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Differential Calculus> Show that if f is in?nitely di?erentiable...

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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]

BR Sharma , 10 Years ago

Grade

1 Answers

Jitender Singh

Last Activity: 10 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 k^thderivative,

$f^{k}(0) = \lim_{h\rightarrow 0}\frac{f^{k-1}(h) - f^{k-1}(0)}{h}$

$f^{k}(0) = \lim_{h\rightarrow 0}\frac{f^{k-1}(h)}{h}$

$f^{k}(0) = \lim_{h\rightarrow 0}\frac{0}{h} = 0$

Thanks & Regards

Jitender Singh

IIT Delhi

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