Problem 17 - Calculus for Biology and Medicine

Question

(a) Find the Taylor polynomial of degree 3 about \(x=0\) for \(f(x)=\sin x\). (b) Use your result in (a) to explain why $$ \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 $$

Step-by-Step Solution

Answer

(a) The Taylor polynomial of degree 3 for \( \sin x \) is \( T_3(x) = x - \frac{x^3}{6} \). (b) This shows \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \) because higher-order terms vanish as \( x \rightarrow 0 \).

Step 1: Understand the Taylor Series Formula

The Taylor series for a function \( f(x) \) centered at \( x = a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] For our exercise, \( a = 0 \) and we need the first three non-zero terms of the Taylor series for \( f(x) = \sin x \).

Step 2: Derivatives of Sine Function

Calculate the necessary derivatives of \( \sin x \): - \( f(x) = \sin x \) - \( f'(x) = \cos x \) - \( f''(x) = -\sin x \) - \( f'''(x) = -\cos x \)

Step 3: Evaluate the Derivatives at \( x = 0 \)

Substitute \( x = 0 \) into the derivatives: - \( f(0) = \sin(0) = 0 \) - \( f'(0) = \cos(0) = 1 \) - \( f''(0) = -\sin(0) = 0 \) - \( f'''(0) = -\cos(0) = -1 \)

Step 4: Construct the Taylor Polynomial

Substitute the derivative values into the Taylor polynomial formula: \[ f(x) = 0 + 1\cdot x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 + \cdots \] This simplifies to: \[ T_3(x) = x - \frac{x^3}{6} \] which is the Taylor polynomial of degree 3 for \( \sin x \) centered at \( x = 0 \).

Step 5: Use the Taylor Polynomial to Approximate \( \frac{\sin x}{x} \)

The Taylor polynomial allows us to approximate \( \sin x \) near \( x = 0 \). \[ \frac{\sin x}{x} \approx \frac{x - \frac{x^3}{6}}{x} = 1 - \frac{x^2}{6} \] As \( x \rightarrow 0 \), the term \( \frac{x^2}{6} \rightarrow 0 \).

Step 6: Explain the Limit

We conclude that as \( x \rightarrow 0 \), \[ \frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots \] This implies \[ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \].

Key Concepts

Taylor series

The Taylor series offers a powerful tool for approximating functions through polynomials. By generating a polynomial of infinite degree, a function can closely approximate its value near a specific point, referred to as the center. This series is expressed with the formula:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]where \( f(x) \) represents the function, and \( a \) is the center of the series. The key is in the derivatives: each derivative is calculated at the point \( a \).

• \( f(a) \), \( f'(a) \), \( f''(a) \), etc., are derivatives at \( x = a \).
• The factorial in the denominator normalizes the contribution of each term.

For the case of \( \sin x \) around \( x = 0 \), the series expansion generates a polynomial that matches well-known trigonometric properties. The series simplifies to a shorter polynomial for ease in calculations, especially for non-linear functions such as sine.

derivatives of trigonometric functions

Trigonometric functions like sine and cosine have smooth, cyclical derivatives, which makes them suitable for Taylor series expansions. Here's how they work:

• The sine function \( \sin x \) starts with the derivative \( f(x) = \sin x \), leading to \( f'(x) = \cos x \), \( f''(x) = -\sin x \), and \( f'''(x) = -\cos x \).
• The cosine function, \( \cos x \), cycles through \( f(x) = \cos x \), leading to \( f'(x) = -\sin x \), \( f''(x) = -\cos x \), and \( f'''(x) = \sin x \).

These derivatives reveal the repetitive nature of trigonometric functions. Applying them at \( x = 0 \), especially for sine, gives values that are frequently 0 or \( \pm1 \). In the Taylor series, these derivatives determine the coefficients of the polynomial terms, giving accurate function approximations near the center point.

limit of a function

When investigating the limit of a function as \( x \) approaches a specific value, the Taylor polynomial becomes instrumental. In cases with \( \frac{\sin x}{x} \), examining the behavior near \( x = 0 \) clarifies the limit. Using the Taylor polynomial:\[ \frac{\sin x}{x} \approx 1 - \frac{x^2}{6} \]As \( x \rightarrow 0 \), \( \frac{x^2}{6} \rightarrow 0 \). This simplifies the expression to tend towards 1. Understanding how tiny changes around \( x = 0 \) affect \( \frac{\sin x}{x} \) explains why:

• The numerator, \( \sin x \) approximates very closely to the input \( x \) when using a Taylor polynomial.
• Hence, the overall function approaches the value of 1, confirming that \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \).

This concept helps solidify the understanding of trigonometric limits, which play a significant role in calculus and differential equations.