Find the remainder term Rₙ(x) for the Taylor series centered at 0 for the following functions. Find an upper bound for the magnitude of the remainder on the given interval for the given value of n. (The bound is not unique.)


ƒ(x) = eˣ; bound R₃(x), for |x| < 1

Verified step by step guidance

1. Identify the function and the center of the Taylor series: here, \( f(x) = e^x \) and the series is centered at 0 (Maclaurin series).
2. Recall the Taylor series expansion of \( e^x \) at 0: \( e^x = \sum_{k=0}^\infty \frac{x^k}{k!} \). The remainder term after \( n \) terms is given by the Lagrange form of the remainder:
3. \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \] where \( c \) is some value between 0 and \( x \).
4. Since all derivatives of \( e^x \) are \( e^x \), we have \( f^{(n+1)}(c) = e^c \). To find an upper bound for \( |R_3(x)| \) on \( |x| < 1 \), note that \( |c| < 1 \) and \( e^c \) is increasing on \( \mathbb{R} \). So the maximum value of \( e^c \) on \( |c| < 1 \) is \( e^1 = e \).
5. Therefore, the upper bound for the magnitude of the remainder term is:
6. \[ |R_3(x)| \leq \frac{e}{4!} |x|^4 \] for all \( |x| < 1 \).

Key Concepts

Taylor Series and Remainder Term

A Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point. The remainder term Rₙ(x) measures the error between the function and its nth-degree Taylor polynomial, indicating how closely the polynomial approximates the function near the center.

Lagrange Form of the Remainder

The Lagrange form expresses the remainder term Rₙ(x) as Rₙ(x) = (f⁽ⁿ⁺¹⁾(ξ) / (n+1)!) * xⁿ⁺¹ for some ξ between 0 and x. This form helps estimate the error by bounding the (n+1)th derivative on the interval, providing a practical way to find an upper bound for the remainder.

Bounding the Remainder on an Interval

To find an upper bound for |Rₙ(x)| on an interval, identify the maximum absolute value of the (n+1)th derivative within that interval. For eˣ, all derivatives are eˣ, which is maximized at the endpoint of the interval. Multiplying this maximum by |x|ⁿ⁺¹/(n+1)! gives a valid upper bound for the remainder.