 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  ABC's Of Greatest Power (Posted on 2008-05-16) Determine the value of the positive integer constant A that satisfies this relationship:

6
{y}[y] dy = 19/20
A

Bonus Question:

What are the possible pair(s) of the positive integer constants (B, C) that satisfy this relationship?

C
[y]{y} * ln [y] dy = 18
B

Notes:

(i) [x] is the greatest integer ≤ x, and {x}= x - [x]

(ii) ln x is the natural logarithm of x.

 See The Solution Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 2

Because of their breaks at integer values, the functions are best integrated piecewise, between integral values of y.

For a given unit interval of y, where [y] = k, the first expression has the value

Integ{0 to 1} x^k dx

where x={y} but in my Integ expression, the {} do not have the given meaning, but merely show the limits of integration.

The evaluation of the above integral is

(x^(k+1))/(k+1) | 0 to 1

= 1/(k+1)

when k (that is, [y]) takes on the following values, the integral for the unit interval is:

k   integral
1     1/2
2     1/3
3     1/4
4     1/5
5     1/6

If A were 1, the integral would be the total of these: 29/20. We need to get rid of 10/20 or 1/2, and can do this be removing the k=1 interval. So the integration starts at k=2, or y=2, and thus A = 2.

Bonus:

In the same way, using unit intervals of y, where [y]=k, we add up, from B to C-1 (as we're dealing with the bottom integer of each unit interval), the evaluations of

Integ{0 to 1} k^x * (ln k) dx

This integral is k^x | 0 to 1

= k - 1

So we need a B and C such that

Sigma{k = B to C-1} (k - 1) = 18

or Sigma{i = B-1 to C-2} i  = 18

The lower and upper limits for the Sigma consistent with evaluating to 18, could be, with the corresponding B and C values:

i values     B    C
18 to 18    19   20
5 to 7       6    9
3 to 6       4    8

so (B,C) is (19,20), (6,9) or (4,8)

 Posted by Charlie on 2008-05-16 16:45:37 Please log in:
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