Problem 18 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The Taylor polynomial is \( T_2(x) = 1 - \frac{1}{2}x^2 \). The limit is 0 because the expression simplifies to \(-\frac{1}{2}x\), which approaches 0 as \(x\) approaches 0.

Step 1: Identify the Function and Point

We are tasked with finding the Taylor polynomial of degree 2 for the function \( f(x) = \cos x \) about \( x = 0 \). The Taylor polynomial is computed using the derivatives of the function at the point \( x = 0 \).

Step 2: Compute Derivatives

Calculate the first and second derivatives of \( f(x) = \cos x \):\[ f'(x) = -\sin x \]\[ f''(x) = -\cos x \] Evaluate these at \( x = 0 \):\[ f(0) = \cos 0 = 1 \]\[ f'(0) = -\sin 0 = 0 \]\[ f''(0) = -\cos 0 = -1 \]

Step 3: Substitute into Taylor Polynomial Formula

The Taylor polynomial of degree 2 is given by:\[ T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 \]Substituting the values from Step 2:\[ T_2(x) = 1 + 0 \cdot x + \frac{-1}{2}x^2 = 1 - \frac{1}{2}x^2 \]

Step 4: Apply Taylor Polynomial to Limit Problem

To solve \( \lim_{x \to 0} \frac{\cos x - 1}{x} \), substitute the Taylor polynomial for \( \cos x \) around \( x = 0 \):\[ \lim_{x \to 0} \frac{(1 - \frac{1}{2}x^2) - 1}{x} = \lim_{x \to 0} \frac{-\frac{1}{2}x^2}{x} \]Simplify the expression:\[ \lim_{x \to 0} \frac{-\frac{1}{2}x^2}{x} = \lim_{x \to 0} -\frac{1}{2}x = 0 \]

Step 5: Conclusion

The limit evaluates to 0 because the dominant term in the numerator \(-\frac{1}{2}x^2\) simplifies and vanishes as \(x\) approaches 0. Thus:\[ \lim_{x \to 0} \frac{\cos x - 1}{x} = 0 \]

Key Concepts

Derivatives

When we talk about derivatives, we are referring to the rate at which a function changes at any point. For a function like

• the first derivative, denoted as \( f'(x) \), tells us the slope of the tangent to the curve at a given point.
• the second derivative, \( f''(x) \), provides us with information on the curvature or concavity of the function at the point.

In our exercise, we have the function \( f(x) = \cos x \). The derivatives for this function are particularly interesting:- The first derivative is \( f'(x) = -\sin x \), because the slope of the \( \cos x \) curve is at its steepest at points that coincide with peaks and valleys of the sine curve.- The second derivative is \( f''(x) = -\cos x \), indicating how the curve's bending changes as it moves along the x-axis.For the Taylor polynomial, we evaluated these derivatives at \( x = 0 \) to construct the polynomial, which will approximate the behavior of \( \cos x \) near this point.

Cosine Function

The cosine function, \( \cos x \), is a fundamental trigonometric function that describes a repeated pattern of peaks and troughs. It is periodic, with a period of \( 2\pi \), meaning its values repeat every \( 2\pi \) units.Some important properties of the cosine function include:

• Symmetry: \( \cos x \) is an even function, meaning it is symmetric about the y-axis. So, \( \cos(-x) = \cos x \).
• Maximum and minimum values: The cosine function reaches its maximum value of 1 at multiples of \( 2\pi \), while its minimum value of -1 at odd multiples of \( \pi \).
• The function is differentiable everywhere, making it convenient for constructing Taylor polynomials.

In the context of the exercise, these properties help in understanding why the approximation through the Taylor polynomial is effective around \( x=0 \). With derivatives evaluated, the polynomial emerges as \( 1 - \frac{1}{2}x^2 \), mirroring the function's behavior near zero.

Limit Evaluation

Evaluating limits involves finding the value that a function approaches as the input (\( x \) here) approaches some value. In the exercise, we look at \[ \lim_{x \to 0} \frac{\cos x - 1}{x} \]To evaluate this, we use the Taylor polynomial. We substitute \( \cos x \) with the polynomial of degree 2:\[ \cos x \approx 1 - \frac{1}{2}x^2 \]Substituting into the limit gives:\[ \lim_{x \to 0} \frac{(1 - \frac{1}{2}x^2) - 1}{x} = \lim_{x \to 0} \frac{-\frac{1}{2}x^2}{x} \]The expression simplifies to:\[ \lim_{x \to 0} -\frac{1}{2}x = 0 \]This occurs because as \( x \to 0 \), the terms in the numerator and denominator simplify, showing that the overall expression tends toward zero. The process illustrates how understanding the polynomial approximation aids in simplifying the evaluation of complex limits.