Problem 29 - Thomas Calculus

Question

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=1 / x^{2}, \quad a=1\)

Step-by-Step Solution

Answer

The Taylor series is \(T(x) = 1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 + \cdots\).

Step 1: Identify the Function and Point

We are given the function \( f(x) = \frac{1}{x^2} \) and we need to find its Taylor series around the point \( a = 1 \). This means the series will be expanded in terms of \( x \) around 1.

Step 2: Find the Derivatives of the Function

Calculate the derivatives of \( f(x) \) at \( x = 1 \).1. \( f(x) = x^{-2} \)2. First derivative: \( f'(x) = -2x^{-3} \)3. Second derivative: \( f''(x) = 6x^{-4} \)4. Third derivative: \( f'''(x) = -24x^{-5} \)5. Continue this pattern to determine higher-order derivatives.Evaluate these at \( x = 1 \): \( f(1) = 1, f'(1) = -2, f''(1) = 6, f'''(1) = -24, \)... etc.

Step 3: Write the General Formula of the Taylor Series

The formula for the Taylor series of a function \( f \) about a point \( a \) is:\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]For our function, this becomes:\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^n\]

Step 4: Substitute Derivatives into the Taylor Series Formula

Substitute the calculated derivatives into the Taylor series:\[T(x) = \frac{1}{0!}(x-1)^0 - \frac{2}{1!}(x-1)^1 + \frac{6}{2!}(x-1)^2 - \frac{24}{3!}(x-1)^3 + \cdots\]This leads to:\[T(x) = 1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 + \cdots\]

Step 5: Summarize the Result

The Taylor series expansion of \( f(x) = \frac{1}{x^2} \) centered at \( x = 1 \) is:\[T(x) = 1 - 2(x-1) + 3(x-1)^2 - 4(x-1)^3 + \cdots\]This series continues indefinitely by applying the same pattern to higher derivatives.

Key Concepts

Derivatives

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

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

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

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.