91Ó°ÊÓ

The Taylor series is a powerful tool in calculus, used to approximate functions by a series of polynomials. A Taylor series for a function \( f(x) \) centered at \( x = a \) is expressed as \[ T_f (x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} \( x - a \)^n \, \] where \( f^{(n)}(a) \) denotes the n-th derivative of \( f \) at \( a \. \) This expansion helps us analyze and compute functions that are difficult to handle otherwise.
For example, consider the function \( f(x) = e^{-1 / x^2} \). When applying the Taylor series expansion at \( x = 0, \) we observe that all the derivatives at this point are zero. Hence, the Taylor series simplifies significantly: \[ T_f (x) = 0. \]
The Taylor series not only provides a way to approximate the function locally but also highlights if higher-order terms contribute, showcasing the behavior of the function around the center point.

Understanding the interval of convergence is essential while working with Taylor series. The interval of convergence refers to the set of \( x \) values for which the Taylor series converges to the actual function.
In general, to determine this interval, one can use the ratio test or other convergence tests. For the function \( f(x) = e^{-1 / x^2} \, \) we previously found the Taylor series \( T_f(x) = 0 \. \)
This means the series is convergent for any \( x \) within the interval \((- \infty, \infty) \. \) Therefore, the interval of convergence for our example is the entire real line: \((- \infty, \infty) \. \) Still, it’s critical to notice that the convergence to the function \( f(x) \) only holds true at \( x = 0 \. \)
This peculiarity occurs because, outside this point, \( f(x) = e^{-1 / x^2} \) is not zero, diverging from the Taylor series result.

Derivatives play a pivotal role in constructing Taylor series, representing how a function changes as its input changes. For our function \( f(x) = e^{-1 / x^2} \, \) the derivative at \( x = 0 \) is calculated using the definition of the derivative: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \. \]
Substituting \( f(0) = 0 \) and \( f(h) = e^{-1 / h^2} \), we find: \[ f'(0) = \lim_{h \to 0} \frac{e^{-1 / h^2}}{h} = 0 \. \]
This results from the fact that \( e^{-1 / h^2} \) decreases to zero much faster than \( h \) approaches zero.
Moreover, higher-order derivatives \( f^{(n)}(0) \) for \( n \geq 2 \) are also zero, simplifying the Taylor series significantly. Thus, by knowing the behavior of derivatives at a specific point, we can determine the series' terms effectively.
Capturing this behavior is what enables the Taylor series to provide a detailed local approximation of the function.