Enter questions about math here.

47 responses total.

I am preparing to take the GRE in the spring in order to apply for graduate
school.  My weak area is math--on the SAT, I got a perfect verbal score and
a 650 on math.  My question is:  can anyone recommend a decent way for me 
to brush up  on my math skills?  I took a formal math course last in 1988.
I'm looking for teach-yourself type material, preferably user-friendly.  To
be honest, I feel about math the way some feel about an abusive parent:  we
have not had a very productive relationship.  Any suggestions?

It's been a while since I took the GRE - what kind of math is on it?

Basic and advanced algebra, geometry, and some trig.  Not really any basic
arithmetic-type-stuff, which I guess I should be proficient at by now.

Brenda is a good person to answer this, since she has been brushing up
on her math lately.

There must be GRE "prep" manuals. See if you can find one of those. People
here can help the most on specific difficulties.

Try Border's for GRE Prep manuals.

This response has been erased.

I ran out and got two prep books--one Arco and one Barrons, and they both look
good.  Wish I had a Border's down here!.  I'll be back with specific questions,
 believe me. 

can someone please help me on this one.

        3 men chech into a very cheap motel.  They decid to share a room.
Each man hands $10 to the person at the front desk.  Later the person realises 
that rooms only cost $25.  He gives $3 back and keeps the extra $2 so there
wouldn't be an argument amongst the 3 men.

So...each man paid $10 minus the $1 recieved back=$9
9x3 men =$27 plus the $2 the person at the desk kept= $29 !
Where is the missing dollar?

You are adding incomensurables. The hotel got $25, the men paid $27,
and the clerk ripped off the difference of $2. 

read the iq conf.  This exact same item is there, with a ton of responses.

Oh! I thought ryan1 really wanted to know! (not).

O.K. You figured me out!  I was playing a trick on you.  I wanted to see how
many responces it would take to figure it out.

Wow! This math section certainly lacks participation!!  Maybe math is
falling into disuse these days?  Anyway, to stir the pot (perhaps): 
Can anyone tell me the origin and the value of the number known as a
google?  I have a general idea of the origin, but cannot recall the
value.

A Googol is the decimal number written as 1 followed by 100 zeroes.
That is, 10^100. The name was invented by the nine-year-old (then)
nephew of one Dr. Kasner.

Define G = 10^100. A larger number is the Googolplex, defined
as 10^G. 

The apparent lack of participation is due to few people seeking
tutoring. But the tutors are hanging around.....

Where would you use numbers as large as a googol or a googolplex?  I read
somewhere (Maybe it was in A Brief History of Time,  
Let me start again.  Where would you use numbers as large as a Googol or
larger?  I read somewhere (I think it was in A Brief History of Time, by
Hawkings) that the mass of the universe was in the range of 10^80 grams.

Accepting that, and assuming the universe is mostly hydrogen, that is
10^80 *gram-atoms*. Each gram-atom contains 2.03E23 atoms, giving
2.03E103 atoms in the universe. That is, 2,030 googols. (Or, is that
2.03 kilogoogols?)  In regard to the googolplex, my book that discusses
this says:

  "One might not believe that such a large number would ever really
  have any application, b ut one who felt that way would not be a
  mathematician. A number as large as the googolplex might be of real
  use in problems of combination."

The article then goes on to describe an example of "combination", or
combinatorial probability, that requires the googolplex. Summarized,
it is that if an object is hung on a string, what is the probability
that it will spontaneously jump above the hanging point (because all
the atoms at that moment vibrated in the same directiion). It can be
shown that this probability lies between (and not outside) the
range (googol)^-1 and (googolplex)^-1. It could, of course, happen
today, as this is probability statement. Watch for it.

Ok, makes sense in terms of probabilities - location of an electron, chemical
reaction, etc.

<sigh> and I had so much hope for this cf.  Well, I have a little math 
problem that has been nagging me for a while now.  I'll post it next 
time I'm on from my room.

Okay, here it is.  I couldn't find the origional sheet it was assigned 
on, but I can reconstruct it.  The problem was to derive the 
reflection/refraction equations that make a rainbow.  It was divided up 
into many parts, but one of them was to use fermat's principal to 
determine the path followed by the light beam.  Thus, an equation was set 
up, and the derivative was taken, and the result set to zero to minimize 
it. So far, so good.  The next part was to solve for x.  Now, in order to 
solve the question, solving for x is not necessary, and they eventually 
took this part out of the project, and all was fine and good.  However, I 
had gotten to that part, and attempted to solve for x, and couldn't do 
it.  I spent many hours and many sheets of paper and much frustration on 
it, and then was rather ticked when they took it out of the problem.  I 
since haven't had time to go back to it, but here it is, for the tutoring 
cf to worry about for a while:

solve for x, assuming that all other letters are constants:

(n_1*x)/(c(a^2+x^2)^(1/2))-(n_2(d-x))/(c(b^2+(d-x)^2)^(1/2)) = 0

Enjoy!

Michael says, the way it's written, the equation = 0 only when x=0.

Are n_1 and n_2 coefficients n1 and n2, and what is the *, which
is not associated with n_2(d-x)? If I take n1 and n2 as coefficients,
and * as multiplication, then I don't agree with #21. c cancels. The
equation expands to forth order polynomial. I do not see a useful way
to factor it. 

You're right, n_1 and n_2 are coefficients n1 and n2 ( was attempting to 
imply subscript...) and the * is multiplication.  The reason it's only 
used there is that's the only place where there isn't a paren to imply 
multiplication.  and, when x = 0, I get -n2*d/(c(b^2 + d^2)^(1/2), not 0. 
I came up with the same thing as in the above response, and couldn't for 
the life of me solve it.  I'm assured there is an answer, but my 
calculator has plugged away at it for an hour, and not given me anything 
useful.

Do you want an analytic factoring, or a numerical answer? The latter 
is easy to find for specific cases. The quartic has an interesting
structure, so I have not *given up* on factoring it. It would be easy
for specific relations between some of the constants. What is your
calculator trying to do with it? 

Solve it for x... Or so it claims. :)  I have an HP48sx, and it has an 
isol function that isolates one variable.  That's what I was trying to 
use.  I'd like a general solution, if that's possible, involving the 
constants.  I know the solutions involving specific constant ratios, as 
that was the origional point of the exercise.

Would you check the expression as you wrote it above? I am suspicious,
because the constant c cancels. Seems silly for it to be there, if it
doesn't affect the solution. 

Well, it looks right to me, and the c doesn't cancel as far as I can 
tell.  There are 2 c's, and they are both in the denometer of an 
expression, and one expression is being subtracted rom the other. I'll 
attempt a more visual translation of it.


        n1 * x                       n2 * (d - x)
----------------------   -   ---------------------------   =   0
 c * (a^2 + x^2)^(1/2)       c * (b^2 - (d - x)^2)^(1/2)


Is that an improvement?

Multiply the whole expression through by c. The c's in the denominator will
cancel out and disappear. So, why are they there in the first place?

I suspect two reasons.  First, this was a minimization problem, so it was 
a derivative set equal to zero, so my physics bood probably didn't cancel 
because it was more interested in the derivative than the solution, and 
second because sin was substuted for some of those quantaties, and the 
c's are necessary for that.

A sign has changed in the denominator of the second term in #27, compared
to that in #20. Typo? Well, I still cannot see a way to an explicit solution.

The one in 27 is a typo, and should read + instead of -.  That's 
basically the way I felt.  I think I'll ask my prof...

I want to place a near-equilateral triangle with its vertices on a
square grid. What is the succession of sizes (or vertex placements)
of a triangle on a square grid such that the vertex angles are all
60 degrees +/- 1 degree?

If one side conincident with a line of the square grid (is there a better
way of saying that), I found the following succession that meets the
criteria:

                base      height     base angle
               ^^^^^^     ^^^^^^     ^^^^^^^^^
                 8          7         60.25
                16         14         60.25
                20         17         59.53
                22         19         59.93

I bogged down, however, when considering near-equilateral triangles tilted
off a grid line. The +/- 1 degree criterion is also arbitrary but I did
not think of a way to generalize it.

Also, where is the "center" of each of these triangles? (Say, the "center"
of the successive triangles formed by the intersection of normal side
bisectors. Do any of these "centers" fall exactly on a grid point?)

So, the chart is a list of isoceles triangles?

Those are, but the general tilted case would be scalenes too. This problem
arose, by the way, because I wanted to put three spacers between two
circular pieces of perf boards (centered on a hole) at as close to an
equilateral placement as possible. In trying to figure it out I saw some
of the inherent mathematical difficulties. 

Is it known that there is no way to do it with a completely equilateral
triangle?  It doesn't seem likely, but I don't see an easy way to prove that
it can't be done; not w/o a piece of paper in front of me, anyways.

I looked at that (with a piece of paper), and realized that all sides
would be roots of integers (on a unit grid). This is a considerable
constraint on the the vertex angles per the law of cosines, viz

               cos(a) = (B^2 + C^2 - A^2)/2BC  [ = 1/2?]

for which the numerator is an integer, while the denominator is twice the
product of two roots of integers. Can this ratio ever be 1/2? The sides
would be related by

                        B^2 + C^2 - A^2 = BC

and the sqares are all integers. 

This is where I bogged down.

Hmmm...  That also gives us:

                     B/C + C/B = 1 + (A^2)/BC

Which maybe isn't helpful, but looks kind of nice.  :-)

Perhaps more helpfully:

                       B^2 - BC + C^2 = A^2

                       (B - C)^2 + BC = A^2

Well, maybe not more helpful.  In any case, all of these are obviously true
when A=B=C, so I'm not sure this line of thought is helpful at all.  *sigh*

The fact that certain of the numbers must be integers or roots of integers
is, I think, the key. I recall the designation of "diaphantine" equations,
which are integer relations - but don't recall enough to do anything
useful.

I've been gone for a few months, but I just discovered this discussion. 
I'm sorry I missed it earlier. The following addresses just one of the 
questions Rane raised back in October:

Theorem: No equilateral triangle can be oriented so that each vertex 
falls at a point on a lattice formed by points in a Euclidean plane 
whose coordinates are all integers (i.e. a square plane grid).

Proof: Assume that you can do this. If there is an equilateral triangle 
that can be oriented so that each vertex falls at a point on a lattice 
formed by points in a Euclidean plane whose coordinates are all 
integers, then we can pick the smallest such triangle, and we can place 
one vertex arbitrarily at the origin. Let the other two vertices be at 
coordinates (a,b) and (c,d).

Setting the sides equal can be expressed with these formulas
(by dropping the square root on each side of the equality, something we 
can legitimately do if we know we always have the positive root):

a^2 + b^2 = c^2 + d^2 = (a-c)^2 + (b-d)^2

                      = a^2 + b^2 + c^2 + d^2 -2ac - 2bd

                      = 2 (ac + bd)

The last one is obtained by subtracting each side of the equation from 
a^2 + b^2 + c^2 + d^2

Now, if all of a,b,c,d are even, then we could have a smaller triangle 
using a/2, b/2, c/2, and d/2, so at least one must be odd.

We now show that if one is odd, it's pair must also be odd. 
If a or b is odd, then the other must also be odd, because if a and b 
had opposite parity, then a^2+b^2 would be odd, but we can see it must 
be even because it is equal to 2(ac+bd). THe same can be said of c and d 
by the same reasoning. So we know that at least one pair (a,b) or (c,d) 
are both odd. 

Now any odd number can be written as 2n+1 for some integer n. So if you 
square it you get 4n^2+4n+1 which is congruent to 1 mod 4. Therefore the 
square of any odd number is congruent to 1 mod 4, so either a^2 + b^2 or 
c^2 + d^2 is congruent to 2 mod 4. They're equal so both must be. 
Therefore all four of a,b,c, and d must be odd.

However, that makes ac even and also bd even, so ac+bd is even, and 
2(ac+bd) must therefore be congruent to 0 mod 4, so it cannot be equal 
to a^2+b^2 as required.

QED - Therefore the initial assumption must be incorrect. You cannot 
place an equilateral triangle on a lattice as assumed.

---

Rane also observed that you can find triangles that are nearly 
equilateral that do work. This is an interesting area to explore, but I 
can't think of any way to express a set of nearly equilateral triangles 
that fit. 

It's "Di*o*phantine Equations", btw. Named after a Greek mathematician. 
In general they are very tough to solve, and usually it is impossible to 
find closed form solutions.

- Backtalk version 1.3.30 - Copyright 1996-2006, Jan Wolter and Steve Weiss