20111115, 04:21  #34  
Apr 2010
233_{8} Posts 
Quote:
What does the following C code do to the binary representation of an unsigned int a? Code:
a &= a  1; 

20111115, 04:42  #35  
"Ben"
Feb 2007
3·17·73 Posts 
Quote:
p.s., I like the new suggestion... although maybe it should be tidbit of the week Last fiddled with by bsquared on 20111115 at 04:44 

20111115, 05:25  #36 
Jun 2003
3^{4}·67 Posts 

20111115, 21:40  #37 
Oct 2011
7·97 Posts 
So I may post again... I was technically wrong when I solved x=4.54 being tired and from lack of practice, but you were in too much of a hurry or something:
Using the quadratic formula. x= (b +/ sqrt(b^2 4ab))/2a x^2 + 3x + 7 = 0 x=(b +/ sqrt(b^24ab))/2a (3 +/ sqrt(3^24(1)(7)))/2*1 (3 +/ sqrt(3^228))/2 (3 +/ sqrt(19))/2 invalid function, cannot sqrt negative number Dunno if you meant x^2+3x7=0 which would be: (3 +/ sqrt(3^2  4(1)(7)))/2*1 (3 +/ sqrt(9+28))/2 (3 +/ sqrt(37))/2 (3 + sqrt(37))/2 = 1.54138127 (3  sqrt(37))/2 = 4.54138127 then x= 1.541 or 4.541 Or if you meant x^23x7=0 (3 +/ sqrt(3^2  4(1)(7)))/2*1 (3 +/ sqrt(9+28))/2 (3 +/ sqrt(37))/2 (3 + sqrt(37))/2 = 1.54138127 (3  sqrt(37))/2 = 4.54138127 then x= 4.541 or 1.541 35 years ago I could 'solve' a rubics cube in under 30 seconds. Today, it'd probably take me 30 minutes. Why? Practice. On a scale of 1 to 10, my mathematical brain is sitting a 1 due to lack of practice while yours is sitting at 10 from constant use. Your flinging useless equations at me while I am trying to explain something from so long ago that I have probably forgotten more than I remember tells me that you fail to see the big picture. I can look at the S0 = 4, SN = (SN12  2) mod (2P1) and figure out that S1=14 mod P, S2=194 mod P, s3=37634 mod P, etc for the exponents currently in use. I can also see how 2kp+1 is used in the P1 factoring. This is the beginning of understanding. Can I write a proof? nope Can I code this? nope Can I appreciate the math involved even though it surpasses me? Yep Last fiddled with by bcp19 on 20111115 at 21:47 
20111115, 22:12  #38 
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
actually you can but in involves the complex number i .
sqrt(x) = sqrt(1)*sqrt(x) = i*sqrt(x). Last fiddled with by science_man_88 on 20111115 at 22:13 
20111115, 22:17  #39 
Oct 2011
7×97 Posts 

20111115, 22:27  #40  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:


20111115, 22:27  #41 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
pinhodecarlos, taunts are not particularly constructive, less so at the present time than usual. Please avoid them.

20111115, 22:28  #42  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×937 Posts 
Quote:
In particular the part about "flinging useless equations" and the part about "fail to see the big picture". I ask bcp19: Are you interested in learning or not???? 

20111115, 23:48  #43 
Mar 2010
11000000_{2} Posts 
Try this problem, but only if you want to waste your time, since it is a very difficult problem:
Let { a_{i} }, where i=1,2...n, be a sequence of n real numbers. Assume that a_{1}^2 +a_{2}^2+...+a_{n}^2 is equal 1. Take all possible sums of the form abs(b_{1}*a_{1} + b_{2}*a_{2} +...+ b_{n}*a_{n}) where b_{i} is sign (is equal 1 or 1) and abs(x) means absolute value of x. This way we construct 2^n real numbers. Show that at least half of these numbers is less or equal 1. 
20111116, 00:20  #44  
Oct 2011
1247_{8} Posts 
Quote:
One of the problems in working with electronics, there is so much you don't deal with. Thus seeing your x^2+3x+7=0 my first reaction is solve for x because that is what I am used to: constants. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
The Month of Double Check effort  Aillas  PrimeNet  48  20120215 19:17 
WhineroftheMonth  cheesehead  Soap Box  0  20090124 07:59 
One month of NPLB  em99010pepe  No Prime Left Behind  5  20080224 14:37 
Best month ever for PSPs prp effort  ltd  Prime Sierpinski Project  22  20060302 17:55 
New Month's Resolution  JuanTutors  Lounge  6  20050228 22:59 