Wednesday, April 23, 2014

We will now consider some other techniques in addition to the previous post. One way to obtain bounds on a difficult summation is to express the series as the sum of two or more series by partitioning the range of the index and then to bound each of the resulting series. For example, suppose we try to find a lower bound on the arithmetic series we had seen earlier. If the number of terms were even, this could comprise of two partitions. The lower bound has to be greater than or equal to that obtained from the upper half. i.e. the lower half can now be ignored because the upper half has similar and is greater in value for the term at each of the corresponding positions. In other words, the initial terms of the summation are all constant.  The sum of the upper half  we can compute from arithmetic series to be equal to (n/2)^2.   This gives us the lower bound of Omega(n^2)  which is an asymptotically tight bound because the arithmetic sum of those n numbers is big-oh(n^2).
The caveat here is that if we had chosen to bound each term by the smallest term, because that term happens to be 1, we may have got a lower bound of n which would not be near the better bound we found.
This technique tells us that when performing an analysis of an algorithm, we can split the summation and ignore a constant number of initial terms and is generally applicable when each term is independent.
Another way this technique can help us is with infinite series.
For example to find an asymptotic upper bound on Sum k  = 0 to infinity of ((k ^ 2) / 2 ^ k ), we observer that the ratio of the consecutive terms is less than or equal to 8/9.
In this case if k > = 3 then the summation can be split into a fixed set of initial terms (in this case upto 2) and the rest ( 3 onwards to infinity) in the other set.
The latter can be said to be less than or equal to 9/8 * Sum k = 0 to infinity of ((8/9) ^ k).
In other words he total of the such a infinite series  is a constant.
This technique can also work for more difficult series such as harmonic series.
In the harmonic series of sum k = 1 to n of (1/k), we split the range into lg N pieces and upper bound the contribution of each piece by 1. Each piece consists of the terms starting at 1/(2^i) and going up to but not including 1/(2^(i+1)). This makes each piece to be upper bounded by 1 and the that for the harmonic series to be lg N + 1. With each of the pieces contributing a constant and there being lg N pieces, we get the upper bound as lg N  + 1.
We will next look at approximation by integrals.
Integrals are another technique for algorithm analysis

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