Name: Partitions - Numberphile
Uploaded: 2020-03-27T18:04:37.000Z
Duration: 11 min 45 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

The superlative

He's known as one of the mathematicians with perhaps the most natural talent for maths

Past simple

It's a man called Ramanujan. He was an Indian mathematician. He started with no formal training

To-infinitive

in mathematics. But just by working alone, isolated, he was able to come up with all these mathematical facts,

these theorems, all by himself. Some were rediscoveries of things that were already known

Some were new! And he sent these to Cambridge mathematicians and they recognized his talent

And he was invited to come to Cambridge and show off what he could do

When we're filming this - a film has just come out all about Ramanujan

this film just concentrated on one achievement. And it was something to do with partitions.

So partitions is about the number of ways that we can break down a number.

We take a number, an integer, and how many ways can we break it down into positive parts?

And to prove that I'm a serious mathematician, I'm going to illustrate this by using a DUPLO here.

Let's say we take the number four, and I've got four units of DUPLO there,

how many ways can I break this down into positive parts? So I could start with this. It's a tower of four.

Or, I could break it down, right? I could have a three and a one. Right? Or, I could have a two, a one, and a one.

Or I could break it down again, I could have one, one, one, one, and there is one I've missed there if you noticed

I have missed out breaking it down into two and two. So in fact, there are five ways to break this down.

So we draw this out. We could either draw it like this: one, two, three, four. We can do a three and a one.

Like that. Two, two. We could have two, one, one. One, one, one, one, like that.

So this is what mathematicians actually do, they do it this this way because by seeing it as a picture,

you can actually make conclusions about partitions which are easier to see.

The French do this slightly differently. The French draw them like this: you can have a row of four like that.

And you can have a row of three with a row of one like that, so it's the other way around.

Or you could it like this, above. And then the last one here: one, one, one, one. So it's a slightly different way of doing it.

My favorite foot note from a textbook - do you have a favorite footnote from a textbook? - oh, I know I do.

Favorite footnote from a textbook was by Ian MacDonald who said

"if you want to do the French notation, you should read this textbook while looking upside down and in a mirror"

So I get to say this for myself for the first time!

If you want to use the French notation, watch this video upside down and in a mirror.

So let's look at the sequence of partitions. So the sequence is this:

one, two, three, five, seven, eleven, fifteen, twenty-two, thirty, forty-two, and so on like this.

Ok, so when we were looking at partitions before, that's this one here, there were five partitions of four.

So if that's breaking down the number four, there are five ways of doing it.

If you wanted to break down the number five, there'd be seven ways of doing it, and so on.

So, these numbers get larger and by the time we get to something like one hundred,

how many ways can we break down the number one hundred. If we did that, it's going to be a large number.

It's going to be one hundred and ninety million, five hundred and sixty-nine thousand, two hundred and ninty two.

So these get really large. Now, it becomes very difficult to calculate these partitions by hand.

Wouldn't it be nice to have a formula that helps us work out these partitions?

So, then if I have a formula, you can give me a number, and I can tell you how many ways to break that down.

Oh, fantastic! And this is the problem that Ramanujan decided to tackle.

So there were some formulas before Ramanujan, which go back to Euler. Not particularly pretty formulas,

but I can show you what they look like. Should I do that?

Ok, so, I'm gonna draw out a polynomial using partitions as coefficients.

It means it's gonna look like this: one plus, I'm gong to have, plus x plus,

now I'm going to have x squared next and I'm going to use how many ways to break down two.

There, which is two ways of doing that. So, if I have x cubed next,

I want how many way to break down three and there are three ways of doing that.

And I would keep going like this. If I had x four, we know how many ways to break that down, right?

There are five ways of doing it. And then it's seven for x five and so on and so on.

So this is a very mathsy looking thing. In other words, if you like this notation, it means "sum".

If I want to sum - these are called partitions of n and then it's times x to the power of n. And you do that all the way up to infinity.

Now, what Euler saw was this was equal to, this is equal to - I'm going to use another symbol here: this symbol,

which is a capital pi, it means product, it means you are multiplying these things together.

It's not three point one four, yeah, we're limited in how many letters we've got to use so we're using pi here for product.

The product of one over one minus x to the k. Yeah, we've got this. It's from k equals 1 to infinity.

So we've got something that we're multiplying together is equal to this polynomial. This is a way of working out partitions.

And it's what you used to do classically. And so if you want to work out the partitions of one hundred,

you could use a formula like this and actually do the calculation. But it's not particularly easy.

But that is what they used to do if they had to do it by hand,

and it's what your computer would do if it wants to work out partitions, it would use a formula like that

Ramanujan had an amazing intuition for numbers.

Another Cambridge mathematician called John Littlewood said that all the positive integers were Ramanujan's personal friends.

Which sounds like an utter nightmare to me, having infinitely many personal friends,

imagine the Christmas card list, sounds terrible. So let's look at some of the things he noticed about partitions.

I guess this is something that attracted him to the subject. Looking at partitions of, let's say,

a number that looks like this: five k plus four. Okay, so if you have a number like that, it ends with a four or a nine.

Four, nine, fourteen, nineteen, so these numbers. So you have a number like that, you notice something.

You notice that the number of ways of breaking it down is divisible by five. So there's a fact that he noticed.

The first example: if we took k equals zero, we're looking at - if it's zero - looking at partitions of four.

Ok, we did that, alright, how many ways were there to do it? There were five ways of doing it.

It's divisible by five. Great! What about partitians of nine? So if you have k equals one,

this would be partitions of nine, that's the next step up. How many ways of doing that?  And if you work it out,

I can tell you that there are thirty ways of doing that and, hey, again it works, it's divisible by five.

It always appears to be the case. Ramanujan was able to show this,

and he was able to show a couple of things that were similar to this as well.

Like if you had partitions of a number that looked like this: seven k plus five, then that is divisible by seven.

Or if you had partitions that look like this: eleven k plus six, then this is divisible by eleven.

So he had these little facts. And if you see that pattern, it does seem to appear like the next one should be something like this

partitions of thirteen k plus seven is divisible by thirteen. And that is not right. That is incorrect.

The pattern doesn't hold. Ramanujan found these three and he found no other formula like this, and he thought

"well that's it, those are the only formula's we've got. The next one doesn't hold,

the others don't appear to hold." Eventually fifty years later, they did solve this. There is one that looks like this,

it's not as simple as what they thought. Seventeen thousand, three-hundred and three k plus two three seven is divisible by thirteen.

Which is nothing like what Ramanujan found before. And it seems crazy but now we know that we have a formula like this for every prime.

But they're just not as nice as the first three that was found by Ramanujan.

But it's still not the big formula they wanted to find. Let's have a look at the big formula, the main punchline.

Ok. Number of partitions is one over four n square root of three multiplied by the exponential of pi.

That's a traditional pi. That's three point one four.  Times the square root of two n over three.

So that's a nice, closed formula for how many partitions there are. That twiddle sign, if you notice,

means it's not equals. It means it's an approximation, but it's an approximation that gets better with larger values.

So this is actually getting better and better as you go along the sequence. So if we did this with an example

Let's do one hundred, right? We know what the one hundred value was, and if we did this with one hundred, put one hundred in here, we would get

one hundered and ninety-nine million, two hundred and eighty thousand, eight hundred and ninety-three

Now this is in this right area for the true value. But it's about four point six percent off. Maybe it's not the best,

but again this formula is getting better with larger values.

But I can do better than that because this formula I gave you here is actually just the first term in a series.

So this can actually become more accurate, in fact it becomes equals.

So what Ramanjun was able to do was use his full formula

and you just needed the first five terms to work out the value for partitions of two hundred to the nearest integer.

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Partitions - Numberphile

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