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

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Theorem

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


Proof

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

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


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

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


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:

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

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


To prove the third relation, we have:

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


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

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


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

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


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

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

$\blacksquare$


Sources

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