Conference rusure::math

What does

(a(n))n 

mean ?  Is it different from

a(n) ?


Is the final n multiplication ?  Or a subscript ?  I don't see its purpose.



Then, "a sequence of non-null real numbers".  Is this the same as
"a sequence of non-zero real numbers" ?  Or do you mean "a non-null sequence
of real numbers" ?

> What does (a(n))n mean ?  Is it different from a(n) ?
> Is the final n multiplication ?  Or a subscript ?  I don't see its purpose.

It's just a classical notation for a generalized sequence, where the family
of indices is mentioned:

		(a(i))
                      i /in I

When this family is  N  = the set of natural numbers, the string "/in N" is
usually skipped, so we get 

		(a(n))
		      n

I wrote this on the same line:  (a(n))n . So it's a subscript.

> Then, "a sequence of non-null real numbers".  Is this the same as
> "a sequence of non-zero real numbers" ?  Or do you mean "a non-null sequence
> of real numbers" ?

Sorry for my bad English. I meant "a sequence of non-zero real numbers".

/Mihai.

Well, let's try an example.

From your wording, it seems you're saying that any a(n) I choose, and any
real number x I choose, we will be able to find a lambda(n) sequence of
integers that work.

So, let's try

	a(n) = 1/2 + 1/4 + 1/8 ...

These converge to 0 so they qualify.

For our real number, let's pick 1/3.

So, our challenge is to find a sequence of integers lambda(n) such that

	1/3 = a * 1/2   +    b * 1/4    + c * 1/8 ...

I'm saying that lambda(n) is {a,b,c...}

This particular example can be solved with the sharing-the-cookie puzzle:

I start with a whole cookie.  I eat half and hand the other half to you.

You eat half of that and hand back the rest to me.

I eat half of that and hand the rest back to you.

We continue this until what's left is just a crumb, which we discard
and pretend doesn't exist.

How much of the cookie did I eat and how much did you eat ?

Well, instead of launching into infinite sequences and series, there's
a fun way of approaching this.

In round 1, I ate 1/2 of the cookie, and handed it to you and you ate
1/4 of the cookie.

So I ate twice as much as you in round 1.

In round 2, I ate 1/8 of the cookie and you ate 1/16.

So I ate twice as much as you in round 2.

Hence every round, I eat twice as much as you !  Therefore, at the end,
we've together consumed 1 whole cookie, and I've eaten twice as much
as you.

Hence I've eaten 2/3 of a cookie and you've eaten 1/3.

In round 1, I ate 1/2 a cookie, and in round 2 I ate 1/8, and in round 3,
I ate 1/32, and we already know I ate 2/3 of a cookie, so we've proven that

	2/3 = 1/2 + 1/8 + 1/32 ...

without doing any fancy math.

Similarly, you ate 1/3 of the cookie, and we know you ate 1/4 of the
cookie in the first round, then 1/16 in the second round, then 1/64 in
third round, so again without fancy math we've proven that

	1/3 = 1/4 + 1/16 + 1/64 ...

which brings us back to the example I posed at the beginning.

We want to find integers a,b,c... such that

	1/3 = 1/2 * a   +   1/4 * b    +       1/8 * c  ...

We can now see that this will work:

	a,b,c... = 0,1,0,1,0,1...


/Eric

For each m in turn, m>=0, choose lambda(m) to minimize the difference between x
and sum(n in 0..m)lambda(n)*a(n).  (Since a(m) isn't 0, the choice is unique
unless the two nearest values are equidistant from x; in this case arbitrarily
choose the larger value for lambda(m).)  The partial sum of the series up to
index m differs from x by at most |a(m)|/2, so the series converges to x.

Yes, this is the right circle of ideas. For completeness's sake, you should
minimize the absolute value of the difference you mentioned.

Please reassure me, you *did* take part in some IMO contests? Congratulations for
the prize(s) that you must have won!

/Mihai.

>Please reassure me, you *did* take part in some IMO contests? Congratulations
for the prize(s) that you must have won!

Not IMO, but I was in the top five of the Putnam exam three times; and with
Arthur Rubin and Bruce Resnick, helped give Caltech the all-time best team
score.  (Lest I seem overly proud, I will note that Arthur was in the top five
all four years he entered, and almost surely outscored me every year.  But the
exact rankings of the top five contestants each year are kept secret.)

    Half a lifetime ago, I was lucky enough to scrape a silver medal as one of
    the British team in IMO 77, in Belgrade. I got a favourable line-call
    after I explicitly assumed Dirichlet's theorem in one question.
    
    The team came third to USA & USSR. The group theorist Richard Borchards
    is the only member of the team who is today a practising research 
    mathematician, so far as I now. Most of the team got involved in
    computing, sooner or later. I have encountered *3* of the 6 questions
    that year solved in Automated Reasoning papers.
    
    However, a lot of beer has gone under the bridge since then, and I'm
    not as sharp as I used to be. I too enjoy Mike's solutions: good to
    have you aboard in these latter Stan-less days.
    
    Cheers,
    Andrew. 

        My third Putnam I ended in the top ten, along with another MIT
        student.  I wasn't on the team then, but my fourth year (the
        Dec. '79 test) we were both on the team, along with someone
        who hadn't taken the test my third year but had been in the
        top five the year before that.  We were all gung-ho to all end
        up in the top five and impress the world with how great MIT
        was.  One of the three did, but the other one and I both
        dropped to fortysomething and we "only" won first place team.
        :-)  The student newspaper quoted one of our professors as
        calling this the equivalent of getting the Nobel prize in
        undergraduate mathematics.
        
        Dan

There is an interesting story associated with my second Putnam.  My first year I
had come in 49th, and a junior named James Shearer had come in below me.  When
it came time to choose people for next year's team, some wanted to pick me
because I had the better score (even though I had been a freshman), and others
wanted to pick James because he would be a senior.  It wasn't strictly a merit
vs. seniority argument: this was going to be his last chance to be on the
Caltech team.  I remember being asked by Dr. Gary Lorden what I thought should
be done, and answering that although it might seem self-serving, I thought the
team ought to be picked to maximize the expected team ranking, and that going
strictly by score was the way to do that.  (I remember saying that the Putnam
test was supposedly designed to try to test raw ability rather than stored lore,
and to the extent it succeeded, going just by score was right.)  Anyway, Dr.
Lorden, whose specialty was statistics, thought the arguments on this side
sounder than the others, and when I made it to the top five Jim was gracious
enough to admit that it had been the right choice.

I ranked 16th in 1974 and 23rd in 1975 at the Romanian "barrage" for the IMO,
but didn't get on the team which had 8 players and 2 substitutes.

Even when young, I was already slow.

/Mihai.

Corollary 1.
------------

Let  f: R --> R  be periodic and have a sequence of periods converging to
zero. If  f  is continuous, then it is constant.


Corollary 2.
------------

Let  a,b  be real numbers, a,b != 0, and let

	Z(a,b) = {alpha * a + beta * b / alpha, beta /in Z}

Then the set  Z(a,b)  is dense in  R  if and only if  a/b  is irrational.


Corollary 2.1.
--------------

Let  a,b  be real numbers,  a,b > 0 , and let

	Z+-(a,b)  = {alpha * a - beta * b / alpha, beta /in N}

Then the set  Z+- (a,b)  is dense in  R  if and only if  a / b  is
irrational.


Corollary 2.2.
--------------

Let  a,b  be real numbers,  a,b > 0 , and let

	F(a,b)  = {a^alpha / b^beta / alpha, beta /in Z}

Then the set  F(a,b)  is dense in  R  if and only if  log a / log b  is
irrational.


Corollary 2.1.1.
----------------

Let  a,b  be real numbers,  a,b > 0 ,  (a - 1) * (b - 1) > 0 , and let

	F+-(a,b)  = {a^alpha / b^beta /  alpha, beta /in N}

Then the set  F+-(a,b)  is dense in  R  if and only if  log a / log b  is
irrational.


Corollary 2.1.1.1.
------------------

Let  p,q  be natural numbers,  p,q >= 2 , and such that  log p / log q  is
irrational. Then for any finite sequence of digits in base  p , there is a
power of  q  whose writing in base  p  begins precisely with  that sequence
of digits.

(This generalizes note 1906.0, as stated in 1906.4).


Corollary 2.1.2.
----------------

Find the limit points of the sequence  (sin n)n .


Mihai.

> Let  f: R --> R  be periodic and have a sequence of periods converging to
> zero. If  f  is continuous, then it is constant.



What about a damped sin wave.  That's periodic, and it converges to 0, but
it's not constant (although becomes increasingly so as it damps).

/Eric

Eric, what do you mean by "damped sin wave"? Can you give its definition?

Mihai

I assume a "damped sin wave" is something like  a*sin(b*x)/x .

I would say this isn't periodic, assuming that periodic with period p
means that f(x)=f(x+p) for all x.

Having a sequence of periods p1, p2, ... means that
	f(x)=f(x+pn) for all x, for each n

To show that f(x)=f(y) for any x,y, we simply show how to express
y-x as a sum of pn's.  Then f(x)=f(x+a1p1) = f(x+a1p1+a2p2) = ...
Along with the assumption that f is continuous, we get f(x)=f(y).

Damped sine wave is what you get if you attach a pen to the end of a spring,
give it a nudge, then drag a long sheet of paper by it.


But I realize now it's not periodic, since although the space between
humps might be constant, each hump is more damped than previous,
and hence isn't strictly periodic.

/Eric

in case anyone wants to follow this diversion, someone did work on it,
long ago.  back in the mid to late fifties i saw a book called
"almost periodic functions" or "nearly periodic functions", or 
something similar. i have no idea of the author or publisher,
but an impression if was one of a rather long series of books.