My Math Forum Formula for n^2 without explicitly performing n*n:
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 September 17th, 2018, 07:22 AM #1 Member   Joined: Feb 2013 From: London, England, UK Posts: 38 Thanks: 1 Formula for n^2 without explicitly performing n*n: I am reposting this here, because it's being shut down on Maths Stack Exchange. There's some snobby people on there who don't like stuff like this. Thanks!! Formula for n^2 without explicitly performing n*n: $$n^2 = [(1 - (-1)^n)/2] + [(1 + (-1)^n)/2] * [4 * ({n + (1 - (-1)^n)/2}/2)] + [4 * ({n + (1 - (-1)^n)/2}/2)] * [({n + (1 - (-1)^n)/2}/2) - 1]$$ That's it. Hopefully, I've not made any errors in it. Here's a simplified version: Let s = (1 - (-1)^n)/2 ; and t = (1 + (-1)^n)/2 Then, expression for n^2 is as follows: n^2 = s + t * 4 * ((n + s)/2) + 4 * ((n + s)/2) * ((n + s)/2 - 1)
 September 17th, 2018, 09:18 AM #2 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Perhaps they are not snobby. Maybe they just don't see the point. Is this supposed to be more efficient for a computer or calculator? I don't have the requisite expertise to answer that. If this is your claim, then post some data so that the computer gurus on this site can assess the efficiency of your claim. If this is supposed to be more efficient for a human, then I would try $\displaystyle n^2=(n-d)(n+d)+d^2$. For a well chosen small value of $\displaystyle d$, you can square $\displaystyle n$ in your head. If it is neither of these, then you should really say what the point is.
 September 17th, 2018, 09:21 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $$n^2=\sum_{k=1}^n(2k-1)$$ Thanks from Sebastian Garth and JeffM1
 September 17th, 2018, 09:27 AM #4 Member   Joined: Feb 2013 From: London, England, UK Posts: 38 Thanks: 1 @mrtwhs The point isn't that there is no point. But equally, need everything have a point? I'm not making any claims, it's a statement of fact. I don't need a disclaimer because nothing is being claimed. I simply chanced upon this as I was doing other research, thought it might be interesting or useful to someone at some point, so posted it. I personally have no real use for it, but felt, perhaps misguidedly, that it might be of utility to others.
September 17th, 2018, 09:37 AM   #5
Senior Member

Joined: Feb 2010

Posts: 711
Thanks: 147

Quote:
 Originally Posted by Jopus @mrtwhs The point isn't that there is no point. But equally, need everything have a point? I'm not making any claims, it's a statement of fact. I don't need a disclaimer because nothing is being claimed. I simply chanced upon this as I was doing other research, thought it might be interesting or useful to someone at some point, so posted it. I personally have no real use for it, but felt, perhaps misguidedly, that it might be of utility to others.
OK! Cool! But I am curious ... what did the other board do that made you think they were snobby?

September 17th, 2018, 09:47 AM   #6
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
Thanks: 1038

Quote:
 Originally Posted by Jopus Formula for n^2 without explicitly performing n*n: $$n^2 = [(1 - (-1)^n)/2] + [(1 + (-1)^n)/2] * [4 * ({n + (1 - (-1)^n)/2}/2)] + [4 * ({n + (1 - (-1)^n)/2}/2)] * [({n + (1 - (-1)^n)/2}/2) - 1]$$
Why do you not simply say: simplify right side ?

Last edited by Denis; September 17th, 2018 at 09:51 AM.

 September 17th, 2018, 10:01 AM #7 Member   Joined: Feb 2013 From: London, England, UK Posts: 38 Thanks: 1 @mrtwhs Sorry, I wasn't trying to be surly or anything. It's just that a few people on the MSE website basically shut the question down. I can't see that there's a good reason. I'm not saying it's all that useful, just that it might be, in case someone is looking for some such thing at some point in the future. Speaking from personal experience, I often encounter interesting things in maths which seem irrelevant to the matter at hand at a particular juncture of research, but the trick is to write them up semi-neatly, and then archive the little result, no matter how trifling or seemingly insignificant. Then invariably, some other avenue opens up down the road, at which point the little finding becomes useful and applicable.
September 17th, 2018, 10:07 AM   #8
Senior Member

Joined: Feb 2010

Posts: 711
Thanks: 147

Quote:
 Originally Posted by Jopus Formula for n^2 without explicitly performing n*n: $$n^2 = [(1 - (-1)^n)/2] + [(1 + (-1)^n)/2] * [4 * ({n + (1 - (-1)^n)/2}/2)] + [4 * ({n + (1 - (-1)^n)/2}/2)] * [({n + (1 - (-1)^n)/2}/2) - 1]$$ Here's a simplified version: Let s = (1 - (-1)^n)/2 ; and t = (1 + (-1)^n)/2 Then, expression for n^2 is as follows: n^2 = s + t * 4 * ((n + s)/2) + 4 * ((n + s)/2) * ((n + s)/2 - 1)
Well these don't match. I substituted $\displaystyle n=3$ into the bottom formula and correctly got $\displaystyle 3^2=9$.

However the top formula gave $\displaystyle 3^2=36$.

I think there is a typo in the top formula.

September 17th, 2018, 10:13 AM   #9
Senior Member

Joined: Aug 2012

Posts: 2,355
Thanks: 737

Quote:
 Originally Posted by mrtwhs Perhaps they are not snobby.
Math.stackexchange is snobby. In a good cause, since there are professional mathematicians there answering technical questions. But snobby for sure. No sense of humor, no discussions, no poorly phrased questions allowed, etc.

September 17th, 2018, 10:18 AM   #10
Math Team

Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 14,597
Thanks: 1038

Quote:
 Originally Posted by Jopus $$n^2 = [(1 - (-1)^n)/2] + [(1 + (-1)^n)/2] * [4 * ({n + (1 - (-1)^n)/2}/2)] + [4 * ({n + (1 - (-1)^n)/2}/2)] * [({n + (1 - (-1)^n)/2}/2) - 1]$$
...may we ask what does this ridiculously long equation represent?
Why stuff like /2/2 and not /4?

 Tags explicitly, formula, performing

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post velasco10 Advanced Statistics 1 November 24th, 2014 07:41 AM LordRaaa Algebra 3 May 15th, 2012 10:01 AM Alex1 Algebra 0 October 29th, 2010 05:38 PM agro Probability and Statistics 3 August 27th, 2009 06:17 AM wmoonw Algebra 1 January 16th, 2009 12:15 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top