Continued Fraction Expansion of Euler's Number/Proof 1/Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

For $n \in \Z , n \ge 0$:
\(\displaystyle A_n\) \(=\) \(\displaystyle q_{3 n} e - p_{3 n}\)
\(\displaystyle B_n\) \(=\) \(\displaystyle p_{3 n + 1} - q_{3 n + 1} e\)
\(\displaystyle C_n\) \(=\) \(\displaystyle p_{3 n + 2} - q_{3 n + 2} e\)


Proof

To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:

\(\displaystyle A_0\) \(=\) \(\displaystyle \int_0^1 e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \bigintlimits {e^x} {x \mathop = 0} {x \mathop = 1}\) Primitive of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle e - 1\)
\(\displaystyle \) \(=\) \(\displaystyle q_0 e - p_0\)
\(\displaystyle B_0\) \(=\) \(\displaystyle \int_0^1 x e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \bigintlimits {x e^x - e^x} {x \mathop = 0} {x \mathop = 1}\) Primitive of x by Exponential of a x
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle p_1 - q_1 e\)
\(\displaystyle C_0\) \(=\) \(\displaystyle \int_0^1 \paren {x - 1} e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 x e^x \rd x - \int_0^1 e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle B_0 - A_0\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - \paren {e - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 - e\)
\(\displaystyle \) \(=\) \(\displaystyle p_2 - q_2 e\)


The final step needed to validate the assertion, we must demonstrate that the following three recurrence relations hold:

\((1):\quad\) \(\displaystyle A_n\) \(=\) \(\displaystyle -B_{n - 1} - C_{n - 1}\)
\((2):\quad\) \(\displaystyle B_n\) \(=\) \(\displaystyle -2 n A_n + C_{n - 1}\)
\((3):\quad\) \(\displaystyle C_n\) \(=\) \(\displaystyle B_n - A_n\)


To prove the first relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n - 1}$ and the integrand of $C_{n - 1}$.

By integrating both sides of the equation, we verify the first recurrence relation:

\(\displaystyle \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n } {n!} e^x}\) \(=\) \(\displaystyle \frac {x^n \paren {x - 1 }^n} {n!} e^x + \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x + \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n} {n!} e^x}\) \(=\) \(\displaystyle \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x + \int_0^1 \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x \rd x + \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x\) Integrating both sides of the equation over the interval from $0$ to $1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \intlimits {\frac {x^n \paren {x - 1}^n} {n!} e^x} {x \mathop = 0} {x \mathop = 1}\) \(=\) \(\displaystyle A_n + B_{n - 1} + C_{n - 1}\) Fundamental Theorem of Calculus
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle A_n + B_{n - 1} + C_{n - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle A_n\) \(=\) \(\displaystyle -B_{n - 1} - C_{n - 1}\) rearranging


To prove the second relation, we note that the derivative of the integrand of $C_n$ is equal to the sum of the integrand of $B_n$ with two times n times the integrand of $A_{n}$ minus the integrand of $C_{n - 1}$.

By integrating both sides of the equation, we verify the second recurrence relation:

\(\displaystyle \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}\) \(=\) \(\displaystyle \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x + 2 n \frac {x^n \paren {x - 1}^n} {n!} e^x - \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}\) \(=\) \(\displaystyle \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x + 2 n \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x\) Integrating both sides of the equation over the interval from $0$ to $1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \intlimits {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x} {x \mathop = 0} {x \mathop = 1}\) \(=\) \(\displaystyle B_n + 2 n A_{n} - C_{n - 1}\) Fundamental Theorem of Calculus
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(=\) \(\displaystyle B_n + 2 n A_n - C_{n - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle B_n\) \(=\) \(\displaystyle -2 n A_n + C_{n - 1}\) rearranging


To prove the third relation, we have:

\(\displaystyle C_n\) \(=\) \(\displaystyle \int_0^1 \frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \frac {x^n \paren {x - 1 }^n} {n!} e^x \paren {x - 1} \rd x\) factoring out $\paren {x - 1}$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren 1 \rd x\) separate integrals
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle B_n - A_n\)


From the first relation, combined with the initial condition at $n = 0$ being satisfied, we have:

\(\displaystyle A_n\) \(=\) \(\displaystyle -B_{n - 1} - C_{n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle -\paren {p_{3 n - 2} - q_{3 n -2 } e} - \paren {p_{3 n - 1} - q_{3 n - 1} e}\)
\(\displaystyle \) \(=\) \(\displaystyle q_{3 n} e - p_{3 n}\)


From the second relation, combined with the initial condition at $n = 0$ being satisfied, we have:

\(\displaystyle B_n\) \(=\) \(\displaystyle -2 n A_n + C_{n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle -2 n \paren {q_{3 n} e - p_{3 n} } + \paren {p_{3 n - 1} - q_{3 n - 1} e}\)
\(\displaystyle \) \(=\) \(\displaystyle p_{3 n + 1} - q_{3 n + 1} e\)


From the third relation, combined with the initial condition at $n = 0$ being satisfied, we have:

\(\displaystyle C_n\) \(=\) \(\displaystyle B_n - A_n\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {p_{3 n + 1} - q_{3 n + 1} e} - \paren {q_{3 n} e - p_{3 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle p_{3 n + 2} - q_{3 n + 2 } e\)

$\blacksquare$


Sources