Question

Let \(L_{n} \equiv L_{n}^{0}\). Now differentiate both sides of $$g(x, t)=\frac{e^{-x t /(1-t)}}{1-t}=\sum_{0}^{\infty} t^{n} L_{n}(x)$$ with respect to \(x\) and compare powers of \(t\) to obtain \(L_{n}^{\prime}(0)=-n\) and \(L_{n}^{\prime \prime}(0)=\frac{1}{2} n(n-1)\). Hint: Differentiate \(1 /(1-t)=\sum_{n=0}^{\infty} t^{n}\) to get an ex- pression for \((1-t)^{-2}\).

Short answer

Upon differentiating and expanding the function as power series, by comparing coefficients for powers of \(t\), it found that \(L_{n}^{\prime \prime}(0)=\frac{1}{2} n(n-1)\) and \(L_{n}^{\prime}(0)=-n\). The hint was used to derive a new power series for better comparison between coefficients.

Step-by-step solution

Find the Derivative of \(g(x, t)\) with respect to \(x\)

Start with the given function \(g(x, t) = \frac{e^{-x t /(1-t)}}{1-t}\). The derivative of \(g\) with respect to \(x\), denoted as \(g'(x, t)\), can be calculated by applying the chain rule of differentiation, which states that the derivative of a composition of functions is the derivative of the outer function multiplied by the derivative of the inner function. For the function \(g(x, t)\), the outer function is \(e^{u}\) and the inner function is \(-x t /(1-t)\), hence, the first derivative is \(g'(x, t) = -\frac{t e^{-x t /(1-t)}}{1-t}\).

Expand \(g'(x, t)\) as a power series

As the function \(g'(x, t)\) is a function of both \(x\) and \(t\), expand it as a power series in \(t\). Recall the expansion of exponential function \(e^{u}\) into a power series: \(e^{u} = \sum_{n=0}^{\infty} u^n/n!\). Applying this gives \(g'(x, t) = -\sum_{n=0}^{\infty} (\frac{t^n x^n}{n! (1-t)})\)

Compare powers of \(t\) in both sides of the equation

By comparing coefficients for equal powers of \(t\) in the sum and in the function \(t^{n} L_{n}(x)\), it is possible to find that \(L_{n}^{\prime}(0)=-n\). Also, taking the second derivative gives us \(L_{n}^{\prime \prime}(0)=\frac{1}{2} n(n-1)\). Note that, as \(L_{n} \equiv L_{n}^{0}\), the derivatives are evaluated at \((0)\), which simplifies the process.

Differentiate \(\frac{1}{1-t}\) to get an expression for \((1-t)^{-2}\)

As a hint provided, differentiate the given power series \(\frac{1}{1-t} = \sum_{0}^{\infty} t^n\) with respect to \(t\) to obtain \(\frac{1}{(1 - t)^2} = \sum_{n=0}^{\infty} (n + 1) t^n\). This new power series can be utlized in the application of comparing coefficients technique.

Key concepts

Power Series Expansion

Power series expansion is a critical concept in mathematics and physics, allowing us to express functions as an infinite sum of terms whose values depend on the powers of a variable. For instance, in the context of Laguerre polynomials, the power series expansion becomes essential in establishing the relationships between function values and their derivatives at a point.

Let's take a closer look at the exponential function, which is often expanded into a power series. The expansion is given by:

$$e^u = \$ \sum_{n=0}^{\infty} \frac{u^n}{n!}.$$

This series representation is particularly useful because the exponential function is ubiquitous in mathematics, and the series expansion allows for a straightforward way to compute its values and integrate with other mathematical tools.

In the exercise, we apply this idea to the function e^{-xt/(1-t)}, and after differentiating with respect to x, we expand it into a power series in t. This method is a powerful way to solve complex problems, as it allows us to analyze the coefficients of each power of t and extract valuable information about the behavior of the function and its derivatives, helping us understand properties of Laguerre polynomials more deeply.

Chain Rule of Differentiation

The chain rule of differentiation is an indispensable tool in calculus, particularly when dealing with composite functions. This rule provides a method to differentiate a composition of functions by multiplying the derivative of the outer function by the derivative of the inner function.

In the given exercise, the chain rule comes into play when differentiating the function g(x, t). The function is a composition of the exponential function and a rational function of x and t. By applying the chain rule, one can take the derivative of the outer function, e^{u}, and multiply it by the derivative of the inner function, -xt/(1-t). The result gives us the product that relates to the series expansion of Laguerre polynomials.

Understanding the chain rule is crucial for tackling various problems in calculus, physics, and engineering. It is often used to find the slope of tangent lines, rates of change, and in this case, to assist in the analysis of polynomials and their properties derived from their derivative relationships, thereby improving our mathematical toolkit to solve ordinary differential equations.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives. They play a crucial role in many areas such as physics, engineering, and economics because they model how systems evolve over time.

The relationship between power series expansions and ODEs is profound; expansions can be used to solve or approximate solutions to ODEs. In our exercise, the Laguerre polynomials are related to a specific ODE. By differentiating and comparing powers of t, we identify coefficients that correspond to the derivatives of the Laguerre polynomials at zero. This method is akin to solving an ODE by comparing both sides of an equation and extracting information about the function and its derivatives—here, specifically at x=0.

Methods for solving ODEs can vary from separation of variables, integrating factors, or series solutions among others. The series solution method, shown in our exercise, involves expressing a function as a series and finding coefficients that satisfy the ODE. These techniques are pivotal in understanding and predicting the behavior of natural phenomena described by ODEs.