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In each of Exercises \(1-10\), express the given function as a power series in \(x\) with base point \(0 .\) Calculate the radius of convergence \(R\). \(\frac{1}{1-2 x}\)

Short Answer

Expert verified
The power series is \( \sum_{n=0}^{\infty} 2^n x^n \) with radius of convergence \( R = \frac{1}{2} \).

Step by step solution

01

Identify the Function Form

Notice that the function \( \frac{1}{1-2x} \) is already in the form \( \frac{1}{1-ax} \), where \( a = 2 \). This resembles the geometric series form \( \sum_{n=0}^{\infty} (ax)^n \).
02

Express as a Power Series

Using the geometric series formula, \( \frac{1}{1-ax} = \sum_{n=0}^{\infty} (ax)^n \), substitute \( a = 2 \) to get the power series representation: \( \frac{1}{1-2x} = \sum_{n=0}^{\infty} (2x)^n = \sum_{n=0}^{\infty} 2^n x^n \).
03

Check for Convergence

A geometric series \( \sum_{n=0}^{\infty} (ax)^n \) converges when \( |ax| < 1 \). For our series, this condition becomes \( |2x| < 1 \).
04

Solve for Radius of Convergence

The inequality \( |2x| < 1 \) simplifies to \( |x| < \frac{1}{2} \). This tells us that the power series converges when \( |x| < \frac{1}{2} \), meaning the radius of convergence \( R \) is \( \frac{1}{2} \).

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Key Concepts

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

Radius of Convergence
The radius of convergence is a key concept when working with power series. It determines the interval around the base point within which the series converges and maintains validity. To calculate this, we examine the condition for convergence in the series: for the series \( \sum_{n=0}^{\infty} (ax)^n \), convergence is achieved when \( |ax| < 1 \).

In our example, the series \( \sum_{n=0}^{\infty} (2x)^n \) converges when \(|2x| < 1\). By solving this inequality, we attain \( |x| < \frac{1}{2} \). Thus, the radius of convergence \( R \) is \( \frac{1}{2} \). This informs us on how far from the base point 0 we can consider the series to still represent the function accurately within a particular range.
Geometric Series
A geometric series is a special kind of series with a constant ratio between successive terms. It forms a crucial foundation in understanding power series and their applications. The general form of a geometric series is \( \sum_{n=0}^{\infty} r^n \), where \( r \) is the common ratio.

For convergence, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). In the context of power series in the form \( \frac{1}{1-ax} \), the geometric series formula \( \sum_{n=0}^{\infty} (ax)^n \) is derived. This formula helps us to transform functions like \( \frac{1}{1-2x} \) into their series representation, making it easier to work with or understand the approximation of complex functions.
Function Representation
Representing functions as power series is an effective method in mathematical analysis and applications, especially when dealing with complex functions or finding approximations. A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n x^n \), where \( c_n \) are coefficients. This representation is highly versatile, allowing for functions to be expressed in a way that can be readily analyzed or manipulated.

In the exercise example, we use the geometric series representation \( \frac{1}{1-2x} = \sum_{n=0}^{\infty} 2^n x^n \). This breakdown into a series makes it easier to approximate the function for values within the radius of convergence \( R = \frac{1}{2} \). Power series also prove useful in solving differential equations, integrating functions, and performing operations like differentiation term by term, all of which leverage the polynomial form that we derive from the series representation.

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Most popular questions from this chapter

Let \(f(x)=(x+1)^{4} /\left(x^{4}+1\right)\). It is known that \(f\) has a power series expansion of the form \(f(x)=1+4 x+6 x^{2}+4 x^{3}-4 x^{5}-6 x^{6}-4 x^{7}+4 x^{9}+\cdots\) a. Plot the central difference quotient approximation \(D_{0} f\left(x, 10^{-5}\right)\) of \(f^{\prime}(x)\) for \(-0.5

In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} $$

Let \(\left\\{a_{n}\right\\}\) be a sequence of positive numbers. In a course on mathematical analysis, one learns that if the two limits \(\lim _{n \rightarrow \infty} a_{n+1} / a_{n}\) and \(\lim _{n \rightarrow \infty} a_{n}^{1 / n}\) exist, then they are equal. In each of Exercises \(65-68\), produce a plot that illustrates the equality of these two limits. Your plot should include a horizontal line that is the asymptote of the points \(\left\\{\left(n, a_{n+1} / a_{n}\right)\right\\}\) and \(\left\\{\left(n, a_{n}^{1 / n}\right)\right\\}\). \(a_{n}=e^{n} / \ln (1+n)\)

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose that the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges on \((-R, R)\) to a function \(f(x)\) and that \(|f(x)| \geq k>0\) on that interval for some positive constant \(k\). Then, \(1 / f(x)\) also has a convergent power series expansion on \((-R, R) .\) Compute its coefficients in terms of the \(a_{n}\) 's. Hint: Set $$ \frac{1}{f(x)}=g(x)=\sum_{n=0}^{\infty} b_{n} x^{n} $$ Use the equation \(f(x) \cdot g(x)=1\) to solve for the \(b_{n}\) 's.

In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{1-\cos (x)}{x^{2}} $$

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