Problem 19 - Calculus for Biology and Medicine

Question

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

Taylor polynomial approximation: 1.4375; Actual value: approximately 1.414.

Step 1: Determine the function and derivatives

We first define the function and its derivatives. The function is given as \(f(x)=\sqrt{x}\). We'll need the first three derivatives since we are finding a third-degree Taylor polynomial. Calculate the derivatives:\[ f(x) = x^{1/2}, \quad f'(x) = \frac{1}{2}x^{-1/2}, \quad f''(x) = -\frac{1}{4}x^{-3/2}, \quad f'''(x) = \frac{3}{8}x^{-5/2} \]

Step 2: Evaluate derivatives at the point a

Next, we evaluate each derivative at the point \(a = 1\):\[ f(1) = 1^{1/2} = 1 \] \[ f'(1) = \frac{1}{2} \times 1^{-1/2} = \frac{1}{2} \] \[ f''(1) = -\frac{1}{4} \times 1^{-3/2} = -\frac{1}{4} \] \[ f'''(1) = \frac{3}{8} \times 1^{-5/2} = \frac{3}{8} \]

Step 3: Write the Taylor polynomial

The Taylor polynomial of degree 3 around \(a = 1\) for the function \(f(x)\) is:\[ P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 \] Substitute the evaluated derivatives:\[ P_3(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{4}\frac{1}{2}(x-1)^2 + \frac{3}{8}\frac{1}{6}(x-1)^3 \] Simplify:\[ P_3(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 \]

Step 4: Calculate the Taylor polynomial at x = 2

Substitute \(x = 2\) into the polynomial \(P_3(x)\):\[ P_3(2) = 1 + \frac{1}{2}(2-1) - \frac{1}{8}(2-1)^2 + \frac{1}{16}(2-1)^3 \] Which simplifies to:\[ P_3(2) = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} = 1 + 0.5 - 0.125 + 0.0625 = 1.4375 \]

Step 5: Compute the actual function value at x = 2

Calculate the actual value of the function \( f(x) \) at \( x = 2 \):\[ f(2) = \sqrt{2} \approx 1.414 \]

Step 6: Compare the approximation with actual value

Now compare the Taylor polynomial approximation with the actual function value. We found:\( P_3(2) = 1.4375 \) and that the actual \( f(2) \approx 1.414 \). The Taylor polynomial approximation is slightly larger than the actual function value.

Key Concepts

Taylor series

The Taylor series is a crucial mathematical concept used to approximate complex functions with a polynomial. It's an infinite sum of terms calculated from the values of a function's derivatives at a single point. This expansion allows us to estimate the value of a function near that point using a finite number of terms. The general formula for the Taylor series of a function \(f(x)\) around a center \(a\) is:

• \( P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \)

Here, \(n!\) denotes the factorial of \(n\), which is the product of all positive integers less than or equal to \(n\). The more terms we include, the more accurate the approximation becomes. When \(n\) approaches infinity, if the series converges, it can exactly represent the function within its radius of convergence. In our problem, the Taylor polynomial is used up to the third degree to approximate the square root function about the point \(a = 1\).

derivative evaluation

Derivative evaluation is an important step in constructing the Taylor polynomial. It involves calculating the derivatives of the function up to the desired degree and then evaluating these derivatives at a specific point, often called the center. This is how we determine the coefficients of the polynomial terms.

• The function: \(f(x) = \sqrt{x} = x^{1/2}\)
• First derivative: \(f'(x) = \frac{1}{2}x^{-1/2}\)
• Second derivative: \(f''(x) = -\frac{1}{4}x^{-3/2}\)
• Third derivative: \(f'''(x) = \frac{3}{8}x^{-5/2}\)

By evaluating these derivatives at \(a = 1\), we simplify the polynomial's coefficients:

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

These values are essential for building the Taylor polynomial up to the desired degree.

error approximation

Error approximation gives us insight into how well our polynomial predicts the function's behavior. For a Taylor polynomial, the error is linked to the next term in its Taylor series expansion that wasn't included. In simpler terms, it shows how accurate our approximation might be by considering the influence of the highest degree term omitted.The remainder or error \(R_n(x)\) of a Taylor polynomial can be evaluated as:

• \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \)

Where \(c\) is some value between \(a\) and \(x\). In our example, since we used a third-degree polynomial, the error would be related to the fourth derivative term, though it is typically small when \(x\) is close to \(a\). For the given problem, it showed that the approximation \( P_3(2) = 1.4375\) was close to \(\sqrt{2} \approx 1.414\). The error is approximately 0.0235, a modest error factor near the center point.

mathematical functions

Mathematical functions are expressions or rules that describe the exact relationship between variables. Functions can be represented in multiple forms such as equations, graphs, or even through real-world phenomena translations. In mathematics, functions like \(f(x) = \sqrt{x}\) can sometimes be cumbersome to compute directly for various values.Here is where approximation comes in handy. Approximating these functions using simpler algebraic expressions, like polynomials, allows for easier computation and understanding:

• Approximations provide an estimate rather than an exact value.
• The complexity of operations like division or square roots can be reduced.
• Polynomials, due to their simple arithmetic operations, are computationally less intensive.

In essence, formulating a Taylor polynomial helps us tackle difficult calculations by aligning them to easier ones while preserving high accuracy. Functions in mathematics are more than equations; they're gateways to modeling any form of complex relations in simpler terms.