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 area of a circle

Hello everyone. Today we will prove the formula for the area of ​​a circle. We will use the method of direct proof as a proof method. 

Theorem: Denote by \(P \) the area of ​​a circle and by \(r \) the radius of a circle. Then the following equation holds: \(P = r ^ 2 \pi \) 

Proof: 

The equation of a circle in Cartesian  coordinate system is \(x ^ 2 + y ^ 2 = r ^ 2 \). Hence we have that \(y = \pm \sqrt {r ^ 2-x ^ 2} \). Based on the geometric interpretation of a certain integral, it follows: \[P = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx \] So, \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx \] \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx \] \[P = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx \] Let us now introduce the substitution \(x = r \cos \theta \). Hence, we have that \(dx = -r \sin \theta d \theta \). So we can write the following equation: \[P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\frac{r^2\cos^2\theta}{r^2}}\sin\theta \, d\theta\] After simplification we get:  \[P=-\int\limits_{\pi}^0 2r^2\sqrt{1-\cos^2\theta}\sin\theta \, d\theta\]\[P=-\int\limits_{\pi}^0 2r^2\sin^2\theta \, d\theta\]\[P=-r^2\int\limits_{\pi}^0 2\sin^2\theta \, d\theta\]\[P=-r^2\int\limits_{\pi}^0 (1-\cos2\theta) \, d\theta\]\[P=-r^2\left(\int\limits_{\pi}^0 d\theta-\int\limits_{\pi}^0 \cos2\theta \, d\theta\right)\]Since \(\int\limits_{\pi}^0 \cos2\theta \, d\theta=0\) we have: \[P=-r^2\int\limits_{\pi}^0 d\theta\]\[P=-r^2(0-\pi)\]\[P=r^2\pi\]

\(\blacksquare\)

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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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