/*! 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 16 Let \(f(x)=(x-1)^{10}, p=1\), an... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)=(x-1)^{10}, p=1\), and \(p_{n}=1+1 / n\). Show that \(\left|f\left(p_{n}\right)\right|<10^{-3}\) whenever \(n>1\) but that \(\left|p-p_{n}\right|<10^{-3}\) requires that \(n>1000\).

Short Answer

Expert verified
After evaluating \(f(p_n)\) and the difference between \(p\) and \(p_n\), it can be concluded that both the properties given in the problem are true. The function \(f(p_n)\) is less than \(10^{-3}\) when \(n > 1\), and the difference between \(p\) and \(p_n\) is less than \(10^{-3}\) when \(n > 1000\).

Step by step solution

01

Calculate \(f(p_n)\) as \(n\) goes to infinity

Start by substituting \(p_n\) into \(f(x)\), resulting in \(f(p_n) = \left(1 + \frac{1}{n} - 1\right)^{10} = \left(\frac{1}{n}\right)^{10}\). As \(n\) goes to infinity, \(\frac{1}{n}\) goes to 0, thus \(\left(\frac{1}{n}\right)^{10}\) is definitely less than \(10^{-3}\) if \(n>1\). This proves the statement that \(\left|f\left(p_{n}\right)\right| < 10^{-3}\) whenever \(n>1\) is correct.
02

Calculate the difference between \(p\) and \(p_n\)

Now we will look at the absolute difference between \(p\) and \(p_n\). Mathematically, it's written as \(\left|p - p_n\right| = \left|1 - \left(1 + \frac{1}{n}\right)\right| = \frac{1}{n}\). By setting this less than \(10^{-3}\), we find that \(n > 1000\). This verifies the statement that \(\left|p-p_{n}\right| < 10^{-3}\) requires that \(n > 1000\).
03

Conclusion

Thus, both the statements given in the problem are true. For the first part, as \(n\) increases, the function \(f(p_n)\) reduces and eventually goes to zero which is certainly less than \(10^{-3}\). For the second part, the difference between \(p\) and \(p_n\) will be less than \(10^{-3}\) only when \(n\) is more than 1000 as we found out from the calculation.

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Key Concepts

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

Convergence
In numerical analysis, convergence is a fundamental concept that describes how a sequence or series approaches a specific value or state. In the context of our original exercise, we studied the sequence \( p_n = 1 + \frac{1}{n} \). As \( n \) becomes very large, this sequence converges to the value \( p = 1 \). Understanding convergence helps us determine how closely the sequence approaches the desired value without necessarily reaching it exactly.
  • As \( n \to \infty \), the term \( \frac{1}{n} \to 0 \), making the expression \( p_n \to 1 \).
  • This convergence assures that for a sufficiently large \( n \), the difference between \( p_n \) and \( p \) becomes negligible.
This is tied to the first part of our exercise, where the convergence of \( f(p_n) = \left(\frac{1}{n}\right)^{10} \) toward zero as \( n \to \infty \) ensures it remains less than \( 10^{-3} \) for \( n > 1 \). Understanding convergence is crucial in numerical analysis as it provides insights into the behavior and reliability of numerical methods.
Error Bound
Error bound provides a quantitative measure of the accuracy of an approximation or numerical solution, indicating how far off our estimate could possibly be from the true value. In our specific case, the error bound is concerned with the inequality \( |p - p_n| < 10^{-3} \). This tells us how close the terms of our sequence come to the limit in absolute terms.
  • The inequality \( |p - p_n| = \left|1 - \left(1 + \frac{1}{n}\right)\right| = \frac{1}{n} \).
  • To ensure \( \frac{1}{n} < 10^{-3} \), it requires \( n > 1000 \).
This threshold of \( n > 1000 \) acts as our error bound, which guarantees a specific degree of precision. Error bounds are essential in computational mathematics as they help quantify and control numerical errors, ensuring our solutions are within acceptable limits.
Mathematical Proof
Mathematical proof is the rigorous demonstration that a particular statement or theorem is true. In this exercise, we used mathematical proof to verify the original conditions provided:- For \( |f(p_n)| < 10^{-3} \), we substituted to find \( f(p_n) = \left(\frac{1}{n}\right)^{10} \) and showed it becomes sufficiently small when \( n > 1 \).- Likewise, for \( |p - p_n| < 10^{-3} \), by evaluating \( \frac{1}{n} < 10^{-3} \), it was proven that \( n > 1000 \).Mathematical proofs use defined steps, logical reasoning, and well-understood properties from mathematics to establish the truth within its scope. It is a vital tool in mathematics and numerical analysis, facilitating a formal method to confirm or refute proposed claims with exactness and clarity.

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Most popular questions from this chapter

A can in the shape of a right circular cylinder is to be constructed to contain \(1000 \mathrm{~cm}^{3}\). The circular top and bottom of the can must have a radius of \(0.25 \mathrm{~cm}\) more than the radius of the can so that the excess can be used to form a seal with the side. The sheet of material being formed into the side of the can must also be \(0.25 \mathrm{~cm}\) longer than the circumference of the can so that a seal can be formed. Find, to within \(10^{-4}\), the minimal amount of material needed to construct the can.

An object falling vertically through the air is subjected to viscous resistance as well as to the force of gravity. Assume that an object with mass \(m\) is dropped from a height \(s_{0}\) and that the height of the object after \(t\) seconds is $$ s(t)=s_{0}-\frac{m g}{k} t+\frac{m^{2} g}{k^{2}}\left(1-e^{-k t / m}\right) $$ where \(g=32.17 \mathrm{ft} / \mathrm{s}^{2}\) and \(k\) represents the coefficient of air resistance in \(\mathrm{lb}-\mathrm{s} / \mathrm{ft}\). Suppose \(s_{0}=300 \mathrm{ft}\), \(m=0.25 \mathrm{lb}\), and \(k=0.1 \mathrm{lb}-\mathrm{s} / \mathrm{ft}\). Find, to within \(0.01 \mathrm{~s}\), the time it takes this quarter-pounder to hit the ground.

Use Newton's method to solve the equation $$ 0=\frac{1}{2}+\frac{1}{4} x^{2}-x \sin x-\frac{1}{2} \cos 2 x, \quad \text { with } p_{0}=\frac{\pi}{2} $$ Iterate using Newton's method until an accuracy of \(10^{-5}\) is obtained. Explain why the result seems unusual for Newton's method. Also, solve the equation with \(p_{0}=5 \pi\) and \(p_{0}=10 \pi\).

Use algebraic manipulation to show that each of the following functions has a fixed point at \(p\) precisely when \(f(p)=0\), where \(f(x)=x^{4}+2 x^{2}-x-3\). a. \(\quad g_{1}(x)=\left(3+x-2 x^{2}\right)^{1 / 4}\) b. \(\quad g_{2}(x)=\left(\frac{x+3-x^{4}}{2}\right)^{1 / 2}\) c. \(\quad g_{3}(x)=\left(\frac{x+3}{x^{2}+2}\right)^{1 / 2}\) d. \(\quad g_{4}(x)=\frac{3 x^{4}+2 x^{2}+3}{4 x^{3}+4 x-1}\)

Use Maple to find a real zero of the polynomial \(f(x)=x^{3}+4 x-4\).

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