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Chapter 10: Problem 12

Write down the Taylor series for \(x e^{x}\) around \(x=0 .\) Integrate and substitute \(x=1\) to find the sum of \(1 / n !(n+2)\).

Short Answer

Expert verified
The sum of \(1 / n!(n+2)\) is the integrated series of \(x e^x\) evaluated at \(x=1\).

Step by step solution

01

Write the Taylor Series for \(e^x\)

The Maclaurin series for \(e^x\) around \(x = 0\) is given by:\[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\] This provides the exponential function written as an infinite sum of power terms.
02

Multiply the Series by \(x\)

To find the series for \(x e^x\), multiply each term in the series for \(e^x\) by \(x\):\[x e^x = x \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}\]This adjusts each power of \(x\) up by one.
03

Integrating Term by Term

Integrate the series obtained for \(x e^x\) term by term:\[\int x e^x \, dx = \int \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \, dx = \sum_{n=0}^{\infty} \frac{1}{n!} \int x^{n+1} \, dx\]The integral of \(x^{n+1}\) is \(\frac{x^{n+2}}{n+2}\). Thus,\[\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)n!}\]
04

Evaluate the Integral at \(x=1\)

Substitute \(x=1\) into the integrated series:\[\sum_{n=0}^{\infty} \frac{1^{n+2}}{(n+2)n!} = \sum_{n=0}^{\infty} \frac{1}{(n+2)n!}\]This represents the evaluated convergent series at \(x=1\).
05

Conclusion

The integrated Taylor series at \(x = 1\) gives the sum of the series:\[\sum_{n=0}^{\infty} \frac{1}{(n+2) n!}\] which coincides with our original target formula for finding the sum. Thus, we can integrate the series of \(x e^x\), substitute \(x=1\), and confirm that the sum is achieved through this series expansion.

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Key Concepts

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

Maclaurin Series
A Maclaurin series is a special case of the more general Taylor series, where the series is expanded around zero. It's a powerful tool that approximates functions through an infinite sum of terms. Each term's coefficients are derived from the derivatives of the function evaluated at zero.
The general form of a Maclaurin series is:
  • The coefficient of each term is the derivative of the function evaluated at zero, divided by the factorial of the term's order.
  • For a function \( f(x) \), its Maclaurin series can be written as \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \).
For instance, the exponential function \( e^x \) has a simple and frequently used Maclaurin series. This series is given by \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
This expansion is significant because it converges to \( e^x \) for all real values of \( x \), showcasing the smoothness and continuity of the exponential function.
Exponential Function
The exponential function \( e^x \) is a crucial mathematical function with unique properties. It is often used in various mathematical, scientific, and engineering applications.
The function \( e^x \) itself is defined as:
  • A constant, \( e \), is the base of natural logarithms, approximately equal to 2.718.
  • The function is its own derivative. This means that the derivative \( \frac{d}{dx} e^x = e^x \).
  • Similarly, the function is its own integral, up to a constant, meaning \( \int e^x \, dx = e^x + C \).
Because \( e^x \) is smooth and continuous, it naturally leads to a straightforward Maclaurin series expansion. The factor of \( n! \) in the denominator of the series terms ensures convergence across the real number line. This property is beneficial when calculating series for practical applications, like solving differential equations or evaluating limits.
Convergent Series
Convergent series are series that approach a specific limit as more terms are added. In mathematics, understanding whether a series converges is crucial because it reveals whether the infinite sum yields a finite result.
When working with series, consider:
  • A series \( \sum_{n=0}^{\infty} a_n \) is said to converge if there exists a number \( L \) such that the partial sums \( s_n = a_0 + a_1 + \cdots + a_n \) approach \( L \) as \( n \) increases.
  • The behavior of the series terms \( a_n \) as \( n \to \infty \) often dictates convergence. For example, rapidly decreasing terms, such as those involving factorial growth in the denominator, suggest convergence.
In the context of our problem, the series \( \sum_{n=0}^{\infty} \frac{1}{(n+2)n!} \) converges because the factorial in the denominator ensures the terms become increasingly smaller. Evaluating at specific points, like substituting \( x=1 \) within the integral, emphasizes the practical applications of convergent series in determining specific numerical sums.

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Most popular questions from this chapter

Find all derivatives of \(f(x)=x^{3}\) at \(x=a\) and write out the Taylor series around that point. Verify that it adds to \(x^{3}\).

For a conditionally convergent series, explain how the terms could be rearranged so that the sum is \(+\infty .\) All terms must eventually be included, even negative terms.

(Composition of series) If \(f=a_{0}+a_{1} x+a_{2} x^{2}+\cdots\) and \(g=b_{1} x+b_{2} x^{2}+\cdots\) find the \(1, x, x^{2}\) cocficicnts of \(f(g(x))\) Test on \(f=1 /(1+x), g=x(1-x),\) with \(f(g(x))=1-x\)

Write down the first three nonzero terms of the Taylor series around \(x=0\), from the series for \(e^{x}, \cos x,\) gnd \(\sin x\). $$ e^{x} \cos x $$

Solve difierential equations by series. Find the infinite series that solves \(y^{\prime}=y\) staring from \(y=\) 1 at \(x=3\) (use powers of \(x-3\) ). Identify \(y\) as a known function.
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