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Evaluating an infinite series Let \(f(x)=\left(e^{x}-1\right) / x,\) for \(x \neq 0\) and \(f(0)=1 .\) Use the Taylor series for \(f\) about 0 and evaluate \(f(1)\) to find the value of \(\sum_{k=0}^{\infty} \frac{1}{(k+1) !}\)

Short Answer

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Question: Determine the value of the infinite series \(\sum_{k=0}^{\infty} \frac{1}{(k+1)!}\) using the function \(f(x)=\frac{e^x-1}{x}\), where \(f(0)=1\). Answer: The value of the infinite series is given by \(f(1)\), or \(\sum_{k=0}^{\infty} \frac{1}{(k+1)!} = f(1)\).

Step by step solution

01

Finding the Taylor series of the given function about 0.

To find the Taylor series of \(f(x)\), we will need to find the derivatives of \(f(x)\) and evaluate them at \(x=0\). Let's first find the first few derivatives of \(f(x)\) and evaluate them at \(x=0\): 1. \(f'(x) = \frac{e^x(x-1)}{x^2}\), and \(f'(0)=\lim_{x\to 0} \frac{e^x(x-1)}{x^2} =-1\) (Using l'Hôpital's rule) 2. \(f''(x) = \frac{e^x(x^2-4x+2)}{x^3}\), and \(f''(0)=\lim_{x\to 0} \frac{e^x(x^2-4x+2)}{x^3} =\frac{1}{2}\) (Using l'Hôpital's rule) Similarly, if we continue, we get all \(f^{(n)}(0)=\frac{1}{(n+1)!}\) for \(n\ge0\). Now we can write the Taylor series about 0: \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\cdot x^n = \sum_{n=0}^{\infty} \frac{\frac{1}{(n+1)!}}{n!} \cdot x^n = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}\)
02

Evaluate f(1)

To evaluate \(f(1)\), substitute \(x=1\) in the Taylor series we obtained: \(f(1) = \sum_{n=0}^{\infty} \frac{1^n}{(n+1)!} = \sum_{n=0}^{\infty} \frac{1}{(n+1)!}\) So, the value of the infinite series is: \(\sum_{k=0}^{\infty} \frac{1}{(k+1)!} = f(1)\)

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Key Concepts

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

Infinite Series
An infinite series is a sum of an infinite sequence of terms. Unlike a finite series that stops after a certain number of terms, an infinite series continues indefinitely. Just because it has infinite terms doesn't mean the sum is infinite. In fact, many infinite series converge to a specific value. This means that as you add more and more terms, the total sum approaches a certain limit.

In the context of functions, infinite series are often used in Taylor series to represent functions as sums of polynomials. When we calculate the value of a function using its infinite series, we're essentially summing up an infinite number of terms to find the value of the function at a certain point. This is exactly what happens with the function \(f(x) = \frac{e^x - 1}{x}\), where its Taylor series representation is an infinite series of terms that allow us to compute \(f(1)\).

The interesting part about infinite series is determining whether they converge or not. Using tests for convergence is necessary to understand the behavior of the series as it tends towards infinity.
Derivatives
Derivatives are fundamental components in calculus, representing the rate of change or slope of a function. When we talk about derivatives in the context of Taylor series, they're used to analyze how each polynomial term in the series contributes to approximating a function.

For \(f(x) = \frac{e^x - 1}{x}\), finding its Taylor series required calculating its derivatives at zero. By doing this:
  • \(f'(x) = \frac{e^x(x-1)}{x^2}\) with \(f'(0) = -1\)
  • \(f''(x) = \frac{e^x(x^2-4x+2)}{x^3}\) with \(f''(0) = \frac{1}{2}\)
we find each derivative results in a value used to form the terms of the Taylor series.

Each derivative gives us information about the function's behavior at \(x = 0\), allowing us to substitute these derivatives into the Taylor polynomial formula effectively making the Taylor series:\[f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}\].

The derivatives, therefore, play a crucial role as the foundation for constructing the series.
L'Hôpital's Rule
L'Hôpital's Rule is a method in calculus used to evaluate indeterminate forms, typically of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When direct substitution in limits returns one of these indeterminate forms, L'Hôpital's Rule offers a viable path forward by differentiating the numerator and the denominator separately.

In the process of computing derivatives of \(f(x) = \frac{e^x - 1}{x}\) for its Taylor series, we often encounter situations where the limit evaluates to an indeterminate form. Here, L'Hôpital's Rule becomes instrumental as it helps to resolve such limits to get the needed values:
  • Using L'Hôpital's Rule, \(f'(0) = \lim_{x\to 0} \frac{e^x(x-1)}{x^2} = -1\).
  • Similarly, for the second derivative, \(f''(0) = \lim_{x\to 0} \frac{e^x(x^2-4x+2)}{x^3} = \frac{1}{2}\).
L'Hôpital's Rule simplifies these complex limits by reducing them to simpler forms that are easily evaluated, proving essential for finding derivatives where normal methods aren't applicable due to their indeterminate nature.

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Most popular questions from this chapter

Let $$f(x)=\sum_{k=0}^{\infty} x^{k}=\frac{1}{1-x} \quad \text { and } \quad S_{n}(x)=\sum_{k=0}^{n-1} x^{k}$$ The remainder in truncating the power series after \(n\) terms is \(R_{n}(x)=f(x)-S_{n}(x),\) which depends on \(x\) a. Show that \(R_{n}(x)=x^{n} /(1-x)\) b. Graph the remainder function on the interval \(|x|<1\) for \(n=1,2,3 .\) Discuss and interpret the graph. Where on the interval is \(\left|R_{n}(x)\right|\) largest? Smallest? c. For fixed \(n,\) minimize \(\left|R_{n}(x)\right|\) with respect to \(x .\) Does the result agree with the observations in part (b)? d. Let \(N(x)\) be the number of terms required to reduce \(\left|R_{n}(x)\right|\) to less than \(10^{-6} .\) Graph the function \(N(x)\) on the interval \(|x|<1 .\) Discuss and interpret the graph.

Consider the following common approximations when \(x\) is near zero. a. Estimate \(f(0.1)\) and give a bound on the error in the approximation. b. Estimate \(f(0.2)\) and give a bound on the error in the approximation. $$f(x)=\cos x \approx 1-x^{2} / 2$$

Consider the following function and its power series: $$f(x)=\frac{1}{(1-x)^{2}}=\sum_{k=1}^{\infty} k x^{k-1}, \quad \text { for }-1

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sin 1$$

Replace \(x\) with \(x-1\) in the series \(\ln (1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^{k}}{k}\) to obtain a power series for \(\ln x\) centered at \(x=1 .\) What is the interval of convergence for the new power series?

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