php – How to add the "x" closing button in the upper right corner of the Surbma popup plugin?

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Non-asymptotic upper limit of the right tail of the Gamma function

I wonder if there is any upper limit not asymptotic for the next Gamma function:
$$ f_a (x) = int_ {x} ^ { infty} t ^ a exp (-t) dt $$
for $ x> 0, a> 0 $? Something like $ x ^ a exp (-x) $?

Algorithms: use of Chebyshev to obtain an upper limit for the problem of the coupon collector

I am taking a course and I have problems solving an exercise.

Leave $ X $ be an RV defined as the number of tests required to collect at least one of each type of coupon (of which there is $ n $). So $ E[X] = nH_n = n ln (n) + O (n) $ Y $ sigma ^ 2_X = n ^ 2 sum_ {i = 1} ^ n frac {1} {i ^ 2} – nH_n $ with $ lim_ {n to infty} frac { sigma ^ 2_X} {n ^ 2} = frac { pi ^ 2} {6} $. The exercise asks how Chebyshev's inequality can be used to limit the probability that $ X> beta n ln (n) $ for some $ beta> 1 $.

I tried this:
begin {align *}
Pr[X>beta n ln(n)] & leq Pr[|X-mu_X| geq (beta – 1)n ln n] \
& leq Pr left[|X-mu_X| geq (beta – 1) left( frac{sqrt 6}{pi} sigma_X right) ln nright] (one) \
& leq frac { pi} {6 ( beta -1) ^ 2 ln ^ 2 n} = O left ( frac {1} { ln ^ 2n} right)
end {align *}

where (1) follows because $ n> frac { sigma_X sqrt 6} { pi} $. However, I think the first inequality is incorrect because I'm discarding the $ – O (n) $ since $ mu_X $ in deriving an upper limit.

I also tried something else:
begin {align *}
Pr[X>beta n ln(n)] & leq Pr[|X-mu_X| geq beta n ln n – nH_n] \
& leq Pr[|X-mu_X| geq (beta -1)nH_n] \
& = Pr left[|X-mu_X| geq frac{6(beta – 1)H_n}{pi^2n} left(n^2 sum_{i=1}^infty frac{1}{i^2} – n H_n right)right]
end {align *}

But I'm not sure where to go from here. Thanks for any help!

Approach theory: adjusted upper limits for a non-linear recurrence that increases monotonically

I have the following non-linear recurrence:

$$ y_ {n + 1} = sqrt { frac {2} {1 + y_n}} y_n, y_0 in[0,1]$$

Some basic thoughts show that $ 0 $ Y $ 1 $ They are fixed points of this, and that $ 0 $ is repelling and $ 1 $ is attracting (I'm only worried about the region $ y_i in[0,1]$).

Non-linear recurrences rarely have closed forms in terms of elementary functions, and I doubt that this does.
Fortunately, I do not particularly care exact solutions, but only close the upper limits.
Keep in mind that there is a trivial upper limit of $ y_i leq 1 $, but this is insufficient for my case.

This sequence tends to converge to $ 1 $ quite quickly (and in the region of interest it seems to be pretty “ regular ''), so you can get a pretty good idea of ​​how it looks by choosing some arbitrary data. $ y_0 $ The values ​​and their layout.
I have included the Python code that does this here. You must be able to write plot (1/2) To see the plot of 1/2 (for example).
Keep in mind that you continue adding lines to the graph — to eliminate them, eliminate plot.png, and reload the page (hacky I know).

For those who just want to see some photos, see below:Plots for several starting points.
Note that this image is slightly misleading, this is technically a discreet sequence.
Even so, this shows that a reasonable continuous interpolation could have arrived.

What techniques exist to obtain upper limits for non-linear recurrence relationships that increase monotonously?
The only thought I have is to approach. $ 2y_n / sqrt {1 + y_n} $ through some kind of Taylor series (which is not difficult to use Newton's binomial theorem), and truncating this.
This gives us a recurrence relation "inequality", which will probably remain non-linear if we do not want to incur much error in the remaining term.

I have investigated things like this document, but I could not make my repetition fit his frame (although it is quite close).
The related series:
$$ z_ {n + 1} = frac {1} { sqrt {1 + z_n}} z_n, z_0 in[0,1]$$
It seems to fit, although the results that I obtained from the application of the technique of that article to this repetition were, at best, irregular (probably due to computational errors on my part, but I do not want to pursue them without being able to apply this technique to my recurrence)

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Trigonometry – Find a trigonometry solution to locate the center of an arc intersecting a given arc in the upper right quadrant

Circles with collinear centers (on the X axis) Working completely in the quadrant where x and y are positive, I am looking for a solution based on trigonometry for something that I can easily build with a compass but I still can not find the elegant solution. Given a circle in normal orientation, then selecting a point somewhere in that circle in that upper right quadrant, I look for a way to locate the center of another circle that intersects the chosen point in the first circle, limited by the center of the circle. second circle. on the x axis

Noobs can not post images, but it would be easier to describe using the tags in this linked photo:
Original circle of center C and radio CA;
point B is the objective of the intersection by new circles with centers in F and G whose radii are FE and GD respectively; AD = AE & AD <AH (in fact, it is not interested in AD approaching AH, which would push GD towards infinity, equally disinterested in the ACB angle approaching zero extremes or pi / 2 rads)

What is the point of closing windows by double clicking on its upper left?

Define "… not intuitive …"

As a PC user with Windows since the mid 90's, this is completely intuitive for me. Me, and thousands of people like me, intuitively Close the windows with the double click of the upper left corner without thinking.

Why? Because we have done this for decades.

A more interesting question would be

"When was this behavior established?"

and, more to the point,

"Why is this behavior no longer intuitive for new users?"

@PatomaS responds well to the first and the latter is simply this: in most cases, the current style of the Windows window no longer shows a visual signal for this useful feature, as highlighted below through three examples (which are show from the top: Microsoft Internet Explorer, Google Chrome and Windows Explorer). Only the lower example (Windows Explorer) has an icon in this area.

Showing some examples of corners above left

By the way, the Windows universal platform applications (such as Edge) do not have this feature. Over time, through the use of these applications, my behavior will be reinforced more and more negatively until the day comes when I no longer double-click in the upper left corner of the window to close it.

… I'll probably never know when the last time I do this is …

hashpower: hash rate of average PC / upper end PC

Hi, I was wondering if anyone had any idea about how the hash rates are with recently released PCs. I'm looking for figures to use for an average user if they wanted to use their PC for mine, and then a kind of better case, for the hash rate of someone who has a good graphics card, etc. Any sources or information from personal experiences would be great, thank you!

control inversion: where lies the responsibility of eliminating upper level caches in a pyrimidal calculation system with multi-level caching?

I'm working on a system that analyzes non-trivial amounts of data. The analysis is pyramidal, since several combine to generate intermediate values, which in turn combine with each other and more inputs to generate intermediate values ​​of higher order, and so on, until the value is produced. As a simple example: the inputs A and B combine to make the intermediate product X, the inputs A and C combine to make the intermediate product Y, the inputs C and D combine to make the intermediate product Z and then combine the intermediate products X, Y and Z With input E to make the final result. See below for a rough diagram that represents this example relationship.

Paint in rough Diagram of pyramidal calculation paint.

Almost all of these steps are expensive enough to have significant performance costs for the user, both to retrieve the inputs and to calculate the intermediate products. On the positive side, all entries change only when requested by the user, so their values ​​can be cached and reused, and the system can be specifically notified when one of the entries has changed. Therefore, worry about when to invalidate caches It is not a problem

The problem that I am struggling with is where is the responsibility to clean the caches of intermediate products; after all, if the input A changes in the previous example, we must recalculate X and Y, but we can keep the values ​​stored in Z cache, as well as B, C, D and E.

I am using the event aggregator pattern, but I can not decide which event the cache will use, say, and to activate its invalidation. I can see a couple of options:

  1. The Y cache can subscribe to the A-data-changes and C-data change events and invalidate when it receives any of them. This requires Y to know that it depends on A & C, and means that high-level caches must subscribe to more and more modified data events to cover all their dependencies (and the dependencies of their dependencies, etc.).
  2. The A & C caches could send a data change event AND each time they receive their own data change events. This will require that they know that And depends on them, and it would also require them to send data change events for all subsequent products that depend on them. In addition, it would generate unnecessary duplicate change events for higher level products (since all dependencies try to transmit the message).
  3. A separate dependency manager / event translator could track all dependencies and know that, for example, if you see a C data change event, you need to send Y data change events and Z data changes, as well as events for anything that depends on Y or Z, up to the pyramid.

I'm inclined to go with # 3 and put the new manager with / near the root DI. Are there advantages / disadvantages to these approaches that I have overlooked? Or alternative approaches, to that?