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Exponential function In Section 3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty < x < \infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-x}$$

Short Answer

Expert verified
Answer: The power series representation of the function \(f(x) = e^{-x}\) is \(\sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}\), and its interval of convergence is \(-\infty < x < \infty\).

Step by step solution

01

Recall the power series of the exponential function

The power series for the exponential function \(e^x\) is given by: $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}, \quad \text{for }-\infty < x < \infty$$
02

Substitute \(-x\) into the power series of \(e^x\)

To find the power series for the function \(f(x) = e^{-x}\), we substitute \(-x\) into the power series of \(e^x\). $$f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!}$$
03

Simplify the power series for \(f(x)\)

Observe that \((-x)^k = (-1)^k x^k\). Thus, we can rewrite the power series for \(f(x)\) as: $$f(x) = e^{-x} = \sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}$$
04

Identify the interval of convergence

The original interval of convergence for the \(e^x\) is \(-\infty < x < \infty\). Since the power series for \(f(x)\) is derived from the power series for \(e^x\), the interval of convergence should remain unchanged. Therefore, the power series for the function \(f(x) = e^{-x}\) is: $$f(x) = e^{-x} = \sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k!}$$ and the interval of convergence for the resulting series is \(-\infty < x < \infty\).

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Key Concepts

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

Exponential Function
The exponential function, denoted as \( e^x \), is one of the most fundamental functions in mathematics. It's especially important because of its unique property of being its own derivative. This means that no matter how many times you differentiate \( e^x \), you will always get \( e^x \) again. The power series representation of the exponential function is incredibly handy, especially when you need to calculate values or solve problems involving \( e^x \). The power series for \( e^x \) is expressed as:
  • \( e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \)
This series is an infinite sum where each term involves powers of \( x \) divided by the factorial of the term index \( k \). Factorials \( k! \) grow very rapidly, which means the terms of the series get smaller very quickly, making the series converge nicely.
Interval of Convergence
The interval of convergence is a crucial part of understanding power series. It tells us the range of values for \( x \), where the series converges to a definite value. For the exponential power series of \( e^x \), the interval of convergence is quite broad:
  • \(-\infty < x < \infty\)
This means that for any real number \( x \), the series will converge. The interval of convergence is vital as it assures us that the series representation is valid across a wide range. This property is particularly useful when extending these concepts to other functions, like \( f(x) = e^{-x} \), because the interval remains the same.
Substitution
Substituting variables within a power series allows us to derive new series for related functions. In this exercise, we substitute \(-x\) into the power series for \( e^x \) to find the series for \( f(x) = e^{-x} \).Here's how the substitution works:
  • Replace every instance of \( x \) in the original power series \( \sum_{k=0}^{\infty} \frac{x^k}{k!} \) with \(-x\).
This yields:
  • \( f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} \)
Substitution is a powerful tool that lets us build series for more complex functions based on simpler ones. It retains the characteristics of the original function, like the interval of convergence.
Simplification
Simplifying power series is often necessary to make them more practical for computation and analysis. In the given expression for \( f(x) = e^{-x} \), simplification involves handling the term \((-x)^k\). This can be done by noting:
  • \((-x)^k = (-1)^k x^k\)
Thus, each term in the series can be written to separate the negative sign:
  • \(f(x) = e^{-x} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k!}\)
This transformation helps to visualize the series as alternating terms between positive and negative, dictated by \((-1)^k\). Simplification refines the power series for use in calculations and theoretical discussions.

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Most popular questions from this chapter

The theory of optics gives rise to the two Fresnel integrals $$S(x)=\int_{0}^{x} \sin t^{2} d t \text { and } C(x)=\int_{0}^{x} \cos t^{2} d t$$ a. Compute \(S^{\prime}(x)\) and \(C^{\prime}(x)\) b. Expand \(\sin t^{2}\) and \(\cos t^{2}\) in a Maclaurin series and then integrate to find the first four nonzero terms of the Maclaurin series for \(S\) and \(C\) c. Use the polynomials in part (b) to approximate \(S(0.05)\) and \(C(-0.25)\) d. How many terms of the Maclaurin series are required to approximate \(S(0.05)\) with an error no greater than \(10^{-4} ?\) e. How many terms of the Maclaurin series are required to approximate \(C(-0.25)\) with an error no greater than \(10^{-6} ?\)

Assume that \(f\) has at least two continuous derivatives on an interval containing \(a\) with \(f^{\prime}(a)=0 .\) Use Taylor's Theorem to prove the following version of the Second Derivative Test: a. If \(f^{\prime \prime}(x) > 0\) on some interval containing \(a,\) then \(f\) has a local minimum at \(a\) b. If \(f^{\prime \prime}(x) < 0\) on some interval containing \(a,\) then \(f\) has a local maximum at \(a\)

Let $$f(x)=\sum_{k=0}^{\infty} c_{k} x^{k} \quad \text { and } \quad g(x)=\sum_{k=0}^{\infty} d_{k} x^{k}$$ a. Multiply the power series together as if they were polynomials, collecting all terms that are multiples of \(1, x,\) and \(x^{2} .\) Write the first three terms of the product \(f(x) g(x)\) b. Find a general expression for the coefficient of \(x^{n}\) in the product series, for \(n=0,1,2, \ldots\)

a. Find a power series for the solution of the following differential equations. b. Identify the function represented by the power series. $$y^{\prime}(t)+4 y(t)=8, y(0)=0$$

Determine whether the following statements are true and give an explanation or counterexample. a. The function \(f(x)=\sqrt{x}\) has a Taylor series centered at 0 . b. The function \(f(x)=\csc x\) has a Taylor series centered at \(\pi / 2\) c. If \(f\) has a Taylor series that converges only on \((-2,2),\) then \(f\left(x^{2}\right)\) has a Taylor series that also converges only on (-2,2) d. If \(p(x)\) is the Taylor series for \(f\) centered at \(0,\) then \(p(x-1)\) is the Taylor series for \(f\) centered at 1 e. The Taylor series for an even function about 0 has only even powers of \(x\)

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