Questions tagged [asymptotics]
For questions involving asymptotic analysis, including function growth, Big-$O$, Big-$\Omega$ and Big-$\Theta$ notations.
9,946 questions
0 votes
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Saddle Point / Steepest Descent for Bessel Functions
I am trying to understand how to approximate integrals with Bessel functions. In particular I have something like: $$I_{\ell} = \int_{0}^{\infty} j_{\ell}(pr) dr = \frac{\sqrt{\pi} \Gamma[(1+\ell)/2] ...
1 vote
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Asymptotic Variance
Let $X_1,X_2,\cdots,X_n$ be an i.i.d. sample from a distribution $p_{\theta}$. Denote $\hat{\theta}_{MLE}$ is the MLE of $\theta$. Then under some regularity conditions $$\sqrt{n}\,(\hat{\theta}_{MLE} ...
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1 vote
1 answer
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Divergence, convergence of the series $\sum_{n=1}^{\infty} \left[ 2^{(n^x - n^x \cos(1/n^2))} - 1 \right]$
I have this series $$\sum_{n=1}^{\infty} \left[2^{\left(n^{x}-n^{x} \cos \frac{1}{n^{2}}\right)}-1\right].$$ let $a_n$ the general term of the series, $$ a_n = 2^{\left(n^x - n^x \cos \frac{1}{n^2}\...
- 8,886
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Approach to solve $\sum_{n=1}^{\infty} \left[ 2^{\frac{1}{n^3}} - 1 - \frac{1}{\sqrt{n}} \right]$
I should calculate the series: $$\sum_{n=1}^{\infty} \left[ 2^{\frac{1}{n^3}} - 1 - \frac{1}{\sqrt{n}} \right]$$ Consider the sequence $$ a_n = 2^{\frac{1}{n^3}} - 1 - \frac{1}{\sqrt{n}}. $$ Let us ...
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Help with asymptotics of a summation
Fix $s_0 > 1$. Suppose $s_1 \in (s_0, s_{\max}]$, where $s_{\max} > s_0$ is arbitrary. Let $K = [s_0, s_{\max}]$. For $n \in \mathbb{N}$, define $D_n : K \times K \rightarrow [0, \infty)$ by \...
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2 votes
1 answer
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Technical step in the proof of Linnik's theorem in Iwaniec-Kowalski (18.82)
Going through the proof of Linnik's theorem in Iwaniec and Kowalski's Analytic number theory, I came across an affirmation I don't really understand. On Page 440, starting from the explicit formula ...
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1 vote
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Asymptotics of eigenvalues of Sturm-Liouville problems
Given a regular Sturm-Liouville problem with smooth, positive, bounded $p(x),q(x),w(x)$, over a finite interval $(a,b)$, we know that the eigenvalues $\lambda_n$ are discrete and grow without bound. ...
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1 vote
2 answers
146 views
Why does $\sum_{1 \le m \le q/2} \frac{1}{m} \le \log q$ hold on page 199 of *Analytic Number Theory* by Iwaniec and Kowalski?
I am reading about exponential sums in Analytic Number Theory by Iwaniec and Kowalski, page 199. At one point, they use the inequality $ \sum_{1 \le m \le q/2} \frac{1}{m} \le \log q. $ I understand ...
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