91Ó°ÊÓ

Derivatives are the building blocks of calculus and play a crucial role in the development of Taylor series. They represent how a function changes as its input changes. In simpler terms, a derivative tells us the slope of the function at any given point.
To construct a Taylor series for a function, we need its derivatives at a specific point. For example, in the series expansion of the function \( f(x) = \frac{1}{x^2} \), we calculate the derivatives at \( x = 1 \):

• The first derivative \( f'(x) = -2x^{-3} \) tells us how \( f(x) \) changes at the point.
• The second derivative \( f''(x) = 6x^{-4} \) gives us information about the curvature of the function.
• Higher-order derivatives provide even deeper insights into the behavior of the function.

By evaluating these at \( x = 1 \), we obtain \( 1, -2, 6, -24 \), and so on, to construct the series.

Series expansion involves expressing a function as an infinite sum of terms, each computed as a derivative of the function. The Taylor series is a type of series expansion that approximates functions using polynomials based on the derivatives calculated at one point.
The general formula for a Taylor series about a point \( a \) is:
\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]
This means each term in the series is derived from the function's derivatives, divided by factorial, and then multiplied by \( (x-a)^n \). In the given exercise, we used \( a = 1 \) to form the series:
\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^n\]
Each term captures more of the function's behavior around \( x=1 \), allowing us to write \( f(x) \) as an infinite sum.

Polynomials are mathematical expressions involving sums of powers of the variable \( x \). They are simple yet powerful tools in mathematics. Taylor series use polynomials to approximate functions based on their derivatives.
Polynomial terms of a Taylor series have specific coefficients defined by the derivatives at the given point. For our function, \( f(x) = \frac{1}{x^2} \), the Taylor series expansion resulted in the polynomial:
\[T(x) = 1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 + \cdots\]

• The constant term (0th degree polynomial) is the function value at the center, \( 1 \).
• Higher degree terms (like \((x-1)\), \((x-1)^2\)) adjust the approximation based on the slope, curvature, and more.

Thus, Taylor series allow us to approximate complex functions with a series of straightforward polynomial computations.

Function evaluation is crucial in understanding how a Taylor series approximates a function. It involves replacing the variable with specific values to see how well the polynomial represents the function.
The expansion derived from \( f(x) = \frac{1}{x^2} \) allows us to estimate \( f(x) \) near \( x=1 \) by evaluating the Taylor series. For example, if we plug in \( x = 1.1 \) into the Taylor series:
\( T(1.1) = 1 - 2(1.1-1) + 3(1.1-1)^2 - 4(1.1-1)^3 + \cdots \)
This evaluation will yield a value close to the precise function \( \frac{1}{x^2} \) at \( x = 1.1 \). The more terms we consider, the better the approximation.
Function evaluation demonstrates the power of Taylor series in finding function values and predicting behavior in real-world applications.