What does this function do?

Question

Does this function means its calculating x=(x-1) + x^2?

Function unknown(x)
  if ( x == 1 )
    return 1
  else
    return unknown(x‐1) + x*x

Rick Sanchez · Answer 1 · 2014-09-06T15:02:55.997

4

Well it recursively calculates x² + (x-1)² + (x-2)² + ⋯ + 2² + 1.

edited Sep 06 '14 at 15:02

answered Sep 06 '14 at 12:52

Rick Sanchez

149
6

score 2 · Answer 2 · answered Sep 06 '14 at 19:48

2

This is the sum of squares of natural numbers. The sum of first n natural numbers can be represented mathematically by:

n*(n+1)*(2n+1)/6

answered Sep 06 '14 at 19:48

Ankur Agarwal

121

score 2 · Accepted Answer · answered Sep 07 '14 at 00:41

The function can be written in a more mathematical form as:

unknown(1) = 1
unknown(x) = unknown(x - 1) + x * x    [if x is not 1]

To see why this is the same as the sum of squares, you can substitute the second line into itself recursively and observe the pattern that emerges:

unknown(x) =                                         unknown(x - 1) + x * x
unknown(x) =                     unknown(x - 2) + (x - 1) * (x - 1) + x * x
unknown(x) = unknown(x - 3) + (x - 2) * (x - 2) + (x - 1) * (x - 1) + x * x
etc ...

After expanding several times, you find that eventually* you will reach the special case of unknown(1) since the argument is steadily decreasing by one at each expansion. Hence,

(after expanding many times ...)
unknown(x) = unknown(1) + 2 * 2 + ... + (x - 1) * (x - 1) + x * x
unknown(x) =         1  + 2 * 2 + ... + (x - 1) * (x - 1) + x * x

Hence, this is simply the sum of squares.

[*] Unless your initial x is less than one, in which case the function diverges (never terminates) since the argument will just keep becoming more and more negative.

What does this function do?

3 Answers3