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 2012-11-27, 18:14 #1 Flatlander I quite division it     "Chris" Feb 2005 England 31·67 Posts Playing with WolframAlpha and musing. As someone who really struggles with maths I am interested in what a person with a 'good mathematical understanding'* sees when they look at something like this. Do they look at this and just understand or is this not obvious to such a person? *Define as appropriate. Attached Thumbnails
 2012-11-27, 18:55 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 32·1,117 Posts WolframAlpha is really good for one-offs: say you don't want to derive (and don't remember by heart) Sum(0<=i<=n)i^3, then it will tell you! However, with a little more obscure requests, it gives you some fluff, i.e. it tries to be helpful and applies every possible diversion, and every once in a while is been almost sarcastic (it would seem ;-) to the point of being absurd.
 2012-11-27, 20:11 #3 NBtarheel_33     "Nathan" Jul 2008 Maryland, USA 100010110112 Posts I have an advanced degree in math(s), if that counts, and I can tell you that the "simplification" provided by Alpha is something that no one (outside of a student trying to be a smart aleck) would ever make serious use of. It would be akin to asking for a glass of "aqueous dihydrogen monoxide" rather than "water". I suspect that it is an example of the computer's AI scheme running amok. On the other hand, at least it didn't reply to 64!/32! by loudly chanting "SIXTY-FOUR! over THIRTY-TWO!". There was a guy in a *college* class of mine that actually saw 3! and pronounced it "THREE!" and wondered why they wanted us to shout the numbers.
 2012-11-27, 20:43 #4 frmky     Jul 2003 So Cal 22×11×59 Posts Breaking it down, I first notice that $-1+\cos(128\pi)$ and $-1+\cos(64\pi)$ are both 0, so those terms drop out. Then, $2^{-64}/2^{-32} = 1/2^{32}$. Next, $\sin(64\pi)$ and $\sin(32\pi)$ are both 0, so $\pi^0$ is simply 1. Thus we conclude that this was constructed by a joker trying to be cute. We are then left with $\frac{64!}{32!}=\frac{128!!}{64!! \times 2^{32}$. This becomes clear one you write out a few terms of the expansion.
2012-11-28, 02:22   #5
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

1027410 Posts

Quote:
 Originally Posted by NBtarheel_33 There was a guy in a *college* class of mine that actually saw 3! and pronounced it "THREE!" and wondered why they wanted us to shout the numbers.
(this is cool, I have to remember it! )

Quote:
 Originally Posted by frmky We are then left with $\frac{64!}{32!}=\frac{128!!}{64!! \times 2^{32}$. This becomes clear one you write out a few terms of the expansion.
According with the former, the last should be ONE HUNDRED TWENTY EIGHT!! etc...

Last fiddled with by LaurV on 2012-11-28 at 02:23

 2012-11-28, 16:51 #6 Flatlander I quite division it     "Chris" Feb 2005 England 31·67 Posts I pronounce these as 'sixty-four pling', like a rather depressed microwave. Don't know where I picked that up.
 2012-11-28, 17:19 #7 bsquared     "Ben" Feb 2007 7·13·41 Posts Many people would maybe use a 'click' (although probably not English speakers), for example, the !Kung.
 2012-11-28, 20:55 #8 NBtarheel_33     "Nathan" Jul 2008 Maryland, USA 111510 Posts So, is there a symbol for "factorial missing the least n terms"? For instance, is there a notation for 64!/32! other than 64!/32!?
2012-11-28, 21:26   #9
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts

Quote:
 Originally Posted by NBtarheel_33 So, is there a symbol for "factorial missing the least n terms"? For instance, is there a notation for 64!/32! other than 64!/32!?
$\prod_{i=33}^{64}i$

 2012-11-28, 21:36 #10 bsquared     "Ben" Feb 2007 7×13×41 Posts Generalizing to any m! with n least terms missing: $\prod_{i=m-n+1}^{m}i$
 2012-11-28, 22:18 #11 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 32×1,117 Posts Relevant to PE 403, try this: Sum(over odd d from d1 to d2) (2 + (d3+5d)/3) and the same for even d values.

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