1
$\begingroup$

Suppose that $f$ is twice continuously differentiable on $(-1,1)$ and that $f(0) = 0$.
Find: $$\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}f(kx)$$

My attempt: I was thinking about the Riemann sum, but I am not able to solve.

Any hints/solution will be appreciated.

Thank you.

$\endgroup$
3
  • 3
    $\begingroup$ "My attempt : i was thinking about the Reimann sum" - With all due respect, thinking about something without doing anything doesn't usually count as an attempt... $\endgroup$
    – Yuriy S
    Nov 3, 2018 at 10:14
  • 1
    $\begingroup$ Is [] the floor function here? $\endgroup$
    – Yuriy S
    Nov 3, 2018 at 10:16
  • $\begingroup$ @YuriyS Greatest Integer Function $\endgroup$
    – jasmine
    Nov 3, 2018 at 10:18

1 Answer 1

7
$\begingroup$

$$f(x)=f'(0)x+\frac{f''(0)}2x^2+o(x^2)$$

and

$$\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}f(kx)=\lim_{x \rightarrow 0^+} \sum_{k=1}^{[\frac{1}{\sqrt x}]}(f'(0)kx+\frac{f''(0)}2k^2x^2+o(k^2x^2)).$$

If we set $x=\dfrac1{n^2}$,

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\left(f'(0)\frac k{n^2}+\frac{f''(0)}2\frac{k^2}{n^4}+o\left(\frac{k^2}{n^4}\right)\right)$$ the only term that does not vanish is

$$f'(0)\frac{(n+1)n}2\to\frac{f'(0)}2.$$


For complete rigor, we must account for the fact that for a given $n$,

$$\frac1{(n+1)^2}<x\le\frac1{n^2}$$ and there will be a corrective term bounded by

$$\sum_{k=1}^{n}k\frac{(n+1)^2-n^2}{n^2(n+1)^2},$$ which is harmless.

$\endgroup$
4
  • $\begingroup$ thanks u @Yves Daoust $\endgroup$
    – jasmine
    Nov 3, 2018 at 10:27
  • $\begingroup$ @YuriyS: oooops, yes. Of course, this doesn't change the answer. $\endgroup$
    – user65203
    Nov 3, 2018 at 10:32
  • 1
    $\begingroup$ Cool. desmos.com/calculator/htvkyihy5p $\endgroup$
    – Mason
    Nov 3, 2018 at 10:33
  • $\begingroup$ @Mason: this confirms the formula for this moderate $n$, $1.01\times 1/2=0.505$. $\endgroup$
    – user65203
    Nov 3, 2018 at 10:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .