CLASS NOTES. Contents 1. The Total Derivative 1 2. The Chain Rule 4 3. Multi-variable Taylor Expansions 7. f(a + h) f(a) f (a)h

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1 CLASS NOTES JOHN B ETNYRE Contents 1 The Total Deivative 1 2 The Chain Rule 4 3 Multi-vaiable Taylo Expansions 7 1 The Total Deivative Recall, fom calculus I, that if f : R R is a function then f f(a + h f(a (a = h 0 h We can ewite this as f(a + h f(a f (ah = 0 h 0 h Witten this way we could then say that f is diffeentiable at a if thee is a numbe λ R such that f(a + h f(a λh = 0 h 0 h Then if such a numbe λ exists we define f (a = λ This might seem a bit of a ound about way to define the deivative, but it genealizes nicely to functions of moe than one vaiable The way to intepet this definition is that f (a is the best linea appoximation to f(x at a (o moe pecisely, the best linea appoximation to f(x + a f(a at 0 Now if f : D R n R m is a function and a D then genealizing the idea above we want the deivative of f at a to be a linea map that appoximates f at a Recall, a linea map fom R n to R m can be given by an m n matix T Indeed we say f : D R n R m is diffeentiable at a in D if thee is an n m matix T, which we think of as a linea map T : R n R m, such that f(a h f(a Th = 0 h 0 h If T exists it is called the total deivative of f at a and we wite Df(a = T 1

2 2 JOHN B ETNYRE Theoem 11 If f : R n R m is a diffeentiable function at a then f is continuous at a Example 12 Suppose f : R n R m is given by f(x = Bx + b, whee B is an m n-matix and b R m Fo example if B = then [ ] f(x, y, z = x [ y 1 + = ] z We claim that Df(a = B fo all a To check this note So indeed, Df(a = B f(a + h f(a Bh = h 0 h [ ] 2x + 3y + 4z + 1 x + y + 2z + 2 (B(a + h b (Ba + b Bh = h 0 h 0 h 0 h = 0 [ ] [ and b = ] Example 13 Let f : R n R be a function and u be a unit vecto Suppose that f is diffeentiable at a Then so Thus f(a + hu f(a Df(a(hu = 0, by definition h 0 hu = h ( f(a + hu f(a Df(a(u = 0 h 0 h f(a + hu f(a [Df(a]u = = D u f(a = [ f(a] u h 0 h Whee D u f(a is the diectional deivative of f in the diection of u Since it is easy to see check that if v and w ae two vectos then v w = v t w, whee v t means the tanspose of v, we have [Df(a]u = [ f(a] t u We fomalize this in a theoem Theoem 14 If f : D R n R has continuous fist patial deivatives in a neighbohood of some point a in D, then Df(a = [ f(a] t

3 CLASS NOTES 3 Example 15 Find the total deivative of f(x, y = sin xy + x 2 y at the point (1, π In geneal we have Df(1, π = [ f f (1, π ] (1, π y Theoem 16 Let f : R n R m be given by = [ (y cosxy + 2xy (x cosxy + x 2 ] (1,π = [ π 0 ] f(x = (f 1 (x,,f m (x, whee f i : R n R, i = 1,,m If f i j is continuous at a fo all i and j, then f is diffeentiable at a and the total deivative is f 1 f 1 (a 1 n (a Df(a = f m 1 (a f m n (a That is, Df(a is the m n-matix with the i th ow being the tanspose of the gadient of f i Example 17 Let f(x, y = (e xy + x 2 y, x 2 + y 2, x Fine the total deivative of f at y (1, 1 Befoe we stat this poblem let s note that f is a function fom R 2 to R 3 so we expect to get a 3 2-matix as ou answe We can think of f as f = (f 1 (x, y, f 2 (x, y, f 3 (x, y, whee f 1 (x, y = e xy + x 2 y, f 2 (x, y = x 2 + y 2, and f 3 (x, y = x Then the matix in y the theoem is f 1 f 1 y (ye xy + 2xy (xe xy + x 2 f 2 f 2 = 2x 2y f 3 y f 3 y 1 y Since all the patial deivatives in this matix ae continuous at (1, 1 we can just evaluate the tems at (1, 1 to compute the total deivative So Df(1, 1 = ( e 2 2 (e Example 18 Let f(x, y = (x cosy, x sin y Find the total deivative We know the patials of the functions x cos y and x sin y ae continuous so [ ] cosy x sin y Df = sin y x cos y x y 2

4 4 JOHN B ETNYRE 2 The Chain Rule Now suppose you have two functions f : R n R p and g : R p R m If you compose these functions you get g f : R n R m We know that Df(a is the best linea appoximation to f at a and that Dg(f(a is the best linea appoximation to g at f(a So what is the best linealy appoximation to g f at a? Well, you would expect it just to be the composition of the othe two linea appoximations and compositions of linea maps coesponding to matices is achieved by matix multiplications This leads to the multi-vaiable chain ule Theoem 21 (Mutli-Vaiable Chain Rule Let f : R n R p and g : R p R m be two functions If f is diffeentiable at a and g is diffeentiable at f(a then g f is diffeentiable at a and D(g f(a = [ Dg(f(a ][ Df(a ] Example 22 Let f(x, y = (x 2 + y 2, xy and g(x, y = (4xy, x y, 3x 2 + 2y 2 Find the total deivative of g f at ( 3, 1 Use the chain ule to compute this D(g f( 3, 1 = [ Dg(f( 3, 1 ][ Df( 3, 1 ] = [Dg(10, 3] [ Df( 3, 1 ] 4y 4x [ ] = 1 1 2x 2y y x 6x 4y ( 3,1 (10, ] = 1 1 [ = Example 23 Hee we will check that a special case of the chain ule used when studying the gadient follows fom the geneal chain ule To this end let f : R R n and g : R n R and wite f(t = (f 1 (t,,f n (t Then we have d dt (g f(t = D(g f(t = [ Dg(f(t ][ Df(t ] = = [ g 1 (f(t ] g n (f(t ( g f1 (f(t 1 t (t + + f 1 (t t f n (t t ( g (f(t n fn t (t Example 24 Hee we show how to use the multi-vaiable chain ule to compute deivatives in diffeent coodinate systems Recall catesian coodinates (x, y ae elated to pola coodinates (, θ by x = cosθ, y = sin θ

5 CLASS NOTES 5 and = x 2 + y 2, θ = tan 1 x y Now suppose we ae given a function expessed in xy-coodinates, like g(x, y = x 2 y 3 We can compute the total deivative of g Dg(x, y = [ 2xy 3 3x 2 y 2] But this is the total deivative with espect to the coodinates x and y What about the deivative with espect to the coodinates and θ? We could plug x = cosθ and y = sin θ into g to expess g in tems of and θ Then we could compute the total deivative of g with espect to and θ It tuns out that it is just as easy (and in othe cases easie! to use the multi-vaiable chain ule To this end conside f(, θ = ( cosθ, sin θ Then g thought of as a function of and θ is just Now the chain ule gives us g(, θ = g f(, θ D(g f(, θ = [Dg(f(, θ][df(, θ] = [ [ ] 2xy 3 3x 2 y 2] cos θ sin θ ( cos θ, sin θ sin θ cos θ = [ 2( cos θ( sin θ 3 3( cosθ 2 ( sin θ 2][ cos θ sin θ sin θ cos θ = [ (2 4 cos 2 θ sin 3 θ cos 2 θ sin 3 θ ( 2 5 cosθ sin 4 θ cos 3 θ sin 2 θ ] (,θ = [ (5 4 cos 2 θ sin 3 θ ( 5 cosθ sin 2 θ(3 cos 2 θ 2 sin 2 θ ] This is the total deivative with espect to and θ Notice, now that we have this woked out we know what g g and ae: θ g = 54 cos 2 θ sin 3 g θ, θ = 5 cosθ sin 2 θ(3 cos 2 θ 2 sin 2 θ Again this poblem could have been woked out by just witing g in tems of and θ and computing the deivative, but you cannot always do this Fo example, in the next poblem we don t know what ou function is, just that it satisfies an equation in xy-coodinates and we want to see what the coesponding equation is in θ-coodinate Example 25 A function u : R 2 R is said to satisfy Laplace s equation if 2 u u y 2 = 0 This equation is used in heat conduction, electodynamics and many othe aeas It is fequently useful to have Laplace s equation witten in pola coodinates To do this we need to figue out how to wite the patial deivatives with espect to x and y ]

6 6 JOHN B ETNYRE in tems of patial deivatives with espect to and θ So again conside the change of coodinates f(, θ = ( cosθ, sin θ Now as above So Du(, θ = D(u f(, θ = [Du] f(,θ [Df(, θ] [ ] [ θ = ] y f(,θ [ ] cosθ sin θ sin θ cosθ Denote the 2 2-matix on the ight by A and notice that [ ] cosθ sin θ A 1 = Thus [ ] y = [ f(,θ θ sin θ cos θ ] [ ] cos θ sin θ sinθ = [ cosθ sinθ θ cos θ sin θ + cos θ θ] So = cos θ sin θ θ, and = sin θ y + cosθ θ So we have the fist deivatives of u in tems of x and y witten in tems of and θ Of couse these same fomulas hold fo a function v as well So if v = we have 2 u = ( = 2 v v θ = cosθ v sin θ = cosθ sin θ θ = cosθ ( cos θ sin θ sin θ θ = cosθ v sin θ θ v ( cosθ θ sin θ θ = cos 2 θ 2 u sin θ 2 u 2cosθ sin2 θ 2 u 2 θ + sin2 θ cosθ + 2sin 2 2 θ Similaly Thus 2 u y = 2 sin2 θ 2 u sin θ 2 u + 2cosθ cos2 θ 2 u 2 θ + cos2 θ cos θ 2sin 2 2 θ 2 u + 2 u 2 y = 2 u u 2 2 θ + 1 2

7 And finally, Laplace s equation becomes CLASS NOTES 7 2 u u 2 2 θ = 0 3 Multi-vaiable Taylo Expansions Recall, fom calculus I, that the single vaiable Taylo polynomial is the following: given a function f : R R that is (k + 1-times diffeentiable nea a then the k th ode Taylo polynomial of f about a is P k f,a(x = f(a + f (a(x a f (a(x a k! f(k (a(x a k which we can wite moe succinctly This polynomial satisfies P k f,a (x = k i=0 ( i 1 d f(a(x a i i! dx f(x P k f,a(x K x a k+1 fo some constant K and x nea a In othe wods, P k f,a (x is the kth ode polynomial that best appoximates f nea a The Taylo seies is just P f,a(x k k We will see that a fomula almost identical to the boxed one above will also give the multi-vaiable Taylo polynomial The difficult pat is intepeting what this should mean in a multi-vaiable context If we ae in dimension n then a mutli-index I is an odeed sequence of n non-negative integes I = (i 1,,i n If x = (x 1,,x n is a point in R n then define x I = x i 1 1 x i 2 2 x in n Example 31 Fo example if x = (x, y, z then x (1,0,0 = x 1 y 0 z 0 = x, x (2,0,3 = x 2 y 0 z 3 = x 2 z 3, x (1,1,1 = xyz The degee of a multi-index I = (i 1,, i n is I = i i n

8 8 JOHN B ETNYRE Example 32 We can wite any multi-vaible polynomial in multi-index notation Fo example xy 2 + xz + 3y + 7 can be witten whee, of couse, x = (x, y, z x (1,2,0 + x (1,0,1 + 3x (0,1,0 + 7x (0,0,0, Example 33 A geneal second ode polynomial in two vaiables can be witten which we can wite (a (2,0 x 2 + a (1,1 xy + a (0,2 y 2 + (a (1,0 x + a (0,1 y + a (0,0, a I x I + a I x I + a I x I, I =2 I =1 whee, of couse, x = (x, y Note all the multi-indices in two dimensions with degee 2 ae (2, 0, (1, 1, (0, 2 So the fist sum in this fomula is ove these thee multiindices and give the fist thee tems in the polynomial Similaly the multi-indices with degee 1 ae (1, 0, (0, 1 and these two tems give us the next two tems in the polynomial Finally the constant tem is given fom them multi-index (0, 0 of degee 0 Now we can wite this last expession fo the polynomial even moe succinctly as 2 a I x I i=0 I =i Note this same fomula gives an expession fo a second ode polynomial in any numbe of vaiables The only diffeence is the multi-indecies which equal a given degee will be diffeent Fo example in 3 dimensions, the multi-indices of degee 2 ae (2, 0, 0, (0, 2, 0, (0, 0, 2, (1, 1, 0, (1, 0,1, (0,1, 1 I =0 In geneal, a k th ode polynomial in the vaiable x = (x 1,, x n is just k a I x I i=0 I =i We ae almost eady fo the multi-vaiable Taylo polynomial, we just need a few moe definitions If I = (i 1,,i n then We abbeviate i by D i, in othe wods Finally, I! = (i 1!(i 2! (i n! D i f = f i ( i1 ( i2 ( in D I f = D i 1 1 D i 2 2 Dn in f = f 1 2 n

9 Example 34 Fo example if I = (2, 1, 1 then CLASS NOTES 9 D I (x 5 y 2 z 3 = ( 2 ( 1 ( 1 (x 5 y 2 z 3 y z = 5 4x 3 2y3z 2 = 120x 3 yz 2 We can finally state: Theoem 35 (Multi-vaiable Taylo Polynomials Let f : D R n R be a function with continuous mixed patial deivatives of ode less than o equal to k +1 nea a Then k Pf,a(x k 1 = I! DI f(a(x a I i=0 i =i is the best k th ode polynomial appoximation of f nea a in the sense that fo some constant K and all x nea a f(x P k f,a (x K x a k+1, Notice that this theoem is almost identical to the one fo single vaiable Taylo polynomials The main diffeence is intepeting eveything in tems of multi-indices Example 36 Let f : R 2 R be a function of two vaiables Wite out the 2 nd ode Taylo polynomial at a = (a, b 2 Pf,(a,b 2 (x, y = 1 I! DI f(a(x a I i=0 i =i = I =0 1 I! DI f(a(x a I + I =1 1 I! DI f(a(x a I + I =2 1 I! DI f(a(x a I

10 10 JOHN B ETNYRE Expanding this out we get Pf,(a,b 2 = 1 0!0! D(0,0 f(a, b((x, y (a, b (0,0 ( 1 + 1!0! D(1,0 f(a, b((x, y (a, b (1, !1! D(0,1 f(a, b((x, y (a, b (0,1 ( 1 + 2!0! D(2,0 f(a, b((x, y (a, b (2, !1! D(1,1 f(a, b((x, y (a, b (1, !2! D(0,2 f(a, b((x, y (a, b (0,2 = f(a(x a 0 (y b 0 ( f + (a, b(x a1 (y b 0 + f y (a, b(x a0 (y b 1 ( 1 2 f + 2 (a, b(x 2 a2 (y b f y (a, b(x 11 (y f 2 y (a, b(x 2 a0 (y b 2 = f(a, b + f f (a, b(x a + (a, b(y b y f 2 (a, b(x 2 a2 + 2 f y (a, b(x a(y b f 2 y (a, b(y 2 b2 OK that took some wok Let s look at a specific function Example 37 Compute the 2 nd ode Taylo polynomial of f(x, y = sin(x + y 2 at (0, 0 We know fom that last example that we need to compute all deivatives up to ode 2 and evaluate them at (0, 0 We have f = cos(x + y2, f y = 2y cos(x + y2, 2 f = sin(x + 2 y2, 2 f y = 2y sin(x + y2, and 2 f y = 2 cos(x + 2 y2 4y 2 sin(x + y 2 When we evaluate at (0, 0 we get f(0, 0 = 0, f f (0, 0 = 1, (0, 0 = 0, y

11 CLASS NOTES 11 and 2 f 2 f 2 f (0, 0 = 0, (0, 0 = 0, (0, 0 = 2 2 y y2 Plugging this into the fomula fom the last execise we get Pf,(0,0 2 (x, y = 0 + 1(x 0 + 0(y 0 + 0(x (x 0(y (y 02 2 = x + y 2 This is a faily tedious way to compute Taylo polynomials Let s find some shotcuts Fist, ecall the standad Taylo polynomials fom single vaiable calculus Moe pecisely we have the Taylo seies: and e x = 1 + x x2 + = sin x = x 1 3! x ! x5 + = cosx = x ! x4 + = ln(1 + x = x 1 2 x2 + = n=0 n=0 x n n!, ( 1 n x 2n+1 (2n + 1!, n=0 n=1 ( 1 n x2n (2n!, ( 1 n+1xn n Now a theoem that helps us put Taylo polynomials togethe is the following Theoem 38 We have the following (1 If f is a polynomial then P k is all the tems in f of degee less than o equal f,0 to k (2 Pf±g,a k = P f,a k ± P g,a k, (3 Pfg,a k is the tems of degee less than o equal to k in P f,a k P g,a, k (4 Pf g,a k is the tems of degee less than o equal to k in P f,g(a k P g,a k Example 39 Compute the ode 4 Taylo polynomial of f(x, y = sin(x + y 2 at (0, 0 Note we can think of f(x, y as g h(x, y whee g(x = sin x and h(x, y = x + y 2 Fom item (1 in the theoem we know the Taylo seies of a polynomial is just that polynomial so P 4 h,(0,0 = x + y2 Also h(0, 0 = 0 so fom item (4 in the theoem we ae inteested in the Taylo polynomial of g at 0 = h(0, 0 Fom above we know the ode 4 Taylo polynomial of g(x at 0 is P 4 g,0(x = x x3 6

12 12 JOHN B ETNYRE Let s compute P 4 g,0 P 4 h,(0,0 P 4 g,0 P 4 h,(0,0 = P 4 h,(0,0 1 6 ( P 4 h,(0,0 3 = (x + y (x + y2 3 = (x + y (x3 + 3x 2 y 2 + 3xy 4 + y 6 We want all the tems of degee less than o equal to 4 So we have P 4 f,(0,0(x, y = x + y x3 1 2 x2 y 2 Example 310 Compute the ode 2 Taylo polynomial of f(x, y = cos(πe x cosy at (0, 0 We will think of f(x, y as the composition g h(x, y whee g(x = cosx and h(x, y = πe x cosy Now Ph,(0,0 2 (x, y is the tems of ode less than o equal to 2 in Pe 2 x,0 P cos 2 y,0 We have Pe 2,0P 2 x cos y,0 = π (1 + x + x2 2 Thus = π (1 y2 2 [1 y2 2 + x xy ( Ph,(0,0 2 (x, y = π 1 + x y2 2 + x2 2 ] (x 2 x2 y 2 Now h(0, 0 = π so we need to find the expansion of g(x at π We know cosx = ( 1 n x2n (2n!, so but Thus and cos(x π = n=0 ( 1 n(x π2n, (2n! n=0 cos(x π = cosxcosπ sin x sin π = cosx cosx = ( 1 n(x π2n, (2n! n=0 P 2 g,π = (x π2 2

13 We now need to conside Pg,π 2 Ph,(0,0 2 : Pg,π 2 Ph,(0,0 2 = (π(1 + x y x2 = 1 + π2 2 CLASS NOTES π (x 2 y 2 x + x 3 + y4 4 x2 y 2 + x4 2 4 Taking the tems of ode less than o equal to 2 we have P 2 f,(0,0(x, y = 1 + π2 2 x2