Problem 42 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Taylor series centered at 0 for the given finction. b. Use the first four terms of the series to approximate the given quantity. $$f(x)=(1+x)^{-3} ; \text {approximate } 1 / 1.331=1 / 1.1^{3}$$

Step-by-Step Solution

Answer

Question: Use the first four nonzero terms of the Taylor series for the function $$f(x) = (1 + x)^{-3}$$ to approximate $$\frac{1}{1.331}$$. Answer: The approximation using the first four nonzero terms of the Taylor series is $$\boxed{0.752}$$.

Step 1: Find the Taylor series expansion of the function

To find the Taylor series expansion, we need to find the nth derivative of the function and evaluate it at 0. The general term of a Taylor series expansion is given by: $$T_n(x) = \frac{f^{(n)}(0)}{n!} x^n$$ We will now find the first four terms of the Taylor series expansion for the function $$f(x) = (1+x)^{-3}$$.

Step 2: Find the nth derivative of the function

To find the nth derivative, we can use the chain rule iteratively. Since our function is of the form $$(1+x)^n$$, we can write the general formula for its nth derivative as: $$f^{(n)}(x) = (-1)^n \cdot n! \cdot (1+x)^{-3-n}$$

Step 3: Evaluate the nth derivative at 0

To find the Taylor expansion centered at 0, we need to evaluate the nth derivative at 0. This yields: $$f^{(n)}(0) = (-1)^n \cdot n! \cdot (1+0)^{-3-n} = (-1)^n \cdot n!$$

Step 4: Find the general term of the Taylor series

Now, we can substitute the value of the nth derivative at 0 into our general term formula: $$T_n(x) = \frac{(-1)^n \cdot n!}{n!} x^n = (-1)^n x^n$$

Step 5: Find the first four nonzero terms of the Taylor series

By substituting the values of n, we can find the first four nonzero terms of the Taylor series: - \(n=1\): \((-1)^1 x^1 = -x\) - \(n=2\): \((-1)^2 x^2 = x^2\) - \(n=3\): \((-1)^3 x^3 = -x^3\) - \(n=4\): \((-1)^4 x^4 = x^4\) So the first four nonzero terms of the Taylor series are $$-x + x^2 - x^3 + x^4$$.

Step 6: Use the Taylor series approximation to estimate the given quantity

We have to approximate $$\frac{1}{1.331}$$ using the first four terms of the Taylor series. We can rewrite the given expression as: $$\frac{1}{1.331} = \frac{1}{1.1^3}$$ So, let $$x = 0.1$$ in our Taylor series approximation: $$f(0.1) \approx -0.1 + (0.1)^2 - (0.1)^3 + (0.1)^4$$ Evaluating this expression, we get: $$f(0.1) \approx -0.1 + 0.01 - 0.001 + 0.0001 = 0.752 \approx 0.752$$ Therefore, our approximation for $$\frac{1}{1.331}$$ using the first four terms of the Taylor series is $$\boxed{0.752}$$.

Key Concepts

nth Derivative

In calculus, the nth derivative of a function represents the derivative taken repeatedly n times. A derivative describes the rate at which a function changes at a given value. When you're looking for higher derivatives, you're essentially studying the pattern of how that rate changes. For Taylor series expansions, finding the nth derivative at a specific point—usually at 0 or another convenient point—is crucial. This is because Taylor series are crafted around points where these derivatives are evaluated.
To better grasp the concept, let's consider the function \( f(x) = (1+x)^{-3} \). Using the power of calculus, particularly the chain rule, you can find any nth derivative. For this function, the nth derivative formula is given by \( f^{(n)}(x) = (-1)^n \cdot n! \cdot (1+x)^{-3-n} \). When you compute it at \( x = 0 \), it simplifies to \( (-1)^n \cdot n! \), making it easier to plug directly into Taylor series expansions.

Taylor Series Expansion

The Taylor series expansion is a method for approximating functions using an infinite sum of terms calculated from the values of a function's derivatives at a single point. This series provides a polynomial representation of a complex function, which makes it easier to analyze and solve problems involving the function.
The general formula for the Taylor series expansion around the point \( x = 0 \) (also called Maclaurin series) is:

• \( T_n(x) = \frac{f^{(n)}(0)}{n!} x^n \)

In our context, for \( f(x) = (1+x)^{-3} \), we computed the Taylor series for the first four non-zero terms. Each term \( (-1)^n x^n \) arises as we insert the function's derivative values into the formula. These terms converge towards the actual function value, offering an efficient way of approximation.
Understanding Taylor series is critical because they not only facilitate easier computation but also provide deep insights into the function behavior near a given point.

Function Approximation

Function approximation is an invaluable tool in mathematics. It allows us to estimate the value of a function close to a point using its Taylor series. \(\frac{1}{1.331}\) is such a quantity that appears tricky to calculate directly. But turning it into \( \frac{1}{1.1^3} \) allows for an elegant simplification using a Taylor series.
By substituting \( x = 0.1 \) into the obtained Taylor series \(-x + x^2 - x^3 + x^4\), we can estimate the function value around this point. This estimation becomes incredibly handy when calculating complicated expressions with ease and speed.

• The first four terms of the Taylor series give us: \(-0.1 + 0.01 - 0.001 + 0.0001 = 0.752\).

Despite the series being technically infinite, just a few terms often yield a good approximation, demonstrating the power and efficiency of Taylor series in practical applications.