Chapter 9: Problem 17
Use sigma notation to write the Taylor series about \(x=x_{0}\) for the function. $$ \ln x ; x_{0}=1 $$
Short Answer
Expert verified
\( \ln x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(x-1)^n}{n} \)
Step by step solution
01
Identify the function and expansion point
The function given is \( \ln x \) and the expansion point is \( x_0 = 1 \). We need to write its Taylor series about this point.
02
Recall the general Taylor series formula
The Taylor series of a function \( f(x) \) about a point \( x = x_0 \) is given by:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\]Our task is to express \( \ln x \) using this formula.
03
Compute derivatives of the function
Let's calculate the first few derivatives of the function \( f(x) = \ln x \):- \( f(x) = \ln x \) - First derivative: \( f'(x) = \frac{1}{x} \) - Second derivative: \( f''(x) = -\frac{1}{x^2} \) - Third derivative: \( f'''(x) = \frac{2}{x^3} \) Continue this pattern to obtain a general expression for the \( n^{\text{th}} \) derivative, \( f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{x^n} \).
04
Calculate derivatives at the expansion point
Evaluate each derivative at \( x = 1 \):- \( f(x_0) = \ln(1) = 0 \)- \( f'(x_0) = \frac{1}{1} = 1 \)- \( f''(x_0) = -\frac{1}{1^2} = -1 \)- \( f'''(x_0) = \frac{2}{1^3} = 2 \)In general, \( f^{(n)}(1) = (-1)^{n-1} (n-1)! \).
05
Write the series in sigma notation
Substitute the calculated derivatives into the Taylor series formula:\[\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} (n-1)!}{n!} (x-1)^n\]Recognizing that \( \frac{(n-1)!}{n!} = \frac{1}{n} \), we can simplify to:\[\ln x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(x-1)^n}{n}\]
06
Final expression
Thus, the Taylor series for \( \ln x \) about \( x = 1 \) is:\[\ln x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(x-1)^n}{n}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sigma Notation
Sigma notation is a powerful mathematical tool used to represent the summation of a sequence of terms. It uses the Greek letter sigma (\(\Sigma\)) to indicate that you are summing over a series. This is particularly helpful in expressing complex sums in a concise form. For example, when writing Taylor series, sigma notation allows us to capture the infinite sum of terms generated by the formula.
Here’s a basic introduction to how sigma works:
Here’s a basic introduction to how sigma works:
- The expression inside the sigma signifies the general form of the term in the series.
- The limits of summation, often written as numbers below and above the sigma, define the range of terms to sum.
- For instance, \(\sum_{n=1}^{\infty} \), shows that you are summing starting from \(n = 1\) to infinity.
Derivatives
Derivatives in calculus measure how a function changes as its input changes. They form the backbone for Taylor series, where derivatives at a specific point help approximate the function nearby.
The process of finding a derivative is called differentiation, which gives us the slope of the function at any point. For the Taylor series of \(\ln x\), we begin by differentiating to find successive derivatives:
The process of finding a derivative is called differentiation, which gives us the slope of the function at any point. For the Taylor series of \(\ln x\), we begin by differentiating to find successive derivatives:
- The first derivative of \(\ln x\) is \(f'(x)= \frac{1}{x}\).
- The second derivative becomes \(f''(x) = -\frac{1}{x^2}\).
- Further differentiation continues in a similar pattern.
Expansion Point
The expansion point in a Taylor series is the value around which we are expanding the function. It serves as the center of the approximation. In our example, the expansion point, \(x_0 = 1\), is essential since all the derivatives of the function \(\ln x\) are evaluated at this point.
Choosing the right expansion point can simplify calculations by:
Choosing the right expansion point can simplify calculations by:
- Making derivative evaluation easy due to the nature of the point, like zeroing certain terms.
- Focusing approximation accuracy around values where the expansion point is close to the values you're interested in.