Recommendation Systems - Mining of Massive Datasets - page 316

(5 − (2.6 y + 1)) 2 + (3 − ( y + 1)) 2 + (2 − ( y + 1)) 2 + (2 − ( y + 1)) 2 + (4 − ( y + 1)) 2
This expression simplifies to
(4 − 2.6 y ) 2 + (2 − y ) 2 + (1 − y ) 2 + (1 − y ) 2 + (3 − y ) 2
As before, we find the minimum value of this expression by differentiating and equating to
0, as:
−2 × (2.6(4 − 2.6 y ) + (2 − y ) + (1 − y ) + (1 − y ) + (3 − y )) = 0
The solution for y is y = 17.4 / 10.76 = 1.617. The improved estimates of U and V are shown
in Fig. 9.14 .
Figure 9.13 v 11 becomes a variable y
Figure 9.14 Replace y by 1.617
We shall do one more change, to illustrate what happens when entries of M are blank.
We shall vary u 31 , calling it z temporarily. The new U and V are shown in Fig. 9.15 . The
value of z affects only the entries in the third row.
Figure 9.15 u 31 becomes a variable z
We can express the sum of the squares of the errors as
(2 − (1.617 z + 1)) 2 + (3 − ( z + 1)) 2 + (1 − ( z + 1)) 2 + (4 − ( z + 1)) 2
Note that there is no contribution from the element in the second column of the third row,
since this element is blank in M . The expression simplifies to
(1 − 1.617 z ) 2 + (2 − z ) 2 + (− z ) 2 + (3 − z ) 2
The usual process of setting the derivative to 0 gives us