Derive a formula for evaluating the following expression in terms of
p and its higher powers, given that
p is a
positive integer.
Σ ([3√m] + <3√m>)
m=1 to p3
Note: [x] is the greatest integer ≤ x, and <x> is the least integer ≥ x
When m is a perfect cube, the two numbers being added into the summation are equal, as the cube root of m. In between, perfect cubes, the two numbers being added in differ by 1. The last m will in fact be a perfect cube, i.e., p^3.
The following table shows, first, the last row to stop on for a given p. The next column shows the m value or range of m values for which it is valid. The next column shows how many m values are in that range. The next column is a sum, showing the two numbers that are added for that m and the total of those values. The next column multiplies that total by the number of m values that have that total (i.e., the third column). And finally the last column shows the cumulative total thus far for valid p values:
p m range how many m's pair sum total value cum. total
1 1 1 1+1=2 2 2
2-7 6 1+2=3 18
2 8 1 2+2=4 4 24
9-26 18 2+3=5 90
3 27 1 3+3=6 6 120
...
The alternate terms are in arithmetic progression and are easy enough to evaluate. The intermediate ones are more difficult.
The overall summations for several values of p are:
1 2
2 24
3 120
4 380
5 930
6 1932
7 3584
8 6120
9 9810
10 14960
11 21912
12 31044
13 42770
14 57540
15 75840
16 98192
17 125154
18 157320
19 195320
20 239820
calculated by
DEFDBL A-Z
CLS
FOR p = 1 TO 20
t = 0
FOR i = 1 TO p * p * p
cr = INT(i ^ (1 / 3) + .5)
IF cr * cr * cr <> i THEN cr = i ^ (1 / 3)
t = t + INT(cr) - INT(-cr)
NEXT
PRINT p, t
NEXT
Note the first three values agree with the cum. totals given in the manually drawn table.
I'd imagine there'd be a polynomial for these but I haven't found it using difference tables.
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Posted by Charlie
on 2008-05-22 17:23:07 |
thoughts
