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Chapter 10: Problem 13

Find the Maclaurin series for the functions \(\frac{1}{1+x}\)

Short Answer

Expert verified
The Maclaurin series is \(1 - x + x^2 - x^3 + \ldots = \sum_{n=0}^{\infty} (-1)^n x^n\).

Step by step solution

01

Understanding the Maclaurin Series

The Maclaurin series is a special case of the Taylor series, where the expansion of the function is performed at the point \(x = 0\). The Maclaurin series for a function \(f(x)\) is given by the formula: \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\]
02

Identify the Function and its Derivatives

We have the function \(f(x) = \frac{1}{1+x}\). Let's compute its derivatives at \(x = 0\) to use them in the series expansion. First, compute \(f(0)\): - \(f(x) = \frac{1}{1+x}\) so \(f(0) = \frac{1}{1+0} = 1\). Next, compute the first few derivatives:- First derivative: \(f'(x) = -\frac{1}{(1+x)^2}\), so \(f'(0) = -1\).- Second derivative: \(f''(x) = \frac{2}{(1+x)^3}\), so \(f''(0) = 2\).- Third derivative: \(f'''(x) = -\frac{6}{(1+x)^4}\), so \(f'''(0) = -6\).
03

Construct the Maclaurin Series

Using the derivatives calculated at \(x = 0\), we construct the Maclaurin Series:\[f(x) = 1 - x + \frac{2}{2!}x^2 - \frac{6}{3!}x^3 + \ldots \]This simplifies to:\[f(x) = 1 - x + x^2 - x^3 + \ldots \]
04

Conclude the Pattern of the Series

The pattern of the Maclaurin series for \(\frac{1}{1+x}\) emerges as:\[f(x) = \sum_{n=0}^{\infty} (-1)^n x^n\]This is the series representation, where the coefficient alternates the sign for each successive power of \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor series
The Taylor series is a powerful mathematical tool used to represent functions as infinite sums of derivatives at a single point. This allows us to understand complex functions in simpler terms.
  • The Taylor series can be expressed using a formula involving function derivatives at a specific point, often denoted by \(a\).
  • The general form of a Taylor series for a function \(f(x)\) centered around \(a\) is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
  • In the case of a Maclaurin series, which is a special case of the Taylor series, the expansion is done around \(a = 0\).
  • This effectively means that the Maclaurin series is just a simplified form of the Taylor series, applicable when the derivatives are evaluated and summed at the origin.
Each term in the Taylor or Maclaurin series involves higher-order derivatives, which provide insight into the behavior of the original function. This is particularly important when the function doesn't have simple algebraic expressions or is difficult to evaluate directly.
function expansion
Function expansion through series is a way to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This approach is crucial in applications like physics, engineering, and other fields where you model systems using mathematical functions.
  • Function expansion captures the essence of the function by using polynomial-like structures to approximate the function values over an interval.
  • The Maclaurin series is an example of this, expanding the function \( \frac{1}{1+x} \) into an infinite series: \[ f(x) = 1 - x + x^2 - x^3 + \ldots \]
  • Expanding a function allows for easier computation, especially when dealing with complex analytical problems.
  • One key advantage is that with Maclaurin series, you can evaluate function values close to zero with higher accuracy and less computational effort.
Function expansion is a crucial concept in calculus that helps in linearizing non-linear equations, making them simpler to work with. This technique is not only foundational in mathematics, but also assists in simplifying real-world problems, making complex systems manageable.
derivative computation
Computing derivatives is a central task in finding series expansions like the Maclaurin series. Derivatives measure how a function changes as its input changes, providing information needed to construct polynomial approximations.
  • For the function \( f(x) = \frac{1}{1+x} \), knowing the first few derivatives and their values at \( x = 0 \) is necessary for its Maclaurin series.
  • The derivatives computed are as follows:
    • First derivative: \( f'(x) = -\frac{1}{(1+x)^2} \)
    • Second derivative: \( f''(x) = \frac{2}{(1+x)^3} \)
    • Third derivative: \( f'''(x) = -\frac{6}{(1+x)^4} \)
  • Evaluating these at \( x = 0 \) determines their contribution to the Maclaurin series:
    • \( f'(0) = -1 \)
    • \( f''(0) = 2 \)
    • \( f'''(0) = -6 \)
By using derivatives, we can approximate a wide range of functions for practical calculations. Understanding derivatives is crucial, not just for series expansion, but for broader mathematical and scientific applications as well.

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Most popular questions from this chapter

Uniqueness of convergent power series a. Show that if two power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(\sum_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) . (Hint: Let \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !) . )\) b. Show that if \(\sum_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n .\)

How many terms of the Taylor series for \(\ln (1+x)\) should you add to be sure of calculating ln \((1.1)\) with an error of magnitude less than \(10^{-8} ?\) Give reasons for your answer.

How many terms of the convergent series \(\sum_{n=1}^{\infty}\left(1 / n^{1.1}\right)\) should be used to estimate its value with error at most 0.00001\(?\)

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Use series to estimate the integrals' values with an error of magnitude less than \(10^{-5}\) . (The answer section gives the integrals' values rounded to seven decimal places.) \begin{equation} \int_{0}^{0.5} \frac{1}{\sqrt{1+x^{4}}} d x \end{equation}
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