Arithmetical Algorithms and "Polyseries"
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The Algebraic Calculus Two course begins by enlarging our arithmetical understanding in a rather general way. We show how to augment our arithmetic from polynumbers to polyseries, which are close to, but distinct from, power series in standard analysis. This introductory lecture motivates our treatment of this important topic, from both the Integral and Differential Calculus points of view, as well as algorithms from arithmetic, including quotients, square roots and cube roots. Please note that the Algebraic Calculus One course is now publicly available at https://www.openlearning.com/courses/.... Why not join? This is an important pre-requisite to learning the Algebraic Calculus Two.
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[Music]
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welcome everyone to the algebraic
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calculus I'm Norman wobbler so this is
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the algebraic calculus to our second
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course where we're going to be slowly
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moving towards differential calculus and
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a good starting point is to extend our
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arithmetic okay so we've been doing
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arithmetic since primary school it's
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time to step up a grade and that's going
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to open a lot of doors for us in our
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understanding of calculus so just to
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remind you we started primary school
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with arithmetic with numbers addition
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and multiplication some rational numbers
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etc and then somewhere in high school we
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learned about polynomial arithmetic what
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we prefer to call poly numbers so the
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way we like to think about things as a
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poly number is a vertical list of
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numbers they could be integers like most
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of my examples but actually they could
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also be rational numbers which are sort
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of the default a number system that we
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adopt so here's a Pollin number to one
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and here's another one minus one three
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and the way we add them is
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component-wise and if we want to
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multiply them it's a little bit more
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sophisticated something called a Koshi
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product so one two times three minus one
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how do we get three five minus two well
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the top entry which is sort of the level
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zero is the product of the two level
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zero fellows one times three is three
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the level one guy down here the five is
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obtained from taking the level zero
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times the level one here and adding the
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level one here times the level zero here
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so 1 times minus 1 plus 2 times 3 that's
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a total of 5 and there's also an
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additional one at level two which is
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attained from the product of the two
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level one terms two times minus one now
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it becomes convenient to connect this
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with polynomial arithmetic we can do
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that by introducing the special poly
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number alpha which is 0-1 and that way
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we can quickly rewrite a poly number
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like this as two plus
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alpha this one is minus one plus three
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alpha and then the addition is just the
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usual addition of polynomials in alpha
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but now crucially the Alpha is not some
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variable for which it's hard for us to
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pin down
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it's rather this very specific poly
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number okay and then the multiplication
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like over here 1 plus 2 alpha times 3
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minus alpha that's 3 plus 5 alpha minus
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2 alpha squared that then corresponds to
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the usual polynomial arithmetic okay so
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that's great that has been a very
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important framework for the algebraic
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calculus 1 course because a lot of the
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basic elementary stuff with calculus
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really rests on polynomial arithmetic
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but now entering the algebraic calculus
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2 course we want to go up a level and
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this is a very big jump sort of taking
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step from the high school arithmetic to
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college-level arithmetic so this next
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step up is in traditional calculus
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courses called power series but we are
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going to call them poly series because
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we want to separate what we're doing
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from the usual story because we're
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approaching things in a much more finite
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specifiable fashion we're not allowed to
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say let's add up a following infinite
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number of things so we have to have
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everything on the page and that means
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that it's a big challenge for us
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actually to bring this into a proper
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focus so today's lecture is not meant to
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actually do that but just to introduce
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these objects and kind of convince you
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that they are actually quite interesting
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and important and that they really do
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enlarge our arithmetic so what are these
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things well from our point of view a
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poly series is like a poly number it's a
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vertical list of numbers but with the
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crucial difference that it doesn't
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necessarily end ok there's these three
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dots that in some sense represent an
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ongoing aspect of this poly number so
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it's a poly number that's ongoing if you
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like but to say precisely what that
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means we'll require a fair
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work for us so it's not at all clear at
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this stage but anyway we can sort of
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visually understand here is a poly
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series that starts 0 1 2 3 and then
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carries on in some fashion and here's
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another one 1 2 4 8 carries on in some
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fashion now you might look at this one
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say well this is looking like the poly
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number with the consecutive integers
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down the column and maybe that's the
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case but maybe that's just a good guess
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and not quite correct because we are not
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saying what exactly the pattern is
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represented by these three dots at this
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point likewise here you might say okay
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that looks like powers of two well that
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could be but there are other sequences
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interesting sequences that start this
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way and then don't carry on like you
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might think okay so adding these things
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is just like with poly numbers we add
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them component wise so in this games we
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get 1 3 6 and 11 and then what happens
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afterwards well we don't really know but
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we represent that with three dots if we
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knew a few more terms here then we could
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augment this some also so here then is
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corresponding multiplication we choose
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some different ones here's the phone
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number or poly series 1 1 1 1 1 and then
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carries on and this one 0 1 3 7 5 and
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there's more and to multiply them we
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still do that Koshi product kind of
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business so 1 times 0 is 0 1 times 1
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plus 1 times 0 that's 1 just I'm gonna
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go through it this time so to make sure
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we all know ok this 4 is obtained by 1
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times 3 plus 1 times 1 plus 1 times 0 so
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it's 3 plus 1 that's 4 this 11 is
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obtained by 1 times 7 plus 1 times 3
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plus 1 times 1 plus 1 times 0 it's 11
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this 26 is obtained by 1 times 15 plus 1
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times 7 plus 1 times 3 plus 1 times 1
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plus 1 times 0 you can see that what
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we're doing here is kind of adding up
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successive elements in this one right
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this is performing like a summation
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really on this sequence so we get to 26
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and if we knew more well then we could
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get more but at this stage we're not
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able to say what the next one is because
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we don't know what the next ones are in
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the two sequences or two poly series
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they're now just as with poly numbers we
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can alternatively represent what's going
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on with alpha notation so here this
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would be alpha plus 2 alpha squared plus
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3 alpha cubed plus something going on
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this is 1 plus 2 alpha plus 4 alpha
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squared plus 8 alpha cube plus something
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further that's going on but we don't
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know what and the sum is well 1 plus 3
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alpha plus 6 alpha squared plus 11 alpha
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cubed and that we can obtain just from
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the information given the higher powers
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of alpha which are sort of implicit
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there but not known are not going to
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affect these ones so whatever these
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things are beyond here it's not going to
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affect this computation that we have
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made at the beginning the first four
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terms are going to be 1 plus 3 alpha
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plus 6 alpha squared plus 11 alpha cubed
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similarity over here this one is 1 plus
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alpha plus alpha square plus alpha Q
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plus alpha the fourth and then it keeps
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going and this one is alpha there's no
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constant term plus 3 alpha squared plus
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7 alpha cubed plus 15 alpha and forth
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and it keeps going again in some fashion
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that we don't know and if you just did
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polynomial arithmetic on this well you
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could calculate everything up to the
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fourth powers of alpha you wouldn't be
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able to say anything about alpha to the
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fifth or higher because there's some
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unknown alpha to the fifth terms in both
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of these initial expressions that we
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don't know so we're not able to
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calculate any higher coefficients but we
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can determine that the initial ones are
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alpha plus 4 alpha squared plus 11 alpha
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cube plus twenty six out of the four and
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at this level of support anout like this
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horizontally in terms of alpha how for
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example could we get this alpha cubed
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with an 11 in front of it well we asked
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how can we get alpha cubed out of this
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product we can get it with 1 times alpha
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cubed
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seven of them or alpha times three alpha
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squared there's three of them or alpha
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squared times alpha there's one of them
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and that's all the possibilities for a
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total of seven plus three plus one which
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is eleven okay so this Koshi product at
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the level of sequences or lists is
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exactly what we are familiar with with
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highschool polynomial algebra but
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crucially now we're recognizing that
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we're only stating things up to a
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certain point and then beyond it there's
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some indeterminacy we don't know what's
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happening after that at this point now
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it turns out just this example here it's
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a little bit interesting so this is
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obviously kind of a an obvious pattern
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the obvious Ben and this year is also
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kind of an obvious pattern you could get
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a good guess for what that is just one
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less than powers of two and then it
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turns out that these numbers here are
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not without interest 1 for 11 26 they
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are the first of the oil area numbers in
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Sloan's
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online encyclopedia of integer sequences
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which is going to be playing a very
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interesting role in this course that
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sequence is numbered a triple zero to
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nine five okay that's that sequence that
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starts one for eleven twenty six maybe
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there's some zeros in front so it turns
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out that these oil area numbers wherever
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they're coming from probably some kind
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of comm unit or XOR algebra can be
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served generated by multiplying these
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two simpler poly series so that's kind
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of interesting and we can hope that more
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interesting things like that will happen
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all right so it looks like there's a
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cute or interesting arithmetic that can
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be developed with these poly series
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which kind of are like poly numbers but
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they carry on but what are the
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connections with calculus so it turns
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out that there's a very good reasons for
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adopting this arithmetic into the
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algebraic calculus so one of them comes
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from the integral calculus point of view
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that we've already been talking about at
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length in the algebra
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calculus one course where we were
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concerned with curves and signed areas
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determined by segments of curves for
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example from gamma of T to gamma u this
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segment of the curve determines a
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certain signed area suitably interpreted
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with orientations perhaps this way and
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then the opposite way for this little
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loop here so if our curve is a very
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pleasant curve for us that means it's a
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de casteljau bezier curve okay those are
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the most pleasant kinds of curves in
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calculus for sure they are the ones for
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which these signed areas can be defined
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and calculated completely correctly
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unambiguously precisely okay so we have
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a very beautiful theory of signed areas
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for these kind of polynomial a
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parameterised curves or DCB curves but a
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lot of curves that come up in practice
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are not of that pleasant kind so we are
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looking for ways of extending what we
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already have developed in this context
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to a wider arena to be able to deal with
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other curves for example even something
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like a circle or a hyperbola or maybe
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some kind of cubic curve so these
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generally are not de casteljau bezier
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curves and so we'd like to access
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integration for them in some way and one
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approach to doing that is to augment our
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arithmetic first by including these poly
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series that's going to be very important
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technology for pushing the integral
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calculus further in fact it's very much
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along the line of Newton's approach to
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the calculus and actually what a lot of
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modern physicists do as well now the
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differential calculus which is going to
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be playing a bigger role in the
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algebraic calculus to course concerns
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itself let's say at least initially with
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the local structure of curves so here's
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a lemniscate
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lemma skater Bernoulli here's its
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equation as an algebraic curve
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sort of like a figure eight or infinity
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sign so this kind of infinity is allowed
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for us okay
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and suppose we have a point on this
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curve suppose we actually can find such
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a point then we might be interested in
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what does this curve look like near this
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point the simplest answer that question
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is to consider the tangent line to the
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curve okay so that's a good
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approximation to the curve but it's not
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really great because it's a straight
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line a better approximation is to take a
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quadratic approximation so one can do
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this so maybe there's a parabola that
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approximates the curve well at that
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point and maybe in fact there are higher
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order occurs that provide better and
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better approximations to the curve near
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the point where we're interested so
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that's a primary kind of motivation in
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the differential calculus is to be able
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to provide finer and finer descriptions
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or understandings of curves near some
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particular point or region so a natural
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question might be sort of how do we
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approximate a curve a near point P well
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another way of approaching this or
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actually kind of related is to try to
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express the the curve in a parametric
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form now it turns out that this kind of
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curve is not a de castilla Bezier curve
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okay so you cannot parameterize it with
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poly numbers that's just a fact of life
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if you could then all integral questions
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become a lot simpler automatically but
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you can't but what you can do it turns
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out often at least is to sort of capture
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this with not Pauli numbers but poly
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series so we can aspire to describing
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the curve gamma at least say near the
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point P by a series like this where we
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have instead of a poly number P and
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another poly number Q we have a poly
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series ease
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to say 1 alpha plus a 2 alpha squared
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plus and then there's more stuff and
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here's another one
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B 0 plus B 1 alpha plus B 2 alpha plus
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more stuff at this stage we're not
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saying what the more stuff is but when
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we allow ourselves this possibility of
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using this augmented arithmetic then the
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possibility emerges that we can describe
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curves like this which cannot be
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described in a DC B fashion but with
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this sort of larger framework that's
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great because a lot of the kinds of
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things that we do with DC B curves based
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on essentially understanding Pauli
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numbers can be applied to this setting
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because these things are almost poly
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numbers okay so this is a really good
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idea and that's gonna be powerful for us
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but we do need to make the step up when
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we're considering general curves from
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poly numbers to poly series so I hope
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I've provided some kind of motivation
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for us why we might be interested in
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poly series in the context of calculus
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but in fact we can also see poly series
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arising naturally in much more
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elementary arithmetic so you just start
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thinking about arithmetic with
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polynomials or poly numbers then it
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doesn't take too much before you start
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realizing that in certain situations
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it'd be really useful to be able to
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extend our poly numbers to have more and
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more terms so to an ongoing kind of poly
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numbers namely poly series so let me
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give you an example or two of those so
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the first one comes from division okay
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we have our four operations addition
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subtraction multiplication those three
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are very simple or relatively simple
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with just poly numbers and don't take us
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outside of poly numbers but division
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which is always the hardest
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Bharath medical operation that that
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gives us problems just as it does for
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arithmetic with
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ordinary numbers what's 3/5 well you
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know what's 17 / 59 these are more
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subtle questions okay and the similar
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kind of thing is happening here
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so let me remind you of a computation
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that we did actually in end videos with
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the algebraic calculus one course we
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were talking about divisibility of paly
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numbers how to divide one pali number by
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another i'm showing you how i prefer you
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to do long division and here was an
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example we were dividing 1 minus alpha
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to the 6 by 1 minus alpha plus alpha
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squared and i want to just quickly
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review that because doing this long
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division is very important ok it's a
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very important skill that we're going to
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need in our development of differential
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calculus and working with poly series so
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what do we do we ask ourselves how many
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times does this polynomial sort of
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divide this one and note that we've
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arranged our poly numbers so that they
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are in increasing order as we read from
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left to right so the constant term first
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and then the linear term and the
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quadratic term as we move to the right
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okay so when we answer this question
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what we do is we say okay let's look at
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the constant term we're sort of reading
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from left to right the constant term one
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okay here there's a constant term one
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how many of these fins do this
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well one of them so we take one we write
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down that one
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and we multiply this by one and we get
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that there so we think that one of these
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will fit inside here because this one is
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dividing that one there or less than
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that one so we then subtract we take
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this thing minus this so that gives us a
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total of alpha minus alpha squared minus
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alpha to the six so this one here is the
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result of taking this minus this so
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that's like the remainder after we've
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removed one of these from what we
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started with okay now we ask again how
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many of these will fit into this thing
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here and it's quite simple because we
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just look at this constant term here
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that's the only thing that we're really
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using we say okay there's an alpha here
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there's a one here we can take alpha of
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these
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okay so we multiply this by alpha get
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alpha minus alpha square plus alpha
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cubed and then we're gonna subtract this
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1 minus this one
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the alphas cancel the alpha Squared's
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cancel and we get this minus is minus
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alpha cubed minus alpha to the 6 now how
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many of these fit into there we just
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look at the first or the lowest term
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minus alpha cube we say minus alpha
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cubed of them will fit in we multiply
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this by minus alpha cubed there it is
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there we do the subtraction we get this
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one here then how many of these fit in
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there when you can't - out through the
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four of them and by the time we've done
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that then we finally get a remainder of
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zero that means that the division has
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been successful we have divided this by
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this and obtained this ok so here's the
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result of that so writing it sort of in
20:51
terms of a quotient 1 minus alpha to the
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6 over 1 minus alpha plus alpha squared
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that equals 1 plus alpha minus alpha
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cubed minus alpha to the 4 and we could
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check that by doing a multiplication so
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we could multiply both sides by this
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denominator here bring that to this side
21:09
here and if we multiply this poly number
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by this poly number expanded it all out
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collect the terms simplified we would
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get exactly 1 minus alpha to the 6 with
21:20
nothing left over but just ask for
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numbers not all divisions work so well
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we can divide 24 by 6 we get an exact
21:30
number 4 if we try to divide 24 by 7
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well then it's a very different story
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okay so that also happens in the poly
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number situation for example suppose we
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just change that a little bit now we
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have 1 minus alpha to the 5 divided by
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the same thing 1 minus alpha plus alpha
21:50
squared so I'm going to do by using the
21:53
same strategy but now I'm going to sort
21:56
of write the the answers the quotients
22:01
vertically because well actually we'll
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see that they keep on going okay so
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let's start how many of these fit into
22:08
here
22:08
again we just look at the left most
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terms here and you're one of them so
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here's one of them now subtract this
22:14
minus this
22:15
get alpha minus alpha squared minus
22:17
alpha v how many of these fit into here
22:20
alpha of them so I'll put the Alpha on
22:22
the same line alpha that so alpha times
22:25
this is powerful minus alpha squared
22:27
plus alpha cubed this - this is those
22:32
things cancel we still get minus alpha
22:33
cubed minus alpha v how many of these
22:37
fit into this minus alpha cubed of them
22:39
here we have myself with cute plus alpha
22:41
fourth myself for the fifth we do the
22:44
subtraction we get myself to the four
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actually so so far it's looking very
22:51
much like what we did in the last
22:53
example and how many of these fit into
22:55
minus alpha the four well minus alpha to
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the four of them well this x minus out
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of the four is minus out of the 4 plus
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alpha v minus alpha the six we subtract
23:04
this minus this we get minus alpha v
23:06
plus alpha to the six last time we got
23:10
zero this time we're not getting zero
23:12
okay so we could carry on okay how many
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of these fitted here well minus alpha to
23:19
the fifth we can see that but further
23:22
once well would require further work now
23:28
is this going to end
23:31
like maybe if we go down 28 steps will
23:33
it find land
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no what won't because so far what we've
23:38
got is we got 1 minus alpha v over 1
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minus alpha plus alpha squared equals 1
23:43
plus alpha minus alpha cubed minus alpha
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fourth minus alpha v and then there's
23:47
some more stuff which we haven't
23:49
computed yet and this is not going to
23:53
end at alpha to the 28th because if it
23:55
did then we would have an identity that
23:58
involves 1 minus alpha to the fifth
24:00
equals the product of this term times
24:03
this big poly number okay this is degree
24:07
2 and this thing is of degree 28 when we
24:09
multiply them we're gonna get something
24:11
of degree 30 not of degree 5 like we
24:14
need so that argument shows that this
24:17
procedure in this case is not going to
24:20
end it's a completely different story
24:22
from the previous division that we did
24:25
it's an algorithm which will not and
24:28
and this is such an important aspect of
24:32
this story right now what we have here
24:34
is an algorithm a well-defined procedure
24:38
that's very clearly laid out and we're
24:42
pressing the Run button and it's
24:43
churning over and over and over giving
24:45
us output excellent more and more and
24:48
more output but we can see that this
24:50
program is not gonna stop it's not going
24:53
to terminate we do not get to the end
24:55
this just keeps on churning out more and
24:59
more terms in this poly series now my
25:08
pure mathematician friends look at
25:11
something like this and say okay well
25:13
why don't we just go to infinity you
25:17
know get our computer and we look at the
25:20
back somewhere and there's somewhere
25:21
there's a go to infinity button big red
25:24
button does your computer have such a
25:25
thing a big red button go to infinity
25:28
and suddenly you've gone to infinity in
25:31
the minds of my friends so this is a
25:34
thought experiment which unfortunately
25:37
is somehow removed from actual
25:39
computation reality because if you do
25:42
this for a few hours you will see that
25:44
you know you're not gonna be able to go
25:46
to infinity you're not going to be able
25:48
to continue this thing to the end of
25:50
time so in a case we're getting
25:54
something that we could informally call
25:56
a poly series we've divided this
25:59
opponent number by this poly number we
26:01
have not yet another poly number we've
26:02
gotten a poly series so this is a very
26:06
good reason why having an arithmetic
26:09
including poly series is a good idea
26:12
because it allows us to do division in
26:15
the world of poly numbers now another
26:20
kind of interesting arithmetic that we
26:22
can access with poly series that we
26:25
can't access so easily with poly numbers
26:27
are square roots so for example suppose
26:32
we want to solve p squared equals 1 plus
26:34
alpha okay so 1 plus alpha is a very
26:38
simple poly number and P is some other
26:40
Pauline umber
26:42
so we're looking for a poly number that
26:43
has the property that if you square it
26:45
you're gonna get 1 plus alpha now it's
26:48
pretty easy to see that you're not going
26:50
to be able to find such a poly number
26:52
but it turns out that if you expand your
26:55
horizon a little bit and allow poly
26:58
series that you can find such a P at
27:02
least kind of okay so let me show you
27:05
how that goes so our poly number P we're
27:08
gonna represent it with unknown entries
27:11
say p0 p1 p2 and it carries on in this
27:16
case the pattern is obvious and here
27:19
we're multiplying P times P and we would
27:22
like the product to be this thing here
27:26
which is 1 plus alpha so what we're
27:29
gonna do and this is a fundamental sort
27:31
of strategy for many such problems is
27:34
we're going to successively go through
27:36
the various pis and solve for them ok so
27:41
we start up at the top at the zero level
27:44
the top level is p0 times p0 should
27:48
equal one so p0 squared equals 1
27:51
well we know what the solutions are
27:53
possibilities for that are so p0
27:55
therefore is going to be equal to plus
27:58
or minus 1 let's say that T 0 is 1 ok so
28:07
suppose it's 1 so we have 1 now and now
28:10
we have this next entry ok how do we get
28:13
this next entry well it's this this is
28:15
not give me a 1 so 1 times P 1 plus P 1
28:20
times 1 that's 2 P 1 should equal 1
28:23
that's gonna be our next thing I'll put
28:25
it maybe here to p 0 equals 1 that tells
28:30
us that P is going to be 1/2 ok so this
28:33
is 1/2 here ok so this is 1 this is 1/2
28:36
this is unknown p2 what's that going to
28:39
be so let's look at the third entry
28:42
which here is actually a 0 but it's
28:44
nothing there but it means it's a 0 so
28:46
that's 0 is going to be the result of p0
28:48
times p2 that's p2 so let's write P 2
28:53
and then plus p1 times p1
28:56
well p1 we said is it was not p0 sorry
29:00
this was p1 pardon me
29:02
p1 was 1/2 so p1 times p1 is 1/4 1/4 and
29:11
then there's another p2 times p0 well
29:13
that's 1 so it'll be another p2 and that
29:17
should be equal to 0 okay so we bring
29:20
the one chord to the other side that's
29:22
minus 1/4 then we divide by 2 we get
29:24
minus 1/8 get minus 1/8 and you could
29:29
see that you could kind of carry on like
29:32
this and in fact if you carry on a
29:35
little bit further we get P equals 1
29:43
plus 1/2 alpha minus 1/8 alpha squared
29:51
plus 1/16 of alpha cubed then minus 5
29:58
over 1 28 alpha to the 4 plus 7 over 256
30:07
alpha to the 5 and then it keeps going
30:12
after the next one is negative and just
30:17
as in the division case it doesn't stop
30:20
we've outlined an algorithm that allows
30:23
us to sequentially calculate more and
30:26
more of these pis with the understanding
30:29
that the algorithm doesn't end we never
30:32
get to the end because if we got to the
30:35
end we had some polynomial a polynomial
30:42
equal to this thing okay so there's no
30:46
way we're gonna get a poly number answer
30:47
out of this this is necessarily gonna
30:49
have to keep on going in some sense the
30:52
algorithm does not stop oh there's going
30:55
to be another solution and that's when
30:57
we take minus P instead of this piece
31:01
which just multiply all of those by a
31:02
negatives and that's gonna have the
31:04
property that when you square it you're
31:05
still gonna get the same thing as you
31:06
did before
31:08
okay this is quite interesting this is
31:10
actually a very important poly series
31:14
although we're just describing the first
31:18
five or six terms here okay keeps on
31:21
going but there will be a lot of
31:23
physicists for example who will who will
31:25
know this off by heart because it's an
31:28
important tool for a lot of calculations
31:31
if you want to calculate a square root
31:33
in a lot of situations it ends up being
31:35
close to close to this all right so
31:39
that's a important example of how poly
31:42
series allows us to solve a kind of
31:45
equation that we have no hope of solving
31:48
originally with just poly numbers okay
31:52
so this is such an important technology
31:54
I want to show you the same kind of
31:56
problem but with a little bit more
31:58
sophisticated right-hand side so let's
32:00
try to solve Q squared equals 1 plus
32:03
alpha minus alpha squared we're looking
32:05
for a poly number or probably a poly
32:07
series that will satisfy this equation
32:10
and we suppose that Q is given by Q 0 Q
32:13
1 Q 2 some coefficients or entries which
32:17
are to be determined and our job is to
32:19
determine them we're going to do so one
32:21
at a time starting for the top starting
32:24
from the top so Q zero times Q 0 equals
32:29
1 okay so I'm gonna put the result here
32:33
so we have Q 0 squared equals 1 that's
32:36
our top line so let's say that Q 0 is 1
32:39
there's 2 possibility is that before I'm
32:41
just gonna show you one possibility
32:42
ok so Q 0 is now 1 okay now what's the
32:46
next line 1 times Q 1 plus 1 times Q 1
32:49
so it's 2 Q 1 is going to be 1 I would
32:52
get the same thing as we did last time Q
32:55
1 is going to be 1/2 okay what's the
32:58
next one so we have Q 2 and there's
33:01
another tooth Q 2 so there's 2 Q 2 I'll
33:04
plus this 1/2 times 1/2 so that's 1/4
33:07
equals and now minus 1 okay so minus 1
33:13
bring this to the other side so that's
33:15
minus 5 quarters / - that's minus 5
33:20
eighths
33:23
so we could carry on right we could
33:26
carry on but what we're gonna get is
33:30
something like this P equals 1 plus 1/2
33:34
alpha minus 5/8 alpha squared and then
33:41
plus 5/16 alpha cubed minus 45 over a
33:51
hundred and twenty eight alpha to the 4
33:55
plus 95 over 256 alpha to the fifth and
34:04
then well we could keep going again we
34:09
have a pretty clear algorithm you could
34:11
program a computer to do this but
34:15
there's lots of software that will do it
34:16
very quickly anyway but you could
34:18
program a computer to do this quite
34:20
pleasantly and then you could churn out
34:22
a few thousand or million entries in
34:26
this poly series and this is how it
34:28
would start there's another one where we
34:31
just multiply this 1 by minus 1 I guess
34:34
we're not talking about P we're talking
34:36
about Q aren't we okay
34:39
pardon me Q so let's put a box around
34:48
here that's our solution to this
34:53
quadratic equation in in poly numbers
34:58
but the solution is this poly series and
35:02
as you can kind of guess it's maybe not
35:05
entirely easy to predict what the
35:09
general coefficient is that's going to
35:12
be in front of alpha to the N all those
35:15
may be docked impossible either so in
35:19
the course proper you'll have a chance
35:21
to practice a lot of different examples
35:23
to strengthen this technology here's
35:26
another sort of step up where we look at
35:28
cube roots instead of square roots
35:31
suppose we're trying to solve this
35:33
equation R cubed equals 1 plus
35:35
alpha well let's let R be the poly
35:40
series starting with R 0 and then R 1 R
35:43
2 R 3 and so on and we have to cube this
35:47
thing and set it equal to this poly
35:51
number 1 plus alpha so it turns out that
35:53
when you cube this you get a poly series
35:58
whose terms successively get bigger and
36:01
bigger but they are polynomials in the
36:06
entries up to this point so like say
36:09
let's say we're going up to this point
36:10
here this number 3 a tree this certain
36:13
degree 0 1 2 3 so this degree 3 entry is
36:19
polynomial combination of all the
36:22
previous ones but has a crucial
36:24
additional property so let's have a look
36:28
at successive one so the the top one is
36:30
our zero cubed we're setting that equal
36:32
to one so one solution is to let that be
36:36
one so suppose our zero is equal to one
36:39
then the next one would be 3 times R 1
36:42
equals this one here so 3 times R 1 that
36:45
means that R 1 has to be one-third so
36:47
there's the 1/3 the next line so we say
36:51
R 0 is 1 and R 1 is 1/3 so this be 3
36:56
times 1/3 squared since our total of 1/3
36:59
and this is just 3 R 2 and that's
37:03
supposed to equal the second entry here
37:05
which is a zero okay so what's that give
37:08
us so I guess we get minus 3 divided by
37:11
3 so that's minus 1/9 that was why we
37:13
get to a minus one night there for the
37:14
alpha squared term and then you could
37:17
check that the next one here the R 3
37:20
term crucially is appearing only
37:24
linearly here is appearing linearly here
37:26
linearly here so that's very pleasant
37:28
because it means that we can recursively
37:30
solve for these various ours ok once we
37:34
have at this stage R 0 R 1 R 2 then this
37:37
is a linear equation set it equal to 0
37:39
for our three linear equations we can
37:42
always solve especially since the
37:44
coefficient is non zero right we're
37:46
saying R 0 is 1 so this
37:49
here is not zero that means we can solve
37:51
for our three and we would get five over
37:54
81 and we can carry on this way we get
37:59
this poly series so this is kind of a
38:05
strange new arithmetic isn't it
38:07
right we're solving an equation like
38:10
this in poly numbers or polynomials and
38:13
we're getting answers which are
38:18
polynomials which carry on which we're
38:20
calling poly series and it's not clear
38:24
what this means like how are we supposed
38:26
to interpret this what is actually the
38:28
definition of this R right now it's in
38:30
quotes because there's this crucial
38:31
three dots there and we haven't said
38:34
what that really means are we obliged to
38:37
spell out what the actual coefficients
38:40
are or is it enough to specify the
38:43
algorithm or can we just write something
38:46
down like this and then add three dots
38:48
and say well this is the object there's
38:51
a very subtle question so we have to
38:52
deal with so that's definitely a
38:55
question about of this what is really
38:57
happening and then another obviously
39:00
interesting question is is there a
39:02
formula for these coefficients you know
39:04
there's some kind of pattern here it
39:06
seems but a little bit of a pattern some
39:09
powers of three involved here
39:12
can we write down a formula for the
39:14
coefficients well Isaac Newton back in
39:17
the 17th century worked this out in one
39:22
of the really great discoveries of
39:26
modern mathematics he he realized what's
39:29
going on with with this equation and
39:31
with the square root equation and fact
39:33
with a much more general class of
39:35
equations and this is really his
39:37
generalized binomial series okay so
39:40
that's that's great we're going to touch
39:43
base with with Newton's work and that's
39:46
going to give us new insight into the
39:50
structure of these poly series that are
39:53
coming up with questions like this so
39:56
there's much to learn
39:57
I'm nowhere la Berger thanks for
40:00
listening
40:01
[Music]
40:09
[Music]