] Euler's identity

[3] This is a modified Version of an alternative formula for the factorial: [math]\displaystyle{ 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.

[4] 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. If [math]\displaystyle{ \beta_i }[/math] is a real number this just boils down to splitting the interval of integration. But if [math]\displaystyle{ \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.

[5] [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(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]\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]. Note that the fractions [math]\displaystyle{ \frac{\left(\rho+k\right)!}{\rho!} }[/math] are also integers.

[6] 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}\equiv b^{M\rho+M}\rho! b_M^{\rho+1} }[/math]

[7] [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]

[8] 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.

[9] See Newton-Girard Formulas

[10] ultiples of [math]\displaystyle{ a\cdot b\cdot b_M }[/math]

[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 6: / Line 6: @@
 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 <math>\left(i^2=-1\right)</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> (Euler's identity) we have
+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>.
@@ Line 25: / Line 26: @@
 and because no <math>\beta_i</math> is equal to zero <math>b_M\neq 0</math> also.
-===First Part===
+====Split====
-First we will multiply our equation with the integral <math>\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz</math>,
+First we will multiply our equation with the integral
-where <math>\rho</math> is some positive integer.
+<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>
-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,
+====First Sum, First Term====
-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
+Then, one can show that
-real axis in the complex plane.
+<math>
-<ref>
+	\int_0^{\infty}
-If <math>\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.
+	=\int_0^{\infty}z^{\rho}\left[g(z)\right]^{\rho + 1}e^{-z}dz
-So let <math>\beta_i</math> be a complex number with imaginary part <math>\text{Im}\left(\beta_i\right)\neq 0</math>.
+</math>
-As above, <math>\int_0^{\beta_i}</math> is the integral along the line segment from <math>0</math> to <math>\beta_i</math>,
+is an integer that is divisible by <math>\rho!</math> by expanding
-<math>\int_{\beta_i}^{R'}</math> along the line segment parallel to the real axis from <math>\beta_i</math> to a number <math>R'</math> with real part <math>\text{Re}\left(R'\right)= R</math>,
+<math>
-<math>\int_{R'}^{R}</math> along the line segment parallel to the imaginary axis from <math>R'</math> to the real number <math>R</math>,
+	\left[g(z)\right]^{\rho+1}=\left[f(z)b^M\right]^{\rho + 1}
-and <math>\int_{R}^0</math> along the real axis from <math>R</math> to <math>0</math>.
+</math>
+to an (integer) polynomial and then using the fact that
-So <math>\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,
+<math>
-which is not self-intersecting if we choose <math>R</math> large enough (<math>R>\text{Re}\left(\beta_i\right)</math>).
+	n!=\int_0^{\infty}z^{n}e^{-z}dz
-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.
+</math>
+for positive integers <math>n</math> to get rid of the <math>e</math>-terms and integrals.
-Note that if we let <math>R\longrightarrow\infty</math> the lenght of the line segment from  <math>R'</math> to  <math>R</math> stays the same
-while the function <math>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>\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>\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>
-</ref>
-===Second Part===
-The reason to do this is that, surprisingly, this turns most of the equation into integers.
-First, one can show that <math>\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>\rho!</math> by expanding <math>\left[f(z)b^M\right]^{\rho + 1}</math>
-to an integer polynomial and then using the fact that <math>\int_0^{\infty}z^{\sigma}e^{-z}dz = \sigma!</math> for positive integers <math>\sigma</math>.
 <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(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>
-:::<math>= 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>
+	\int_0^{\infty}z^{\rho}\left[f(z)b^M\right]^{\rho + 1}e^{-z}dz
-:::<math>=b^{M\rho+M}\rho!\left(c\frac{\left(M\rho+M+\rho\right)!}{\rho!}+\text{ . . . }+c_{M\rho+M}\right)</math>
+	=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
-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>.
+	= b^{M\left(\rho+1\right)}\left(c\left(M\rho+M+\rho\right)!+\text{ . . . }+c_{M\rho+M}\rho!\right)
-The fractions are also integers because <math>\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>.
+	=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 get's the remainder <math> b^{\left(M\rho+M\right)}\cdot b_M^{\left(\rho+1\right)}\cdot\rho! </math>.
+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>.
+	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}</math>,
+	Hence, modulo <math>(\rho+1)</math> we have the congruence
-and since
+<math>
-<math>\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>.
+	\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>
-===Third Part===
+====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[f(z)b^M\right]^{\rho + 1}e^{-z}dz</math>.
+Something slightly weaker can be shown for the terms
-Using the substitution <math>\omega = z+\beta_i</math> we can set the bounds of integration to <math>0</math> and <math>\infty</math>,
+<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>
+<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>\int_0^{\infty}z^{\sigma}e^{-z}dz = \sigma!</math> again,
+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
+	We can expand the first factor using the binomial theorem
+<math>
-<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{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>
+	\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>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{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>
+<math>
+	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>\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{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>
+<math>
+	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
+:	We can "distribute" the integral into the sum
-<math>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>
+<math>
+	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>\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{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>
+<math>
+	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>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
-<math>\left(\rho+1\right)!\cdot G\left(\beta_i\right)</math>,
+<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.
-===Fourth Part===
+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

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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