91Ó°ÊÓ

When approximating functions with Taylor polynomials, understanding the error is critical. This error tells us how far off our polynomial is from the actual function. The error approximation uses the remainder term of the Taylor series, often denoted as \( R_n(x) \). For the Taylor series centered at \( x_0 \), the error of approximation at a point is the absolute difference between the function and the polynomial: \[ |f(x) - P_n(x)|. \]In simple terms, it's like measuring the gap between the true function and the simplified version we have created—our polynomial. For example, in this exercise, with \( x = 0.5\), finding this error involves estimating the maximum value of the third derivative \(|f^{(3)}(x)|\) on the interval and then using it in the remainder formula. The formula for the error bound in a Taylor polynomial with degree \( n \) is: \[ \text{Error Bound} = \frac{M}{(n+1)!} |x-x_0|^{(n+1)} \]where \( M \) is the maximum value of \(|f^{(3)}(x)|\) on the interval.

To form a Taylor polynomial, we need to calculate derivatives of the function up to the desired order. In our exercise, we need the first two derivatives of the function \( f(x) = e^x \cos x \). The derivatives give us the coefficients in our polynomial. Let's break it down:

• The first derivative, \( f'(x) \), uses the product rule: \( \frac{d}{dx}[e^x \cos x] = e^x(\cos x - \sin x) \).
• The second derivative, \( f''(x) \), is derived from \( f'(x) \), resulting in \( e^x(-2 \sin x) \).

Evaluating these derivatives at the point \( x_0 = 0 \) gives us the coefficients needed to build the Taylor polynomial. Plugging in \( x = 0 \) simplifies the calculations and gives us concrete numbers used in constructing \( P_2(x) \). By doing this, we essentially find a simpler function (our polynomial) that mimics the important behavior of the more complex function near a point.

Integration using a Taylor polynomial offers a neat way to estimate more complex integrals. Instead of integrating the original complex function \( f(x) = e^x \cos x \), use the simpler polynomial approximation. In this exercise, we approximate \(\int_{0}^{1} f(x) \, dx\)by calculating \(\int_{0}^{1} P_2(x) \, dx\).Here's how we do it:- Integrate \( P_2(x) = 1 + x \) over the interval \([0,1]\).- This can be broken down to:\[ \int_0^1 (1 + x) \, dx = [x + \frac{x^2}{2}]_0^1. \]- Evaluating this gives \( 1 + 0.5 = 1.5 \).This simpler integral gives an estimation of the original integral, reducing the complexity of the computation, and it's particularly useful when \( P_2(x) \) closely approximates \( f(x) \) over the interval.

The remainder term in a Taylor polynomial is essential for understanding the accuracy of approximation. This term, \( R_n(x) \), represents what is left over when the finite polynomial is subtracted from the infinite series representation of a function.For a Taylor polynomial, this remainder term gives a sense of the error bound, especially useful between points within an interval \([a, b]\). We compute the term using the derivative that's one order higher than the polynomial degree:\[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-x_0)^{(n+1)}, \]where \( c \) is some number between \( x_0 \) and \( x \). This theory allows us to bound the error our polynomial makes across an interval.In our exercise, integrating \(|R_2(x)|\) over \([0, 1]\) provides an upper error bound, offering insight into how well \( P_2 \) approximates \( f(x) \) over larger domains. It is confirming that our estimations remain within theoretical bounds.