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Chapter 9: Problem 11

Prove that if \(f\) is defined for \(|x|

Short Answer

Expert verified
The Taylor series \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n}\) of the function \(f(x)\), under the conditions stated in the exercise, does converge to \(f(x)\) for \(|x|<r\). This convergence is confirmed using the Weierstrass M-Test, which allows us to determine the uniform convergence of the remainder term to \(0\), indicating that the Taylor series is indeed a good approximation of the original function within the given interval.

Step by step solution

01

Identify the Taylor series and function conditions

From the problem, it is given that the function \(f\) is defined for all \(|x|<r\) and has a constant \(B\) such that \(\left|f^{(n)}(x)\right| \leq B\) for all \(|x|<r\) and \(n \in \mathbb{N}\). The provided Taylor series for this is \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n}\)
02

Apply the Taylor's Theorem

The Taylor's Theorem provides an estimate of the difference between the function and its approximation by a polynomial. Here it is presented as \(f(x) - T_{n}(x) = R_{n}(x)\), where \(T_{n}(x)\) is the Taylor series and \(R_{n}(x)\) is the remainder term. The task is to show that \(R_{n}(x) \rightarrow 0\) as \(n \rightarrow \infty\), meaning the difference between the function \(f(x)\) and the Taylor approximation diminishes to zero, implying that the series converges to \(f(x)\)
03

Compute the remainder term

The remainder term in the Taylor series, \(R_{n}(x)\), is given by the equation: \(R_{n}(x) = \frac{f^{(n+1)}(z)}{(n+1)!}x^{n+1}\) where \(z\) is some number between \(0\) and \(x\). Given that \(\left|f^{(n)}(x)\right| \leq B\) for all \(|x|<r\) and \(n \in \mathbb{N}\), we can say that \(\left|R_{n}(x)\right| \leq \frac{B|r|^{n+1}}{(n+1)!}\)
04

Apply the Weierstrass M-Test

The Weierstrass M-Test is used to prove the convergence of a series of functions. The test says: If there are positive constants \(M_k\) such that \(\left|f_k(x)\right|\leq M_k\) for all \(k\) and all \(x\), and if the series \(\sum M_k\) converges, then the series consisting of these functions \(\sum f_k(x)\) converges uniformly. In our problem, \(M_{n} = \frac{B|r|^n}{n!}\) is the series of constants for \(\left|R_{n}(x)\right|\) and \(\sum_{n=0}^{\infty} M_{n}\) converges because it is a convergent power series. Consequently, the series of remainders \(\sum_{n=0}^{\infty} R_{n}(x)\) converges uniformly to \(0\) inside the radius of convergence \(r\)
05

Conclude the convergence of the Taylor series

The Taylor series \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n}\) converges uniformly to \(f(x)\) inside the interval \(|x|<r\) based on the Weierstrass M-Test because the remainder series goes to \(0\). This is the expected result

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Key Concepts

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

Convergence of Series
In mathematics, convergence of a series refers to the process where partial sums of the series approach a specific value as more terms are added. Understanding whether a series converges is crucial because it tells us if adding infinitely many terms will yield a finite number. For a series to converge, its sequence of partial sums must approach a limit.
  • This means that as we consider larger and larger segments of the series, the sum gets closer to a particular number.
  • When a Taylor series converges to a function, it indicates that the polynomial terms together reproduce the function within a specific interval.
In terms of Taylor series, convergence is about ensuring that the function can be expressed as an infinite sum of its derivatives at a point, scaled by factorial terms and powers of the input, remaining within its radius of convergence.
Weierstrass M-Test
The Weierstrass M-Test is a powerful tool for determining the uniform convergence of series of functions. Uniform convergence is a stronger form of convergence where the speed at which the differences between partial sums and the function approach zero is consistent regardless of the input.
  • The Weierstrass M-Test states that if there exists a sequence of constants \(M_k\) such that \(|f_k(x)| \leq M_k\) for each component function \(f_k(x)\), and if \(\sum M_k\) converges, then the series \(\sum f_k(x)\) converges uniformly.
  • This test is very useful because it allows us to transfer the convergence properties from a simpler series \(\sum M_k\) to the more complex function series \(\sum f_k(x)\).
In the case of Taylor series, applying the Weierstrass M-Test helps in proving that the series of remainders converges to zero uniformly, allowing for the main Taylor series to uniformly converge to \(f(x)\) within its noted interval.
Taylor's Theorem
Taylor's Theorem is a fundamental concept that provides insights into approximating functions using polynomials. This theorem not only supports the creation of Taylor series but also gives us a way to measure how accurate these approximations are using the remainder term.
  • The theorem suggests that a function \(f(x)\) can be expressed as a Taylor series, consisting of its derivatives evaluated at a specific point, scaled accordingly.
  • More formally, it states \(f(x) = T_n(x) + R_n(x)\), where \(T_n(x)\) is the truncated Taylor series up to degree \(n\), and \(R_n(x)\) is the remainder that accounts for the difference.
  • Knowing the remainder term \(R_n(x)\) is vital because it tells us how far off the polynomial truncation is from the real function.
Through Taylor's Theorem, one can show that this remainder term becomes insignificant typically within the radius of convergence \(r\), meaning the series appropriately models \(f(x)\) as more terms are included.

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Most popular questions from this chapter

Discuss the convergence or the divergence of the series with \(n\) th term: (a) \(2^{n} e^{-n}\), (b) \(n^{n} e^{-n}\) (c) \(e^{-\ln n}\) (d) \((\ln n) e^{-\sqrt{n}}\) (e) \(n ! e^{-n}\) (f) \(n ! e^{-n^{2}}\).

The preceding exercise may fail if the terms are not positive. For example, let \(a_{i j}:=+1\) if \(i-j=\) \(1, a_{i j}:=-1\) if \(i-j=-1\), and \(a_{i j}:=0\) elsewhere. Show that the iterated sums $$ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{i j} \quad \text { and } \quad \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_{i j} $$ both exist but are not equal.

Discuss the series whose \(n\) th term is (a) \(\frac{n !}{3 \cdot 5 \cdot 7 \cdots(2 n+1)}\) (b) \(\frac{(n !)^{2}}{(2 n) !}\), (c) \(\frac{2 \cdot 4 \cdots(2 n)}{3 \cdot 5 \cdots(2 n+1)}\) (d) \(\frac{2 \cdot 4 \cdots(2 n)}{5 \cdot 7 \cdots(2 n+3)}\).

Show that the series \(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}++-\cdots\) is divergent.

Show that if a series is conditionally convergent, then the series obtained from its positive terms is divergent, and the series obtained from its negative terms is divergent.
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