/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 In Problems \(55-57\), find the ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Problems \(55-57\), find the described value from the series $$C(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{4 n+1}}{(2 n) !(4 n+1)}$$ The integer \(a\) where \(x^{k} / a\) is the term of the series corresponding to \(n=2\).

Short Answer

Expert verified
The integer \(a\) is 216.

Step by step solution

01

Identify the Term for n=2

First, we need to know the form of the series: \[C(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{4n+1}}{(2n)! (4n+1)}\]We substitute \(n = 2\) into the series.This becomes the term: \[\frac{(-1)^{2} x^{4 \times 2 + 1}}{(2 \times 2)! (4 \times 2 + 1)} = \frac{x^{9}}{4! \times 9}\]
02

Calculate Factorials and Simplify

Next, simplify the term by calculating the factorial and the multiplication:Calculate \(4!\):\[4! = 4 \times 3 \times 2 \times 1 = 24\]Substitute back into the fraction:\[\frac{x^{9}}{24 \times 9} = \frac{x^{9}}{216}\]
03

Identify and Solve for Integer a

We found that the term in the series for \(n=2\) is \(\frac{x^{9}}{216}\).The given term was asked in the form \(\frac{x^k}{a}\).Here, \(k = 9\) and \(a = 216\).

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Key Concepts

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

Factorials
Factorials are a key part of understanding many mathematical series and sequences, including power series. The factorial of a non-negative integer, denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For instance, \( 4! \) (read as "four factorial") is calculated as \( 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow at a rapid pace since each term multiplies by successively larger numbers.Factorials are used in permutations, combinations, and various other mathematical concepts. In the context of power series, factorials often appear in the denominators to help scale down large products. This becomes especially evident in series like the Taylor series, where each term includes a factorial in the denominator as part of the general formula. To summarize the key points on factorials:
  • Factorial \( n! \) is the product of integers from 1 to \( n \).
  • Factorials rapidly increase, as they involve cumulative multiplication.
  • Common in series, factorials help in transforming sequences to converge.
Summation Notation
Summation notation, often represented by the Greek letter sigma (\( \Sigma \)), is a concise way to represent the sum of a sequence of terms. The notation \( \sum_{n=0}^{\infty} a_n \) indicates that we are summing terms \( a_n \) from \( n = 0 \) to infinity. Each term in this format can be seen as a part of a larger series where each component follows a specific formula.This notation simplifies the expression of long sequences, allowing mathematicians to work efficiently with complex series. It is crucial in the analysis of series and is widely used in calculus, statistics, and other mathematical fields. Key elements in summation notation include:
  • \( n \): The index, usually starting from an integer and increasing by 1.
  • \( a_n \): The general term or formula expressing each element in the series.
  • \( \infty \): Typically signifies the series continues indefinitely.
Understanding summation notation is essential to working with series like the Taylor series, as it allows readers to interpret and manipulate entire sequences compactly.
Taylor Series
The Taylor series is a powerful tool in calculus that represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It essentially expresses complex functions as power series, where the terms can be computed using factorials and powers of variables.The general formula for a Taylor series about the point \( a \) for a function \( f(x) \) is:\[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]In this formula:
  • \( f^n(a) \) is the \( n \)-th derivative of \( f \) at \( a \).
  • The denominator includes \( n! \), linking factorials to each term's calculation.
  • The series extends infinitely, improving approximation within its radius of convergence.
Taylor series are particularly useful in approximating functions that are otherwise difficult to calculate manually. Since each function is expanded about a point \( a \), it allows us to estimate values of the function near \( a \) with increasing accuracy as more terms are included. Understanding the way Taylor series work aids in grasping both theoretical and applied aspects of calculus.

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Most popular questions from this chapter

Decide if the statements in Problems \(65-71\) are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If the Taylor series for \(f(x)\) around \(x=0\) has a finite number of terms and an infinite radius of convergence, then \(f(x)\) is a polynomial.

Show that for positive integers \(k\), the periodic function \(f(x)=a_{k} \cos k x+b_{k} \sin k x\) of period \(2 \pi\) has energy \(a_{k}^{2}+b_{k}^{2}\).

The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass, \(m\) of the object when it is moving at speed \(v\) is given by the formula $$m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}$$, where \(c\) is the speed of light and \(m_{0}\) is the mass of the object when it is at rest. (a) Use the formula for \(m\) to decide what values of \(v\) are possible. (b) Sketch a rough graph of \(m\) against \(v,\) labeling intercepts and asymptotes. (c) Write the first three nonzero terms of the Taylor series for \(m\) in terms of \(v\) (d) For what values of \(v\) do you expect the series to converge?

(a) Find and multiply the Taylor polynomials of degree 1 near \(x=0\) for the two functions \(f(x)=\) \(1 /(1-x)\) and \(g(x)=1 /(1-2 x)\) (b) Find the Taylor polynomial of degree 2 near \(x=0\) for the function \(h(x)=f(x) g(x)\) (c) Is the product of the Taylor polynomials for \(f(x)\) and \(g(x)\) equal to the Taylor polynomial for the function \(h(x) ?\)

Decide if the statements are true or false. Give an explanation for your answer. If \(f(x)\) and \(g(x)\) have the same Taylor polynomial of degree 2 near \(x=0,\) then \(f(x)=g(x)\)

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