(iii)
The first digit of x^3 goes through its entire cycle infinitely many times as x approaches 0 from the right. So lets consider a point where none of the cycles is cut off (a fraction of a cycle), as each cycle will have the same set of probabilities. The hex value 0.1 (1/16 in decimal) is at the end of the last (rightmost, highest) cycle in this infinite set. All the cycles will have the same probabilities. And further, 3^x will all begin with hex digit 1 in that range, as was previously said in part (i) for all x lower than .A1849C... in hex.
So what amount of the number line below .1 hex has x^3 with 1 as its first non-zero digit?
In every range of x = 1 to x = 10 hex, (and its images .1 to 1, etc.), cuberoot(2)/15 decimal is the fraction occupied by numbers whose cube root starts with 1 in hex. This fraction is .08399473665965822.... The portion of the number line in question is 1/16 of 1 unit, so that occupied by these numbers whose cube begins with 1 is .00524967104122864. Remember that since this is the actual portion of number line occupied, it still has to be divided by 16 at the end, together with all the other addends, to get the probability sought.
So we've found the portion of the number line below .1 hex that fits the criterion. The rest of the line from .1 to 10 hex now has to be evaluated and added in. It starts with a segment in which there is indeed a match, ending where x^3 no longer starts with 1 or where 3^x no longer starts with 1, whichever comes first.
It sounds like this calls for a computer program to keep track of the first digit changes up to 10 hex = 16 decimal and add in the portions where the digits match.
DefDbl A-Z
Dim crlf$, pwrref
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
digx3 = 1: dig3x = 1
pwrx3 = -3: pwr3x = 0
x = 1 / 16
valx3 = x ^ 3: val3x = 3 ^ x
hexx3$ = hexreal(valx3): hex3x$ = hexreal(val3x)
Text1.Text = Text1.Text & hexx3 & " " & hex3x & crlf
Text1.Text = Text1.Text & hexreal(digx3 * 16 ^ pwrx3) & " " & hexreal(dig3x * 16 ^ pwr3x) & crlf & crlf
Do
DoEvents
newpwrx3 = pwrx3
newdigx3 = digx3 + 1: If newdigx3 > 15 Then newdigx3 = newdigx3 - 15: newpwrx3 = newpwrx3 + 1
newpwr3x = pwr3x
newdig3x = dig3x + 1: If newdig3x > 15 Then newdig3x = newdig3x - 15: newpwr3x = newpwr3x + 1
newvalx3 = newdigx3 * 16 ^ newpwrx3
newval3x = newdig3x * 16 ^ newpwr3x
newxForx3 = newvalx3 ^ (1 / 3)
newxFor3x = Log(newval3x) / Log(3)
nextx = newxForx3: used$ = "x3"
If newxFor3x < nextx Then nextx = newxFor3x: used$ = "3x"
If newxFor3x = newxForx3 Then used$ = "x33x"
If 16 < nextx Then nextx = 16: used$ = "mx"
span = nextx - x
If digx3 = dig3x Then
tot = tot + span: ct = ct + 1
Text1.Text = Text1.Text & x & " " & hexreal(x) & " "
Text1.Text = Text1.Text & nextx & " " & hexreal(nextx)
Text1.Text = Text1.Text & " " & span & crlf
Text1.Text = Text1.Text & val3x & " " & hexreal(val3x) & " "
Text1.Text = Text1.Text & valx3 & " " & hexreal(valx3) & crlf
Text1.Text = Text1.Text & 3 ^ nextx & " " & hexreal(3 ^ nextx) & " "
Text1.Text = Text1.Text & nextx ^ 3 & " " & hexreal(nextx ^ 3) & crlf & crlf
End If
x = nextx
valx3 = x ^ 3: val3x = 3 ^ x
hexx3$ = hexreal(valx3)
digx3 = InStr("123456789abcdef", firstdig(valx3))
pwrx3 = pwrref
If InStr(used, "x3") > 0 Then
digx3 = newdigx3
pwrx3 = newpwrx3
End If
hex3x$ = hexreal(val3x)
dig3x = InStr("123456789abcdef", firstdig(val3x))
pwr3x = pwrref
If InStr(used, "3x") > 0 Then
dig3x = newdig3x
pwr3x = newpwr3x
End If
Loop Until Abs(nextx - 16) < 0.000000000001
Text1.Text = Text1.Text & crlf & ct & crlf & tot & crlf
End Sub
Function firstdig$(n)
s$ = hexreal$(n)
hexpt = InStr(s, ".")
For i = 1 To Len(s$)
If InStr("123456789abcdef", Mid(s, i, 1)) > 0 Then Exit For
Next i
firstdig$ = Mid(s, i, 1)
pwrref = hexpt - i - 1: If pwrref < 0 Then pwrref = pwrref + 1
End Function
Function hexreal$(n)
intpart = Int(n)
fracpart = n - intpart
h$ = "."
While intpart > 0
q = Int(intpart / 16)
r = intpart - q * 16
intpart = q
h = Mid("0123456789abcdef", r + 1, 1) + h
Wend
While fracpart > 0
m = fracpart * 16
d = Int(m)
fracpart = m - d
h = h + Mid("0123456789abcdef", d + 1, 1)
Wend
hexreal$ = h
End Function
Function mform$(x, t$)
a$ = Format$(x, t$)
If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
mform$ = a$
End Function
The results take some explaining. There are three lines to each segment so we'll take one at a time. Let's take the first segment below:
In decimal the first valid segment extends from .0625 to 0.078745... (note E-02 means x10^-2). In hex that's .1 to .1428a....
The last number on line one is the span in decimal, i.e., the space occupied on the number line: 0.078745... - .0625 = 0.016245.... There turn out to be 18 such segments and the total span is shown at the bottom.
Line 2 shows the values at the start of the span: 3^x = 1.07107548307291 decimal or 1.123200bb58774 hex and x^3 = 0.000244140625 decimal or .001 hex.
The last of the three lines shows the values that terminate the segment. In this case it's that the value of x^3 becomes .002 in hex at the end of the interval. Note that there are rounding problems throughout. Here, the extraneous zeros and digit 6 in .0020000000000006 don't upset the programming too much. Programming had to be done to work around in such cases as the last of the segments below, where d00000 is converted as cfffff.ffffff78 and e00000 as dfffff.ffffff88: the first a seeming mismatch to d0f.957ffd96468, and the second a seeming match to d3d.09c7e740fb.
0.0625 .1 7.87450656184296E-02 .1428a2f98d728c 1.62450656184296E-02
1.07107548307291 1.123200bb58774 0.000244140625 .001
1.09036259486235 1.172200c5b729e 0.00048828125 .0020000000000006
0.157490131236859 .285145f31ae516 0.198425131496025 .32cbfd4a7adc7a 4.09350002591658E-02
1.18889058827495 1.305b223329c27 0.00390625 .01
1.24357747691624 1.3e5b17f165bef 0.0078125 .020000000000002
0.39685026299205 .6597fa94f5b8f4 0.5 .8 0.10314973700795
1.54648494109336 1.8be66fe5c0a1f 0.0625 .10000000000001
1.73205080756888 1.bb67ae8584caa 0.125 .2
2.0800838230519 2.14805f98f253 2.09590327428938 2.188d1df2a6f82 1.58194512374803E-02
9.82770707891505 9.d3e49c729b9a 9 9.
10 9.ffffffffffff8 9.2069059840103 9.34f7ca62abc2
2.15443469003188 2.278908270e09e 2.18265833864414 2.2ec2b266d179e 2.82236486122542E-02
10.6641582839993 a.aa0646fd24fd 10 a.0000000000008
11 b. 10.3981787011419 a.65ef0a135e4
2.22398009056932 2.3956c25bf348a 2.26185950714291 2.430939835353c 3.78794165735994E-02
11.510870586235 b.82c86a2c5e228 11 a.ffffffffffff
12 b.ffffffffffff 11.5716923061382 b.925a6d4e3d06
2.28942848510666 2.4a17fc36105ea 2.33471751947279 2.55b00c1f88ace 0.045289034366129
12.3690114459331 c.5e7788bc21db 12 c.
13 d. 12.7263254917658 c.b9f077a962a28
2.35133468772076 2.59f111f1b605c 2.40217350273288 2.66f4d7b98e344 5.08388150121224E-02
13.2395052594758 d.3d503778ab278 13 c.ffffffffffff
14 e. 13.8615921511138 d.dc914d9f86208
2.41014226417523 2.68ff155b570a8 2.46497352071793 2.77088130fd4e4 5.48312565426974E-02
14.1231021770879 e.1f839fd0a8b88 14 d.fffffffffffe8
15 e.ffffffffffff8 14.9774119479579 e.fa37ab5f32dc8
2.46621207433047 2.7759acac3feba 2.51984209978975 2.85145f31ae516 5.36300254592765E-02
15.0204242456681 f.053a85fb30a3 15 e.fffffffffffe8
15.9319973076973 f.ee9760248517 16 10.
2.52371901428583 2.86127306a6a7a 3.15464876785729 3.27970fc850518 0.630929753571457
16 f.ffffffffffff8 16.0739642992686 10.12ef53066e23
32 1f.fffffffffffe 31.394461522009 1f.64fb6e288f1c
3.1748021039364 3.2cbfd4a7adc78 3.52371901428583 3.86127306a6a7a 0.348916910349431
32.7164040647631 20.b76641bce0fc 32 1f.fffffffffffe
48 2f.fffffffffffe 43.7525943313296 2b.c0aa05a83738
3.63424118566428 3.a25da15e344fc 3.78557852142874 3.c91bac89f9fb6 0.151337335764465
54.1968086954604 36.32620dfe9216 48 2f.fffffffffffc
64 3f.fffffffffffa 54.2496295100315 36.3fe7b835b3b4
7.57115704285749 7.92375913f3f6c 8 7.ffffffffffffc 0.428842957142511
4096 fff.fffffffffc8 433.997036080252 1b1.ff3dc1ad9da
6560.99999999999 19a0.fffffffff9 512 1ff.ffffffffffd
8.20208679642895 8.33bbf5d59da08 8.57115704285749 8.92375913f3f68 0.369070246428542
8191.99999999999 1fff.fffffffff4 551.78905571083 227.c9ff8e18bca
12288 2fff.ffffffffe8 629.677764111656 275.ad81f2e5f74
11.7258058107148 b.b9ce68dc4448 11.8661198063047 b.ddba07127f288 0.140313995589965
393215.999999999 5ffff.fffffffcc 1612.23406396067 64c.3beb9da0444
458751.999999999 6ffff.fffffffcc 1670.8066172931 686.ce7e788e46
14.8804545785721 e.e16578a494998 14.930044627269 e.ee176799f0da 4.95900486969294E-02
12582912 bfffff.ffffff9 3294.94823194994 cde.f2bf543dff8
13287449.4577502 cac019.752f1e08 3328 cff.fffffffffe8
14.9533125909019 e.f40c4b40c9f28 15.020768574162 f.055116dacf7a 6.74559832600874E-02
13631488 cfffff.ffffff78 3343.58398423117 d0f.957ffd96468
14680064 dfffff.ffffff88 3389.03820653283 d3d.09c7e740fb
There are 18 segments totalling 2.63329868149249 on the number line.
We need to add in that .00524967104122864 calculated at the beginning for the stretches before .1 hex. The total is 2.638548352533718, which is to be divided by 16, giving .1649092720333574 as the probability sought for part (iii).
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Posted by Charlie
on 2016-06-20 09:01:51 |
part (iii) solution
