/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Let \(c \in \mathbb{R}, r>0, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 53

Let \(c \in \mathbb{R}, r>0, f:(c-r, c+r) \rightarrow \mathbb{R}\), and \(n \in \mathbb{N}\) be such that \(f^{(n)}(c)\) exists. Show that \(\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1) !\right]}{h^{n}}=\frac{f^{(n)}(c)}{n !}\)

Short Answer

Expert verified
Using Taylor's theorem, we rewrite the expression inside the limit and evaluate it at \(x=c+h\). After subtracting the given expression and simplifying, we divide both sides by \(h^n\) and apply the limit as \(h \rightarrow 0\). Due to the properties of Taylor's theorem, the limit becomes \(\frac{f^{(n)}(c)}{n!}\), which proves the given statement.

Step by step solution

01

Rewrite the expression inside the limit using Taylor's theorem

Taylor's theorem states that any function f(x) can be approximated by a polynomial function centered at a point c, with a radius of convergence r. In this exercise, we are considering the function \(f:(c-r, c+r) \rightarrow \mathbb{R}\). Therefore, we can write \[f(x) = f(c) + f^{(1)}(c)(x-c) + \frac{f^{(2)}(c)}{2!}(x-c)^2 + \cdots + \frac{f^{(n)}(c)}{n!}(x-c)^n + R_n(x)\] where R_n(x) is the remainder term. This remainder term estimates the error when approximating the function by the nth-degree Taylor polynomial centered at c.
02

Evaluate the expression inside the limit at \(x=c+h\)

Now, we need to evaluate this expression at \(x=c+h\), which gives, \[f(c+h) = f(c) + f^{(1)}(c)h + \frac{f^{(2)}(c)}{2!}h^2 + \cdots + \frac{f^{(n)}(c)}{n!}h^n + R_n(c+h)\]
03

Subtract from the given expression and simplify

Now, we will subtract the given expression from the result obtained in step 2. After doing so, we obtain the following: \[\frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1)! \right]}{h^{n}} = \frac{f^{(n)}(c)}{n!}h^n + R_n(c+h)\]
04

Divide both sides by \(h^n\)

We need to divide both sides by \(h^n\) to isolate the limit term: \[\frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1)! \right]}{h^{n}} = \frac{f^{(n)}(c)}{n!} + \frac{R_n(c+h)}{h^n}\]
05

Apply the limit as \(h \rightarrow 0\)

Take the limit of both sides of the equation as \(h \rightarrow 0\): \[\lim_{h\to 0} \frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1)! \right]}{h^{n}} = \lim_{h\to 0} \left(\frac{f^{(n)}(c)}{n!} + \frac{R_n(c+h)}{h^n}\right)\] From the properties of Taylor's theorem, we know that as \(h \rightarrow 0\), \(R_n(c+h) \rightarrow 0\). Therefore, the limit becomes: \[\lim_{h\to 0} \frac{f(c+h)-f(c)-h f^{\prime}(c)-\cdots-h^{n-1}\left[f^{(n-1)}(c) /(n-1)! \right]}{h^{n}} = \frac{f^{(n)}(c)}{n!}\] Hence, the given limit is proven.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Limits in Calculus
The concept of limits is foundational to calculus, representing the value that a function approaches as the input (or the variable) approaches some value. Limits are essential for defining both derivatives and integrals, the core operations of calculus. To understand limits, one must grasp the behavior of functions as they get close to a particular point鈥攅ven if they never actually reach that point.

For instance, in the problem provided, we are examining the limit as the variable, denoted by 'h,' approaches zero. This concept helps us to deduce what value the function tends towards, giving us insight into its behavior at a minute scale that cannot be achieved by simple observation. By establishing the limit, we can confirm the accuracy of our approximations like the Taylor polynomial in representing the behavior of a function around a specific point.
Calculus Derivatives
Calculus derivatives are the centerpiece of differential calculus, quantifying how a function changes as its input changes. Loosely speaking, a derivative represents the slope of the tangent line to the function at any given point. It's a measure of instantaneous rate of change, and calculating derivatives is a way to predict the behavior of a function.

Derivatives are notated by the prime symbol (') or the fraction involving 'd,' e.g., 'f'(x) or \( \frac{df}{dx} \). They play a crucial role in various branches of science and engineering, allowing us to find rates of change, maximum and minimum values of functions, and other characteristics of functional behavior.vIn our exercise, derivatives are pivotal because the Taylor polynomial is built using derivatives of the function at a specific point, up to the nth degree, to predict the function's behavior in a neighborhood around that point.
nth Derivative Test
The nth derivative test is a method used to analyze the behavior of functions based on higher-order derivatives. It is less common than the first and second derivative tests, which are frequently used to find local extremum points (maxima and minima) and concavity, respectively. However, the nth derivative test can offer insights into the function's trends and possible inflection points when considering higher degrees of change.

In the context of Taylor's theorem, we are concerned with the nth derivative because it features in the approximation polynomial of the function. The nth derivative at a point 'c' provides a coefficient for the nth term of the Taylor series. The exercise demonstrates how we involve this nth derivative in evaluating the limit as 'h' approaches zero, showcasing its importance in confirming the validity of a Taylor polynomial's approximation at a certain point.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the MVT to prove that for all \(n \in \mathbb{N}\) and \(a, b \in \mathbb{R}\) such that \(0

Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\). If there is \(\alpha \in \mathbb{R}\) such that \(\left|f^{\prime}(x)\right| \leq \alpha\left|g^{\prime}(x)\right|\) for all \(x \in(a, b)\) and if \(g^{\prime}(x) \neq 0\) for all \(x \in(a, b)\), then show that \(|f(b)-f(a)| \leq \alpha|g(b)-g(a)|\). Is the conclusion valid if the condition " \(g^{\prime}(x) \neq 0\) for all \(x \in(a, b) "\) is omitted?

Let \(I\) be an interval and \(f: I \rightarrow \mathbb{R}\) be continuous on \(I\) and differentiable at every interior point of \(I\). If there is a constant \(\alpha\) such that \(\left|f^{\prime}(x)\right| \leq \alpha\) for all interior points \(x\) of \(I\), then show that \(f\) is uniformly continuous on \(I\). Is the converse true? In other words, is it true that if \(f: I \rightarrow \mathbb{R}\) is uniformly continuous on \(I\) and differentiable at every interior point of \(I\), then there is a constant \(\alpha\) such that \(\left|f^{\prime}(x)\right| \leq \alpha\) for all interior points \(x\) of \(I ?\)

Let \(f:(0, \infty) \rightarrow \mathbb{R}\) satisfy \(f(x y)=f(x)+f(y)\) for all \(x, y \in(0, \infty) .\) If \(f\) is differentiable at 1 , show that \(f\) is differentiable at every \(c \in(0, \infty)\) and \(f^{\prime}(c)=f^{\prime}(1) / c .\) In fact, show that \(f\) is infinitely differentiable. If \(f^{\prime}(1)=2\), find \(f^{(n)}(3) .\)

Let \(f, g, h:[a, b] \rightarrow \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\). Show that there is \(c \in(a, b)\) such that the \(3 \times 3\) determinant $$ \left|\begin{array}{lll} f(a) & f(b) & f^{\prime}(c) \\ g(a) & g(b) & g^{\prime}(c) \\ h(a) & h(b) & h^{\prime}(c) \end{array}\right| $$ is zero, that is, \(f(a)\left[g(b) h^{\prime}(c)-h(b) g^{\prime}(c)\right]-f(b)\left[g(a) h^{\prime}(c)-h(a) g^{\prime}(c)\right]+\) \(f^{\prime}(c)[g(a) h(b)-h(a) g(b)]=0 .\) Deduce that if \(h(x)=1\) for all \(x \in[a, b]\) we obtain the conclusion of Cauchy's Mean Value Theorem (Proposition \(4.36\) ). What does the result say if \(g(x)=x\) and \(h(x)=1\) for all \(x \in[a, b] ?\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.