Binomial to Multinomial to Multiseries
We want to move from the familiar Binomial theorem to the more general Multinomial theorem, and then on to less familiar territory involving what we call the Multiseries theorem. But we will have to be rather careful in making this transition. A first step is to think more generally about binomial coefficients to facilitate the step to multinomial coefficients. We will work our way up slowly ---leaving you with a nice challenge to determine the important pattern here!
And along the way, we will have to say some critical things about the modern treatment of "infinite sums", which is of course seriously misguided.
Video Contents (thanks to Lucas Lofaro)
0:00 Intro
0:13 Overview
1:03 A More General View of the Binomial Theorem
2:45 The Trinomial Theorem
5:55 Multinomial Coefficients
9:14 Multinomial Theorem
13:28 Example of Multinomial Expansion
17:38 Towards the Multiseries Theorem
22:18 Problem with Infinite Sums
25:37 Motivation from Newton's Polyseries Theorem
29:36 Multiseries Theorem
33:40 A Preview of the General Pattern
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0:13
[Music] Hello everyone and welcome to the algebraic calculus 2 course i'm Norman Wildberger
0:19
we'd like to move towards understanding infinite series or better yet ongoing series
0:26
which we are tending to call poly series and the exact nature of these things is a big big
0:34
question mark that we are trying to gently move towards today we're going to take some steps in
0:39
that direction by starting out with a core theorem in algebra the binomial theorem and reminding you
0:47
how to generalize that to the multinomial case the multinomial theorem and then we're going to
0:53
move beyond that towards a much bigger picture which is in the direction of power series or
1:01
poly series okay so let me remind you that the usual binomial theorem can be expressed this way
1:08
i'll use a1 plus a2 instead of a and b or x and y because i'm interested in in increasing the number
1:16
of constituents here in the bracket so if we have a1 plus a2 to the n and n is here a natural number
1:23
then that's the sum from k equals 0 to n n choose k a 1 to the k a 2 to the n minus k so in this
1:29
case increasing powers of a1 decreasing powers of a2 and these familiar binomial coefficients
1:35
now there's an interesting alternate way to write this which is actually a little bit more conducive to generalization which is to introduce a
1:46
kind of a different kind of binomial coefficient where we have n choose k1 k2
1:53
and we can define n choose k1 k2 as being n factorial over k1 factorial k2 factorial
2:00
and then we can re-express this by having a sum over two quantities k1 and k2 which are
2:08
natural numbers or starting from zero but that have the property that k1 plus k2 equals n
2:15
then we can write it in this form where this is this new kind of binomial coefficient times a1 to
2:21
the k1 a2 to the k2 so what we've really done is we've replaced the k that was sitting here with k1
2:27
and replace the n minus k with k2 so now k1 plus k2 symmetrically add up to n okay that makes it
2:36
a little bit more symmetrical and uh it's also a step towards expressing the multinomial theorem
2:45
so first step in that direction is the trinomial theorem where we have a1 plus a2 plus a3 to
2:50
some power if we square that quantity then it's pretty familiar most of you will know that you
2:56
get the sum of the squares of the individual constituents a1 squared plus a2 squared plus a3 squared plus twice the possible products a1 times a2 plus a2 times a3 plus a1 times a3
3:10
a little bit less familiar is what happens when you cube a1 plus a2 plus a3 well then you get
3:16
the sum of the individual cubes you also get a um let's go to this term here a term here that's
3:25
just sort of like this where we're taking a whole bunch of um products involving two of these three
3:32
quantities and a third one so a1 squared a2 so we have two a1 factors and one a2 factor
3:40
and there's six possible such terms here they all are okay and this combination is multiplied by
3:47
three so that's sort of analogous to this and then there's an additional term which is six a one a
3:54
two a three six times the product of all three of them and what is the the general situation for the
4:03
trinomial theorem if you have a 1 plus a 2 plus a 3 to the n then you can write it as a sum of terms
4:11
each is a power of a1 times the power of a2 times the power of a3 say a1 to the k1 a2 to the k2
4:17
a3 to the k3 the total degree had better be n so k1 plus k2 plus k3 equals n
4:25
and otherwise k1 and k2 and k3 are allowed to be arbitrary natural numbers so starting with zero
4:33
and the coefficient is a trinomial coefficient okay it's n choose k1 k2 k3
4:41
which is well it's n factorial divided by k1 factorial k2 factorial k3 factorial so for example
4:49
say we're interested in this term here that would be the term when k1 equals k2 equals k3 equals 1.
4:56
then this trinomial coefficient is going to be 3 choose 1 comma 1 comma 1. what's that
5:02
well that's 3 factorial in the numerator with a product 1 factorial 1 factorial 1 factorial
5:07
in the denominator so that indeed is 6. on the other hand if we have say 2 1 0 if k 1 is 2 and
5:18
k 2 is 1 and k 3 is a 0. so they still sum up to two to three then what's the
5:29
what's the coefficient well then it's three choose two comma one comma zero so that's
5:35
gonna be three factorial six in the numerator and in the denominator we have a 2 factorial of 1 factorial and a 0 factorial that reminds you 0 factorial is the same as 1 so there's a total of 2
5:44
in the denominator so 6 over 2 that's 3. that's where that 3 a1 squared a2 term is coming from
5:52
okay so that's the trinomial theorem okay so let's introduce these multinomial coefficients
5:58
more generally if n is k1 plus k2 all the way up to say plus k sub m then n choose k1 comma k2
6:08
comma all the way up to k sub m is n factorial in the numerator the product k1 factorial k2
6:15
factorial up to km factorial in the denominator now actually this condition that the sum of these
6:23
terms here is equal to the n on the top is not really necessary in order to define this
6:29
but it's this particular case that's going to come up in the application to the multinomial theorem
6:36
okay so in this case what does this number represent first of all it's not entirely obvious that this is a natural number but it is and we can see that by interpreting it as
6:48
a count of something so this is a count of the number of ways of choosing
6:53
a certain number of objects from n objects starting out with and putting them in in m
7:00
boxes so we have a bunch of objects n of them and we have m boxes box one box two up to box m
7:07
and we're interested in putting k one of these ends into the first box k two of them into the
7:13
second box and all together then km of them into the m box so that all of the objects here get
7:20
put into boxes so we're putting all the objects here into these boxes that's this condition here and we want to know how many ways are there of doing this if we know n and we know m and we know
7:30
all these numbers k1 k2 and up to km and i talk about this uh this issue in in my math foundation
7:38
series so earlier on in the math foundation talk about the binomial theorem the trinomial theorem the multinomial theorem so you might want to remind yourself about that but here's sort of
7:47
a quick and easy way of understanding where this formula comes from what you do is you you tackle this problem in the following way you say okay well the number of ways of doing
7:55
this is first let's concentrate on putting k1 things in the first box so there's n choose k1
8:03
ways of doing that okay so they're in there and now there's n minus k1 things left over and we're
8:09
going to choose k2 of them to put in the second box so then we have to multiply by n minus k1
8:17
choose k2 okay and then the third one okay then there'll be n minus k1 minus k2 left and we're
8:26
choosing k3 of them to put in the third box so we have to multiply by n minus k1 minus k2 choose 3.
8:34
and we carry on like that all the way to the last box by the time we get to the last box there will
8:40
be exactly a case of m things left and so the number of ways of putting them in here is just km
8:47
choose km or if you like n minus k1 minus k2 minus k3 minus all the way up to km minus 1.
8:56
okay and so then if you write down those those m binomial coefficients
9:02
and you sort of expand them out into factorials and do enough cancellation then you'll see that
9:07
you end up with this expression here so that that really is an integer because it really is counting
9:13
the number of ways of doing this okay so now we come to the multinomial theorem
9:19
which tells us how to expand a power of a general sum a1 plus a2 all the way up to plus am
9:29
as a sum of individual terms which are all products of powers of a1 through am
9:38
with some coefficients which are multinomial coefficients so the theorem says that we're getting a sum of terms of form a1 to the k1a2 to the k2 all the
9:49
way up to am to the km where k1 plus k2 plus km the sum of these things here is going to be equal
9:58
to n because the whole thing has to be of degree n and furthermore each of these k i's have to be
10:06
natural numbers that is either 0 1 2 3 etc so we've got positive natural numbers that sum
10:15
total to n and here's that multinomial coefficient that we've just been talking about
10:22
okay so why is this true well let me try to explain it to you here so when you're taking a
10:28
product like this you're really you know expanding out this product of n terms each of which is a1
10:36
up to am the sum okay so we have the sum sum sum with n of these sums multiplied together
10:44
when we expand all of this out okay what's a typical term going to look like well a
10:50
typical term involves taking one of the terms from this expression maybe a 3. we take an a3
10:57
from that expression and we multiply it by one of the terms here maybe a5 and then we take one
11:02
from the third one maybe another a5 and so on so we go through each of these n terms and pick
11:08
out a single a sub i so suppose we end up with a situation where we've all together picked out
11:16
uh two a1s maybe an a1 from this box and an a1 from from this box and all together we've
11:22
picked out three a2s and maybe one a3 and maybe five a sub m's and some other stuff
11:30
okay so that's one of many possible terms that that will appear so we'd like to know what's
11:36
the coefficient of this term going to be in this big sum once we expand it all out how many ways
11:42
are there of getting exactly this term when we do all this well see what that really amounts to it's
11:50
it's like you you have these m boxes representing the a1s the a2s the a3s all the way up to the ams
11:58
and we're deciding in this individual case to put two uh two of them in here and three of them
12:05
in here and one of them in here and something else in these guys and five of them in there
12:11
and what actually are we putting in here well we're putting in the the the constituent uh
12:20
terms in this product so there's n of these brackets okay n of these brackets
12:26
and let's say this a1 squared it's coming from two out of those n brackets maybe from this one
12:33
and and from from this one okay so that's the case we have n things and we're choosing two of them to
12:40
be in that number one box and then we're choosing three of them to be in the two box corresponding
12:46
to the places where the a2 is uh is is being taken from the factors that a2 is being taken from etc
12:54
so if you think about it a little bit okay this is exactly the situation that i described
12:59
previous slide which is counting the number of ways of making this distribution and that's
13:05
exactly captured by this multinomial coefficient n choose k1 k2 up to km so that's a kind of a
13:14
an argument of why the multinomial theorem is true it's just a question of of looking carefully at
13:20
what it means to expand a product of n terms like this where each of them has m constituents now in
13:28
our mathematical careers we end up having lots of practice with binomial expansions
13:35
not so much with more advanced or more complicated multinomial expansions
13:41
and i think it's an important skill to to become familiar with because after all this is the algebraic calculus and algebra and algebraic manipulations are in some sense the heart of
13:51
what's going on and a lot of things ultimately reduce to these algebraic manipulations so having
13:57
a good sense of things is really important and this multinomial theorem is a key kind of central
14:03
a theoretical result it's really good to have some some understanding of it so in the course um which i hope you'll all take at some point um we'll have lots of opportunities to investigate
14:13
examples in this direction some of them quite interesting okay so but let's do a specific case here let's look at a1 plus a2 plus a3 plus a4 to the fifth we want to expand this thing out
14:27
okay well there's going gonna be a lot of terms right this can be four to the fifth terms but
14:32
you know they're um some of them are gonna be the same they're gonna get grouped together
14:37
so i'm expressing what we get here in this fashion here it's a little bit incomplete but
14:43
let me explain so this first little bracket here refers to a1 to the fifth and so on so that means
14:50
you have to kind of symmetrically extend that so that all the variables are playing the same role
14:55
so you should think about this bracket as representing a1 to the fifth plus a2 to the fifth plus a3 to the fifth plus a4 to the fifth okay then it's a nice symmetric expression
15:07
the next term has um a1 to the fourth a2 so it's a product of two of these things only two are being
15:15
involved and one of them is being involved to the fourth power and the other one to the first power
15:20
so there's a lot of those there's a1 to the fourth times a2 a1 4 times a 3 a 1 4 times a 4 a 2
15:27
to the fourth times a 1 a 2 to the fourth times a 3 et cetera okay so that's what's all contained
15:33
in the sum here and we're saying that each one of those occurs with a multiplicity 5.
15:39
and then there are terms of form a1 cubed a2 squared take any two of these terms and one of
15:46
them gets raised to the third power the other one gets raised to the second power there's 10 of them and here we have a product of three terms an a1 cubed times an a2 times an a3 so one of them
16:01
gets raised to the third power and the other two are just single powers okay there's 20 of
16:06
each of those here we have a1 squared a2 squared a3 and all symmetric combinations of that there's
16:13
30 of them and here we have a1 squared a2 a3 a4 we have 60 of them okay so let's um
16:24
verify some of these some of these coefficients with the multinomial theorem so let's say this
16:31
thing here okay so what's involved here in the coefficient of this term this is the case when
16:37
k1 is 4 and k2 is 1. okay so we're talking about 5 factorial over 4 factorial 1 factorial which is 5.
16:46
this one here is 5 factorial over three factorial times two factorial or five choose three comma
16:53
two that's ten this one here five choose three three comma one comma one that's five factorial
17:00
over three factorial one factorial one factorial that's 20. this one is 30 that's 5 factorial over
17:06
2 factorial 2 factorial 1 factorial and this last one what would that be that would be 5 factorial
17:13
over 2 factorial times 1 factorial 1 factorial 1 factorial so 120 over 2 which is indeed also 60.
17:23
so that's the multinomial theorem and it's it's useful to to stare at something like this and
17:29
sort of convince yourself do you understand how how it's happening it's complicated but it's
17:34
not that complicated there's a nice sort of structure to it so we've expanded our view from the binomial theorem to the multinomial theorem so a natural question is can we go further
17:47
what would that mean well can we consider a power of a sum like this where there's not
17:54
two or three or four or five you know not a finite number of constituents in the sum but
18:01
an infinite number like a sum that keeps on going so a1 plus a2 plus a3 plus and so on
18:10
well let's think about that i mean if such a thing did make sense what are we likely to get
18:17
well we know that if we have a finite sum what we get is we get the sum of the squares of the individual constituents plus twice the sum of all possible products so you know maybe
18:28
it makes sense to say yeah this is still going to give us something similar we're going to get an infinite sum involving a1 squared plus a2 squared the sum of the squares of all of these
18:38
things and then there's another term with a 2 in front that consists of all possible products and
18:43
they'll of course be you know sort of a double infinity in some sense of all possible products
18:49
so this is obviously in quotes and we would like to understand you can we make sense of this does
18:54
this does this work in some sense i mean sort of intuitively we feel that something like this should exist but what what actually are we talking about so does it make sense well first of all
19:06
if we are thinking of a1 a2 and a3 as rational numbers okay which is often what we're doing then
19:13
the answer is no it doesn't really make sense and the reason is because even this initial sum is not
19:21
defined okay so a1 plus a2 plus a3 plus and so on is not um a not a defined object in our arithmetic
19:31
you know it would be like me introducing some weird new symbol and asking does this make sense and you say wait a minute that's a new symbol that you haven't defined yet so so
19:39
the question is meaningless so that's the way i want you to think about it i want you to think of um a sum like this involving a an infinite number of rational numbers that's just something
19:49
that is not defined it's not a question of whether it exists or not it's a question of that that addition is a binary operation you know at every stage you're allowed to add
20:01
one rational number to another rational number okay that's what you're allowed to do so
20:08
um i know that modern calculus is keen to try to circumvent this okay and so we're gonna have a lot
20:16
to say about that but let's let's just from our point of view our position is going to be that this is not defined however there's an alternate way of thinking about what this might mean
20:27
and that's in terms of variables okay so we could look at this and say okay wait a minute what if
20:33
a1 and a2 and a3 are not numbers but are rather variables well for us variables are kind of a
20:42
dubious metaphysical term and we prefer to think about poly numbers and i remind you that we have
20:50
two poly numbers that we depend on quite a lot alpha and beta and we know exactly what they are
20:55
right alpha is like a vertical poly number and beta is a horizontal poly number and we have a
21:02
nice algebra of bipolar numbers built up from alpha and beta there's only two of them so we
21:08
could maybe re relabel them and call alpha alpha one and and beta is now alpha two
21:15
okay but then we would have to in order to make something like this work we'd have to introduce a hierarchy of additional variables or poly numbers what would that involve well that would involve
21:26
taking our framework of bipolar numbers and sort of making a higher dimensional extension of it
21:33
now that's actually a pretty interesting idea which is one that we are going to ultimately talk
21:38
a lot about that's actually quite an important development but it it does take some doing so one
21:45
has to first establish uh this theory of um not just poly numbers and bi-poly numbers but sort of
21:52
multi-poly numbers in an ongoing uh setting and so that's a major expansion of our algebra
21:59
in order to sort of support this kind of statement but it is possible uh given that
22:05
to eventually look at something like this and try to interpret it in that kind of framework where
22:12
both sides are representing some some arithmetic with these extended poly numbers now this
22:20
algebraic calculus course is quite novel uh it's a major departure from the usual story and i'm
22:26
not trying to spend too much time criticizing the usual story here i do that in my math foundations
22:32
series and elsewhere to to a great extent but it is important for me to just say a few words about
22:38
the inadequacies of the current framework towards infinite sums okay so modern mathematics would
22:45
like there to be the possibility of having infinite sums but this is really a problematic
22:50
thing of trying to define an infinite sum a1 plus a2 plus a3 et cetera of rational numbers and one
22:56
of the key problems with this is that if you try to do this in a kind of a consistent way you'll find that the usual commutative and associative rules break down and uh and here's a great example
23:06
which goes back to the the 17th century so you know it's already known well known to Leibniz and the Bernoullis et cetera so suppose you have a relatively simple sum where each of these
23:16
constituents is just one and minus one so one plus minus one plus one plus minus y etc okay so if
23:22
you're going to treat this infinite sum something like you treat ordinary numbers right then
23:29
associativity is it should be freely available you should be able to group these things in
23:34
different ways so in particular you could group those first two together one plus minus one and then the next two together one plus minus one and if you do that then you see that each one
23:42
of these pairs is giving you zero so you end up getting zero plus zero plus zero and and so on so you have to get zero for the value of this sum however if you take the same sum and you sort of
23:54
use associativity and regroup them in a slightly different way so you you peel that first one off
23:59
and then you start grouping after that so minus one plus one plus minus 1 plus 1. well then you're going to get 1 plus 0 plus 0 plus 0 et cetera et cetera and so you're obviously going to get 1.
24:10
okay so how does this work so this is an issue a very fundamental issue with you know trying to add
24:18
infinite numbers or some sets of numbers together okay you you run the problem that the usual laws
24:25
of arithmetic break down so modern modern calculus tries to get around this and it has a whole
24:32
elaborate framework for trying to do this in terms of restricting oneself to so-called convergent
24:38
sums okay and if you do that the right way then you can kind of circumvent this kind of thing but the price that you pay for that is that you have to um impose a a theory of real numbers which
24:49
are sort of the the pots of goals at the end of these rainbows uh of um of these convergent uh
24:56
series that you you cook up okay so you you have to um sort of imagine this vastly more complicated
25:03
and ultimately intractable arithmetic in order to support a theory of of convergence and and
25:10
therefore definition of infinite sums so it's very quick summary okay of why we are are very um very
25:18
sensitive to this issue okay this is a central topic we want to address this squarely on and and
25:25
hold our gaze and really come to terms with the problem is and try to find a solution okay
25:32
i think nobody has has really spent the effort in doing this and we're going to attempt it
25:38
and our motivation is going to be those aspects of ongoing series that we've already had a look at
25:45
and that we are thinking has a certain amount of consistency namely newton's generalized binomial
25:52
series or we're saying newton's poly series theorem let me remind you that we considered
25:58
things of the form 1 plus alpha to the r some ongoing sum like a poly number but going ongoing
26:05
it's a poly series with coefficients h sub k of r so r is a rational number we're summing k
26:12
equals 0 and onward and we're getting a sum like this 1 plus r alpha plus r times r minus 1 over
26:18
2 alpha squared and so on so the coefficients are like a sort of generalized binomial coefficients
26:24
and we saw that with this kind of series there was a sense in which 1 plus alpha to the r
26:32
times 1 plus alpha to the s was consistently equal to 1 plus alpha to the r plus s this is
26:38
this is the key result of newton where r and s are arbitrary rational numbers
26:45
so as a special case um if r equals s here okay so we're just basically squaring 1 plus alpha
26:51
to the r so we're squaring this thing here so if we square this here then according to this we should get 1 plus alpha to the 2r okay so we should get the same kind of series where
27:02
the r is replaced with 2r so this is a kind of an example right of
27:07
of something which is in the direction of what we're we're talking about we're we're taking some
27:12
ongoing series with an ongoing number of terms and we're squaring it
27:19
it's not a multinomial expression it's not a finite number of terms and we're squaring it we have a an ongoing number of terms and we're squaring it and we're seemingly getting something
27:28
so how did we interpret this well we interpreted in the following way we're saying okay what does
27:35
this mean it means that if you pick some natural number n and you look up to degree n in alpha on
27:43
both sides then you get going to get equality so what does that mean so it means say on this side
27:49
here we go up to terms of alpha to the n and we truncate or drop everything beyond that that's
27:56
going to give us a finite polynomial over here if we go up to terms of degree n inside the bracket
28:04
okay and then square that okay then we're going to get additional terms but if we drop those as well
28:10
beyond alpha to the n so we end up with just an expression involving powers of alpha up to n
28:18
then those two poly numbers are going to be equal and that should be the case for any n
28:25
so it's sort of like the case that we're we're agreeing that we're going to have a window a truncation window we're only going to look at this window and ignore everything beyond okay
28:33
same thing here we're going to look at this thing here and evaluate it up to a window and ignore everything beyond that and if what we see in these two finite computable windows is the same and
28:44
that's independent of the size of the window okay even if the window is big that same is
28:49
going to happen then that's what we mean by this ongoing series is equal to this ongoing series
28:57
okay so it's a it's a quite a departure from thinking about things in terms of arithmetic of
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numbers so it's not that we're computing this and we're thinking this as being a number you know this is a real number and we're squaring it and over here there's another real number
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that's not the way we're thinking about it all right we're thinking in this algebraic sense of in terms of truncating going up to degree n or actually since our polynomials
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are generally going down so we're going down to degree n okay and truncating everything below that
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that's how we are going to approach the arithmetic of these infinite or ongoing series
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so here is what i propose is a good framework for thinking about the extension of the multinomial
29:43
theorem to having an ongoing number of
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constituents the multi-series theorem we're going to aim for a statement of the following kind
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we're going to propose to find a formula that for an n a natural number if we look at this thing
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here there's c 0 plus c 1 alpha plus c 2 alpha squared and so on i'm using c's instead of a so
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that we don't get the a's and the alphas mixed up okay alpha is our favorite polynomial okay
30:14
it's a familiar polynomial so this is an ongoing polynomial or a poly series with coefficients
30:21
which are c0 c1 c2 c3 etc and it keeps on going and we're proposing to raise this to the nth power
30:30
and the idea is that this is going to be another poly series say with coefficients d0 d1
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d2 d3 etc and we are going to want to establish a relationship like that where these coefficients
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over here the d's say d sub k which really depends on k and also on n because we're starting with n
30:55
is some kind of function of the variables or the numbers c 0 c 1 up to c k so we're proposing that
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we are looking for a formula where if we go up to a certain degree here maybe up to you
31:10
know degree two okay and look at that d2 that what should happen is that that d2 should be a
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function just of c0 c1 and c2 right which kind of makes sense because uh if you know we're looking
31:22
just for quadratic terms then there's no way that cubic and quartic terms involved here are going to contribute to the amount of alpha squareds on the right hand side right they're not and so we
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want to determine what kind of relationship is there going to be between these d's and the c's
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how do we get these d sub k's as functions of the c's and as i said before equality here is going
31:48
to be interpreted um via this truncations idea in alpha we're going to think about truncating both
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sides up to some arbitrary degree m that's like our window okay we're looking up to degree two
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and on this side degree two on this side comparing what we get uh it works good so tick off that
32:07
works for two then we have to demonstrate it for three for four for five et cetera and hopefully we have some kind of you know inductive proof perhaps or or some kind of formula or some kind of
32:17
way of proving it for a general m so if we can prove the equality for these finite windows
32:23
of arbitrary size and that's what we're going to mean by a statement like this
32:30
and this is going to be a great kind of thing because we're going to be able to recover the usual multinomial theorem as a kind of a special case that you get by
32:41
essentially dropping a lot of the higher coefficients so let's say we're you know up to cls
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and we say we drop c sub l and c sub l plus one that would make them all zero after that point we
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agree that all the c's are going to be zero well then that suddenly becomes then just a multinomial
32:58
theorem and then we should be able to recover the multinomial theorem by setting alpha equals one
33:04
okay when we set alpha equals one after we've truncated then we have finite sums in in both
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cases and then we hopefully will get the multinomial theorem so we have a prospect
33:15
of getting equations like this that sort of capture all of the multinomial theorems the
33:23
binomial and the trinomial and the quadronomial or whatever it is um the quinomial you know
33:30
with that we're capturing all of those essentially with with one with one blow
33:35
so this is a really interesting thing to do and and the question is well can't we do this
33:41
all right so i suggest that there is actually cause for optimism here because in some sense
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if we're just looking at one level at a time you're going down to a certain degree
33:55
then it's sort of reducing uh things to to the multinomial case and we sort of know what happens
34:00
in the multinomial case so we should be able to exploit our understanding of the multinomial case
34:05
to try to come up with something and here's what happens with n equals three at least initially
34:10
so suppose we have c zero plus c one alpha plus c two alpha squared plus and it keeps on going just keeps on going and we wanna cube this thing
34:21
well here's what you're going to get initially in other words here is sort of the top of this poly series that you're going to get the constant term is c0 cubed
34:30
okay so you have imagine having these three big uh things that you're multiplying the only way of getting a constant term is to take the constant term here times the constant term here times the
34:39
constant term here so we're getting c zero cubed that's sort of obvious and what about the alpha terms well it's not hard to see that there's not a lot of ways of getting an alpha you know you could
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take this alpha here and then that constant term and that constant term and there's sort of three different ways of of doing that so you can get 3 c 0 squared c 1 times alpha and the quadratic
35:01
term the coefficient of alpha squared is going to be this 3 c 0 squared c 2 plus 3 c 0 c 1 squared
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and the coefficient of alpha cubed is going to be 3 c 0 squared c 3 plus 6 c0 c1 c2 plus c1 cubed
35:19
now we can sort of feel something of the trinomial theorem there but maybe something in addition
35:27
also and then for alpha to the fourth well we get these four terms alpha the fifth these five terms
35:37
and perhaps there's enough information here so it's a good little exercise for you into
35:42
a pattern solving and um you know just using your imagination can you find what the pattern is here
35:51
okay so in particular for example could you write down what you think the coefficient of alpha to
35:56
the sixth is going to be and this is optimistic because it looks like there is there's a pattern
36:01
here to be found and then we can hopefully jack that up instead of having a three there we could
36:06
maybe have a general n and maybe write down some multi-series theorem that goes far beyond the
36:14
binomial theorem so there's much to learn i'm Norman Wildberger thanks for listening. [Music]