Chapter 16: Problem 15
Find the power series in \(x\) of \(f(x)\) if \(f^{\prime \prime}(x)=-f(x), f(0)=0\), and \(f^{\prime}(0)=1\). Also, find the radius of convergence of the resulting series.
Short Answer
Expert verified
The power series for \(f(x)\) is \(f(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\). The radius of convergence is \(\infty\).
Step by step solution
01
Write the general form of the power series
The general form of a power series for a function can be written as \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]
02
Find the second derivative of the power series
We need to find the second derivative of the power series:\[ f''(x) = \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} \]
03
Use the given differential equation
According to the problem, we have \(f''(x) = -f(x)\). Substitute the expressions for \(f(x)\) and \(f''(x)\) from the power series:\[ \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} = -\sum_{n=0}^{\infty} a_n x^n \]
04
Adjust indices to match powers of x
Adjust the indices so the powers of \(x\) match:\[ \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} = -a_0 - a_1 x - \sum_{n=2}^{\infty} a_n x^n \]Shift the index of the left-hand side:\[ \sum_{k=0}^{\infty} a_{k+2}(k+2)(k+1)x^k = -a_0 - a_1 x - \sum_{k=0}^{\infty} a_k x^k \]
05
Equate coefficients of powers of x
Match the coefficients of \(x^k\) on both sides:For the constant term: \(a_2 \cdot 2 \cdot 1 = -a_0 \rightarrow a_2 = -\frac{a_0}{2} = 0\) since \(a_0 = 0\).For the linear term: \(a_3 \cdot 3 \cdot 2 = -a_1 \rightarrow a_3 = -\frac{a_1}{6} = -\frac{1}{6}\) since \(a_1 = 1\).Continuing this pattern, the recurrence relation is given by:\[a_{n+2} = -\frac{a_n}{(n+2)(n+1)}\]
06
Identify the non-zero coefficients
Using the recurrence:\[a_2 = a_0 = 0\, a_3 = -\frac{1}{6},\, a_4 = 0,\, a_5 = -\frac{1}{120},\,\ldots\]This results in terms where only odd powers are non-zero.
07
Write the power series solution
The power series for \(f(x)\) is therefore:\[f(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots = x + \sum_{k=1}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!} \]
08
Determine the radius of convergence
The radius of convergence for a power series can be found using the ratio test:\[\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \]For this series:\[\left| \frac{(-1)^{k+1} \frac{1}{(2k+3)!}}{(-1)^k \frac{1}{(2k+1)!}} \right| = \left| \frac{1}{(2k+3)(2k+2)} \right| \rightarrow 0 \text{ as } k \rightarrow \infty \]Hence, the radius of convergence is infinite.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a way to represent a function as an infinite sum of terms, each of which is a power of the variable multiplied by a coefficient. The general form of a power series for a function is given by:
\[ f(x) = \sum_{n=0}^{\infty}a_nx^n \].
In this expression, \( a_n \) are the coefficients of the series and 'n' is the index that runs from 0 to infinity. Each term in the series involves a power of 'x' and the corresponding coefficient.
Power series are incredibly useful because they allow us to work with very complicated functions in a much simpler, polynomial-like form.
\[ f(x) = \sum_{n=0}^{\infty}a_nx^n \].
In this expression, \( a_n \) are the coefficients of the series and 'n' is the index that runs from 0 to infinity. Each term in the series involves a power of 'x' and the corresponding coefficient.
Power series are incredibly useful because they allow us to work with very complicated functions in a much simpler, polynomial-like form.
- Example: The exponential function \( e^x \) can be expanded as a power series: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \].
Radius of Convergence
When dealing with power series, one important concept is the 'radius of convergence'. This term refers to the distance within which the power series converges to a finite value as you move away from the center of the series (usually around \( x=0 \)).
The radius of convergence helps us understand where the series representation of a function is valid. To find this radius, we often use the ratio test. For a power series \( \sum_{n=0}^{\infty} a_n x^n \), the ratio test involves finding:
\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \].
The radius of convergence helps us understand where the series representation of a function is valid. To find this radius, we often use the ratio test. For a power series \( \sum_{n=0}^{\infty} a_n x^n \), the ratio test involves finding:
\[ \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \].
- If this limit is 'L', then the radius of convergence 'R' is given by \( R=\frac{1}{L} \).
- If \( L=0 \), the series converges everywhere, thus having an infinite radius of convergence.
- If \( L \to \infty \), the series converges only at the center point.
Analytic Solutions
An analytic solution of a differential equation is one that can be expressed in terms of power series. We call a function 'analytic' at a point if it can be represented by a convergent power series in some interval around that point. Let's break down why this is important:
- Exact Results: Analytic solutions give us exact results within the radius of convergence, unlike numerical methods which provide approximate results.
- Explanation of Behavior: Power series solutions can provide insight into the behavior of the function near specific points.