How to resolve the algorithm Averages/Pythagorean means step by step in the Julia programming language

Published on 22 June 2024 08:30 PM

How to resolve the algorithm Averages/Pythagorean means step by step in the Julia programming language

Table of Contents

Problem Statement

Compute all three of the Pythagorean means of the set of integers 1 through 10 (inclusive). Show that

A (

x

1

, … ,

x

n

) ≥ G (

x

1

, … ,

x

n

) ≥ H (

x

1

, … ,

x

n

)

{\displaystyle A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})}

for this set of positive integers.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Averages/Pythagorean means step by step in the Julia programming language

The provided Julia code defines three functions: amean, gmean, and hmean for calculating the arithmetic mean, geometric mean, and harmonic mean of an array A, respectively. Here's a detailed breakdown of each function:

  1. amean(A):

    • This function calculates the arithmetic mean or average value of the elements in array A.
    • It sums all the elements using the sum(A) function and then divides the result by the number of elements in A using length(A).
    • The result is the arithmetic mean of the array elements.
  2. gmean(A):

    • This function calculates the geometric mean of the elements in array A.
    • It multiplies all the elements together using the prod(A) function and then takes the 1/length(A)th root of the product using the ^ operator.
    • The result is the geometric mean of the array elements.
  3. hmean(A):

    • This function calculates the harmonic mean of the elements in array A.
    • It calculates the sum of the reciprocals (1 divided by each element) of the array elements using sum(1./A) and then divides the number of elements in A by this sum.
    • The result is the harmonic mean of the array elements.

Source code in the julia programming language

amean(A) = sum(A)/length(A)

gmean(A) = prod(A)^(1/length(A))

hmean(A) = length(A)/sum(1./A)


  

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