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From: chris@questrel.questrel.com (Chris Cole)
Subject: rec.puzzles Archive (analysis), part 02 of 35
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Followup-To: rec.puzzles
Summary: This is part of an archive of questions
and answers that may be of interest to
puzzle enthusiasts.
Part 1 contains the index to the archive.
Read the rec.puzzles FAQ for more information.
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Date: Wed, 18 Aug 1993 06:04:22 GMT
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Archive-name: puzzles/archive/analysis
Last-modified: 17 Aug 1993
Version: 4
==> analysis/bicycle.p <==
A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
walk at 2 mph and the dog can trot at 4 mph. They also have a bicycle
which only one of them (including the dog!) can use at a time. When
riding, the boy and girl can travel at 12 mph while the dog can pedal
at 16 mph. What is the shortest time in which all three can complete
the trip?
==> analysis/bicycle.s <==
First note that there's no apparent way to benefit from letting either the
boy or girl ride the bike longer than the other. Any solution which gets the
boy there faster, must involve him using the bike (forward) more; similarly
for the girl. Thus the bike must go backwards more for it to remain within
the 10-mile route. Thus the dog won't make it there in time. So the solution
assumes they ride the bike for the same amount of time.
Also note that there's no apparent way to benefit from letting any of the three
arrive at the finish ahead of the others. If they do, they can probably take
time out to help the others. So the solution assumes they all finish at the
same time.
The boy starts off on the bike, and travels 5.4 miles. At this
point, he drops the bike and completes the rest of the trip on foot. The
dog eventually reaches the bike, and takes it *backward* .8 miles (so the
girl gets to it sooner) and then returns to trotting. Finally, the girl makes
it to the bike and rides it to the end. The answer is 2.75 hours.
The puzzle is in Vasek Chvatal, Linear Programming, W. H. Freeman & Co.
The generalized problem (n people, 1 bike, different walking and riding speeds)
is known as "The Bicycle Problem". A couple references are
Masuda, S. (1970). "The bicycle problem," University of California, Berkeley:
Operations Research Center Technical Report ORC 70-35.
Chvatal, V. (1983). "On the bicycle problem," Discrete Applied Mathematics 5:
pp. 165 - 173.
As for the linear program which gives the lower bound of 2.75 hours, let
t[person, mode, direction] by the amount of time "person" (boy, girl or dog)
is travelling by "mode" (walk or bike) in "direction" (forward or backwards).
Define Time[person] to be the total time spent by person doing each of these
four activities. The objective is to minimize the maximum of T[person], for
person = boy, girl, dog, e.g.
minimize T
subject to T >= T[boy], T >= T[girl], T >= T[dog].
Now just think of all the other linear constraints on the variables t[x,y,z],
such as everyone has to travel 10 miles, etc. In all, there are 8 contraints
in 18 variables (including slack variables). Solving this program yields the
lower bound.
==> analysis/boy.girl.dog.p <==
A boy, a girl and a dog are standing together on a long, straight road.
Simulataneously, they all start walking in the same direction:
The boy at 4 mph, the girl at 3 mph, and the dog trots back and forth
between them at 10 mph. Assume all reversals of direction instantaneous.
In one hour, where is the dog and in which direction is he facing?
==> analysis/boy.girl.dog.s <==
The dog's position and direction are indeterminate, other than that the
dog must be between the boy and girl (endpoints included). To see this,
simply time reverse the problem. No matter where the dog starts out,
the three of them wind up together in one hour.
This argument is not quite adequate. It is possible to construct problems
where the orientation changes an infinite number of times initially, but for
which there can be a definite result. This would be the case if the positions
at time t are uniformly continuous in the positions at time s, s small.
But suppose that at time a the dog is with the girl. Then the boy is at
4a, and the time it takes the dog to reach the boy is a/6, because
the relative speed is 6 mph. So the time b at which the dog reaches the
boy is proportional to a. A similar argument shows that the time the
dog next reaches the girl is b + b/13, and is hence proportional to b.
This makes the position of the dog at time (t > a) a periodic function of
the logarithm of a, and thus does not approach a limit as a -> 0.
==> analysis/bugs.p <==
Four bugs are placed at the corners of a square. Each bug walks always
directly toward the next bug in the clockwise direction. How far do
the bugs walk before they meet?
==> analysis/bugs.s <==
Since the bugs start out walking perpendicularly, and there is nothing
in the problem to alter this symmetry, the bugs are always walking
perpendicularly. Since each bug is walking perpendicularly to the line
separating it from the bug chasing it, the gap is closing at the speed
of the chasing bug. Therefore, each bug walks a distance equal to the
side of the square before it meets the next bug.
In order to conveniently express the equation of the bugs' motion,
use standard polar coordinates, and let the first bug's position at
any instant be (r(t), theta(t)). Assume the initial square is
centered at the origin.
Then by symmetry the bugs are always at four corners of some square
centered at the origin, and if they meet they meet at the origin.
Also, each bug is always walking in a direction pi/4 (45 degrees) away
from the radial line to the origin. This means that in the limit as
the time step goes to zero, the bug travels sec(pi/4) = sqrt(2) units
along its path for every unit of progress made good toward the center.
Since the corners are initially d/sqrt(2) distance from the center,
each bug travels distance d before they meet, assuming they meet at
all.
Since a bug's path always crosses radial lines at angle psi = pi/4,
the path is a logarithmic spiral with angle psi = pi/4 and equation
r(t) = e^(a*theta(t) + b)
Moreover since the bugs walk clockwise, both r(t) and theta(t)
decrease as t increases, in other words r increases as theta
increases, hence a is positive. Also, psi = pi/4 gives us
d(r)/d(theta) = r
(this is easiest to see by drawing the path for a small time step
delta-t and taking the limit as delta-t goes to 0). The solution is
r(t) = e^(theta(t) + b)
(that is, a = 1). As we know, this spiral makes infinitely many
"wraps" around the origin as the radius approaches zero, but it does
have finite length from any point inward and its limit point is the
origin, where the bugs will meet (unless one wants to quibble about
the behavior at the exact limit point).
How much closer is the bug to the origin after its first complete cycle
around the origin? Recall that r(0) = d/sqrt(2). As the bug walks
clockwise around the origin, after one full circuit its angle decreases
from theta(0) to theta(t1), where t1 (the time at which full circuit
occurs) is defined by
theta(t1) = theta(0) - 2*pi
Hence
r(0) = e^(theta(0) + b)
r(t1) = e^(theta(0) - 2*pi + b)
r(t1)/r(0) = e^(-2*pi)
The quantity we want is
r(0) - r(t1) = r(0)*(1 - e^(-2*pi))
= d * (1 - e^(-2*pi))/sqrt(2)
==> analysis/c.infinity.p <==
What function is zero at zero, strictly positive elsewhere, infinitely
differentiable at zero and has all zero derivatives at zero?
==> analysis/c.infinity.s <==
exp(-1/x^2)
There are infinitely many other such functions.
This tells us why Taylor Series are a more limited device than they might be.
We form a Taylor series by looking at the derivatives of a function at a given
point; but this example shows us that the derivatives at a point may tell us
almost nothing about its behavior away from that point.
==> analysis/cache.p <==
Cache and Ferry (How far can a truck go in a desert?)
A pick-up truck is in the desert beside N 50-gallon gas drums, all full.
The truck's gas tank holds 10 gallons and is empty. The truck can carry
one drum, whether full or empty, in its bed. It gets 10 miles to the gallon.
How far away from the starting point can you drive the truck?
==> analysis/cache.s <==
If the truck can siphon gas out of its tank and leave it in the cache,
the answer is:
{ 1/1 + 1/3 + ... + 1/(2 * N - 1) } x 500 miles.
A much harder problem occurs when the truck can siphon gas, but if it does,
it must siphon the gas into one of the initial barrels.
Now, remarkably, for initial gas values of 50, 100, 150, 200, 250, 300,
and 350 gallons, the two problems give identical optimal distances, viz.
gal dist
--- ----
50 500
100 733.3333
150 860
200 948.5714
250 1016.8254
300 1072.3810
350 1117.8355
However, for the 8 barrel case (400 gallons), the unlimited cache optimal
distance is 1157.6957, but limited cache is only 1157.2294.
What happened is that the unlimited cache optimal solution has started to
siphon out more than 50 gallons (60-80/13 to be exact) of gas on trips
to the first depot. With a limited cache, the truck can only leave a
maximum of 50 gallons (one barrel worth), and does not have a big
enough gas tank (only ten gallons) to carry around the excess.
The limited cache problem is the same as the "Desert Fox" problem
described, but not solved, by Dewdney, July '87 "Scientific American".
Dewdney's Oct. '87 Sci. Am. article gives for N=2, the optimal distance
of 733.33 miles.
In the Nov. issue, Dewdney lists the optimal distance of 860 miles for
N=3, and gives a better, but not optimal, general distance formula.
Westbrook, in Vol 74, #467, pp 49-50, March '90 "Mathematical Gazette",
gives an even better formula, for which he incorrectly claims optimality:
For N = 2,3,4,5,6:
Dist = (600/1 + 600/3 + ... + 600/(2N-3)) + (600-100N)/(2N-1)
For N > 6:
Dist = (600/1 + 600/3 + ... + 600/9) + (500/11 + ... + 500/(2N-3))
The following shows that Westbrook's formula is not optimal for N=8:
Ferry 7 drums forward 33.3333 miles (356.6667 gallons remain)
Ferry 6 drums forward 51.5151 miles (300.0000 gallons remain)
Ferry 5 drums forward 66.6667 miles (240.0000 gallons remain)
Ferry 4 drums forward 85.7143 miles (180.0000 gallons remain)
Ferry 3 drums forward 120.0000 miles (120.0000 gallons remain)
Ferry 2 drums forward 200.0000 miles ( 60.0000 gallons remain)
Ferry 1 drums forward 600.0000 miles
---------------
Total distance = 1157.2294 miles
(Westbrook's formula = 1156.2970 miles)
["Ferrying n drums forward x miles" involves (2*n-1) trips,
each of distance x.]
Other attainable values I've found:
N Distance
--- --------- (Ferry distances for each N are omitted for brevity.)
5 1016.8254
7 1117.8355
11 1249.2749
13 1296.8939
17 1372.8577
19 1404.1136 (The N <= 19 distances could be optimal.)
31 1541.1550 (I doubt that this N = 31 distance is optimal.)
139 1955.5702 (Attainable and probably optimal.)
So...where's MY formula?
I haven't found one, and believe me, I've looked.
I would be most grateful if someone would end my misery by mailing me
a formula, a literature reference, or even an efficient algorithm that
computes the optimal distance.
If you do come up with the solution, you might want to first check it
against the attainable distances listed above, before sending it out.
(Not because you might be wrong, but just as a mere formality to check
your work.)
[Warning: the Mathematician General has determined that
this problem is as addicting as Twinkies.]
Myron P. Souris
EDS/Unigraphics
souris@ug.eds.com
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
The following output comes from some hack programs that I've used to
empirically verify some proofs I've been working on.
Initial barrels: 12 (600 gallons)
Attainable distance= 1274.175211
Barrels Distance Gas
Moved covered left
>From depot 1: 10 63.1579 480.0000
>From depot 2: 8 50.0000 405.0000
>From depot 3: 7 37.5000 356.2500
>From depot 4: 6 51.1364 300.0000
>From depot 5: 5 66.6667 240.0000
>From depot 6: 4 85.7143 180.0000
>From depot 7: 3 120.0000 120.0000
>From depot 8: 2 200.0000 60.0000
>From depot 9: 1 600.0000 0.0000
Initial barrels: 40 (2000 gallons)
Attainable distance= 1611.591484
Barrels Distance Gas
Moved covered left
>From depot 1: 40 2.5316 1980.0000
>From depot 2: 33 50.0000 1655.0000
>From depot 3: 28 50.0000 1380.0000
>From depot 4: 23 53.3333 1140.0000
>From depot 5: 19 50.0000 955.0000
>From depot 6: 16 56.4516 780.0000
>From depot 7: 13 50.0000 655.0000
>From depot 8: 11 54.7619 540.0000
>From depot 9: 9 50.0000 455.0000
>From depot 10: 8 32.1429 406.7857
>From depot 11: 7 38.9881 356.1012
>From depot 12: 6 51.0011 300.0000
>From depot 13: 5 66.6667 240.0000
>From depot 14: 4 85.7143 180.0000
>From depot 15: 3 120.0000 120.0000
>From depot 16: 2 200.0000 60.0000
>From depot 17: 1 600.0000 0.0000
==> analysis/calculate.pi.p <==
How can I calculate many digits of pi?
==> analysis/calculate.pi.s <==
long a=10000,
b,
c=2800,
d,e,
f[2801],
g;
main()
{
for (;b-c;) f[b++]=a/5;
for (;
d=0,g=c*2;
c-=14, printf("%.4d",e+d/a), e=d%a)
for (b=c;
d+=f[b]*a,f[b]=d%--g,d/=g--,--b;
d*=b);
}
==> analysis/cats.and.rats.p <==
If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
kill one rat in one minute?
==> analysis/cats.and.rats.s <==
The following piece by Lewis Carroll first appeared in ``The Monthly
Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_,
edited by John Fisher, Bramhall House, 1973.
/Larry Denenberg
larry@bbn.com
larry@harvard.edu
Cats and Rats
If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100
rats in 50 minutes?
This is a good example of a phenomenon that often occurs in working
problems in double proportion; the answer looks all right at first, but,
when we come to test it, we find that, owing to peculiar circumstances in
the case, the solution is either impossible or else indefinite, and needing
further data. The 'peculiar circumstance' here is that fractional cats or
rats are excluded from consideration, and in consequence of this the
solution is, as we shall see, indefinite.
The solution, by the ordinary rules of Double Proportion, is as follows:
6 rats : 100 rats \
> :: 6 cats : ans.
50 min. : 6 min. /
.
. . ans. = (100)(6)(6)/(50)(6) = 12
But when we come to trace the history of this sanguinary scene through all
its horrid details, we find that at the end of 48 minutes 96 rats are dead,
and that there remain 4 live rats and 2 minutes to kill them in: the
question is, can this be done?
Now there are at least *four* different ways in which the original feat,
of 6 cats killing 6 rats in 6 minutes, may be achieved. For the sake of
clearness let us tabulate them:
A. All 6 cats are needed to kill a rat; and this they do in one minute,
the other rats standing meekly by, waiting for their turn.
B. 3 cats are needed to kill a rat, and they do it in 2 minutes.
C. 2 cats are needed, and do it in 3 minutes.
D. Each cat kills a rat all by itself, and take 6 minutes to do it.
In cases A and B it is clear that the 12 cats (who are assumed to come
quite fresh from their 48 minutes of slaughter) can finish the affair in
the required time; but, in case C, it can only be done by supposing that 2
cats could kill two-thirds of a rat in 2 minutes; and in case D, by
supposing that a cat could kill one-third of a rat in two minutes. Neither
supposition is warranted by the data; nor could the fractional rats (even
if endowed with equal vitality) be fairly assigned to the different cats.
For my part, if I were a cat in case D, and did not find my claws in good
working order, I should certainly prefer to have my one-third-rat cut off
from the tail end.
In cases C and D, then, it is clear that we must provide extra cat-power.
In case C *less* than 2 extra cats would be of no use. If 2 were supplied,
and if they began killing their 4 rats at the beginning of the time, they
would finish them in 12 minutes, and have 36 minutes to spare, during which
they might weep, like Alexander, because there were not 12 more rats to
kill. In case D, one extra cat would suffice; it would kill its 4 rats in
24 minutes, and have 24 minutes to spare, during which it could have killed
another 4. But in neither case could any use be made of the last 2
minutes, except to half-kill rats---a barbarity we need not take into
consideration.
To sum up our results. If the 6 cats kill the 6 rats by method A or B,
the answer is 12; if by method C, 14; if by method D, 13.
This, then, is an instance of a solution made `indefinite' by the
circumstances of the case. If an instance of the `impossible' be desired,
take the following: `If a cat can kill a rat in a minute, how many would be
needed to kill it in the thousandth part of a second?' The *mathematical*
answer, of course, is `60,000,' and no doubt less than this would *not*
suffice; but would 60,000 suffice? I doubt it very much. I fancy that at
least 50,000 of the cats would never even see the rat, or have any idea of
what was going on.
Or take this: `If a cat can kill a rat in a minute, how long would it be
killing 60,000 rats?' Ah, how long, indeed! My private opinion is that
the rats would kill the cat.
==> analysis/dog.p <==
A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
In a unit of time, they march forward 50m in formation to take up the
position DCEF. The army's mascot, a small dog, is standing next to its
handler at location A. When the
B----C----E soldiers start marching, the dog
| | | forward--> begins to run around the moving
A----D----F body in a clockwise direction,
keeping as close to it as possible.
When one unit of time has elapsed, the dog has made one complete
circuit and has got back to its handler, who is now at location D. (We
can assume the dog runs at a constant speed and does not delay when
turning the corners.)
How far does the dog travel?
==> analysis/dog.s <==
Let L be the side of the square, 50m, and let D be the distance the
dog travels.
Let v1 be the soldiers' marching speed and v2 be the speed of the dog.
Then v1 = L / (1 time unit) and v2 = v1*D/L.
Let t1, t2, t3, t4 be the time the dog takes to traverse each side of
the square, in order. Find t1 through t4 in terms of L and D and solve
t1+t2+t3+t4 = 1 time unit.
While the dog runs along the back edge of the square in time t1, the
soldiers advance a distance d=t1*v1, so the dog has to cover a distance
sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2.
Solving for t1 gives t1=L/sqrt(v2^2 - v1^2).
The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1).
In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by
using v1=L/(1 time unit), obtaining
2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1
which can be turned into
D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0
which has a root D = 4.18113L = 209.056m.
==> analysis/e.and.pi.p <==
Without finding their numerical values, which is greater, e^(pi) or (pi)^e?
==> analysis/e.and.pi.s <==
e^(pi). Put x = pi/e - 1 in the inequality e^x > 1+x (x>0).
==> analysis/functional/distributed.p <==
Find all f: R -> R, f not identically zero, such that
(*) f( (x+y)/(x-y) ) = ( f(x)+f(y) )/( f(x)-f(y) ).
==> analysis/functional/distributed.s <==
1) Assuming f finite everywhere, (*) ==> x<>y ==> f(x)<>f(y)
2) Exchanging x and y in (*) we see that f(-x) = -f(x).
3) a <> 0 ==> f((a-a)/(a+a)) = (f(a)-f(a))/(f(a)+f(a)) ==> f(0) = 0.
4) a <> 0 ==> f((a+0)/(a-0)) = f(a)/f(a) ==> f(1) = 1.
5) x<>y, y<>0 ==> f(x/y) =
f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) = f(x)/f(y)
==> f(xy) = f(x)f(y) by replacing x with xy and by noting that
f(x*1) = f(x)*1 and f(x*0) = f(x)*f(0).
6) f(x*x) = f(x)*f(x) ==> f(x) > 0 if x>0.
7) Let a=1+\/2, b=1-\/2; a,b satisfy (x+1)/(x-1) = x ==>
f(x) = (f(x)+1)/(f(x)-1) ==> f(a)=a, f(b)=b. f(1/\/2) = f((a+b)/(a-b))
= (a+b)/(a-b) = 1/\/2 ==> f(2) = 2.
8) By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1)
we get that f(n)=n for all integer n. #5 now implies that f fixes
the rationals.
9) If x>y>0 (*) ==> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6.
Thus f is order-preserving.
Since f fixes the rationals *and* f is order-preserving, f must be the
identity function.
This was E2176 in _The American Mathematical Monthly_ (the proposer was
R. S. Luthar).
==> analysis/functional/linear.p <==
Suppose f is non-decreasing with
f(x+y) = f(x) + f(y) + C for all real x, y.
Prove: there is a constant A such that f(x) = Ax - C for all x.
(Note: continuity of f is not assumed in advance.)
==> analysis/functional/linear.s <==
By induction f(mx) = m(f(x)+C)-C. Let x=1/n, m=n and find that
f(1/n) = (1/n)(f(1)+C)-C. Now let x=1/n and find that f(m/n) =
(m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x)
(since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C. Since f is
monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem).
==> analysis/integral.p <==
If f is integrable on (0,inf) and differentiable at 0, and a > 0, and:
inf f(x)
Int ---------------- dx is defined
0 x
show:
inf ( f(x) - f(ax) )
Int ---------------- dx = f(0) ln(a)
0 x
==> analysis/integral.s <==
First, note that if f(0) is 0, then by substituting u=ax in
the integral of f(x)/x, our integral is the difference of two
equal integrals and so is 0 (the integrals are finite because f is
0 at 0 and differentiable there. Note I make no requirement of
continuity).
Second, note that if f is the characteristic function of the
interval [0, 1]--- i.e.
1, 0<=x<=1
f (x) =
0 otherwise
then a little arithmetic reduces our integral to that of
1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar),
which is ln(a) = f(0)ln(a) as required. Call this function g.
Finally, note that the operator which takes the function f to the
value of our integral is linear, and that every function meeting the
hypotheses (incidentally, I should have said `differentiable from the right',
or else replaced the characteristic function of [0,1] above by that of
(-infinity, 1]; but it really doesn't matter) is a linear combination of
one which is 0 at 0 and g, to wit
f(x) = f(0)g(x) + (f(x) - g(x)f(0)).
==> analysis/irrational.stamp.p <==
You have an ink stamp which is so amazingly precise that, when inked
and pressed down on the plane, it makes every circle of irrational
radius (centered at the center of the stamp) black.
Question: Can one use the stamp three times and make every point
in the plane black? [assume plane was white to begin with, and
ignore the fact that no such stamp is physically possible]
==> analysis/irrational.stamp.s <==
Yes. Center the stamp at (0,0), (1,0), and (0,pi).
Suppose there is a point (x,y) which is not covered.
Then there are rational numbers a,b,c satisfying the following equations:
(1) x^2 + y^2 = a
(2) (x-1)^2 + y^2 = b
(3) x^2 + (y-pi)^2 = c
Subtract (2) from (1) and solve for x. Thus x is rational.
From equation (2), y is algebraic. But equation (3) implies
that y is transcendental, contradiction.
==> analysis/minimum.time.p <==
N people can walk or drive in a two-seater to go from city A to city B. What
is the minimum time required to do so?
==> analysis/minimum.time.s <==
Let the speed of the car be v, the speed of a walking person be u, and
the distance between cities by L. We want to minimize T, the time t
at which all persons are at displacement x=L (city B), when they all
start at displacement x=0 (city A) at time t=0.
I'll assume that the solution has everyone starting out from city A at
the same time t=0 arriving in city B at the same time t=T so nobody is
standing around idly waiting. Let's plot everyone's movements on a
graph showing coordinates (t,x). Then at time t just after 0, (N-2)
walkers are on the line L0 through (0,0) with slope dx/dt = u, and 2
in the car are on a line through (0,0) with slope v, and at t just
before T, (N-2) walkers are on the line L1 through (T,L) with slope u,
and 2 in the car are on a line through (T,L) with slope v. Obviously
L1 lies "above" L0 (greater x coordinate given the same t coordinate).
In between t=0 and t=T, the car zigzags between L0 and L1 along lines
of slope v and -v, picking up people from L0 and dropping them off at
L1. I will not prove that this is the optimal strategy; in fact you
can make an infinite number of variations on it which all come up with
the same elapsed time.
Now examine the graph again. Say the car travels distance r between
picking someone up and dropping that person off, and distance s back
to pick up the next person. The car makes (N-1) trips forward and
(N-2) trips back to pick up and ferry everyone, so its total travel is
vT = (N - 1)r + (N - 2)s = (N - 2)(r + s) + r
Moreover the car makes (r-s) net displacement on each round trip, plus
r displacement on the extra forward trip, so
L = (N - 2)(r - s) + r
Note that a person walks distance (r-s) in the time it takes the car to
go (r+s), so
r - s = (u/v)(r + s)
A little algebraic manipulation of this equation shows us that
r - s = r * 2u/(v + u)
r + s = r * 2v/(v + u)
Plug this into the equation for L, and we get the first important
piece of information, how far the car should drive before dropping off
the passenger (once you know this, you tell it to the driver and this
guarantees the people get to B in minimum time):
L = r + (N - 2) r * 2u/(v + u)
= r * (v + u + (N - 2)*2u)/(v + u)
v + u
r = ------------ L
2uN + v - 3u
We can also find out what the elapsed time T will be:
vT = (N - 2)r*2v/(v + u) + r
= r * ((N - 2)*2v + v + u)/(v + u)
1 2vN - 3v + u
T = - * ------------ r
v v + u
Therefore
2vN - (3v - u)
T = -------------- (L/v)
2uN + v - 3u
-- David Karr (karr@cs.cornell.edu)
==> analysis/particle.p <==
What is the longest time that a particle can take in travelling between two
points if it never increases its acceleration along the way and reaches the
second point with speed V?
==> analysis/particle.s <==
Assumptions:
1. x(0) = 0; x(T) = X
2. v(0) = 0; v(T) = V
3. d(a)/dt <= 0
Solution:
a(t) = constant = A = V^2/2X which implies T = 2X/V.
Proof:
Consider assumptions as they apply to f(t) = A * t - v(t):
1. integral from 0 to T of f = 0
2. f(0) = f(T) = 0
3. d^2(f)/dt^2 <= 0
From the mean value theorem, f(t) = 0. QED.
==> analysis/period.p <==
What is the least possible integral period of the sum of functions
of periods 3 and 6?
==> analysis/period.s <==
Period 2. Clearly, the sum of periodic functions of periods 2 and
three is 6. So take the function which is the sum of that function of
period six and the negative of the function of period three and you
have a function of period 2.
This proves that a period-2 solution exists, but not that it is minimal.
Since we're talking about integers, the only lower possibility is 1.
But the sum or difference of a period-1 and a period-3 function must have
period 3, not 6, therefore 1 is indeed impossible.
==> analysis/rubberband.p <==
A bug walks down a rubber band which is attached to a wall at one end and a car
moving away from the wall at the other end. The car is moving at 1 m/sec while
the bug is only moving at 1 cm/sec. Assuming the rubber band is uniformly and
infinitely elastic, will the bug ever reach the car?
==> analysis/rubberband.s <==
Let w = speed of bug and N = ratio of car speed/bug speed = 100. Paint N+1
equally spaced stripes on the rubberband. When the bug is standing on one
stripe, the next stripe is moving away from him at a speed slightly < w
(relative to him). Since he is walking at w, clearly the bug can reach
the next stripe. But once he reaches that stripe, the next one is only
receeding at < w. So he walks on down to the car, one stripe at a time.
The bug starts gaining on the car when he is at the next to last stripe.
==> analysis/sequence.p <==
Show that in the sequence: x, 2x, 3x, .... (n-1)x (x can be any real number)
there is at least one number which is within 1/n of an integer.
==> analysis/sequence.s <==
Throw 0 into the sequence; there are now n numbers, so some pair must
have fractional parts within 1/n of each other; their difference is
then within 1/n of an integer.
==> analysis/snow.p <==
Snow starts falling before noon on a cold December day. At noon a
snow plow starts plowing a street. It travels 1 mile in the first hour,
and 1/2 mile in the second hour. What time did the snow start
falling??
You may assume that the plow's rate of travel is inversely proportional
to the height of the snow, and that the snow falls at a uniform rate.
==> analysis/snow.s <==
11:22:55.077 am.
Method:
Let b = the depth of the snow at noon, a = the rate of increase in the
depth. Then the depth at time t (where noon is t=0) is at+b, the
snowfall started at t_0=-b/a, and the snowplow's rate of progress is
ds/dt = k/(at+b).
If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that
s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where
A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for
x and t_0 = -(1 hour)/x.
The exact answer is 11:(90-30 Sqrt[5]).
_American Mathematics Monthly_, April 1937, page 245
E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa.
The solution appears, appropriately, in the December 1937 issue,
pp. 666-667. Also solved by William Douglas, C. E. Springer,
E. P. Starke, W. J. Taylor, and the proposer.
See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff.
==> analysis/tower.p <==
R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
==> analysis/tower.s <==
ANSWER: e^(1/e)
Let N be the number in question and R the result of the process. Then
R can be defined recursively by the equation:
(1) R = N^R
Taking the logarithm of both sides of (1):
(2) ln(R) = R * ln(N)
Dividing (2) by R and rearranging:
(3) ln(N) = ln(R) / R
Exponentiating (3):
(4) N = R^(1/R)
We wish to find the maximum value of N with respect to R. Find the
derivative of N with respect to R and set it equal to zero:
(5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0
For finite values of R, (5) is satisfied by R = e. This is a maximum of
N if the second derivative of N at R = e is less than zero.
(6) d2(N)/d2(R) | R=e = (2 * ln(R) - 3) / R^3 | R=e = -1 / e^3 < 0
The solution therefore is (4) at R = e:
(7) Nmax = e^(1/e)