Number System and Simplification

Using the Möbius Inversion Formula with ( f(n) = d ) and ( g(n) = \sum_{d|n} \mu(d) d ), we get ( d = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = \sigma(\frac{n}{d}) ) for all ( d | n ), we have ( \sum_{d|n} d = \sigma(n) ).

Since ( n ) is square-free, all of its divisors are also square-free. Therefore, ( \mu(d) = 1 ) for all ( d | n ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 1 ) for all ( d | n ), we have ( \sum_{d|n} \mu(d) = 1 ).

If ( m ) is divisible by ( n ), then all of the divisors of ( n ) that are divisible by ( m ) are also divisible by ( \frac{n}{m} ). Therefore, ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 1 & \text{if } m = 1 \ 0 & \text{if } m > 1 \end{cases} ).

If ( m ) is not divisible by ( n ), then there is at least one prime factor of ( n ) that is not a prime factor of ( m ). Therefore, there is at least one divisor of ( n ) that is not divisible by ( m ). Since ( \mu(d) = 0 ) for all ( d | n ) that are divisible by ( m ), we have ( \sum_{d|n} \mu(d) = -1 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = -1 ) for all ( d | n ) that are not divisible by ( m ), we have ( \sum_{d|n} \mu(d) = \begin{cases} 0 & \text{if } m = 1 \ -1 & \text{if } m > 1 \end{cases} ).

If ( d ) is a divisor of ( n ) that is not relatively prime to ( n ), then there is at least one prime factor of ( n ) that is also a prime factor of ( d ). Therefore, ( \mu(d) = 0 ). Using the Möbius Inversion Formula with ( f(n) = 1 ) and ( g(n) = \sum_{d|n} \mu(d) ), we get ( 1 = \sum_{d|n} \mu(d) g(\frac{n}{d}) ). Since ( g(\frac{n}{d}) = 0 ) for all ( d | n ) that are not relatively prime to ( n ), we have ( \sum_{d|n} \mu(d) = 0 ).

The Brahmagupta formula is given by the equation $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

The general formula for the Madhava series for the arctangent function is $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$.

The general formula for the Madhava series for pi is $$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$$.

To find the value of the expression, we can add the fractions as follows: (\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = \frac{11}{6}).

The general formula for the Madhava series for the arctangent function is $$arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$. This series converges for all values of x between -1 and 1.

The general formula for the Madhava series for pi is $$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$$. This series converges very slowly, but it was the first series expansion for pi to be discovered.

The equation (\frac{a}{b} = \frac{c}{d}) is known as the cross-multiplication rule, which is used to solve proportions.

Modular arithmetic involves calculations with integers modulo a fixed integer. In this case, we are working modulo 5. To find 7^3 mod 5, we can calculate 7^3 = 343, and then divide it by 5 to get the remainder. 343 ÷ 5 = 68 remainder 3. Therefore, 7^3 mod 5 = 3.

The Gamma function at (\frac{1}{2}) is equal to (\sqrt{\pi}).

The Gamma function at (\frac{3}{2}) is equal to (\frac{\sqrt{\pi}}{2}).