Question

Prove that, for positive integral \(n:\) $$ \Gamma\left(n+\frac{1}{2}\right)=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2^{n}} \sqrt{\pi}=\frac{(2 n) !}{4^{n} n !} \sqrt{\pi}. $$

Short answer

Using the recurrence relation for \( \text{Γ}(x) \), \( \text{Γ}\big(n+\frac{1}{2}\big) = \frac{(2n)!}{4^n n!}\text{√π} \).

Step-by-step solution

Recall the Gamma Function

The Gamma function \(\text{Γ}(n)\) is a generalization of the factorial. For a positive integer \(n\), \( \text{Γ}(n) = (n-1)! \). When \(n = k + \frac{1}{2}\), the Gamma function can be expressed as \( \text{Γ}\big(k + \frac{1}{2}\big) = \frac{(2k)!}{4^k k!}\frac{\text{Γ}\big(\frac{1}{2}\big)}{\text{Γ}\big(1\big)} = \frac{(2k)!}{4^k k!}\text{Γ}\big(\frac{1}{2}\big).\)

Evaluate \( \text{Γ}\big(\frac{1}{2}\big) \)

It is known that \( \text{Γ}\big(\frac{1}{2}\big) = \text{Γ}\big(1-0.5\big) = \text{Γ}(0.5) \). This is equal to \(\text{√π}\).

Substitute known values

Now, substitute \( \text{Γ}\big(\frac{1}{2}\big)= \text{√π} \) back into the expression from Step 1: \[ \text{Γ}\big( k + \frac{1}{2} \big) = \frac{(2k)!}{4^k \text{k}!} \text{√π}. \]

Rewrite in terms of \( n \)

Given that \( n = k+0.5 \, \text{thus} \ k = n-0.5 \). Therefore: \[ \text{Γ}(n+0.5) = \frac{(2(n-0.5))!}{4^{n-0.5} (n-0.5)!} \text{√π} = \frac{(2n-1)!}{4^{n-0.5} (n-0.5)!} \text{√π}. \]

Use recurrence relation

Utilize the recurrence relation of \( \text{Γ}(x+1) = x \text{Γ}(x) \). We know: \(\text{Γ}(n+0.5) = (n-0.5)! (2n-1)!! \text{Γ}(0.5) \). Therefore: \[ \text{Γ}(n + 0.5) = \frac{(2n-1)!}{2^n} \text{√π}. \]

Simplify and derive final formula

Rewriting the simplified expression: \[ \text{Γ}\big(n+\frac{1}{2}\big) = \frac{1 · 3 · 5 · . . . (2n-1)}{2^n} \text{√π} = \frac{(2n)!}{4^n n!}\text{√π}, \]

Key concepts

Gamma Function properties

The Gamma Function, denoted as \( \Gamma(x) \), is a fundamental concept in mathematics, extending the idea of factorials to non-integer values. It's defined for all complex numbers except non-positive integers. For positive integers, the Gamma Function is linked directly to factorials: \( \Gamma(n) = (n-1)! \).Some key properties of the Gamma Function include:

• \( \Gamma(x+1) = x \Gamma(x) \) (recurrence relation)
• \( \Gamma(1) = 1 \)
• \( \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \)

Understanding these properties helps in manipulating and simplifying expressions involving the Gamma Function.

Factorials

Factorials, typically denoted as \( n! \, \), play a crucial role in various mathematical areas, including combinatorics, calculus, and more. The factorial of a positive integer \( n \) is the product of all positive integers less than or equal to \( n \):
\[ n! = n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 \cdot 1 \]
By definition, \( 0! = 1 \). For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factorials grow rapidly with larger \( n \), which is why they are useful in describing combinations and permutations in larger datasets. In relation to the Gamma function, we see that \( \Gamma(n) = (n-1)! \), tying it to some useful properties like the recurrence relation.

Recurrence Relation

The recurrence relation is a fundamental property in both Gamma Functions and Factorials. For the Gamma function, it's expressed as:
\( \Gamma(x+1) = x \Gamma(x)) \)
This property signifies that the value of the Gamma Function at \( x+1 \) can be derived by multiplying \( x \) with the Gamma Function at \( x \). This property simplifies calculations and derivations involving the Gamma Function substantially.
For instance, in our exercise, we use the property to express gamma values and ultimately simplify our problem statement. Additionally, in factorials, the notion of recurrence is seen when defining \( n! = n \cdot (n-1)! \). This ties back to the Gamma Function, making both areas interconnected and crucial for deeper mathematical understanding.