proof of Taylor expansion

Proof of Taylor's Expansion Theorem PhL 8.24.08 1. The Long Proof. I didn't realize this was non-trivial until today. I am reading on page 124 of Ahlfors book. Let's start with what is called Cauchy's Formula F(z) = (1/2πi) ∫C dy F(y) /(y-z) where C is a contour which goes clockwise around the pole in the y plane. This thing is only true as long as the function F(y) is itself not singular at the point y = z. In this case, we get that the contour integral picks up the normal pole residue without any derivatives required. Notice that the integrand F(y) /(y-z) has a pole at the point y = z, but the contour stays away from the pole, so nothing dramatic happens during the integration, it is all well defined. Now consider for our function F the following function F(z) = [ f(z) - f(a) ] / (z-a) where f(z) is some function which is analytic in our entire region of interest. We are concerned that this function F(z) is not analytic right at the point z = a because we have a 0/0 situation there. But suppose we know that F(z) is at least finite at this point z = a, limz→a F(z) = f '(a) = some finite number since f(z) is analytic. Then the above Cauchy Formula is valid for the function F. Again, it says F(z) = (1/2πi) ∫C dy F(y) /(y-z) = (1/2πi) ∫C dy [ f(y) - f(a) ] / [ (y-a)(y-z) ] and here is a picture showing the contour and the two poles Because the contour stays away from both poles, nothing singular happens during the integration over C. If the two poles coincide, fine, nothing special happens, the integral is still well defined. The function of z represented by the integral is analytic at z = a. Alfors wants to call this integral f1(z). It is identical with F(z) everywhere other than z = a, and it is well defined and analytic also at z=a. Therefore, the technical point he is making is this: f1(z) is an "extension" of the function F(z) which was undefined at z=0 and this extension is defined at z=a and is in fact analytic there. So we write the above Cauchy formula as [ f(z) - f(a) ] / (z-a) = f1(z) = analytic at z=a. We then have, given that f(z) is continuous and differentiable at point a, ie, f '(a) exists and is finite, we can write f(z) = f(a) + (z-a)f1(z) This seems like a stupid trivial result but here is what it says: if f(z) is analytic in our disk, then it can be written as shown where f1(z) is also analytic in the disk. Now do this again making the replacement f → f1. After all, f1 is analytic just as f was. So: f1(z) = f1(a) + (z-a)f2(z) where f2 is the name we give to the integral in this case, and of course f2 is analytic in our disk. Now combine these to get f(z) = f(a) + (z-a)[ f1(a) + (z-a)f2(z)] = f(a) + (z-a) f1(a) + (z-a)2 f2(z) Obviously we can keep going, and we get this series f(z) = f(a) + (z-a) f1(a) + (z-a)2 f2(a) + (z-a)3 f3(a) + (z-a)4 f3(a) + ..... = Σn=0,∞ (z-a)n fn(a) where we now call f = f0 just so we can make this sum. Then f'(z) = Σn=1,∞ n(z-a)n-1 fn(a) = 1 f1(a) + Σn=2,∞ n(z-a)n-1 fn(a) f"(z) = Σn=2,∞ n(n-1)(z-a)n-2 fn(a) = 2*1 f2(a) + Σn=3,∞ n(n-1)(z-a)n-2 fn(a) f(m)(z) = Σn=m,∞ n(n-1)...(n-m+1)(z-a)n-m fn(a) = m! fm(a) + Σn=m+1,∞ n(n-1)...(n-m+1)(z-a)n-m fn(a) Therefore, when we go to z=a, the residual series vanishes in each case and we then get f(m)(a) = m! fm(a) => fm(a) = f(m)(a)/ m! and then we can write our series as f(z) = Σn=0,∞ (z-a)n f(n)(a)/ n! and this is the Taylor series expansion around the point z = a. 2. The Short Proof. Just expand f(z) in a power series around f(a) f(z) = f(a) + k1 (z-a) + k2 (z-a)2 + k3(z-a)3 + .... which we know we can do because f(z) is analytic Now do m derivatives of f(z) to find this pattern: f(m)(a) = m! km and then we get our Taylor expansion. 3. Do this in two variables f(x,y) = f(a,b) + k10 (x-a) + k01(y-b) + k20(x-a)2 + k11(x-a) (y-b) + k02(y-b)2 + k30(x-a)3 + k21(x-a)2 (y-b) + k12(x-a) (y-b)2 + k03(y-b)3 Comment: Motivation for assuming the above form: First, treat y as a bystander variable and expand f(x,y) around x = a using 1D Taylor series, In this little comment, the kij functions are not necessarily the same as above, they are just nameless functions. f(x,y) = f(a,y) + k1(y) (x-a) + k2(y) (x-a)2 + k3(y)(x-a)3 + .... or f(x,y) = f(a,y) + k1(y) (x-a) + k2(y) (x-a)2 + k3(y) (x-a)3 + .... Now expand each function of y in a 1D Taylor series about y = b f(x,y) = [ f(a,b) + s1(b) (y-b) + s2(b) (y-b)2 + s3(y)(y-b)3 + ....] + [ k1(a,b) + k11(b) (y-b) + k12(y) (y-b)2 + k13(y)(y-b)3 + ...] (x-a) + [ k2(a,b) + k21(b) (y-b) + k22(y) (y-b)2 + k23(y)(y-b)3 + ...] (x-a)2 + [ k3(a,b) + k31(b) (y-b) + k32(y) (y-b)2 + k33(y)(y-b)3 + ...](x-a)3 + .... + ... Now assume appropriate convergence so can reorder terms: f(x,y) = f(a,b) + k1(a,b)(x-a) + s1(b) (y-b) // lead term plus linear terms + k2(a,b) (x-a)2 + k11(b) (x-a) (y-b) + s2(b) (y-b)2 + k3(a,b) (x-a)3 + k21(b) (x-a)2 (y-b) + k12(y) (x-a) (y-b)2 + s3(y)(y-b)3 + ... This then shows why you might get all those different combinations at each power level. It just comes from doing this double expansion. So far we don't know how to evaluate the coefficients. So we then assume this general form (new k's), f(x,y) = f(a,b) + k10 (x-a) + k01(y-b) + k20(x-a)2 + k11(x-a) (y-b) + k02(y-b)2 + k30(x-a)3 + k21(x-a)2 (y-b) + k12(x-a) (y-b)2 + k03(y-b)3 Notice that {∂2x ∂1y f}|a,b = 2 k21 A general term in the series will have this form Tnm = knm (x-a)n (y-b)m and notice that {∂nx ∂my f}|a,b = n! m! knm Any lesser derivative acting Tnm gives 0 when we evaluate at x=a and y=b. Any greater derivative acting Tnm gives 0 because derivatives above ∂nx ∂my act on constant. Thus, you can say {∂nx ∂my f}|a,b Tn'm' = δn,n' δm,m' {∂nx ∂my f}|a,b = δn,n' δm,m'( n! m! knm ) Now rewrite the original series as a sum of its Tnm terms f(x,y) = Σn=0∞ Σm=0∞ Tnm(x,y) Then we find that -- assuming we can do term-by-term differentiation -- {∂n'x ∂m'y f}|a,b = Σn=0∞ Σm=0∞ {∂n'x ∂m'y f}|a,b Tnm(x,y) = Σn=0∞ Σm=0∞ δn,n' δm,m'( n! m! knm ) = ( n'! m'! kn'm' ) Thus we have proven that (removing primes), {∂nx ∂my f}|a,b = n! m! knm , and our conclusion is this for the double Taylor series : f(x,y) = Σn=0∞ Σm=0∞ Tnm(x,y) = Σn=0∞ Σm=0∞ knm (x-a)n (y-b)m = Σn=0∞ Σm=0∞ (x-a)n (y-b)m Comment. Suppose in the above we have n=1 and m=1. That term is supposed to be { (∂2f/∂x∂y)|a,b /( 1! 1!) } (x-a)1 (y-b)1 = (∂2f/∂x∂y)|a,b (x-a)(y-b) and this is correct because there is only one such term. Now suppose we regard x = x1 and y = x2. Then the above expansion is written f(x1, x2) = Σn,m=0,∞ knm(x1-a)n (x2-b)m = Σn,m=0,∞ { ∂n+mf/∂nx1∂mx2|a,b /( n! m!) } (x1-a)n (x2-b)m = Σn,m=0,∞ { (∂1n∂2mf) |a,b /( n! m!) } (x1-a)n (x2-b)m The n=m=1 term still gives a single term. In the derivative denominator, we never vary the order of the variables. For example, we never have ∂n+mf/∂nx2∂mx1 and we never have ∂2n∂1mf. It seems clear how you would extend this to three 3 variables: f(x1, x2, x3) = Σn1,n2,n3=0,∞ kn1,n2,n3 (x1-x10)n1 (x2 - x20)n2 (x3 - x30)n3 = Σn1,n2,n3=0,∞ { ∂n1+n2+n3f/∂n1x1∂n2x2∂n3x3|x10,x20,x30 /( n1! n2! n3!) } (x1-x10)n1 (x2 - x20)n2 (x3 - x30)n3 which I alternately write as f(x,y,z) = Σn=0∞ Σm=0∞ Σs=0∞ (x-a)n (y-b)m (z-c)s 4. Do it for a vector function of multiple variables We just do the above for each component of the vector, fi(x,y) = Σn,m=0,∞ kinm(x-a)n (y-b)m = Σn,m=0,∞ { ∂n+mfi/∂nx∂my|a,b /( n! m!) } (x-a)n (y-b)m 5. Do it for a matrix function of multiple variables Again, just do it for each component: fij(x,y) = Σn,m=0,∞ kijnm(x-a)n (y-b)m = Σn,m=0,∞ { ∂n+mfij/∂nx∂my|a,b /( n! m!) } (x-a)n (y-b)m 6. Do it for a matrix function of a matrix The variables are now just the elements of the matrix which you list in some order. Here for a 2x2 we could say fij (x11,x12; x21, x22) = Σn11,n12,n21,n22 = 0,∞ { ∂n11+n12+n21+n22 fij/∂n11x11∂n12x12∂n21x21∂n21x22|x110,x210,x120,x220 /( n11! n12! n21! n22!) } (x11-a11)n11 (x12-a12)n12 (x21-a21)n21 (x22-a22)n22 = Σn11,n12,n21,n22 = 0,∞ kijn11,n12.... (δx11)n11 (δx12)n12..... Now suppose all the displacements are of the same general small magnitude perhaps λ, what would be the order λ term in the above expansion? We can list off the terms: { n11, n12, n21,n22} = {1000} + {0100}+ {0010}+ {0001} = ∂fij/∂x11 (x11-a11) + ∂fij/∂x12 (x12-a12) + ∂fij/∂x11 (x21-a21) + ∂f/∂x22 (x22-a22) = Σrs=1,2 (∂fij/∂xrs) (xrs-ars) = Σrs=1,2 kijrs δxrs Now for fun, let's go for the λ2 term: { n11, n12, n21,n22} = {1100} + {1010}+ {1001}+ {0110} + {0101} + {0011} + {2000} + {0200}+ {0020}+ {0002} = ( ∂2fij/∂x11∂x12) δx11δx12 +5 similar + (∂2fij/∂2x11)/2!(δx11)2 + 3 similar Try to state in more general terms, for an NxN matrix: fij (x11,x12... .) = Σn11,n12... = 0,∞ ( kijn11,n12....) (δx11)n11 (δx12)n12..... Now let's say that (δx11) = λ(δy11) just to expose the smallness. Then we get fij (x11,x12... .) = Σn11,n12... = 0,∞ ( kijn11,n12....) (δy11)n11 (δy12)n12..... λn11+n12... So the point is that if we want the order M contribution in this sum, we need to do this fij (x11,x12... .) = Σn11,n12.. ( kijn11,n12....) (δx11)n11 (δx12)n12..... where we restrict the sum to values such that n11+n12 + ... = M. Conclusion: it is not easy to write the terms higher than order λ=1 in some kind of closed compact form. However, it is easy to do it for the λ=1 terms. For example, with a 3x3 matrix there will be 9 terms which we could indicated by {100 000 000} and so on, sliding the 1. There are nine possible δx objects and we can write fij (x11,x12...) = Σrs=1,3 (∂fij/∂xrs) (xrs-ars) = Σrs=1,3 kijrs δxrs and we can even put this into a sort of matrix notation F(X)first order = Σrs=1,3 (∂F/∂Xrs) (Xrs-Ars) = Σrs=1,3 Krs δXrs where F, X and each of the Krs objects is a 3x3 matrix, and δXrs are just numbers. Don't make the mistake of thinking Krs is an element of a matrix, it is a whole matrix. The form on the right is the sum of nine matrices Krs each weighted by the scalar δXrs . We can generalize to an NxN matrix F(X)|first order = Σrs=1,N (∂F/∂Xrs) (Xrs-Ars) = Σrs=1,N Krs δXrs where we are expanding X near that matrix A. Suppose A = 1, the identity matrix, then F(X)|first order = Σrs=1,N (∂F/∂Xrs)|X=1 (Xrs-δrs) = Σrs=1,N Krs (Xrs-δrs) or we write through first order: F(X) = F(1) + Σrs=1,N Krs (Xrs-δrs) F, X and K are all NxN matrices as appears, for example, in equation 2.14 page 20 of B&D where F(1) = 1 and Krs = (-i/4) σrs and we have the case N=4. We even know something about the matrices Krs= (∂F/∂Xrs)|X=1 = the matrix obtained by differentiating matrix F with respect to the single variable Xrs which is one element of the X matrix 8. Taylor expansion in N variables (added 3.3.12) The expansion around the point (x,y) = (a,b) given above is the following fi(x,y) = Σn,m=0,∞ kinm(x-a)n (y-b)m = Σn,m=0,∞ { ∂n+mfi/∂nx∂my|a,b /( n! m!) } (x-a)n (y-b)m Now write a = x0 and then x-x0 = dx etc to get fi(x,y) = Σn,m=0,∞ kinm(dx)n (dy)m = Σn,m=0,∞ { (∂n+mfi/∂nx∂my)|x0,y0 /( n! m!) } (dx)n (dy)m Is there some way to put this into vector notation where x = (x,y) ? First rewrite as fi(x1,x2) = Σn,m=0,∞ kinm(dx1)n (dx2)m = Σn,m=0,∞ { (∂n+mfi/∂nx1∂mx2)|x10,x20 /( n! m!) } (dx1)n (dx2)m Probably the adding a third variable gives fi(x1,x2,x3) = Σn,m,k=0,∞ kinmk (dx1)n (dx2)m (dx3)k = Σn,m,k=0,∞ { (∂n+m+kfi/∂nx1∂mx2∂kx3)|x10,x20,x30 /( n! m! k!) } (dx1)n (dx2)m (dx3)k Does this derivative object have a name? (∂n+m+kfi/∂nx1∂mx2∂kx3) = [∂1n∂2m∂3k fi(x)] Not so bad. So then we have fi(x1,x2,x3) = Σn,m,k=0,∞ [∂1n∂2m∂3k fi(x)]x=x0 /( n! m! k!) } (dx1)n (dx2)m (dx3)k fi(x1,x2,x3) = Σn1,n2,n3=0,∞ [∂1n1∂2n2∂3n3 fi(x)]x=x0 /( n1! n2! n3!) } (dx1)n1 (dx2)n2 (dx3)n3 Then our general formula might be written fi(x) = Σn [∂1n1∂2n2∂3n3..... fi(x)]x=x0 / (n1! n2! n3! ...) } (dx1)n1 (dx2)n2 (dx3)n3 ..... fi(x) = Σn [∂1n1∂2n2∂3n3..... fi(x)]x=x0 / Πs=1N (ns)! } Πs=1N (dxs)ns fi(x) = Σn [∂1n1∂2n2∂3n3..... fi(x)]x=x0 Πs=1N [(dxs)ns/ns!] fi(x) = Σk [∂1k1∂2k2∂3k3..... fi(x)]x=x0 Πs=1N [(dxs)ks/ks!] NB: The order of variables in the ∂'s are always monotonic increasing!! In Stakgold volume 2 page 2 we had the concept of a vector "multi-index" k such that k = k1 + k2+ .. + kN The derivative is just called Dk so the above becomes fi(x) = Σk [Dkfi(x)]x=x0 Πs=1N [(dxs)ks/ks!] Dk = ∂1k1∂2k2∂3k3..... Good, that is a fairly compact result I think. Let's now write the terms of various smallness orders. zeroth term = fi(x0) first order term =(∂1fi)dx1 + etc = (∂jfi) dxj = (fi) dx = [(f) x]i = (f)ijxj And then so far we have, in vector notation, f(x) = f(x0) + (f)|x0 x and we see the utility of the f notation. second order term = (∂1∂1fi)dx1dx1/2! + (∂2∂2fi)dx2dx2/2! + .... + (∂1∂2fi ) dx1dx2 // there IS no term (∂2∂1fi ) dx2dx1 !!! Example: fi(x) = xy fi(x+dx) = (x+dx)(y+dy) = xy + ydx + xdy + dxdy So how can we put this into some reasonable notation? second order term (N=2) = (∂1∂1fi)dx1dx1/2! + (∂2∂2fi)dx2dx2/2! + (∂1∂2fi ) dx1dx2 = (1/2!) Σn+m=2 (∂n∂mfi) dxndxm In this double sum, we have two terms appearing (∂1∂2fi) dx1dx2 and (∂2∂1fi) dx2dx1 which is different from the mindset of the general equation. Only one of these terms should count, and that is corrected for by the outside factor (which provides the 1/2! for the other two terms). Is there some vector or tensor way to write this second order term? I guess we could define (f)inm ≡ (∂n∂mfi) but that doesn't help much with vector notation. Notice that n,m are coordinate index labels, whereas the ki integers are powers. fi(x)|1st = Σk=1 [∂1k1∂2k2∂3k3..... fi(x)]x=x0 Πs=1N [(dxs)ks/ks!] = N terms fi(x)|2nd = Σk=2 [∂1k1∂2k2∂3k3..... fi(x)]x=x0 Πs=1N [(dxs)ks/ks!] = N + (N,2) terms In writing this, I was trying to find a way to associate Γcab with these second order terms.