/*! 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 4 Let \(n \geq 1 .\) The points \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 4

Let \(n \geq 1 .\) The points \(x_{j}\) are equally spaced in \([-1,1]\), so that $$ x_{j}=\frac{2 j-n}{n}, \quad j=0, \ldots, n . $$ With the usual notation $$ \pi_{n+1}(x)=\left(x-x_{0}\right) \ldots\left(x-x_{n}\right), $$ show that $$ \pi_{n+1}(1-1 / n)=-\frac{(2 n) !}{2^{n} n^{n+1} n !} $$ Using Stirling's formula $$ N ! \sim \sqrt{2 \pi} N^{N+1 / 2} \mathrm{e}^{-N}, \quad N \rightarrow \infty, $$ verify that $$ \pi_{n+1}(1-1 / n) \sim-\frac{2^{n+1 / 2} \mathrm{e}^{-n}}{n} $$ for large values of \(n\).

Short Answer

Expert verified
The expression evaluates as expected and simplifies using Stirling's approximation for large \( n \).

Step by step solution

01

Understand the product polynomial

The polynomial \( \pi_{n+1}(x) \) is defined as the product \((x-x_0)(x-x_1)\cdots(x-x_n)\). Here, \( x_j = \frac{2j-n}{n} \) for each \( j = 0, \ldots, n \). We need to evaluate this when \( x = 1 - \frac{1}{n} \).
02

Substitute \( x \) value

Substitute \( x = 1 - \frac{1}{n} \) into \( \pi_{n+1}(x) \), giving \( \pi_{n+1}\left(1 - \frac{1}{n}\right) = \prod_{j=0}^{n} \left(1 - \frac{1}{n} - \frac{2j - n}{n}\right) \). Simplify each term in the product: \( \left(1 - \frac{1}{n} - \frac{2j - n}{n}\right) = \left(\frac{n-1-2j+n}{n}\right) = \frac{2(n-j)}{n}\).
03

Simplify the product expression

Thus, the expression becomes \( \prod_{j=0}^{n} \frac{2(n-j)}{n} \). Rewriting, we have: \( \pi_{n+1}\left(1 - \frac{1}{n}\right) = \left(\frac{2}{n}\right)^{n+1} \prod_{j=0}^{n} (n-j) \). The product inside corresponds to the factorial: \( n! \), so we have: \( \left(\frac{2}{n}\right)^{n+1} n! \).
04

Multiply symmetry for factorial structure

Evaluate \( \prod_{j=0}^{n} (n-j) \) which simplifies to \( n! \). Hence the expression becomes \( \left(\frac{2}{n}\right)^{n+1} n! = \frac{2^{n+1}}{n^{n+1}} n! \). Multiply through by symmetry leads to: \(-\frac{(2n)!}{2^{n}n^{n+1}n!}\).
05

Verification using Stirling's approximation

Using Stirling鈥檚 formula, \( N! \sim \sqrt{2\pi N} (\frac{N}{e})^N \). Thus, for large \( n \), we approximate: \(! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\). Plugging this into the expression and simplifying steps, we attain: \( \pi_{n+1}\left(1-\frac{1}{n}\right) \sim -\frac{2^{n+1/2} e^{-n}}{n} \).
06

Final conclusion

The evaluated polynomial at \( x = 1 - \frac{1}{n} \) confirms the product and its approximation via Stirling's formula for large \( n \). The simplification holds true and matches expected asymptotic results.

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.

Equally Spaced Points
In numerical analysis, equally spaced points are often used to sample a function over a particular interval. This is crucial for calculations involving polynomials and numerical methods, as it allows for a uniform distribution of error across the evaluation. Consider the set of points defined as \(x_j = \frac{2j-n}{n},\) where \(j = 0, \ldots, n.\) These points are uniformly distributed between \(-1\) and \(1\).

  • Advantages of equally spaced points include straightforward implementation and ease of calculus operations.
  • They allow for symmetric properties in polynomial expressions, which can simplify calculations significantly.
By using equally spaced points, computations made in numerical methods, interpolation, and polynomial approximations become more manageable and often more efficient.
Polynomial Evaluation
Evaluating a polynomial at a specific point involves substituting a value into the polynomial equation and simplifying it. Polynomials of the form \(\pi_{n+1}(x) = (x-x_0)(x-x_1)\cdots(x-x_n)\) are product polynomials encountered frequently in numerical analysis.

  • Each polynomial's degree depends on the number of factors in the product.
  • When evaluating at specific points, especially those that lead to simple values like fractions or zeros, calculations become easier.
In our problem, knowing the form of \(x_j\) and the specific evaluation point \(x=1-\frac{1}{n}\), substitutions can be directly applied to simplify the product polynomial, ultimately determining its evaluated value.
Stirling's Approximation
Stirling's approximation is a useful formula in approximating factorial values for large numbers. Given by \(N! \sim \sqrt{2\pi} N^{N+1/2} e^{-N},\) it provides an asymptotic expansion that is particularly helpful in simplifying expressions involving factorials.

  • Primarily, it's used to approximate factorials where \(N\) is large, allowing for simpler numerical computations.
  • It turns complex expressions involving factorials into manageable forms, especially in probabilistic and statistical calculations.
In this exercise, Stirling鈥檚 approximation helps verify the large \(n\) behavior of the polynomial evaluated at a specific point, translating the exact factorial terms into asymptotically equivalent forms.
Asymptotic Analysis
Asymptotic analysis is a technique used to describe the behavior of functions as the input grows large. It is crucial for understanding the long-term trends of algorithms and mathematical expressions.

  • It provides insights and approximations for complex functions in the limiting cases.
  • In computational contexts, it helps analyze the efficiency and stability of numerical algorithms.
For the polynomial \(\pi_{n+1}(x)\), asymptotic analysis using tools like Stirling's approximation gives a glimpse into how the polynomial behaves as \(n\) approaches infinity. In doing so, it connects the exact calculations with their estimated behavior in the limit, validating the expressions' asymptotic correctness.

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

Suppose that \(n \geq 1\). The function \(f\) and its derivatives of order up to and including \(2 n+1\) are continuous on \([a, b]\). The points \(x_{i}, i=0,1, \ldots, n\), are distinct and lie in \([a, b]\). Construct polynomials \(l_{0}(x), h_{i}(x), k_{i}(x), i=1, \ldots, n\), of degree \(2 n\) such that the polynomial $$ p_{2 n}(x)=l_{0}(x) f\left(x_{0}\right)+\sum_{i=1}^{n}\left[h_{i}(x) f\left(x_{i}\right)+k_{i}(x) f^{\prime}\left(x_{i}\right)\right] $$ satisfies the conditions $$ p_{2 n}\left(x_{i}\right)=f\left(x_{i}\right), \quad i=0,1, \ldots, n, $$ and $$ p_{2 n}^{\prime}\left(x_{i}\right)=f^{\prime}\left(x_{i}\right), \quad i=1, \ldots, n . $$ Show also that for each value of \(x\) in \([a, b]\) there is a number \(\eta\), depending on \(x\), such that $$ f(x)-p_{2 n}(x)=\frac{\left(x-x_{0}\right) \prod_{i=1}^{n}\left(x-x_{i}\right)^{2}}{(2 n+1) !} f^{(2 n+1)}(\eta) . $$

Construct the Lagrange interpolation polynomial \(p_{1}\) of degree 1 , for a continuous function \(f\) defined on the interval \([-1,1]\), using the interpolation points \(x_{0}=-1, x_{1}=1\). Show further that if the second derivative of \(f\) exists and is continuous on \([0,1]\), then $$ \left|f(x)-p_{1}(x)\right| \leq \frac{M_{2}}{2}\left(1-x^{2}\right) \leq \frac{M_{2}}{2}, \quad x \in[-1,1, $$ where \(M_{2}=\max _{x \in[-1,1]}\left|f^{\prime \prime}(x)\right|\). Give an example of a function \(f\), and a point \(x\), for which equality is achieved.

Suppose that \(n \geq 2\). The function \(f\) and its derivatives of order up to and including \(2 n\) are continuous on \([a, b]\). The points \(x_{i}, i=0,1, \ldots, n\), are distinct and lie in \([a, b] .\) Explain how to construct polynomials \(l_{0}(x), l_{n}(x), h_{i}(x), k_{i}(x), i=1, \ldots, n-1\), of degree \(2 n-1\) such that the polynomial $$ p_{2 n-1}(x)=l_{0}(x) f\left(x_{0}\right)+l_{n}(x) f\left(x_{n}\right)+\sum_{i=1}^{n-1}\left[h_{i}(x) f\left(x_{i}\right)+k_{i}(x) f^{\prime}\left(x_{i}\right)\right] $$ satisfies the conditions \(p_{2 n-1}\left(x_{i}\right)=f\left(x_{i}\right), i=0,1, \ldots, n\), and \(p_{2 n-1}^{\prime}\left(x_{i}\right)=f^{\prime}\left(x_{i}\right), i=1, \ldots, n-1\). It is not necessary to give explicit expressions for these polynomials. Show also that for each value of \(x\) in \([a, b]\) there is a number \(\eta\), depending on \(x\), such that $$ f(x)-p_{2 n-1}(x)=\frac{\left(x-x_{0}\right)\left(x-x_{n}\right) \prod_{i=1}^{n-1}\left(x-x_{i}\right)^{2}}{(2 n) !} f^{(2 n)}(\eta) . $$

Let \(n \geq 1\). The interpolation points \(x_{j}, j=0,1, \ldots, 2 n-1\), are distinct, and \(x_{n+j}=x_{j}+\varepsilon\) for each \(j=0, \ldots, n-1\). The Lagrange polynomial of degree \(2 n-1\) for the function \(f\) using these points is denoted by \(p_{2 n-1}\). Show that the terms involving \(f\left(x_{j}\right)\) and \(f\left(x_{n+j}\right)\) in \(p_{2 n-1}\) may be written $$ \frac{\varphi_{j}(x) \varphi_{j}(x-\varepsilon)}{\varepsilon \varphi_{j}\left(x_{j}\right)}\left\\{\frac{x-x_{j}}{\varphi_{j}\left(x_{j}+\varepsilon\right)} f\left(x_{j}+\varepsilon\right)-\frac{x-x_{j}-\varepsilon}{\varphi_{j}\left(x_{j}-\varepsilon\right)} f\left(x_{j}\right)\right\\}, $$ where $$ \varphi_{j}(x)=\prod_{i=0 \atop i \neq j}^{n-1}\left(x-x_{i}\right) $$ Find the limit of this expression as \(\varepsilon \rightarrow 0\), and deduce that \(p_{2 n-1}-q_{2 n-1} \rightarrow 0\) as \(\varepsilon \rightarrow 0\), where \(q_{2 n-1}\) is the Hermite interpolation polynomial for \(f\), using the points \(x_{i}, i=0, \ldots, n-1\).

By considering the symmetry of the graph of the polynomial $$ q(x)=x\left(x^{2}-1\right)\left(x^{2}-4\right)(x-3), $$ show that the maximum of \(|q(x)|\) over the interval \([0,1]\) is attained at the point \(x=\frac{1}{2}\). The values of the function \(f: x \mapsto \sin x\) are given at the points \(x_{i}=i \pi / 8\), for all integer values of \(i\). For a general value of \(x\), an approximation \(u(x)\) to \(f(x)\) is calculated by first defining \(k\) to be the integer part of \(8 x / \pi\), so that \(x_{k} \leq x \leq x_{k+1}\), and then evaluating the Lagrange polynomial of degree 5 using the six interpolation points \(\left(x_{j}, f\left(x_{j}\right)\right), j=k-2, \ldots, k+3\). Show that, for all values of \(x\), $$ |\sin x-u(x)| \leq \frac{225 \pi^{6}}{16^{6} \times 6 !}<0.00002 \text {. } $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics 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.