Problem 26 - Calculus Early Transcendentals

Question

a. Differentiate the Taylor series about 0 for the following functions. b. Identify the function represented by the differentiated series. c. Give the interval of convergence of the power series for the derivative. $$f(x)=\cos x$$

Step-by-Step Solution

Answer

Answer: The function represented by the differentiated series is \(-\sin x\), and its interval of convergence is \((-\infty, \infty)\).

Step 1: Recall the Taylor series formula for a function f at x = 0 (Maclaurin series)

The Taylor series formula for a function f at x = 0 is given by: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$ Step 2: Find the Taylor series for \(f(x) = \cos x\)

Step 2: Write down the derivatives and their values at x = 0

To find the Taylor series for \(f(x) = \cos x\), we need to find the derivatives of f and their values at x = 0. The first few derivatives and their values at x=0 are: 1. \(f(x) = \cos x \Rightarrow f(0) = 1\) 2. \(f'(x) = -\sin x \Rightarrow f'(0) = 0\) 3. \(f''(x) = -\cos x \Rightarrow f''(0) = -1\) 4. \(f'''(x) = \sin x \Rightarrow f'''(0) = 0\) Notice a pattern here: the even derivatives are either cosine or negative cosine while odd derivatives are sine or negative sine. At x = 0, cosine evaluates to 1, and sine evaluates to 0. This pattern continues. Step 3: Write down the Taylor series of \(f(x) = \cos x\)

Step 3: Write down the Taylor series using the derivatives and their values

Using the derivatives and their values, we can now write the Taylor series of \(f(x) = \cos x\): $$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ Step 4: Differentiate the Taylor series

Step 4: Differentiate the Taylor series to find the Taylor series of the derivative

Now, we need to differentiate the Taylor series: $$\frac{d}{dx} \cos x = \frac{d}{dx} \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) = -\frac{2x}{2!} + \frac{4x^3}{4!} - \frac{6x^5}{6!} + \cdots$$ Step 5: Identify the function represented by the differentiated series

Step 5: Compare the differentiated series to the sine function

Comparing the differentiated series with the sine function, we can see that: $$-\frac{d}{dx} \cos x = -\sin x = -\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = -x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots$$ Step 6: Interval of convergence

Step 6: State the interval of convergence for the derivative

The interval of convergence for the power series of both the cosine and sine functions is \((-\infty, \infty)\). Since our derived series represents the function \(-\sin{x}\), the interval of convergence remains \((-\infty, \infty)\).

Key Concepts

Maclaurin series

The Maclaurin series is a special case of the Taylor series centered at zero. It is used to express a function as an infinite sum of terms calculated from the values of its derivatives at zero. This technique is particularly useful for approximating functions that might be difficult to calculate directly. For any function \( f(x) \), the Maclaurin series is given by:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \)

In our context, the function \( f(x) = \cos x \) can be expressed using its Maclaurin series:

• The derivatives \( f'(x), f''(x), \dots \) are evaluated at \( x = 0 \).
• This yields the series \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \)

This series allows us to approximate \( \cos(x) \) for values of \( x \) close to zero with increasing accuracy as more terms are included. It is a powerful tool in calculus that enables the study and application of functions in a simpler algebraic form.

Power series convergence

Understanding the convergence of a power series is crucial. It tells us the values of \( x \) for which the series sums to the actual function. In mathematical terms, a power series converges if the infinite sum approaches a finite limit. The radius of convergence is a measure that helps to determine the interval around zero where the series converges.

• For the cosine function \( \cos x \), the power series is convergent for all real \( x \).
• This means the radius of convergence is infinite, indicated as \(( -\infty, \infty )\).

This property reflects the fact that trigonometric functions like sine and cosine are entire functions, meaning they are defined and differentiable at all complex numbers, not just real numbers. Hence, their power series are valid across the entire real line. Understanding convergence is key in applying these series accurately in real-world problems or computational methods.

Differentiation of series

Differentiating a power series term by term is quite straightforward and follows the same rules as regular differentiation. When differentiating a series, each term of the series is differentiated individually, which yields a series that represents the derivative of the original function.

• For \( \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} \), the derivative is: \( \frac{d}{dx}\left(\cos x\right) = \sum_{n=1}^{\infty} \frac{d}{dx}\left(\frac{(-1)^n}{(2n)!} x^{2n}\right) \)
• Upon differentiation, this results in: \(-\sin x = -\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)\)

The differentiation process shifts the structure of the series slightly. Instead of even powers, sin terms are in odd powers. This differentiation process also preserves the interval of convergence, remaining \((-\infty, \infty)\) in this case. Differentiation of power series is an essential method that helps in finding derivatives of complex functions and performing more advanced calculus manipulations.