Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function at $x_0=0$ and $f(0)=0$ and let $k \in \mathbb{N}$. Evaluate the limit:
$$\lim_{x\rightarrow 0}\frac{1}{x}\left [ f(x)+ f\left ( \frac{x}{2} \right )+ \cdots+ f\left ( \frac{x}{k-1} \right )+ f\left ( \frac{x}{k} \right ) \right ]$$
Solution
We are evaluating the limit according to the limit derivative definition:
$$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}= f'(0)$$
So , according to the above limit we split the limit accordingly:
$$\begin{align*}
\lim_{x\rightarrow 0}\left [ \frac{f(x)-f(0)}{x}+\frac{1}{2}\cdot \frac{f\left (\frac{x}{2} \right )-f(0)}{\frac{x}{2}-0} +\cdots + \frac{1}{k}\frac{f\left ( \frac{x}{k} \right )-f(0)}{\frac{x}{k}-0}\right ] &=f'(0)\left ( 1+ \frac{1}{2} + \cdots + \frac{1}{k}\right ) \\
&= \mathcal{H}_k f'(0)
\end{align*}$$
$$\lim_{x\rightarrow 0}\frac{1}{x}\left [ f(x)+ f\left ( \frac{x}{2} \right )+ \cdots+ f\left ( \frac{x}{k-1} \right )+ f\left ( \frac{x}{k} \right ) \right ]$$
Solution
We are evaluating the limit according to the limit derivative definition:
$$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}= f'(0)$$
So , according to the above limit we split the limit accordingly:
$$\begin{align*}
\lim_{x\rightarrow 0}\left [ \frac{f(x)-f(0)}{x}+\frac{1}{2}\cdot \frac{f\left (\frac{x}{2} \right )-f(0)}{\frac{x}{2}-0} +\cdots + \frac{1}{k}\frac{f\left ( \frac{x}{k} \right )-f(0)}{\frac{x}{k}-0}\right ] &=f'(0)\left ( 1+ \frac{1}{2} + \cdots + \frac{1}{k}\right ) \\
&= \mathcal{H}_k f'(0)
\end{align*}$$
Well, I have found a nice result, let's check it.
ReplyDeleteIf $f:\mathbb{R}\to\mathbb{R}$ is a differentiable at $x_0=0$ function with $f(0)=0$ and $f'(0)>0$ then compute the following limit
$$\lim_{k\to\frac{1}{e-1}}\lim_{x\to 0}\left[\frac{1}{x}\left(f(x)+f\left(\frac{x}{2}\right)+\cdots+f\left(\frac{x}{k}\right)\right)\right].$$
Hello George. Very interesting. Well the inner limit is easy , for we have seen it above. The outer limit actually appears to be quite difficult:
ReplyDeleteIt is actually to compute the limit:
$$\mathscr{L}=f'(0)\lim_{k\rightarrow 1/(e-1)}\mathcal{H}_k$$
where $\mathcal{H}_k$ is the harmonic number. Hmm , my first reaction was to use something of the generalized harmonic number , namely the identities:
$$\mathcal{H}_z= \mathcal{H}_{z-1} + \frac{1}{z} \tag{1}$$
and
$$\mathcal{H}_z=\gamma+\psi(z+1) \tag{2}$$
where $\gamma$ is the Εuler - Mascheroni constant.. I reached a dead- end.
Using the Inequality
Delete$$\ln(k+1)<\mathcal{H}_k\leq 1+\ln k$$
along with the squeeze theorem yields the result.
I should have imagined that you were bounding it using a double inequality and then you applied the squeeze theorem, Well done George. Nicely executed.
DeleteMy hat is off to you. :-h
Under the same assumptions one can eventually also evaluate the following limit:
ReplyDelete$$\mathscr{L}= \lim_{x \rightarrow 0} \frac{1}{x} \left[f(x)+f(2x)+\cdots + f(kx) \right]$$