Question

. Show why Lagrange's power series representation fails for the case \(f(x)=e^{-1 / x^{2}}\).

Short answer

Short Answer: The Lagrange's power series representation fails for the function \(f(x)=e^{-\frac{1}{x^2}}\) because the derivatives of the function are not defined at \(x=0\), and therefore, a Taylor series centered at \(x=0\) cannot be constructed. Moreover, the function is not analytic at \(x=0\), meaning it cannot be represented by a convergent power series in a neighborhood of that point.

Step-by-step solution

Find the Taylor series representation of the given function

Let's find the Maclaurin series (Taylor series centered at \(x=0\)) for the function \(f(x) = e^{-\frac{1}{x^2}}\). We will find its derivatives and evaluate them at \(x=0\): 1. \(f(x) = e^{-\frac{1}{x^2}}\) 2. \(f'(x) = \frac{d}{dx}(e^{-\frac{1}{x^2}}) = 2x e^{-\frac{1}{x^2}}\frac{1}{x^3} = \frac{2}{x} e^{-\frac{1}{x^2}}\) 3. \(f''(x) = -\frac{4}{x^3} e^{-\frac{1}{x^2}} + \frac{4} {x^5} e^{-\frac{1}{x^2}} = \frac{4}{x^3} e^{-\frac{1}{x^2}} \left( 1 - \frac{1}{x^2} \right)\) 4. And so on... Now evaluating each derivative at \(x=0\): 1. \(f(0) = e^{-\frac{1}{0^2}}\) is not defined. 2. \(f'(0) = \frac{2}{0} e^{-\frac{1}{0^2}}\) is not defined. 3. \(f''(0) = \frac{4}{0^3} e^{-\frac{1}{0^2}} \left( 1 - \frac{1}{0^2} \right)\) is not defined. 4. And so on... Since the derivatives of the given function are not defined at \(x=0\), we cannot construct a Taylor series representation centered at \(x=0\). This indicates that the Lagrange's power series representation fails for the given function.

Confirm that the function is not analytic at x = 0

The fact that the derivatives are not defined at \(x=0\) suggests that the function is not analytic at \(x=0\). A function is said to be analytic at a point if it can be represented by a convergent power series in a neighborhood of that point. Since there are no convergent power series representations centered at \(x=0\) for the given function, it is not analytic at \(x=0\). This is another reason why the Lagrange's power series representation fails for the given function.

Key concepts

Taylor series

The Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. It is a powerful tool in mathematical analysis that allows for approximating complex functions with polynomials, making calculations and other manipulations more feasible.

For a function to have a Taylor series expansion at a point, it must be infinitely differentiable at that point. The Taylor series of a function f(x) about a point a is given by the formula: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \cdots \]
When the point a is zero, the series is also referred to as a Maclaurin series. In our exercise, attempting to find a Maclaurin series for f(x) = e^{-1/x^2} fails because the derivatives of this function are not defined at x=0, rendering the Taylor series non-existent at that point.

Analytic functions

Analytic functions are a class of functions that are locally given by convergent power series. In other words, around any point in their domain, an analytic function can be represented as a power series. This property allows for the functions to be manipulated algebraically and, under certain conditions, differentiated and integrated term by term.

It's important to recognize that being differentiable at a point does not necessarily make a function analytic at that point. The function must be differentiable in some interval containing the point and it must be representable as a power series within that interval.

In the provided exercise, we determined that f(x) = e^{-1/x^2} is not analytic at x=0 because its power series representation doesn't converge. This insight further supports the finding that Lagrange's power series representation cannot be applied to this function at x=0.

Maclaurin series

A Maclaurin series is a special case of the Taylor series, centered at x=0. It has the form: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots \]
This series makes it easy to approximate functions near the origin if the function and all its derivatives are defined there.

However, as with our exercise, not all functions can be expanded into a Maclaurin series. If the necessary derivatives do not exist (such as at singular points or discontinuities), then the function cannot be represented by a Maclaurin series at that point, emphasizing the importance of a function's differentiability at the center point of expansion.

Mathematical analysis

Mathematical analysis is the branch of mathematics dealing with limits and related theories, such as differentiation, integration, measure, infinite series, and analytic functions. It enables mathematicians and scientists to describe and analyze the behavior and properties of functions and the abstract mathematical concepts they represent.

Through analysis, one can formalize concepts like continuity, convergence of sequences and series, and the rigorous definition of limits. The field is fundamental for proving the existence and uniqueness of solutions to differential equations, optimizing functions, and modeling real-world phenomena. The challenge presented by the function f(x) = e^{-1/x^2} is an example of the nuance and depth found in mathematical analysis, as it reveals the limitations of certain representations and the importance of a function's behavior at specific points for analysis.