Question

Show that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\), and that \((2 k-1) ! ! \equiv(2 k-1)(2 k-3) \cdots 5 \cdot 3 \cdot 1=\frac{2^{k}}{\sqrt{\pi}} \Gamma\left(\frac{2 k+1}{2}\right)\).

Short answer

\(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\), and \( (2 k-1)!!=\frac{2^{k}}{\sqrt{\pi}}\Gamma\left(\frac{2 k+1}{2}\right)\)

Step-by-step solution

Proof of First Part

Using the integral definition of Gamma function, can write \(\Gamma\left(\frac{1}{2}\right)\) as \(\Gamma\left(\frac{1}{2}\right)=\int_0^\infty t^{-1/2}e^{-t}dt\). Using the substitution \(u=\sqrt{t}\), \(du=\frac{1}{2\sqrt{t}}dt\). Thus, \(\Gamma\left(\frac{1}{2}\right)=2\int_0^\infty e^{-u^2}du\). Now, use polar coordinates to evaluate the integral, denote as \(I\). \(I^2=(2\int_0^\infty e^{-u^2}du)^2=4\int_0^\infty\int_0^\infty e^{-x^2-y^2}dxdy\). This integral can be evaluated in polar coordinates to get \(I=\sqrt{\pi}\), hence \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Proof of Second Part

Begin by noting that the double factorial is for odd numbers only, and can be written as \( (2n-1)!!=(2n-1)\cdot(2n-3)\cdot\ldots\cdot3\cdot1\), which holds for both positive integers and zero. The double factorial of \( (2k-1)!!=(2k-1)(2k-3)\cdots5\cdot3\cdot1\). Using the properties of the Gamma function and factorials, it can be shown that \( (2k-1)!!=(2k-1)(2k-3)\cdots5\cdot3\cdot1=\frac{2^{k}}{\sqrt{\pi}}\Gamma\left(\frac{2 k+1}{2}\right)\). Since for \(n>0\), \( (2n)!!=2^n\cdot n!\). Then, \( (2k-1)!!=\frac{(2k)!}{(2^k\cdot k!)\cdot 2^k}=\frac{2^{k}}{\sqrt{\pi}}\Gamma\left(\frac{2 k+1}{2}\right)\), which completes the proof.

Key concepts

Double Factorial

The double factorial is a mathematical operation denoted by two exclamation points (!!); it's not the same as taking the factorial of a factorial. For a positive integer, the double factorial of an odd number is the product of all the odd integers from 1 up to that number. For example, the double factorial of 9, denoted as 9!!, would be calculated as follows:

\( 9!! = 9 \times 7 \times 5 \times 3 \times 1 = 945 \)

This operation is helpful in various branches of mathematics and physics, such as quantum mechanics and combinatorics. It can be especially useful in simplifying expressions involving factorials in combinatorial problems.

Integral Definition of Gamma Function

The Gamma function (denoted as \(\Gamma\)) is a concept that extends the factorial function into the real and complex number domains, except for non-positive integers. Its integral definition, for a complex number s with a real part greater than 0, is:

\(\Gamma(s) = \int_0^\infty t^{s-1}e^{-t}dt \)

This function allows us to calculate factorials of non-integer values. It has numerous properties and plays a vital role in various areas of mathematics including probability and statistics, combinatorics, and complex analysis. An interesting fact is that for positive integers \(n\), the Gamma function satisfies \(\Gamma(n)=(n-1)!\), which beautifully ties it to the classic factorial for whole numbers.

Polar Coordinates Integration

Integration in polar coordinates is a powerful technique when dealing with problems that exhibit radial symmetry. In such a case, converting a double integral from Cartesian coordinates (x, y) to polar coordinates (r, \(\theta\)) can simplify the process. The conversion is obtained by the relations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), with the Jacobian of the transformation being \(r\). Hence, the double integral would be expressed as:

\( \int_\alpha^\beta \int_a^b f(r, \theta) r \, dr \, d\theta \)

In the context of the Gamma function, using polar coordinates allows us to evaluate otherwise challenging integrals by exploiting symmetry, particularly that of the exponential function involving a square of variables.

Mathematical Proof Techniques

Mathematical proofs are logical arguments that demonstrate the truth of a mathematical statement. There are various techniques to prove a theorem or an equation:

• Direct Proof: Begins with known facts and uses logical deductions to arrive at the statement to be proved.
• Proof by Contradiction: Assumes the opposite of what needs to be proved and shows that this leads to a contradiction, thus proving the original statement.
• Inductive Proof: Proves the base case and then demonstrates that if the statement holds for one case, it must hold for the next.
• Proof by Contrapositive: Instead of proving a statement directly, proves that if the conclusion is false, then the hypothesis must also be false.

Each of these techniques can be appropriately applied depending on the nature of the problem. Understanding these methods is fundamental to progressing in advanced mathematics and tackling complex proofs.