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Grex > Tutoring > #14: Algebra, Geometry, Calculus, Trig, all that good stuff |  |
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| 25 new of 132 responses total. |
drew
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response 100 of 132:
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Aug 31 16:06 UTC 1997 |
Someone recently came out with a 100-sided die. I'm not sure whether hexagons
could be circumscribed about the faces, but if a 100 sided die is possible,
then maybe something like hexagon-world could be done if you don't mind some
irregularity and mismatch.
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i
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response 101 of 132:
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Aug 31 22:26 UTC 1997 |
If you're just interested in generating random numbers from 1 to n, then a
fair n-sided die can easily be made. Besides the common 6-siders, 4-, 8-
12-, and 20-sided dice (based on the regular polyhedra) are quite common.
(Dungeons & Dragons uses all these, as do a number of other games.) The
results are often (mathematically and visually) uglier for other values
of n. My guess is that commercial 100-sided dice would either split the
faces of a 20-sided die into 5 pieces (each) or follow an apple-peeler
pattern. But I wouldn't be surprised by something really odd like a little
10-sided die inside a large, clear 10-sided die (read as 1's and 10's).
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remmers
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response 102 of 132:
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Aug 31 23:41 UTC 1997 |
Okay, change-of-pace time! Here's a reasonably straightforward
geometry problem:
Three billiard balls of diameter 1 are resting on a flat
tabletop. Each ball is touching the other two. A fourth ball of
diameter 1 rests on top of the first three. How high above the
surface of the table is the apex of the fourth ball?
(Supply reasoning to justify your answer.)
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rcurl
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response 103 of 132:
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Sep 1 01:55 UTC 1997 |
It is the diameter of a ball plus the height of a tetrahedron with sides
equal to the diameter of a ball. (Reduced to previously solved problem....)
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toking
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response 104 of 132:
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Sep 1 03:20 UTC 1997 |
re 100: 100 sided die have been out for a while...I picked one
up in like 5th grade to use with D&D..
I recently finished "Flatland" by Edwin Abbott Abbott, is the sequel
"spehere Land" and good?
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rcurl
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response 105 of 132:
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Sep 1 06:12 UTC 1997 |
I have a "die" that was made by a statistician which is an aluminum
ten-sided prism with the sides numbered (on the end of the prism) from
0 to 9. It is for generating random decimal numbers.
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remmers
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response 106 of 132:
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Sep 1 12:15 UTC 1997 |
Re #103: Yeah, but what's the ANSWER? :)
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i
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response 107 of 132:
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Sep 1 15:20 UTC 1997 |
Okay, let's roll out the old HS geometry and take a closer look at that
tetrahedron. It's regular with edges of length 1, so all 4 faces are
equilateral triangles with sides of length 1. Bisect the 3 angles in the
base triangle and run the lines through the opposite sides. The result
just oozes symmetries, similarities, and 30-60-90 triangles. Note that
by easy symmetry arguments, the 3 bisectors meet at the center directly
under the vertex. Using the old 1|1/2|sqtr(3)/2 sides rule for those
30-60-90 trianges gives us a distance of 1/sqrt(3) from a corner of the
base to the center. This distance and the (unknown) height are legs of
a right triangle with an edge as hypotenuse (1). Pythagoras says the
height is sqrt(2/3), so (per #103), the answer is 1+sqrt(2/3).
On to a probability problem. It's not hard, but I've seen grad. students
flub it and I'm told that the government of India once made a major policy
decision based on a wrong answer to it.
Assume that the gender ratio at birth is 50-50 and that all births are
independent events. (In other words, having babies is like flipping a
fair coin - heads it's a boy, tails it's a girl.) The following policy
is proposed: Each couple may have two children. If both are girls, then
they may have a third, if that's a girl, a 4th is permitted, etc. (Two
very different ways you could interpret this sexism! The intent was to
guarantee everyone a son who wasn't an only child.) Assume that couples
will have as many children as permitted. Ignore divorce, birth out of
wedlock, fertility running out before a son is born, etc.
What will the gender ratio of children born under this policy be?
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drew
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response 108 of 132:
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Sep 1 21:53 UTC 1997 |
Okay, I'll take a crack at it.
If you stop at two kids, then the ratio is 50/50. However, 25% of these
families will be two girls, and thus can have another kid. Of these, 50% will
have yet another girl, and of these 50% yet another... Thus we have:
M M 1/4
M F 1/4
F M 1/4
F F M 1/8
F F F M 1/16
F F F F M 1/32
.........
(n) F M 2^(-n-1)
In the first two cases, the ratio is one female to three males, multiplied
by 1/2. The rest can be generalized as n females to one male, multiplied by
(1/2)^(n+1). Expressing it as a sum:
R = 1/6 + sum(i=1 to inf) { i / 2^(i+1) }.
The series works out as 1/6 + 1/4 + 1/4 + 3/16 + 1/8 + 5/64 + 3/64 + ...
By experiment, the series seems to converge at 1.166666667 or 7/6. Shouldn't
be too much of a disaster in itself.
But the real problem is the population *growth*, the increase per generation.
This is best state, per couple, as
2/4 + 2/4 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
or
1/2 + sum(i=1 to inf) { i / 2^i }.
This series converges, by experiment, at 2.5 children per male-female pair,
or 1.25 multiplication of the population every generation.
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aruba
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response 109 of 132:
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Sep 2 00:59 UTC 1997 |
Where did you get that 1/6 from, Drew?
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drew
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response 110 of 132:
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Sep 3 00:04 UTC 1997 |
The first two cases are taken as the first term in the series - actually not
really a part of the series itself, but a special case. The female to male
ratio among these first two cases is 1:3, and they take up half the outcomes,
for a product of 1/6.
I suspect my answer may not be right. I'll have to think this one over.
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srw
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response 111 of 132:
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Sep 4 05:21 UTC 1997 |
There is a simple proof that the series sum(i=1 to inf) (i / 2^i) = 2
based on the method of differentiating the series...to wit
Start with y= sum[i=0,inf](x^i) = 1/(1-x) in the region 0<x<1
differentiate w.r.t. x - (you can differentiate each term of the series
separately). so it yields
dy/dx = sum[i=0,inf] (i * x^(i-1) ) = (1 - x)^(-2)
so if we substitute x=2^(-1) (another name for 0.5) we have this
sum[i=0,inf] (i / 2^(i-1)) = 4
now divide each side by 2 and note that the i=0 term is 0, so drop it
sum[i=1,inf] (i / 2^i ) = 2 QED
H O W E V E R .....
If you wish to determine the gender ratio you do not need to evaluate a
series. It is 1:1 because each birth is independent. I don't see any
need to argue further.
For skeptics, if you really must do it by the series approach, then
evaluate the expected number of males per family and the expected number
of females. Both evaluate to 1.25 so the ratio really is 1:1
case probability E(M) E(F)
MM 1/4 2/4 0/4
MF 1/4 1/4 1/4
FM 1/4 1/4 1/4
FFM 1/8 1/8 2/8
FFFM 1/16 1/16 3/16
FFFFM 1/32 1/32 4/32
The E(M) series is 1/4 + sum[i=1,inf](1/2^i) = 1/4 + 1
The E(F) series is 1/4 + sum[i=1,inf](i/2^(i+1)) = 1/4 + 1
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aruba
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response 112 of 132:
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Sep 4 17:42 UTC 1997 |
Another way to find the expected number of girls G (which doesn't require
any calculus) is:
G = 0/4 + 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
= 1/4 + 1/4 + 2/8 + 3/16 + 4/32 + ...
==> 2G = 2/4 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...
==> 2G-G = 1/4 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
==> G = 1/4 + (1/2)/(1-1/2)
= 1/4 + 1
= 5/4
Drew's interesting observation about the increase in population is quite
right, each generation will be 25% bigger than the last if everyone has
children. That would be kind of a problem.
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i
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response 113 of 132:
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Sep 4 22:52 UTC 1997 |
(Compared to the actual growth rates, 25% per generation [less early
mortality, infertility, etc.] would have been wonderfully low, and
India would be in much better shape today.)
Steve's "H O W E V E R..." in #111 is the correct and "best" solution.
There are simpler "add things up" solutions, but this is one of those
problems where the student's approach tells the teacher more about his/her
understanding than anything else.
(The instructor threw this problem at a class of PhD-track math (not stats)
grad students I was in. The initial class consensus was for a wrong answer
[and they almost agreed on which one].)
On to a more familiar mathematical topic - Sets. Apples and oranges are
elements of the set of fruits, Mr. Figston's 3rd grade is a subset of the
set of students at Washington Elementary School, the intersection of the
set of wet thing with the set of dry things is the empty set, and all that
fun.
One very popular set is the Universal Set, which contains EVERYTHING - real,
abstract, imagined or undiscovered, simple or complex, it's all there. But
the idea that such a set can exist suffers from a fatal logical flaw. It
is not that the Universal Set can't contain itself as a subset - but that's
a good hint on where to start looking.
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remmers
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response 114 of 132:
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Sep 5 00:34 UTC 1997 |
Did you mean to say "contain itself as a subset" or "contain
itself as an element"?
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srw
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response 115 of 132:
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Sep 5 02:38 UTC 1997 |
resp:112 Hey, that's cool, Mark. I learned how to solve that silly
series using differentiation. Your way is much better.
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tpryan
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response 116 of 132:
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Sep 5 22:37 UTC 1997 |
Would the Universal Set, then try to contain 'null' and 'infinity'
at the same time?
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i
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response 117 of 132:
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Sep 5 22:59 UTC 1997 |
The Universal Set (U for short), by "definition", contains null, infinity,
and anything else you can think up. Yourself included.
Re #114: U must do both. I believe that a contradiction can be derived
either way, but the "nontraditional" one requires minimal knowledge of
power sets.
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remmers
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response 118 of 132:
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Sep 6 13:15 UTC 1997 |
Well, every set contains itself as a subset, so I figured you
must have meant "contain itself as an element".
The most familiar contradiction based on the notion of sets
containing themselves as elements is the Russell Paradox, which
goes as follows: Let S be the set of all sets that are not
elements of themselves. Then if S is an element of itself, then
by definition of S, S is not an element of itself. Conversely,
if S is not an element of itself, then again by definition of
S, S is an element of itself.
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rcurl
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response 119 of 132:
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Sep 6 16:09 UTC 1997 |
Mike shaves all men that do not shave themselves. Does Mike shave himself?
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mcnally
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response 120 of 132:
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Sep 6 16:40 UTC 1997 |
re #119: Your statement of Russell's "Barber" paradox is insufficient
to indicate the paradox since it says nothing about whether or not the
barber shaves some of those who shave themselves (which would only include
the barber..) or whether or not the restriction applies to the barber
(e.g. is the barber a woman?) You need a restriction more like "the barber
shaves all those and *only* those who don't shave themselves.."
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rcurl
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response 121 of 132:
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Sep 6 17:42 UTC 1997 |
Ok, fine. But if you are going to be particular, remember that the
defining relative pronoun is *that*, so it has to be stated as "the barber
shaves all those and only those *that* don't shave themselves."
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i
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response 122 of 132:
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Sep 7 14:06 UTC 1997 |
Re: #118 - my meaning was that a contradiction can be obtained by looking
either at sets which are elements of U or at sets which are subsets of U
(the power sets are used in the latter). I did't want to give too big a hint.
This Statement Is False.
(Is the above statement a paradox? Is it not a paradox only for "poetic"
reasons - the contradiction is too poorly hidden, insufficiently interesting,
etc.? What is a paradox?)
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tpryan
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response 123 of 132:
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Sep 12 23:00 UTC 1997 |
Hey, I know Barry & Sally Childs-Helton, they both have Phds,
they are a Paradox.
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dang
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response 124 of 132:
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Sep 29 17:16 UTC 1997 |
re 122: two places to moor ships.
Speaking of Russel's paradox, Greg (my roommate, flem) was just reading his
autobiography and mentioned it to me last night. Interesting coincidance.
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