## Improved search of XenForo 2.x – | NulledTeam UnderGround

Today, just over a year after we presented the new version of XenForo for the first time to the public, we are delighted to be able to launch XenForo 2.0.0 and the attached 2.0.0 versions of our official add-ons. These are the first supported versions of the 2.0 series.

XenForo 2.0 is an important update since version 1. It includes many new features and significant changes under the hood, with the central code completely redesigned.

Some of the new features include:

## raw – The best software to sample up to approximately 2x?

My 2 cents here.

I have some high quality RAW files.

This is important. Let's go

I need to have them to print approximately twice the size I get with the ppi set at 240.

I guess 240 ppi is a preset camera setting … Well, you can modify that value to whatever value you need.

Normal people will not see a difference in a 100ppi photo and a 200ppi to 30cm photo, except in some very small or very small details (less than a foot).

The question is how far away normal viewers will be. If you double the distance, you can cut to half the resolution.

(These images have an error, says dpi instead of ppi)
http://www.otake.com.mx/Foros/Resolution1.jpg

Now. If we start with a 300ppi file seen at 30 cm, we can start making some numbers.

In my experience, the expected viewing distance is somehow equal to the largest size of your photo.

http://otake.com.mx/Foros/Grafica1Eng.png

But suppose you really need to resample an image. That is a matter of taste.

This test takes time but you can compare different programs. The first image is a resize without resampling. The second with a Lanczos algorithm, which is similar to the bicubic. And the other two are from the popular remodeling software. Benvista and Reshade.

http://www.otake.com.mx/Foros/Zoom200.jpg

In my opinion, a simple "bicubic sharper" in Photoshop is good enough. You can use additional sharpening if you need it.

My recommendation is that you should only use round numbers. Resample 200%, not 187%, for example.

I would never sample more than 200%. For a blured background it could go to 300%.

## sequences and series: \$ 1 + 2x + 3x ^ 2 + points \$ converge to what?

Assuming the following terms are the following, you must have:

$$1 + 2x + 2x ^ 2 + cdots + 2x ^ n$$
$$= left (1 + x + x ^ 2 + cdots + x ^ n right) + left (x + x ^ 2 + cdots + x ^ n right)$$

instead of $$Sigma_ {n = 0} ^ infty nx ^ n$$.

Now you can use the formula for an infinite sum of a geometric series.

## integration – Why does derivation -2 / (x + 1) or 2x / (x + 1) result in the same function 2 / (x + 1) ^ 2?

I have some problems to understand the integration of 2 / (x + 1) ^ 2.

When integrating using u-subition I get:

-2 / (x + 1)

However, I also know that:

2x / (x + 1)

It is also a solution, because when you take the derivative From either of the two solutions, you end up with the same answer:

2 / (x + 1) ^ 2

I want to understand how the relationship between the two integrated solutions works and why. I can not find any proof or example that can guide me in the right way. Any help to understand this phenomenon would be greatly appreciated.

Thank you.

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## Actual analysis: Prove that \$ sum_ {n = 1} ^ infty int _ {- infty} ^ infty cos (n ^ 2x) I (x) \$ dx converges absolutely.

Leave $$I$$ be a measurable subset of $$mathbb R$$. We define
$$I (x) = int_I frac { chi _ {(- 1 le x-y le 1)}} {1 + y ^ 2} dy.$$
by $$n ge 1$$ we define
$$a_n = int _ {- infty} ^ infty cos (n ^ 2x) I (x) dx.$$
Test it $$sum_ {n = 1} ^ infty a_n$$ it converges absolutely.

I know that $$I (x)$$ it is a non-negative subject $$L ^ 1$$ work in $$mathbb R$$. To demonstrate that $$sum_ {n = 1} ^ infty a_n$$ converges absolutely, I tried to estimate the integral:
begin {align} sum_ {n = 1} ^ infty | a_n | & le sum_ {n = 1} ^ infty int _ {- infty} ^ infty left | cos (n ^ 2x)[arctan(x+1)-arctan(x-1)] right | dx \ end {align}

However, I do not know how to proceed without sacrificing the term. $$cos (n ^ 2x)$$. I have the intuition that $$arctan (x + 1) – arctan (x-1)$$ it goes to zero like $$x$$ goes to infinity. Meanwhile, $$| cos (n ^ 2x) |$$ should be magnified to something with respect to $$n$$ from where we compare each term with a convergent series to conclude. But I'm not sure how to carry it out explicitly. Some help?

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## Find all ordered pairs (x, y) that satisfy both (3x-4y) / xy = -8 and (2x + 7y) / xy = 43

This is listed in the "multiple variable" category of my Alg 1 task.

Find all ordered pairs (x, y) that satisfy both (3x-4y) / xy = -8 and (2x + 7y) / xy = 43