The Design Expert software gives me the picture of resolution for the fractional factorial design, in which Resolution III, IV and V are most critical.  But can I ask, how can I determine the resolution by myself, based on the number of factors?
120814
In design of experiments, is there an intuitive way to understand (and explain) the idea of the "resolution" of the design?
124267
Does anyone know where the factorial "!" symbol came from?  I can't decide if it is my favorite or least favorite notation in mathematics...
102527
If  $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$  $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$  could some help me with this, thanks
17369
For a very specific reason I need to check the computer form factor(Desktop, Notebook or All-in-one) remotelly, using a command  through the command-prompt, powershell or vbs(Windows 7/8 systems).   There is any way to do such thing?  Thanks in advance.
98
Here's the very dumb way:  def divisorGenerator(n):     for i in xrange(1,n/2+1):         if n%i == 0: yield i     yield n   The result I'd like to get is similar to this one, but I'd like a smarter algorithm (this one it's too much slow and dumb :-)  I can find prime factors and their multiplicity fast enough.  I've an generator that generates factor in this way:  (factor1, multiplicity1) (factor2, multiplicity2) (factor3, multiplicity3) and so on...  i.e. the output of   for i in factorGenerator(100):     print i   is:  (2, 2) (5, 2)   I don't know how much is this useful for what I want to do (I coded it for other problems), anyway I'd like a smarter way to make  for i in divisorGen(100):     print i   output this:  1 2 4 5 10 20 25 50 100     UPDATE: Many thanks to Greg Hewgill and his "smart way" :) Calculating all divisors of 100000000 took 0.01s with his way against the 39s that the dumb way took on my machine, very cool :D  UPDATE 2: Stop saying this is a duplicate of  post. Calculating the number of divisor of a given number doesn't need to calculate all the divisors. It's a different problem, if you think it's not then look for "Divisor function" on wikipedia. Read the questions and the answer before posting, if you do not understand what is the topic just don't add not useful and already given answers.
5200
In the Pottermore article on  it's mentioned that other fractional platforms exist (7 1/2, for instance). Is it ever stated elsewhere where these platforms lead? Perhaps you could get a connection from your hometown by accessing one of these platforms?
59048
I saw this approximation for the factorial given by Ramanujan as $$\log(n!) \approx n \log n - n + \frac{\log(n(1+4n(1+2n)))}{6} + \frac{\log(\pi)}{2}$$ in , which claims the approximation is superior to Stirling's approximation. I tried to locate the reference but unfortunately I could not. I would appreciate if someone can throw light on how this asymptotic is obtained and the order of the error.
98444
I understand that the factorial gives the number of arrangements. For example: the factorial of zero i.e. an empty set ( doesn't occur) is 1.   As the empty set can be arranged only in 1 way - i.e. by filling nothing. Now, let's take an example: 5 distinct seats. How many ways 5 distinct seats can be arranged? - 5! ways i.e. 120.  So, basically, factorial gives us the arrangements. Now, the question is why do we need to know the factorial of a negative number?, let's say -5. How can we imagine that there are -5 seats, and we need to arrange it? Something, which doesn't exist shouldn't have an arrangement right? Can someone please throw some light on it?. I saw some research done on the formula -
8402
I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.  My attempt at the program was as follows:   A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.   LONG and TEDIOUS way: $$(8k)^2 = 0 (mod 8)$$ $$(8k+1)^2 = 1 (mod 8)$$ $$(8k+2)^2 = 4 (mod 8)$$ $$(8k+3)^2 = 1 (mod 8)$$ $$(8k+4)^2 = 0 (mod 8)$$ $$(8k+5)^2 = 1 (mod 8)$$ $$(8k+6)^2 = 4 (mod 8)$$ $$(8k+7)^2 = 1 (mod 8)$$  That is, using three of these modulo there is no way to arrive at  $$a^2 + b^2 + c^2 = 7 (mod 8)$$
81954
Is there any way I can 'undo' the factorial operation? JUst like you can do squares and square roots, can you do factorials and factorial roots (for lack of a better term)?   Here is an example: 5! = 120.  Is there a way I can work out the number that must be factorialed (??) to give the answer 120?
25345
Does it exist some tikz register (let say \zoomfactor) that provides the curent scale factor. For example can I do something like :   \begin{tikzpicture}[scale=2]     \begin{scope}[scale=1.5]         \draw (0,0) circle (\zoomfactor);     \end{scope} \end{tikzpicture}   and to obtain circle of size 3=2*1.5 ?
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The number of different ways to choose $k$ out of $n$ unique items:   Without repetition and without order-significance: $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ Without repetition and with order-significance: $\binom{n}{k}\times k!=\frac{n!}{(n-k)!}$ With repetition and with order-significance: $n^k$ With repetition and without order-significance: $?$   Thank you!
2387
I'm trying to solve hard combinatorics that involve complicated factorials with large values.  In a simple case such as $8Pr = 336$, find the value of $r$, it is easy to say it equals to this: $$\frac{8!}{(8-r)!} = 336.$$  Then $(8-r)! = 336$ and by inspection, clearly $8-r = 5$ and $r = 3$.  Now this is all and good and I know an inverse function to a factorial doesn't exist as there is for functions like sin, cos and tan etc. but how would you possibly solve an equation that involves very large values compared to the above problem without the tedious guess and checking for right values.  Edit: For e.g. if you wanted to calculate a problem like this (it's simple I know but a good starting out problem) Let's say 10 colored marbles are placed in a row, what is the minimum number of colors needed to guarantee at least $10000$ different patterns? WITHOUT GUESS AND CHECKING  Any method or explanation is appreciated!
90946
My task is to factorize $n$ knowing it has two factors and $\varphi(n)$. How can I do it?  Thanks for any advice.
38130
Is there an equation to determine the Ability Modifier?  I'm not talking about the chart that lists   1 = −5 2-3 = −4 4-5 = −3 6-7 = −2 etc.   I mean a legit equation that, if I somehow misplaced this information, I could determine what the modifier is based on the score?
65612
I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, etc.  In general how do we compute $_nC_k$ when $n$ is less than $k$? Do we still use the formula $\frac{n!}{(n-k)!k!}$ and use the gamma function to compute the negative factorial?  Thanks!
114686
How to deal with $m+\frac{4}{m^2}\geq3$ for every $m &gt; 0$ ? I multiplied both sides by $m^2$ and got $m^3+4-3m^2\geq0$ and have no idea how to continue with this