Concrete mathematics : a foundation for computer science

Please, don’t remind
me of the
midterm.
5.2 BASIC PRACTICE 175
The more complicated the derivation, the more important it is to check
the answer. This one wasn’t too complicated but we’ll check anyway. In the
small case m = 2 and n = 4 we have
(g/(40) + (f)/(Y) + ($yJ = l +i+i = :;
yes, this agrees perfectly with our closed form (4 + 1)/(4 + 1 - 2).
Problem 2: From the literature of sorting.
Our next sum appeared way back in ancient times (the early 1970s)
before people were fluent with binomial coefficients. A paper that introduced
an improved merging technique [165] concludes with the following remarks:
“It can be shown that the expected number of saved transfers . . is given by
the expression
Here m and n are as defined above, and mCn is the symbol for the number
of combinations of m objects taken n at a time. . . . The author is grateful to
the referee for reducing a more complex equation for expected transfers saved
to the form given here.”
We’ll see that this is definitely not a final answer to the author’s problem.
It’s not even a midterm answer.
First we should translate the sum into something we can work with; the
ghastly notation m-rPICm-n-l is enough to stop anybody, save the enthusiastic
referee (please). In our language we’d write
T = gk(zI:I:>/(:)) integers m > n 3 0.
The binomial coefficient in the denominator doesn’t involve the index of summation,
so we can remove it and work with the new sum
What next? The index of summation appears in the upper index of the
binomial coefficient but not in the lower index. So if the other k weren’t there,
we could massage the sum and apply summation on the upper index (5.10).
With the extra k, though, we can’t. If we could somehow absorb that k into
the binomial coefficient, using one of our absorption identities, we could then