Question

a.Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b.Determine the radius of convergence of the series. $$f(x)=\cos 2 x+2 \sin x$$.

Short answer

Question: Find the first four nonzero terms of the Taylor series centered at 0 for the function $$f(x)=\cos 2x+2\sin x$$ and determine the radius of convergence. Answer: The first four nonzero terms of the Taylor series are $$f(x) \approx 1 + 2x - 4x^2 - \frac{2x^3}{3} + 4x^4 + \frac{2x^5}{15}$$, and the radius of convergence is infinite ($$R = \infty$$).

Step-by-step solution

1. Expand the Taylor series for Cosine Function

Recall the general Taylor series expansion for the cosine function: $$\cos (x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ Substitute \(2x\) into the above formula: $$\cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n x^{2n}}{(2n)!}$$

2. Expand the Taylor series for Sine Function

Recall the general Taylor series expansion for the sine function: $$\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$ Substitute \(x\) with itself and apply the constant 2: $$2\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 2x^{2n+1}}{(2n+1)!}$$

3. Combine the expansions

Add the two expansions obtained in steps 1 and 2 to find the Taylor series of the given function: $$\cos(2x) + 2\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n 4^n x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{(-1)^n 2x^{2n+1}}{(2n+1)!}$$ Write the sums as a single sum: $$f(x) = \sum_{n=0}^{\infty} \left( \frac{(-1)^n 4^n x^{2n}}{(2n)!} + \frac{(-1)^n 2x^{2n+1}}{(2n+1)!} \right)$$

4. Determine the first four nonzero terms

We will write out the first four nonzero terms of the Taylor series expansion: 1. For \(n = 0\): $$\frac{(-1)^0 4^0 x^{2\cdot 0}}{(2\cdot 0)!} + \frac{(-1)^0 2x^{2\cdot 0+1}}{(2\cdot 0+1)!} = 1 + 2x$$ 2. For \(n = 1\): $$\frac{(-1)^1 4^1 x^{2\cdot 1}}{(2\cdot 1)!} + \frac{(-1)^1 2x^{2\cdot 1+1}}{(2\cdot 1+1)!} = -4x^2 - \frac{2x^3}{3}$$ 3. For \(n = 2\): $$\frac{(-1)^2 4^2 x^{2\cdot 2}}{(2\cdot 2)!} + \frac{(-1)^2 2x^{2\cdot 2+1}}{(2\cdot 2+1)!} = 4x^4 + \frac{2x^5}{15}$$ The first four nonzero terms of the Taylor series are: $$f(x) \approx 1 + 2x - 4x^2 - \frac{2x^3}{3} + 4x^4 + \frac{2x^5}{15}$$

5. Determine the radius of convergence

For both trigonometric functions, the radius of convergence is infinite, since Taylor series expansions for sine and cosine functions converge for all real values of x. This means that their sum also has an infinite radius of convergence: $$R = \infty$$

Summary

We found the first four nonzero terms of the Taylor series centered at 0 for the given function: $$f(x) \approx 1 + 2x - 4x^2 - \frac{2x^3}{3} + 4x^4 + \frac{2x^5}{15}$$ The radius of convergence of the series is infinite: $$R = \infty$$

Key concepts

Radius of Convergence

When dealing with Taylor series, understanding the radius of convergence is crucial. The radius of convergence tells us over what range of values the series converges to the function it's representing. For the series to be useful, it must converge over the interval that interests us.

In the case of the trigonometric functions like sine and cosine, their Taylor series are unique. They have the interesting property of converging for all values of x. This means that their radius of convergence is infinite, symbolized as \( R = \infty \).

Why is this important? When we sum the series for \( \cos(2x) \) and \( 2\sin(x) \) to create the Taylor series for \( f(x) = \cos(2x) + 2\sin(x) \), the resulting series also has an infinite radius of convergence.

• This means the series is valid for all real numbers.
• No matter what value of x you choose, the series will converge to the function's true value.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in understanding periodic phenomena. Their repetitive patterns make them perfect candidates for expansion using Taylor series.

The Taylor series provides a way to approximate these functions as an infinitely long polynomial. For instance:

• \( \cos(x) \) series is \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \) which includes terms like 1, \(-x^2/2!\), and \(x^4/4!\).
• \( \sin(x) \) series is \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \) with terms like \( x \), \(-x^3/3!\), and \( x^5/5! \).

These expansions are particularly useful because they allow us to compute sine and cosine to any desired level of accuracy by controlling the number of terms in the series. For the function \( f(x) = \cos(2x) + 2\sin(x) \), the expansions leverage the Taylor series of each individual trigonometric function. We perform algebraic manipulations to find the series that best represents our desired function, cementing the power of analyzing periodic behavior with simple polynomials.

Analytical Methods

Analytical methods in calculus allow us to break down complex functions into simpler components. This process helps us understand functions deeply and solve problems across mathematics and engineering.

When deriving the Taylor series for a complex function like \( f(x) = \cos(2x) + 2\sin(x) \), we use analytical methods to combine the known series expansions of \( \cos(x) \) and \( \sin(x) \). These methods involve:

• Substitution: Replacing variables like \( x \) with \( 2x \) to adapt the series to our needs.
• Summation: Adding individual series to form a new, unified expansion for the entire function.

By understanding the individual series, we piece together the overall series representation.

This structured approach provides us with a coherent way to handle the complexities of trigonometric combinations. We end up with an expression like \( f(x) \approx 1 + 2x - 4x^2 - \frac{2x^3}{3} + 4x^4 + \frac{2x^5}{15}\), highlighting the practical power of analytical methods in calculus.