Problem 21 - 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 \(x\). $$ f(x)=\cos x, a=\frac{\pi}{6}, n=3 ; x=\frac{\pi}{7} $$

Step-by-Step Solution

Answer

Use a degree 3 Taylor polynomial centered at \( \frac{\pi}{6} \) to approximate \( \cos(\frac{\pi}{7}) \) and compare with the cosine value at that point for accuracy.

Step 1: Determine the Taylor Series Formula

The Taylor series for a function \( f(x) \) centered at \( a \) is given by:\[ P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \] For degree \( n=3 \), this formula includes up to the third derivative of \( f(x) \).

Step 2: Calculate the Function Derivatives

The derivatives of \( f(x) = \cos x \) are:- \( f(x) = \cos x \)- \( f'(x) = -\sin x \)- \( f''(x) = -\cos x \)- \( f'''(x) = \sin x \)

Step 3: Evaluate Derivatives at \( a = \frac{\pi}{6} \)

Compute the derivatives at \( a = \frac{\pi}{6} \):- \( f(a) = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)- \( f'(a) = -\sin \frac{\pi}{6} = -\frac{1}{2} \)- \( f''(a) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2} \)- \( f'''(a) = \sin \frac{\pi}{6} = \frac{1}{2} \)

Step 4: Construct the Taylor Polynomial

Substitute the values into the Taylor polynomial formula:\[P_3(x) = \frac{\sqrt{3}}{2} - \frac{1}{2}(x - \frac{\pi}{6}) - \frac{\sqrt{3}}{4}(x - \frac{\pi}{6})^2 + \frac{1}{12}(x - \frac{\pi}{6})^3\]

Step 5: Approximate \( f(x) \) At \( x = \frac{\pi}{7} \) Using the Polynomial

Substitute \( x = \frac{\pi}{7} \) into the polynomial:\[P_3(\frac{\pi}{7}) = \frac{\sqrt{3}}{2} - \frac{1}{2}(\frac{\pi}{7} - \frac{\pi}{6}) - \frac{\sqrt{3}}{4}(\frac{\pi}{7} - \frac{\pi}{6})^2 + \frac{1}{12}(\frac{\pi}{7} - \frac{\pi}{6})^3\]

Step 6: Compare with the Function Value at \( x = \frac{\pi}{7} \)

Calculate \( f(\frac{\pi}{7}) = \cos(\frac{\pi}{7}) \). Compare this value with the polynomial approximation from the previous step to determine the accuracy of the approximation.

Key Concepts

Taylor polynomial

Imagine constructing a precise map of your neighborhood. A Taylor polynomial does something similar by mapping out a function around a specific point, capturing the essence of the function in a finite and manageable form.

The Taylor polynomial of degree \( n \) is a pivotal mathematical concept to approximate functions closely at a given point \( a \). It does this by summing up terms involving the function and its derivatives, evaluated at \( a \). For a function \( f(x) \) and its derivatives, the polynomial is expressed as:

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

Each term in the summation becomes increasingly insignificant with higher powers of \( (x-a) \), which makes the Taylor polynomial a true reflection of the function near \( a \). Thus, it provides a simpler form of the function by effectively using a limited number of terms, particularly useful in computations where an infinite series is not feasible.

approximation

An approximation is like finding a middle ground when the complete picture is too complex to handle directly. With Taylor polynomials, we approximate the function to understand its behavior near a certain point better.

In our worked problem, we estimated the function \( f(x) = \cos x \) at \( x = \frac{\pi}{7} \) using a Taylor polynomial around \( a = \frac{\pi}{6} \). By using a Taylor polynomial of degree \( n=3 \), we constructed:

• \( P_3(x) = \frac{\sqrt{3}}{2} - \frac{1}{2}(x - \frac{\pi}{6}) - \frac{\sqrt{3}}{4}(x - \frac{\pi}{6})^2 + \frac{1}{12}(x - \frac{\pi}{6})^3 \)

This approximation method allows us to replace the need for calculating \( \cos \frac{\pi}{7} \) directly with a simpler polynomial evaluation. The polynomial gives us a good estimate, particularly effective since \( x \) is close to \( a \). The closer \( x \) is to \( a \), the more reliable the approximation is.

function derivatives

Think of function derivatives as the secret ingredients sprinkled into a recipe to finesse the overall flavor. Similarly, derivatives give Taylor polynomials their depth and precision by incorporating the rate of change of the function.

Derivatives of a function measure how the function value changes as \( x \) shifts infinitesimally. For the function \( f(x) = \cos x \), the derivatives needed for building the Taylor polynomial up to the third degree were:

• First derivative, \( f'(x) = -\sin x \)
• Second derivative, \( f''(x) = -\cos x \)
• Third derivative, \( f'''(x) = \sin x \)

These derivatives were evaluated at \( a = \frac{\pi}{6} \), resulting in values \( f(a) = \frac{\sqrt{3}}{2} \), \( f'(a) = -\frac{1}{2} \), among others. Each successive derivative feeds into higher-degree terms of the polynomial, adjusting the curve of the polynomial to better match \( f(x) \). By understanding these derivatives and their values, we can craft a Taylor polynomial that beautifully mimics the behavior of \( f(x) \) around \( a \).