Question

Consider the function \(f(z)=(1+z)^{\alpha}\). (a) Show that \(d^{n} f /\left.d z^{n}\right|_{z=0}=\Gamma(\alpha+1) / \Gamma(\alpha-n+1)\), and use it to derive the relation $$ \begin{array}{l} (1+z)^{\alpha}=\sum_{n=0}^{\infty}\left(\begin{array}{l} \alpha \\ n \end{array}\right) z^{n}, \quad \text { where } \\ \left(\begin{array}{l} \alpha \\ n \end{array}\right) \equiv \frac{\alpha !}{n !(\alpha-n) !} \equiv \frac{\Gamma(\alpha+1)}{n ! \Gamma(\alpha-n+1)} . \end{array} $$ (b) Show that for general complex numbers \(a\) and \(b\) we can formally write $$ (a+b)^{\alpha}=\sum_{n=0}^{\infty}\left(\begin{array}{l} \alpha \\ n \end{array}\right) a^{n} b^{\alpha-n} $$ (c) Show that if \(\alpha\) is a positive integer \(m\), the series in part (b) truncates at \(n=m .\)

Short answer

Yes, \(f(z)=(1+z)^α\) can indeed be expanded as a power series to \(\sum_{n=0}^{\infty} \frac{\Gamma (α+1)}{n! \Gamma(α - n + 1)} z^n\). This applies to complex numbers \(a\) and \(b\) too – \((a+b)^α =\sum_{n=0}^{\infty} \frac{\Gamma(α+1)}{n! \Gamma(α - n + 1)} a^n b^{α-n}\). When α is a positive integer, the series truncates at \(n = α\).

Step-by-step solution

Compute Derivative

Let's first compute the derivative of \(f(z)\) at \(z=0\). Using the chain rule of differentiation, we find that \(f'(z) = α(1+z)^{α-1}\). This implies \(f'(0) = α\), hence \(\frac{{\alpha!}}{{(α-n)!}}\), which is the \(n\)th derivative of \(f(z)\) at \(z=0\).

Expand \(f(z)\) into Infinite Series

Substitute the derivative of \(f(z)\) from step 1 into the power series expansion. We get \(f(z)=(1+z)^α=\sum_{n=0}^{\infty} \frac{\Gamma (α+1)}{n! \Gamma(α - n + 1)} z^n\). This concludes the expansion of the function \(f(z)\), which fulfills the first part of the problem.

Apply the Binomial Theorem

To generalize part (a) to complex numbers \(a\) and \(b\), we make use of the binomial theorem. Replace \(z\) in the above equation with \(ab\), we get \((a+b)^α =\sum_{n=0}^{\infty} \frac{\Gamma(α+1)}{n! \Gamma(α - n + 1)} a^n b^{α-n}\), which is what we wanted to show in the second part of the question.

Truncate the Series for Positive Integer α

In the case where α is a positive integer – say \(m\) – the term \(\Gamma(α - n + 1)\) will become undefined (because it becomes \(\Gamma(0)\)) after \(n = m\). Because of this, the series truncates at \(n = m\). This completes the third part.

Key concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in mathematics, allowing us to expand expressions that have been raised to a power. In classical terms, this is usually of the form

• \((x + y)^n\)

The theorem provides a formula for expanding these expressions into a sum involving terms of the form \(x^{n-k} y^k\). In our context of complex analysis, we deal with complex exponents and variables, such as expressing \((1+z)^\alpha\) in terms of a series.

This center around the idea of a generalized binomial coefficient or term. This coefficient is given by the expression:

• \[\left(\begin{array}{c}\alpha \end{array}\right) = \frac{\Gamma(\alpha+1)}{n! \Gamma(\alpha-n+1)} \quad \text{for non-integer}\ \alpha\]

Here, the Gamma function helps us extend the factorial concept, which is normally not defined for non-integers, to non-integer values. This is essential for working with series in complex analysis. Understanding the binomial theorem in this broader context allows us to tackle more sophisticated problems involving complex numbers and power series.

Gamma Function

The Gamma Function is a remarkable concept in mathematical analysis, often considered an extension of the factorial function to complex numbers. While the factorial, denoted as \(n!\), is defined only for non-negative integers, the Gamma Function \(\Gamma(n)\) extends this definition to include complex numbers. In fact, for a positive integer \(n\), we have

• \(\Gamma(n) = (n-1)!\)

The Gamma Function takes the form:

• \[\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt\]

This integral definition allows it to handle non-integer and even complex \(z\).

In the context of the exercise, the Gamma Function is crucial for determining the coefficients in the power series expansion. When using a general exponent \(\alpha\) in expressions like \((1+z)^\alpha\), the coefficients \(\frac{\Gamma(\alpha+1)}{n! \Gamma(\alpha-n+1)}\) ensure that the series makes sense even for complex or fractional exponents. Thus, the inclusion of the Gamma Function in these expansions provides a seamless transition from integer to non-integer scenarios, enhancing the scope and applicability of our analyses.

Power Series Expansion

A Power Series Expansion is a way to represent a function using an infinite sum of terms involving powers of a variable. This concept is invaluable in complex analysis for approximating functions and solving equations.

For a function such as \( (1+z)^\alpha \), the power series expansion enables us to express the function as an infinite sum:

• \[(1+z)^\alpha = \sum_{n=0}^{\infty} \left(\begin{array}{c}\alpha \end{array}\right) z^n\]

Each term in this series is composed of the binomial coefficient, which involves the Gamma Function, and the corresponding power of \(z\). This expansion is not just theoretical but practical, allowing computations to be carried out with ease.

When \(\alpha\) is a positive integer, the power series becomes a finite polynomial, which simplifies calculations substantially. This truncation occurs because the coefficients become zero when \(n\) exceeds \(\alpha\). Power series expansions are thus fundamentally important for dealing with complex variables, offering both depth and flexibility in mathematical analysis.