Find all possible values of a real number x that satisfy this equation:
9x floor(x)
---- = --------------
10 x – floor(x)
From the
graph (click to see), one looks for solutions between 1 and 2, between 2 and 3, etc., probably ending around 10 or 11.
First one:
9(1+f) / 10 = 1/f
where f is the fractional part.
9 + 9f = 10/f
9f + 9f^2 -10 = 0
f = (-9+sqrt(81+360))/18
= 2/3
first solution: 5/3
Then
9(2+f) / 10 = 2/f
18 + 9f = 20/f
18f + 9f^2 -20 = 0
9f^2 + 18f - 20 = 0
f = (-18 + sqrt(324+720))/18
~= .7950549357115
second solution: approx. 2.7950549357115
In general
k + (sqrt(81k^2 + 360k) - 9k)/18
The valid values of k are 1 through 8, as the table shows:
1 1.666666666666667
2 2.795054935711502
3 3.862907813126304
4 4.905932629027116
5 5.935921354681383
6 6.95811402901264
7 7.975240527365851
8 8.988876515698589
9 10
10 11.00925212577332
11 12.01707159867238
12 13.02376916856849
13 14.02957133205101
14 15.03464719546263
15 16.03912563829967
16 17.04310664416703
as 10 is not valid as 9.99999... in this context, i.e., floor(10) <> 9; and, the more so, floor(11+) <> 10, etc.
table from:
DEFDBL A-Z
FOR k = 1 TO 16
PRINT k,
f = (SQR(81 * k * k + 360 * k) - 9 * k) / 18
PRINT k + f
NEXT
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Posted by Charlie
on 2015-01-30 14:36:09 |
solution
