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Show that if f is in?nitely di?erentiable on (-1, 1) and f(1/n)= 0 for all n > 1 then f^(k)(0) = 0 for all k > 0. [f^(k)(x) is differentiating f(x) k times]

Jitender Singh IIT Delhi
6 years ago
Ans:
f is infinitely differentiable on (-1, 1). So it would be continuous in (-1, 1).
$f(\frac{1}{n}) = 0, n> 1$
$\lim_{n\rightarrow \infty }f(\frac{1}{n}) = f(0) = 0$
$f'(x) = \lim_{h\rightarrow 0}\frac{f(h) - f(0)}{h}$
$f'(0) = \lim_{h\rightarrow 0}\frac{f(h) }{h}$
h is the delta neibhbourhood of ‘0’. So its value is zero.
$f'(0) = \lim_{h\rightarrow 0}\frac{0 }{h} = 0$
$f''(0) = \lim_{h\rightarrow 0}\frac{f'(h) - f'(0)}{h}$
$f''(0) = \lim_{h\rightarrow 0}\frac{f'(h)}{h}$
$f''(0) = \lim_{h\rightarrow 0}\frac{0}{h} = 0$
Similarly for the kthderivative,
$f^{k}(0) = \lim_{h\rightarrow 0}\frac{f^{k-1}(h) - f^{k-1}(0)}{h}$
$f^{k}(0) = \lim_{h\rightarrow 0}\frac{f^{k-1}(h)}{h}$
$f^{k}(0) = \lim_{h\rightarrow 0}\frac{0}{h} = 0$
Thanks & Regards
Jitender Singh
IIT Delhi