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Chapter 5: Problem 15

Integrating Power Series For each of the following functions \(f\), find a power series representation for \(f\) by integrating the power series for \(f^{\prime}\) and find its interval of convergence. $$ \begin{aligned} &\text { a. } f(x)=\ln (1+x) \\ &\text { b. } f(x)=\tan ^{-1} x \end{aligned} $$

Short Answer

Expert verified
a. \(\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^{n+1}\), \((-1,1)\); b. \(\tan^{-1}x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}\), \((-1,1)\).

Step by step solution

01

Find the derivative of the function

For both functions, we start by identifying their derivatives.For \( f(x) = \ln(1+x) \), the derivative is \( f'(x) = \frac{1}{1+x} \).For \( f(x) = \tan^{-1} x \), the derivative is \( f'(x) = \frac{1}{1+x^2} \).
02

Express the derivative as a power series

We recognize \( f'(x) \) as a geometric series where applicable. For \( f'(x) = \frac{1}{1+x} \), it can be expanded as:\[\sum_{n=0}^{\infty} (-1)^n x^n, \quad |x|<1.\]For \( f'(x) = \frac{1}{1+x^2} \), it becomes:\[\sum_{n=0}^{\infty} (-1)^n x^{2n}, \quad |x|<1.\]
03

Integrate the power series to find the function's power series

Integrating term-by-term the power series representation of \( f'(x) \) gives us the function \( f(x) \).For \( f(x) = \ln(1+x) \), integrate each term:\[\int \sum_{n=0}^{\infty} (-1)^n x^n \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^{n+1} + C.\]Since \( \ln(1+0) = 0 \), the constant \( C = 0 \), so:\[f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^{n+1}, \quad |x|<1.\]For \( f(x) = \tan^{-1} x \), integrate each term:\[\int \sum_{n=0}^{\infty} (-1)^n x^{2n} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} + C.\]Since \( \tan^{-1}(0) = 0 \), the constant \( C = 0 \), so:\[f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, \quad |x|<1.\]
04

Identify the interval of convergence

The interval of convergence for both series is derived from the geometric series used in the derivatives:For \( \ln(1+x) \), the series \( \sum_{n=0}^{\infty} (-1)^n x^n \) converges for \(|x| < 1\), so the interval of convergence is \((-1, 1)\).For \( \tan^{-1} x \), the series \( \sum_{n=0}^{\infty} (-1)^n x^{2n} \) also converges for \(|x| < 1\), so its interval of convergence is also \((-1, 1)\).

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Key Concepts

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

Interval of Convergence
When dealing with power series, one crucial aspect is determining where the series actually converges. This is known as the "interval of convergence." For a power series in the form of \[ \sum_{n=0}^{\infty} a_n (x - c)^n, \] we determine for which values of \( x \) this series converges.
This involves checking the values of \( x \) for which the series converges absolutely.
  • A common method to find the interval of convergence is using the ratio test. This test helps to determine where the series converges and also if the endpoints \( x = c \pm R \) (where R is the radius of convergence) are included.
  • In some cases, like the geometric series, the convergence might be straightforward. But remember, the interval \( |x| < 1 \) is typical. After finding \( R \), check the endpoints separately.
  • For the functions \( \ln(1+x) \) and \( \tan^{-1} x \), both have an interval of convergence \((-1, 1)\). The reason for this is that they derive from the geometric series expansion.
Geometric Series
The geometric series is a type of series in the form:\[ \sum_{n=0}^{\infty} ar^n, \]where \( a \) is the first term, and \( r \) is the common ratio. A geometric series converges when the absolute value of \( r \) is less than 1. This means that the values for which the series converges or diverges depend heavily on this ratio.
  • The fundamental formula for its sum is \( S = \frac{a}{1-r} \) when \( |r| < 1 \). This is where the magic of geometric series lies, as it allows us to switch the infinite series to a finite expression easily.
  • In the exercise, both derivatives \( \frac{1}{1+x} \) and \( \frac{1}{1+x^2} \) can be represented by geometric series with \( r = -x \) and \( r = -x^2 \) respectively. These attack problems by converting seemingly complex functions into simple infinite series expressions we can manipulate, especially for integration.
  • By finding such representations, it becomes easier to integrate and determine other functions as shown in our derivatives conversion. This simplifies finding the series form of complex logarithmic or arctangent expressions.
Term-by-term Integration
This method is used when integrating power series to find the antiderivative in terms of a series expansion. Since power series act somewhat like polynomials, they allow us to perform integration term-by-term. However, this technique needs treatment with care:
  • Integrate each term separately as if you would a polynomial term. So for \( \sum a_n x^n \), its integral would be \( \sum \frac{a_n}{n+1} x^{n+1} + C \).
  • The key is in knowing the radius and interval of convergence for your series. Ensure the process remains valid within this interval.
  • The constant of integration, \( C \), plays a critical role. You solve for \( C \) by using initial conditions or known values, ensuring your function matches the series outside of its infinite behavior.
  • For instances like \( \ln(1+x) \) and \( \tan^{-1} x \), we saw how by starting with their derivatives we could integrate term-by-term and arrive at these well-known function series.

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Most popular questions from this chapter

The following exercises make use of the functions \(S_{5}(x)=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}\) and \(C_{4}(x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\) on \([-\pi, \pi]\). [T] Plot \(e^{x}-e_{4}(x)\) where \(e_{4}(x)=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{24}\) on \([0,2] .\) Compare the maximum error with the Taylor remainder estimate.

In the following exercises, given that \(\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}\), use term-by-term differentiation or integration to find power series for each function centered at the given point. Subtract the infinite series of \(\ln (1-x)\) from \(\ln (1+x)\) to get a power series for \(\ln \left(\frac{1+x}{1-x}\right)\). Evaluate at \(x=\frac{1}{3}\). What is the smallest \(N\) such that the Nth partial sum of this series approximates \(\ln (2)\) with an error less than \(0.001 ?\)

In the following exercises, compute the Taylor series of each function around \(x=1\). $$ f(x)=e^{-x} $$

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum. Suppose that the coefficients \(a_{n}\) of the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) are defined by the recurrence relation \(a_{n}=\frac{a_{n-1}}{n}+\frac{a_{n-2}}{n(n-1)}\). For \(a_{0}=0\) and \(a_{1}=1\), compute and plot the sums \(S_{N}=\sum_{n=0}^{N} a_{n} x^{n}\) for \(N=2,3,4,5\) on \([-1,1]\).

In the following exercises, find the radius of convergence and the interval of convergence for the given series. $$ \sum_{n=0}^{\infty} \frac{2^{n}}{e^{n}}(x-e)^{n} $$
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