Algebraic Calculus One from Wild Egg
Wild Egg Maths
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The very first and arguably most important calculation in Calculus was Archimedes' determination of the slice area of a parabola in terms of the area of a suitably inscribed triangle, involving the ratio 4/3. Remarkably, Archimedes' formula extends to the cubic case once we identify the right class of cubic curves. These are the de Casteljau Bezier cubic curves with an additional Archimedean property, characterized either by the nature of the point at infinity on the curve, or alternatively by the geometry of the quadrilateral of control points.
This is a very pleasant situation, and shows the power of the Algebraic Calculus to not only explain current theories more carefully and correctly, but also to discover novel results and open new directions.
I should have mentioned in the video that this Archimedean situation covers also the special case of a cubic function of one variable, that is a curve with equation y=a+bx+cx2+dx3.
Note: The projective coordinates we are now using identifies the affine point [x,y] with the projective point [1 : x : y]. Correspondingly the line with projective coordinates (a : b : c) has equation a+bx+cy=0. This is in keeping with our move to express polynumbers or polynomials with the lowest degree terms appearing first.
Screenshot PDFs for my videos are available at the website http://wildegg.com. These give you a concise overview of the contents of the lectures for various Playlists: great for review, study and summary.
My research papers can be found at my Research Gate page, at https://www.researchgate.net/profile/...
My blog is at http://njwildberger.com/, where I will discuss lots of foundational issues, along with other things.
Online courses will be developed at openlearning.com. The first one, already underway is Algebraic Calculus One at https://www.openlearning.com/courses/... Please join us for an exciting new approach to one of mathematics' most important subjects!
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00:01
[Music]
00:13
hello everyone I'm Norman wahlburger and
00:15
welcome to the algebraic calculus
00:17
today's video is something of a
00:19
highlight of this first course in the
00:22
algebraic calculus because we're
00:24
extending a classical result going back
00:27
to Archimedes for the parabolic area
00:29
formula to the much more interesting and
00:33
complicated case of cubic curves and
00:36
it's a little bit of a subtle situation
00:39
but there's all kinds of really
00:40
beautiful new aspects that are going to
00:43
be rising so we're really exploring new
00:46
territory opening new doors here one of
00:49
the advantages that we get by this
00:52
particular approach to calculus very
00:54
rational very algebraic very concrete so
01:00
the objects of interest are the castle
01:02
Yale Bezier cubic curves so these are
01:05
formed with control points p0 p1 p2 and
01:09
p3 and we've talked a lot about them in
01:12
our course they are fundamental building
01:14
blocks for calculus and we've seen that
01:18
such a curve say gamma can be expressed
01:21
in terms of parameterization involving
01:24
say lambda and some convex combination
01:27
of the points p0 p1 p2 and p3 with these
01:31
cubic polynomials or paly numbers which
01:35
are the Bernstein polynomials of degree
01:37
three now if we substitute in here the
01:42
actual coordinates of some particular
01:44
set of four points then we can
01:46
re-express this in our more standard
01:49
fashion as a polynomial a parameterised
01:51
curve so that we have an x component or
01:54
a first component involving alpha up to
01:58
alpha cubed terms and here are the
01:59
coefficients a0 a1 a2 a3 and
02:02
correspondingly the y-coordinate B 0
02:05
plus B 1 alpha plus B 2 alpha squared
02:07
plus B 3 alpha cubed so that's going to
02:09
be our algebraic form of the de castilla
02:13
a curve and these two things are
02:15
equivalent so having it too costly a
02:17
Bezier cubic curve or having a
02:19
polynomial a parameterized curve of this
02:21
form they're basically the same we can
02:23
go from one to the other now there's a
02:26
lot of numbers here but there are these
02:28
secondary quantities that are going to
02:30
play a very important role and so let's
02:32
define them they are the Delta IJ so for
02:36
I and J indices between 0 & 3 let's
02:40
assume that I is less than J will define
02:43
Delta IJ to be AI BJ minus a JB I so for
02:50
example a sub 2 3 would be a 2 times B 3
02:55
minus a 3 times B 2 okay so that's our
03:00
starting point and now we're going to
03:02
try to apply Archimedes area formula to
03:06
this situation and so now we're going to
03:10
identify a particular kind of such de
03:14
castilla Bezier cubic curve we're going
03:16
to call such a curve Archimedean
03:19
precisely when this quantity Delta - 3
03:22
which is a 2 B 3 minus a 3 B 2 is zero
03:28
so this is the critical definition here
03:31
identifying a subfamily of the family of
03:34
all cubic to costly a Bezier curves now
03:39
one other thing that we're going to have
03:42
to mention is the point at infinity so
03:45
if we shift to projective coordinates
03:47
and I remind you that sort of involves
03:50
augmenting the affine coordinates x and
03:52
y with a third coordinate and then
03:55
looking at things in a scale invariant
03:59
fashion then there's a unique point at
04:01
infinity on this curve and it's 0 a 3 B
04:06
3 so you can think about that is what
04:08
happens when the parameters say alpha
04:10
gets large when the parameter gets large
04:13
the alpha cubed terms dominate the other
04:15
ones and more or less the ratios of the
04:19
entries are a 3 2 B 3 and so when you
04:23
actually sort of go to the limit and the
04:25
projective sense then you
04:26
get that pair there together with zero
04:29
in the first coordinate and we'll call
04:31
this point I so it turns out that this
04:35
Archimedean condition which is just
04:38
nicely expressed in terms of these these
04:40
coefficients here has an interpretation
04:43
in terms of the nature of this point I
04:46
at infinity so the theorem is that gamma
04:50
is Archimedean precisely when the point
04:53
at infinity namely I is singular so what
04:59
does it mean to be singular for a point
05:02
to be singular on a curve well here are
05:04
some examples so this point here here's
05:08
a curve and this point here is a
05:10
singular point one way of saying that is
05:12
that the tangent is not really
05:14
well-defined there are two possible
05:16
tangents to the curve at this particular
05:18
point this is often called a node here's
05:21
another kind of situation that can
05:22
happen you have two branches sort of
05:24
coming in at the same time that's called
05:27
a cusp here there's arguably there's a
05:31
tangent but actually it's a kind of a
05:33
double tangent because it's a tangent of
05:34
this branch and of this branch so it
05:36
really ought to be counted twice so it's
05:38
not really an honest tangent and here's
05:41
another kind of an example of a singular
05:44
point you have an isolated point where
05:46
most of the curve is over here and then
05:47
there's this one point which is off by
05:50
itself again here obviously there's no
05:52
tangent to the curve this theorem is
05:56
saying that this Archimedean condition
05:58
geometrically is equivalent to the point
06:01
at infinity actually being singular now
06:06
that does involve some discussion of
06:08
projective coordinates and how we
06:10
analyze points at infinity and in the
06:13
course proper and the work problems
06:14
we'll go over that and explain why this
06:18
is and why it's true okay but it's
06:21
important to understand to at least to
06:23
appreciate that this algebraic condition
06:26
has a geometrical interpretation
06:28
involving this point at infinity which
06:30
plays quite an important role in the
06:32
story now here is a very different
06:35
characterization of that same
06:36
Archimedean property this theorem says
06:39
that
06:40
as our comedian precisely when a certain
06:44
condition is satisfied for the four
06:48
points the four control points of the de
06:52
casteljau Bezier curve
06:54
so here's p0 p1 p2 p3 these are the
06:57
control points R curve gamma something
07:00
perhaps like that in red and this curve
07:04
is Archimedean precisely when the signed
07:08
area of the cyclic spline p0 p1 p2 p3 in
07:13
other words this quadrilateral oriented
07:16
quadrilateral like that if that side
07:18
area is twice the signed area of the
07:21
oriented cyclic spline p 0 p2 p1 p3
07:28
so the spy that we get by going like
07:30
this ok so if the first one is twice the
07:36
second
07:36
that's exactly equivalent to it being
07:38
Archimedean now this is actually
07:41
relatively easily checked because the
07:44
condition on Archimedean is just a
07:48
certain algebraic combination of
07:51
coefficients being zero and if we expand
07:53
that out in the context of the four
07:56
control points that define the curve
07:58
then this formula kind of just falls out
08:03
by a little bit of rearrangement now
08:07
there's an alternate way of thinking
08:09
about this because well we can think
08:13
about these sort of intermediate or
08:16
interior triangles so if s0 is the
08:19
triangle formed by p0 on this
08:21
intersection point here in p3 so that
08:23
triangle there and s1 is this signed
08:25
area and s2 is this signed area and s3
08:27
is this signed area all oriented in the
08:29
same way then this condition turns out
08:32
to be equivalent to perhaps the simpler
08:36
condition that s 0 is equal to s1 plus 3
08:41
times s2 plus s3 interesting that
08:45
there's a 3 here that 3 will figure in
08:47
other places in this theory ok so that
08:50
condition is a local condition
08:53
purely in terms of the four control
08:55
points and the shape of the
08:57
quadrilateral that they form involving
08:59
the standard signed areas of cyclic
09:03
splines and and that's equivalent to the
09:06
Archimedean condition which we've seen
09:08
is equivalent to having a singular point
09:10
at infinity now if we put these two
09:13
conditions together and stare at them we
09:16
see that there's an interesting
09:17
consequence because let's say this
09:21
condition is satisfied for these four
09:23
control points that means that the point
09:25
at infinity on the curve is going to be
09:28
singular but that means that if we take
09:30
any other set of control points for the
09:34
same curve any other set of controllers
09:38
then that other quadrilateral will
09:41
satisfy exactly the same condition
09:43
because it's equivalent to the point at
09:45
infinity being singular so automatically
09:49
we get as a consequence that this
09:51
particular property if it holds for one
09:55
control quadrilateral that it holds for
09:59
all of them that's quite interesting
10:02
already by itself okay and so now we
10:06
come to our committees area formula for
10:09
cubics which is in my view a really
10:13
beautiful result and I think firmly
10:17
establishes the power of the algebraic
10:20
calculus and it really is hard to argue
10:22
with a theory which after one sort of
10:27
course can discover and easily prove and
10:31
come to grips with something so
10:33
ultimately fundamental and and powerful
10:36
and interesting and historically
10:38
important because what we're doing is
10:39
we're going back to really the first
10:42
computation in the history of calculus
10:45
and certainly one of the most important
10:47
calculations done by Archimedes in the
10:49
context of the parabola or quadratic
10:52
curves and extending that to the cubic
10:54
situation it's a very natural thing to
10:57
do and the way it works out is so
11:01
beautiful that it really ought to be in
11:03
every first-year calculus text
11:07
so what is this lovely hero suppose that
11:10
gamma is first of all described or you
11:13
know written down as in our first
11:15
formula and it's an Archimedean DCB
11:18
cubic curve and that we have two points
11:21
on it a which is gamma of t and b which
11:25
is gamma of U then we have two
11:28
statements so the first is a kind of
11:29
preliminary statement the first
11:32
statement is that if M is the midpoint
11:34
of a B so here's point A and point B on
11:37
our curve gamma and there's the midpoint
11:39
M then the line I am I is the point at
11:44
infinity on this curve but actually at
11:47
infinity but I'm drawing it here and
11:48
actually the curve kind of would go
11:50
through here so really we should put
11:52
some some blue curve going through there
11:54
but this is that infinity so okay anyway
11:57
this line I am there is there meets
12:01
gamma at a unique point see there is
12:06
there which in fact is just gamma of t
12:10
plus u over to the curve gamma at the
12:14
average of the parameter values defining
12:19
a and B now a cubic curve typically
12:25
meets a line in in three points that's
12:28
sort of the generic kind of thing that
12:30
can happen although other things can
12:32
happen depending on the situation but
12:34
typically we expect three points of
12:37
intersection and so you might ask well
12:39
why is it appear in here that there's
12:41
only two points of intersection this
12:43
Point C and this point I at infinity the
12:47
reason is because I is a singular point
12:49
and so in some sense the curve is going
12:52
through there perhaps like twice so you
12:54
can think about it that way all right so
12:55
the curve is really crossing through
12:57
there twice there's two points of
12:58
intersection of the curve at I and this
13:01
is then the third point of intersection
13:03
and it's unique and this is a formula
13:05
for it okay and now we come to the
13:08
Archimedes area formula in this cubic
13:11
context which is that if we look at the
13:15
signed slice area from T to U of this
13:21
of gamma so that's the signed area of
13:24
what we get when we take the curve
13:26
together with the oriented segment from
13:30
B to a so we're going around the curve
13:32
from A to B and then straight back from
13:34
B to a and it's this green area there
13:36
that we're interested in the statement
13:40
is that that's 4/3 the signed area of
13:43
the triangle a C B so when we put it
13:50
this way it's exactly the same
13:53
formulation as Archimedes parabolic area
13:57
formula they're also the Point C could
14:01
be defined in terms of the unique meet
14:03
of the the line through the midpoint and
14:05
the point at infinity on the parabola
14:07
with the curve in that case the point C
14:11
had some additional properties which are
14:13
not necessarily so applying here at
14:18
least in the in the naive way but this
14:21
particular property that I've emphasized
14:24
in the quadratic case is the one that
14:26
generalizes in the cubic situation and
14:28
remarkably the 4/3 factor is exactly the
14:31
same factor that appears in the
14:33
quadratic situation so this is very much
14:36
an extension of Archimedes parabolic
14:41
area formula to cubics alright so let me
14:46
give you the sketch of the proof so you
14:48
can see the logical structure and the
14:50
details involve some computation
14:53
sometimes rather considerable
14:54
computations and we'll leave that for
14:56
the online course so details will be
15:00
laid out there more fully so let's start
15:02
with a general DCB curve here's our form
15:05
for it gamma in terms of A's and B's x
15:09
and y-coordinates and we'll assume that
15:11
we have a which is gamma of T and B
15:14
which is gamma of you these are two
15:16
points on the curve now the line I am I
15:20
is the point at infinity that we've
15:22
already talked about and M is the
15:24
midpoint of a and B okay so those two
15:28
points determine a line and this is the
15:31
equation for that line in projective
15:33
chord
15:34
and it obviously involves tea and you
15:38
but it also involves of course these
15:40
coefficients but it does so in a way
15:43
through these Delta objects so Delta 0 3
15:48
that's a particular cross-product of the
15:51
A's a 0 B 3 and a 3 B 0 and here's a
15:56
delta 1 3 times T Plus u over 2 and
15:59
here's the Delta 2 3 times T squared
16:02
plus u squared over 2 and the other
16:05
coefficients are minus B 3 2 a 3 now
16:10
this is the general case where we're not
16:13
necessarily assuming an Archimedean
16:14
curve so it's interesting to try to
16:17
compute things in the general case as
16:19
well okay so notice that here there is a
16:21
delta 2 3 appearing in the Archimedean
16:23
case that term will disappear that would
16:26
be important for us in any case in
16:28
general this thing here meets the curve
16:31
gamma of V so if you let V be a third
16:34
parameter and as when does this line
16:36
meet to this point well then you see you
16:39
have a quadratic equation in V just sort
16:43
of write down what it means to be
16:45
incident this point being incident with
16:47
this line then you get this equation
16:49
here involving Delta 1 3 again Delta 2 3
16:52
some T Plus u over 2 T squared plus u
16:55
squared over 2 and minus Delta 1 3 V and
16:59
minus Delta 2 3 v squared equals 0
17:02
that's a quadratic equation for V now
17:06
when Delta 2 3 equals 0 in other words
17:10
when we're in the Archimedean situation
17:13
then this term happily is gone and then
17:17
we have a linear equation in V and this
17:21
term here is also gone so it becomes
17:23
actually very simple that Delta 1 3 is
17:26
canceled and we immediately see that V
17:28
is T Plus u over 2 so in the Archimedean
17:33
case this simplifies dramatically and
17:35
there is indeed only one solution so
17:38
there'll be one point on the curve which
17:41
is on this line aside from the point I
17:45
at infinity so that's the first half of
17:49
the theorem that that I wrote down okay
17:54
so now let's look at the second part of
17:56
the proof where we are starting with
17:58
let's say a general cubic curve gamma
18:01
and here is the expression for its
18:05
differential or the area differential D
18:07
of gamma so if gamma is the curve P
18:12
comma Q then this is the quantity P DQ
18:15
minus QD P that's the quantity that we
18:17
have to follow up or integrate in order
18:20
to get a signed area so it turns out to
18:23
be a recently Pleasant expression
18:24
especially when you use these Delta
18:26
objects
18:27
so there's Delta 0 1 plus 2 Delta 0 to
18:30
alpha plus 3 Delta 0 3 plus Delta 1 2
18:34
times alpha squared plus 2 Delta 1 3
18:38
times alpha cubed plus Delta 2 3 times
18:42
alpha to the fourth and then we're in a
18:45
position to find the signed slice area
18:48
so I remind you the sign slice area is
18:50
obtained by taking the signed area of a
18:52
segment and adding to it the signed area
18:55
of the oriented segment coming back from
18:57
the end point to the beginning okay so
19:01
we can work all of that out and then we
19:03
can simplify and here is the formula for
19:07
the signed slice area from tdu of gamma
19:09
so it's an expression times u minus T
19:12
cubed and in here is Delta 1 2 over 6
19:15
plus Delta 1 3 over 4 times T Plus u
19:17
plus Delta 2 3 over 10 times T squared
19:21
plus 3t u Plus u squared now when we're
19:27
in the Archimedean situation the delta 2
19:31
3 equals 0 then what happens is that
19:34
first of all we can introduce this other
19:38
point C which is gamma of T Plus u over
19:41
2 and we can calculate the signed area
19:45
of the triangle a Seabee because we have
19:49
three points we have their coefficients
19:51
exactly we can compute the signed area
19:52
of that triangle and it turns out that
19:54
this is the expression that we get
19:56
there's a u minus T cubed
19:58
term as before and now multiplying that
20:01
Delta 1 2 over 8 plus 3 Delta 1 3 over
20:06
16 times t plus u now how does that
20:13
compare with the sign slice area in the
20:16
case when Delta 2 3 equals 0 well when
20:19
Delta 2 3 equals 0 then this quadratic
20:22
part is not there and we just have these
20:25
two terms multiplying this and when you
20:28
look at these two terms and compare them
20:30
with these two terms pleasently there's
20:32
a nice ratio in fact the sign slice area
20:36
is exactly 4/3 the side area of ACB and
20:41
we can see that because if we multiply
20:43
this by 4/3
20:44
we'll get 1/6 there and we multiply this
20:47
by 4/3 we get 1/4 which is exactly what
20:50
we have there and otherwise these two
20:51
expressions are exactly the same and the
20:54
factor here is exactly the same so
20:57
that's a proof that given this important
21:00
this crucial condition that Delta 2/3
21:03
equals 0 so we have to have this
21:04
argument in property around otherwise
21:06
says it's not going to work
21:07
but if we do have that then the sine
21:10
slice area is exactly 4/3 the signed
21:14
area of this inscribed triangle that's
21:18
exactly the same form as the classical
21:24
formula of Archimedes for the slice area
21:29
of a parabola same 4/3 same kind of
21:34
triangle same slice area it's exactly
21:38
the same now there's actually quite a
21:42
lot of additional really beautiful stuff
21:44
that's happening in this situation sort
21:46
of around this Archimedean formula and I
21:49
especially want to show you this kind of
21:52
Restatement of it where we are making
21:55
contact with the de castilla bezzie a
21:58
framework of having four control points
22:00
okay so we'll assume that our points a
22:04
and B are the outside of these four
22:08
control points so p0 is a and there is
22:10
p1 and there's p2 and there's p3 which
22:12
be so that's the segment that we've been
22:14
considering up till now and we're
22:16
considering that portion of this DCB
22:19
cubic curve so the theorem now is that
22:23
if gamma is our Q median and I remind
22:27
you that that condition can be expressed
22:30
just in terms of areas associated to
22:32
this quadrilateral formed by these four
22:35
control points okay we don't have to
22:37
look at infinity we can just look at
22:39
some areas and make some relatively easy
22:41
calculations to determine exactly when
22:43
this condition is happening so if we
22:46
have gamma Archimedean then we have
22:49
another formula for the sine slice area
22:52
of the curve so again the same quantity
22:54
there is the curve there's the segment
22:56
coming back so we're talking to this
22:57
green area the area sliced off by this
23:02
segment a-b
23:04
from the curve that sine slice area is
23:07
in fact three quarters of the signed
23:11
area of the cyclic quadrilateral p0 p1
23:15
p2 p3 this quadrilateral here you take
23:19
three quarters of its side area and you
23:23
get the sine slice area so that means
23:28
actually in practice that you don't need
23:31
to find this point C okay you don't need
23:34
to make this line law this line I am and
23:37
see where it intersects the curve if you
23:40
want to know what the area of this slice
23:41
is then all what you need to do is
23:44
calculate the area of this quadrilateral
23:47
oriented quadrilateral and take three
23:50
quarters of it and I also want to point
23:53
out that the the number 4/3 is then
23:56
appearing at really twice in this
23:58
picture so here's the triangle ACB if we
24:01
multiply that by 4/3 then we get the
24:04
signed slice area of the curve and if we
24:07
multiply the signed slice area of the
24:09
curve by 4/3 let's put that on that side
24:13
then we get the area of the big control
24:17
quadrilateral so from triangle to curve
24:21
slice areas factor 4/3 and from the
24:24
sliced
24:26
area of the curve to the big
24:28
quadrilateral is another factor of 4/3
24:31
and finally just to suggest that there
24:35
is actually more interesting stuff
24:37
happening here and in the course there
24:39
will be a wider investigation there's
24:42
actually quite a lot of additional
24:43
fairly fascinating things that are
24:45
happening here in this situation they
24:48
will be investigated there but I just
24:50
wanted to end with this thing here which
24:52
shows that this point see that we were
24:55
talking about this inscribed point it
24:58
can actually be obtained in a very nice
25:00
way so if n is the midpoint of p1 and p2
25:03
m m was the midpoint of this segment and
25:07
if now n is the midpoint over this
25:09
assignment ok then it turns out that
25:12
this point C is actually on this line
25:15
between N and M and it's a nice affine
25:18
combination it's 3/4 of n plus 1/4 of M
25:23
so that means it's in the ratio of 1 2 3
25:26
the segment to this segment is in 1 2 3
25:28
and so we're connecting with the convex
25:31
linear combination so that we talked
25:34
about earlier in the course and because
25:37
N and M are themselves just combinations
25:40
of P 1 P 2 and P Zero P 3 then it's not
25:44
hard to just rewrite this as a fine
25:47
combination of the four control points
25:50
of de casteljau and Messier and we get
25:53
that the Point C here is 1/8 of p0 plus
25:57
3/8 of p1 plus 3/8 of p2 plus 1/8 of p3
26:04
and of course C is actually on this line
26:08
so this line to infinity can also be
26:10
found by taking the midpoint here in the
26:12
midpoint here and joining up so this is
26:14
all happening in the Archimedean case
26:17
important to note that right this is not
26:19
for a general cubic it's just for the
26:23
argument D in cubics for which our
26:25
Archimedean formula works so I think
26:30
this is a highlight of the course and
26:32
it's I think solid evidence that yes
26:36
this algebraic calculus point of view
26:38
not only clarify
26:40
a lot of things not only provides a
26:42
logical and rational and completely
26:45
computationally sound approach to
26:47
calculus it also opens the doors for new
26:50
discoveries and beautiful new vistas so
26:56
there's much to learn I'm Norman
26:58
wahlburger thanks for listening
27:02
[Music]