Problem 36 - Calculus Early Transcendentals

Question

a. Find the nth-order Taylor polynomials for the given function centered at the given point a, for \(n=0,1,\) and 2 b. Graph the Taylor polynomials and the function. $$f(x)=\sqrt[4]{x}, a=16$$

Step-by-Step Solution

Answer

Answer: The Taylor polynomials for the function \(f(x) = \sqrt[4]{x}\) centered at \(a = 16\) are: Order 0: \(P_0(x) = 2\) Order 1: \(P_1(x) = 2 + \frac{1}{32}(x - 16)\) Order 2: \(P_2(x) = 2 + \frac{1}{32}(x - 16) - \frac{3}{2048 \cdot 2}(x - 16)^2\)

Step 1: 1. Find the nth derivative of the given function

To find the n-order Taylor polynomial, we need to find the nth derivative of the function. In this case, we will find the 0th, 1st, and 2nd derivatives of the function \(f(x) =\sqrt[4]{x}\) to accommodate the values of n specified in the exercise. 0th Derivative: $$f(x)=\sqrt[4]{x}$$ 1st Derivative: $$f'(x)=\frac{1}{4}\cdot x^{-\frac{3}{4}}$$ 2nd Derivative: $$f''(x)=-\frac{3}{16} \cdot x^{-\frac{7}{4}}$$

Step 2: 2. Evaluate the nth derivatives at the given point a

Now, we will evaluate the derivatives at the given point \(a=16\): 0th Derivative at a: $$f(16)=\sqrt[4]{16}=2$$ 1st Derivative at a: $$f'(16)=\frac{1}{4}\cdot 16^{-\frac{3}{4}}=\frac{1}{32}$$ 2nd Derivative at a: $$f''(16)=-\frac{3}{16}\cdot 16^{-\frac{7}{4}}=-\frac{3}{2048}$$

Step 3: 3. Construct the nth-order Taylor polynomial formula

To construct the nth-order Taylor polynomial formula, we'll use the following general formula: $$P_n(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$$ In our case, we have: \(n=0\) Taylor Polynomial: $$P_0(x)=\frac{f(16)}{0!}(x-16)^0=2$$ \(n=1\) Taylor Polynomial: $$P_1(x)=2+\frac{1}{32}(x-16)$$ \(n=2\) Taylor Polynomial: $$P_2(x)=2+\frac{1}{32}(x-16)-\frac{3}{2048\cdot 2}(x-16)^2$$

Step 4: 4. Graph the Taylor polynomials and the function

To graph the Taylor polynomials, along with the original function, plot them on a coordinate plane, using technology or graphing software. We'll represent the function \(f(x)=\sqrt[4]{x}\) with a solid line, and the Taylor polynomials \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\) with different dashed lines. On the graph, you'll observe that as the order of the Taylor polynomials increases, they become better approximations of the original function around the given point \(a = 16\). The polynomial with \(n=2\) will be a closer approximation to the original function as compared to the polynomials with \(n=0\) and \(n=1\).

Key Concepts

Derivative of a Function

When studying Taylor polynomials, derivatives play a crucial role. The derivative of a function, which is a fundamental concept in calculus, gives us the rate at which the function's value changes with respect to change in the input value. For Taylor polynomials, we calculate the derivatives of the function at a specific point. This helps to build the polynomial approximation.
In our example, we were given the function \(f(x) = \sqrt[4]{x}\) centered at \(a = 16\). We calculated the 0th, 1st, and 2nd derivatives:

• The 0th derivative is simply the function itself: \(f(x) = \sqrt[4]{x}\).
• The 1st derivative, found using basic differentiation rules, is \(f'(x) = \frac{1}{4}x^{-\frac{3}{4}}\).
• The 2nd derivative is given by \(f''(x) = -\frac{3}{16}x^{-\frac{7}{4}}\).

This process of finding derivatives is central to forming a Taylor polynomial, as each derivative corresponds to a different term in the polynomial.

Approximation

Taylor polynomials are all about approximation. They provide a way to estimate the value of complex functions using simpler polynomial expressions. By taking derivatives at a point, we capture the behavior of the function locally around that point. For instance, in our exercise, we approximated \(f(x) = \sqrt[4]{x}\) near \(x = 16\) using Taylor polynomials of orders 0, 1, and 2.
Each Taylor polynomial adds layers of accuracy:

• The 0th-order polynomial is just a constant function equal to the function value at the center of expansion, providing a basic approximation.
• The 1st-order includes the first derivative term, which allows the approximation to "tilt," adjusting for the function's slope.
• The 2nd-order adds a curvature component, refining the approximation further.

Thus, as the order of the polynomial increases, the approximation becomes more precise around the center point \(x = 16\). This demonstrates how Taylor polynomials serve as powerful tools in simplifying complex mathematical concepts.

Center of Expansion

The center of expansion is an essential part of Taylor polynomial construction. It is the point \(a\) around which the polynomial is constructed and essentially where the function is approximated most accurately. For our example with \(f(x) = \sqrt[4]{x}\), the center of expansion is \(a = 16\).
The Taylor polynomial uses this center to calculate each derivative's value and incorporate them into the polynomial's terms. The expression \((x-a)\) in the Taylor polynomial formula reflects the distance from the center, dictating how the approximation behaves as you move away from \(x = a\).
If you were to move the center of expansion to a different value, say \(a = 4\) or \(a = 25\), the polynomial would change. The closer the x-values are to the center of expansion, the better the approximation becomes. Thus, selecting the appropriate center is key to obtaining an effective Taylor polynomial approximation.