#Ok to post homework #Quentin Dubroff, Assignment 21, April 14 2019 with(linalg): ############################THINGS FROM C21.txt######################################## #C21.txt, April 11, 2019, Functional chains (and networks) Help:=proc(): print(` EvalFC1(L,x,0), EvalFC(L,x,x0) , RP(d,K,x) , RC(d,K,n,x) `): print(`EvalFCdiff(L,x,x0), SumCoeff(A) `): end: #y=a*x+b #Data-Set [x1,y1], ..., [xn,yn] L(a,b):-=Sum(a*xi+b-yi)^2,i=1..n); #diff(L,a)=0, diff(L,b)=0 a*x_{n+1}+b # [x1,..., xm; y]: Loss(a1, ..., am,a0):=Sum((a1*x1+...+am*xm+a0-y)^2 , data points in the set) #[m1,m2,m3, ..., mk] m1xm2 +m2xm3+.... #CNN #functional chain x->f1(x)->f2(f1(x)) #Def: A functional chain is a list of expressions [f1, ..., fn] in x #EvalFC1(L,x,x0): inputs L=[f1,..., fn] expressions in x and a number (or symbol) x0 #outputs f1(f2(...fn(x0))))) EvalFC1:=proc(L,x,x0) local i,L1: if nops(L)=1 then RETURN(subs(x=x0,L[1])): fi: L1:=[op(1..nops(L)-1,L)]: expand(subs( x=EvalFC1(L1,x,x0),L[nops(L)])): end: #RP(d,K,x): a random polynomial of degree d in x with coeff. between -K and K RP:=proc(d,K,x) local ra,i: ra:=rand(-K..K): add(ra()*x^i,i=0..d): end: #RC(d,K,n,x): a random functional chain of length n RC:=proc(d,K,n,x) local i: [seq(RP(d,K,x),i=1..n)]: end: #f(g(x))'= f'(g(x))*g'(x) #f1(f2(f3(x))'=f1'(f2(f3(x))*f2'(f3(x))*f3'(x) #(fk(f(k-1)(f(k-2)))'= fk'(f(k-1)....) [(f(k-1)(f(k-2)....) f1'(x0)] #x0->f1(x0)->f2(f1(x0)->.... fn(f_{n-1}(x)) ... #EvalFC(L,x,x0): inputs L=[f1,..., fn] expressions in x and a number (or symbol) x0 #outputs [f1(x0),f2(f1(x0)), f3(f2(f1(x0)), ..., fn(....)]: EvalFC:=proc(L,x,x0) local i,L1: if nops(L)=1 then RETURN([subs(x=x0,L[1])]): fi: L1:=[op(1..nops(L)-1,L)]: L1:=EvalFC(L1,x,x0): [op(L1),subs(x=L1[nops(L1)] , L[nops(L)])]: end: #EvalFCdiff(L,x,x0): inputs a functional chain L, variable x, and number or symbol x0 #outputs the derivative of EvalFC(L,x,y) w.r.t y and y=x0 # EvalFCdiff:=proc(L,x,x0) local i,Ld,M,L0: L0:=EvalFC(L,x,x0): Ld:=[seq(diff(L[i],x),i=1..nops(L))]: M:=[subs(x=x0,Ld[1])]: for i from 2 to nops(L) do M:=[op(M), subs(x=L0[i-1] ,Ld[i]) ]: od: convert(M,`*`): end: #SumCoeff(A): the sum of the coefficient in a + expressin SumCoeff:=proc(A) local su,i, lauren: su:=0: if not type(A,`+`) then RETURN(FAIL): fi: for i from 1 to nops(A) do lauren:=op(1,op(i,A)): if type(lauren,integer) then su:=su+lauren: else su:=su+1: fi: od: su: end: ##########################END C21.txt############################################## #EvalFCdiffk(L,x,x0,k) inputs a functional chain L, variable x, and number or symbol x0 and outputs the #k-th derivative of EvalFC(L,x,y) w.r.t. y and y=x0. #I am not sure how to do this without calling EvalFC(L,x,y) and differentiating # w.r.t. y k times. #Terms of diff(f(g(x)),x\$i) in the OEIS is sequence #A000041: the number of partitions of n. #It is not mentioned in the notes that this is the number of terms for arbitrary f and g # but instead says terms of n-th derivative of 1/f or of log(f). #As found in class, the sum of the coefficients in the above is the sequence of Bell numbers #A000110 #Terms of diff(f(g(h(x))),x\$i) in the OEIS is sequence #A022811: Number of terms in n-th derivative of a function composed with itself 3 times. #We find that if we add the coefficients of the above expansion we get the sequence #A000258 e.g.f.: exp(exp(exp(x)-1)-1) #EvalNR2(L,c1,c2,x0,y0)inputs a list L of pairs [A1,B1],[A2,B2], ..., [An,Bn] and ouputs #the value in going from left to right and applying to the current vector the transformation #[x,y] -> [max(a11*x+a12*y+b1,0), max(a21*x+a22*y+b2,0)] for the current #A=[[a11,a12],[a21,a22]], and B=[b1,b2], max(c1*x+c2*y,0) to the final [x,y]. #NOTE It is unclear to me what the role of c1 and c2 are. EvalNR2:=proc(L,x0,y0) local x,y,A,B,i: if nops(L)<2 then RETURN([x0,y0]): fi: A:=L[1][1]: B:=L[1][2]: x:=max(dotprod(A[1],[x0,y0])+B[1],0): y:=max(dotprod(A[2],[x0,y0])+B[2],0): for i from 2 to nops(L) do A:=L[i][1]: B:=L[i][2]: x:=max(dotprod(A[1],[x,y])+B[1],0): y:=max(dotprod(A[2],[x,y])+B[2],0): od: [x,y]: end: #Using plot3d, we find that the graph looks like a piece of folded paper.