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In each exercise, functions \(f(t)\) and \(g(t)\) are given. The functions \(f(t)\) and \(g(t)\) are defined by a power series that converges in \(-R

Short Answer

Expert verified
Answer: The largest value of R for which both functions \(f(t)\) and \(g(t)\) converge is \(R=1\). The first four terms of the power series for each function are as follows: \(f(t) + g(t) = 4t^2\) \(f(t) - g(t) = 2t+6t^3\) \(f'(t) = 1 + 4t + 9t^2 + 16t^3 + \cdots\) \(f''(t) = 2 + 6t + 12t^2 + \cdots\)

Step by step solution

01

Convergence of the Power Series

To find the largest value of \(R\) for which both \(f(t)\) and \(g(t)\) converge, we use the ratio test. This states that if \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\), the series converges, and if the limit is greater than 1, it diverges. For \(f(t)\), we have \(a_n=nt^n\). The ratio is: $$ \lim_{n\to\infty}\left|\frac{(n+1)t^{n+1}}{nt^n}\right| = \lim_{n\to\infty}|t|\left|\frac{n+1}{n}\right| = |t|. $$ Thus, \(f(t)\) converges if \(|t|<1\). For \(g(t)\), we have \(a_n=(-1)^n nt^n\). The ratio is: $$ \lim_{n\to\infty}\left|\frac{(-1)^{n+1}(n+1)t^{n+1}}{(-1)^n nt^n}\right| = \lim_{n\to\infty}|t|\left|\frac{n+1}{n}\right| = |t|. $$ Thus, \(g(t)\) also converges if \(|t|<1\). So, the largest value of \(R\) is \(R=1\). Both functions converge when \(-R<t-t_0<R\), which is \(-1<t-t_0<1\).
02

Finding the Terms for \(f(t)\) and \(g(t)\)

For the first four terms of \(f(t)\), we find: $$ f(t) = \sum_{n=0}^{\infty} n t^{n} = 0 + 1t + 2t^2 + 3t^3 + \cdots $$ For the first four terms of \(g(t)\), we find: $$ g(t) = \sum_{n=0}^{\infty}(-1)^{n} nt^{n} = 0 - 1t + 2t^2 - 3t^3 + \cdots $$
03

Finding the Terms for \(f(t)+g(t)\) and \(f(t)-g(t)\)

Adding the first four terms of \(f(t)\) and \(g(t)\) (part b), we get their sum: $$ f(t) + g(t) = (0+0) + (1t-1t) + (2t^2+2t^2) + (3t^3-3t^3) = 4t^2 $$ Subtracting the first four terms of \(f(t)\) and \(g(t)\) (part c), we get their difference: $$ f(t) - g(t) = (0-0) + (1t+1t) + (2t^2-2t^2) + (3t^3+3t^3) = 2t+6t^3 $$
04

Finding the Terms for \(f'(t)\) and \(f''(t)\)

Taking the first derivative of \(f(t)\), we have: $$ f'(t) = \frac{d}{dt}\sum_{n=0}^{\infty} n t^{n} = \sum_{n=0}^{\infty} n^2 t^{n-1} = 1 + 4t + 9t^2 + 16t^3 + \cdots $$ Taking the second derivative of \(f(t)\), we have: $$ f''(t) = \frac{d^2}{dt^2}\sum_{n=0}^{\infty} n t^{n} = \sum_{n=0}^{\infty} n(n-1) t^{n-2} = 2 + 6t + 12t^2 + \cdots $$
05

Summary

The largest value of R for which both functions \(f(t)\) and \(g(t)\) converge is \(R=1\). The first four terms of the power series for each function are as follows: (a) \(f(t) = \sum_{n=0}^{\infty} n t^{n} = 0 + 1t + 2t^2 + 3t^3 + \cdots\) and \(g(t) = \sum_{n=0}^{\infty}(-1)^{n} nt^{n} = 0 - 1t + 2t^2 - 3t^3 + \cdots\) (b) \(f(t) + g(t) = 4t^2\) (c) \(f(t) - g(t) = 2t+6t^3\) (d) \(f'(t) = 1 + 4t + 9t^2 + 16t^3 + \cdots\) (e) \(f''(t) = 2 + 6t + 12t^2 + \cdots\)

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Key Concepts

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

Ratio Test
The ratio test is a powerful tool for determining the convergence of infinite series, particularly when the series involves factorials, exponentials, or other terms that grow quickly with n. It involves taking the limit of the absolute value of the ratio of consecutive terms in the series.

When applying the ratio test to a power series, we typically express the terms of the series as \(a_n\text{ and }a_{n+1}\). Then, we evaluate the following limit:\
\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\]
If this limit is less than 1, the series converges absolutely. If it's more than 1, the series diverges, and if it equals 1, the test is inconclusive. In our case, both the functions \(f(t)\) and \(g(t)\) have shown through the ratio test that they converge for \(R=1\), which simplifies decision-making regarding convergence for a given value of t.

It's crucial for students to practice working through the ratio test with various series to gain fluency in recognizing when and how it can be applied effectively.
Series Convergence
Understanding series convergence is at the heart of working with infinite series and sequences. Series convergence refers to whether the sum of the infinite series approaches a finite number as more terms are added. An important aspect of working with power series is identifying the interval, often called the radius of convergence (R), within which the series converges.

Here, both \(f(t)\) and \(g(t)\), represented with power series, converge for |t|<1. This means within the interval (-1,1), which is a vital piece of information when studying functions defined by power series. From a practical standpoint, recognizing convergence patterns helps predict behaviors of more complex functions in advanced calculus and various applications in physics and engineering.

Students should practice by solving numerous types of series to get comfortable with identifying convergence or divergence quickly.
Power Series Differentiation
Differentiating power series term by term is a process known as power series differentiation. Because power series converge to a differentiable function within their interval of convergence, we can differentiate them just like polynomials, applying standard differentiation rules to each term individually.

For example, to find \(f'(t)\) for our function \(f(t)\), we differentiate each term \(nt^n\) to get \(n^2t^{n-1}\). Similarly, for \(f''(t)\), we apply differentiation again to find \(n(n-1)t^{n-2}\). It is crucial to ensure that the radius of convergence remains unchanged or is recalculated after differentiation. This concept reinforces our understanding of the relationship between power series and their derivatives, highlighting the utility of power series in solving differential equations.
Power Series Terms
Breaking down a power series into its component power series terms is fundamental to working with these infinite expansions. Each term of a power series has the form \(a_n(t-t_0)^n\), where \(t_0\) is the center of the series, and \(a_n\) is the coefficient for the nth term.

In the exercises we've considered, the terms for \(f(t)\) are \(0, t, 2t^2, 3t^3, \ldots\) and for \(g(t)\) are \(0, -t, 2t^2, -3t^3, \ldots\). Recognizing and writing out these terms aids in understanding the behavior and properties of the function the series represents. Students should become adept at identifying and manipulating these terms, which includes operations like addition and subtraction of series, as well as series differentiation and integration, to get a comprehensive grasp of the topic.

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Most popular questions from this chapter

For the special case \(v=\frac{1}{2}\), consider the modified Bessel equation for \(t>0\), \(t^{2} y^{\prime \prime}+t y^{\prime}-\left(t^{2}+\frac{1}{4}\right) y=0\). (a) Define a new dependent variable \(u(t)\) by the relation \(y(t)=t^{-1 / 2} u(t)\). Show that \(u(t)\) satisfies the differential equation \(u^{\prime \prime}-u=0\). (b) Show that the differential equation has a fundamental set of solutions $$ \frac{\sinh t}{\sqrt{t}}, \quad \frac{\cosh t}{\sqrt{t}}, \quad t>0 . $$

In each exercise, (a) Verify that the given differential equation has a regular singular point at \(t=0\). (b) Determine the indicial equation and its two roots. (These roots are often called the exponents at the singularity.) (c) Determine the recurrence relation for the series coefficients. (d) Consider the interval \(t>0\). If the two exponents obtained in (c) are unequal and do not differ by an integer, determine the first two nonzero terms in the series for each of the two linearly independent solutions. If the exponents are equal or differ by an integer, obtain the first two nonzero terms in the series for the solution having the larger exponent. (e) When the given differential equation is put in the form \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), note that \(t p(t)\) and \(t^{2} q(t)\) are polynomials. Do the series, whose initial terms were found in part (d), converge for all \(t, 0

Obtain the general solution of the given differential equation for \(0

As in Example 1, use the ratio test to find the radius of convergence \(R\) for the given power series. $$ \sum_{n=0}^{\infty} n^{3}(t-1)^{n} $$

Identify all the singular points of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) in the interval \(-10

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