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Taylor Series offers a way to represent a function as an infinite sum of terms. This series approximates a function using polynomials, making it easier to compute and understand functions that might initially seem complex. When approximating a function like the exponential function, the Taylor series' main power is its ability to provide accurate estimates.

To develop a Taylor polynomial for a function at a point, we sum over the derivatives of the function evaluated at that point. With each increment, the series becomes more precise. The Taylor polynomial of order \( n \), denoted as \( P_n(x) \), around the point \( x_0 \) is given by:

• \( P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k \)

For the function \( f(x) = e^x \) around \( x_0 = 0 \), each derivative \( f^{(k)}(0) \) equals \( e^0 = 1 \), resulting in the series:

• \( P_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!} \)

The Lagrange Error Bound provides a way to understand how good a polynomial approximation is. It gives a maximum amount that the Taylor polynomial \( P_n(x) \) is off by when approximating a function \( f(x) \). Knowing this helps us determine the necessary degree \( n \) of a Taylor polynomial to achieve a desired level of accuracy.

This error is calculated as follows:

• \( |R_n(x)| \le \frac{M |x - x_0|^{n+1}}{(n+1)!} \)

"\( M \)" represents the largest value of the \( (n+1) \)-th derivative of \( f(x) \) on the interval. For \( e^x \), every derivative is \( e^x \), so for an interval \( [0, 0.5] \), \( M \) is \( e^{0.5} \).

To ensure the error does not exceed \( 10^{-6} \), we set the right side of the error bound formula to \( 10^{-6} \) and solve for \( n \). When solving this numerically, we find that \( n = 5 \) meets the criteria.

The exponential function \( e^x \) is one of the fundamental functions in calculus. It describes continuous growth and is defined with the base \( e \), an irrational number approximately equal to 2.71828. Understanding \( e^x \) is crucial because of its unique properties, like having the same derivative as the function itself:

• \( \frac{d}{dx}e^x = e^x \)

This property simplifies Taylor series calculations, as we see with the example:

• The Taylor polynomial for \( e^x \) is formed by taking derivatives that remain \( e^x \).
• When evaluated at \( x_0 = 0 \), these derivatives become 1, streamlining the process.
• The series becomes \( P_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!} \).

Recognizing the simplicity of calculating derivatives helps in applying the Taylor series efficiently to tackle many calculus problems involving exponential growth.