The differential calculus for curves, via Lagrange! | Differential Geometry 4 | NJ Wildberger

[Music] [Applause] [Music] hello everyone i'm norman weillberger we're here  at the university of new south wales and this is a   course on differential geometry today we're going  to look at a very important sort of elementary   issue of the the role of the differential  calculus for curves so this particular lecture   is probably going to be a broader interest it  will probably appeal to anyone who's studying   calculus or or teaching calculus because i'm  basically going to present a an alternative   kind of view to calculus different from the  one that one finds in the standard texts   this is a powerful alternative that has a very  strong algebraic focus and was in fact initiated   not recently but quite a long time ago  it was really an outgrowth of the work   of euler and lagrange so euler  and lagrange where they're both pioneering figures of course  in mathematics generally   the towering figures of 18th century  mathematics first oiler and then   he passed on the the torch to lagrange and they  had a rather algebraic approach to analysis   that that was i suppose enunciated very clearly  by an important work of lagrange and the work   is called theory de function analytic or theory of  analytic functions so that was written in 1797 and the subtitle is very interesting here's the  english translation of the subtitles very long   so theory of analytic functions containing the  principles of the differential calculus disengaged   from all considerations of infinitesimals  vanishing limits or flexions and reduced to the   algebraic analysis of finite quantities so that  was the subtitle of lagrange's important 1797 work   theory of analytic functions so in that that  work he embraces euler's thinking and uh and   lays out an approach to calculus which we're going  to follow today okay and it'll probably be novel   to most of you and you may easily wonder why  are we not learning calculus this way rather   than the more complicated way that we generally  learn calculus today that's a very good question   and it's an important question that educators  really ought to give some serious thought to all   right so let me start by reviewing the standard  approach to calculus and let me do it in the   context of a particular polynomial  so let's take the polynomial   say p of x equals eight minus five x  plus four x cubed minus x to the fourth   that's a quartic polynomial that looks something like this okay maybe here's 20 and  minus 20 and one two three   four minus one minus two minus three and  so on okay it looks a little bit like this okay all right so that's just a random polynomial   and suppose we choose a point on it say a  point -1 then what are we doing when we do   the differential calculus well the first  object of interest is the tangent line to   the curve at that point so we're looking for  a line passing through that point which is in   some sense a good approximation to the function  and we all know how to do that what we do is we   choose a nearby point and we draw a secant line  between the point in question and the nearby point   and we then let the nearby point approach  the given point and then that secant line   will approach the tangent okay so once we have  that tangent what's of interest to the tangent   or about the tangent is its slope so we  say s is the slope of this tangent line and that's usually in terms of a ratio of changes   of y and changes of x but we have to  sort of take a limiting in a limiting way all right so we get a number okay we  get a number so in this case this thing   here well maybe a change of 2 change of 10 or  something like that so maybe there's a slope of   5 or something like that all right so what we  do then is we plot that value okay so we define   p prime of x to be equal to  the slope of the tangent line at x and that gives us okay another value here  and then we can do the same thing for various uh   points on the curve and we get the the derivative  okay so the derivative will be uh what well   maybe it goes zero here a little bit negative  through the zero there will be a zero here okay so the derivative is not something  like this okay and that's then so the basic object of differential calculus  is usually presented to go from the function   p of x to its derivative p prime of x and  we also might call this say just d p of x   d standing for differentiation operator  okay now in in practice students learn   what this actually is in this case the  derivative is minus 5 plus 12 x squared   minus 4 x cubed and that's a consequence of a  couple of different formulas the one of them is   a very fundamental formula that the derivative  of x to the n is n to the x to the n minus one that's a fundamental fact and then another kind  of fundamental fact is that if you have two   polynomials the the derivative acts linearly so that we can break up a derivative like  this into a sum of lots of little derivatives   and then we just use this formula over and  over again we can write down the derivative   okay and then we have the notion of the  second derivative so then the second   derivative of p is by definition d of d of p  and we can keep going with higher derivatives okay so this is fine there are some  requirements here that are a little bit subtle   one is this limiting quantity this limiting idea  another is there's an underlying assumption of   sort of real numbers that we assume are  is underlying this limiting uh business   in the 18th century mathematicians previously  had approached these limits in rather dodgy ways   using so-called infinitesimals and a bishop irish  bishop berkeley wrote a scathing denunciation of   the techniques that were being thrown around  at the beginning of the 18th century and this   motivated mathematicians to start saying well what  are we actually doing here can we be a little bit   more precise about this limiting uh process  and it was partly in response to that that   lagrange and euler would have tried to frame  this this discussion in a different way all   right so what i want to do is i want to give you a  completely alternate or different approach to the   differential calculus now and i'll stick with this  same example so we can sort of see what happens okay so what i'm going to outline now  is really the approach of lagrange which rather remarkably has uh almost  completely dropped off the radar screen   here at the beginning of this new millennium  okay so what is lagrange's approach lagrange   says all right let's start with this polynomial  p and let's consider p of x plus y okay he may   not say it exactly this is my reformulation of  lagrange's idea but it's basically lagrange's idea   all right so we're going to look at p of x plus y  we introduce another variable and we're going to   just write this out so this is 8 minus 5 x plus y  plus 4 x plus y cubed minus x plus y to the fourth and we're going to expand this out  because we know the binomial theorem   so this is 8 minus 5x minus 5y plus 4 times x cubed plus 3x squared y plus 3xy squared  plus y cubed for that term there and then minus   x plus y to the fourth would be x to  the fourth plus four x to the fourth y   plus six x squared y squared plus  four x y cubed plus y to the fourth so correct me if i make this mistake there okay and now what we can do is we can rewrite this by sort of separating the x's and y's  so we can write this as a another polynomial in x   with coefficients in y so we get eight minus  five y plus four y cubed minus y to the fourth   that's sort of the constant term so i've done  this one this one uh this one and this one   and then there will be the terms linear  in x so that will be plus minus 5 plus 12   y squared minus 4 y cubed times x so now i've picked up minus 5x 12y squared x and minus 4y cubed x oh there's a minus sign here yet so the minus  sign so that should be a minus sign that's fine   okay and then the x squared terms so then there'll  be plus 12 y minus 6 y squared times x squared   and then there will be four minus four y times  x cubed and a single minus x to the fourth left over all right now i'm going to rewrite that formula slightly and now i'm what i'm going  to do is i'm going to replace now replace x with x minus y in this formula all right so   the if i if i replace x with x minus y  then i'll just have x here so p of x equals uh right here so i'm only replacing the x's so  i still have eight minus five y plus 4 y cubed   minus y to the fourth plus minus 5 plus 12 y squared  minus 4y cubed times x minus y plus 12 y minus 6y squared times x minus y squared plus 4 minus 4y x minus y cubed minus x minus y to the fourth so this is an identity you can think of it  as a way of rewriting the original polynomial   the original polynomial is now also equal  to this so if we expanded this all out again   we would just get the  original polynomial which was actually we can read it from here eight minus  five x plus four x cubed minus x to the fourth all right so now it's perhaps  more convenient or more usual to   replace y which is up to now just  another abstract variable with a number okay so let's replace a y with a number oh we could have done this  at the beginning but okay so then we would get well let me write it p of x equals eight  minus five r oh the number will be r   five r plus four r cubed minus r to the fourth  plus minus five plus twelve r squared minus 4r   cubed times x minus r and do i have room in here  plus 12y not 12r minus 6r squared times x minus   r squared plus four minus four r times x minus  four r to the cubed minus x minus r to the fourth i've just taken this thing and replaced  y with r i'm thinking of r is a number okay so what does this do for us well this identifies  certain we are certain polynomials are arising   here that uh that are occurring in the in  the coefficients of these various increasing   powers of x minus r okay and what are  those polynomials that are appearing [Applause] well let me write it without writing them  all down let me write this as d0 of p of r   plus d1 of d1 of p of r times x minus r plus  d2 p of r x minus r squared plus d 3 p of r   x minus r cubed plus d 4 p  of r x minus r to the fourth so what are these polynomials well we can see that  d zero of p is just really exactly the same as the   original polynomial p polynomial was eight minus  five x plus four x cubed minus x to the fourth   that's exactly the same thing there and this  next one what we're calling d sub one of p   is what's usually called the derivative that's  the derivative all right the derivative of   this is minus five plus twelve x squared  minus 4 x to the cubed so that's the   like p prime that's the derivative we also have these other ones so d to d sub  2 of p is not exactly the second derivative if you took the second derivative  you would take the derivative of this   you would get 24y minus 12 12y squared  here we're only getting half of that   so it's what we might say the second derivative  if you know what that is over of p divided   by two and then the third p is the the third  derivative of p over 3 factorial and and so on okay i like to call these  functions sub derivatives sub derivatives of p yes yes that's the taylor expansion of the  polynomial which is something that usually comes   quite a way a little ways into a calculus course  after you've done a whole bunch of other things   so what lagrange is saying is that we can go  directly right to the what we call the taylor   expansion basically lagrange knew this we can do  do it directly using just high school algebra and   once we once we look at things that way then not  only do we get the first derivative popping up for   free but up to this little bit normalization we  get all the derivatives listed right there for us okay in fact there's a a nice way of of  getting these uh all these coefficients   so but what we're really doing is just using the  binomial theorem so if we look at pascal's array which i'll write like this so one and then one  one and then one two one and one three three   one and one four six four one and so  on one five ten ten five one and so on   then what we're doing is we're  just multiplying these diagonals by the appropriate coefficients in our original polynomial  eight minus five zero four and minus one so i'll put maybe a eight here and then minus  five and then the x squared coefficient is zero   and then there's a four and then there's a minus  one all right so if we multiply each of these sort   of diagonals in this direction by these numbers  we get the array eight minus five minus five to   multiply that by l by zero one three three one  gets multiplied by four so four twelve twelve four no oh yeah that's right and then the  one four six four one gets multiplied   by minus one so that's minus one minus  four minus six minus four minus one and then that array there we have  immediately all the sub derivatives   ready for our reading off there's eight minus  five zero four minus one that's the original one   uh there is the first derivative there is   uh the second sub derivative 12 y in the right  columns and then 4 minus 4y and then minus 1 there so if you didn't feel like doing all this  calculation that i originally did once   you've done it once or twice you realize all  what you really need to do is you take this   pascal's array multiply the columns by  the coefficients of the matrix and boom   all the information is right  there right in front of you all right so that's good so  what what can we do with that well one pleasant thing to do is to follow  lagrange and say all right suppose that we have   two polynomials so we have one  polynomial p which we'll write as d0 of p   of r like everything so r plus d1 of p of  r times x minus r plus d2 p of r times x   minus r squared and so on and then we'll  have another polynomial let's call it q so we'll have a similar expansion d zero of q of r   plus d one q of r x minus r plus d  2 q of r x minus r squared and so on then if we multiply these two  polynomials p times q will then be well the zero term will be d d zero  of p times d zero of q okay at r and then the coefficient of x minus r will be  d zero of p times d one of q plus d zero of q   times d1 of p all of that at r times x minus r and then the next one will be the coefficients of  x minus r squared will be there'll be a d0 of p   times this d2q plus a d1 p times  a d1q and also a a d2p times a d0q okay all of that actually evaluated r  times x minus r squared and we can easily   see what the pattern is now of course that has to also be equal to the the corresponding series for p times q so  that has to be d zero of p times q at r plus   d one of p times q of r at x minus r plus d two  [Music] of p times q x minus r squared and so on   okay so what can we conclude well we  conclude uh it's obvious that d zero   of p q is equal to z d zero of p times d  zero of q because that's just p times q   but what's maybe not so obvious is that we now  know that the derivative of p times q is well   d zero p is just p so p times d one of q plus  q times d1 of p a familiar product rule for   derivatives but we can read off more we can read  off that the second derivative of the product is p times the second derivative of q plus the  derivative of p times the derivative of q   plus q times the derivative second derivative of p and we could just as easily write  down corresponding formulas for   third the third derivative of a product or  the fourth derivative of product and so on okay so that's uh that's good but we're interested in in geometry here for us what we're interested in in differential  geometry is to study a curve or a surface uh near a point okay we're interested in the  shape of things near points so we're interested in   you know what what's how do you describe  the shape of that curve near that point   or how do you describe the shape of  this surface near near that point so for this this lagrange kind of expansion  is very useful for the following reason   so we're going to imagine well we'll just take this is the basic formula there and   what we can do is we can we can truncate  that okay so to find approximations to the curve p of x at the point r we truncate our all-important formula star what that  means is we're going to consider a number   of approximations and we're going to give  them names so the first the first one   this will all be at the point r the first one  or the zeroth one we'll put a zero up here   of p is just by definition uh d0 of p  which is just at r which is just p of r okay t sub r one of p will be what we get when we  take i might just write okay d0 of p of r plus d1   p of r times x minus r and d two p is the sum of the first three  terms all quantities up to degree two and in general we take t r k if  p is the sum of all terms in star up 2 and including degree a k in x minus r and we'll call this thing we'll call this the kth tangent the kth tangent to p at r okay let's go back to our example and all right so for our original curve which  was p of x equals eight minus five x   plus four x cubed minus x to the fourth  for that example the star looked like well i'll write it again so p of x equals eight minus five r plus four  r cubed minus r to the fourth   plus minus five plus twelve r  squared minus four r cubed x minus r plus four minus four r  times x minus r cubed and uh well there was a there was a square  term here let me squeeze it in here um   12 r minus 6 r squared  times x minus r squared plus   4 minus 4 r times alpha minus x minus r  cubed and minus x minus r to the fourth all right so what do these various  tangents to p and r look like   so the zeroth tangent at some point r is just this it's just a number that's just p of r it's just a number the  first tangent is well we have to take the   sum of the first two terms so we have to take  eight minus five r plus four r cubed minus r   to the fourth plus minus five plus twelve r  squared minus four r cubed times x minus r that's an equation that's linear in x actually we can if we want to sort  of simplify it because we could now   expand this out if we wanted to and we would get eight minus eight r cubed plus  three r to the fourth plus minus   five minus four r cubed plus 12 r squared times x that's then an equation of a line  whose coefficients depend on r okay what about d sub r two the second tangent to the curve well  we have to take the sum of the first   terms up to degree two including degree  two so we have to take eight minus five   r plus four r cubed minus r to the fourth plus  minus five plus twelve r squared minus four   r cubed times x minus r plus 12 r minus  six r squared times x minus r squared that's going to be a quadratic and you  can expand that out too if you want to i'm going to just tell you what  it is okay it turns out to be   8 plus 4 r cubed minus three r to the  fourth plus minus five minus twelve   r squared plus eight r cubed times x plus  twelve r minus six r squared times x squared and we could keep going if we went to the next  one since uh well we would get a cubic maybe i'll   tell you what that is i'll write just uh write  down the simplified form so the third one is   eight plus r to the fourth minus five plus four  r cubed times x plus six r squared x squared   plus four minus four r [Applause] times x cubed  and the the fourth one well that would just be   p itself that would just be p of  x because that would be everything so what we have now is a hierarchy of  approximations to the original curve   the hierarchy obtained simply by truncating  this taylor or taylor lagrange polynomial   just truncating it up to whatever number of  degree terms we want if we go just one term we're   just getting a constant function which is the  value of the polynomial at the point r if we go   to the first terms we get an equation  of a line and that's the tangent line here it is that's the tangent line to the polynomial p at the point r and the second degree one is the tangent conic we'll give it this name we'll  call it the tangent conic to p at r and this third one we'll call  it well we can call it the   three tangent or we'll call  it the tangent cubic to p at r okay so this is very very important  for differential geometry but   we're interested in in curves  and surfaces we're interested in   understanding what things look like near  a point here we have a succession of   increasingly sophisticated approximations to  a curve at a given point starting with the   function's value then going to the tangent plane  and then introducing this perhaps new idea to you   it's almost surely new to most of us the idea  of the tangent conic to the curve at a point and then one step further tangent  cubic and we can keep going if we want   getting better and better approximations  to a curve at a point let's have a look   to see what it actually looks  like for our running example so okay okay so here's uh this kind of thing can be done of  course very pleasantly with geogebra all right so maybe here's 20   minus 20 one two three four this one minus  two minus three all right so the curve itself watch there is p of x i remind you that was  eight minus five x plus four x cubed minus x   to the fourth that fourth degree polynomial  let's choose the point r equals minus one just as an example so r equals minus 1. okay so up here at -1 the function's value  is uh eight so the value actually is eight   okay so the the zeroth at minus one the  zeroth approximation to the polynomial   it's just the number eight represented by  just that that dot or if you like you can   think of it as a constant function if  you like okay then at the tangent line   if we take the equation for the tangent  line and plug in r equals minus 1 we get this tangent line which looks like this all right so that's t that's the tangent  minus 1 the first tangent of p that is then   has equation 19 plus 11x that's the equation of the line  there it is there the tangent conic okay here's the tangent conic looks like approximates it better okay it is so here's the tangent conic which  is t at minus one second degree of p and that   is equal to right in this color  so you can see one minus 25 x minus 18 x squared that's the tangent conic that's the tangent line there's also a tangent cubic which is uh  over here if i plug in r equals minus 1   to that i'm going to get an equation  of a cubic what's that going to be it will have the form something like that so that's  the tangent cubic of p which is 9 minus x plus 16 x squared plus eight x cubed all right so here's a little problem for you problem calculate and graph for the tangent curves 2p the point r equals 3. well that point over there okay so so why is this uh interesting well  admittedly there was a bit of work there   was a bit of algebra right we had to do  some fiddling around with polynomials but   in some sense that's completely trivial kind of  arithmetic these days a computer algebra system   does that stuff just like your calculator does  you know multiplies big numbers simplifying   polynomials expanding them out grouping  terms these are all ultimately almost   trivial operations these days with a with  a calculator with a computer algebra system   like maple or mathematica or matlab or even  geogebra will do some of that for us right so all   yeah so previous generations would have found some  of this owners but these days you can get your   computer to do this almost instantaneously you can  generate these kinds of pictures very very easily   the point is we are not using limits we've  avoided any discussion of limits we've   avoided any prior understanding of real numbers  everything's just been over the rational numbers   and it's all really been with high school algebra  and what have we done we've established well the   derivative essentially the higher derivatives  and we've it's been easier for us to get at these   these curves in in the classical setup  to get the tangent line first you have to   compute the derivative and then you have  to do some analytic geometry to get the   tangent line here the tangent line is just just  what you get by truncating the taylor expansion   lagrange taylor expansion the tangent  conic is just what you get by truncating   the crucial operation is truncation okay but  there are some other very important consequences [Applause]   and we'll we'll talk a little bit about these   in further lectures so one important  consequence is that everything is algebraic basically we're only using addition  multiplication subtraction and division   so it really works over a general field so the discussion holds over a general field in particular it holds over a finite field  okay so we are now in the position of being   able to do calculus over a finite field and i'll  show you some examples of that perhaps next time   even even looking at this over finite  field makes perfectly good sense secondly this generalizes to   more dimensions eat more readily than the  usual business so it generalizes more smoothly to not just one dimensional calculus  but two dimensional calculus to   to calculus of several variables so calculus of several variables is something  that you usually only do in a second course   second university course in calculus but we'll see  that this approach means that you can basically   almost immediately start thinking about  cal functions of more than one variable another is that the importance of the tangent  conic i think comes out okay so these tangent   conics are remarkable things obviously if you're interested in  approximating a curve you shouldn't   just restrict yourself to the tangent line yes  the tangent line is a useful first approximation   but for finer more refined work you  may very well prefer working with   this tangent conic which approximates the  curve better than the tangent line does that's going to be a very very important idea  for us in differential geometry because in   some sense a lot of differential geometry  is really built on the idea that we should   approximate curves and surfaces quadratically  not linearly but quadratically how we do that   and with the consequences of that that's  really a core issue in differential geometry and next time i'm going to show you what i  consider the the most beautiful cal most beautiful   theorem in in calculus okay which is a direct  um a direct consequence of looking at this idea   of tangent conics and will actually um it's just  an absolutely remarkable thing that was uh first   observed by atn geese so we'll look at that  in the context of oscillating curves so   lots of interesting potential for this students  of calculus give it some thought okay and purely   algebraic approach going back to the masters  oiler and lagrange see you tomorrow [Music] you

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Length: 48min 27sec (2907 seconds)

Published: Thu Aug 15 2013