Pi Squared is Irrational/Proof 3/Lemma
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Pi Squared is Irrational: Lemma
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let it be supposed that $\pi^2$ is irrational, so that:
- $\pi^2 = \dfrac p q$
where $p$ and $q$ are integers and $q \ne 0$.
Let $A_n$ be defined as:
- $\ds A_n = \frac \pi 2 \frac {p^n} {n!} \int_0^1 \paren {1 - x^2 }^n \map \cos {\dfrac {\pi x} 2} \rd x$
Then:
- $A_n = \paren {16 n - 8} q A_{n - 1} - 16 p q A_{n - 2}$
is a reduction formula for $A_n$.
Proof
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \paren {1 - x^2 }^n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds n \paren {1 - x^2 }^{n - 1} \paren {- 2 x}\) | Power Rule for Derivatives and Chain Rule for Derivatives |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \map \cos {\dfrac {\pi x} 2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac 2 \pi \map \sin {\dfrac {\pi x} 2}\) | Primitive of Cosine Function |
Then:
\(\ds A_n\) | \(=\) | \(\ds \frac \pi 2 \frac {p^n} {n!} \int_0^1 \paren {1 - x^2 }^n \map \cos {\dfrac {\pi x} 2} \rd x\) | by definition | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2 \frac {p^n} {n!} \paren {\frac 2 \pi \bigintlimits {\paren {1 - x^2 }^n \paren {\map \sin {\dfrac {\pi x} 2} } } 0 1 - \frac 2 \pi \int_0^1 \paren {\map \sin {\dfrac {\pi x} 2} } n \paren {1 - x^2 }^{n - 1} \paren {- 2 x} \rd x }\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p^n} {n!} \int_0^1 n \paren {1 - x^2 }^{n - 1} \paren {2 x} \map \sin {\dfrac {\pi x} 2} \rd x\) | as $\dfrac 2 \pi \bigintlimits {\paren {1 - x^2 }^n \paren {\map \sin {\dfrac {\pi x} 2} } } 0 1$ trivially evaluates to $0$ |
$\Box$
Let:
\(\ds u\) | \(=\) | \(\ds n \paren {1 - x^2 }^{n - 1} \paren {2 x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds -n \paren {n - 1} \paren {1 - x^2 }^{n - 2} \paren {2 x}^2 + 2 n \paren {1 - x^2 }^{n - 1}\) | Power Rule for Derivatives, Product Rule for Derivatives and Chain Rule for Derivatives |
Now let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \map \sin {\dfrac {\pi x} 2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds -\frac 2 \pi \map \cos {\dfrac {\pi x} 2}\) | Primitive of Sine Function |
Then:
\(\ds A_n\) | \(=\) | \(\ds \frac {p^n} {n!} \int_0^1 n \paren {1 - x^2 }^{n - 1} \paren {2 x} \map \sin {\dfrac {\pi x} 2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p^n} {n!} \paren {-\frac 2 \pi \bigintlimits {n \paren {1 - x^2 }^{n - 1} \paren {2 x} \paren {\map \cos {\dfrac {\pi x} 2} } } 0 1 - \paren {-\frac 2 \pi } \int_0^1 \paren {\map \cos {\dfrac {\pi x} 2} } \paren {-n \paren {n - 1} \paren {1 - x^2 }^{n - 2} \paren {2 x}^2 + 2 n \paren {1 - x^2 }^{n - 1} } \rd x}\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p^n} {n!} \paren {\frac 2 \pi \int_0^1 \paren {\map \cos {\dfrac {\pi x} 2} } \paren {-n \paren {n - 1} \paren {1 - x^2 }^{n - 2} \paren {2 x}^2 + 2 n \paren {1 - x^2 }^{n - 1} } \rd x}\) | as $\bigintlimits {n \paren {1 - x^2 }^{n - 1} \paren {2 x} \paren {\map \cos {\dfrac {\pi x} 2} } } 0 1$ trivially evaluates to $0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p^n} {n!} \frac 2 \pi \int_0^1 \paren {2 n \paren {1 - x^2 }^{n - 1} - n \paren {n - 1} \paren {1 - x^2 }^{n - 2} \paren {2 x}^2 } \map \cos {\dfrac {\pi x} 2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {p^n} {n!} \frac {4 n} \pi \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {p^n} {n!} \frac {8 n \paren {n - 1} } \pi \int_0^1 \paren {1 - x^2 }^{n - 2} \paren {x}^2 \map \cos {\dfrac {\pi x} 2} \rd x\) | Linear Combination of Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 p} \pi \frac {p^{n - 1} } {\paren {n - 1}!} \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \paren {x}^2 \map \cos {\dfrac {\pi x} 2} \rd x\) | Definition of Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \paren {x^2} \map \cos {\dfrac {\pi x} 2} \rd x\) | recalling $\ds A_{n - 1} = \frac \pi 2 \frac {p^{n - 1} } {\paren {n - 1}!} \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \paren {x^2} \map \cos {\dfrac {\pi x} 2} \rd x + \paren {\frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \map \cos {\dfrac {\pi x} 2} \rd x }\) | adding $0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} + \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \map \cos {\dfrac {\pi x} 2} \rd x\) | Linear Combination of Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} + \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \times \frac {\paren {n - 1} } {\paren {n - 1} } \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {8 p^2} \pi \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \map \cos {\dfrac {\pi x} 2} \rd x\) | multiplying top and bottom by $\paren {n - 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} + \frac {\paren {8 n - 8} p} \pi \frac {p^{n - 1} } {\paren {n - 1}!} \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x - \frac {16 p^2} {\pi^2} A_{n - 2}\) | recalling $\ds A_{n - 2} = \frac \pi 2 \frac {p^{n - 2} } {\paren {n - 2}!} \int_0^1 \paren {1 - x^2 }^{n - 2} \map \cos {\dfrac {\pi x} 2} \rd x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 p} {\pi^2} A_{n - 1} + \frac {\paren {16 n - 16} p} {\pi^2} A_{n - 1} - \frac {16 p^2} {\pi^2} A_{n - 2}\) | recalling $\ds A_{n - 1} = \frac \pi 2 \frac {p^{n - 1} } {\paren {n - 1}!} \int_0^1 \paren {1 - x^2 }^{n - 1} \map \cos {\dfrac {\pi x} 2} \rd x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {16 n - 8} p} {p / q} A_{n - 1} - \frac {16 p^2} {p / q} A_{n - 2}\) | assuming $\pi^2 = \dfrac p q$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {16 n - 8} q A_{n - 1} - 16 p q A_{n - 2}\) |
$\blacksquare$