welcome to another mathologer video.
many of you will have heard of the
indian mathematical genius srinivasa
ramanujan.
largely self-taught, astonishingly
original,
died way too young 100 years ago and so
on. an incredible story.
one of the things ramanujan is famous
for are these totally insane
infinite continued fraction equations
like the one over there.
a completely unexpected connection
between the three super constants
pi phi and e.
and just like most of our ramanujan's
mathematics this equation is super hard
to make sense of
even if you are a professional
mathematician.
luckily there is still some more
accessible ramanujan magic to be
mathologerised
before i have to tackle that part of
ramanujan's legacy.
in particular there is this great
anecdote featuring ramanujan
and an infinite continued fraction and
today's mission is to explain what's
really going on in the story
and what the infinite fraction is doing
there.
okay in 1914.
ramanujan's countrymen and fellow
mathematician prasantha
mahalanobis was visiting ramanujan at his
rooms in cambridge in england.
ramanujan was cooking a meal and mahalanobis
was passing the time reading the
1914 december
issue of The Strand a very popular
illustrated magazine at the time.
fun fact: a lot of sherlock holmes
stories were first published in this
magazine. for example, this particular
issue featured the latest sherlock
holmes story: the valley of fear.
the strand also featured a regular
puzzle column written by one of the
greatest puzzle inventors of all time, i'm a real
fan, henry dudeney. it's 1914 and the first
world war had broken out
earlier that year dudeney's puzzles in
this issue of the strand are woven into
a story of some of the patrons of a
village inn
discussing the war. for example, one
puzzle asks
how this red cross here can be cut into
five pieces that can be reassembled into
the two smaller crosses.
nope no continued fractions in sight yet.
soon promise :)
but can you do the red cross puzzle? if
you can
post a link to your solution in the
comments.
but now let's look at the other puzzle
the one that ramanujan's friend was
pondering while ramanujan is cooking
away.
this puzzle is set in Louvain a belgian
city
which had just made the war headlines.
there's a man who lives on a long street
numbered on his side
one two three and so on and that all the
numbers on one side of him
added up exactly the same as all the
numbers on the other side of him.
so all the purple house numbers add up
to the same sum
as the green house numbers. the puzzle
consists in finding out how many houses
were on the street
and the number of the house in which the
man lived. all clear?
it's a tricky puzzle but after a while
mahalanobis figured it out and then
challenged ramanujan to do the same.
what happened?
yep ramanujan waved his magic wand, or
his magic brain
or whatever had the answer straight away.
pretty damn
impressive, right? but what impressed
mahalanobis even more was that
ramanujan gave his answer in the form of
one of these
infinite fractions. and this infinite
fraction not only gave the answer to the
strand puzzle
but also gave the infinitely many
answers
of a closely related and important super
puzzle a so-called Pell equation.
okay so today we'll have a go at solving
the strand puzzle,
understanding ramanujan's solution,
understanding why on earth people would
bother with these infinite fractions in
the first place
and at the end come up with a solution
that even
beats ramanujan's. believe it or not.
and the key to all this is the fact that
root 2 is an irrational number
and a curious property of A size pieces
of paper.
lots of ancient, beautiful and deep math(s)
to look forward to.
okay are you ready to solve the strand
puzzle together with me?
yes? well that's great :) then let's go.
okay a street and a house with a special
property. as usual,
to get a feel for this problem let's
have a look at some small
examples. so can the street be five
houses long?
let's first focus on house number three
the house in the middle.
purple one plus two is three and green
four plus five is nine.
since the sums aren't equal our man is
definitely not living in house number
three on this street.
of course, with hindsight that's
never going to work, right? the house will
definitely have to be past the middle of
the street
since the numbers on the left are
smaller than those on the right.
so how about house number four?
close but no banana. one plus two plus
three is six
which is not equal to five and that
means that the street we are puzzling
over is definitely not
five houses long. what's next? well i
leave it to you to rule out the tiny
streets with two,
three or four houses. so let's go bigger
and try a street with
six houses. we still have a sum of
six on the left and now five plus six is
eleven on the right. doesn't work
and this time one plus two plus three
plus four is ten on the left
and six on the right. too small, no dice :)
okay so forget about streets with six
houses.
on to seven. left sum
is still ten and on the right we have
six plus seven
is thirteen. nope. okay
fifteen on the left now and seven on the
right.
but wait a minute. can you see it?
15 on the left and 7 plus what is 15?
well 8 of course! 15 on both sides, so
8 works? great, problem solved.
not so fast!
remember, when you used to do word
problems in school and your teacher
would demand that you:
READ EVERY WORD. well,
welcome back to school. let's read the
puzzle again. it says
there's a man who lived on a long street.
yep it's a long straight and i think we
can agree that eight houses
is not long enough to be called long. so
apparently dudeney was after a different
solution.
well if you keep on searching, you
actually do find another solution,
this one here and that's the solution
dudeney was after?
well, no. confession time.
there's actually a bit more to dudeney's
puzzle than i showed you.
so not only did you have to read every
word like in school,
you had to read some words you weren't
actually given like in a really
mean school :) here they are.
there, it also says that there are more
than 50 houses in our street and less
than 500. this means our solution up
there
is not dudney's. there must be yet another
solution.
but wait a minute, why did the maestro
specify
that the solution he's looking for is
less than 500. maybe there's an
even larger solution? well it turns out
that there are infinitely many solutions
and by stipulating more than 50 and less
than 500
dudeney simply picked out one of these
infinitely many solutions of just the
right difficulty.
so how do we find it? you reckon it's a
good idea to keep checking streets of
increasing length: 50 then 51 then 52
and so until we hit the next solution?
without a computer?
by hand? promise, not a good idea :)
of course there's a better way which
mahalanobis would have figured out. but
now comes the punchline. ramanujan's
infinite fraction method
picks out all the infinitely many
solutions,
all at once. intrigued?
well then it's time for a little algebra.
so we're looking for the number of
houses and the number of the special
house.
well let's call those unknowns, wait
let's be really original,
let's call them x and y.
what a surprise. so the purple sum on
the left is
1 plus 2 plus 3 up to x minus 1. easy
right
we've already encountered the formula
for this kind of sum a couple of times
in previous videos. remember
it's this. cool and this sum is supposed
to be equal to the sum
of the greens. but the green sum
is just a difference of two of those one
plus two plus three sums.
can you see it? the green sum is
one plus two plus three all the way up
to y
minus one plus two plus three all the
way up to
x. and so our formula for this sort of
sum
strikes again, twice. okay
now some algebra autopilot to simplify
this mess.
very nice :) as a quick check that we
haven't
messed up, let's sub in one of the
solutions we found earlier.
okay 6 squared is 36 times 2 is 72 and 8
squared
is 64 plus 8 is also 72, works.
great. okay what's next? well maybe play
with the equation a bit.
right, written this way the equation
looks even nicer and that x squared plus
x squared
equals y squared and a bit is somewhat
reminiscent of pythagoras.
is there any point to this? well
reasonable to try but
sadly it doesn't lead anywhere. what else?
well it's definitely tempting to solve
for x or y and then start plotting and
see where that gets us.
but there's something else that's crying
out to be done here
maybe some of you will have guessed
already. what we can do is some good old
completing the square on the right.
for this let's just hit the algebra
autopilot button again.
hmm, did that help? at first glance this
looks more complicated than the equation
we started with
right? but what's nicer about the new
equation is that there is
only one x and only one y,
with a bit of noise around each.
to suppress the noise, let's abbreviate
the red by capital
x and the green by capital y.
now our new capitals equation is
definitely much nicer.
and, importantly, if we find solutions to
this new equation, then it's easy to
translate them
into the solutions of our original
equation. how?
well by substituting back for a little x
and little y.
like that. okay
so this is the equation we want to solve.
just a little shuffle to put it in its
most useful form.
there. okay someone like ramanujan would
see
everything we've done so far in a flash.
but even more so,
ramanujan also saw immediately that
all solutions to this equation are
contained in this
special infinite fraction.
what does that even mean? well, that means
he must be a genius, right?
well believe it or not that's not a
rhetorical question.
before i really address this question,
let's extract all the solutions
out of the infinite fraction, without
worrying about the why, at least
not to start with.
to extract the solutions of our equation
from the infinite fraction we have to
calculate the so-called
partial fractions. again, please just run
with it,
we won't worry about the why just yet.
the partial fractions are the finite
fractions that you get
when you stop at one of the plus signs
and zap everything from there on.
so to get things going chop things off
at the first plus.
okay not much left so the first partial
fraction is
one, which we can also write as one over
one.
nothing impressive yet but let's keep
going.
chop off at the second plus
one plus one half that's three halves.
that's the second partial fraction. the
third one is this
and that's equal to seven over five, and
so on
there and the denominator and numerators
they grow in size.
a quick challenge for you: can you guess
a simple rule that relates consecutive
fractions?
anyway choose one of the fractions, let's
go for a really simple one,
3 over 2. let's make 3 the y
and 2 the x. 3 squared minus 2 times
2 squared that's 9 minus 8. that's one.
works :)
the numerator and denominator solve our
equation.
let's try the next fraction: 7 over 5.
there
7 squared that's 49 minus 2 times 25, so
49 minus 50
that's minus one. that didn't work
but we plow on regardless and the next
one works again.
289 minus two times 144 that's one.
actually what turns out to be the case
is that every second fraction works.
in fact these fractions correspond to
all the solutions to our equation in
positive integers.
and what about the other fractions? well
they correspond to all the solutions of
the companion equation, with a minus 1
on the right. pretty amazing, isn't it?
now all that remains to be done is to
translate these orange fractions
into the solutions of the strand puzzle.
the translation rules were what? well
these here.
so to get the house numbers, the little
x's
on the left, we simply have to divide by
two. like
this. and to get the corresponding
numbers of houses in the streets,
we subtract 1 and then divide by 2. like
this. nice.
here's our first solution. there's the
second one
and of course next is the solution between
50 and 500 that dudeney actually had in
mind.
imagine finding this solution by a trial
and error. scary.
i guess even just checking by hand that
our street equation is satisfied by
these two numbers
would be way too taxing for most people
who grew up punching buttons on a
calculator.
before we move on, maybe ponder the one-one solution at the top.
can you make sense of this? easy right? or
not?
anyway to wrap up, here is a summary of
all the stuff that ramanujan saw
in a flash. whoa.
now there's no doubt that  ramanujan was
an absolute genius.
and there's definitely some genius
mathematics in all this, especially that
infinite fraction.
and seeing all this at a glance is super
impressive.
however that genius mathematics is
actually not due to ramanujan
and although continued fractions are a
little forgotten
many mathematicians a century ago would
have been familiar with this approach
and would have followed something like
ramanujan's path to a solution of the
problem.
and so great story that it is,  ramanujan's
solution
is not as amazing a mathematical feat as
it may appear at first glance
and as it is often made out to be in the
telling of the story.
in particular, the equation that we are
solving at the end
is super famous and all its
solutions were first found a couple of
thousand years ago.
also any mathematician who has studied
continued fractions
will be aware that our equation is the
simplest
example of what is called a pell
equation, that is an equation
with some non-square positive integer in
the place of the two over there.
and these mathematicians would also know
that all the solutions to the pell
equation are contained in the simple
continued
infinite fraction of root k. and indeed
our infinite fraction
is that simple continued fraction of
root 2.
well that's all great but i'm sure
everybody else who has made it so far
you all
won't be able to sleep tonight if i
don't explain the connection with root
2
and where on earth this infinite
fraction comes from. well, don't worry
that's what i'll do next. so catch your
breath, grab yourself a cup of coffee and
then
let's get to it.
so what's the natural way to come up
with that infinite fraction
as a way to solve this equation and
what's the connection with root 2?
well that root 2 is involved is actually
easy to see.
let's tweak our Pell equation a little
and replace the 1 by 0. then autopilot.
what this tells us is that if x and y
were integer solutions of our tweaked
equation
then the fraction y over x is equal to
root 2.
cool. but of course that's not going to
happen. since root 2 is irrational
such integer solutions cannot exist. on
the other hand,
going back to the original Pell equation,
it's now
also clear that if x and y solve that
equation,
then y over x will be approximately
equal to
root two. and the larger the x and y the
better this approximation will be.
let's see how that works in practice. we
already know the solution to the pell
equation.
so let's compare the largest fraction
here with root two.
okay in decimals and that's really
approximately equal to root 2.
five decimals already. pretty impressive,
don't you think.
now if you are a long-time mathologer
viewer,
this all may ring a bell or two or
three on forever.
we already stumbled across the Pell
equation and its solutions while
proving that root two is irrational in
the marching squares video a couple of
years ago.
here are a couple of screenshots from
that video.
uh well that was a younger me. do you see
a difference?
and we weren't just lucky then. it's
actually quite natural
to stumble across solutions of our Pell
equation
while playing around with certain proofs
of the irrationality of rule 2.
another one of these irrationality
proofs is powered
by the so-called euclidean algorithm, an
ancient mathematical superweapon
with which many of you will be familiar.
and now it turns out, and i'm guessing
even most of you who are familiar with
the euclidean algorithm won't know this,
infinite fractions like ours for root 2
are really just the euclidean algorithm
in disguise.
got it? not only does the euclidean
algorithm lead to an irrationality
proof of root two it is also at the
heart of our
infinite fraction. so there in a nutshell
is one possible natural connection to
the infinite fraction.
in the rest of the video i'd like to
explain how someone playing with the
euclidean algorithm could naturally
make these connections. if you think
you've got what it takes to also survive
this more challenging, sort of
master class part of the video, now is
the time to don your mathematical crash
helmets and seat belts.
and those of you who make it to the very
end will be rewarded with a solution of
the strand puzzle
that i think even beats ramanujan's.
i'll now take you on a wonderful journey
of discovery. we'll start with some
school math(s)
the euclidean algorithm, head on to the
irrationality of root 2,
and wind up with ramanujan's super-duper
solution of the strand problem.
ok you're all familiar with the
euclidean algorithm?
no maybe not. it's not taught as much now
as it used to be. so
just to be safe here is a refresher.
take any two positive integers, let's say
38 and 16.
divide the larger number by the
smaller with remainder. so in this
example
16 goes into 38 two times and we're left
with a remainder of
six. 38 equals two times sixteen plus six.
now repeat the process with the new
numbers 16 and six.
the 16 equals two times six that's
twelve plus
4. do the same with 6 and 4.
okay once more with  4 and 2.
okay again. well, actually,
2 divided by 0 doesn't make any sense
unless you want to enter the land of
infinities.
well i absolutely love infinities but
there is no need for them here and so
this is where things stop.
all easy right? but what's the point? well
there are lots and lots of points.
the point you will most likely have come
across is that the euclidean algorithm
captures the highest common factor of
our two starting numbers.
there that final two is the highest
common factor of 38 and 16. and sure
in this case the highest common factor
is just staring at us.
but what about two more intimidating
numbers such as
these guys? not so easy to spot the
highest common factor now right?  (Easter egg hidden on this slide!)
but the euclidian algorithm can easily click
through those two numbers. there five
easy steps
and the highest common factor 153 pops
out
at the end. it's also super easy to prove
that the euclidean algorithm always
captures the highest common factor in
this way.
anyway back to our earlier example and
back to work.
let's play around a bit with those four
equations over there.
divide the equations by the blue numbers
there
and watch the magic unfold.
very cool don't you think? a really
beautiful fractional
capturing of what the euclidean
algorithm is doing.
there's also a really nice visual
representation of the euclidean
algorithm that is lurking just around
the corner.
to see this let's start with a 38 times
16 rectangle.
now watch this.
this shows that you can perform the
euclidean algorithm by repeatedly
chopping off as many squares as possible
alternating between chopping
horizontally and vertically.
and the highlighted numbers are just the
numbers of the different squares you
chop off.
two big blue squares, two aqua ones, and
so on.
and the side length of the final,
smallest two by two square
is the highest common factor of the
starting numbers 38 and 16.
brilliant isn't it? so next time you have
to find the highest common factor of two
integers you know what to do right?
mathematical masterchef: just cook up the
corresponding rectangle and
chop it into squares :)
okay so there's the result of the
euclidean algorithm on top and the
continued fraction that we distilled
from it. now
what if instead of 38 and 16 we had
started with
half those numbers. 38 divided by 2 is 19
and 16 over 2 is 8. then,
except for the change in grid size, the
top picture will look the same
and since the top picture stays the same
so will the continued fraction.
so the continued fraction for 19 over 8
is the same as the one for 38 over 16.
and of course the same would have been
true if we had multiplied both the top
and bottom of
19 over 8 by 2 or by any other number
right?
multiplying the top and the bottom by
the same number
just corresponds to scaling the
rectangle and so nothing changes in
terms of the continued fraction.
in other words the continued fraction is
really
the continued fraction of the number 19
over 8
and not just that of a particular
fractional representation of that number.
this means that if we remove the grid on
top
what's left is a complete visual
representation
of the continued fraction. really
incredibly
beautiful :) from all this
it's also pretty obvious that if, instead
of 38 and 16,
we start with any two numbers whose
ratio is a rational number
then the euclidean algorithm will
terminate after a finite number of steps
and the corresponding continued fraction
will also be finite.
but wait a minute. what
if for some starting numbers a and b
the algorithm does not terminate and we
end up with an
infinite continued fraction? well
this would mean that a over b, the ratio
of our starting numbers,
cannot be rational, that this ratio must
be an irrational number.
so let's apply this trick to the obvious
guinea pig, root two.
we need to think of root two as a ratio
of two numbers, so let's start with a
rectangle with aspect ratio root two to
one
and get the euclidean algorithm going.
fun fact and it's no coincidence,
A4 and other A size paper has these
proportions.
okay let's start cutting off squares,
calculate the corresponding continuous
fraction and see where we end up.
well that's the only square of this size
that fits. this means the continued
fraction for root two starts out with a
one.
next square. here we go, another one.
okay two of those squares and so our
fraction continues with a two, like this.
next. okay.
another two and another two
in fact it looks like twos will keep
popping up forever.
if that was true, we'd end up with the
infinite continued fraction for root 2
that we encountered before
and we would have proved that root 2
is irrational. but how do we know that
there will only
be twos in this infinite fraction and
infinitely many of them.
well that turns out to be pretty easy. to
begin
after cutting off a number of squares
it's really hard not to notice that
apart from the twos other things repeat
as well.
for example, after we've cut off the
first two squares
the blue one and one aqua one
have a look at the left over white
rectangle. there,
doesn't it look very much like the
starting rectangle, just rotated and
scaled
down. and in fact that's true
and it's not too hard to prove. i'll
leave the justification for this as a
little challenge for you.
write up your proofs in the comments. now
since the same
shape rectangles appear over and over
the 2s they spit out keep appearing
as well,
on and on forever. right? cut off the next
two squares
and you're left with another white root
two rectangle, and so on
ad infinitum.
okay that's a beautiful proof but still
this fraction monster equaling root 2
still leaves me scratching my head a bit.
maybe you too?
for example just looking at this thing
how would you go about calculating its
value? not clear at all right?
to get a better idea of what's going on
let's have a closer look at some of
these
partial fractions. as we've already seen
these partial fractions are really
harmless and easy to evaluate?
okay so let's chop things off right
behind the yellow two.
to find the corresponding diagram up on
top, we first throw away all the squares
smaller than the yellow ones.
there. infinitely many little squares
gone. now
we just rescale all the squares in sight a
tiny little bit
so that the little indent at the bottom
right goes away like this.
there goes away. then the value of the
partial fraction
is just the aspect ratio of this new
rectangle.
and, as well, we can just read this aspect
ratio
off the diagram. if the yellow square is
one unit wide then the green is one plus
one
equals two units wide. how wide are the
aqua squares?
well two plus two plus one that's five.
and finally the blue square. one plus one
plus five plus five is twelve. this means
the rectangle is
12 units high and 12 plus 5
equals 17 units wide. and so our partial
fraction is equal to
17 over 12.. there.
of course you can also just do the
algebra and get the same answer.
but isn't this absolutely marvellous?
thinking about continued fraction in
this way really helps me and hopefully
helps you too.
it provides real insight into the true
nature of these mysterious mathematical
creatures.
here's another such insight. when we
first form a partial fraction by
truncating at a plus sign
it's not at all clear what the effect of
that truncation will be
on the value of the fraction. however
when we recognize that this truncation
just amounts to smoothing out the little
indent resulting from discarding some
tiny squares,
it's clear that the resulting rectangle
will be very similar in shape to the
original rectangle.
well there's a smoothing action again
for our example. there,
smooth out. and just one more time. okay
there
smooth out. so nice. so it's really clear
at a glance that 17 over 12 is a good
approximation of root two.
it's also intuitively clear that the
further down the infinite fraction we go
before chopping off,
the closer we'll get to root 2 the value
of the complete
infinite fraction. well life's good and
we can stop
scratching our heads. saying that the
infinite fraction is equal to root 2
really makes sense. now
depending on what you include, this is
either the third or fourth mathologer
video
dealing with aspects of continued
fractions.
and if you are a regular mathologerer,
you may have heard me mention that those
partial fractions
are in some sense the best possible
rational approximations to the full
fractions.
but that seems a very strange statement
doesn't it? how can
any one fraction such as 17/12
possibly be a best approximation of root
2
since all the partial fractions further
down the track are much better?
well let me explain. the
obvious measure of how good a fraction
approximates root 2 is simply the
difference of the two numbers. right, the
smaller the difference
the better the approximation? it's a
no-brainer isn't it? and then because
root 2 is irrational
there is no best approximation in this
sense. but we can approximate
as well as we wish with a difference as
close to 0 as we wish.
but really good approximations of this
type come with a price:
both x and y will be huge numbers.
so there's this other simple measure, a
sort of
value for money estimate for determining
how good an approximation y over x
is to root 2. that quantity down there
the left side of our pell equation.
that at least makes some sense right? if
y over x was exactly equal to root 2,
this would make this expression equal to
zero.
that's not possible and so since x and y
are integers,
the best way we can do by this measure
is to score one.
and i've already told you that scoring
one is actually possible and is achieved
by our partial fractions
in lowest terms. remember half of our
partial fractions have
y squared minus two x squared evaluate
to one.
and the fraction in the other half
return the value
-1.
in fact there are no other
superfractions of this type and by the
end of this video
the keen among you shouldn't have much
trouble actually proving this.
anyway if we were to dig down and tidy
up more,
more than we have time for today, we
would discover that our partial
fractions being best in the second weird
way
translates into the partial fractions
also being best
in a simpler and more obvious sense.
among all the fractions with denominators
 less than a certain number,
it turns out our partial fractions are
closest to root 2. for example
17 over 12 is a better approximation to
root 2
than any other fraction with the
denominator
x in the range from 1 to 12.
maybe the clearest way to describe this
measure of 17/12's bestness
is to say that 17/12
is to root 2 what a super famous
fraction
22 over 7 is to pi. as i said,
there's a lot more to all this than we
have time for today, but i hope this
provides you with some intuition.
anyway our partial fractions are really
super nice and super important
and so let me finish up by proving to
you some of the nice things that I've
claimed to be true so far and
then
use all this to make a formula for the
strand puzzle that
best ramanujan's. okay are you
ready to outramanujan ramanujan.
all right, let's first pin down an easy
way to generate all these partial
fractions.
to get the earlier partial fractions
from the one of 17/12
just remove the two smallest squares,
smooth out,
and repeat. here we go. remove,
smooth out, remove, smooth out,
okay there remove, well not much to
smooth out :) there but anyway.
wonderful. so in this way, starting with
one of the partial fractions of the
infinite fraction
we've constructed all the earlier ones.
and we've done so by successfully
removing the two smallest squares
and smoothing out the resulting indents.
now because of the nice periodic
structure of the continued fraction of
root 2
there is a second way of constructing
those early partial fractions.
instead of removing the smaller squares
we can also remove the largest two
squares. here we go.
get rid of the largest square and the
second largest square
and what's left is the partial fraction
seven over five
placed on its side. get rid of the
largest two squares
and that's three over two. one more time
one over one. what's
extra nice about the second way is that
you really just have to get rid of the
squares
and no extra adjustment is needed. also,
running this construction in reverse,
that is adding two squares at a time,
we can construct absolutely all partial
fractions starting with
just this one square.
another thing that's very easy to glean
from this is an
easy relationship between successive
partial fractions.
there if the sides here are
L and S then the sides of the next
larger rectangle are, well,
S plus L and
S plus S plus L which is 2S plus L.
so this means that if the first partial
fraction is L over S
then the next one is 2S plus L over
S plus L. i guess that's the relationship
that a lot of you will have guessed
previously
when i challenged you to do so. now we
have a proof.
now starting with the first partial sum
1 over 1 we can use
this formula to also successfully
generate all the other partial fractions
algebraically.
there feed that one in the front
splits out the next one and do it again
spits out the next one.
and great challenge for you: use this
relationship to prove
that our measure for bestness of
approximation
alternates between one and minus one.
it's an easy one but a nice one.
okay this is a simple relationship that
will get us
all those partial fraction easily. but we
can do
even better. our relationship consists of
two consecutive fractions. what's the
next fraction?
easy just iterate this relationship, that
is plug the second fraction into itself.
wow can you see it? there
so there in the numerators you've got
something like fibonacci in action
right? to get the numerator of the third
fraction
you simply have to add 2 times the
numerator of the previous fraction
to the numerator of the fraction before
that.
and the same is true for the
denominators.
wow there two times the orange
three is six plus the green one is seven.
two times seven is fourteen plus three
is seventeen,
etc. and at the bottom
2 times 12 is 24 plus 5 is 29.
brilliant. so both the numerators and the
denominators grow in exactly the same
fibonacci-like fashion. the only
difference is that the numerators start
with
1 and 3, whereas the denominators start
with 1 and 2.
there. for the fibonacci numbers
highlighted in blue
there's this amazing general formula on
the right, Binet's formula.
plug-in one it spits out one, plug-in two
it splits out one plug in
three it spits out two, and so on, one
fibonacci number after the other.
i actually showed you how to derive this
formula in the video on the
tribonacci sequence. remember that video?
well the hardcore ones among you will
remember.
now in pretty much exactly the same way
we can also find general formulas for
our denominator and numerator sequences.
here they are. pretty :)
and these formulas now straightaway
translate into general formulas
for the infinitely many solutions of the
strand puzzle.
what a wonderful solution isn't it. beats
ramanujan's infinite fraction.
what do you think? anyway lots more
amazing stuff i could mention
and i'm tempted to go on for a couple
more hours but
maybe that's a good time to stop. until
next time :)
