Problem 19 - Calculus for Biology and Medicine

Question

In Problems 19-23, compute the Taylor polynomial of degree \(n\) about a and compare the value of the approximation with the value of the function at the given point \(x\). $$ f(x)=\sqrt{x}, a=1, n=3 ; x=2 $$

Step-by-Step Solution

Answer

The Taylor polynomial approximation at $x=2$ is 1.4375, compared to $f(2) = \sqrt{2} \approx 1.4142$.

Step 1: Understanding the Taylor Polynomial

A Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(a\) is given by: \[ P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \]In this problem, we need to find the Taylor polynomial of degree 3 for \(f(x) = \sqrt{x}\) about \(a = 1\).

Step 2: Finding the Derivatives

Find the first three derivatives of \(f(x) = \sqrt{x}\):1. \(f(x) = x^{1/2}\)2. \(f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\)3. \(f''(x) = -\frac{1}{4}x^{-3/2} = -\frac{1}{4x^{3/2}}\)4. \(f'''(x) = \frac{3}{8}x^{-5/2} = \frac{3}{8x^{5/2}}\).

Step 3: Evaluating Derivatives at a=1

Evaluate these derivatives at \(a = 1\):1. \(f(1)=1\)2. \(f'(1)=\frac{1}{2}\)3. \(f''(1)=-\frac{1}{4}\)4. \(f'''(1)=\frac{3}{8}\).

Step 4: Writing the Taylor Polynomial

Using the derivatives evaluated at \(a = 1\), write the Taylor polynomial of degree 3:\[ P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2}(x-1)^2 + \frac{f'''(1)}{6}(x-1)^3 \]Substitute the values:\[ P_3(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 \].

Step 5: Compare Values at x=2

Find the approximation \(P_3(2)\):1. \(P_3(2) = 1 + \frac{1}{2}(2-1) - \frac{1}{8}(2-1)^2 + \frac{1}{16}(2-1)^3\)2. \(= 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16}\)3. Resolve the fractions: - Convert to a common denominator: \[ = \frac{16}{16} + \frac{8}{16} - \frac{2}{16} + \frac{1}{16} \] - Total = \[\frac{23}{16} = 1.4375 \].Then, compare with \(f(2) = \sqrt{2} \approx 1.4142\).

Key Concepts

Taylor series

The Taylor series is a mathematical concept used to approximate complex functions through polynomials. This technique involves expressing a function as an infinite sum of terms that are calculated from its derivatives at a specific point, usually denoted as \( a \). For a number \( x \) close to \( a \), the Taylor series provides an accurate approximation of the function

• The general form of a Taylor series centered at \( a \) for a function \( f(x) \) is given by:

\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots \]This series can be truncated to produce a Taylor polynomial of a certain degree \( n \), providing a simpler polynomial approximation. This series becomes more effective for approximation when enough terms are included. Each term in the series becomes progressively smaller and contributes less to the accuracy of the series approximation.

Application in Approximating Functions

When computing the Taylor polynomial for \( f(x) = \sqrt{x} \) about \( a = 1 \) up to degree 3, we sum the first four terms (from degree 0 to degree 3). This polynomial gives us a convenient way to approximate the function \( f(x) \) near \( a \). In real applications, Taylor series help solve problems where exact solutions are difficult or impossible to obtain by other means, offering an approximation that is almost indistinguishable from the actual value when calculated properly.

derivatives

Derivatives play a crucial role in determining the coefficients of the Taylor polynomial. The derivative of a function gives us the rate at which the function's value changes. In the context of the Taylor series, derivatives help us determine how the function behaves near a particular point.When computing a Taylor polynomial, we calculate several derivatives of the function:

• The first derivative \( f'(x) \) gives us the slope of \( f(x) \).
• Higher-order derivatives such as \( f''(x) \), \( f'''(x) \), and so on, inform us about the function's curvature and changes beyond the slope.

For our specific problem with \( f(x) = \sqrt{x} \), derivatives are:

• \( f'(x) = \frac{1}{2\sqrt{x}} \)
• \( f''(x) = -\frac{1}{4x^{3/2}} \)
• \( f'''(x) = \frac{3}{8x^{5/2}} \)

Evaluating these derivatives at \( a = 1 \) simplifies them into constants:

• \( f'(1)=\frac{1}{2} \)
• \( f''(1)=-\frac{1}{4} \)
• \( f'''(1)=\frac{3}{8} \)

These values are crucial for forming the Taylor polynomial. By multiplying these derivative terms with their respective polynomial powers, we build an accurate approximation for \( f(x) \). Each derivative contributes essential information about the function's behavior, enabling our polynomial to mimic the function closely.

approximation

Approximation involves creating a simpler representation that is close to the original value. In calculus, we use Taylor polynomials for approximating functions like \( \sqrt{x} \). A Taylor polynomial uses a finite number of terms from the Taylor series to estimate the value of the function.How does approximation work?Understanding approximation involves recognizing that:

• The aim is to create a polynomial that matches the function closely around a chosen point \( a \).
• We base our polynomial on the derivatives of the function at \( a \).
• With more terms, the polynomial becomes a better approximation of the function.

Applying this to our example, the degree 3 Taylor polynomial for \( f(x) = \sqrt{x} \) centered at \( a = 1 \) is \[ P_3(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 \]. At \( x = 2 \), this provides an approximate value of \( f(x) \).
Evaluation of the ApproximationTo gauge the approximation:

• Compute \( P_3(2) = 1.4375 \).
• Compare this with the actual \( \sqrt{2} \approx 1.4142 \).

This shows the polynomial is a good approximation since it is quite close to the actual function value. Approximation becomes important in situations where exact calculation is cumbersome. The ability to approximate functions reliably allows us to handle complex calculations in physics, engineering, and economics with greater ease.