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In Problems \(1-4\), find the radius of convergence and interval of convergence for the given power series. $$ \sum_{k=0}^{\infty} k !(x-1)^{k} $$

Short Answer

Expert verified
The radius of convergence is 0, and the interval of convergence is [1, 1].

Step by step solution

01

Identify the given power series

The power series given is \( \sum_{k=0}^{\infty} k! (x-1)^k \). This is a series of the form \( \sum_{k=0}^{\infty} a_k (x-c)^k \) where \( a_k = k! \) and \( c = 1 \).
02

Apply the Ratio Test

To find the radius of convergence, apply the ratio test. The ratio test involves finding \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Substituting \( a_k = k! \), we have: \( \frac{a_{k+1}}{a_k} = \frac{(k+1)!}{k!} = (k+1) \).
03

Evaluate the Ratio Test

Evaluate the limit: \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} (k+1) = \infty \). Because the series's limit is infinite, the radius of convergence \( R = 0 \).
04

Determine the interval of convergence

Since the radius of convergence is 0, the series converges only at \( x - 1 = 0 \), which means \( x = 1 \). So, the interval of convergence is \([1, 1]\).

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Key Concepts

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

Interval of Convergence
The interval of convergence is crucial in understanding where a power series converges. In the context of a power series, it refers to the range of values for which the series converges to a limit. To find this interval, it's essential to determine both the radius of convergence and evaluate the behavior of the series at the endpoints of this interval.

Since the series given has a radius of convergence of zero, as calculated from the previous solution, the series only converges at a single point: where the center of the series is located. In this specific example, the center is at \( x = 1 \). Hence, the interval of convergence is rather trivial—just \([1, 1]\). This means the series only converges when \( x \) equals exactly 1 and not at any other point.
Ratio Test
The Ratio Test is a powerful tool for finding the radius of convergence of a power series. When applied, it allows us to assess the behavior of the series as it extends towards infinity. The main idea involves evaluating the limit of the absolute value of the ratio of consecutive terms as \( k \) approaches infinity.

In mathematical form, the Ratio Test considers \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). For the given series, with \( a_k = k! \), this simplifies to \( (k+1) \), since \( \frac{(k+1)!}{k!} = (k+1) \).

Because this limit is infinite, the Ratio Test tells us that the series diverges for any \( x eq 1 \). Thus, the radius of convergence is zero, indicating convergence at one specific point: \( x = 1 \). The Ratio Test, in this context, confirms that for any departure from this point, the series will not converge.
Power Series
A power series is a series of the form \( \sum_{k=0}^{\infty} a_k (x-c)^k \) where \( a_k \) are coefficients and \( c \) is the center of the series. Such series are fundamental in representing functions in calculus because they allow complex functions to be expressed as an infinite polynomial.

In this exercise, the power series under discussion is \( \sum_{k=0}^{fty} k! (x-1)^k \), with the center \( c = 1 \) and coefficients \( a_k = k! \). This expansion means the series is expressed in terms of \( (x-1)^k \).

Power series can closely approximate functions within their interval of convergence, making them useful in practical applications such as solving differential equations, modeling in physics, and many areas of engineering. However, it's important to calculate the interval where they converge to prevent misuse or misinterpretation of the series beyond its valid range.
Convergence Analysis
Conducting convergence analysis involves determining where and how a series converges. This process often taps into theorems and tests like the Ratio Test or Root Test. Each test provides insights on different aspects of the series' behavior.

For this power series \( \sum_{k=0}^{\infty} k! (x-1)^k \), convergence analysis helped establish that the series has a radius of convergence of zero, thanks to the infinite result of the ratio test. This indicates that the only value of \( x \) for which the series converges is at the center \( x = 1 \).

Convergence analysis is crucial not only for understanding single point convergence but also for situating the series functionality and its limit behavior in broader mathematical contexts. It helps in learning where predictions and models using power series are applicable, ensuring accurate scientific or mathematical outcomes.

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Most popular questions from this chapter

Use a root-finding application to find the zeros of \(P_{1}(x)\), \(P_{2}(x), \ldots, P_{7}(x)\). If the Legendre polynomials are built-in functions of your CAS, find the zeros of Legendre polynomials of higher degree. Form a conjecture about the location of the zeros of any Legendre polynomial \(P_{n}(x)\), and then investigate to see whether it is true.

\(x=0\) is a regular singular point of the given differential equation. Show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain two linearly independent series solutions about \(x=0\). Form the general solution on the interval \((0, \infty)\). $$ 9 x^{2} y^{\prime \prime}+9 x^{2} y^{\prime}+2 y=0 $$

(a) Find two power series solutions for \(y^{\prime \prime}+x y^{\prime}+y=0\) and express the solutions \(y_{1}(x)\) and \(y_{2}(x)\) in terms of summation notation. (b) Use a CAS to graph the partial sums \(S_{N}(x)\) for \(y_{1}(x)\). Use \(N=2,3,5,6,8,10 .\) Repeat using the partial sums \(S_{N}(x)\) for \(y_{2}(x)\). (c) Compare the graphs obtained in part (b) with the curve obtained using a numerical solver. Use the initial conditions \(y_{1}(0)=1, y_{1}^{\prime}(0)=0\), and \(y_{2}(0)=0, y_{2}^{\prime}(0)=1\). (d) Rexamine the solution \(y_{1}(x)\) in part (a). Express this series as an elementary function. Then use (5) of Section \(3.2\) to find a second solution of the equation. Verify that this second solution is the same as the power series solution \(y_{2}(x)\).

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. $$ \left(x^{3}-2 x^{2}+3 x\right)^{2} y^{\prime \prime}+x(x-3)^{2} y^{\prime}-(x+1) y=0 $$

In Problems 13 and 14 , verify by direct substitution that the given power series is a particular solution of the indicated differential equation. $$ y=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{2 n}(n !)^{2}} x^{2 n}, \quad x y^{\prime \prime}+y^{\prime}+x y=0 $$

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