Chapter 9: Problem 11
Use sigma notation to write the Taylor series about \(x=x_{0}\) for the function. $$ e^{x} ; x_{0}=1 $$
Short Answer
Expert verified
\( e \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!} \)
Step by step solution
01
Understand the Taylor Series
The Taylor series of a function \( f(x) \) around a point \( x_0 \) is given by:\[f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots\]Or in sigma notation:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\]To write the Taylor series for \( e^x \) about \( x=1 \), we need the derivatives of \( e^x \).
02
Compute Derivatives at \(x_0=1\)
The function \( e^x \) has a simple derivative pattern: its derivative is \( e^x \) itself. Therefore, all derivatives at any point are \( e^x \). At \( x_0 = 1 \), we have:\[f(x_0) = e^1 = e \]Thus, every derivative evaluated at \( x_0 = 1 \) yields \( e \).
03
Write the Taylor Series
Now substitute the derivatives into the Taylor series formula centered at \( x_0 = 1 \):\[\sum_{n=0}^{\infty} \frac{e}{n!}(x-1)^n\]This is the sigma notation for the Taylor series of \( e^x \) around \( x=1 \).
04
Simplify the Expression
Since \( e \) is constant, it can be factored out of the summation:\[e \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!}\]This represents the Taylor series in sigma notation, clearly indicating the structure of the series.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sigma Notation
Sigma notation, represented by the Greek letter \( \Sigma \), is a way to concisely express the sum of a series of terms. This notation is particularly useful for representing series where there is a clear repetition in the formation of terms, such as with Taylor series.
When using sigma notation, the general format is:
When using sigma notation, the general format is:
- \( \sum_{n=a}^{b} \text{expression} \)
- "\(n\)" is the index of summation, which takes on integer values starting from \(a\) and ending at \(b\).
- The expression is what is being summed, often involving \(n\).
Derivatives
When dealing with Taylor series, derivatives play a crucial role. They determine the coefficients in the series expansion. The \(n\)-th derivative of a function \(f\), evaluated at a specific point \(x_0\), determines how much that term \((x-x_0)^n\) contributes to the overall function approximation.
The beautiful aspect of the exponential function \(e^x\) is its simplicity:
The beautiful aspect of the exponential function \(e^x\) is its simplicity:
- The derivative of \(e^x\) with respect to \(x\) is \(e^x\).
- Consequently, the higher-order derivatives \(f'(x), f''(x),\ldots\) are also \(e^{x}\).
Exponential Function
The exponential function, denoted as \( e^x \), is a fundamental function in mathematics. It has unique properties that make it crucial in various fields, such as calculus, differential equations, and complex analysis. Some of its remarkable characteristics include:
- It is the only function whose derivative is the same as the function itself, \( \frac{d}{dx}e^x = e^x \).
- The function is always positive, \( e^x > 0 \) for all \( x \).
- As \( x \to \infty \), \( e^x \to \infty \), and as \( x \to -\infty \), \( e^x \to 0 \).