Problem 63 - Calculus Early Transcendentals

Question

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\frac{e^{x}+e^{-x}}{2}$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series are \(f(x) = 1 + \frac{x^2}{2}+\frac{x^4}{24}+\cdots\), and the radius of convergence is infinity.

Step 1: Find the first four derivatives of the function#

We want to find the first four derivatives of the function: $$f(x) = \frac{e^{x} + e^{-x}}{2}$$ Let's start by finding f'(x), f''(x), f'''(x), and f^{(4)}(x): $$f'(x) = \frac{1}{2}(\frac{d(e^x)}{dx} + \frac{d(e^{-x})}{dx}) = \frac{e^x - e^{-x}}{2}$$ $$f''(x) = \frac{1}{2}((e^x) - (-e^{-x})) = \frac{e^x + e^{-x}}{2}$$ $$f'''(x) = \frac{1}{2}((e^x) - (-e^{-x})) = \frac{e^x - e^{-x}}{2}$$ $$f^{(4)}(x) = \frac{1}{2}((e^x) + (-e^{-x})) = \frac{e^x + e^{-x}}{2}$$

Step 2: Evaluate the derivatives at x = 0#

Now that we have the first four derivatives, we will evaluate them at x = 0 to find the coefficients for our Taylor series: $$f(0) = \frac{e^0 + e^0}{2} = 1$$ $$f'(0) = \frac{e^0 - e^0}{2} = 0$$ $$f''(0) = \frac{e^0 + e^0}{2} = 1$$ $$f'''(0) = \frac{e^0 - e^0}{2} = 0$$ $$f^{(4)}(0) = \frac{e^0 + e^0}{2} = 1$$

Step 3: Find the Taylor series using the coefficients#

Using the coefficients we found in step 2, we can write the first four nonzero terms of the Taylor series centered at 0: $$f(x) = \frac{1}{0!}x^0 + \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \cdots = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \cdots$$

Step 4: Determine the radius of convergence#

The function is given by: $$f(x) = \frac{e^x + e^{-x}}{2}$$ Since this function is the sum of two exponential functions, it converges for all x. Therefore, the radius of convergence is infinity. That means the Taylor series converges for all real x. In conclusion, the first four nonzero terms of the Taylor series centered at 0 for the given function are: $$f(x) = 1 + \frac{x^2}{2}+\frac{x^4}{24}+\cdots$$ And the radius of convergence for the series is infinity.

Key Concepts

Derivatives

When exploring Taylor series, finding derivatives is a crucial first step. A Taylor series is an infinite sum of terms that are expressed as derivatives of a function at a single point.
To build the series, you need to calculate higher-order derivatives and evaluate them at the center value, often zero.
In our case, we are focusing on the function:

• \( f(x) = \frac{e^x + e^{-x}}{2} \)
• By finding the derivatives \( f'(x), f''(x), f'''(x), \text{ and } f^{(4)}(x) \), we can assess the pattern they form.
  These derivatives give us coefficients for the Taylor series.

A simple trick with derivatives of exponential functions is observing predictable patterns:

• First derivative: Shows how the function changes.
• Second derivative: Explores the curvature of the function.
• Higher-order derivatives: Continue to capture more complex behavior.

Evaluating these at \( x = 0 \) provides the needed coefficients. These coefficients help us construct each term in the Taylor series, building up the overall approximation of the function.

Radius of Convergence

The radius of convergence determines the interval where a Taylor series is a valid approximation of the function. For many mathematical functions, especially exponentials, this radius is crucial.

This concept answers the question: "For which x values will the series converge to the function?"
Understanding this allows us to confidently use a Taylor series approximation.
For the function \( \frac{e^x + e^{-x}}{2} \), here's why the radius of convergence is infinity:

• The function is essentially a sum of two exponential functions.
• Exponential functions, \( e^x \) and \( e^{-x} \), converge for all real numbers.

Since the series is derived from such functions, it remains valid over all real x values without any restrictions. This universal convergence is a powerful property that simplifies the analysis and application of the Taylor series.

Exponential Function

The exponential function \( e^x \) is fascinating due to its unique properties. Here, the function \( f(x) = \frac{e^x + e^{-x}}{2} \) engages these properties effectively.

To start, let's look at some core features:

• Exponentials grow or decay rapidly, depending on the sign of x.
• Their derivatives are particularly simple: the derivative of \( e^x \) is \( e^x \) itself.
• This leads to precise calculations and appreciations of change.

In the context of Taylor series:

• The exponential function allows operations like differentiation and integration to align beautifully with series representations.
• Due to these propitious characteristics, exponentials make an excellent choice for constructing Taylor series.

Understanding exponential functions sets the stage for accurately approximating more complex behaviors, opening further exploration into mathematics.