Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

  Hello everyone.  Today we will prove the product formula for  \(\dfrac{\pi}{2\sqrt{3}}\).  We will use the method of direct proof as a proof method.  Theorem 1:  \[\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\]\[\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}\] Theorem 2:  We have\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]expression whose numerators are the sequence of the odd prime numbers greater than \(3\) and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that \[\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots\] we will have \[\frac{1}{5} \cdo...
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Proof of the formula for the nth derivative of the natural logarithm

Hello everyone. Today we will prove the formula for the nth derivative of the natural logarithm. We will use the method of mathematical induction as a proof method. 

Theorem: The nth derivative of the function \(\ln (x) \) for \(n \ge 1 \) is given by the formula: \[\frac {\mathrm {d} ^ n} { \mathrm {d} x ^ n} \ln (x) = \frac {(n-1)! (-1) ^ {n-1}} {x ^ n} \] 

Proof: 

1. Base case (n = 1) \[\frac {\mathrm {d}} {\mathrm {d} x} \ln (x) = \frac {(1-1)! (-1) ^ {1-1}} {x ^ 1} \] \[ \frac {1} {x} = \frac {(0)! (-1) ^ {0}} {x} \] \[\frac {1} {x} = \frac {1} {x} \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\frac{\mathrm{d}^m}{\mathrm{d}x^m}\ln(x)=\frac{(m-1)!(-1)^{m-1}}{x^m}\]

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\frac {\mathrm {d} ^ {m + 1}} {\mathrm {d} x ^ {m + 1}} \ln (x) = \frac {m! (-1) ^ {m}} {x ^ {m + 1}} \]

So,

\[\frac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}\ln(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}^m}{\mathrm{d}x^m}\ln(x)\right)=\]\[\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{(m-1)!(-1)^{m-1}}{x^m}\right)=\]\[\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left((m-1)!(-1)^{m-1}\right)\cdot x^m-\left((m-1)!(-1)^{m-1}\right)\cdot \frac{\mathrm{d}}{\mathrm{d}x}x^m}{x^{2m}}=\]\[\frac{-(m-1)!(-1)^{m-1}mx^{m-1}}{x^{2m}}=\]\[\frac{m(m-1)!(-1)^mx^{m-1}}{x^{2m}}=\]\[\frac{m!(-1)^m}{x^{2m-m+1}}=\]\[\frac{m!(-1)^m}{x^{m+1}}\]

\(\blacksquare\)

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Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

  Hello everyone.  Today we will prove the product formula for  \(\dfrac{\pi}{2\sqrt{3}}\).  We will use the method of direct proof as a proof method.  Theorem 1:  \[\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\]\[\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}\] Theorem 2:  We have\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]expression whose numerators are the sequence of the odd prime numbers greater than \(3\) and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that \[\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots\] we will have \[\frac{1}{5} \cdo...
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