Problem 9 - A Primer of Real Analysis

Question

Find the Taylor polynomial \(P_{9}\) of order 9 for \(f(x)=\sin (x)\) at 0. Note that this is equal to the Taylor polynomial of order 10 for \(f\) at \(0 .\) Is \(P_{9}\left(\frac{1}{2}\right)\) an overestimate or an underestimate for \(\sin \left(\frac{1}{2}\right) ?\) Find an upper bound for the error in this approximation.

Step-by-Step Solution

Answer

\(P_9(1/2)\) is an underestimate with an error less than \(2.76 \times 10^{-8}\).

Step 1: Understand the Taylor Series for Sine Function

The Taylor series expansion for a function \( f(x) \) around the point \( a \) is given by \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \). For \( f(x) = \sin(x) \), which is centered at \( a = 0 \), the series is \( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \).

Step 2: Calculate the Taylor Polynomial P_9

For the sine function, the Taylor polynomial \( P_n \) is formed by taking terms up to the \((2n+1)\)-th derivative. Hence, for order 9, we take terms for \( n=0 \) to \( n=4 \) in the series: \[ P_9(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} \].

Step 3: Evaluate P_9 at x = 1/2

Substitute \( x = \frac{1}{2} \) into the polynomial: \[ P_9\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{(\frac{1}{2})^3}{3!} + \frac{(\frac{1}{2})^5}{5!} - \frac{(\frac{1}{2})^7}{7!} + \frac{(\frac{1}{2})^9}{9!} \].After calculating, we find: \[ P_9\left(\frac{1}{2}\right) \approx 0.479425 \].

Step 4: Determine if P_9 is an Overestimate or Underestimate

The error sign depends on the next term in the series, which would be negative \((x^{11}/11!)\). Since the sine function is alternating and increasing in magnitude up to \( x = \pi/2 \), and since \( \sin(1/2) \approx 0.479426 \), \( P_9(1/2) \) is an underestimate because adding the next negative term would make it decrease further.

Step 5: Calculate an Upper Bound on Error

For alternating series with magnitude decreasing terms, the maximum error is less than the absolute value of the first omitted term. So, calculate:\[ \left|\frac{(1/2)^{11}}{11!}\right| < \frac{(1/2)^{11}}{4\times 10^7} \approx 2.76 \times 10^{-8} \].

Key Concepts

Taylor Series

The Taylor series is an essential concept in calculus, providing a way to approximate complex functions with simpler polynomials. This idea is grounded in the expansion of a function into an infinite sum of terms, each calculated from the derivatives of the function at a single point. For a function \( f(x) \) centered around the point \( a \), the Taylor series is expressed as:

• \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \).

This series effectively represents the function as the sum of its derivative values at that point. It's like building a staircase using small steps to reach the height of a tall building. By selecting more terms from the series, we can achieve a closer approximation to the function's actual behavior. In this exercise, the sine function is expressed in terms of the Taylor series around zero (also known as the Maclaurin series).What makes the Taylor series particularly useful is its ability to approximate functions that are otherwise difficult to evaluate, especially around a specific point.

Sine Function

The sine function, denoted \( \sin(x) \), is one of the fundamental trigonometric functions, typically used to describe the oscillation in waves or circular motion. When expanded into a Taylor series around zero, its simplicity manifests elegantly:

• \( \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \).

This series alternates between positive and negative values, showing how the function swings above and below the x-axis like a pendulum. Each term in the series corresponds to the odd-powered derivatives of sine at zero because even-powered derivatives are zero at the origin.In our exercise, we are interested in the Taylor polynomial \( P_9(x) \) of degree 9 for the sine function—comprising terms up to the eleventh power of x. This is derived by taking the first few terms of the series, up to the ninth derivative. The polynomial therefore approximates the sine function near zero and provides a useful way of calculating sine for small angles.

Approximation Error

In mathematical approximations such as the Taylor polynomial, the concept of approximation error is crucial to understanding the accuracy of the approximation. When using a Taylor series to approximate a function, we don't use all infinite terms, which results in an 'error' or 'remainder'. This error, known as the approximation error, is the difference between the actual function value and its Taylor polynomial approximation.In the case of a Taylor series for a sine function, an alternating series, the error (or the remainder) can be controlled. The approximation error is often estimated as being less than the magnitude of the first omitted term. Specifically, for our function \( f(x) = \sin(x) \), the approximation error when \( n = 9 \) is less than:

• \( \left|\frac{(x)^{11}}{11!}\right| \).

For \( x = 0.5 \), this results in an approximation error of less than \( 2.76 \times 10^{-8} \). Consequently, this tiny error shows that our polynomial approximation is extremely close to the actual value of \( \sin(0.5) \), making it a powerful tool when precision is required.