How to resolve the algorithm Faulhaber's triangle step by step in the Racket programming language

Published on 12 May 2024 09:40 PM
#Racket

How to resolve the algorithm Faulhaber's triangle step by step in the Racket programming language

Table of Contents

Problem Statement

Named after Johann Faulhaber, the rows of Faulhaber's triangle are the coefficients of polynomials that represent sums of integer powers, which are extracted from Faulhaber's formula:

where

B

n

{\displaystyle B_{n}}

is the nth-Bernoulli number.

The first 5 rows of Faulhaber's triangle, are:

Using the third row of the triangle, we have:

∑

k

1

n

k

2

=

1 6

n +

1 2

n

2

1 3

n

3

{\displaystyle \sum _{k=1}^{n}k^{2}={1 \over 6}n+{1 \over 2}n^{2}+{1 \over 3}n^{3}}

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Faulhaber's triangle step by step in the Racket programming language

Source code in the racket programming language

#lang racket
(require math/number-theory)

(define (second-bernoulli-number n)
  (if (= n 1) 1/2 (bernoulli-number n)))

(define (faulhaber-row:formulaic p)
  (let ((p+1 (+ p 1)))
    (reverse
     (for/list ((j (in-range p+1)))
       (* (/ p+1) (second-bernoulli-number j) (binomial p+1 j))))))

(define (sum-k^p:formulaic p n)
  (for/sum ((f (faulhaber-row:formulaic p)) (i (in-naturals 1)))
    (* f (expt n i))))

(module+ main
  (map faulhaber-row:formulaic (range 10))
  (sum-k^p:formulaic 17 1000))

(module+ test
  (require rackunit)
  (check-equal? (sum-k^p:formulaic 17 1000)
                (for/sum ((k (in-range 1 (add1 1000)))) (expt k 17))))

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