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Magazine Chapter 11: Problem 20
Find a Taylor series for \(f(x)\) at c. (Do not verify that \(\left.\lim _{n \rightarrow \infty} R_{n}(x)=0 .\right)\) \(f(x)=e^{x}: \quad c=-3\)
Short Answer
Expert verified
The Taylor series for \(e^x\) at \(c = -3\) is \(\sum_{n=0}^{\infty} \frac{(x + 3)^n}{n!}\).
Step by step solution
01
Identify the Function and Center
The given function is \(f(x) = e^x\), and we need to find the Taylor series centered at \(c = -3\).
02
Recall the Taylor Series Formula
The Taylor series for a function \(f(x)\) centered at \(c\) is given by:\[T(x) = \sum_{n=0}^{fty} \frac{f^n(c)}{n!} (x - c)^n\] where \(f^n(c)\) is the \(n\)-th derivative of \(f\) evaluated at \(c\).
03
Calculate the Derivatives at Center
Since the function is \(f(x) = e^x\), all derivatives \(f^n(x) = e^x\). Evaluated at \(c = -3\), we have:\(f^n(-3) = e^{-3}\) for all \(n\).
04
Substitute Into the Taylor Series Formula
Substitute \(f^n(-3) = e^{-3}\) and \(c = -3\) into the Taylor series formula:\[T(x) = \sum_{n=0}^{fty} \frac{e^{-3}}{n!} (x + 3)^n\]
05
Simplify the Expression
The Taylor series can now be written as:\[T(x) = e^{-3} \sum_{n=0}^{fty} \frac{(x + 3)^n}{n!}\], which resembles the series expansion for \(e^{u}\) where \(u = x+3\).
06
Final Series Representation
Thus, the Taylor series of \(f(x) = e^x\) centered at \(c = -3\) is:\[T(x) = e^{-3} e^{x + 3}\] Simplifying gives \(T(x) = e^x\). However, in Taylor series form, it is \[T(x) = \sum_{n=0}^{fty} \frac{(x + 3)^n}{n!}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Function
The exponential function, denoted as \( e^x \), is a fundamental mathematical function with numerous applications in calculus and beyond. It is defined as the power series:
- \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
- \( \frac{d}{dx} e^x = e^x \)
Derivatives
Derivatives provide information about the rate of change of a function. For a smooth function \( f(x) \), its derivative \( f'(x) \) represents the function's slope at any given point. In the case of the exponential function \( f(x) = e^x \), every derivative is strikingly simple:
- \( f^n(x) = e^x \) for all \( n \)
- \( f^n(-3) = e^{-3} \)
Series Expansion
A series expansion, like the Taylor series, enables us to approximate functions near a point \( c \). The Taylor series provides a polynomial approximation to a function, allowing for easier computation and analysis. The general form is:
- \( T(x) = \sum_{n=0}^{\infty} \frac{f^n(c)}{n!} (x - c)^n \)
- Thus, \( e^x \) expanded about \( c = -3 \) is \( T(x) = e^{-3} \sum_{n=0}^{\infty} \frac{(x+3)^n}{n!} \)
Function Evaluation
Evaluating a function within its Taylor series framework means reconstructing the function using its derivatives at a specific center point. By evaluating \( e^x \) at \( c = -3 \), we employ:
- \( T(x) = e^{-3} \sum_{n=0}^{\infty} \frac{(x+3)^n}{n!} \)
- \( e^{a+b} = e^a e^b \)
