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Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule $$ \frac{d\left(u^{n}\right)}{d x}=n u^{n-1}\left(\frac{d u}{d x}\right) $$ by adding a new clause to the deriv program and defining appropriate procedures exponentiation?, base, exponent, and make-exponentiation. (You may use the symbol \(* *\) to denote exponentiation.) Build in the rules that anything raised to the power 0 is 1 and anything raised to the power 1 is the thing itself.

Short Answer

Expert verified
Add a differentiation rule handling u^n with special cases for exponents 0 and 1.

Step by step solution

01

Modify the Differentiator Structure

To extend the differentiator, you begin by adding a new clause to the `deriv` procedure to handle exponentiation. Check if the expression is an exponentiation and apply the new differentiation rule: \( \frac{d(u^n)}{dx} = n \, u^{n-1} \frac{du}{dx} \).
02

Implement the Function exponentiation?

Define the procedure `exponentiation?` that checks if the given expression is of the form \(u^n\). This will determine if part of the expression should be differentiated using the exponentiation rule.
03

Define base, exponent, and make-exponentiation

The `base` function should extract the base \(u\) from the expression \(u^n\), while the `exponent` function retrieves the exponent \(n\). The function `make-exponentiation` should return the expression \(u^n\) given a base \(u\) and an exponent \(n\).
04

Apply Differentiation Rule for Exponentiation

Within the added clause, use the differentiation rule for exponentiation. Calculate \(n \, u^{n-1}\) using the `base` and `exponent` functions. Then recursively derive `du/dx` using the core differentiation program. Multiply these to find the derivative of \(u^n\).
05

Build in Special Cases for Exponents 0 and 1

Add special cases: when the exponent is 0, return 1; when the exponent is 1, return the base itself, effectively bypassing further calculations for these simplified scenarios.

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Key Concepts

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

Differentiation Rules
Differentiation rules are the foundational guidelines used to find the derivative of mathematical expressions. These rules help us determine how functions change with respect to a variable. In calculus, derivatives represent the rate of change, and therefore, understanding the standard differentiation rules is essential.
  • **Power Rule**: This is one of the most common rules. For an expression of the form \(x^n\), the derivative is \(n \cdot x^{n-1}\).
  • **Product Rule**: Used when differentiating products of functions. If \(f(x)\) and \(g(x)\) are functions, then the derivative of their product is \(f'(x)g(x) + f(x)g'(x)\).
  • **Chain Rule**: Applied to composite functions. If \(f(g(x))\) is a composite function, its derivative is \(f'(g(x)) \cdot g'(x)\).
Differentiation rules allow us to handle a variety of mathematical problems and are critical to proficiencies in calculus and other fields.
Exponentiation
Exponentiation involves raising a number or expression to a power, representing repeated multiplication. For example, in the expression \(u^n\), \(u\) is multiplied by itself \(n\) times. It's crucial to handle this operation properly when dealing with differentiation.
  • **Basic Understanding**: For any expression \(x^n\), if \(n = 0\), the result is 1. If \(n = 1\), the result is \(x\).
  • **Differentiation of Exponential Functions**: The differentiation rule specific to exponentiation states that \(\frac{d(u^n)}{d x} = n \cdot u^{n-1} \cdot \frac{d u}{d x}\). This is applied when the base of the exponent \(u\) is a function itself.
Exponentiation is a vital concept in both algebra and calculus, often requiring special attention when programming symbolic differentiation.
Programming Recursion
Programming recursion is a technique where a function calls itself to solve sub-problems. It is particularly useful in the context of symbolic differentiation, where expressions can be broken down recursively until they reach a base case. Recursion simplifies the process of differentiating complex functions by handling smaller parts of the problem first.
  • **Base Case**: Every recursive function needs a base case to avoid infinite loops. This could be simple forms of expressions that can be solved without further recursion.
  • **Recursive Case**: The part of the function where the method calls itself to work on smaller or simpler sub-problems. In differentiation, this might involve breaking down expressions into their components and applying differentiation rules recursively.
Recursive solutions are elegant and powerful, often simplifying code when dealing with nested or repeated patterns in data or calculations.
Mathematical Expressions
Mathematical expressions are combinations of symbols representing numbers, variables, operations, and sometimes functions. They are the basic building blocks in mathematics, used to convey calculations and equations.
  • **Components**: Standard elements of a mathematical expression can include constants, variables, coefficients, operators (like +, -, *, /), and functions (like sin, cos, etc.).
  • **Expression Trees**: In computer science, expression trees are used to represent expressions in a structured way. Each node is an operation or operand, allowing expressions to be evaluated or transformed systematically, which is useful for symbolic differentiation.
Mathematical expressions are central to the understanding and execution of programming tasks involving calculus and algebra, especially when translating written formulas into functional code.

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Most popular questions from this chapter

Suppose we want to modify the differentiation program so that it works with ordinary mathematical notation, in which \(+\) and \(*\) are infix rather than prefix operators. Since the differentiation program is defined in terms of abstract data, we can modify it to work with different representations of expressions solely by changing the predicates, selectors, and constructors that define the representation of the algebraic expressions on which the differentiator is to operate. a. Show how to do this in order to differentiate algebraic expressions presented in infix form, such as \((x+(3 *(x+(y+2))))\). To simplify the task, assume that \(+\) and \(*\) always take two arguments and that expressions are fully parenthesized. b. The problem becomes substantially harder if we allow standard algebraic notation, such as \((\mathrm{x}+3 *(\mathrm{x}+\mathrm{y}+2))\), which drops unnecessary parentheses and assumes that multiplication is done before addition. Can you design appropriate predicates, selectors, and constructors for this notation such that our derivative program still works?

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?

The internal procedures in the scheme-number package are essentially nothing more than calls to the primitive procedures \(+,-\), etc. It was not possible to use the primitives of the language directly because our type-tag system requires that each data object have a type attached to it. In fact, however, all Lisp implementations do have a type system, which they use internally. Primitive predicates such as symbol? and number? determine whether data objects have particular types. Modify the definitions of type-tag, contents, and attach-tag from section 2.4.2 so that our generic system takes advantage of Scheme's internal type system. That is to say, the system should work as before except that ordinary numbers should be represented simply as Scheme numbers rather than as pairs whose car is the symbol scheme-number.

Louis Reasoner has noticed that apply-generic may try to coerce the arguments to each other's type even if they already have the same type. Therefore, he reasons, we need to put procedures in the coercion table to "coerce" arguments of each type to their own type. For example, in addition to the scheme- number->complex coercion shown above, he would do: (define (scheme-number->scheme-number n) n) (define (complex->complex z) z) (put-coercion'scheme-number 'scheme-number scheme-number->scheme-number) (put-coercion 'complex 'complex complex->complex) a. With Louis's coercion procedures installed, what happens if apply-generic is called with two arguments of type scheme-number or two arguments of type complex for an operation that is not found in the table for those types? For example, assume that we've defined a generic exponentiation operation: (define (exp x y) (apply-generic' \(\exp x y\) )) and have put a procedure for exponentiation in the Scheme-number package but not in any other package: "following added to Scheme-number package (put'exp' '(scheme-number scheme-number) (lambda (x y) (tag (expt \(x y)\) ))) ;using primitive expt What happens if we call exp with two complex numbers as arguments? b. Is Louis correct that something had to be done about coercion with arguments of the same type, or does apply-generic work correctly as is? c. Modify apply-generic so that it doesn't try coercion if the two arguments have the same type.

Two lists are said to be equal? if they contain equal elements arranged in the same order. For example, (equal? '(this is a list) '(this is a list)) is true, but (equal? '(this is a list)' (this (is a) list)) is false. To be more precise, we can define equal? recursively in terms of the basic eq? equality of symbols by saying that a and b are equal? if they are both symbols and the symbols are eq?, or if they are both lists such that (car a) is equal? to (car b) and (cdr a) is equal? to (cdr b). Using this idea, implement equal? as a procedure. \({ }^{36}\)

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