Cancel common factors in symbolic product

Suppose I have the following function

G[n_, k_] := Product[g[n + i]/g[i], {i, 0, k}] 

where $ n,k$ are assumed to be natural numbers and $ g$ is a nonzero symbolic function. I want to simplify the expression G[n,k+1]/G[n,k] and cancel all common terms to get g[n+k+1]/g[k+1]. However, Simplify, FullSimplify, Expand and Cancel all yield the symbolic (not cancelled) product.

Turn around Cell Phone Database – 3 Important Factors

As you are very much aware, there are truly many catalogs that you can browse on the web that will permit you to do a converse telephone query. Some look fancier than others, some give you more data, and some give you quality and security. In the event that you are in the chase for an opposite telephone query, there are the things that you have to consider. 
The main thing that you have to consider is the amount you are happy to spend. In the event that you  phone database are simply searching for land line telephone numbers, it should come at not cost to you. On the off chance that you are searching for unlisted and mobile phone numbers, it will include some significant pitfalls. This is because of PDAs not being openly enlisted, hence you need to pay for the administration. A normal registry should cost you around $4.00 every month which ought to incorporate boundless looks for a year.
On the off chance that you have an inclination that you don’t have to pay for the administration, you can take a stab at finding a free one. Be careful, a large portion of these catalogs just give you restricted data and are not refreshed consistently. To guarantee quality, the catalog ought to be refreshed consistently and give quality data. This implies you ought to get data on names, addresses, criminal records, warrant look, business look, neighborhood watches, and significantly more.
The exact opposite thing you need to search for in a converse assistance is the no hit, no charge strategy. This will permit you to look for your number and in the event that it isn’t in the registry, you basically don’t pay. It is a strategy that ought to consistently be remembered for the administration, and without it, you may not comprehend what you are accepting.

What factors of the integer dataset being sorted can I change, in order to compare two sorting algorithms?

I am comparing two comparison and binary data structure based sorting algorithms, the Tree Sort, and the Heap Sort. I am measuring the time taken for both algorithms to sort an increasing size of an integer dataset. However, I am wondering if there are any other variables which I can modify, for example standard deviation, in the integer dataset itself that would be of any benefit to my comparison.

What factors led to the poor reception and early cancellation of Wraith: The Oblivion?

Wraith: The Oblivion was published by White Wolf Games in 1994, but cancelled in 1999. Many of the books during the run show signs of being compressed together. While White Wolf typically published one book for each clan, bloodline, creed, etc., for Wraith two guilds were often smashed into one book. Wikipedia says that more books were planned, but never published.

Why was Wraith cancelled early? Wikipedia also mentions it was nominated for some awards, and it seems like a fairly novel game in some regards. It has a unique character that really sets it apart. I haven’t played it, but reading the books doesn’t reveal anything telling – it seems to use the same system as other White Wolf games with some reskinning.

Things I’ve tried:

  • I’ve tried asking the proprieters of my local game stores. They either don’t know anything, or remember that it sold poorly with no real explanation.
  • I’ve asked around my local World of Darkness tables. Many of the oldbies were playing in the 1990’s, but none admit to having played the game. Their responses mostly came down to "everyone knows this game sucks", with no explanation of why.
  • White Wolf advertised some listservs in its game books, but I haven’t been able to find archives of them.
  • I’ve read some reviews of Wraith. Most of them are rather artistic or critical in nature, rather than something that I suspect reflects the real played experience of actual tables. Additionally, none of them describe the game’s failure or why it was cancelled.

So what happened? Why was Wraith cancelled?

Finding the twiddle factors for FFT algorithm

I am trying to calculate the twiddle factors for the FFT algorithm and the iFFT algorithm, but i am unsure if i have it correctly calculated and was hoping some one could tell me if i have gone wrong as currently i get the wrong output for my FFT and i believe the twiddle factors might be the reason.

This is my code (in C#) to calculate them:

For _N = 4 and _passes = log(_N)/log(2) = 2

        //twiddle factor buffer creation         _twiddlesR = new Vector2[_N*_passes]; //inverse FFT twiddles         _twiddlesF = new Vector2[_N*_passes]; //forward FFT twiddles                  for (int stage = 0; stage < _passes; stage++)         {             int span = (int)Math.Pow(2, stage); // 2^n              for (int k = 0; k < _N; k++) // for each index in series             {                 int arrIndex = stage * _N + k; // get index for 1D array                                  // not 100% sure if this is correct for theta ???                 float a = pi2 * k / Math.Pow(2,stage+1);                  //inverse FFT has exp(i * 2 * pi * k / N )                 Vector2 twiddle = new Vector2(Math.Cos(a), Math.Sin(a));                  //forward FFT has exp(-i * 2 * pi * k/ N ) which is the conjugate                 Vector2 twiddleConj = twiddle.ComplexConjugate();                  /*this ternary checks if the k index is top wing or bottom wing                 the bottom wing requires -T top wing requires +T*/                  float coefficient = k % Math.Pow(2, stage + 1) < span ? 1 : -1;                  _twiddlesR[arrIndex] = coefficient * twiddle;                 _twiddlesF[arrIndex] = coefficient * twiddleConj;             }         } 

My debug data:

For inverse FFT twiddles:

First pass 1 + 0i 1 + 0i 1 + 0i 1 + 01 Second pass: 1 + 0i 0 + i 1 + 0i 0 + i 

For forward FFT twiddles:

First pass 1 + 0i 1 + 0i 1 + 0i 1 + 01 Second pass 1 + 0i 0 - i 1 + 0i 0 - i 

I am not convinced i have it right, but i am unsure what i have got wrong. Hoping some one who has a better understanding of this algorithm can spot my math error.

Canceling factors from numerator and denominator with complex number

I have the following expression:

$ \text{expr} = i \frac{\sqrt{a-b}}{\sqrt{b-a}}x$ .

Clearly simplifying expression gives $ \text{expr}=x$ by cancelling the numerator and denominator. However, I am not able to perform this simplification in Mathematica. I have tried PowerExpand, Cancel, Factor, FullSimplify and Simplify. Nothing works. Can anyone suggest any method?

External user gets ‘Access Required’ could it be down to these two factors?

I have an external user who’s trying to access a SharePoint online site but keeps getting the Access Request page.

I’ve been doing some rooting around and found that when I checked their permissions I get this:

enter image description here

There’s a lot of Deny in that list! I read in another post that I should go to Application Management in Site Collections and select Configure quotas but can’t find Application Management in Online

Also when I was in Active Sites and clicked on the site in question I saw a message on the right hand side saying ‘We couldn’t find the Office 365 group connected to this site’.

Could either or both be causing the access issue and how would I resolve?

Thanks in advance

What are the success factors to security and performance testing?

I’m performing a research regarding security and performance verifiction (testing and reviews). I would like to know the opinion of practitioners regarding my findings. Could you answer my survey at ?

Besides contributing to the evolving of knowledge, you will be helping those who need it – We will be donating R$ 1 to Red Cross (Brazil) for each of valid survey response.

Thank you very much!