0

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
  • Why is this tagged [array] and what language is it supposed to be? – JensG Sep 6 '14 at 12:45
  • opps sorry. It's pseudocode. Once I know what is the function is doing, I will then know how to write it into C++/C#. – Zainu Sep 6 '14 at 12:53
  • 3
    Have a look at this: programmers.stackexchange.com/questions/25052/… – 11684 Sep 6 '14 at 12:54
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2

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.

4

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

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

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