EDIT: I was offline for the last few hours, so I couldn't look at the other question of which this one here was a duplicate.  It is not a duplicate, because the linked question mentions Stack&nbsp;Exchange links explicitly. The problem I face has to do with the fact, that links in Stack&nbsp;Exchange mailings are not exactly what the Stack&nbsp;Exchange app would expect an Stack&nbsp;Exchange link to look like. I would not even call this a Stack&nbsp;Exchange link, since it doesn't lead to a site within the Stack&nbsp;Exchange network you would normally link to.  I guess they are some sort of click-tracker link, or something like that, but that's not the important part, here. Here's what one of the links looks like:  http://sg-links.stackoverflow.email/wf/click?upn=pDzA4vSaPmKy14MynsLvnZrGP7ewVN5UzFsPjLNsxf37lv5vH-2FL7YDbAQsAtLugYNp-2BJiZcu65iHCdCmWwcPykni0h3G-2BW3a5DJQvmvpqO5YYk0exrZkOZd357pJfsQtX23gZixSuNBvjUBdJW0S-2FYJxGOjASUTK1FgJWcFKhWk-3D_AOPg-2F-2BQL2BN1Z1adM2aHoJZemfJJPhUU1r33PmBub3XkdttNEsj9eAI3iXegmVDgNR1kPnNqBKQmCk2dsyn9uQ5g2T8fxWFcEgLgWVEljU9HeN8V3E9CiWFKeHBN3AsBll3Ol5ooWPcw1zAzDtJBxhi9y23WRvLODAGnxo8GjN2JvPOgbDchbKP4fGLU5bdnPtYYvcBBziE3leh3q-2FXLhm6oVzcczCqFM3tXWWWXJuY-3D   If this question is not attracting any more attention and doesn't loose the dupe tag, I guess I will have waited too long with my edit. In this case, I will delete it and submit a "new" question.    Yeah, so, the title says it. Because the links in the emails are not specifically from Stack&nbsp;Exchange, the app doesn't recognize it as something it's supposed to want to open and each link I tap in an email just opens my mobile browser.   App Version: 1.0.95 Device Manufacturer: Samsung Device Model: SM-A520F   OS Version: 8.0.0 (A520FXXU8CSC1)
I clicked on  to a comment in my email and it prompted a list of apps to open link. All browsers (Chrome, Firefox and etc) are in the list, but not SE app.  Most apps that can open site specific links get prompted for that app to open  the link, but not Stack Exchange links for some reason.    App Version: 1.0.95 Device Manufacturer: Asus Device Model: ASUS_Z010D   OS Version: 6.0.1 (WW_Phone-13.8.26.76-20161230)
What steps must an educational institution wanting to recruit good faculty take? There is a cycle in the academia:    Good doctorates from top universities invariably join other top universities as faculty members.  Good students aspire to study in top universities under excellent faculty members.   Suppose a new educational institution or department springs up and desires to set up a top-quality group, how does it break into this cycle? I suppose this could take some years, but is money-power the only thing that could be wielded by the institute to achieve its end? Suppose $x&gt;0$ and we have function $f(x)=\frac{\sin x}{x}$ how can we show $$\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}$$ I need a hint to show this property .Thanks in advance . I tried for $n=1 ,2$  by finding maximum of $|f'| ,|f''|$ but I get stuck to show for $n$ How do I space quads evenly? 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Please find all given information in the following:  Given information:  $m$ countries each with a population of $n_i$; $y_i$ is the number of people having a disease; $y_i$ is a poisson random variable with mean $\frac{n_i*\lambda}{100}$  The prior distribution for $\lambda$: $\lambda \sim Gamma(a_0,b_0)$  Task:  Show that the posterior distribution is also a gamma distribution with $a_m = a_0 + \sum_{i = 1}^{m}y_i$ and $b_m = b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  My solution:  $Posterior \propto Likelihood \times Prior$  $p(\theta|y) \propto \prod_{i=1}^{m}(\theta^{y_i}e^{-\theta})\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i}e^{-m\theta}\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i+a_0-1}e^{-\theta(m+b_0)}$  $a_m = a_0 + \sum_{i = 1}^{m}y_i$  $b_m = b_0 + m \neq b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  I think I am doing a mistake, because I don't know how to incorporate the information that the mean of $y_i$ is given as $\frac{n_i*\lambda}{100}$. 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