mersenneforum.org Problem of the Month
 Register FAQ Search Today's Posts Mark Forums Read

2011-11-15, 04:21   #34
ccorn

Apr 2010

2338 Posts

Quote:
 Originally Posted by bcp19 It is obvious that I do not have the jargon for what I am trying to say, and your taking my words 10x more literal than I am causes a gulf of difference and an attitude than I am no longer going to deal with.
Let us redeclare the thread's topic to "tidbit of the month". For the following, you might have the jargon:

What does the following C code do to the binary representation of an unsigned int a?
Code:
a &= a - 1;

2011-11-15, 04:42   #35
bsquared

"Ben"
Feb 2007

3·17·73 Posts

Quote:
 Originally Posted by ccorn Let us redeclare the thread's topic to "tidbit of the month". For the following, you might have the jargon: What does the following C code do to the binary representation of an unsigned int a? Code: a &= a - 1;
clears the lowest set bit

p.s., I like the new suggestion... although maybe it should be tidbit of the week

Last fiddled with by bsquared on 2011-11-15 at 04:44

2011-11-15, 05:25   #36
axn

Jun 2003

34·67 Posts

Quote:
 Originally Posted by ccorn Let us redeclare the thread's topic to "tidbit of the month". For the following, you might have the jargon: What does the following C code do to the binary representation of an unsigned int a? Code: a &= a - 1;
Clears the least significant lit bit

2011-11-15, 21:40   #37
bcp19

Oct 2011

7·97 Posts

Quote:
 Originally Posted by R.D. Silverman Show me the 'factors' of: x^2 + 3x + 7 = 0
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:

x= (-b +/- sqrt(b^2 -4ab))/2a

x^2 + 3x + 7 = 0
x=(-b +/- sqrt(b^2-4ab))/2a
(-3 +/- sqrt(3^2-4(1)(7)))/2*1
(-3 +/- sqrt(3^2-28))/2
(-3 +/- sqrt(-19))/2
invalid function, cannot sqrt negative number

Dunno if you meant x^2+3x-7=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^2-3x-7=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 = (SN-12 - 2) mod (2P-1) 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 P-1 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 2011-11-15 at 21:47

2011-11-15, 22:12   #38
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by bcp19 invalid function, cannot sqrt negative number
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 2011-11-15 at 22:13

2011-11-15, 22:17   #39
bcp19

Oct 2011

7×97 Posts

Quote:
 Originally Posted by science_man_88 actually you can but in involves the complex number i . sqrt(-x) = sqrt(-1)*sqrt(x) = i*sqrt(x).
Obviously my brain did not pick up on imaginary numbers... the rust will come out eventually.

2011-11-15, 22:27   #40
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by bcp19 Obviously my brain did not pick up on imaginary numbers... the rust will come out eventually.
found this a while back ( and again since I wanted to show you): http://mathforum.org/library/drmath/view/61911.html

 2011-11-15, 22:27 #41 akruppa     "Nancy" Aug 2002 Alexandria 2,467 Posts pinhodecarlos, taunts are not particularly constructive, less so at the present time than usual. Please avoid them.
2011-11-15, 22:28   #42
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23×937 Posts

Quote:
 Originally Posted by bcp19 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.
Would anyone else care to comment on the attitude displayed above?
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????

 2011-11-15, 23:48 #43 literka     Mar 2010 110000002 Posts Try this problem, but only if you want to waste your time, since it is a very difficult problem: Let { ai }, where i=1,2...n, be a sequence of n real numbers. Assume that a1^2 +a2^2+...+an^2 is equal 1. Take all possible sums of the form abs(b1*a1 + b2*a2 +...+ bn*an) where bi 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.
2011-11-16, 00:20   #44
bcp19

Oct 2011

12478 Posts

Quote:
 Originally Posted by R.D. Silverman Would anyone else care to comment on the attitude displayed above? 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????
Yes, but, I need to figure out what I remember before I can go further, otherwise it is a repeat of high school. I know I worked through geometry, algebra, probability and statistics, can vaguely remember something about trigonometry and calculus. That said, I know I am in a sad state due to lack of 'activity'.

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.

 Similar Threads Thread Thread Starter Forum Replies Last Post Aillas PrimeNet 48 2012-02-15 19:17 cheesehead Soap Box 0 2009-01-24 07:59 em99010pepe No Prime Left Behind 5 2008-02-24 14:37 ltd Prime Sierpinski Project 22 2006-03-02 17:55 JuanTutors Lounge 6 2005-02-28 22:59

All times are UTC. The time now is 22:26.

Mon Dec 5 22:26:07 UTC 2022 up 109 days, 19:54, 0 users, load averages: 0.63, 0.93, 1.08