/*! 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 10 Complete al-Samaw'al's procedure... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Complete al-Samaw'al's procedure of dividing \(20 x^{2}+30 x\) by \(6 x^{2}+12\) to get the result stated in the text. Prove that the coefficients of the quotient satisfy the rule \(a_{n+2}=-2 a_{n}\) where \(a_{n}\) is the coefficient of \(\frac{1}{n}\)

Short Answer

Expert verified
Based on the given division problem, we found the quotient expression to be \(Q(x) = \frac{10}{3} - \frac{5}{6x}\). When we tried to verify the rule \(a_{n+2} = -2a_n\) for the non-zero coefficients, we found that it did not hold for both n=0 and n=1. Therefore, the coefficients of the quotient do not satisfy the given rule.

Step by step solution

01

Divide the given polynomials using al-Samaw'al's procedure

First, we'll set up the division like this: \(\frac{20x^2 + 30x}{6x^2 + 12}\) To proceed with al-Samaw'al's procedure, divide the first term of the numerator by the first term of the denominator: \(\frac{20x^2}{6x^2} = \frac{10}{3}\) Next, multiply the entire denominator by the result: \(\frac{10}{3}(6x^2 + 12) = 20x^2 + 40x.\) Subtract this new expression from the original numerator: \((20x^2 + 30x) - (20x^2 + 40x) = -10x.\) Now, divide the result by the same \(6x^2\) we used before: \(\frac{-10x}{6x^{2}} = -\frac{5}{6x}\) Now, our expression for the quotient is: \(\frac{10}{3} - \frac{5}{6x}\)
02

Write the result of the division as a series

We can view the quotient expression as a series, so: \(Q(x) = \frac{10}{3} - \frac{5}{6x}\) As a series, we have: \(Q(x) = a_0 + a_1x + a_2x^2 + ...\) where \(a_0 = \frac{10}{3}\), \(a_1 = -\frac{5}{6}\), and the rest of the coefficients are zero, since there are no other terms in the quotient.
03

Verify that the coefficients of the quotient satisfy the given rule \(a_{n+2} = -2a_n\)

To verify that the coefficients of the quotient \(Q(x)\) satisfy the rule \(a_{n+2} = -2a_n\), let's check the rule for the non-zero coefficients: For \(n = 0\): \(a_{0} = \frac{10}{3}\), \(a_{2} = 0\) \(a_{n+2} = -2a_{n} \iff a_2 = -2a_0 \iff 0 = -2 \cdot \frac{10}{3} \iff 0 = -\frac{20}{3}\) This is not true, so the rule doesn't hold for n=0. For \(n = 1\): \(a_{1} = -\frac{5}{6}\), \(a_{3} = 0\) \(a_{n+2} = -2a_{n} \iff a_3 = -2a_1 \iff 0 = -2 \cdot -\frac{5}{6} \iff 0 = \frac{10}{6}\) This is also not true, so the rule doesn't hold for n=1. Thus, the coefficients of the quotient do not satisfy the rule \(a_{n+2} = -2a_n\).

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 91Ó°ÊÓ!

Key Concepts

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

al-Samaw'al's procedure
Al-Samaw'al's procedure is a method of polynomial division named after the medieval mathematician al-Samaw'al. It simplifies polynomial division, providing a straightforward approach. To understand it, think of it like long division but for polynomials. This method is especially useful in ensuring that each step of the division accounts for the powers of the variables involved. When dividing the polynomials \(20x^2 + 30x\) by \(6x^2 + 12\), the process begins by focusing on the leading terms.
  • First, divide the first term of the numerator \(20x^2\) by the first term of the denominator \(6x^2\).
  • This gives the coefficient \(\frac{10}{3}\).
  • Next, multiply the entire denominator by this coefficient and subtract from the original numerator.
  • Continue this process until no further division can be made without resulting in a fraction that includes the variable as the denominator.
Al-Samaw'al's procedure allows us to derive a quotient of polynomials that are part integer, part fractional powers of variables. This gives a clearer understanding of polynomial behavior.
Coefficient Rule
In polynomial division, coefficients of the resulting quotient often follow specific patterns or rules. One such rule proposed in this exercise is that the coefficients \(a_{n+2}\) of the quotient should satisfy \(-2a_n\). However, upon closer examination through the division of the given polynomials, it turns out that this rule does not apply in this case. Let’s break it down.
  • The quotient derived was \(\frac{10}{3} - \frac{5}{6x}\); hence coefficients are \(a_0 = \frac{10}{3}\), \(a_1 = -\frac{5}{6}\), and all other higher-degree coefficients are zero.
  • Testing this against the rule, we see that \(a_2\) should be equal to \(-2\cdot a_0\) which does not hold true as \(a_2 = 0\).
  • Similarly, \(a_3\) should be \(-2\cdot a_1\) but this is also not satisfied as \(a_3 = 0\).
This discrepancy highlights the importance of verifying such rules, as they may not universally apply or require specific polynomial conditions to hold.
Mathematical Proof
Mathematical proof is a logical argument demonstrating the truth of a mathematical statement. In polynomial division, attempting to prove coefficient rules requires careful verification. Let's use the given problem as an example to illustrate the proof process.Attempting to apply and prove the rule \(a_{n+2} = -2a_n\) involves evaluating the coefficients from the derived polynomial quotient \(Q(x) = \frac{10}{3} - \frac{5}{6x}\). Here's the step-by-step reasoning:
  • Start by identifying the coefficients. For \(a_0 = \frac{10}{3}\) and \(a_1 = -\frac{5}{6}\), assess whether \(a_2 = a_3 = 0\) works with the rule.
  • For \(n = 0\), calculate \(a_2 = -2 \times a_0\) which fails since \(a_2 = 0\) does not equal \(-\frac{20}{3}\).
  • Similarly, for \(n = 1\), \(a_3\) should equal \(-2 \times a_1\), but it also does not satisfy the equation as \(a_3 = 0\) is not \(\frac{10}{6}\).
Such proofs necessitate methodical checking of every step. They ensure that any derived rules or patterns accurately reflect the relationships between polynomial coefficients. Even when a rule appears mathematically sound, empirical verification, like the one in this exercise, clarifies applicability and 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

Show that 1184 and 1210 are amicable numbers that are not a consequence of the theorem of Th?bit ibn Qurra.

Show that the radius \(r_{\alpha}\) of a latitude circle on the earth at \(\alpha^{\circ}\) is given by \(r_{\alpha}=R \cos \alpha\), where \(R\) is the radius of the -earth.

Analyze the possibilities of positive solutions to \(x^{3}+d=\) \(c x\) by first showing that the maximum of the function \(x(c-\) \(x^{2}\) ) occurs at \(x_{0}=\sqrt{\frac{c}{3}}\). Use calculus to consider the graph of \(y=x^{3}-c x+d\) and determine the conditions on the coefficients giving it zero, one, or two positive solutions.

Use ibn al-Haytham's procedure to derive the formula for the sum of the fifth powers of the integers: $$ 1^{5}+2^{5}+\cdots+n^{5}=\frac{1}{6} n^{6}+\frac{1}{2} n^{5}+\frac{5}{12} n^{4}-\frac{1}{12} n^{2} $$

Ab? Sahl al-K?hì knew from his own work on centers of gravity and the work of his predecessors that the center of gravity divides the axis of certain plane and solid figures in the following ratios: Tetrahedron: \(\frac{1}{4}\) Segment of a parabola: \(\frac{2}{3} \quad\) Paraboloid of revolution: \(\frac{2}{6}\) Hemisphere: \(\frac{3}{8}\) Noting the pattern, he guessed that the corresponding value for a semicircle was \(3 / 7\). Show that al-K?hi's first five results are correct, but that his guess for the semicircle\\} implies that \(\pi=31 / 9\). (Al-K?h? realized that this value contradicted Archimedes' bounds of \(310 / 71\) and \(31 / 7\), but concluded that there was an error in the transmission of Archimedes' work.)

See all solutions

Recommended explanations on Math Textbooks

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.