Difference between revisions of "Pi is transcendental"
Jump to navigation
Jump to search
(moved the calculations to the references to make it more readable) |
|||
| Line 37: | Line 37: | ||
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 | 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. | real axis in the complex plane. | ||
<ref> | |||
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. | 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. | ||
So let <math>\beta_i</math> be a complex number with imaginary part <math>\text{Im}\left(\beta_i\right)\neq 0</math>. | So let <math>\beta_i</math> be a complex number with imaginary part <math>\text{Im}\left(\beta_i\right)\neq 0</math>. | ||
| Line 55: | Line 55: | ||
:<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> | :<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=== | ===Second Part=== | ||
The reason to do this is that, surprisingly, this turns most of the equation into integers. | |||
that | |||
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>\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>= 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>= 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> | ||
| Line 65: | Line 69: | ||
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>. | 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>. | ||
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>. | 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>. | ||
</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>. | |||
Note that in the parentheses every term is divisible by <math>(\rho+1)</math> except the last one <math>c_{M\rho+M}</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>. | |||
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 <math>\int_0^{\infty}\equiv b^{M\rho+M}\rho! c_{M\rho+M}</math>, | ||
and since | and since | ||
<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>. | <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>. | ||
</ref> | |||
===Third Part=== | ===Third Part=== | ||
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>. | |||
Using the substitution <math>\omega = z+\beta_i</math> we can set the bounds of integration to <math>0</math> and <math>\infty</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, | |||
<ref> | |||
We can expand the first factor using the binomial theorem | |||
<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> | |||
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> | |||
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> | |||
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 | ||
where <math>G(\beta_i)</math> is a polynomial in <math>\beta_i</math> with integer coefficients | <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. | |||
</ref> | |||
which tells us that the integral is equal to | |||
<math>\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. | |||
===Fourth Part=== | |||
... | |||
{{Claim | {{Claim | ||
Revision as of 20:28, 24 January 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.
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
...
| Statement of the claim | Pi is transcendental |
| Level of certainty | Proven |
| Nature | Theoretical |
| Counterclaim | Pi is algebraic |
| Dependent on |
|
| Dependency of |
|
References
- ↑
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]
- ↑
- [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]
- [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]
- ↑ 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].
- ↑
- [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.