Chapter 1: Problem 17
Find the Taylor series for \(f(x)=e^{x}\) about the point \(c=3\). Then simplify the series and show how it could have been obtained directly from the series for \(f\) about \(c=0 .\)
Short Answer
Expert verified
The Taylor series for \(f(x)=e^x\) about \(c=3\) is \(e^3 \sum_{n=0}^{\infty} \frac{(x-3)^n}{n!}\).
Step by step solution
01
Define the Taylor Series Formula
The Taylor series of a function \(f(x)\) centered at \(c\) is given by: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \] where \(f^{(n)}(c)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x = c\).
02
Derivatives of the Function
For \(f(x) = e^x\), all the derivatives are equal to \(e^x\). Therefore, \(f^{(n)}(x) = e^x\) for all \(n\). Thus, \(f^{(n)}(3) = e^3\) for all \(n\).
03
Substitute into Taylor Series Formula
Substitute \(f^{(n)}(3) = e^3\) into the Taylor series formula to get: \[ f(x) = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n \] which is the Taylor series for \(f(x)=e^x\) about \(c=3\).
04
Simplify the Derived Series
Factor out \(e^3\) from the series as it is constant with respect to the summation: \[ f(x) = e^3 \sum_{n=0}^{\infty} \frac{1}{n!} (x-3)^n \] This simplifies to \(e^3 \cdot e^{(x-3)} = e^x\), since the series inside the summation \(\sum_{n=0}^{\infty} \frac{(x-3)^n}{n!}\) is the familiar Taylor series expansion of \(e^{x-3}\).
05
Connect to Series at c=0
The series for \(f(x) = e^x\) about \(c=0\) is \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\). By substituting \(x\) with \(x-3\), this directly gives \(\sum_{n=0}^{\infty} \frac{(x-3)^n}{n!}\), showing how the series centered at 3 can be derived from the series centered at 0.
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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, often denoted by \( e^x \), is a fundamental mathematical function with a variety of important properties and applications. The base \( e \) is an irrational number approximately equal to 2.71828, and its unique property makes the exponential function extremely useful in calculus and analysis. The function \( e^x \) is unique because:
- Its derivative is the same as the function itself, \( \frac{d}{dx}(e^x) = e^x \).
- Its value is positive for all real numbers \( x \).
- It grows faster than any polynomial function as \( x \) increases.
Series Expansion
Series expansion is a powerful technique used in mathematics to represent functions as an infinite sum of terms. One popular type is the Taylor series, which expresses a function as an infinite sum of derivative terms evaluated at a specific point. For many functions, finding a series expansion provides a way to approximate the function with a finite number of terms for practical computation.The Taylor series for a function \( f(x) \) about a point \( c \) is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \]This expansion allows us to understand functions by breaking them down into simpler polynomial terms, making complex functions easier to work with and analyze.
- Taylor series provide approximations of functions that are useful for calculations where exact values are hard to come by.
- Through substitution of terms, series expansions centered at different points can be derived from one another.
Function Derivatives
Function derivatives play a crucial role in the formation of a Taylor series expansion. Essentially, the derivatives of a function provide the coefficients in the Taylor series, revealing how the function changes: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \]For the exponential function \( f(x) = e^x \), all derivatives are equal to \( e^x \), making it unique and extremely regular.
- The consistency of derivatives simplifies series expansions, particularly Taylor series, because each successive term in the series has the same form.
- For the exponential function, this means each term beyond the base \( e^x \) is a scaled version of itself, facilitating simple and effective computation.