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By shifting the index of summation as in equation \((9)\) or \((13)\), rewrite the given power series so that the general term involves \(t^{n}\). $$ \sum_{n=0}^{\infty}(-1)^{n+1}(n+1) a_{n} t^{n+2} $$

Short Answer

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Question: Rewrite the given power series so that the general term involves \(t^{n}\): \(\sum_{n=0}^{\infty}(-1)^{n+1}(n+2) a_nt^{n+2}\) Answer: \(\sum_{m=2}^{\infty}(-1)^{m-1}(m-1)a_{m-2}t^{m}\)

Step by step solution

01

Define a new variable.

Let's define a new variable \(m\) such that \(m = n + 2\). This will help us change the exponents of \(t\) in the general term to \(t^{n}\).
02

Express original variables in terms of the new variable.

Now in terms of \(m\), we have \(n = m - 2\), and the exponent of \(t\) becomes \(t^{m}\).
03

Adjust the limits of summation.

As \(n=0\) initially, for the lower limit, we have \(m=2\). For the upper limit, as \(n\) goes to infinity, we see that \(m\) will also go to infinity.
04

Rewrite the power series using the new variable and adjusted summation limits.

Now we can write the power series with the new variable \(m\) and adjusted limits: $$ \sum_{m=2}^{\infty}(-1)^{(m-2)+1}(m-1)a_{m-2}t^{m} $$
05

Simplify the expression and obtain the final result.

We can further simplify the expression inside the summation by combining the terms and simplifying the exponents. The final expression for the power series with the general term involving \(t^{n}\) is: $$ \sum_{m=2}^{\infty}(-1)^{m-1}(m-1)a_{m-2}t^{m} $$

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Key Concepts

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

Index of Summation
When we work with power series, the index of summation is crucial as it tracks which term of the series we're currently summing. Think of it as a placeholder representing each consecutive number in the series. In the given exercise, the index was initially expressed with the variable 'n' and the series involved terms like \(t^{n+2}\). To make the power series easier to work with, the problem suggests shifting the index of summation.
This 'shift' is a strategic move that simplifies the expression, and it's like a translation: if you picture the series on a number line, it's as if you're sliding it over to align with another point. Here, by redefining our index with a different base starting point, we are able to rewrite the series terms to involve \(t^{n}\) instead. It's a matter of convenience and readability—having \(t^{n}\) simply makes the series neater and often paves the way to compare it with known series or apply standard theorems.
Variable Substitution
The technique of variable substitution is like a secret passage in mathematics, allowing us to navigate complex expressions with greater ease.
In our exercise, a simple yet powerful trick was applied. A new variable \(m\) was introduced, which is essentially 'stepping into the shoes' of the original variable \(n\) plus 2. This new variable helps to reframe the problem, bridging the gap between the initial expression and a more digestible form.
With \(m = n + 2\), we sort of re-center our viewpoint. This not only affects the exponent of \(t\) but also the limits of the series—where our summation begins and ends. Variable substitution can be a best friend for anyone diving into power series; it brings clarity to potentially convoluted expressions and often reveals patterns that were not immediately observable.
Infinite Series
An infinite series is a sum that goes on forever—it's like a gift that keeps on giving, but in the world of math. It's formed by adding up an infinite sequence of numbers, and such series can converge to a finite value or diverge to infinity.
In the context of our exercise, we are dealing with an infinite power series, a particular type of series where each term is a power of a variable multiplied by a coefficient. These series are incredibly important in various fields, from physics to finance, because they can represent functions in a form that's often easier to work with.
Determining what an infinite series converges to, if it does converge, is a pivotal concept in calculus. Some series are fairly tame and will calmly settle down to a single value, while others will wildly race off toward infinity. It's like a saga with each term adding to the storyline; understanding the nature of these series opens up a new lens through which we can examine functions and model real-world phenomena.

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Most popular questions from this chapter

By shifting the index of summation as in equation \((9)\) or \((13)\), rewrite the given power series so that the general term involves \(t^{n}\). $$ \sum_{n=0}^{\infty} 2^{n} t^{n+2} $$

In each exercise, determine the polynomial \(P(t)\) of smallest degree that causes the given differential equation to have the stated properties. \(\begin{array}{ll}y^{\prime \prime}+\frac{1}{t} y^{\prime}+\frac{1}{P(t)} y=0 & \begin{array}{l}\text { There is a regular singular point at } t=0 . \text { All other } \\ \text { points are ordinary points. }\end{array}\end{array}\)

Exercises 31 and 32 outline the proof of parts (a) and (b) of Theorem 8.2, respectively. In each exercise, consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), where \(p\) and \(q\) are continuous on the domain \((-b,-a) \cup(a, b), a \geq 0\). Now let \(p\) and \(q\) be analytic at \(t=0\) with a common radius of convergence \(R>0\), where \(p\) is an odd function and \(q\) is an even function. (a) Let \(f_{1}(t)\) and \(f_{2}(t)\) be solutions of the given differential equation, satisfying initial conditions \(f_{1}(0)=1, f_{1}^{\prime}(0)=0, f_{2}(0)=0, f_{2}^{\prime}(0)=1\). What does Theorem \(8.1\) say about the solutions \(f_{1}(t)\) and \(f_{2}(t)\) ? (b) Use the results of Exercise 31 to show that \(f_{1}(-t)\) and \(f_{2}(-t)\) are also solutions on the interval \(-R

By shifting the index of summation as in equation \((9)\) or \((13)\), rewrite the given power series so that the general term involves \(t^{n}\). $$ \sum_{n=2}^{\infty} n(n-1) a_{n} t^{n-2} $$

In each exercise, find the singular points (if any) and classify them as regular or irregular. $$ \left(t^{2}-1\right) y^{\prime \prime}+(t-1) y^{\prime}+y=0 $$

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