Problem 18 - Calculus for Biology and Medicine

Question

(a) Find the Taylor polynomial of degree 2 about \(a=0\) for \(f(x)=\cos x\) (b) Use your result in (a) to give an intuitive explanation why $$ \lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0 $$

Step-by-Step Solution

Answer

The Taylor polynomial for \(\cos x\) is \(1 - \frac{x^2}{2}\); this shows \(\lim_{x \rightarrow 0} \frac{\cos x-1}{x} = 0\) because \(\cos x - 1 \approx -\frac{x^2}{2}\) near zero.

Step 1: Identify the Function and Derivatives

We begin by identifying the function we are analyzing, which is given as \(f(x) = \cos x\). To find the Taylor polynomial, we'll need the derivatives of the function at \(a = 0\). The first derivative is \(f'(x) = -\sin x\), and the second derivative is \(f''(x) = -\cos x\).

Step 2: Evaluate Function and Derivatives at \(x = 0\)

Next, we evaluate the function and its derivatives at \(a = 0\): \(f(0) = \cos(0) = 1\), \(f'(0) = -\sin(0) = 0\), and \(f''(0) = -\cos(0) = -1\).

Step 3: Construct the Degree 2 Taylor Polynomial

The Taylor polynomial of degree 2 about \(a = 0\) for function \(f(x)\) is given by: \[ P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 \]. Plugging in the evaluated derivatives, we get: \[ P_2(x) = 1 + 0 \times x + \frac{-1}{2}x^2 = 1 - \frac{x^2}{2} \].

Step 4: Analyze the Limit

Now, use the Taylor polynomial to explain the limit. Near \(x = 0\), \(\cos x \approx 1 - \frac{x^2}{2}\). Therefore, \(\cos x - 1 \approx -\frac{x^2}{2}\). Substituting this approximation into the limit, \(\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} \approx \lim_{x \rightarrow 0} \frac{-\frac{x^2}{2}}{x} = \lim_{x \rightarrow 0} -\frac{x}{2} = 0\), since \(x\) approaches 0.

Key Concepts

Limit of a Function

The limit of a function helps us understand the behavior of a function as its input approaches a certain point. In the context of the exercise, we are interested in the limit:

• \(\lim_{x \to 0} \frac{\cos x - 1}{x} = 0\)

The goal is to find out what the function value gets closer to as the input \(x\) gets closer to zero. As seen in the solution, using the Taylor polynomial simplifies the trigonometric function \(\cos x\) around \(x = 0\) to \(1 - \frac{x^2}{2}\). This simplification allows us to approximate complex expressions, making the limit easier to evaluate.
When we substitute this into the limit and simplify, we find that the value approaches zero. This intuitive analysis demonstrates how calculus provides tools for understanding function behavior in a precise manner.

Trigonometric Functions

Trigonometric functions such as \(\cos x\) are periodic functions that have unique properties. In this exercise, the function \(f(x) = \cos x\) is at the heart of the problem.

• Trigonometric functions are crucial in calculus because of their applications in modeling waves, oscillations, and more.
• The Taylor polynomial transforms these functions into polynomial expressions for simpler analysis near a point.

The trigonometric function \(\cos x\) is known for its repetitive pattern which can be effectively analyzed using derivatives. By calculating the first few derivatives at a chosen point, you gain a polynomial approximation with manageable terms like \(1 - \frac{x^2}{2}\). This is highly useful in understanding how these functions behave for small angles \(x\) (i.e., close to zero) where such approximations are most accurate.

Derivatives

Derivatives are fundamental in calculus and involve the rate at which a function changes. They are core components used to construct the Taylor polynomial, as they measure how a function behaves locally around a certain point.

• For \(f(x) = \cos x\), the first derivative \(f'(x) = -\sin x\) indicates how \(\cos x\) changes as \(x\) changes.
• The second derivative \(f''(x) = -\cos x\) shows us the rate of change of the rate of change of \(\cos x\).

Each derivative is used in the polynomial expression to reflect the local behavior around zero. For our specific exercise, evaluating these at \(x = 0\) provides coefficients for each term in our Taylor polynomial: \(1\) for the constant term, and \(-\frac{1}{2}\) for the \(x^2\) term. This polynomial offers a simplified model to observe how the original function behaves near the target point.