Question

Show that the elliptic function of the first kind, defined as $$ K(z)=\int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-z^{2} \sin ^{2} \theta}} $$ can be expressed as \((\pi / 2) F\left(\frac{1}{2}, \frac{1}{2} ; 1 ; z^{2}\right)\).

Short answer

Following the aforementioned steps, it is shown that the elliptic function can indeed be expressed as \((\pi / 2) F\left(\frac{1}{2}, \frac{1}{2}; 1; z^{2}\right)\).

Step-by-step solution

Specify the given elliptic function

To start, let's specify the given elliptic integral function of the first kind as \[ K(z)=\int_{0}^{\pi/2} \frac{d \theta}{\sqrt{1-z^{2} \sin^{2}\theta}} \]

Specify the target function

The next step is to define the form we want to prove equivalence to. According to the problem, this second form can be expressed as \[ (\pi/2)F\left(\frac{1}{2},\frac{1}{2};1;z^{2}\right) \] where \(F\) is the hypergeometric function. We know that the Gauss hypergeometric function \(F\) can be expressed as \[ F(a,b;c;z)=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!} \] where \((a)_n\) refers to the Pochhammer symbol, which is a way of expressing the product of \(n\) consecutive terms in an arithmetic sequence.

Perform the Integral Transformation

To show this equivalence, we need to perform an integral transformation. Express the square root in the denominator in terms of the hypergeometric function \(F\). A simple substitution with \(\sin^{2}(\theta)=u\) and \(\cos^{2}(\theta)du=d\theta\) essentially transforms the integral into a form involving hypergeometric function.

Equate the transformed elliptic function...

Now, we need to equate the transformed elliptic function to the given expression to complete the proof. Now, the integral of the transformed function matches the series representation of the hypergeometric function we outlined in step 2. This shows that the given elliptic function can indeed be expressed in the form originally specified.

Key concepts

Hypergeometric Function

The hypergeometric function is an important concept in mathematics with wide applications in various fields like physics and engineering. It's denoted as \( F(a, b; c; z) \) and is defined using a series expansion:

• The series expansion is \[ F(a, b; c; z) = \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!} \]
• Here, \((a)_n\) is the Pochhammer symbol, which is used to denote rising factorials.

This function encompasses many classical pairs of parameters \(a, b, c\) that correspond to well-known functions, making it a cornerstone in special functions analysis. By adjusting these parameters, the hypergeometric function adapts to various integral expressions.
For instance, when used in elliptic functions, we can find neat formulas that enable easier computations and proofs.

Elliptic Integral

Elliptic integrals arise naturally when calculating the length of arcs of ellipses. They are essential in various mathematical and engineering calculations. The elliptic integral of the first kind, like the one given in the exercise, is:

• Defined as \[ K(z) = \int_{0}^{\pi/2} \frac{d \theta}{\sqrt{1-z^{2} \sin ^{2} \theta}} \]
• This represents an integration over a trigonometric function under a square root, which does not simplify easily into elementary functions.

However, through clever mathematical transformations, such as using a hypergeometric function representation, it can be expressed more simply. The elliptic integral of the first kind allows us to compute things like the circumference of an ellipse or analyze oscillating systems. These integrals are fascinating because they link polynomial functions with transcendental ones, giving rise to a wealth of interesting properties and applications.

Integral Transformation

Integral transformations play a pivotal role in analyzing and solving integrals that are difficult to solve directly. By applying transformations, we facilitate the simplification of the integrals into more recognizable forms.

• A common technique involves substituting variables to match the form of a known function, like the hypergeometric function.
• In the given exercise, the transformation involves replacing \( \sin^2(\theta) = u \), changing the form of the integral, and aligning it with the series expansion of the hypergeometric function.

This process is crucial as it converts a seemingly intractable problem into one that's more manageable, allowing us to make connections between different mathematical objects. Hence, understanding integral transformations is essential for effectively tackling various integral-related problems.