Question

3. (a) Let f be an infinitely differentiable function on R and define х F(x) = e-y f(y) dy. Find and prove a formula for F(n), the nth derivative of F. (b) Show that if f is a polynomial then there exists a constant C such that F(n)(x) = Cem for sufficiently large n. Find the least n for which it is true.

Answer 1

on IR. 3 o = y=x OC) Let f be infinitely differenliable function define F(x) = sexy.fo) dy. :: f'(30) en (ex9-f68]dy [exy-f%)]dy +[ex-946); 1 [By Leibnitz rule for = 5*e**FOX4+ 100) differentiation under integral sign. FG) + foc) F"00) F' GB) + f '60) = F20) + f60+ f'(6) 115 p'(X)= F(0) + f(x) F"60 = 766) + f'@l) + f" = F4) + f() + f'(x)+f" so ON). D Thus (n) + 2.) = F(0) + f@) ++'&l) + f(1) Ge]

MONA F(%) = $ *) fejdy x ty parts my f03) se f'@)dy by integration, bu e e XC 0 X 6-X. - XC x9f"(t)dy e no fosex - f(x) + [*e*if'0) [-ex?fo)) + Sexst"Ceely f@)e-f60) + of'ne*- f'@). + (continuing in this way) = fee*- f@)+f@)e*.4'®)+----+ floroje vyroben X X-yrim) + sex?) dy =e*Fo+fear to fres= f'on... holy X- en dy. ot se e

© so we can also can also write F(x) = e*{{@)+80)+ - + f("%} - {f(x)+f'*+-- + frontline)} 24 e-> وا(يون exe / + f e m = n. farina in previous case) i pln) (x) = f(x) + f«) +4'60)+ -- + f(*)(x) =e* {f®)+f'@)+ -- + f(40) (0)3 + fm (4) dy x ent (6) Let f be a polynomial of dogree m. Then f(n) (c) = 0 for n>m taking n= mtl, we will get ie. f (m+) () = 0 so from previous part, taking {f@+10)+:--+ x F (M+1) (x) ( Se fontos? + sexy from the رو (9)

- Ei e*{f@+f'@)+-- + femmin )} +0 (As flert)) = 0). cietWhere C= f®) +f69) +--+f6m) (0) Also this is the least possible n. for which + F max) = = C ex . is Hence, if f is a polynomicil, then for' n = dkg 6f3+1, where c FM) 6) = ce a constant Also it is dear that F) (C)= cex for all na deg ef)+)