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Chapter 2: Problem 88

Extend the polynomial system to include subtraction of polynomials. (Hint: You may find it helpful to define a generic negation operation.)

Short Answer

Expert verified
Subtracting polynomials involves adding the negation of the subtracted polynomial.

Step by step solution

01

Understanding Polynomial Subtraction

To subtract polynomials, recognize it as adding the negation of a polynomial. For example, to subtract polynomial \( Q(x) \) from polynomial \( P(x) \), compute \( P(x) + (-Q(x)) \).
02

Defining the Negation Operation

The negation of a polynomial changes the sign of each term in the polynomial. If \( Q(x) = q_nx^n + q_{n-1}x^{n-1} + ... + q_1x + q_0 \), then \(-Q(x) = -q_nx^n - q_{n-1}x^{n-1} - ... - q_1x - q_0 \).
03

Applying Negation to the Polynomial

Given a polynomial \( Q(x) = 3x^2 + 2x - 5 \), its negation is \(-Q(x) = -3x^2 - 2x + 5 \).
04

Performing the Subtraction

To subtract \( Q(x) \) from \( P(x) \), compute \( P(x) + (-Q(x)) \). If \( P(x) = 5x^2 + x + 4 \), then \( P(x) - Q(x) = (5x^2 + x + 4) + (-3x^2 - 2x + 5) = 2x^2 - x + 9 \).

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Key Concepts

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

Negation Operation
The negation operation is a crucial step in polynomial subtraction. When you negate a polynomial, you change the sign of each of its terms. This means converting positive terms into negative and vice versa. This operation allows us to transform a subtraction problem into an addition problem, which can be easier to handle.
For instance, consider the polynomial \( Q(x) = 3x^2 + 2x - 5 \). The negation of this polynomial is achieved by changing the sign of each term, resulting in \(-Q(x) = -3x^2 - 2x + 5 \). Notice how each term in \( Q(x) \) has had its sign flipped to produce \(-Q(x) \). By handling subtraction in this manner, we can use the familiar rules of polynomial addition instead.
Polynomial Addition
Once you have negated one of the polynomials in a subtraction problem, the task becomes a matter of polynomial addition. Adding polynomials involves combining like terms, which are terms that have the same variables raised to the same power. The coefficients of these terms are added together.
For example, if we are to subtract \( Q(x) = 3x^2 + 2x - 5 \) from \( P(x) = 5x^2 + x + 4 \), we rewrite it as an addition problem: \[ P(x) - Q(x) = P(x) + (-Q(x)) \]. After negating \( Q(x) \) as \(-3x^2 - 2x + 5 \), add them up:
  • Combine the \( x^2 \) terms: \( 5x^2 - 3x^2 = 2x^2 \)
  • Combine the \( x \) terms: \( x - 2x = -x \)
  • Combine the constant terms: \( 4 + 5 = 9 \)
This results in \( P(x) - Q(x) = 2x^2 - x + 9 \). Therefore, what seemed challenging initially is simplified into straightforward arithmetic.
Algebraic Expressions
Algebraic expressions like polynomials are mathematical sentences that include constants, variables, and exponents. Understanding these components is essential when performing operations such as addition and subtraction. Each term in an algebraic expression consists of a coefficient (which is a constant or a number), a variable, and an exponent.
For example, in the algebraic expression \( 3x^2 + 2x - 5 \), the terms are \( 3x^2 \), \( 2x \), and \(-5 \). Here, \( 3 \), \( 2 \), and \(-5 \) are coefficients; \( x \) is the variable; and \( 2 \), \( 1 \) (implied for \( 2x \)), and \( 0 \) (since \(-5 \) is a constant term without a variable) are the exponents.
When performing operations on algebraic expressions, like subtraction by negation and then addition, it's important to identify and correctly manipulate these terms. This kind of clarity ensures accuracy in simplifying or rearranging these expressions for any mathematical purpose.

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Most popular questions from this chapter

A two-dimensional vector \(\mathbf{v}\) running from the origin to a point can be represented as a pair consisting of an \(x\)-coordinate and a \(y\)-coordinate. Implement a data abstraction for vectors by giving a constructor make-vect and corresponding selectors xcor-vect and ycor-vect. In terms of your selectors and constructor, implement procedures add-vect, sub-vect, and scale-vect that perform the operations vector addition, vector subtraction, and multiplying a vector by a scalar: $$ \begin{aligned} \left(x_{1}, y_{1}\right)+\left(x_{2}, y_{2}\right) &=\left(x_{1}+x_{2}, y_{1}+y_{2}\right) \\ \left(x_{1}, y_{1}\right)-\left(x_{2}, y_{2}\right) &=\left(x_{1}-x_{2}, y_{1}-y_{2}\right) \\ s \cdot(x, y) &=(s x, s y) \end{aligned} $$

As a large system with generic operations evolves, new types of data objects or new operations may be needed. For each of the three strategies-generic operations with explicit dispatch, data-directed style, and message-passing- styledescribe the changes that must be made to a system in order to add new types or new operations. Which organization would be most appropriate for a system in which new types must often be added? Which would be most appropriate for a system in which new operations must often be added?

In case representing pairs as procedures wasn't mind-boggling enough, consider that, in a language that can manipulate procedures, we can get by without numbers (at least insofar as nonnegative integers are concerned) by implementing 0 and the operation of adding 1 as (define zero (lambda (f) (lambda ( \(x\) ) \(x\) ))) (define (add-1 \({n}\) ) (lambda (f) (lambda ( \(x\) ) (f ( \((\) n \(f) x))\) )) This representation is known as Church numerals, after its inventor, Alonzo Church, the logician who invented the \(\lambda\) calculus. Define one and two directly (not in terms of zero and add-1). (Hint: Use substitution to evaluate (add-1 zero)). Give a direct definition of the addition procedure + (not in terms of repeated application of add-1).

Define a better version of make-rat that handles both positive and negative arguments. Make-rat should normalize the sign so that if the rational number is positive, both the numerator and denominator are positive, and if the rational number is negative, only the numerator is negative.

Show that under the assumption of small percentage tolerances there is a simple formula for the approximate percentage tolerance of the product of two intervals in terms of the tolerances of the factors. You may simplify the problem by assuming that all numbers are positive.
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