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Let \(f\) and \(g\) be two one-argument functions. The composition \(f\) after \(g\) is defined to be the function \(x \mapsto f(g(x))\). Define a procedure compose that implements composition. For example, if inc is a procedure that adds 1 to its argument, ((compose square inc) 6) 49

Short Answer

Expert verified
The result is 49.

Step by step solution

01

Understand Function Composition

The composition of two functions, denoted as \((f \circ g)(x)\), is a process where function \(g\) is applied to the input \(x\), and function \(f\) is applied to the result of \(g(x)\). Mathematically, this is expressed as \(f(g(x))\). In this instance, \(f(x) = x^2\) and \(g(x) = x + 1\), so the composed function \((f \circ g)(x) = f(g(x)) = (g(x))^2 = (x+1)^2\).
02

Implement Composition Function

To create a function that performs composition, take two functions, \(f\) and \(g\), and return a new function. In pseudocode, this new function returns \(f(g(x))\) when called with an argument \(x\). Thus, the pseudocode could resemble: `compose(f, g)` that outputs `h(x)`, where \(h(x) = f(g(x))\).
03

Define Individual Functions

Define the increment function \(\text{inc}(x) = x + 1\) and the square function \(\text{square}(x) = x^2\). These will serve as \(g\) and \(f\) respectively in the composition \((f \circ g)(x) = f(g(x))\).
04

Compose Square and Inc

Using the `compose` function from Step 2, plug in the `square` and `inc` functions. This returns a new function that, when given an argument \(x\), will compute \((x+1)^2\).
05

Evaluate Composite Function with Input

Pass the value \(6\) to the composed function. Evaluate \((6 + 1)^2\) as follows: First compute the increment \(6+1=7\), then square the result: \(7^2 = 49\). The result of \((\text{compose square inc})(6)\) is therefore \(49\).

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Key Concepts

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

Programming Concepts
In programming, we often need to perform complex operations by combining simpler functions. This is especially true when utilizing function composition. Function composition is a powerful concept where you create a new function by combining two or more functions. In essence, you pass the output of one function as the input to another.

By doing this, you can build more advanced functionality without modifying the original functions. Think of it like creating a pipeline where each stage represents a function. This approach boosts modularity and reusability, allowing programmers to handle complex problems more efficiently.
  • Create robust programs through modular code.
  • Enhance program reusability and maintainability.
  • Simplify complex operations by breaking them into smaller tasks.
Mathematical Functions
Mathematical functions are at the core of many programming tasks. A function, in mathematics, is a relation that uniquely associates an input to an output. The idea of function composition also originates from mathematics. When you compose two functions, say function \( f \) and \( g \), you apply them in a sequence. This process is denoted mathematically as \((f \circ g)(x) = f(g(x))\).

If \( g(x) = x + 1 \) and \( f(x) = x^2 \), composing them results in: \[ (f \circ g)(x) = f(g(x)) = (x+1)^2 \]

Through this, you combine operations into a single expression, streamlining calculations. Function composition also serves as a helpful abstraction tool in computer science and occurs frequently in areas such as calculus and algebra.
  • Combine multiple operations into a single expression.
  • Simplify calculations with transformative steps.
  • Act as a foundational concept in calculus and algebra.
Procedure Definition
In programming, defining a procedure allows us to encapsulate logic into a function, making our code more organized and maintainable. A procedure definition usually involves naming the procedure, specifying parameters, and detailing the operations it performs.

To define a procedure for function composition, you need to combine two functions \( f \) and \( g \). You can create a new function \( h \), such that \( h(x) = f(g(x)) \). This promotes the use of higher-order functions, which are an advanced feature of many programming languages, including those discussed in the "Structure and Interpretation of Computer Programs" (SICP).
  • Increase code clarity and organization.
  • Enable code reusability with defined logic components.
  • Illustrate the use of higher-order functions effectively.
SICP
"Structure and Interpretation of Computer Programs" (SICP) is a classic textbook that explores programming principles through the lens of computation. One of its focal points is the abstraction of procedures, such as function composition, to demonstrate elegant ways of solving problems.

SICP encourages thinking about programs not just as a set of instructions but as expressions and procedures that form a robust framework for problem-solving. This includes creating functions that operate on other functions, as seen with the "compose" procedure.

The text demonstrates how composing procedures can simplify complex tasks by breaking them down into more manageable parts, showing why it's influential in computer science education.
  • Promote deeper understanding of programming concepts.
  • Emphasize the use of abstraction in code design.
  • Encourage building systems with scalable, composable components.

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Most popular questions from this chapter

Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.

Alyssa P. Hacker doesn't see why if needs to be provided as a special form. "Why can't I just define it as an ordinary procedure in terms of cond?" she asks. Alyssa's friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if: (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause))) Eva demonstrates the program for Alyssa: (new-if (=2 3) 0 5) 5 (new-if (= 1 1) 0 5) 0 Delighted, Alyssa uses new-if to rewrite the s (define (sqrt-iter guess \(x\) ) (new-if (good-enough? guess \(x\) ) $$ \begin{array}{l}\text { guess } \\ \text { (sqrt-iter (improve guess } x \text { ) } \\ x) \text { ) }\end{array} $$ What happens when Alyssa attempts to use this to compute square roots? Explain.

Define a procedure double that takes a procedure of one argument as argument and returns a procedure that applies the original procedure twice. For example, if inc is a procedure that adds 1 to its argument, then (double inc) should be a procedure that adds 2 . What value is returned by (((double (double double)) inc) 5)

Newton's method for cube roots is based on the fact that if \(y\) is an approximation to the cube root of \(x\), then a better approximation is given by the value \(\frac{x / y^{2}+2 y}{3}\) Use this formula to implement a cube-root procedure analogous to the squareroot procedure. (In section \(1.3 .4\) we will see how to implement Newton's method in general as an abstraction of these square-root and cube- root procedures.)

The exponentiation algorithms in this section are based on performing exponentiation by means of repeated multiplication. In a similar way, one can perform integer multiplication by means of repeated addition. The following multiplication procedure (in which it is assumed that our language can only add, not multiply) is analogous to the expt procedure: (define (* a b) (if \((=b \quad 0)\) 0 \((+a(* a(-b \quad 1)))))\)

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