September 17th, 2018, 12:49 PM  #21 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Look at the third summand. To evaluate that, we shall have to multiply n by n. So this complex formula does not avoid the necessity to multiply n times n.

September 17th, 2018, 12:56 PM  #22 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Right! Missed that...to the corner I go...

September 17th, 2018, 01:15 PM  #23 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Missing that is hardly relevant. If n is even, the third summand is $2(n)(0.5n  1) = n^2  2n$, which essentially requires evaluating $n^2.$ If n is odd, the third summand is $2(n + 1)(0.5n  0.5) = n^2  1$, which essentialy requires evaluating $n^2.$ The whole thing is utterly pointless. Last edited by JeffM1; September 17th, 2018 at 01:23 PM. 
September 18th, 2018, 08:01 AM  #24 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 
OK. I am going to admit that I was guilty of slight hyperbole. You CAN calculate $n^2$ without TECHNICALLY multiplying n by itself. There are probably an infinite number of ways to do so. The formula given will involve complex numbers if for example n = 1/2, but we safely stay in the rational numbers if n is an integer. Under the formula, if n is even, the formula simplifies to $\dfrac{1  1}{2} + \dfrac{1 + 1}{2} * 2 \left (n + \dfrac{1  1}{2} \right ) + 2 \left (n + \dfrac{1  1}{2} \right )\left ( \dfrac{n + \dfrac{1  1}{2}}{2}  1 \right ) =$ $2n + 2n(0.5n  1).$ Now obviously $2n + 2n(0.5 n  1) = 2n + n^2  2n = n^2$, but, given any even integer n, you can compute 2n and (0.5n  1), multiply them together, and add the product to the previously calculated 2n to compute the square of n without actually ever multiplying n by itself. Similarly if n is odd, the formula simplies to $\dfrac{1 + 1}{2} + \dfrac{1  1}{2} * 2 \left (n + \dfrac{1 + 1}{2} \right ) + 2 \left (n + \dfrac{1 + 1}{2} \right )\left ( \dfrac{n + \dfrac{1 + 1}{2}}{2}  1 \right ) =$ $1 + 2(n + 1)(0.5n  0.5) = 1 + (n + 1)(n  1).$ Now obviously $1 + (n + 1)(n  1) = 1 + n^2  1 = n^2$, but, given any odd integer n, you can compute (n + 1) and (n  1), multiply them together, and add 1 to that product to compute the square of n without actually ever multiplying n by itself. As I mentioned before, if you do not restrict the formula to integers, you risk having to work with complex numbers. If you do restrict the formula to integers, it is confusingly and unnecessarily complex. It reduces to: $n \text { is an even integer } \implies n^2 = 2n + 2n(0.5n  1); \text { and }$ $n \text { is an odd integer } \implies n^2 = 1 + (n + 1)(n  1).$ It is true, but it does not have any specified practical use or theoretical interest. There are an infinite number of ways to express a given number. 
September 18th, 2018, 08:59 AM  #25 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 
Ya'll wanna have a good laugh at me? I just realised that the problem was a way to square a number without multiplying that number by itself. Duh! I thought that Jopus was simply trying to SIMPLIFY... To make up, here's the way to handle this: n = number to be squared r = random number Are you ready? Here: n^2 = nr  n(r  n) NO Jeff, I ain't multiplying n by n: brackets are evaluated first 
September 18th, 2018, 11:16 AM  #26  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Quote:
Still, I am not sure why the stack exchange folks shut him down. I work at one of the stack exchanges (not the math one obviously), and I am voting not to shut things down all the time. Some people seem to think that the purpose of the stack exchange is to suppress questions. I actually got a question reopened on Saturday. It was a minor triumph.  
September 18th, 2018, 05:15 PM  #27  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Quote:
 

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