AC and dCB 12| Difference operators, the Harriot/Pascal matrix and the velocity ladder of dCB curves
Wild Egg Maths
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We go back to the beginning of Discrete Calculus with the work of Thomas Harriot on the Difference operator on sequences, now expressed conveniently in matrix form and with the language of polynumbers. We connect with well-known properties of Pascal matrices and the Catalan triangle, and then connect with de Casteljau Bezier curves.
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hello everyone i'm norman wahlberger
00:02
thanks for joining me
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today i want to talk about how linear
00:06
algebra can help us
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significantly in
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analyzing
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aspects of the algebraic calculus in
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particular in the direction of the
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castle albezia curves
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this is a
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nice opportunity for amateur
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mathematicians to engage in a wide
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variety of really interesting questions
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so with linear algebra we're working in
00:29
a linear space and the algebra of
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matrices looms as sort of the central
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algebraic object
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and you should think about the algebra
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of matrices as being a kind of an
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arithmetic that goes beyond our usual
00:42
arithmetic with numbers and also is sort
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of a step up from the arithmetic with
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polynomials or polynomials so there's a
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sort of a hierarchy sort of a basic
00:52
number theory where you talk about
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numbers and then there's poly numbers
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where you're doing arithmetic with
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polynomials and then one step up you're
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talking about arithmetic with matrices
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so we have this hierarchy of number
01:05
systems and this matrix
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arithmetic is really powerful and
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for some profound reason seems to be the
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way
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nature expresses a lot of what's going
01:17
on in physics and chemistry and other
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things
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so
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um
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for amateurs this is a particularly
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fruitful area of investigation and i
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hope you'll see some of that today so
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this is an introduction to a vast topic
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which will actually play a major role in
01:35
the algebraic calculus 2 course although
01:38
already in the algebraic calculus one
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course there's a fair amount of this
01:41
stuff okay so what we're going to do is
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we're going to try to
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frame or reframe a lot of the things
01:48
that we've been talking about in matrix
01:50
language
01:51
now i'm going to be showing you things
01:53
in scientific workplace but creating
01:55
similar environments in other systems is
01:57
a good idea so if you're inclined with
02:00
mathematica or maple or mupad or matlab
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or whatever
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it's nice to think about how to set up
02:07
such a system so that you have sort of
02:09
tools that you can readily use okay
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so that might be an interesting thing
02:13
for you to think about
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so we're starting with matrices as
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fundamental objects and hopefully all
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have some understanding of the
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arithmetic with matrices if not go
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through my wild linalge
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series
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now
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an important
02:29
statement however is that here we're
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looking at applications in calculus
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where there's generally an unbounded
02:35
aspect so we're thinking about applying
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things to not necessarily just finite
02:40
vectors but to to vectors that keep on
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going like poly numbers which are
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potentially uh you know of our
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arbitrarily large size in fact even ones
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which are potentially sort of ongoing
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with respect to some sequence or some
02:54
some generating pattern
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so this
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forces us to touch base with a whole
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issue of how do we
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how do we finitely and concretely and
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clearly uh tackle things which have an
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ongoing nature okay so this is a really
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important part of this question which we
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cannot finesse
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but i'm going to
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put that largely aside for now and later
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on we'll talk a lot more about about
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that
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so the way you should think about things
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here is that we're talking about
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matrices which are
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ongoing so we write them down
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in square objects like this but you
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should think of them as extending to the
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right and down okay so we're starting up
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here in the left
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top corner that's the
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zero zero spot so our indexing starts
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with zero very important to remember
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that okay throughout this discussion
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indexing starts with zero so this is the
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zeroth column this is the zeroth row
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etcetera
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okay so um
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also if you really want to do this
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really well you have to
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embrace the recent theory of maxills and
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vexils which i've laid out in my math
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foundation series starting roughly
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around math foundations 166 and
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following
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so at some point i'm going to touch base
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with that and we're going to up the ante
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and move to that better
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situation but because this is
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introductory i'm going to still stick
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with the standard story with matrices
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okay so the fundamental objects that
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we're interested in are sequences which
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will be written vertically okay in the
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same fashion as poly numbers so there's
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there's a sequence and typically these
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sequences are
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continuing they're ongoing but
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i'm going to represent them all with
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just six entries okay so we're gonna
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sort of truncate the all the stories uh
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sort of at level six actually level five
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actually specifically since we're
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starting with zero okay so everything
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that i'm going to
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be showing you actually has an ongoing
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aspect and we're just looking at sort of
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the top
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left
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corner block of things
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okay so here is a typical
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sequence or poly number p
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entries s0 s1 s2
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etc
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and matrices act on sequences by
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multiplication on the left
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okay
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the simplest matrix by the way let's uh
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make some
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some definitions okay so first of all
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let's clear our definitions
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and let's define okay this is now p so
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we've defined p to be this sequence here
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and here's the identity matrix okay so
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let's define that that's called i
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and here is the shift
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matrix which we're going to call s
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okay so now let's check just to see how
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things go so so for example if we take i
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times p
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okay and evaluate that well what we see
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is we get p back again
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and if we take the shift matrix and
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evaluate it at p by multiplying then we
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get this thing here which shifts the
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sequence up
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now you see that it introduces this 0
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here which is not exactly what we really
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want because
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we recognize that this sequence actually
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keeps on going so really we should sort
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of think of replacing the zero with an s
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sub six so we have to be a little bit
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careful that we've lost something in
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this truncation especially when shifts
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are involved okay so let's flag that
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okay
06:30
so this is um only sort of a partial
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truncated representation of what this
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real the real shift matrix which just
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keeps on going does to this real
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sequence which also keeps on going
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okay
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but nevertheless that gives us
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an initial kind of orientation and now
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we can
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define the difference operator so the
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difference
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operator
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is
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minus i
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plus s
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okay let me show you what that is that's
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that
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thing there
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and i'm going to give it a name
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let's call it
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uh well delta because delta is our name
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for the difference operation
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oh
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let's give me that back okay
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there it is there
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so
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delta equals this and let's make a
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definition
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let's just check that that works so if i
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take
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delta and i apply it to the sequence by
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left multiplication
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then i'm going to get
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this and you can see here this really is
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like
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a difference operator we're taking the
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difference between the first element and
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the zeroth element and then between the
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second element and the first element and
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so on
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with again this proviso that because
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we've truncated we don't actually have
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something that's really correct down
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here
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okay so what happens if we do
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the
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difference operator a number of times
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then we can
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easily
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calculate powers
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just by
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matrix
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multiplication
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so for example
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if we want say delta squared we just do
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that and evaluate
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and then we see um
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this thing here this is the effect of
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taking the difference of the difference
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so for example if we wanted to see what
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that actually does on a given sequence
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we would put del squared and then
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there's our generic sequence p and we
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would evaluate and we would see that
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we're getting
08:43
this kind of pattern here
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and we see you know the pattern is quite
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clear here but now we're running into
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the situation where the last two entries
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are not really reflecting the pattern
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and that's a result of our truncation so
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we always have to be conscious of the
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fact that we're only looking at a finite
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portion of this ongoing sequence and so
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that introduces some errors into things
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and these errors will increase as we we
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keep on taking powers so if we look at
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the next one
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del
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the cubed times p
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well then we're getting this thing here
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okay
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and
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one more
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delta to the fourth
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p
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is this one here
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so you can see we're getting these
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binomial
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um
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expressions um s1 minus s0 and then s0
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minus 2s1 plus s 2 and then this is
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really like s
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minus s 0 plus 3 s 1 minus 3 s 2 plus s
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3 so the binomial coefficients involving
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3 and the binomial coefficients
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involving 4 etc ok
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so
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we can put these together we can
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put
09:56
these together
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to
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form
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the
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difference
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transform
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okay and what does that do which sends
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our
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initial sequence p 2
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to what well it's going to be a a new
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sequence
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which
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whose first
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element is
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just s0 whose next element is
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s1 minus s0
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got like the the top of the first
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difference
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and then the next one is the top of the
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second difference
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and then the next one is the top of the
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next difference
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um and maybe i will just
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organize this a little bit cleaner put
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that there
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plus so minus s0 plus
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3s1 minus 3s2 plus s3 and to be
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consistent i should probably rewrite
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this one also
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how's that
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and then the next one
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is that one there
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and uh will be the next one as well
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would be one more let me just calculate
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that one
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fifth power
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on p is going to be this thing here so
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we have to grab this one here that's the
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fifth one
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and i will also organize that by
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re-positioning this one
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there
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minus s05 s1 minus 10 s2 10 s3 minus 5
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s4 plus s5 that's good okay so we don't
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need this
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okay so this is this uh difference
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transform which is kind of the key point
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in the gregory newton
12:02
identity which i told you about
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last time
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so
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we can think about
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just going from this p
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to this this thing here in in one step
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okay so there's a certain matrix
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involved here and what matrix is that
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well so i'm claiming this is a certain
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matrix and what matrix is it well it's
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this matrix right here
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and if i multiply that by p let's just
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check that we're getting what we want is
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that the same
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yes that's the
12:36
same all right so it's this matrix now
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which is encoding the the difference
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transform
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okay which is um essentially sort of
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capturing what we're really doing here
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is we're taking powers of of this delta
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and looking at the top rows so delta
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itself just had a single one and then
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i think sorry delta to the zero had a
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single one that's just the identity and
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here's delta its top row is minus one
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one here's delta squared's top row was
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one minus two one which is what you see
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there
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and delta cubed i'll just show you what
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delta cubed is we already seen it maybe
13:14
i'll delta how about delta five
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delta to the five that matrix
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was that one there and its top row is
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down here
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okay so the various rows that are
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occurring here are the top rows of the
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various powers of
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our
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delta difference
13:34
matrix
13:36
okay so um
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this is
13:40
the the difference transform
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okay so this matrix here is sort of
13:44
capturing the difference
13:47
and
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so we should give it a name
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so let's call it something
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let's
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call
13:59
let's give this thing a name
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how about jay
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all right so that's the matrix j and
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in this story whenever we have a matrix
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its inverse is um
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quite uh
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interesting usually so if we take
14:28
j and we look at its inverse then we get
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this matrix here
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and so and remarkably
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we get this beautiful fact that the
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inverse of this matrix here is this
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matrix here which we recognize as the
14:44
the pascal
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matrix
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sometimes called p
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but we are going to call it h
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okay because that's our harriet
14:55
matrix
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that is essentially the
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matrix whose rows are the binomial
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coefficients
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but more crucially for us it's the
15:06
matrix whose columns are the harriet
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sequences
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okay so let's make that definition
15:15
so we also define
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so also define
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this
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the harriet
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okay
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traditionally called pascal but i think
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harriet deserves a lot more press
15:38
uh harriet or pascal matrix
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so this j is
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so then this j is just
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the inverse matrix
15:53
of
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h
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so you see one of the advantages in
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framing things in terms of linear
16:00
algebra is that the crucial operation
16:04
in in linear algebra is the inversion of
16:06
matrices you know that is that is the
16:08
central computational thing in linear
16:10
algebra okay is to calculate an inverse
16:13
of a matrix it's like at the heart of
16:15
the subject
16:17
so whenever you have an interesting
16:18
matrix you should ask
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first of all is this invertible
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that of course depends on whether it's
16:24
determined it is zero or not if it's
16:25
determined it is non-zero then it is
16:27
invertible and then you ask okay if it
16:30
is invertible what is its inverse
16:33
that's a central problem in linear
16:35
algebra okay and
16:38
that's a question that will be coming up
16:41
often when we do these kinds of things
16:43
when we try to frame
16:45
the uh the manipulations that we're
16:47
making in terms of linear algebra to
16:50
help us
16:52
employ matrix algebra to make things
16:54
smoother
16:55
okay
16:56
so that's the the harriet pascal matrix
16:59
and we then see that it's
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it's some kind of uh you know outgrowth
17:04
of this difference calculus and then the
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um the statement of
17:09
gregory newton becomes the following
17:13
so then the gregory newton inversion
17:14
amounts to the following that if we
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start
17:17
with a sequence
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p
17:23
say that one there
17:24
and apply
17:25
the
17:28
the difference
17:30
transform
17:33
and i might start calling this the
17:35
harriet transform actually i think
17:37
harriet should get a lot more um
17:39
recognition so
17:40
we might call this the harriet or
17:42
difference transform what is that that's
17:44
just multiplying this
17:47
sequence in matrix language with this j
17:49
okay so that's that's what we got when
17:52
we applied
17:54
j
17:55
to p
17:56
these are the successive differences
17:59
then
18:02
then
18:03
this
18:04
then p
18:07
can be recovered
18:12
from j
18:14
p
18:15
by
18:16
well
18:17
if we take
18:19
j p
18:22
and multiply by
18:24
h
18:26
well
18:28
oh having to find h have i okay so
18:30
there's eight i better define that
18:32
let's define that all right now let's
18:34
try this one again
18:37
i'm getting p back and that's not a
18:39
surprise because h and j are inverses so
18:43
this is matrix multiplication it's
18:44
associative we multiply by j and then we
18:46
multiply by its inverse we're getting
18:48
back p again
18:50
but what what this
18:52
really is is if i write this out
18:55
okay so here's the matrix
18:59
h
19:00
so let me just
19:01
i want to just make sure you understand
19:03
gregory newton okay which is a key
19:06
formula in calculus so what i'm saying
19:09
is is the following see
19:11
here's how you should think about this
19:13
we're interested in this sequence p okay
19:16
we want to express this sequence p as a
19:19
linear combination of these sequences
19:21
these are the harriet sequences this is
19:23
what we call h0 this is h1 this is h2
19:26
this h3 is h4 this is h5 okay when you
19:30
multiply
19:32
the matrix like this by a sequence one
19:34
way of interpreting what you're doing in
19:37
the linear algebra term you're taking a
19:39
linear combination of the columns here
19:42
okay so when you do this what you're
19:43
really doing is you're taking this first
19:45
sequence this zero sequence h0 and
19:47
you're multiplying it by s0
19:49
and then you're taking this second one
19:51
you're multiplying that by s1 minus s0
19:53
and then the third one
19:55
by
19:56
uh
19:58
by this difference so the fact that this
19:59
thing is equal to
20:02
well
20:03
i'll show you what it's equal to the
20:04
fact that this is equal to the original
20:05
sequence this is exactly
20:07
a statement of gregory newton
20:10
okay
20:11
so so
20:13
then
20:14
the
20:15
gregory
20:19
newton formula
20:22
is the following
20:23
okay it's expressing
20:26
which i might write which expresses
20:29
the original
20:31
sequence in this case p as a linear
20:34
combination
20:36
of
20:38
the
20:39
columns
20:41
h sub 0 h sub 1 h sub
20:45
2.
20:48
etc
20:52
of the harriet
20:55
matrix
20:56
h
20:58
okay and the coefficients the
21:00
coefficients
21:03
are just
21:05
the
21:06
um
21:07
appropriate
21:10
uh
21:12
difference powers
21:14
difference difference powers of p
21:18
evaluated
21:20
at
21:21
zero
21:22
in other words this thing here
21:24
okay is just
21:26
i.e for example if we look at uh the
21:30
the second difference of pay
21:32
okay there it is there and the
21:34
evaluating at zero means taking the top
21:36
entry
21:37
okay
21:39
has
21:40
zero entry
21:43
this thing here and that's why that um
21:46
that's the coefficient of h2 in the
21:49
expression for p
21:54
okay so please make sure that you you
21:57
understand this that gregory newton from
21:59
this point of view and once we've framed
22:01
it in terms of linear algebra it becomes
22:03
almost like a triviality we see exactly
22:06
why it's happening it's not magical at
22:08
all it really comes down to the fact
22:10
that we have an inverse relationship
22:12
well there is some magic the magic is
22:14
contained in this formula here that the
22:16
inverse of this j
22:19
is h or that the inverse of h is j
22:22
that these two matrices this j here and
22:24
this h which are um very
22:26
similar to each other just with um
22:29
difference of signs are actually inverse
22:31
matrices
22:33
so from a linear algebra point of view
22:35
that's where the magic is that's the
22:37
it's having to do with the inverse
22:38
relationship between these two matrices
22:41
okay so you see the the intimate
22:44
connection between
22:46
the calculus and the linear algebra
22:49
okay so i want to tell you just a little
22:51
bit about further powers h this is an
22:53
interesting thing to um
22:55
for you to
22:56
to think about
22:58
let me
22:59
show you
23:01
a
23:02
a paper
23:03
or well at least just
23:05
beginnings of a paper so this is a paper
23:07
by gregory call and daniel vellman
23:10
it was in 1993 in the american
23:13
mathematical monthly so it's a paper
23:15
about
23:16
powers of pascal matrices and
23:20
they're talking about the powers of
23:21
these matrices okay so um
23:25
the area that we're
23:26
investigating here has seen a lot of
23:28
attention in in the last few decades a
23:30
lot of people have been quite interested
23:31
in the beautiful connection between
23:33
linear algebra and
23:35
sequences
23:37
that are generated in
23:40
in calculus and sort of associated
23:41
combinatorics
23:43
so we can
23:45
get a sense of what's happening here if
23:46
we just take our little
23:49
matrix h okay let's just look at powers
23:51
of h okay here's h
23:53
remind you what it is there it is there
23:55
pascal
23:58
entries along the rows so what is h
24:01
squared let's look at h squared what's
24:03
that
24:06
so you may have to stare at this for a
24:08
little while you stare at it for a
24:09
little while you realize that
24:11
it's actually quite closely connected
24:13
with h what's the pattern well
24:16
the the diagonal is the same but this
24:19
next sub diagonal here it's one two
24:20
three four five over here it's two four
24:22
six eight ten well that's just twice
24:25
the entries here and the next sub
24:27
diagonal here is one three six ten the
24:29
next one here is 4 12 24 40. well that's
24:32
just four times this one
24:34
and the next one is eight times this one
24:36
instead of having one four ten we have
24:38
eight thirty two eighty
24:40
okay well that's cool so what happens if
24:42
you look at uh h cubed
24:45
of course the beauty of of working in a
24:47
system like this is you could just
24:48
generate these matrices instantly and
24:50
then you can study them and
24:52
and observe interesting things
24:55
so here you see we were looking at h
24:57
cubed and the diagonal is still one but
24:59
now the sub-diagonal is three times the
25:02
original sub-diagonal
25:04
and the next one
25:06
nine twenty-seven fifty-four that's nine
25:08
times the next sub-diagonal
25:11
and the next one is 27 times the next
25:13
one so we're just taking increasing
25:15
powers of three and multiplying by these
25:17
cross diagonals by powers of three
25:20
how about how is that consistent with h
25:23
uh
25:25
inverse well we know what h inverse is
25:26
right that was j
25:28
so if we were going to apply the same
25:30
rule here we should multiply these
25:34
sub diagonals with powers of minus one
25:36
and indeed yes here we multiply by minus
25:39
one and here we multiply by minus one
25:41
squared which is one so it doesn't do
25:42
anything and here we multiply by
25:45
minus one cubed et cetera
25:47
okay so if we took uh say h to the
25:51
uh say minus three
25:53
okay
25:54
um well not a surprise it's the same
25:56
pattern okay you have to multiply this
25:59
sub diagonal by minus three you multiply
26:01
the next one by minus three squared
26:03
which is nine the next one by minus
26:04
three cubed which is minus 27 etc
26:08
so
26:10
so there's all kinds of interesting uh
26:12
potential things to
26:14
um to make contact with here
26:18
and sort of investigations that you can
26:20
make it's a very very rich subject to
26:22
matrix arithmetic is really something
26:24
that's
26:25
in some sense really in its infancy and
26:28
it's a
26:29
very fertile ground for amateurs because
26:32
you can explore and you can discover
26:34
things and once you've discovered
26:36
something the proofs are usually not
26:38
that onerous
26:40
some some pleasant induction or some
26:42
some interesting factorizations
26:45
can be quite nice
26:46
and easy to discover
26:51
now these pascal or harriet matrices and
26:54
related things are full of interesting
26:56
content and i just want to
26:58
point out another nice discovery uh this
27:00
is due to robert brayer and magnus
27:03
pirovino a paper that
27:07
appeared
27:09
called the linear algebra of the pascal
27:11
matrix okay and they observe the
27:14
following let me describe it down here
27:17
it's uh given by essentially this this
27:19
formula here so here is our
27:22
harriet matrix h
27:24
okay okay it's done to the five by five
27:27
kind of case or
27:28
from zero to five and here's a
27:31
factorization of it
27:33
into simpler matrices which just consist
27:36
of ones and zeros so they're all lower
27:38
triangular as is the original h
27:41
and here we have a block of entire ones
27:44
below that diagonal and then in the next
27:46
one we sort of remove the first column
27:49
of ones and then now we remove the
27:50
second column of ones and so on still
27:53
maintaining uh ones along the diagonal
27:56
and i guess we could go even one further
27:58
if we went then that would just be the
28:00
identity and it's not really necessary
28:01
so let's just check this by evaluating
28:03
this thing here and we do see that we
28:05
get the harriet matrix
28:07
so this is a
28:09
a nice thing you might
28:12
see if you can prove this and
28:14
and maybe discover some other things so
28:17
for example another interesting thing to
28:19
uh
28:20
to do is with these this five product
28:23
here um suppose that we take these
28:25
things and we've reversed the order just
28:27
for something to do okay so we'll take
28:29
that first one and put it in the back
28:31
and then take the second one and put it
28:34
out
28:34
over here
28:36
and then this one here
28:38
put it back in the second place and the
28:40
third one over here
28:42
so we've reversed the order now and now
28:45
if we evaluate that we get an
28:47
interesting matrix
28:49
um it's diagonal it's got 1 1 1 2 2 1 5
28:53
5 3 14 14 9 4 1 42 actually uh recognize
28:58
the catalan numbers those of you
29:00
watching my other series on solving
29:02
polynomial equations
29:03
we'll have seen this sequence um many
29:06
times
29:07
so we can actually kind of look to see
29:09
whether this is a
29:12
sequence or a matrix that has appeared
29:15
before so we could we could look for
29:18
the sequence on the online encyclopedia
29:20
one one one
29:22
two two sort of just a reading of the
29:24
rows two one
29:27
uh five five three
29:30
fourteen fourteen nine four one etc
29:36
and what do we find
29:37
well we find
29:40
this entry
29:42
there's the sequence right there and
29:44
this is called the catalan triangle
29:47
transpose so it's the transpose of
29:48
what's called the catalan triangle and
29:50
it's
29:52
a033184 and there's
29:55
quite a lot of information about it so
29:57
that's actually quite an interesting
29:58
combinatorial object that's sort of just
30:01
sort of a surprise it appears
30:04
and you might check is there an
30:06
understanding of this as a product of
30:08
these matrices i haven't looked to see
30:10
but you can you can look and if it's not
30:13
there well you can put in uh an
30:15
additional
30:16
entry and make this observation maybe
30:18
prove it first
30:20
all right great okay so now i want to uh
30:23
return to our main story the cast leo
30:27
messier curves and i want to
30:29
show you how these linear algebraic
30:32
ideas simple though they are can be
30:34
quite an effective tool in in helping us
30:36
understand these things so here's the
30:38
quartic curve that we've already seen a
30:40
couple of times control points p0 p1 p2
30:43
p3 p4
30:46
there is the expression in terms of
30:49
the bergstein polynomials the affine
30:52
combination of the points and here is
30:54
this sort of alternate
30:56
uh form
30:57
where we've
30:59
seen that this is to be thought of as a
31:00
point plus a bunch of vectors okay and
31:04
we can interpret these vectors that are
31:06
appearing here as sort of velocities or
31:08
or higher velocities we might call this
31:10
the velocity ladder okay because they're
31:12
like velocity acceleration jerk snap
31:15
crackle pop etc right
31:18
so these are interesting vectors and and
31:20
we have this particular combination of
31:22
these vectors where we're multiplying by
31:24
these
31:25
powers of t and binomial coefficients
31:29
so i want to show you how this
31:31
linear algebraic matrix point of view
31:33
allows us to to deal with this kind of
31:35
thing rather effectively
31:37
all right so let's um let's go down and
31:39
and
31:40
instead of having a sequence of numbers
31:43
s 0 s well let's consider a p to be the
31:46
sequence of points although we don't
31:47
need to say they're points they're just
31:49
a capital p 0
31:52
p 1 etc so we'll make that definition
31:55
okay so then
31:58
the difference transform
32:01
remember that was
32:03
j
32:04
times p
32:05
okay so we get those those values there
32:08
and you can see that these objects here
32:11
are just the objects in the velocity
32:13
ladder okay
32:14
we get of course
32:16
the velocity
32:18
ladder
32:20
put it in quotes
32:25
for the dcb curve
32:30
okay that's great now
32:33
how could we recover the curve so
32:35
suppose we wanted to get at this
32:37
expression
32:38
okay we've got our points we can very
32:41
quickly generate the velocity
32:45
ladder just by multiplying this vector
32:48
by j
32:49
now to get the actual form of the
32:52
curve
32:53
to get the
32:55
form
32:56
of the curve well what we could do is we
32:59
could use h right if we take h
33:02
capital h and multiply by j
33:06
p well we know what we're going to get
33:07
we're going to get the original thing
33:09
back but what we want to do is something
33:11
a little bit um more sophisticated
33:13
because
33:14
i remind you what h is
33:17
h is this thing here okay so if we
33:21
if let me just put this thing here
33:22
beside it and let's let's stare what
33:25
this products this would be h times p
33:30
h times um
33:31
h times j p
33:35
so we know that this is just the
33:36
original p but what we are doing here is
33:38
we're taking when you perform a product
33:41
like this you're taking a linear
33:42
combination of those columns okay with
33:45
coefficients which are
33:46
these entries here so this is p0 times
33:49
that first one plus this one times the
33:51
second etc
33:53
now that's a lot like um
33:55
what we're aiming for up here except
33:58
that
33:59
in addition we want powers of t
34:03
so we can introduce powers of t by an
34:06
axillary matrix
34:09
so let's
34:11
let
34:12
capital t be this diagonal matrices
34:15
consisting of powers of t
34:18
okay so let's define that
34:21
then if i put this in here so if i
34:24
replace this with h
34:26
t j p
34:28
put that in there
34:30
well let me put it in there
34:33
maybe i should tell you what this
34:34
product is first so let's let's just
34:36
look at then h times t what is that
34:40
h times t
34:42
see if you if you multiply h here is h
34:47
by t okay maybe i will put that even
34:49
there as a preliminary
34:51
thing so you can see that the matrix
34:53
multiplication how it works
34:56
is elementary but it's useful
34:59
okay there's h and i'm going to multiply
35:01
by t so when you multiply a matrix with
35:04
this diagonal matrix on the right okay
35:07
what effectively happens is say this t
35:10
squared this t squared is going to
35:11
multiply the second column here so all
35:15
the entries in here are going to get
35:17
multiplied by t squared see that's
35:18
what's happened here that entire column
35:20
has been multiplied by t squared that
35:22
entire column has been multiplied by t
35:24
cubed et cetera
35:25
if we multiplied by this matrix on the
35:27
left then that would multiply the
35:29
corresponding rows by the various powers
35:31
of t
35:32
okay but we want to multiply on the
35:34
right this way because now we're getting
35:35
these these combinations of
35:38
basically powers of of t
35:41
with appropriate binomial coefficients
35:44
so those are exactly the kinds of things
35:46
that are appearing in in the um the
35:48
castle alvesier
35:49
expression up here right 4t 6t squared
35:53
40 cubed so we're basically multiplying
35:56
this vector of uh
35:59
binomial coefficients times t times the
36:02
the jp
36:04
matrix so if we take this thing here and
36:07
we multiply by by this thing here that's
36:09
the same as doing this what are we going
36:11
to get
36:14
well if you just look at this entry over
36:15
here
36:16
okay
36:18
it's a little bit different from let me
36:19
try to move that around just so you can
36:21
see it more clearly i'll put that over
36:23
there
36:24
okay so what you you're getting is
36:27
the um what we're getting
36:30
when we do this product we're getting
36:31
this p0
36:33
times that one and then this p1 minus p0
36:37
times the four t and then this one here
36:39
times a 6t squared etc so we're getting
36:42
exactly
36:43
the expression for the dcb curve
36:47
let me just check that okay so let's
36:49
check that
36:50
this thing here actually is the same as
36:56
this thing here
37:02
okay put it there okay and i'm gonna ask
37:05
my computer please tell me are these
37:08
things the same yes this is true
37:12
but notice that what we're doing here is
37:13
we're kind of doing this all at once
37:15
right um
37:16
this is just the first six things of an
37:19
ongoing pattern but
37:21
any one of these entries is the
37:23
corresponding
37:25
expression for a dcb curve with
37:28
that many control points
37:30
so for example this is the
37:32
expression for the quadratic
37:35
dcb curve with control points p0 p1 and
37:37
p2 only use those first three here is
37:41
the form for the cubic dcb curve with
37:44
control points p0 p1 p2 p3
37:48
so this ongoing sequence of when we
37:50
jacked up the dimensions of all of these
37:53
matrices we would get this this sequence
37:55
and basically we've got all the decastly
37:58
how bezier curves you know sort of
38:01
contained
38:03
ultimately in these um in the entries of
38:05
this sequence
38:07
so
38:08
in some sense here we
38:10
are here
38:11
obtaining
38:13
all the dcb
38:16
uh
38:18
curves
38:19
at once
38:24
okay so this is a very powerful
38:26
way of thinking it's a it's a way of
38:28
organizing one's understanding and
38:31
in harnessing the power of linear
38:34
algebra in particular matrix
38:35
multiplication
38:37
to to help us organize things
38:40
so
38:41
these are really powerful techniques
38:44
very useful in a wide variety of
38:47
different situations okay so
38:50
mastering this
38:51
elementary linear algebra the arithmetic
38:55
of matrix
38:56
theory
38:57
that's a really great tool
38:59
not just for linear algebra but also for
39:01
calculus