Problem 3 - Essential Calculus

Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\cos x, \quad c=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The Taylor series centered at \(c = \frac{\pi}{4}\) for \(f(x) = \cos x\) is \(\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) -\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^2}{2!} + \frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^3}{3!} +\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^4}{4!}-...\).

Step 1: Understand the pattern of derivatives

Find the first few derivatives of the cosine function and observe the pattern:1st derivative (f'): \(-\sin(x)\),2nd derivative (f''): \(-\cos(x)\),3rd derivative (f'''): \(\sin(x)\),4th derivative (f''''): \(\cos(x)\).The pattern then begins to repeat.

Step 2: Calculate the derivatives at \(c = \frac{\pi}{4}\)

Now, calculate the values of each derivative at \(c = \frac{\pi}{4}\) using the known values of sine and cosine at \(\frac{\pi}{4}\), which are equivalent to \(\frac{\sqrt{2}}{2}\):Derivative 0: \(f(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\),1st derivative: \(f'(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\),2nd derivative: \(f''(\frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\),3rd derivative: \(f'''(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\),4th derivative: \(f''''(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). It's clear that the pattern repeats after the 4th derivative.

Step 3: Incorporate the terms into the Taylor series

Add the terms to the general Taylor series formula, dividing by the factorial of the term's degree: \(f(x) = f(a)+f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} +...\) This would yield the series centered at \(\frac{\pi}{4}\):\(f(x) = \frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) -\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^2}{2!} + \frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^3}{3!} +\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^4}{4!}-...\)

Key Concepts

Cosine Function

The cosine function, expressed as \( \cos(x) \), is a fundamental trigonometric function. It describes the ratio of the adjacent side to the hypotenuse in a right-angled triangle. This function is periodic with a cycle of \( 2\pi \), meaning it repeats its values every \( 2\pi \) units. The graph of \( \cos(x) \) is a wave-like curve that oscillates between -1 and 1, peaking at 1 when \( x = 0 \), \( 2\pi \), etc., and dipping to -1 at \( \pi \). Understanding the cosine function is key to exploring its derivatives and making sense of the start of any Taylor series involving \( \cos(x) \).

In this exercise, you're asked to find the Taylor series expansion of \( \cos(x) \), particularly focused on the center point \( c = \frac{\pi}{4} \). This center is important because it dictates where our series approximation revolves around. The value of the cosine at this point is \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), a crucial starting point for the series.

Derivatives

Derivatives are vital in finding series expansions. They tell us how a function changes as its inputs change. Essentially, a derivative provides the slope of the tangent line to the function at any given point.

The derivatives of \( \cos(x) \) follow a predictable pattern that repeats every four steps:

• The first derivative \( f'(x) = -\sin(x) \)
• The second derivative \( f''(x) = -\cos(x) \)
• The third derivative \( f'''(x) = \sin(x) \)
• The fourth derivative \( f''''(x) = \cos(x) \)
• and back to the starting point.

The cycle nature of these derivatives becomes evident.

For any Taylor series about a specific point \( c \), the derivatives at that point are needed. For \( c = \frac{\pi}{4} \), knowing the specific values:

• \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
• \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)

becomes essential for calculating each term of the series.

Series Expansion

A Taylor series provides a polynomial approximation of a function near a specific point \( c \). The more terms included, the closer it approximates the function. For \( \cos(x) \) centered at \( \frac{\pi}{4} \), the Taylor series formula is:\[f(x) = f(c) + \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \dotsb\]Each term in this series is calculated using the function's derivatives at the point \( c \), effectively incorporating the change of \( \cos(x) \) around that point.

Using the known values and derivatives:

• Start with \( \frac{\sqrt{2}}{2} \), the cosine value at \( \pi/4 \).
• Add successive terms, such as \( -\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) \) by incorporating the first derivative with factorial adjustments.

The series helps us approximate \( \cos(x) \) behavior near \( \frac{\pi}{4} \) with a simple polynomial that conveys complex trigonometric behavior.