This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 3 based on use of L'hospital Rule . Let's give it a try !!
Let f be a function such that \(f(0)=0\) and f has derivatives of all order. Show that \( \lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}}=f''(0) \)
where \( f''(0)\) is the second derivative of f at 0.
Differentiability
Continuity
L'hospital rule
Let L= \( \lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}} \) it's a \( \frac{0}{0} \) form as f(0)=0 .
So , here we can use L'hospital rule as f is differentiable .
We get L= \( \lim _{h \to 0} \frac{f'(h)-f'(-h)}{2h} = \lim _{h \to 0} \frac{(f'(h)-f'(0)) -(f'(-h)-f'(0))}{2h} \)
= \( \lim _{h \to 0} \frac{f'(h)-f'(0)}{2h} + \lim _{k \to 0} \frac{f'(k)-f'(0)}{2k} \) , taking -h=k .
= \( \frac{f''(0)}{2} + \frac{f''(0)}{2} \) = \( f''(0) \) . Hence done!
Let \( f:[0,1] \rightarrow[0,1] \) be a continuous function such \( f^{(n)} := f ( f ( \cdots ( f(n \text{ times} )) \) and assume that there exists a positive integer m such that \( f^{(m)}(x)=x\) for all \( x \in[0,1] .\) Prove that \( f(x)=x \) for all \( x \in[0,1] \)
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