Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Rewrite the given expression as a sum whose generic term involves \(x^{n} .\) $$ \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2} $$

Short Answer

Expert verified
Question: Rewrite the expression \(\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}\) as a sum with a generic term that involves \(x^{n}\). Answer: The given expression can be rewritten as \(\sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^{m}\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Adjust index to rewrite generic term

To rewrite the given sum with a generic term involving \(x^n\), let's apply a change of variables to the indices. We replace the index \(n\) with \(m+2\) where \(m = n-2\). As a result, the index will start from \(m=0\) and the sum will be written with respect to \(m\): $$ \sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^{m}. $$
02

Present the final expression

Having replaced the index \(n\) by \(m+2\) and changed the limits of the sum accordingly, we can present the final expression of the sum involving the generic term \(x^{n}\): $$ \sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2} x^{m}. $$

Key Concepts

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

Summation Notation
Summation notation, also known as sigma notation, is a convenient and compact way to write long sums. When dealing with a summation, the symbol \(\sum\) is used to represent the sum of a sequence of terms. The variable right below this symbol is the index of summation, which in our exercise is \(n\), and it changes from a lower bound to an upper bound. In the original expression \(\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}\), the index of summation \(n\) starts at 2 and goes to infinity, indicating that this is an infinite series.

Each term in the sequence is defined as a function of \(n\), which in this case is \(n(n-1) a_{n} x^{n-2}\). Summation notation helps in simplifying and understanding the structure of series, especially when working with complex terms or a large number of terms.
Power Series
A power series is a type of infinite series that can be thought of as a polynomial with infinitely many terms, where each term involves a power of the variable \(x\). It is usually expressed in the form \(\sum_{n=0}^{\infty} a_n x^n\). Power series are used in various branches of mathematics to represent functions and to solve differential equations.

The main advantage of power series is that they can converge to a wide range of functions within their interval of convergence and can be easily manipulated through differentiation and integration. In the exercise, we are dealing with a series that we'd like to express in terms of \(x^n\), leading us to adjust the series to a power series format.
Change of Index in Series
Changing the index of a series refers to the process where we replace the existing index of summation with a new variable. This technique is useful when the goal is to manipulate the series into a more familiar or useful form. In the step-by-step solution, the index \(n\) is replaced with \(m+2\), effectively shifting the entire series.

As a result, \(m = n - 2\), and when \(n = 2\), \(m\) will be 0, thereby changing the series so that it starts at \(m = 0\) instead of \(n = 2\). This re-indexing is particularly helpful in this context because it aligns the powers of \(x\) with the coefficients \(a_{m+2}\), simplifying the analysis and calculation of the series.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. When we talk about an infinite series, such as \(\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}\) from our exercise, it means that the series has an endless number of terms.

Convergence is a critical concept when dealing with infinite series, which means that as more and more terms are added, the sum approaches a certain finite limit. However, some infinite series do not converge to a limit; these are called divergent. The transformed series in our problem is also an infinite series, now starting with \(m = 0\), and it is representative of the original problem in a way that emphasizes each term involving a power of \(x\) as \(x^m\). It's important for students to understand the behavior and nature of infinite series within the context of convergence and divergence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

First Order Equations. The series methods discussed in this section are directly applicable to the first order linear differential equation \(P(x) y^{\prime}+Q(x) y=0\) at a point \(x_{0}\), if the function \(p=Q / P\) has a Taylor series expansion about that point. Such a point is called an ordinary point, and further, the radius of convergence of the series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) is at least as large as the radius of convergence of the series for \(Q / P .\) In each of Problems 16 through 21 solve the given differential equation by a series in powers of \(x\) and verify that \(a_{0}\) is arbitrary in each case. Problems 20 and 21 involve nonhomogeneous differential equations to which series methods can be easily extended. Where possible, compare the series solution with the solution obtained by using the methods of Chapter 2 . $$ y^{\prime}=e^{x^{2}} y, \quad \text { three terms only } $$

Determine the general solution of the given differential equation that is valid in any interval not including the singular point. \(x^{2} y^{\prime \prime}-5 x y^{\prime}+9 y=0\)

Find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point. \(2 x(x+2) y^{\prime \prime}+y^{\prime}-x y=0\)

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that two linearly independent solutions of the Legendre equation for \(|x|<1\) are $$ \begin{aligned} y_{1}(x)=& 1+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{\alpha(\alpha-2)(\alpha-4) \cdots(\alpha-2 m+2)(\alpha+1)(\alpha+3) \cdots(\alpha+2 m-1)}{(2 m) !} x^{2 m} \\ y_{2}(x)=& x+\sum_{m=1}^{\infty}(-1)^{m} \\ & \times \frac{(\alpha-1)(\alpha-3) \cdots(\alpha-2 m+1)(\alpha+2)(\alpha+4) \cdots(\alpha+2 m)}{(2 m+1) !} x^{2 m+1} \end{aligned} $$

Find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as \(x \rightarrow 0\). \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0, \quad y(1)=1, \quad y^{\prime}(1)=-1\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free