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Differential Calculus

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]

Profile image of BR Sharma
12 Years agoGrade
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1 Answer

Profile image of Jitender Singh
12 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
askIITians Faculty

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