91Ó°ÊÓ

Understanding function derivatives is crucial when dealing with Taylor series. A derivative indicates how a function changes as its input changes. In essence, the derivative of a function gives us the slope of the function at any point.
For example, let's take the function given in the exercise: \( f(x) = x \ln x \). First, we find the first derivative, which helps us understand the function's initial rate of change:

• \( f'(x) = \frac{d}{dx} [x \ln x] = \ln x + 1 \)

Next, the second derivative tells us how this rate of change itself is changing:

• \( f''(x) = \frac{d}{dx} [\ln x + 1] = \frac{1}{x} \)

The third derivative continues this process:

• \( f'''(x) = \frac{d}{dx} [\frac{1}{x}] = -\frac{1}{x^2} \)

By calculating these derivatives, we prepare the necessary components to write out the Taylor series.

A special case of the Taylor series is the Maclaurin series, which is centered at \(a=0\). Essentially, a Maclaurin series is just a Taylor series expanded around zero. The formula for a Maclaurin series is:
\[ T(0) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
If the point specified in the problem was \(a=0\), we would be looking to calculate the derivatives of the function at zero. However, since our point is \(a=1\), we follow a similar approach but around \(x=1\).

Series expansion is a powerful tool in calculus that allows us to approximate functions with polynomials. By expanding a function into a series, we can more easily understand its behavior and make calculations.
Using Taylor series, we express a function \(f(x)\) as an infinite sum of its derivatives at a point \(a\). This representation is particularly useful for functions that are difficult to evaluate directly.
For example, in our exercise, after computing the necessary derivatives, we constructed the series:
\[ T(x) = 0 + 1(x-1) + \frac{1}{2!}(x-1)^2 - \frac{1}{3!}(x-1)^3 + \cdots \]
Simplifying it gives us:
\[ T(x) = (x-1) + \frac{(x-1)^2}{2} - \frac{(x-1)^3}{6} + \cdots \]
This polynomial representation makes it easier to analyze the function near \(x=1\).