1
$\begingroup$

Let $f$ be an arithmetic function. If $$\lim_{x \rightarrow \infty}\frac{\sum_{n \leq x} f(n)}{x} = M \neq 0, $$ deduce the asymtotic formula for $$\sum_{n \leq x} f(n)n^{ia}$$ where $a$ is a real number constant.

I am not sure about what the question is asking. We have notation of asymptotic $f \sim g := \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1.$

So I have to find $g$ such that $$\sum_{n \leq x} f(n)n^{ia} \sim g.$$

Is it equivalent to $$\sum_{n \leq x} f(n)n^{ia} = h_1(x) + O(h_2(x)) ?$$ I usually see many theorem in analytic number theory write the terms in the way $$\sum_{n \leq x} f(x) = mainterm + O(something).$$ Is this thing also included as asymptotic formular ?

For the prove, I plan to use Abel summation formula (https://en.wikipedia.org/wiki/Abel%27s_summation_formula) to write

$$\sum_{n \leq x} f(n)n^{ia} = h(x) + O(something)$$ by setting $$a_n = f(n), A(t) = \sum_{n\leq t}a_n, \phi(t) = t^{ia}.$$

But I am not sure if Abel formular valid for complex function, and more over what is the (complex) derivative of $t^{ia} ?$

$\endgroup$
  • $\begingroup$ Abel formula is valid for complex-valued functions as well. And of course the derivative of $\phi : (0, \infty) \to \Bbb{C}$ given by $\phi(t) = t^{ia}$ is $ia t^{ia-1}$ as you expect. Finally, if I computed correctly, Abel formula leads to $$\sum_{n \leq x} f(n) n^{ia} \quad \sim \quad \frac{M}{ia+1} x^{ia+1} $$ as $x \to \infty$. $\endgroup$ – Sangchul Lee Oct 3 '16 at 13:06
  • $\begingroup$ @SangchulLee Thank you for your response. Could you please write what you did ? Just some steps is fine. I will do Abel summation, but it is good if I have something to check when I do it. Actually, I only use Abel and get to $mainterm + O(something)$, not sure how to obtain $\sim$ there. $\endgroup$ – Both Htob Oct 3 '16 at 13:12
5
$\begingroup$

Abel's formula gives you

$$ \sum_{n \leq x} f(n) n^{ia} = A(x)x^{ia} - ia \int_{1}^{x} A(u) u^{ia-1} \, du. \tag{*}$$

Now using the assumption, let us write $A(u) = (M + \epsilon(u))u$ where $\epsilon(u) \to 0$ as $u \to \infty$. Then

\begin{align*} \int_{1}^{x} A(u) u^{ia-1} \, du &= \int_{1}^{x} (M + \epsilon(u)) u^{ia} \, du \\ &= \frac{M}{ia+1}(x^{ia+1} - 1) + \int_{1}^{x} \epsilon(u) u^{ia} \, du \end{align*}

We claim that the last integral is $o(x)$ as $x \to \infty$. Indeed, one can adopt the idea of the Cesaro-mean to check that

$$ \frac{1}{x} \left|\int_{1}^{x} \epsilon(u) u^{ia} \, du\right| \leq \frac{1}{x} \int_{1}^{x} |\epsilon(u)| \, du \to 0 \quad \text{as} \quad x \to \infty. $$

Thus we have

$$\int_{1}^{x} A(u) u^{ia-1} \, du = \frac{M}{ia+1}x^{ia+1} + o(x). $$

Plugging this back to $\text{(1)}$ and using the relation $A(x) = Mx + o(x)$ gives

$$ \sum_{n \leq x} f(n) n^{ia} = \frac{M}{ia+1}x^{ia+1} + o(x), $$

which is essentially equivalent to

$$\sum_{n \leq x} f(n) n^{ia} \sim \frac{M}{ia+1}x^{ia+1}. $$

$\endgroup$
  • $\begingroup$ Thank you very much !! So $\sim$ is connected to little o ? I do not know that. Is it also connect to big O ? Might be not since the definition of Big O and little o is considerably different. $\endgroup$ – Both Htob Oct 3 '16 at 13:35
  • $\begingroup$ @BothHtob, The connection comes from the fact that $f(x) \sim g(x)$ can be written as $f(x)/g(x) = 1 + o(1)$. So you can freely move from $f(x) \sim g(x)$ to $f(x) = g(x) + o(g(x))$. The other direction is also guaranteed given that $g(x)$ never vanishes as $x \to \infty$. In our case, since $M/(ia+1)$ is non-zero and $|x^{ia+1}| = x$, we have $o(\frac{M}{ia+1}x^{ia+1}) = o(x)$. $\endgroup$ – Sangchul Lee Oct 3 '16 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.