(i)∣fk(x)∣≤β,k∈(0,1,..,n−1) and x∈(−∞,∞)
Take f(x)=2
∣f(x)∣=2≤2
f′(x)=0≤2f′′(x)=0≤2∣fk(x)∣≤2,k∈(0,1,..,n−1) and x∈(−∞,∞)
(ii) fk(0)=0, for k∈(0,1,..,n−1)
Take f(x)=0
fk(x)=0∀k∈(0,1,..,n−1
and f(x)=xn
Clearly fk(0)=0∀k∈(0,1,..,n−1)
(iii)fn(0)=α
Take f(x)=ex
f0(0)=e0=1
f′(x)=ex
f′(0)=e0=1
Similarly fk(0)=1∀k∈(0,1,..,n−1)
So, f(x)=ex is used here.
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