CODEWITHSUNDEEP
racket
bluepanda33
bluepanda33

Reputation: 1

How do I compute the summation for a function from 1 to n using Racket?

Summation sum(2x-1, x=1...n). I have to write a recursive function to solve the sum.

(define (sum1 n)
  (if (= n 0)
      -1
      (+ n (- 1 (sum1 (* n 1))))))

This is what I have so far, I am so lost.

Upvotes: 0

Views: 766

Answers (2)

JSells
JSells

Reputation: 95

If you just want to output the sum, this is how I would do it:


#lang racket

(define (sum n)
  (if (= n 1)
     1
     (+ (-(* 2 n) 1) (sum (- n 1)))))

Upvotes: 0

Atharva Shukla
Atharva Shukla

Reputation: 2137

Your base case is 1 not 0 because you're only defining it from 1 to n. Therefore, the function is only defined over N. So we don't need to consider the 0 case for the function (it only consumes Ns).

Furthermore, the recursive call should add n applied to 2x-1 to the summation of the rest of the sequence. Currently the function does not terminate because you are calling it on the same input (* n 1) = n. (sum1 n) => (sum1 n) => ...

#lang racket

(require rackunit)

;; N is one of:
;; - 1
;; - (+ 1 N)
;; interpretation: 1, 2, 3 ...

;; N -> Number
;; sums `(- (* 2 x) 1)` from 1 to n
(define (sum1 n)
  (if (= n 1)
      (- (* 2 1) 1) ;; = 1
      (+ (- (* 2 n) 1) (sum1 (- n 1)))))

(check-equal? (sum1 1) 1)
(check-equal? (sum1 10) 100)

Abstracting out the function:

;; [Number -> Number] N -> Number
;; sums of f from 1 to n
(define (summation f n)
  (if (= n 1)
      (f 1)
      (+ (f n) (summation f (- n 1)))))

(check-equal? (summation identity 5) (+ 1 2 3 4 5))
(check-equal? (summation sqr 3) (+ (sqr 1) (sqr 2) (sqr 3)))

(check-equal? (summation (λ (x) (- (* 2 x) 1)) 1) 1)
(check-equal? (summation (λ (x) (- (* 2 x) 1)) 10) 100)

Upvotes: 1

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