Question #181568

Show that if n is a natural number and α, β are real numbers with β > 0 then there exists a real function f with derivatives of all orders such that: (i) |f(k)(x)| ≤ β for k ∈ {0, 1, ..., n − 1} and x ∈ (−∞, ∞); (ii) f(k)(0) = 0 for k ∈ {0, 1, ..., n − 1}; (iii) f (n)(0) = α.


1
Expert's answer
2021-05-12T03:48:02-0400

(i)fk(x)β,k(0,1,..,n1) and x(,)(i) |f^k(x)|\le \beta, k\in (0,1,..,n-1) \text{ and }x\in (-\infty,\infty)


Take f(x)=2f(x)=2

   f(x)=22|f(x)|=2\le 2


f(x)=02f(x)=02fk(x)2,k(0,1,..,n1) and x(,)f'(x)=0\le 2 \\[9pt] f''(x)=0\le 2 \\[9pt] |f^k(x)|\le 2, k\in (0,1,..,n-1) \text{ and } x\in (-\infty,\infty)


(ii) fk(0)=0, for k(0,1,..,n1)f^k(0)=0, \text{ for } k\in (0,1,..,n-1)


 Take f(x)=0f(x)=0

    fk(x)=0k(0,1,..,n1f^k(x)=0 \forall k\in (0,1,..,n-1


   and f(x)=xnf(x)=x^n

     Clearly fk(0)=0k(0,1,..,n1)f^k(0)=0\forall k\in (0,1,..,n-1)


(iii)fn(0)=αf^n(0)=\alpha


   Take f(x)=exf(x)=e^x


  f0(0)=e0=1f^0(0)=e^0=1


f(x)=exf'(x)=e^x


f(0)=e0=1f'(0)=e^0=1


Similarly fk(0)=1k(0,1,..,n1)f^k(0)=1 \forall k\in (0,1,..,n-1)


So, f(x)=exf(x)=e^x is used here.



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