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Chapter 4: Problem 14

Let $$ f(x)=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}, \quad\left|x-x_{0}\right|

Short Answer

Expert verified
Question: Show that if \(f(t_n) = 0\) for a sequence \(t_n\) converging to \(x_0\) and \(t_n \neq x_0\) for a power series function \(f(x)\), then \(f(x) \equiv 0\) for \(|x - x_0| < R\). Answer: Using Rolle's theorem and analyzing the coefficients of the power series, we found that all coefficients \(a_n\) of powers of \((x - x_0)\) are zero for \(n \geq 1\), and consequently, \(f(x) = a_0(x - x_0)^0 = 0\) for \(|x - x_0| < R\). Therefore, the power series function \(f(x)\) is identically zero in the interval \((x_0 - R, x_0 + R)\).

Step by step solution

01

Rewrite the condition for Rolle's Theorem to apply

Since we are given that \(f(t_n) = 0\) for each \(t_n\), we can conclude that \(f(x)\) is continuous in the interval \((x_0, t_n)\) for all \(n\). We also know that the function \(f(x)\) is differentiable in the interval since it is a power series. Therefore, applying Rolle's theorem, it must be that there exists some \(c_n \in (x_0, t_n)\) such that \(f'(c_n) = 0\).
02

Find the derivative of f(x)

First, let's find the derivative of the power series function \(f(x)\): $$ f'(x) = \sum_{n=1}^{\infty} n a_{n}(x - x_0)^{n - 1}. $$
03

Apply the derivative to c_n

Now let's evaluate \(f'(c_n)\) using the fact that \(f'(c_n) = 0\): $$ 0 = f'(c_n) = \sum_{n=1}^{\infty} n a_{n}(c_n - x_0)^{n - 1}. $$
04

Take the limit as n approaches infinity

Taking the limit as \(n \rightarrow \infty\), we have: $$ 0 = \lim_{n \rightarrow \infty} f'(c_n) = \lim_{n \rightarrow \infty} \sum_{n=1}^{\infty} n a_{n} (c_n - x_0)^{n - 1}. $$ As \(t_n\) converges to \(x_0\), we know \(c_n\) will also converge to \(x_0\) since \(c_n \in (x_0, t_n)\).
05

Replace c_n with x_0 in the limit and use the derivative definition

Thus, we can replace \(c_n\) with \(x_0\) to obtain: $$ 0 = \lim_{n \rightarrow \infty} \sum_{n=1}^{\infty} n a_{n} (x_0 - x_0)^{n - 1} = \sum_{n=1}^{\infty} n a_{n} (0)^{n - 1}. $$
06

Simplify and show f(x) is identically zero

Now let's simplify the final expression: $$ 0 = a_1 + 0 + 0 + \dots = a_1. $$ Thus, the coefficients \(a_n\) of all powers of \((x-x_0)\) in the power series are zero for \(n \geq 1\), and we have: $$ f(x) = a_0(x - x_0)^0 = 0, $$ for \(|x-x_0| < R\). Therefore, \(f(x) \equiv 0\) in the interval \((x_0 - R, x_0 + R)\).

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Key Concepts

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

Power Series
A power series is a type of infinite series that is composed of terms that include powers of a variable, typically written as
  • \( \sum_{n=0}^{\infty} a_n (x - x_0)^n \)
Here, \((x - x_0)^n\) is the variable raised to the \(n^{th}\) power, and \(a_n\) are the coefficients for each term. A power series is centered at \(x_0\), meaning the series behaves in some specific way near that point.
Key points about power series include:
  • They converge within a radius of convergence \(R\), within which they behave like a standard polynomial function and are infinitely differentiable.
  • Outside this radius, the behavior of the series can be unpredictable or divergent.
  • They are very useful in calculus and analysis as they provide approximations for functions, especially for computations and analytical studies.
Understanding power series is crucial for solving problems involving infinite series and polynomials in calculus, such as the problem discussed here, where power series representation of a function and its properties, like convergence, are central themes.
Convergence
Convergence is an essential concept when working with power series. Simply put, a series converges if the sum of its terms approaches a specific value as the number of terms goes to infinity. For power series,
  • The series converges within a specific interval, known as the radius of convergence \((|x-x_0|
This radius defines where the series will behave like a well-defined and smooth function.
The properties of convergence are:
  • Inside the interval, the power series sum is finite and you can perform calculus operations like differentiation and integration term by term.
  • At the ends of the interval, the series may or may not converge, thus requiring specific checks depending on the series itself.
In our exercise, understanding convergence helps in employing Rolle's Theorem effectively. As points \(t_n\) converge to \(x_0\), it aids in proving that the function represented by the power series is zero across the interval of convergence.
Derivative of a Function
The derivative of a function is a core concept in calculus that measures how a function changes as its input changes. When dealing with power series, finding the derivative is straightforward because you can differentiate the series term by term:
  • For the function \( f(x) = \sum_{n=0}^{\infty} a_n (x - x_0)^n \), the derivative \( f'(x) \) is given by: \[ f'(x) = \sum_{n=1}^{\infty} n a_n (x - x_0)^{n-1}. \]
When we apply Rolle's Theorem in our problem, the derivative helps identify points within the interval where the function's rate of change is zero, implying a horizontal tangent line or a local extremum.
Key facts about derivatives in this context are:
  • They indicate how the function behaves locally around a point \(x_0\).
  • They are used to show that all the coefficients \(a_n\) of terms in the series must be zero when conditions are met, proving that \(f(x)\) is identically zero.
This use of derivatives not only validates assumptions made about our function but also highlights how calculus concepts are pivotal in analyzing and solving complex mathematical problems.

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Most popular questions from this chapter

Define (if necessary) the given function so as to be continuous at \(x_{0}=0,\) and find the first four nonzero terms of its Maclaurin series. (a) \(\frac{x e^{x}}{\sin x}\) (b) \(\frac{\cos x}{1+x+x^{2}}\) (c) \(\sec x\) (d) \(x \csc x\) (e) \(\frac{\sin 2 x}{\sin x}\)

Suppose that \(a_{n} \geq 0\) for \(n \geq m\) and \(\sum a_{n}=\infty .\) Prove: If \(N\) is an arbitrary integer \(\geq m\) and \(J\) is an arbitrary positive number, then \(\sum_{n=N}^{N+k} a_{n}>J\) for some positive integer \(k\).

Prove: If \(\sum a_{n}^{2}<\infty\) and \(\sum b_{n}^{2}<\infty,\) then \(\sum a_{n} b_{n}\) converges absolutely.

Determine convergence or divergence. (a) \(\sum \frac{(2 n) !}{2^{2 n}(n !)^{2}}\) (b) \(\sum \frac{(3 n) !}{3^{3 n} n !(n+1) !(n+3) !}\) (c) \(\sum \frac{2^{n} n !}{5 \cdots 7 \cdot(2 n+3)}\) (d) \(\sum \frac{\alpha(\alpha+1) \cdots(\alpha+n-1)}{\beta(\beta+1) \cdots(\beta+n-1)} \quad(\alpha, \beta>0)\)

(a) Prove: If \(\left\\{a_{n} r^{n}\right\\}\) is bounded and \(\left|x_{1}-x_{0}\right|R\), then \(\left\\{a_{n}\left(x_{1}-x_{0}\right)^{n}\right\\}\) is unbounded.
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