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 :)
mathologer once again proving he's the best math youtuber
Mathologer really cannot make a bad video. What a channel.