So what we're going to try to talk about
today is the arithmetic of dynamical
sequences. Let's do an example; so here's
an example maybe you'll recognize. Let's
let f of x be the function 2x plus 1, and
what I mean by iterating something under
this function is let's start at a point
say x equals 0. When we plug x equals 0
into this function we get 1 and then
let's keep doing that; so when we plug 1
into the function we get 2 plus 1 which
is 3, when we plug 3 into the function we
get 6 plus 1 which is 7. So you look at
the last number - so 15, 31. So maybe I'll
do two more just for a demonstration
here, so 62 plus 1 that's 63 and 126 plus
1 it's 127. So this sequence is special,
so the sequence- you might recognize that
each of these numbers in the sequence
are off from a power of 2 by exactly 1.
so this is known as the Mersenne
sequence. Well people have heard of the
Mersenne sequence because there's a
famous question that we don't know how
to answer is are there infinitely many
prime numbers in this sequence? Since
this question is so hard, one easier
question we could ask would be what are
the prime divisors of each element of
this sequence look like? - (Brady: That seems like)
(a harder question!) - Well it might be a
harder question depending on how
specific you want to get. But so what I
mean by the prime divisors of elements
of the sequence is we have these factors,
so each number factors into prime
numbers as long as we're ignoring 0 here.
So 3 is already prime, 7 is already prime,
but 15 this is 3 times 5.
31 that's already prime, 63 well let's
see - so 7 is a prime divisor and 9 is
a divisor but it's not prime, it's
actually 3 times 3 times 7. And 127 - let's
see I think that's prime; and so on.
Anyway so this is what I mean by prime
divisors. And so instead of asking, okay,
do we have infinitely many elements in
this sequence which are prime? We could
at least ask as we go along in the
sequence do we get a new prime at each
step? So that's the question I'm interested
in answering. So here, okay we don't have
any prime devisers, but here 3 is a
prime divisor that never appeared before,
here 7 is a prime divisor that never
appeared before, here 5 never appeared,
31 never appeared - but here we run into a
problem.
(We broke the streak.) - No new primes.
(And is that bad?) - Well depends on what you're
interested in. But if the question is, do
we always have a new prime divisor? For
the Mersenne sequence the answer is no
because look at the sixth element -
doesn't have a new prime divisor. But the
interesting fact is after that all
Mersenne numbers have a new prime divisor.
(Every single one?) - Every single one.
(What happened with 63?) - Well it's just- when
you're at the beginning of a sequence
the numbers are so close and there's so
many primes happening that it can occur
that you sort of fail to have this new
prime divisor. And it also matters that
this was the sixth element of the
sequence, but that's a little bit
complicated. So the point is that after
that after that sixth element of the
sequence, after 63, we always have the new
prime divisors. So this is an interesting
thing and I think the next natural
question once you notice that a sequence
like this has some interesting property
is what's special about this sequence?
What other sequences have this property?
So instead of choosing a function like
2x plus 1 to generate my sequence what
if I chose a nonlinear function? So if I
had an x squared in there. So let's say I
chose x squared plus 1. But we can still
start generating these sequences; so if
we start with 0, if we plug it into this
function, it is going to map to 1 again but
when we plug 1 in we get 1 plus 1
which is 2. When we plug 2 in we have 2
squared plus 1 which is 5. We plug 5 in
we have 25 plus 1 which is 26; and when
we plug 26 in I happen to know we end up
with 677. So because I'm dealing with an
x squared instead of an x here the
sequence gets larger much faster but still
we get this sequence of integers, still we
can ask when do we start getting new
prime divisors? So if we look at what we
have here- so notice again 2 & 5 are
already prime; 26 is 2 times 13 and 677
is also prime. And it turns out that for
this sequence it's even better than
Mersenne sequence: all numbers after 2
have a new prime divisor. So this has
proven that after 2 you always get a new
prime divisor in this particular
sequence. - (You showed me an amazing)
(sequence; then you showed me a more)
(amazing sequence!) - Ok well then get ready!
So let's do the same thing, except that
there's no reason that we're only
talking about positive integers here,
right? So let's look instead at something
like this and see what happens. So again
we start with 0; 0 maps to negative 1,
negative 1 maps to 1 squared minus 1
which is 0. But we already know where 0
maps - 0 maps to negative 1 and so the
sequence we get is completely not interesting.
- (I think that's cool! How can you say that's not interesting?)
Well it's definitely not
going to have this property that we have
new prime divisors at least. So this is
bad news, right? As far as this question
is concerned this is a weird sequence. So
let's let's try a different negative
number, let's try- I don't want to try
minus 2 and maybe you can try it and
find out why, let's try instead minus 3.
All right so let's go; so 0, and 0 maps to 0
minus 3, minus 3 maps to 9 minus 3 which
is 6, 6 maps to 36 minus 3 - 33, 33 maps to
something messy that I don't want to
calculate. But now we'll get an infinite
sequence again; instead of having this
this repetition like we had in the minus
1 case we'll get an infinite sequence
and again this sequence will have new
prime divisors at each point in the sequence.
Every time. - (I'm beginning to think this isn't)
(such a special thing anymore.) - I know; so let
me tell you you're sort of right. So here
I'll write down a general fact: if say c
is an integer - so a whole number or the
negative of a whole number - and c is not
0 minus 1 or minus 2, mostly because we
want to avoid this problem, then every
element of the sequence that's generated
from x squared plus c has a new prime
divisor. - (So the only exceptions are 0,)
(minus 1, minus 2?) - This works for x
squareds but it works for x squareds
plus an integer. If you wanted you could
ask this for a fraction c instead of a
whole number c. So if c was 1/2 let's
say, so if f of x is x squared plus 1/2
well then what starts happening? So 0
maps to 1/2; now the computation is a
little less pleasant but 1/2 is 1/4 plus
1/2 which is 3/4. 3/4 when we put it into
this function we have 9 over 16 plus 1/2
which I guess is 17 over 16? Does that
sound right? And so on. So even though
we're getting these fractions in the
sequence the denominators are just
powers of 2. And so when you try and
factor them the only prime that you get
there is 2 so we can still ask the same
question about the numerators in the
sequence. - (So we're just going to- we're)
(just going to ignore the denominators?) - Just ignore
the denominators, there's not interesting
stuff there happening there anyways. So
if you ignore the denominators then we
look and we have this sequence 0, 1, 3, 17
and so on; and it turns out that again
you will have this property. It seems
sort of fundamental and part of the
reason why is because I've been hiding
an example from you. So- so maybe one last
example of computation that I'll show
you is this kind of tricky little thing
here: x squared minus 7 over 4. So what does
0 map to? 0 maps to minus 7 over 4, 21
over 16, minus 7 over 16 squared. All of a
sudden everything comes to a stop
because look at our numerators - 0,
7, 7 times 3 - so 3 is a
new prime divisor there - but here we
don't have a new prime divisor. So
there's something different between 1/2
and minus 7/4; for some reason there's an
element in this sequence where the
numerator doesn't have a new prime
divisor. - (And this is- is this the only- this)
(is the only number, the only fraction)
(that throws the spanner in the works?)
We don't know, we don't know. So just to give
you a hint of why this is true: so this
is the only one that we know of, but we
don't know if it's the only time it can
possibly happen. So what's really going
on behind the scenes here has something
to do with a particular set of numbers
known as the Mandelbrot set which I
won't go into but it's this interesting
fractal where things go kind of crazy,
and it's hard to draw a fractal so I'll
just do my best. - (Not a bad effort.)
Thank you! So let me
put this into perspective for you, so
this is a picture of the complex numbers
and I'm interested in the numbers that
are inside of this set. And let's look at
the values of c that we've done this
computation for: so 1, we first did x
squared plus 1, 1 is over here. We
also did minus 1, minus 1 is over
here. We did minus 3 which is well
outside of this set. We did one half
which is also not inside this set. But
minus 7 over 4, it is inside of
this set. So minus 7 over 4 it
turns out is right inside this piece of
the Mandelbrot set. - (This line of the real)
(of the real numbers there)
(seems to run right through a whole swath of)
(fractions that would be- is that- aren't)
(there going to be loads of fractions)
(that are going to cause this problem then?)
That's a really good point. So
you're absolutely right, there's there's
tons and tons of fractions on the real
line inside of the Mandelbrot set, but
it's not enough it turns out to just be
inside the Mandelbrot set you have to be
particularly very far inside of the
Mandelbrot set. And so this is really
saying something special about where the
number c can lie inside of the
Mandelbrot set. - (So if there are other)
(numbers like)
(minus 7 over 4 that are going to)
(throw a spanner in our works they)
(definitely have to be in here?) - That's
right. So right away you get for free
that if you take a large enough fraction,
say bigger than 1 for example although
you could get closer, for free you get
that there will always be new primes in
those sequences. But yeah, if we were to
look for other examples of where this
might occur they would have to lie
inside of the Mandelbrot set and they
would have to lie well inside the
Mandelbrot set actually it turns out.
(But it looks like there are numbers here)
(that would be pretty well inside the)
(Mandelbrot set, or is it a particular like-)
It's a particular thing. Inside each of
these bulbs here that I've drawn sort of
half half heartedly there's a special
point. And what might happen, the only
reason why this might happen, is if your
rational number is very very close to
that special point in a technical way
that is is hard to formulate. I think at
heart people don't believe that there's
one beyond minus 7 over 4. If I had
to make a guess I would say probably not,
but it would certainly be nice to know
that for sure.