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Magazine Chapter 11: Problem 29
Find the Taylor series [Eq. (16)] of the given function at the indicated point \(a\). \(f(x)=\ln (1+x), \quad a=0\)
Short Answer
Expert verified
The Taylor series for \(\ln(1+x)\) at \(a=0\) is: \(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\).
Step by step solution
01
Understand the Taylor Series Formula
The Taylor series for a function \(f(x)\) about a point \(a\) is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]We need to find the series at \(a = 0\) for the function \(f(x) = \ln(1+x)\). Thus, \(x-a\) simplifies to \(x-0 = x\).
02
Evaluate the Function and Derivatives at \(a=0\)
Evaluate the function and its derivatives at \(a=0\):- \(f(x) = \ln(1+x)\) implies \(f(0) = \ln(1+0) = 0\).- First derivative: \(f'(x) = \frac{1}{1+x}\) gives \(f'(0) = 1\).- Second derivative: \(f''(x) = -\frac{1}{(1+x)^2}\) gives \(f''(0) = -1\).- Third derivative: \(f'''(x) = \frac{2}{(1+x)^3}\) gives \(f'''(0) = 2\).- Continue finding higher-order derivatives similarly.
03
Substitute into the Taylor Series Formula
Using the evaluations from Step 2, substitute into the Taylor series formula:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]This becomes:\[ \ln(1+x) = 0 + 1x + \frac{-1}{2!}x^2 + \frac{2}{3!}x^3 + \cdots \]
04
Simplify the Taylor Series
Simplify the expression:- \(f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \)This can be written as the infinite series:\[ \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \]
05
Conclusion: Taylor Series for \(f(x) = \ln(1+x)\)
The Taylor series of \(f(x) = \ln(1+x)\) about \(a=0\) is:\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]This series is valid for \(|x| < 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
A derivative tells us how a function changes as its input changes. Think of it as the function's rate of change or its slope at a certain point. In the context of the Taylor series, we need to find multiple derivatives of the function at a specific point. Here, we looked at the first, second, third, and higher derivatives of the function \( f(x) = \ln(1+x) \) evaluated at \( a=0 \).
- The first derivative, \( f'(x) = \frac{1}{1+x} \), gives us how sharply the logarithmic function ascends or descends from \( a = 0 \).
- The second derivative, \( f''(x) = -\frac{1}{(1+x)^2} \), tells us about the curvature of the logarithm function's graph.
- Higher-order derivatives continue to inform us about the more subtle changes in the function's graph.
Function Series
A function series is a sum of functions that represents another function. Taylor series is a type of function series that approximates functions using polynomial expressions. It allows complex functions like \( \ln(1+x) \) to be represented using infinite polynomial terms. By finding each term's derivative and substituting them into the Taylor formula, we approximate \( \ln(1+x) \) with an infinite series.
- The first term is the value of the function at point \( a \).
- The second and subsequent terms integrate the derivatives found at \( a =0 \), getting multiplied by their respective powers of \( x \).
- Each term has coefficients derived from factorial calculations, like \( \, \frac{f''(0)}{2!} \) or \( \, \frac{2}{3!} \) in our given function.
ln(1+x)
\( \ln(1+x) \) is a logarithmic function and one of the classic examples to illustrate Taylor series. The natural logarithm function has a unique property: it converges absolutely for \(|x| < 1\), allowing a Taylor series representation that provides a precise approximation around the point \( a=0 \).
- At \( x = 0 \), the value of \( \ln(1+x) \) is zero, which becomes the starting point of the series.
- Each term in the Taylor series brings more accuracy beyond \( x=0 \).
- This series can be extended as much as needed, giving the essence of a function's behavior in a neighborhood around zero.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. For functions like \( \ln(1+x) \), representing it with an infinite series allows for remarkable computational flexibility. In our context, we use the Taylor series:\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\] These values keep adding up infinitely, adjusting the function's approximation as we move further from \( a=0 \).
- This approximation improves with more terms, providing better estimates and behaviors around \( a \).
- The price of this infinite reach is convergence; it is only valid within \(|x| < 1\). Beyond this range, the series no longer represents the function accurately.
- Concepts like radius and interval of convergence become relevant to determine where this series is usable.
