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Chapter 30: Problem 23

Show that if \(f(x)=\sum_{k=0}^{\infty} a_{k} x^{k}\) is a power series solution to \(f^{\prime}(x)=-f(x)\), then \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{k}}{k !} .\) What function does this series represent?

Short Answer

Expert verified
The power series solution of the differential equation \(f'(x) = -f(x)\) is \(f(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}\), which is the Taylor Series representation of \(f(x) = e^{-x}\).

Step by step solution

01

Differentiate the Power Series

Differentiate \(f(x) = \sum_{k=0}^{\infty} a_k x^k\) to find \(f'(x)\). As the derivative of \(x^k\) is \(kx^{k-1}\), the derivative \(f'(x)\) will be \(\sum_{k=1}^{\infty} k a_k x^{k-1}\). Notice that we are starting at \(k=1\) because the \(k=0\) term vanishes in the derivative.
02

Change of Index for \(f'(x)\)

For ease, shift index \(k\) to \(k+1\) in \(f'(x)\) so that it looks like \(f(x)\). So now, \(f'(x) = \sum_{k=0}^{\infty}(k+1) a_{k+1}x^k\).
03

Substitute \(f(x)\) and \(f'(x)\) in the Differential Equation

Rewrite the initial differential equation \(f'(x)=-f(x)\), substituting \(f(x)\) and \(f'(x)\), we get \(\sum_{k=0}^{\infty}(k+1) a_{k+1}x^k = -\sum_{k=0}^{\infty}a_k x^k\).
04

Equating Coefficients

On comparing the coefficients of the power series, we get \((k+1) a_{k+1} = -a_k\). This gives us a recurrent relationship among the coefficients - the coefficient at index \(k+1\) is equivalent to the negative coefficient at index \(k\) divided by \(k+1\). Which leads to \(a_{k+1} = \frac{-a_k}{k+1}\).
05

Bracket Conventionality

By conventionality, we set \(a_0 = 1\), hence, using the recurrence relationship to generate the coefficients, we obtain \(a_k = (-1)^k \frac{1}{k!}\).
06

Replacement of Coefficients in \(f(x)\)

Replace values of \(a_k\) in \(f(x)\) and you get \(f(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}\). This is as was required.
07

Identify the Function

The function \(f(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}\) is the Taylor Series representation of the function \(f(x) = e^{-x}\).

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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 an infinite series of the form \( f(x) = \sum_{k=0}^{\infty} a_k x^k \). This expression represents a function as an infinite sum of terms, each involving a coefficient \( a_k \) and a variable \( x \) raised to an increasing power.
Power series are useful because they can represent complex functions as simple sums of powers of \( x \).
  • The series converges when \( x \) is in a certain range where the infinite series adds up to a finite number.
  • If \( x \) is outside that range, the series diverges.

The range of \( x \) that makes the series converge is called the "radius of convergence." In the exercise, the power series \( f(x)=\sum_{k=0}^{\infty} a_k x^k\) was used to solve a differential equation, demonstrating the flexibility and power of this mathematical concept.
Taylor Series
The Taylor series is a particular kind of power series that expands functions into an infinite sum centered around a specific point, often \( x = 0 \). It is expressed as \( f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k \), where \( f^{(k)}(0) \) denotes the \( k \)-th derivative of \( f \) evaluated at 0.
The beauty of the Taylor series lies in its ability to approximate functions that might otherwise be difficult to express.
  • Taylor series are used in calculus to simplify complex operations such as integration and differentiation.
  • In many cases, only a few terms of the series are needed to get a very close approximation of the function.

In our exercise, it is shown that the power series solution converges to \( f(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!} \), which is recognized as the Taylor series for the exponential decay function \( e^{-x} \).
Recurrent Relationship
The recurrent or recurrence relationship is a method used to find the coefficients of a power series, and it often emerges from comparing coefficients of like terms. In the solution given, the recurrence relation \( (k+1) a_{k+1} = -a_k \) was derived.
This simplifies to \( a_{k+1} = \frac{-a_k}{k+1} \). This relationship shows how each term \( a_{k+1} \) in the series is related to the preceding term \( a_k \).
  • It's a step-by-step rule used to calculate subsequent terms based on the previous terms.
  • By starting with a known initial condition, such as \( a_0 = 1 \), all other \( a_k \) coefficients can be determined.

Through this approach, we constructively solve complex differential equations by focusing on pattern repetition in the coefficients, leading to coherent functions like \( e^{-x} \) in our example.

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Most popular questions from this chapter

Find the Maclaurin series for \(\arcsin x\) using the fact that \(\int \frac{1}{\sqrt{1-x^{2}}} d x=\sin ^{-1} x+C\). What is the radius of convergence of the series?

Determine whether the series converges or diverges. In this set of problems knowledge of the Limit Comparison Test is assumed. \(\sum_{k=1}^{\infty} \frac{1}{e^{k}-1}\)

Determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. \(\sum_{k=1}^{\infty} \frac{k+2}{3 k^{2}}\)

Determine whether the series converges absolutely, converges conditionally, or diverges. Explain your reasoning carefully. \(\sum_{k=1}^{\infty} \frac{(-1)^{k} 5^{k}}{k !}\)

Use your knowledge of improper integrals to give an upper and lower bound for \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}\)
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