mm-939
===
Subject: Re: How many branches to non-integer roots?
So does this same logic hold for taking the ith root of i?
That is,
i ** (1/i)
which would also render real values?
>Then for fun, we decided to compute i**i (i to the i
power). At first,
>we thought this would be difficult if not impossible. But
using x=pi/2
>we had
>e**(i*pi/2) = cos(pi/2) + i*sin(pi/2) = i
>So i**i = [e**(i*pi/2)]**i = e**(-pi/2)
> Yes. Since exp(it) = cos(t) + i sin(t), it follows that
> i = exp[i*(pi/2 + 2N pi)], for integer N.
>Granted, this seems to be taking a few liberties with the
properties,
>but if we did this right, i**i would actually have a real
value.
> Right.
>Now of course, we could have used (2N + pi/2) in the
substitution,
>giving us an infinite number of values ...
>e**(-2N -pi/2) for integer values of N
> Close, but not quite.
> Rather, i^i = exp[-2Npi - pi/2].
> ^^
>Again .. all real values.
> That's correct. All branches have real values.
>I would have thought that we would get branches for this
power, much
>like we did for fractional values. And of course, all of our
>calculations could be faulty anyway.
> There is branching--infinitely many branches, in fact. It
just
> so happens that they all real give real numbers.
> ---
> Stan Liou
===
Subject: Re: How many branches to non-integer roots?
>So does this same logic hold for taking the ith root of i?
That is,
>i ** (1/i)
>which would also render real values?
Of course. Since i*i = -1, it follows that 1/i = -i,
and therefore i^(1/i) = 1/(i^i).
---
Stan Liou
===
Subject: Re: How many branches to non-integer roots?
CORRECTION
In the calculation of g, you only need to worry about the
positive
square root and the real value of the Log. (The argument is
always >=
0, so there is always a real value.)
What I meant to say, of course, is that the formula fails if
a = b = 0
(no big deal). Otherwise the argument is always > 0 (taking
the
positive square root), so there is always a real value for
the log.
===
Subject: Re: How many branches to non-integer roots?
>So does this same logic hold for taking the ith root of i?
That is,
>i ** (1/i)
>which would also render real values?
> Of course. Since i*i = -1, it follows that 1/i = -i,
> and therefore i^(1/i) = 1/(i^i).
> ---
> Stan Liou
Coincidentally, I have recently also been looking at exactly
this same
question (see Question about powers posting a while back in
alt.math.undergrad). Using Euler's formula I arrived, after 
a
lot of
playing about and head-scratching, at the general formula
(a + b*i)^(c + d*i) = exp(c*g - d*h) * [cos(t) + i*sin(t)]
where
g = Log[Sqrt(a^2 + b^2)]
h = Arctan(b/a) + 2*n*pi
t = d*g + c*h
n is any integer (positive, negative or zero)
In the calculation of g, you only need to worry about the
positive
square root and the real value of the Log. (The argument is
always >=
0, so there is always a real value.)
In the calculation of h, you need to use the
quadrant-sensitive
version of the Arctan function. For example, Arctan(-1/-1) =
5*pi/4
(not pi/4). Also, Arctan(1/0) = pi/2, Arctan(-1/0) = 3*pi/2
etc.
(Depending on your point of view, the + 2*n*pi is already
implicit
in the Arctan function, but I put it in for clarity.)
I'm not totally confident with the result, but 
it's reassuring
that it
does give i^i = exp(-pi/2 + 2*n*pi) which, with the reversal
of the
sign of n, is the same as previous posters have found.
Does anyone agree/disagree with my general formula?
===
Subject: Re: How many branches to non-integer roots?
> Of course. Since i*i = -1, it follows that 1/i = -i,
> and therefore i^(1/i) = 1/(i^i).
>Coincidentally, I have recently also been looking at exactly
this same
>question (see Question about powers posting a while back in
>alt.math.undergrad). Using Euler's formula I arrived, after
a lot of
>playing about and head-scratching, at the general formula
>(a + b*i)^(c + d*i) = exp(c*g - d*h) * [cos(t) + i*sin(t)]
>where
>g = Log[Sqrt(a^2 + b^2)]
>h = Arctan(b/a) + 2*n*pi
>t = d*g + c*h
>n is any integer (positive, negative or zero)
Transforming into polar coordinates,
z = a+bi = r exp[i(T+2npi)], any integer n.
z^w = exp[Log r + ih]^w
= exp[(c+d*i)(Log r + ih)]
= exp[cg-dh]exp[i(dg + ch)]
You are correct.
>In the calculation of g, you only need to worry about the
positive
>square root and the real value of the Log. (The argument is
always >=
0, so there is always a real value.)
Right.
>In the calculation of h, you need to use the
quadrant-sensitive
>version of the Arctan function. For example, Arctan(-1/-1) =
5*pi/4
>(not pi/4). Also, Arctan(1/0) = pi/2, Arctan(-1/0) = 3*pi/2
etc.
>(Depending on your point of view, the + 2*n*pi is already
implicit
>in the Arctan function, but I put it in for clarity.)
Or simply h = Arg(a+bi) + 2npi, the complex argument.
Interestingly, computers often provide atan2(x,y) function for
use in exactly this situation (part of the ANSI C standard,
but
also at the hardware level--e.g., FATAN2 opcode in x86/7
chipsets).
>I'm not totally confident with the result, but 
it's
reassuring that it
>does give i^i = exp(-pi/2 + 2*n*pi) which, with the reversal
of the
>sign of n, is the same as previous posters have found.
Since n is varies over all of the integers, positive, zero,
and
negative, the solution set specified by exp(-pi/2+2npi) and
exp(-pi/2-2npi) are exactly equivalent.
>Does anyone agree/disagree with my general formula?
Not I.
---
Stan Liou
===
Subject: Re: An Improved Number System, Weights & Measures
> I agree that the current system for measuring time is a
mess, but you'll
never
> metricise the number of days in a year (without some serious
planetary
> engineering). This makes any system of metric time
inherently
unsatisfactory,
> and goes some way to explaining why none of the various
attempts to
introduce
> metric time have been successful.
> While day and year are intrinsic physical units, why is a
day divided into
a
> series of 24, 60, and 60 units, instead of, say, 100,000?
> Here's a question I don't know the answer 
to. Were hours,
minutes, and
> seconds a worldwide standard long before metrification was
introduced?
If
> so, that might explain why time wasn't 
metrified.
A 24-hour day goes back at least as far as the Romans, who
stretched hours
so that there were 12 hours of daylight and 12 hours of
night. I suspect
that the fact we use 60's goes back to Babylonians (who used
base 60 for
their numbers).
As to 10 being natural for finger-counting, maybe Edgar del
Ray has 6
fingers on each hand. (This is a dominant genetic trait, btw.)
-- Christopher Heckman
===
Subject: Re: An Improved Number System, Weights & Measures
X-RFC2646:  Original
> A 24-hour day goes back at least as far as the Romans, who
stretched
hours
> so that there were 12 hours of daylight and 12 hours of
night. I suspect
> that the fact we use 60's goes back to Babylonians (who
used base 60 for
> their numbers).
Actually, Babylonians preferred counting things out in groups
of 12. The
rest of the world, however, had standardized on base 10. 60
became a popular
unit of counting in the ancient world because it is the least
common
multiple of 10 and 12.
This accounts for both the measuring of time in units of 12
and 60. This is
also the reason why a degree (as a unit of angle measurment)
is 1 / 360 of a
full circle.
> As to 10 being natural for finger-counting, maybe Edgar del
Ray has 6
> fingers on each hand. (This is a dominant genetic trait,
btw.)
> -- Christopher Heckman
===
Subject: Re: An Improved Number System, Weights & Measures
> Actually, Babylonians preferred counting things out in
groups of 12. The
> rest of the world, however, had standardized on base 10. 60
became a
popular
> unit of counting in the ancient world because it is the
least common
> multiple of 10 and 12.
> This accounts for both the measuring of time in units of 12
and 60. This
is also the reason why a degree (as a unit of angle
measurment) is 1 / 360 of a full circle.
I humbly submit another hypothesis. The Babylonians were
quite good at
astronomical observations, so they noticed a repeating
pattern of the
motion between the Sun and the Earth in some 36* number of
days.
Possibly their measurement was inaccurate, to get at the
number 360.
But I much rather suspect they 
Ôdefined' it to be 360, as this
is such
a wonderful harmonious number (it's the smallest common
multiple of all
numbers 1 through 10, except for that pesky 7).
It then comes perfectly natural to draw this heavenly motion
in a
1-year circle, subdivided in 360 segments, a day each.
That 360-day period is too large to be practical, so they
needed
a subdivision, which also presents itself quite naturally.
Those astronomical observations showed another celestial body
to
display a periodic pattern 12 times within one year. Those
periods,
aptly called Ômo(o)n-ths' had 30 days each. So 
within a year,
we
have 3x4 months of 5x6 days ! Isn't that perfect ? It Had To
Be !
Then they had to subdivide a day further and further to get to
the second, the period of the clock that ticks incessantly in
every human body. I'm not quite sure where 24/60/60 came 
from,
but I always thought the Romans had 12 hours in a full day,
not some stretching 24 hours as stated by Christopher.
--
__________
~ALCATEL/~~~~Walter Baeck
~~BELL~/~~~~~Alcatel Belgium
~~~~~~/~~~~~~DSL Microelectronics Design
~~~~~/~~~~~~~E-mail : walter.baeck@alcatel.be
~~~~/~~~~~~~~Phone : +32 3 240 73 86
===
Subject: Re: An Improved Number System, Weights & Measures
> As to 10 being natural for finger-counting, maybe Edgar del
Ray has 6
> fingers on each hand. (This is a dominant genetic trait,
btw.)
Presumably the trait is dominant because in a eye-poking
fight, when the
other guy block five of your fingers, you still 
have one left
to poke him
in
the eye with.
Boy, we've drifted a ways from the topic, 
haven't we.
===
Subject: Re: An Improved Number System, Weights & Measures
> Here's a question I don't know the answer 
to. Were hours,
minutes, and
> seconds a worldwide standard long before metrification was
introduced?
If
> so, that might explain why time wasn't 
metrified.
and Proginoskes replied:
> A 24-hour day goes back at least as far as the Romans, who
stretched hours
so
> that there were 12 hours of daylight and 12 hours of night.
I suspect that
the
> fact we use 60's goes back to Babylonians (who used base 
60
for their
> numbers).
Similar statements can be made about other non-metric units.
For example,
the relationship between foot and mile goes back to the
romans, but was
replaced by metrification. Presumably 
metrification of length
was easier
to
swallow (than of time) because every country had different
units for
length.
So what I'm wondering is, was the 24-60-60 time standard
universally
accepted (meaning, used in standard practice by every country
in the
world),
thus making it more difficult to replace it with a decimalized
version?
===
Subject: Re: An Improved Number System, Weights & Measures
> For me, it was enlightening to see the digit patterns of
fractions in
other
> bases. For example in base twelve, 1/7 = .186X35,
repeating (X is
do, the
> digit for ten), which has some properties similar to those
of the digits
of
> 1/7 in base ten.
and Phil Carmody replied:
> I would hope that it have all the same properties of 1/7 in
base 10.
Which
> property is it lacking?
I'm talking about properties of the patterns of the
duodecimal and decimal
expansions of 1/7. There's no reason (I can think of) to
expect they'd
have
all the same properties. But there are some they have that
are similar.
What are they?
Bob H
===
Subject: Re: An Improved Number System, Weights & Measures
> For me, it was enlightening to see the digit patterns of
fractions in
other
> bases. For example in base twelve, 1/7 = .186X35,
repeating (X is
do, the
> digit for ten), which has some properties similar to those
of the
digits of
> 1/7 in base ten.
> and Phil Carmody replied:
> I would hope that it have all the same properties of 1/7 in
base 10.
Which
> property is it lacking?
> I'm talking about properties of the patterns of the
duodecimal and
decimal
> expansions of 1/7.
So am I.
> There's no reason (I can think of) to expect 
they'd have
> all the same properties.
Do you have any reason to expect they'd not have all the
similar
properties?
> But there are some they have that are similar.
> What are they?
All of them.
Back to my original point - which property do you believe is
lacking.
Phil
--
They no longer do my traditional winks tournament lunch -
liver and bacon.
It's just what you need during a winks tournament lunchtime
to replace lost
===
Subject: Re: An Improved Number System, Weights & Measures
> For me, it was enlightening to see the digit patterns of
fractions in
other
> bases. For example in base twelve, 1/7 = .186X35,
repeating (X is
do,
> the
> digit for ten), which has some properties similar to
those of the
digits of
> 1/7 in base ten.
> and Phil Carmody replied:
> I would hope that it have all the same properties of 1/7
in base 10.
Which
> property is it lacking?
> I'm talking about properties of the patterns of the
duodecimal and
decimal
> expansions of 1/7.
> So am I.
> There's no reason (I can think of) to expect 
they'd have
> all the same properties.
> Do you have any reason to expect they'd not have all the
similar
properties?
I would not expect the repeat period to be the same, though
for 1/7 it is.
> But there are some they have that are similar.
> What are they?
> All of them.
> Back to my original point - which property do you believe
is lacking.
In decimal, the digit pairs of 1/7 form a doubling sequence:
14 28 57 (it is 57, not 56, because of a carry from the next
term)
This property is lacking in duodecimal. But you have it in
reverse:
35 x2 -> 6X
6X x2 -> (1)18
18 x2 -> 34 (which becomes 35 with the carry)
> They no longer do my traditional winks tournament lunch -
liver and
bacon.
> It's just what you need during a winks tournament 
lunchtime
to replace
lost
> ... liver.
I have been trying to figure out the meaning of your sig for a
while. Is
Ôwinks' what would be called in the U.S. 
Ôtiddlywinks' or is
it something
else?
Bob H
===
Subject: Re: An Improved Number System, Weights & Measures
> For me, it was enlightening to see the digit patterns of
fractions in
other
> bases. For example in base twelve, 1/7 = .186X35,
repeating (X is
do,
> the
> digit for ten), which has some properties similar to
those of the
digits of
> 1/7 in base ten.
 and Phil Carmody replied:
> I would hope that it have all the same properties of 1/7
in base 10.
Which
> property is it lacking?
 I'm talking about properties of the patterns of the
duodecimal and
decimal
> expansions of 1/7.
 So am I.
 There's no reason (I can think of) to expect 
they'd have
> all the same properties.
 Do you have any reason to expect they'd not have all the
similar
properties?
> I would not expect the repeat period to be the same, though
for 1/7 it
is.
> But there are some they have that are similar.
> What are they?
 All of them.
 Back to my original point - which property do you believe
is lacking.
> In decimal, the digit pairs of 1/7 form a doubling sequence:
> 14 28 57 (it is 57, not 56, because of a carry from the
next term)
In accordance with the following sum:
14
+ 28
+ 56
+ 112
+ 224
+ 448
+ 896
+ 1792
...
=1428571428571392,
which is approximately 1/7 (if you start with a decimal). The
infinite
sum is 1/7, because the sequence is geometric, with the first
term
being 14/100, and the common ratio 2/100. Since |2/100| < 1,
the series
converges to (14/100) / [1 - (2/100)] = 14 / (100 - 2) =
14/98 = 1/7.
There's also a similar trick for checking for divisibility 
by
7 that
I found on my own. You break off the last digit, multiply the
rest by
3, then add the digit you've broken off. The new number has
the same
remainder when you divide by 7. Continue until you can do the
division
by 7 mentally.
For instance,
142857 is broken into 14285;7 -> 3 * 14285 + 7 = 42862.
42862 -> 4286;2 -> 3 * 4286 + 2 = 12860
12860 -> 1286;0 -> 3 * 1286 + 0 = 3858
3858 -> 385;8 -> 3 * 385 + 8 = 1163
1163 -> 116;3 -> 3 * 116 + 3 = 351 (351 / 7 = 50 R 1, so you
could
stop)
351 -> 35;1 -> 3 * 35 + 1 = 106
106 -> 10;6 -> 3 * 10 + 6 = 36 (36 / 7 = 5 R 1, so you could
stop)
36 -> 3;6 -> 3 * 3 + 6 = 15
15 -> 1;5 -> 3 * 1 + 5 = 8 (and you MUST stop here, because
you
only have one digit left)
This works because 10^1 - 3 is a multiple of 7. If you pull
off the last
TWO digits, you get a divisibility test for any factor of
10^2 - 3, which
is 97:
142857 -> 1428;57 -> 3 * 1428 + 57 = 4341
4341 -> 43;41 -> 3 * 43 + 41 = 170
170 -> 1;70 -> 3 * 1 + 70 = 73,
so the remainder when you divide 142857 by 97 is 73.
BTW, you can use reverse division (another useless discovery
of mine)
to get the quotient. First, 142857 divided by 97 has a
remainder of 73,
so the quotient is the same as that of (142857 - 73) = 142784.
To divide 142784 by 97, note that the units digit of 142784 /
97
must be a 2. This is the only digit you can multiply by 7 to
get the
one's digit equal to 4. Now, we subtract 97 * 2 = 194 from
the 784
part of 142784, move over a column, and repeat. Since 142784
is
divisible by 97, we will eventually get rid of all the digits:
2
______
97 142784
-194
14259
The next digit must be a 7, so 142784/97 ends in 72.
72
______
97 142784
-194
14259
-679
1358
472
______
97 142784
-194
14259
-679
1358
-388
97
(Isn't cut-and-paste great? 8-) )
1472
______
97 142784
-194
14259
-679
1358
-388
97
-97
so 142784 / 97 = 1472.
Note that the final digit of 97 -- 7 -- is relatively prime to
10. If the
divisor ends in an even digit or a 5, I would be stuck on the
first digit.
So reverse division could be used as an argument AGAINST base
12. Base
13
(or any prime number) would be best, since you could always
use reverse
division when doing division. (Unless the number is a
multiple of 13,
that is. I'll leave it as an exercise to y'all 
to figure out
how to divide
a multiple of 13 by a multiple of 13, in the base 13 system.)
> This property is lacking in duodecimal.
But you do have it for fractions like (A/144) / [1 - (2/144)]
= A / (144 - 2) = A / 142. For instance, 1/142 =
.010204081428...
-- Christopher Heckman
> [...]
===
Subject: Re: An Improved Number System, Weights & Measures
> In decimal, the digit pairs of 1/7 form a doubling sequence:
 14 28 57 (it is 57, not 56, because of a carry from the next
term)
> In accordance with the following sum:
> 14
> + 28
> + 56
> + 112
> + 224
> + 448
> + 896
> + 1792
> ...
> =1428571428571392,
> which is approximately 1/7 (if you start with a decimal).
The infinite
> sum is 1/7, because the sequence is geometric, with the
first term
> being 14/100, and the common ratio 2/100. Since |2/100| <
1, the series
> converges to (14/100) / [1 - (2/100)] = 14 / (100 - 2) =
14/98 = 1/7.
> There's also a similar trick for checking for divisibility
by 7 that
> I found on my own. You break off the last digit, multiply
the rest by
> 3, then add the digit you've broken off. The new number 
has
the same
> remainder when you divide by 7. Continue until you can do
the division
> by 7 mentally.
> For instance,
> 142857 is broken into 14285;7 -> 3 * 14285 + 7 = 42862.
> 42862 -> 4286;2 -> 3 * 4286 + 2 = 12860
> 12860 -> 1286;0 -> 3 * 1286 + 0 = 3858
> 3858 -> 385;8 -> 3 * 385 + 8 = 1163
> 1163 -> 116;3 -> 3 * 116 + 3 = 351 (351 / 7 = 50 R 1, so
you could
stop)
> 351 -> 35;1 -> 3 * 35 + 1 = 106
> 106 -> 10;6 -> 3 * 10 + 6 = 36 (36 / 7 = 5 R 1, so you could
stop)
> 36 -> 3;6 -> 3 * 3 + 6 = 15
> 15 -> 1;5 -> 3 * 1 + 5 = 8 (and you MUST stop here, because
you
> only have one digit left)
> This works because 10^1 - 3 is a multiple of 7. If you pull
off the last
> TWO digits, you get a divisibility test for any factor of
10^2 - 3, which
> is 97:
However, if you pull off 2 digits and multiply by _2_, then
you get
divisibility by 98. When you're down to 2 digits, then you
can turn
that into a divisibility by 7 test.
If you pull off 3 digits, and multiply by 1 :-), and
_subtract_ then
you get divisibility by 1001. When you're down to 3 digits,
then
use the residue to simply work out the divisibility by 7, 11
and 13.
Phil
--
They no longer do my traditional winks tournament lunch -
liver and bacon.
It's just what you need during a winks tournament lunchtime
to replace lost
===
Subject: Re: An Improved Number System, Weights & Measures
> There's also a similar trick for checking for divisibility
by 7 that
> I found on my own. You break off the last digit, multiply
the rest by
> 3, then add the digit you've broken off. The new number 
has
the same
> remainder when you divide by 7. Continue until you can do
the division
> by 7 mentally.
You must be kidding. Multiplying by 3 all those times is
harder than the
original problem.
> If you pull off 3 digits, and multiply by 1 :-), and
_subtract_ then
> you get divisibility by 1001. When you're down to 3 
digits,
then
> use the residue to simply work out the divisibility by 7,
11 and 13.
Nice.
Another method with more but simpler steps, using 7 divides
21:
Subtract twice the last digit from the rest of the digits.
Example: 952 -> 95-2*2=91 -> 9-1*1 = 7.
--
Jens Kruse Andersen
===
Subject: Re: An Improved Number System, Weights & Measures
> For me, it was enlightening to see the digit patterns of
fractions in
other
> bases. For example in base twelve, 1/7 = .186X35,
repeating (X is
do,
> the
> digit for ten), which has some properties similar to
those of the
digits of
> 1/7 in base ten.
> I would not expect the repeat period to be the same, though
for 1/7 it
is.
As it was the repeating 6-digit sequence whose properties were
being compared, I didn't consider having 6 digits to be one 
of
those properties. However, you're right, on average 1 in 3
bases
will have a period of 6 for 1/7. 1 in 3 have a period 3, 1 in
6
have period 2, and 1 in 6 have period 1.
> But there are some they have that are similar.
> What are they?
 All of them.
 Back to my original point - which property do you believe
is lacking.
> In decimal, the digit pairs of 1/7 form a doubling sequence:
> 14 28 57 (it is 57, not 56, because of a carry from the
next term)
> This property is lacking in duodecimal. But you have it in
reverse:
> 35 x2 -> 6X
> 6X x2 -> (1)18
> 18 x2 -> 34 (which becomes 35 with the carry)
Which is a property that is similar.
Order_7(2)=3, so any base b which has Order_7(b)=6, i.e. a
repeating
digit pattern of length 6 will have either digits that either
double
or halve three times before they loop.
Similarly, because Order_7(6)=2, when viewed in triplets the
two sets of 3 digits will be a ratio of ~6.
857_10 A35_12
------ ~= ------ ~= 6
142_10 186_12
(But this time the relationship is in the same direction.)
> They no longer do my traditional winks tournament lunch -
liver and
bacon.
> It's just what you need during a winks tournament 
lunchtime
to replace
lost
> ... liver.
> I have been trying to figure out the meaning of your sig for
a while. Is
> Ôwinks' what would be called in the U.S. 
Ôtiddlywinks' or
is it something
> else?
Absolutely. I believe the Americans use different, less
liquid,
performance-enhancing drugs than us Brits, but the underlying
game's the same. (Trust me - beer steadies the hand.)
Phil
--
They no longer do my traditional winks tournament lunch -
liver and bacon.
It's just what you need during a winks tournament lunchtime
to replace lost
===
Subject: Different kind of meeting problem.
Bill and Ted agree to meet between 3:00 P.M. and 4:00 P.M..
Each can
wait for 15 minutes. What is the
probability that they will meet?
This is an easy problem to calculate if you know the formula.
Let d be
the waiting time which is 0.25 because
15 minutes/60 minutes = .25. n = 2 because there are two
people.
n(d^(n-1)-(n-1)*d^(n-1)=2*.25-(1/16)=0.4375.
But what if there is a 90% probability that Bill will show
up, and a 70%
probability that Ted will show up?
My thinking is that it is ..9*.7*.4375=.63*.4375=0.275625.
Does anyone agree?
--
Patrick D. Rockwell
prockwell@thegrid.net
hnhc85a@prodigy.net
patri48975@aol.com
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00 P.M..
Each can
> wait for 15 minutes. What is the
> probability that they will meet?
that Bill and Ted's ARRIVAL times are (i) uniformly
distributed
between 3:00 and 4:00 and (ii) independent?
> This is an easy problem to calculate if you know the
formula. Let d be
> the waiting time which is 0.25 because
> 15 minutes/60 minutes = .25. n = 2 because there are two
people.
> n(d^(n-1)-(n-1)*d^(n-1)=2*.25-(1/16)=0.4375.
I don't understand this formula. There is a missing bracket
somewhere,
or something else has gone wrong? However, the answer of
0.4375 does
agree with the answer I get by working it out from first
principles...
> But what if there is a 90% probability that Bill will show
up, and a 70%
> probability that Ted will show up?
I think you are assuming that the event of Bill showing up is
independent of the event of Ted showing up, as well as
independent of
the times of arrival? (For example, if both would be deterred
by rain
then the showing up events would no longer be independent and
this
would inßuence the result.)
> My thinking is that it is ..9*.7*.4375=.63*.4375=0.275625.
> Does anyone agree?
With the above caveats, yes.
===
Subject: Re: Different kind of meeting problem.
X-RFC2646:  Original
> Bill and Ted agree to meet between 3:00 P.M. and 4:00
P.M.. Each can
> wait for 15 minutes. What is the
> probability that they will meet?
> that Bill and Ted's ARRIVAL times are (i) uniformly
distributed
> between 3:00 and 4:00 and (ii) independent?
> This is an easy problem to calculate if you know the
formula. Let d be
> the waiting time which is 0.25 because
> 15 minutes/60 minutes = .25. n = 2 because there are two
people.
> n(d^(n-1)-(n-1)*d^(n-1)=2*.25-(1/16)=0.4375.
> I don't understand this formula. There is a missing 
bracket
somewhere,
> or something else has gone wrong? However, the answer of
0.4375 does
> agree with the answer I get by working it out from first
principles...
Your right. There is a missing bracket. It should be
P=n*(d^(n-1))-(d^(n-1))=2*0.25-(1/16)=0.5-0.0625=0.4375.
> But what if there is a 90% probability that Bill will show
up, and a 70%
> probability that Ted will show up?
> I think you are assuming that the event of Bill showing up
is
> independent of the event of Ted showing up, as well as
independent of
> the times of arrival? (For example, if both would be
deterred by rain
> then the showing up events would no longer be independent
and this
> would inßuence the result.)
> My thinking is that it is ..9*.7*.4375=.63*.4375=0.275625.
> Does anyone agree?
> With the above caveats, yes.
--
-------------------------------
Patrick D. Rockwell
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00
P.M.. Each can
> wait for 15 minutes. What is the
> probability that they will meet?
>
...
 But what if there is a 90% probability that Bill will show
up, and a
70%
> probability that Ted will show up?
> I think you are assuming that the event of Bill showing up
is
> independent of the event of Ted showing up...
I had an interesting thought about this. I think it is
IMPOSSIBLE for
these two events to be independent. There is always some
non-zero
possibility that a natural or man-made catastrophe will
prevent both
Bill AND Ted from showing up. For example, destruction of the
city in
which Bill and Ted live, destruction of the earth by an
asteroid
impact, etc. (sorry about the less-than-pleasant scenarios
here
folks). Therefore, there must always be some correlation,
however
small?
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00
P.M.. Each can
> wait for 15 minutes. What is the
> probability that they will meet?
 ...
 But what if there is a 90% probability that Bill will
show up, and a
70%
> probability that Ted will show up?
 I think you are assuming that the event of Bill showing
up is
> independent of the event of Ted showing up...
> I had an interesting thought about this. I think it is
IMPOSSIBLE for
> these two events to be independent. There is always some
non-zero
> possibility that a natural or man-made catastrophe will
prevent both
> Bill AND Ted from showing up.
The dependence will be so small as to be within the margin of
error, I
believe. For the problem in question, this means less than 5%.
How many people were actually affected by something like this
last year,
in a given country/location? I think this might include major
traffic
jams, subway failures, transport union strikes, and
presidential
campaign appearances?
I think the likelihood that the bar they are to meet in 
isn't
open is
higher.
Michael
--
Still an attentive ear he lent Her speech hath caused this
pain
But could not fathom what she meant Easier I count it to
explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls
===
Subject: Re: Different kind of meeting problem.
* matt271828-news@yahoo.co.uk
> I had an interesting thought about this. I think it is
IMPOSSIBLE for
> these two events to be independent. There is always some
non-zero
> possibility that a natural or man-made catastrophe will
prevent both
> Bill AND Ted from showing up. For example, destruction of
the city in
> which Bill and Ted live, destruction of the earth by an
asteroid
> impact, etc. (sorry about the less-than-pleasant scenarios
here
> folks). Therefore, there must always be some correlation,
however
> small?
If such considerations should be standard to include in the
puzzle
world, you might as well start a discussion for the removal of
rec.puzzles and friends.
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00
P.M.. Each can
> wait for 15 minutes. What is the
> probability that they will meet?
 that Bill and Ted's ARRIVAL times are (i) uniformly
distributed
> between 3:00 and 4:00 and (ii) independent?
> This is an easy problem to calculate if you know the
formula. Let d be
> the waiting time which is 0.25 because
> 15 minutes/60 minutes = .25. n = 2 because there are two
people.
 n(d^(n-1)-(n-1)*d^(n-1)=2*.25-(1/16)=0.4375.
> I don't understand this formula. There is a missing 
bracket
somewhere,
> or something else has gone wrong? However, the answer of
0.4375 does
> agree with the answer I get by working it out from first
principles...
> Your right. There is a missing bracket. It should be
> P=n*(d^(n-1))-(d^(n-1))=2*0.25-(1/16)=0.5-0.0625=0.4375.
P = n*(d^(n-1))-(d^(n-1))=2*0.25-(1/4)= (wrong answer)
Thinking about this geometrically, assuming arrival times are
independent
and evenly distributed over [0,1], I think the answer should
be:
P = n (d^(n-1) - (d^n - Integral[0 to d : x^(n-1)dx]))
= n (d^(n-1) - (d^n - (d^n)/n))
= n d^(n-1) - (n-1) d^n
I got this from considering the volume of a sheared prism of
area
d^(n-1)
and length 1, then subtracting the bit which falls outside
the unit
cube,
then multiplying the whole thing by n, as there is one of
these each side
of
the cubes diagonal for each dimension.
Mike.
 But what if there is a 90% probability that Bill will show
up, and a
70%
> probability that Ted will show up?
> I think you are assuming that the event of Bill showing up
is
> independent of the event of Ted showing up, as well as
independent of
> the times of arrival? (For example, if both would be
deterred by rain
> then the showing up events would no longer be independent
and this
> would inßuence the result.)
 My thinking is that it is ..9*.7*.4375=.63*.4375=0.275625.
 Does anyone agree?
> With the above caveats, yes.
> --
> -------------------------------
> Patrick D. Rockwell
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00
P.M.. Each can
> wait for 15 minutes. What is the
> probability that they will meet?
> This is an easy problem to calculate if you know the
formula. Let d
be
> the waiting time which is 0.25 because
> 15 minutes/60 minutes = .25. n = 2 because there are two
people.
> P=n*(d^(n-1))-(d^(n-1))=2*0.25-(1/16)=0.5-0.0625=0.4375.
> P = n*(d^(n-1))-(d^(n-1))=2*0.25-(1/4)= (wrong answer)
The formula simplifies to
P = (n-1) * (d^(n-1)), and since n=2, P=d.
In effect, that's the probability that Bill arrives while 
Ted
waits if
Ted arrives at a fixed time between 3 and 3:45 (or vice 
versa).
Assuming uniform independent arrivals between 3:00 and 4:00.
1) Chance Ted arrives between 3:15 and 3:45= 1/2
Chance Bill arrives max. 15 minutes before or after Ted: 1/2
Total chance: 1/4
2) Chance Ted arrives between 3:00 and 3:15= 1/4
Chance Bill arrives max. 15 minutes later: 1/4
Chance Bill arrives between 3:00 and Ted's arrival:
Average of 0-1/4 = 1/8
Total Chance: 1/4 * (1/4 + 1/8)
3) Chance Ted arrives between 3:45 and 4:00 = 1/4
The situation is 2) reversed.
Total Chance: 1/4 * (1/4 + 1/8)
Total overall:
1/4 + 2* 1/4 * (1/4 + 1/8) = 1/4 * (1 + 3/4) = 7 / 16
Michael
--
Still an attentive ear he lent Her speech hath caused this
pain
But could not fathom what she meant Easier I count it to
explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls
===
Subject: Re: Different kind of meeting problem.
> Bill and Ted agree to meet between 3:00 P.M. and 4:00 P.M..
Each can
> wait for 15 minutes. What is the
> probability that they will meet?
This depends entirely on the distribution of the arrival
times for Bill and
Ted.
> This is an easy problem to calculate if you know the
formula. Let d be
> the waiting time which is 0.25 because
> 15 minutes/60 minutes = .25. n = 2 because there are two
people.
> n(d^(n-1)-(n-1)*d^(n-1)=2*.25-(1/16)=0.4375.
Your brackets above aren't matched. Also, your arithmetic is
incorrect,
regardless of how you complete the brackets...
> But what if there is a 90% probability that Bill will show
up, and a 70%
> probability that Ted will show up?
> My thinking is that it is ..9*.7*.4375=.63*.4375=0.275625.
Assuming your answer of 0.4375 were correct, and assuming the
probability
of
showing up is independent of the distribution of arrival
times, then your
reasoning here is correct.
Mike.
> Does anyone agree?
> --
> Patrick D. Rockwell
> prockwell@thegrid.net
> hnhc85a@prodigy.net
> patri48975@aol.com
===
Subject: Root Finder vi.
Root Finder vi.
by Jon Giffen
The polynomial
a[0]+a[1]t+a[2]t^2+a[3]t^3+ ... + a[n] = 0
may be expressed as the dot product of the two vectors,
T = (t,t^2,t^3,...,t^n) and
N = (a[1],a[2],a[3],...,a[n])
N may be considered the normal to the more general case
of the plane,
a[0] + a[1]x[1] + a[2]x[2] + ... + a[n]x[n] = 0
where x[1]=t x[2]=t^2 x[3]=t^3 ... x[n]=t^n
T*N+a[0]=0 is the nth degree polynomial since
(t,t^2,t^3,..,t^n)*(a[1],a[2],a[3],..,a[n])+a[0]=0
Following T up from the origin, R is the vector parallel
to N and orthogonal to T-R, as T varies along its curve.
then
R*(T-R)=0
taking the derivative with respect to t,
R'*(T-R)+R*(T'-R')=0
or
2R'*R = R'*T + R*T'
The projection of T' onto N/|N| is R'
((T'*N)/|N|)N/|N|=R'
define
Q = (T*N)N/|N|^2 = -a[0]N/|N|^2
Q = (-a[0]/|N|^2)(a[1],a[2].a[3],..a[n])
or
Q = (q[1],q[2],q[3],...,q[4])
Since R' is in the same direction as Q, it holds that
since Q and T'-(T'*N)N/|N|^2 are orthogonal,
Q*(T'-(T'*N)N/|N|^2) = 0
where Q is the shortest vector from the origin to the plane.
T = (t,t^2,t^3,...,t^n)
T'= (1,2t,3t^2,..,nt^(n-1) )
T'*Q =q[1]+2q[2]t+3q[3]t^2+....+ nq[n]t^(n-1)
=(1/t)T*(q[1],2q[2],3q[3],...,nq[n])
T'*N = a[1]+2a[2]t+3a[3]t^2+....+ na[n]t^(n-1)
=(1/t)T*(a[1],2a[2],3a[3],...,na[n])
Suppose the vectors D,S,U are defined,
D=( 1 , 2 , 3 ,...,n )
S=(a[1],2a[2],3a[3],...,na[n])
U=( 1 , 1 , 1 ,...,1 )
Then
T'* U =(1/t)T*D
T'* N =(1/t)T*S
for instance,
(1,2t,3t^2,..,nt^(n-1) )*(1,1,1,..)
=(1/t)(t,t^2,t^3,...,t^n)*(1,2,3,..)
and
(1,2t,3t^2,..,nt^(n-1) )*(a[1],a[2],a[3],..)
=(1/t)(t,t^2,t^3,...,t^n)*(a[1],2a[2],3a[3],..)
dividing the below equations with each other,
T'* U =(1/t)T*D
T'* N =(1/t)T*S
(T'*U)(T*S) - (T'*N)(T*D) = 0
factoring out T,
T*((T'*U)S - (T'*N)D) = 0
so T is orthogonal with (T'*U)S - (T'*N)D
or T is orthogonal with (1/t)((T*D)S - (T*S)D)
or T is orthogonal with (T*D)S - (T*S)D
this is true, since
T*((T*D)S - (T*S)D) = 0
Let,
G = (T*D)S - (T*S)D
If m is some ratio of G such that,
(mG-Q)*N = 0
mG*N - Q*N = 0 since Q*N=T*N,
m[(T*D)(S*N)-(T*S)(D*N)]-(T*N) = 0
or
T*[m((S*N)D-(D*N)S )-N] = 0
T is orthogonal with m((S*N)D-(D*N)S)-N
N is orthogonal with (S*N)D-(D*N)S
OY=m((S*N)D-(D*N)S)-N
Y
*------------ Q T
| /
| /
| /
| /
| /
| /
Z _________|/
|O
OZ=C |
(S*N)D-(D*N)S |
|
|
|
-N
Let C=(S*N)D-(D*N)S
(Q - (m|N|^2 /-a[0])C )*T = 0
L
*
Q
__________*___________T
F | /
| /
| /
| / E
| /
|/
O
OQ = Q
OT = T
FQ = (m|N|^2 /-a[0])C
OF = Q - (m|N|^2 /-a[0])C
LE and FO are parallel
LE and OT are perpendicular
T*Q -a[0]
angle QOT = arccos(------) = arccos(-------)
|T||Q| |T||N|
|FQ| = |FO|cos(OFQ)=|FO|cos(QOT)
|OQ| = |OF|cos(QOF)=|OF|sin(QOT)
|OQ|
---- = tan(QOT)
|FQ|
|OQ| |Q| (a[0])^2
---- = ------------------ = ------------ eqn i.
|FQ| (m|N|^2 /-a[0])|C| m|N|^3 |C|
sin(QOT) |T||N|
tan(QOT) = --------- = -{ (-------)^2 -1 }^(1/2) eqn ii.
cos(QOT) a[0]
equating eqn i. and eqn ii. and squaring both sides,
(a[0])^4 |T||N|
---------------- = (---------)^2 - 1 eqn iii
m^2 |N|^6 |C|^2 a[0]
(mC-N)*T = 0 or
T*C=(T*N)/m = -a[0]/m eqn iv.
T*C=|T||C|sin(QOT) eqn v.
a[0]
sin(QOT)={1 - (--------)^2 }^(1/2) eqn vi.
|T||N|
substituting eqn iv. and eqn vi. into eqn v. and
squaring both sides,
(a[0])^2 a[0]
--------- = 1 - (--------)^2 eqn vii.
m^2 |T||N|
substituting eqn vii. into eqn iii.,
(a[0])^2 a[0] |T||N|
(-------------)(1 - (-------)^2 ) = (---------)^2 - 1
|N|^6 |C|^2 |T||N| a[0]
Let
|N|
a = -(-----)^2
a[0]
(a[0])^2
b = ----------- + 1
|N|^6 |C|^2
(a[0])^4
c = - --------------
|N|^8 |C|^2
then
a|T|^2 + b + c/|T|^2 = 0 or
a|T|^4 + b|T|^2 + c = 0 and
-b +/-{b^2-4ac}^(1/2)
|T|^2 = ----------------------
2a
Then
the solution to T is on the plane
(T-Q)*N=0
somewhere around the circle on the plane
with center at Q and radius r such that
r^2 = |T|^2 - |Q|^2
The vector C in the plane, starting from Q, intersects
the circle at some ratio z of C:
|zC|^2 = |T|^2 - |Q|^2 or
|T|^2 - |Q|^2
z = +/-{ --------------- }^(1/2)
|C|^2
Then,
T = Q + zC
t^3+2t^2+t-4=0
N=(1,2,1)
|N|^2=(1,2,1)*(1,2,1)=1+4+1=6
a[0]=-4
Q=(4/6)(1,2,1)=(2/3)(1,2,1)=(2/3,4/3,2/3)
|Q|^2 =(4/9)(6)=4(2/3)=8/3
D=(1,2,3)
S=(1,4,3)
C=(S*N)D-(D*N)S=12(1,2,3)-8(1,4,3)=(4,-8,12)=4(1,-2,3)
|C|^2=16(1,-2,3)*(1,-2,3)=16(1+4+9)=16(14)=224
|N| |N|^2 6
a = -(-----)^2 = - --------- = - ----- = -3/8
a[0] (a[0])^2 16
(a[0])^2 16 1 3025
b = ----------- + 1 = ---------- + 1 = ------- + 1 = ------
|N|^6 |C|^2 224(6^3) 14(216) 3024
(a[0])^4 4^4 256 1
c = - -------------- = - -------- = - -------- = - ------
|N|^8 |C|^2 224(6^4) 290304 1134
Then
a = -3/8
b ~ 1
c ~ 0 and
-b +/-{b^2-4ac}^(1/2) -1+/-1 2
|T|^2 = ---------------------- = ------- = ----= 8/3 ,0
2a -3/4 3/4
|T|^2 - |Q|^2
z = +/-{ --------------- }^(1/2) ~ 0
|C|^2
T = Q + zC
T ~ Q
Q =(2/3,4/3,2/3)
t = 2/3 t = 0.6666 ~ 1
t^2 = 4/3 t = 1.1547 ~ 1
t^3 = 2/3 t = 0.8736 ~ 1
The correct answer is t=1.
end
===
Subject: Re: Root Finder vi.
> Root Finder vi.
> by Jon Giffen
[cut]
> t = 2/3 t = 0.6666 ~ 1
> t^2 = 4/3 t = 1.1547 ~ 1
> t^3 = 2/3 t = 0.8736 ~ 1
> The correct answer is t=1.
> end
Again, horribly bad accuracy. 33%, 15%, 13% error
respectively. Lets
say I need two places after the decimal point and the answer
is not a
round integer. Your extremely inaccurate results are total
crap if
they are 13% to 33% off.
===
Subject: Re: Root Finder vi.
> Root Finder vi.
> by Jon Giffen
[Derivation clipped -- reason below.]
> t^3+2t^2+t-4=0
Hooray! An example!
[Giffen's procedure clipped -- reason below.]
> t = 2/3 t = 0.6666 ~ 1
> t^2 = 4/3 t = 1.1547 ~ 1
> t^3 = 2/3 t = 0.8736 ~ 1
> The correct answer is t=1.
So your method is WRONG.
(Sigh.) We've been over this dozens of times, Jon. There are
already
methods to APPROXIMATE roots of a polynomial equation. A root
FINDER, by
definition, must provide EXACT solutions.
(Note that if you had gotten a vector of (2/3, 4/9, 8/27),
you would
have found a root, 2/3, since t^1 = 2/3 AND t^2 = 4/9 AND t^3
= 8/27 are
simultaneously satisfied by t = 2/3. The equations you found
have no
solution, so you haven't done anything.)
-- Christopher Heckman
===
Subject: What are the rules for fractional exponents?
If you compute something like 5**(3/11), you can either
compute the 11
roots of 5**(1/11) and then cube them all, or you can cube 5
and then
compute the 11 roots of 125**(1/11). The order is invariant.
This seems obvious when you take X**(m/n) where m/n is
reduced to
simplest terms. But what happens when m/n isn't simply
reduced?
For example, 4**(4/2) would be a good example of what goes
wrong.
4**2 = 16. Simple, right? But if we allow the computation as
follows:
4**2 = 4**(4/2) = (4**4)**(1/2) = 256**1/2 = +/- 16
Of course, this isn't true.
SO ... what ARE the rules for fractional exponents? Is it
just that the
m/n must be reduced to simplest terms? Or are there
counterexamples to
that too?
Op
===
Subject: Re: What are the rules for fractional exponents?
> If you compute something like 5**(3/11), you can either
compute the 11
> roots of 5**(1/11) and then cube them all, or you can cube
5 and then
> compute the 11 roots of 125**(1/11). The order is invariant.
> This seems obvious when you take X**(m/n) where m/n is
reduced to
> simplest terms. But what happens when m/n isn't simply
reduced?
> For example, 4**(4/2) would be a good example of what goes
wrong.
> 4**2 = 16. Simple, right? But if we allow the computation
as follows:
> 4**2 = 4**(4/2) = (4**4)**(1/2) = 256**1/2 = +/- 16
> Of course, this isn't true.
> SO ... what ARE the rules for fractional exponents? Is it
just that the
> m/n must be reduced to simplest terms? Or are there
counterexamples to
> that too?
> Op
Spookily enough, I too have been struggling with this same
problem
(see Question about powers posting at alt.math.undergrad).
This is
something like the blind leading the blind, but what I
gleaned out of
the responses could be summarised as follows. I'd be more
than happy
for somebody to correct me (again!).
The simplest explanation is that the law of exponents p^(q*r)
=
(p^q)^r just plain does not work for non-integers. That's 
all
there is
to it. For example, as you demonstrate, 4^(4*1/2) is NOT the
same
thing as (4^4)^(1/2). The equation only works for (for want
of the
correct technical term) the principal values (in your case 16
= 16).
But on top of this, and muddying the waters considerably,
there seems
to be some ambiguity about what the expression a^(b/c) should
actually
mean. One view (mine) is that a^(b/c) is a function of the two
quantities a and b/c, namely
a^(b/c) = exp[(b/c)*log(a)]
Thus, the meaning of a^(b/c) is independent of the way the
quantity
b/c is represented. In your example, 4^(4/2) is identical to
4^2.
Both have the unique value 16.
However, some people apparently define a^(b/c) to be equal to
(a^b)^(1/c). (Maybe this is more understandable if superscript
notation is used?) With this interpretation, 4^(4/2) is
DEFINED as
(4^4)^(1/2), and therefore IS equal to +/-16.
Not sure if that helped either of us... maybe someone else
will give
you a better explanation!
===
Subject: Re: What are the rules for fractional exponents?
>The simplest explanation is that the law of exponents
p^(q*r) =
>(p^q)^r just plain does not work for non-integers. That's
all there is
>to it. For example, as you demonstrate, 4^(4*1/2) is NOT the
same
>thing as (4^4)^(1/2). The equation only works for (for want
of the
>correct technical term) the principal values (in your case
16 = 16).
The law of exponents still works on the principal branch,
however. This is easily soon from the definition you give
below.
>But on top of this, and muddying the waters considerably,
there seems
>to be some ambiguity about what the expression a^(b/c)
should actually
>mean. One view (mine) is that a^(b/c) is a function of the
two
>quantities a and b/c, namely
> a^(b/c) = exp[(b/c)*log(a)]
Precisely. Hence, for a^p*a^q = exp[(p+q)log(a)] = a^(p+q).
There problem is that there is ambiguity in the expression
log(a): which branch is to be used? As long as the principal
branch is used (or actually, any branch as long as the same
one
is used consistently), there is no problem.
Since ix = log[cos(x)+isin(x)], the logarithm of x is actually
a whole family of values Log(x) + 2pi i, where Log is the
principal branch of the logarithm. This is what gives rise to
the many numbers of roots, which is the source of the above
problem: x^y = exp[y log(R)] = exp[yLog(x)]exp[i(2*n*y*pi)],
so
that, for example, for y = 1/2, the second term is a rotation
by
a multiple of pi radians, which happens to be precisely the
negative. For y = 1/c, there are therefore exactly c roots,
each
rotated 2npi/c radians from each other.
>Thus, the meaning of a^(b/c) is independent of the way the
quantity
>b/c is represented. In your example, 4^(4/2) is identical to
4^2.
>Both have the unique value 16.
>However, some people apparently define a^(b/c) to be equal to
>(a^b)^(1/c). (Maybe this is more understandable if
superscript
>notation is used?) With this interpretation, 4^(4/2) is
DEFINED as
>(4^4)^(1/2), and therefore IS equal to +/-16.
For fractions, It would make more sense to define a^(b/c) as
[a^(1/c)]^b instead. This avoids these kinds of problems--for
example, 4^(4/2) = [4^(1/2)]^4 = [+/- 2]^4 = 16.
---
Stan Liou
===
Subject: Re: What are the rules for fractional exponents?
> 256**1/2 = +/- 16
If
y = x^(1/2)
is to be a function, it cannot have two outcomes (the +/-
sign). So
usually we take 256^(1/2) = 16.
Nevertheless, -16 is a root of x^2 = 256 of course. However,
y = x^2 is
not invertible over the domain of real numbers.
> SO ... what ARE the rules for fractional exponents? Is it
just that the
> m/n must be reduced to simplest terms? Or are there
counterexamples to
> that too?
You can take the pragmatic approach that you should not
consider
negative numbers, then it works out fine. Or, if you wish to
include
negative numbers, the rule x^(a*b) = (x^a)^b is no longer
valid.
--
M.vr.gr.
Dave
(d-dot-langers-at-wxs-dot-nl)
===
Subject: Re: What are the rules for fractional exponents?
>If you compute something like 5**(3/11), you can either
compute the 11
>roots of 5**(1/11) and then cube them all, or you can cube 5
and then
>compute the 11 roots of 125**(1/11). The order is invariant.
>4**2 = 4**(4/2) = (4**4)**(1/2) = 256**1/2 = +/- 16
>Of course, this isn't true.
>SO ... what ARE the rules for fractional exponents? Is it
just that the
>m/n must be reduced to simplest terms? Or are there
counterexamples to
>that too?
Either requiring that the fraction be in lowest terms
or that the root (denominator) be taken first. In the
above, 4^(4/2) = [4^(1/2)]^4 = [+-2]^4 = 16, which is
unproblematic.
---
Stan Liou
===
Subject: Re: Counting Problem
X-RFC2646:  Original
--
-------------------------------
Patrick D. Rockwell
> A few more terms produced by program,
> 1 1
> 2 1
> 3 3
> 4 7
> 5 35
> 6 143
> 7 1001
> 8 5901
> 9 53109
> 10 418661
> 11 4605271
> 12 45369999
> 13 589809987
> 14 ...
> before we gave up on each other.
> For 6, I got 139 patterns rather than 143. Using the
original
> formulation, these were:
> 123456, 123465, 123546, 123564, 123645, 123654, 124356,
124365,
> 124536, 124635, 125346, 125364, 125436, 125634, 126345,
126354,
> 126435, 126534, 132456, 132465, 132546, 132564, 132645,
132654,
> 134256, 134265, 134526, 134625, 135246, 135264, 135426,
136245,
> 136254, 136425, 142356, 142365, 142536, 142635, 143256,
143265,
> 143526, 143625, 145236, 145326, 152346, 152364, 152436,
152634,
> 153246, 153264, 153426, 154236, 154326, 162345, 162354,
162435,
> 162534, 163245, 163254, 163425, 213456, 213465, 213546,
213564,
> 213645, 213654, 214356, 214365, 214536, 214635, 215346,
215364,
> 215436, 215634, 216345, 216354, 216435, 216534, 231456,
231465,
> 231546, 231564, 231645, 231654, 234156, 234165, 234516,
235146,
> 235164, 235416, 241356, 241365, 241536, 241635, 243156,
243165,
> 243516, 251346, 251364, 251436, 251634, 253146, 253164,
253416,
> 312456, 312465, 312546, 312564, 312645, 312654, 314256,
314265,
> 314526, 314625, 315246, 315264, 315426, 316245, 316254,
316425,
> 321456, 321465, 321546, 321564, 321645, 321654, 324156,
324165,
> 324516, 325146, 325164, 325416, 341256, 341265, 341526,
341625,
> 342156, 342165, 342516
Well, I went through it pain stakingly by hand on a text file,
and I think
that you may be right.
Unfortunately, when I enter 1 1 3 7 35 139 into the
Encyclopedia Of Integer
Sequences Look-Up, I don't
get any hits.
===
Subject: Re: Counting Problem
> A few more terms produced by program,
> 1 1
> 2 1
> 3 3
> 4 7
> 5 35
> 6 143
> 7 1001
> 8 5901
> 9 53109
> 10 418661
> 11 4605271
> 12 45369999
> 13 589809987
> 14 ...
 before we gave up on each other.
 For 6, I got 139 patterns rather than 143.
> Yes, I got 139 by counting invalid patterns
> 6! -3*5! -3*4! -12*3! -15*2! -47 = 139
> For 8, I get 5701 instead of 5901
> For 10, I get 402985 instead of 418661
I agree, I get:
1 1
2 1
3 3
4 7
5 35
6 139
7 1001
8 5701
9 53109
10 402985
11 4605271
> This must mean the recursion formula f(2n+1) =(2n+1)*f(2n)
> is incorrect too.
I agree, f(2n+1) = (2n+1)*f(2n) doesn't do it for me either
===
Subject: Famous British author reinvents geometry
Day of the Jackal, The Odessa File, and other
thrillers. His latest book is Avenger, which came
out last year. On page 305 is the passage:
The triangle jutting out to sea was larger than
he had imagined. ...
The base, on which he now looked down from
his mountain hideout, was about two miles from
side to side. It ran, as his aerial photos had shown,
from sea to sea and at each end the mountain
range dropped to the water in vertical cliffs.
The sides of the isosceles triangle he estimated
at about three miles, giving a total land area of
almost six square miles.
-- Mark Spahn
===
Subject: Re: Famous British author reinvents geometry
zrT`*_`>]a//w'-[5]Ien;9)XV(v<R/q-x?
> Day of the Jackal, The Odessa File, and other
> thrillers. His latest book is Avenger, which came
> out last year. On page 305 is the passage:
> The triangle jutting out to sea was larger than
> he had imagined. ...
> The base, on which he now looked down from
> his mountain hideout, was about two miles from
> side to side. It ran, as his aerial photos had shown,
> from sea to sea and at each end the mountain
> range dropped to the water in vertical cliffs.
> The sides of the isosceles triangle he estimated
> at about three miles, giving a total land area of
> almost six square miles.
Perhaps the explanation is not unrelated to the old joke
about land
that's so hilly, real estate agents could sell both sides of
the same
acre.
===
Subject: Re: Lim_{n-->infty} sin(x*P^n)= ??
X-RFC2646:  Original
> Let P , P >= 2 , be an integer and
> f_n(x)= sin(x*P^n) , x in (-infty ,infty) .
> Find all $ x $ for which the sequence (f_n(x))_{n>=1} is
convergent.
> In case of convergence, find the limit .
Well, let us start with the obvious: if x is of the form n *
pi / ( P ^ k ),
where n is an integer and K is a natural number, then f_n(x)
converges to
0.
If f_n(x) converges for any other value of x, I'd be very
surprized.
===
Subject: Re: Lim_{n-->infty} sin(x*P^n)= ??
> Let P , P >= 2 , be an integer and
> f_n(x)= sin(x*P^n) , x in (-infty ,infty) .
> Find all $ x $ for which the sequence (f_n(x))_{n>=1} is
convergent.
> In case of convergence, find the limit .
>Well, let us start with the obvious: if x is of the form n *
pi / ( P ^ k
),
>where n is an integer and K is a natural number, then f_n(x)
converges to
0.
>If f_n(x) converges for any other value of x, I'd be very
surprized.
There are infinitely many counterexamples to that. For 
example,
sin(3^n*pi/4) = sqrt(2)/2 for all n.
Of course, for any converging x, adding any number of
multiples
of 2pi/P^m, for any integer m, will not change convergence,
since in the sequence P^nx, the additional term will
eventually
become a multiple of 2pi.
For odd P, there seem to be exactly P possible converging
limits. In this case, I can show that the limits are the fixed
points of a certain Pth degree polynomial, but I lack decisive
proof that there are exactly P of them.
---
Stan Liou
===
Subject: Re: Lim_{n-->infty} sin(x*P^n)= ??
>Let P , P >= 2 , be an integer and
> f_n(x)= sin(x*P^n) , x in (-infty ,infty) .
>Find all $ x $ for which the sequence (f_n(x))_{n>=1} is
convergent.
For a set S of limits, any x in B = {x:sin(x) in S} gives
convegence. An x not from S cannot give convergence unless
the sequence (P^nx) itself converges to some point in B.
The real question is then what is the set of limits S.
The set of convergents X can be specified explicitly by
X = {x: Sx + 2pi Sum{i>0,j>0}[a(i)/P^{b(j)}] },
where sin(Sx) is in S, a(n) and b(n) are sequences of
integers,
and b(j) is monotone increasing, as long as that series
converges.
>In case of convergence, find the limit .
For now, I only have a complete answer for odd P.
Consider the integral
I(n,m) = Int{-pi,pi}[ sin^n(t) sin(mt) dt ].
The trivial case is I(0,m) = 0, and using integration
by parts on the sin(mt) term gives I(n,m)
I(n,m) = [n/m]Int{-pi,pi}[ sin^{n-1}(t)cos(t)cos(mt) dt ]
and again on the cos(mt) term
I(n,m) = [-n/m]Int{-pi,pi}[ sin(mt)/m *
[(n-1)sin^{n-2}(t)cos^2(t)-sin^{n-1}(t)sin(t)] dt ],
which simplifies to give
I(n,m) = (n/m)^2 I(n,m) - n(n-1)/m^2 I(n-2,m),
or even simpler
I(n,m) = -[n(n-1)/(m^2-n^2)] I(n-2,m), for n!=m.
The significance of this will be made clear in a moment.
Let A[n,0] = 0, A[n,1] = n, with the recurrence relation
A[n,k+2] = -A[n,k][n^2-k^2]/[(k+1)(k+2)],
and for some fixed odd n, define the polynomial
F[n,t] = Sum{k>0}[ A[n,k]t^k ],
and let G(x) = F[n,sin(x)]. G(x) is an odd function with
G(x) = G(x+2pi), so its Fourier series has the form
G(x) = Sum{m>0}[ B(m)sin(mx) ], where
B(m) = [1/pi] Int{-pi,pi}[ G(x) sin(mx) dx ].
Using the above information on integrals, it can be
easily proven that B(m) = 0 except if m = n; in fact,
G(x) = F[n,sin(x)] = sin(nx);
that B(n) = 1 instead of some other constant is somewhat
harder to prove, but can be done via a method not much
different than the above.
The consequences for the problem are immediate.
For an odd P, if L = lim[ sin(P^nx) ] exists, it
must be the case that L = lim[ F[P,sin(P^nx)] ], and
since it is a simple Pth degree polynomial, the limit
must be a fixed point of F[P,t], i.e, L = F[P,L], with
only L in the interval [-1,1] allowed.
These form the limit set S, up to a difference of 2Npi.
--
On the other hand, sin(2t) = 2sin(t)cos(t) and
cos(2t) = 1-2sin^2(t), so that sin(2^a*t) will have
precisely one factor of cos(t). In general, then,
sin(Pt) = F[b,sin(2^at)], where P = (2^a)b, b odd,
and since F is odd, only odd powers of cosine are
produced, all of which can be reduced through cos^2(t)
= 1-sin^2(t). Therefore, for even P (a>0), the
relationship between sin(t) and sin(Pt) is:
sin(Pt) = cos(t)*G[P,sin(t)],
for some polynomial G[P,t]. The corresponding fixed
points would be harder to compute, but should not be
impossible (I have not done so).
---
Stan Liou
===
Subject: Re: Favorite equation?
> snip
> or to show off
> e^(xi) = sin x + i.cos x
Sooo e^0 = i