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} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5
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Algebraic proof of Pythagoras' theorem

Hello everyone. Today we will prove Pythagoras' theorem using elementary algebra. This proof was first appeared in China more than 2000 years ago. 

Theorem: Let \(a \) and \(b \) be the legs, and \(c \) the hypotenuse of a right triangle. Then the equation \(c ^ 2 = a ^ 2 + b ^ 2 \) holds. 

Proof: 

Consider the following diagram consisting of a large square with side length \(a + b \) and a small square with side length \(c \) and notice four right triangles with legs \(a, b \) and hypotenuse \( c \).

Let us denote the area of ​​the large square by \(P \). Then \[P = (a + b)(a + b) \] Let \(P_1 \) be the area of ​​one of the four right triangles, and \(P_2 \) the area of ​​a small square with side length \(c \). Then we have that: \[P_1 = \frac {ab} {2} \] \[P_2 = c ^ 2 \] Since \(P = 4P_1 + P_2 \) we can write the following equation:
\[(a + b) (a + b) = 4 \cdot \frac {ab} {2} + c ^ 2 \] From here we have that: \[a ^ 2 + 2ab + b ^ 2 = 2ab + c ^ 2 \] 
That is, when we subtract \(2ab \) from both sides of the equation we get: \[c ^ 2 = a ^ 2 + b ^ 2 \] 
\(\blacksquare \)



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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} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5
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