Exploring the Limitations of Taylor Series Expansions for f(x) e^x / (x-1)^2 at x1

Understanding the intricacies of Taylor series expansions is a crucial skill for any mathematician or engineer. However, not all functions can be expressed in this form, especially in cases where the function is discontinuous. In this article, we will delve into a specific example that highlights the limitations of Taylor series expansions when the function is not defined in a certain point.

Introduction to Taylor Series Expansions

A Taylor series expansion is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. Given a function f(x), its Taylor series expansion about a point x a is expressed as:

f(x) f(a) f'(a)(x-a) frac{f''(a)}{2!}(x-a)^2 frac{f'''(a)}{3!}(x-a)^3 ...

The Function in Question: f(x) e^x / (x-1)^2

Let's consider the function f(x) e^x / (x-1)^2. At first glance, this function might seem straightforward, but upon closer inspection, we encounter a significant problem. The function is not defined at x 1. Specifically, the denominator becomes zero, resulting in a division by zero error.

Discontinuity and Removability

The discontinuity at x 1 is not a removable discontinuity. A removable discontinuity would be a point where the function is undefined but can be made continuous by redefining the function at that point. In this case, the discontinuity is more severe and cannot be resolved by simply assigning a value to f(1).

Exploring Taylor Series Expansion for f(x)

Let's attempt to see if the Taylor series expansion of the function around x 1 is possible. For a function to have a Taylor series expansion around a point, it needs to be infinitely differentiable at that point. However, the function f(x) e^x / (x-1)^2 is not defined at x 1, and thus its derivatives are also undefined at this point.

Attempting to compute the Taylor series expansion:

Find the derivatives of f(x) at x 1.

Compute the value of the function and its derivatives at x 1.

Construct the Taylor series expansion using the above values.

Note: The function f(x) e^x / (x-1)^2 is not well-behaved at x 1. Therefore, calculating the derivatives at this point will lead to indeterminate forms or undefined expressions, such as 0/0 or ∞/∞.

Alternative Representation: Laurent Series

Given the discontinuity at x 1, the function f(x) e^x / (x-1)^2 is better represented using a Laurent series. A Laurent series is an extension of the Taylor series that allows for negative powers of (x-a), where a is the expansion point. In this case, a Laurent series around x 1 would be:

f(x) sum_{n-infty}^{infty} a_n (x-1)^n

To find the coefficients a_n, we need to perform a complex analysis or use other advanced mathematical techniques, but the key point is that the Laurent series can handle the singularity at x 1 more effectively than the Taylor series.

Conclusion

In summary, the function f(x) e^x / (x-1)^2 does not have a Taylor series expansion around x 1 due to its discontinuity. However, by using a Laurent series, we can represent the function in a way that accounts for the singularity at x 1. This example underscores the importance of understanding the properties of functions and the appropriate mathematical tools for different types of discontinuities.

References

1. Arfken, G. B., Weber, H. J. (2005). Mathematical methods for physicists (6th ed.). Academic Press.

2. Press, W. H., Teukolsky, S. A., Vetterling, W. T., Flannery, B. P. (2007). Numerical recipes: The art of scientific computing (3rd ed.). Cambridge University Press.