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
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
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 :)