Averages/Pythagorean means

Averages/Pythagorean means
You are encouraged to solve this task according to the task description, using any language you may know.

Compute all three of the Pythagorean means of the numbers 1..10.

• The most common, Arithmetic mean is the sum of the numbers divided by their count, n:

$A(x_{1},\ldots ,x_{n})={\frac {1}{n}}(x_{1}+\cdots +x_{n})$

• The Geometric mean is the n'th root of the multiplication of all the numbers:

$G(x_{1},\ldots ,x_{n})={\sqrt[{n}]{x_{1}\cdots x_{n}}}$

• The Harmonic mean is n divided by the sum of their reciprocals:

$H(x_{1},\ldots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\cdots +{\frac {1}{x_{n}}}}}$

Show that $A(x_{1},\ldots ,x_{n})\geq G(x_{1},\ldots ,x_{n})\geq H(x_{1},\ldots ,x_{n})$ for this set of positive numbers.

Lua

<lang lua>function fsum(f, a, ...) return a and f(a) + fsum(f, ...) or 0 end function pymean(t, f, finv) return finv(fsum(f, unpack(t)) / #t) end nums = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

--arithmetic print(pymean(nums, function(n) return n end, function(n) return n end)) --geometric print(pymean(nums, math.log, math.exp)) --harmonic print(pymean(nums, function(n) return 1/n end, function(n) return 1/n end))</lang>

Python

<lang Python>>>> from operator import mul >>> from functools import reduce >>> def amean(num): return sum(num)/len(num)

>>> def gmean(num): return reduce(mul, num, 1)**(1/len(num))

>>> def hmean(num): return len(num)/sum(1/n for n in num)

>>> numbers = range(1,11) # 1..10 >>> amean(numbers), gmean(numbers), hmean(numbers) (5.5, 4.528728688116765, 3.414171521474055) >>> assert( amean(numbers) >= gmean(numbers) >= hmean(numbers) ) >>> </lang>