Problem 3. Let $ \{a_n\}$ and $ \{b_n\}$ be two sequences such that $ \lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=c$, suppose $ f'(c)$ exists. Show that $ \displaystyle \lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)$.
My way.
I have to put some restictions to ensure my deduction is correct. If $ f'(x)$ is continuous near $ c$, then we are done with the Mean Value Theorem. But it, first and foremost, requires $ f(x)$ to be differentiable.
To avoid continuity of derivative, we inevitably set Mean Value Theorem aside, then differentiability of $ f(x)$ other than $ c$ doesn't carry weight any more. But I have to suppose we always have $ \boxed{a_n<c<b_n}$ or $ \boxed{b_n<c<a_n}$ for sufficiently large $ n$, this assumption will be used later on. By the fact that the existence of first order derivative implies the existance of linear approximation near $ c$, we have for any $ x$ near $ c$, \[f(x)=f'(c)(x-c)+f(c)+o(x-c),
\] Here for any $ \epsilon >0$, there exists a $ \delta > 0$ such that $ \displaystyle 0<|x-c|<\delta \implies \left|\frac{o(x-c)}{x-c}\right|<\frac{\epsilon}{2}$.
While for any $\delta_0 >0$, there exists an $ N$ such that $ n>N\implies |b_n-c|, |a_n-c|<\delta_0$, we take $ \delta_0<\delta$, now for $ n>N$,\begin{align*}
&{\color{white}=}\left|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(c)\right|\\
&=\left|\frac{f'(c)(b_n-c-(a_n-c))+o(b_n-c)+o(a_n-c)}{b_n-a_n}-f'(c)\right|\\
&=\left|\frac{o(b_n-c)+o(a_n-c)}{b_n-a_n}\right| \\
&\leq \left|\frac{o(b_n-c)}{b_n-c}\right|\left|\frac{b_n-c}{b_n-a_n}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\left|\frac{a_n-c}{b_n-a_n}\right| \\
&<\left|\frac{o(b_n-c)}{b_n-c}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\text{(by the assumption)}\\
&<\epsilon.
\end{align*}
Redo example 10. Show that $ \displaystyle 2<65^{1/6}<2+\frac{1}{192}$.
My way.
The left hand side is obvious. For the right hand side, observe that \[
65=2\left(1+\frac{1}{2^6}\right)^{\frac{1}{6}}<2\left(1+\frac{1}{6\cdot 2^6}\right)=2+\frac{1}{192}.
\] We have used $ (1+x)^\alpha \leq 1+\alpha x,\forall x\ge 0,\forall \alpha \in[0,1]$.
Problem 6. Let $ f:[0,\infty)\to\mathbb{R}$ be continuous and $ f(0)=0$. If $ |f'(x)|<|f(x)|$ for very $ x>0$, show that $ f(x)=0$, for every $ x\in[0,\infty)$.
My way.
Instead of considering $ [0,\frac{1}{2}]$, we can consider any interval $ [0,p]$, $ p<1$ first. It can be shown that by mean value theorem,\begin{align*}
|f(x)-f(0)|&<|f(y_n)||x|\\
&=|f(y_n)-f(0)||x|\\
&<|f(y_{n-1})||x||y_n|\\
&<\cdots \\
&<|x||y_n|\cdots |y_2||f(y_1)|\\
&<p^n|f(y_1)|,\forall x\in [0,p].
\end{align*} Letting $ n\to\infty$, we get $ f(x)=\inf\{p^n|f(y_1)|:n\in\mathbb{N}\}=0$. We can show by induction that $ f(x)=0$, for all $ x\in [(n-1)p,np],n\in\mathbb{N}$ in the same manner.
Problem 9. Give an example of a function that is differentiable on $ (0,1)$ and $ f(0)=f(1)$ but does not satisfy the Rolle's Theorem.
My way.
Define $ f(x)=\sqrt{x},\forall x\in[0,1)$ and $ f(1) = 0$.
Additional problem. Let $ f(x)$ be a differentiable on $ [a,b]$, $ f'(a)<f'(b),$ for any $ y_0\in (f'(a),f'(b))$, prove that there exists $ c\in(a,b)$ such that $ f'(c)=y_0$.
My way.
Since that $ f$ is differentiable cannot imply the continuity of $ f'$. Intermediate value theorem fails to work here. We on the contrary define a new continuous function, $ g(x) = f(x)-y_0 x$. It can be easily varified that $ g'(a)=f'(a)-y_0<0$ and $ g'(b)=f'(b)-y_0>0$, then extreme value cannot be attained at the end point, the minima must lie somewhere else in the interval $ (a,b)$, hence there exists $ c\in(a,b)$ such that $ g'(c)=0\implies f'(c)=y_0$.
No comments:
Post a Comment