Come on all you students if you wanna be free
from stupid thoughts, then you listen to me
Let f be continuous in [a, b],
sign(f(a)) - sign(f(b))
So we know that f has at least one root
and we're thinking to ourselves "This is really good!"
but how to find a root is what we ask
so let's go ahead and deal with this task
To your right is b and to your left is a
and in between a sequence tries to wind its way
so you're standing in the middle halfway in between
(a+b)/2 if you know what I mean
to your left a segment half the big one's size
to your right 'nother the same way lies
Let's slide to a side where f changes signs
let's do this, isn't that fine?
this interval that's half as long,
has a zero if my logic ain't wrong
now you do it again, divide the line in two,
and if you paid attention, you know what to do
you do it k times, things are gettin' small
(1/2)k's your interval
with this little space, this little bound
a root of f can still be found
You do this forever until Tisha B'Av
'cause infinite recursion is what we love
Generating intervals on and on,
only one point's in every one,
all left endpoints they have a supremum
and the right ones have an infimum
this sup and this inf, they lie in each of these sets
though the distance that's between them's as small as it gets
They are both the same, so "What the hell!
I guess it's a limit, le'me call it L!"
Interval k and all points that it spans,
I only need one, 'cause that's how bad I am
xk's my name, it's in interval k
quite close to L, see I planned it that way!
Pick epsilon as small as you want it to be
nested intervals are working for me
I'll come back with an M so big it'll do:
b - a divided by eps' log base 2
The thing with that M, I picked it so good,
after M xk's in L's hood
L's hood's epsilon sized, all intervals lie in it
QED, we have a sequence converging to that limit