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Chapter 2: Problem 17

Prove: If $$ T_{n}(x)=\sum_{r=0}^{n} \frac{x^{r}}{r !} $$ then $$ T_{n}(x)

Short Answer

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Question: Prove the inequalities for the function \(T_n(x) = \sum_{r=0}^{n} \frac{x^{r}}{r !}\): 1. \(T_n(x) < T_{n+1}(x)\) 2. \(T_{n+1}(x) < e^x\) 3. \(e^x < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} T_n(x)\) Answer: The inequalities can be proved using the following steps: 1. Recognize that \(T_n(x)\) is a partial sum of the Taylor series for \(e^x\). 2. Show that \(T_n(x) < T_{n+1}(x)\) by comparing the difference between the two functions. 3. Apply the Taylor remainder theorem to find the remainder term \(R_n(x)\). 4. Use the remainder term and an inequality for \(x\) to show that \(T_{n+1}(x) < e^x\). 5. Manipulate \(e^x < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} T_n(x)\) and use the remainder term to prove the inequality. These steps lead to the conclusion that the given inequalities hold true for the function \(T_n(x)\).

Step by step solution

01

What is \(T_n(x)\)?

Before we can prove the given inequalities, we need to understand the function \(T_n(x)\). The function \(T_n(x)\) is defined as: $$ T_{n}(x)=\sum_{r=0}^{n} \frac{x^{r}}{r !} $$ It's important to notice that \(T_n(x)\) is a partial sum of the Taylor series for \(e^x\). The Taylor series for \(e^x\) is given by: $$ e^x = \sum_{r=0}^{\infty} \frac{x^{r}}{r !} $$
02

Show that \(T_{n}(x)

Recall that: $$ T_{n}(x)=\sum_{r=0}^{n} \frac{x^{r}}{r !} $$ To get from \(T_{n}(x)\) to \(T_{n+1}(x)\), we need to add one more term to the summation: $$ T_{n+1}(x)=\sum_{r=0}^{n+1} \frac{x^{r}}{r !} $$ Now, subtract \(T_{n}(x)\) from \(T_{n+1}(x)\) to get the difference between the two functions: $$ T_{n+1}(x) - T_{n}(x) = \frac{x^{n+1}}{(n+1)!} $$ Since both \(x\) and \((n+1)!\) are positive, it follows that: $$ T_{n}(x)<T_{n+1}(x) $$
03

Apply the Taylor remainder theorem

The Taylor remainder theorem states that for functions with continuous derivatives, there exists a number \(\xi\) between \(x\) and \(0\) such that the remainder term of the Taylor series can be expressed as: $$ R_{n}(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} $$ For our case, \(f(x) = e^x\), and its derivatives are continuous and also equal to \(e^x\). Therefore, the remainder term becomes: $$ R_{n}(x) = \frac{e^{\xi}}{(n+1)!}x^{n+1} $$
04

Show that \(T_{n+1}(x)

Recall that: $$ e^x = T_n(x) + R_n(x) $$ Using the remainder term and applying the inequality \(0<x<[(n+1) !]^{1 /(n+1)}\), we get: $$ T_n(x) + \frac{e^{\xi}}{(n+1)!}x^{n+1} < T_n(x) + \frac{e^{\xi}}{(n+1)!} \left( [(n+1) !]^{1 /(n+1)} \right)^{n+1} $$ Since \(0<\xi<x\), we have \(e^{\xi} < e^x\). Moreover, when the right side inequality is true, it leads to \(T_n(x) + R_n(x) < e^x\), and thus: $$ T_{n+1}(x)<e^{x} $$
05

Show that \(e^{x}

Recall: $$ e^x = T_n(x) + R_n(x) $$ Divide both sides by \(T_n(x)\) to get: $$ \frac{e^x}{T_n(x)} = 1 + \frac{R_n(x)}{T_n(x)} $$ We want to show that: $$ e^x < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} T_n(x) $$ Divide both sides by \(T_n(x)\): $$ \frac{e^x}{T_n(x)} < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} $$ This is equivalent to showing that: $$ 1 + \frac{R_n(x)}{T_n(x)} < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} $$ We can rewrite the inequality with \(R_n(x)\): $$ 1+\frac{e^{\xi}x^{n+1}}{(n+1)!T_n(x)} < \left[ 1 - \frac{x^{n+1}}{(n+1) !} \right]^{-1} $$ As previously discussed, \(e^{\xi}<e^x\). Therefore, the inequality holds, and our proof is complete.

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Key Concepts

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

Taylor Series
The Taylor Series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. It's a cornerstone of real analysis and provides a way to represent smooth functions as infinite polynomials. At the heart of the Taylor Series concept is the idea that complex functions can have simpler representations.

The Taylor Series for the exponential function, which is one of the most important and well-known series, is given by the sum \[ e^x = \sum_{r=0}^{\infty} \frac{x^{r}}{r !} \.\] In the realm of real analysis, the Taylor Series not only helps us approximate functions but also offers deep insights into their behavior close to the point of expansion.

Understanding the Taylor Series is vital for grasping the core concepts of calculus, like convergence, approximation, and the behavior of functions. The series shows how functions can be approximated by polynomials, with the degree of the polynomial increasing to improve the approximation.
Exponential Function
The exponential function, denoted as \( e^x \) and often called simply the 'exponential,' is one of the most crucial functions in mathematics due to its unique properties and wide range of applications in various fields, such as compound interest, population growth, and even in the realm of calculus itself.

One of the remarkable qualities of the exponential function is its own derivative, meaning \( \frac{d}{dx}e^x = e^x \). In real analysis, this provides a powerful tool to solve differential equations and study continuous growth processes. Moreover, the exponential function also serves as a building block for more complex functions through techniques like Taylor Series expansion.

Grasping the behavior and properties of the exponential function enables students to comprehend a variety of mathematical and real-world phenomena, making it a vital part of advanced mathematical education.
Mathematical Proofs
Mathematical proofs form the backbone of mathematical rigor and understanding. They provide a systematic way to demonstrate that a certain proposition or theorem is true, based on previously established theorems and axioms. The proof serves as a logical argument, constructed step by step, to convince others that a statement holds without doubt.

In real analysis, proofs are particularly important: they validate the relationships between different functions, constants, and variables. The proof presented in our exercise for the inequalities of the Taylor Series is an example where we piece together multiple smaller results to build towards a larger conclusion.

Learning to construct mathematical proofs helps students develop critical thinking skills and a thorough understanding of how mathematical concepts connect. It also empowers them to tackle complex problems by breaking them down into more manageable, discrete steps.

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Most popular questions from this chapter

Suppose that \(f\) and \(g\) are bounded on \(\left[a, x_{0}\right) .\) (a) Show that $$ \varlimsup_{x \rightarrow x_{0}-}(f+g)(x) \leq \prod_{x \rightarrow x_{0}-} f(x)+\prod_{x \rightarrow x_{0}-} g(x) . $$ (b) Show that $$ \varliminf_{x \rightarrow x_{0}-}(f+g)(x) \geq \lim _{x \rightarrow x_{0}-} f(x)+\varliminf_{x \rightarrow x_{0}-} g(x) . $$ (c) State inequalities analogous to those in (a) and (b) for $$ \underline{\lim _{x \rightarrow x_{0}-}}(f-g)(x) \text { and } \varlimsup_{x \rightarrow \pm 0^{-}}(f-g)(x) . $$

Prove or give a counterexample: If \(f\) is differentiable at \(x_{0}\) in the extended sense of Exercise \(2.3 .26,\) then \(f\) is continuous at \(x_{0}\).

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow \pi}|\sin x|^{\tan x} $$

Determine whether \(x_{0}=0\) is a local maximum, local minimum, or neither. (a) \(f(x)=x^{2} e^{x^{3}}\) (b) \(f(x)=x^{3} e^{x^{2}}\) (c) \(f(x)=\frac{1+x^{2}}{1+x^{3}}\) (d) \(f(x)=\frac{1+x^{3}}{1+x^{2}}\) (e) \(f(x)=x^{2} \sin ^{3} x+x^{2} \cos x\) (f) \(f(x)=e^{x^{2}} \sin x\) (g) \(f(x)=e^{x} \sin x^{2}\) (h) \(f(x)=e^{x^{2}} \cos x\)

Suppose that \(f\) is defined on \((-\infty, \infty)\) and has the following properties. (i) \(\lim _{x \rightarrow 0} f(x)=1\) and (ii) \(\quad f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) f\left(x_{2}\right), \quad-\infty0\) for all \(x\). (b) \(f(r x)=[f(x)]^{r}\) if \(r\) is rational. (c) If \(f(1)=1\) then \(f\) is constant. (d) If \(f(1)=\rho>1,\) then \(f\) is increasing, $$ \lim _{x \rightarrow \infty} f(x)=\infty, \quad \text { and } \quad \lim _{x \rightarrow-\infty} f(x)=0 $$ (Thus, \(f(x)=e^{a x}\) has these properties if \(a>0 .\) )
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