Question

Find the Taylor polynomial of degree \(n\), at \(x=c\), for the given function. $$f(x)=\frac{1}{x^{2}}, \quad n=8, \quad c=1$$

Short answer

The Taylor polynomial is \( P_8(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \ldots + (x-1)^8 \).

Step-by-step solution

Understand the Taylor Polynomial

A Taylor polynomial is an approximation of a function at a point using derivatives of that function at the point. For the function \( f(x) = x^{-2} \), the Taylor polynomial of degree \( n \) at \( c \) is given by \[ P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(c)}{k!}(x-c)^k. \] We need to find derivatives of \( f(x) \) up to the 8th order and evaluate them at \( c = 1 \).

Compute the Derivatives

The first derivative of \( f(x) \) is \( f'(x) = -2x^{-3} \). The second derivative is \( f''(x) = 6x^{-4} \).Continuing this process will yield derivatives of the form: \( f^{(k)}(x) = (-1)^k k! x^{-(k+2)} \).

Evaluate Derivatives at x=c

Evaluate the derivatives at \( x = 1 \):\( f^{(k)}(1) = (-1)^k k! \, \text{for all}\, k.\)This is because \( x^{-(k+2)} \) becomes \( 1^-(k+2) \) which is 1.

Construct the Taylor Polynomial

Now substitute the evaluated derivatives into the Taylor polynomial formula:\[ P_8(x) = \sum_{k=0}^8 \frac{(-1)^k k!}{k!} (x-1)^k = \sum_{k=0}^8 (-1)^k (x-1)^k. \]Which simplifies to:\[ P_8(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \ldots + (-1)^8 (x-1)^8. \]

Key concepts

Derivative

A derivative gives us the rate at which a function is changing at any point. For a more intuitive understanding, imagine driving a car: the derivative of your distance function with respect to time is your speed. In math, particularly in calculus, derivatives are fundamental for finding Taylor polynomials.

In the Taylor polynomial process, derivatives of the function at a specific point allow us to construct polynomial approximations. Derivatives are represented by "primes" or the notation \( f'(x) \), \( f''(x) \), \( f^{(3)}(x) \), and so on.

For the function given, \( f(x) = \frac{1}{x^2} \), we compute the first few derivatives. The pattern we observe in these derivatives can be generalized as \( f^{(k)}(x) = (-1)^k k! x^{-(k+2)} \). This formula is crucial for building our Taylor polynomial, expressing how the function's behavior continuously shifts.

Function Evaluation

Once derivatives are computed, evaluating them at a specific point is essential. Consider it as checking how steep a hill is if you plan to climb it from a particular spot.

In the exercise, we evaluate the derivatives at \( x = 1 \). Why 1? Because the Taylor polynomial is centered at \( c = 1 \). Evaluating at this point provides us derivatives that reflect changes in the function precisely where our polynomial is centered.

As a result, calculating \( f^{(k)}(1) \) for \( k = 0 \) to 8, provides constants that we then use in constructing our polynomial. Since every derivative at \( x=1 \) simplifies nicely to \((-1)^k \), it shows a consistent pattern that greatly simplifies the polynomial expression.

Polynomial Approximation

Polynomial approximation with Taylor polynomials allows us to represent complex functions simply and accurately near a point. Imagine replacing a sophisticated curve with a few straight lines and curves near a section of a rollercoaster.

By constructing the Taylor polynomial \( P_8(x) \), we approximate \( f(x) = \frac{1}{x^2} \) to the 8th degree near \( x=1 \). This means the polynomial will reflect the behavior of \( f(x) \) closely in this neighborhood.

• First, derivatives are computed and evaluated.
• Next, these results plug into the polynomial formula.

The final expression, \[ P_8(x) = 1 - (x-1) + (x-1)^2 - \ldots + (-1)^8(x-1)^8 \], is the approximation we are aiming for to represent the function locally.

Series Expansion

Series expansion, particularly Taylor series, enables us to express functions as an infinite sum of terms. It is like growing your vocabulary bit by bit to better express a story.

In a Taylor series, each added term comes from the function's derivatives. Each partial sum of the series, like our Taylor polynomial, approximates the function near the chosen center \( c \).

The Taylor series for \( f(x) = \frac{1}{x^2} \) evaluated and centered at \( x = 1 \), yields terms: \((-1)^k(x-1)^k\). While we've stopped at the 8th term for this exercise, the process could continue for an even better approximation. Understanding series expansion is key to analyzing functions' behavior in math and science, offering a method to work with complex functions more easily.