quick number pattern - please help quick! it's easy!

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[ARCHIVED THREAD] - quick number pattern - please help quick! it's easy!

Posted: 9/20/2004 4:00:03 PM EDT

256, 16, 4, 2, then what?

i see that 2x2=4, 4x4=16, etc,

but what times itself is 2?

Posted: 9/20/2004 4:01:54 PM EDT

[#1]

square root of 2, or 2^(1/2)

Posted: 9/20/2004 4:02:37 PM EDT

[#2]

Quoted:
256, 16, 4, 2, then what?

i see that 2x2=4, 4x4=16, etc,

but what times itself is 2?

The square root of 2, approximately 1.414

Posted: 9/20/2004 4:03:07 PM EDT

[#3]

It's a "complex" number. (Seriously, that's the real name) Denoted as Square Root of 2, with the little Square root bar.

Posted: 9/20/2004 4:05:03 PM EDT

[#4]

thanks

Posted: 9/20/2004 4:21:45 PM EDT

[#5]

Quoted:
It's a "complex" number. (Seriously, that's the real name) Denoted as Square Root of 2, with the little Square root bar.

y = sqrt(x), where x is real, is by definition the positive (and thus real) number with the property that y*y = x.

Posted: 9/20/2004 4:26:24 PM EDT

[#6]

Mmmk, so what do we get for helping you with your homework?

Posted: 9/20/2004 4:27:03 PM EDT

[#7]

note to self, check math before posting

Posted: 9/20/2004 5:10:31 PM EDT

[#8]

Seems obvious to me :

256 = 2^8
16 = 2^4
4 = 2^2
2 = 2^1
1 = 2^0

But then I'm a software engineer, and we tend to think in binary anyhow. And, obviously, I work in integer math

.

Posted: 9/20/2004 5:15:25 PM EDT

[#9]

Quoted:
Seems obvious to me :

256 = 2^8
16 = 2^4
4 = 2^2
2 = 2^1
1 = 2^0

But then I'm a software engineer, and we tend to think in binary anyhow. And, obviously, I work in integer math .

what he said!

Posted: 9/20/2004 5:36:05 PM EDT

[#10]

wolffie,

It's not a complex number, it's an irrational number. Complex numbers contain a factor of the square root of negative one which this number does not have. An irrational number has no repeating sequence, and an infinite number of digits to define it.

Posted: 9/20/2004 6:32:27 PM EDT

[#11]

but the pattern does not work...you are missing all of the binary digits 2^3, 2^5, 2^6, 2^7 of the 9 bits
Also, if you are trying to pattern the 8,4,2,1,0 then what is the pattern?
divide the exponent by 2? then explain the zero... it would be 1/2, as in the square root...

Quoted:
Seems obvious to me :

256 = 2^8
16 = 2^4
4 = 2^2
2 = 2^1
1 = 2^0

But then I'm a software engineer, and we tend to think in binary anyhow. And, obviously, I work in integer math .

Posted: 9/20/2004 7:31:50 PM EDT

[#12]

but the pattern does not work...you are missing all of the binary digits 2^3, 2^5, 2^6, 2^7 of the 9 bits
Also, if you are trying to pattern the 8,4,2,1,0 then what is the pattern?
divide the exponent by 2? then explain the zero... it would be 1/2, as in the square root...

Think bitshifting.

Posted: 9/20/2004 9:20:16 PM EDT

[#13]

Quoted:
but the pattern does not work...you are missing all of the binary digits 2^3, 2^5, 2^6, 2^7 of the 9 bits
Also, if you are trying to pattern the 8,4,2,1,0 then what is the pattern?
divide the exponent by 2? then explain the zero... it would be 1/2, as in the square root...

Think bitshifting.

I considered this, but what is the pattern of the bitshift? 1, 1, 2, 4? It is not a well defined series in that sense. 1, 1, 2, 3 I would give you (Fibonacci).

Posted: 9/20/2004 9:22:28 PM EDT

[#14]

I hate math.

Posted: 9/20/2004 9:31:30 PM EDT

[#15]

Quoted:
Seems obvious to me :

256 = 2^8
16 = 2^4
4 = 2^2
2 = 2^1
1 = 2^0

But then I'm a software engineer, and we tend to think in binary anyhow. And, obviously, I work in integer math .

thats what I was thinking, but it fell apart due to the abovementioned missing 2^3 2^5 2^6 and 2^7

Posted: 9/20/2004 9:55:34 PM EDT

[#16]

What the heck are you guys doing here???

The formula is Xn = SQRT(Xn-1).

Written another way:

Xn = (Xn-1)^1/2

It was seeded (initialized) with a value (Xo = 65536) such that X1 = SQRT (65536) = 256.

n=1, X = 256
n=2, X = 16
n=3, X = 4
n=4, X = 2
n=5, X = SQRT(2)
n=6, X = (2)^1/4
n=7, X = (2)^1/8

n..........Xn-1............................Xn
0..................................................65536
1..............65536..............................256
2.............. 256..................................16
3..............16.....................................4
4.............. 4......................................2
5.............. 2..............................1.414213562
6.............. 1.414213562..............1.189207115
7.............. 1.189207115..............1.090507733
8.............. 1.090507733..............1.044273782
9.............. 1.044273782..............1.021897149
10............ 1.021897149..............1.010889286
11............ 1.010889286..............1.005429901
12............ 1.005429901..............1.002711275
13............ 1.002711275..............1.00135472
14............ 1.00135472..............1.000677131

Posted: 9/21/2004 10:47:08 AM EDT

[#17]

CFII,

If you hate math so much, what is your profession????

Posted: 9/21/2004 10:57:31 AM EDT

[#18]

Ooh ooh ooh, I got it!

The answer is Pi!

... as in I like pi.

Posted: 9/21/2004 11:10:14 AM EDT

[#19]

Quoted:
Quoted:
but the pattern does not work...you are missing all of the binary digits 2^3, 2^5, 2^6, 2^7 of the 9 bits
Also, if you are trying to pattern the 8,4,2,1,0 then what is the pattern?
divide the exponent by 2? then explain the zero... it would be 1/2, as in the square root...

Think bitshifting.

I considered this, but what is the pattern of the bitshift? 1, 1, 2, 4? It is not a well defined series in that sense. 1, 1, 2, 3 I would give you (Fibonacci).

Bitshift is moving all digits left one and adding a zero, dropping off anything greater than 8 bits (or 16, or double, or whatever).


00000001 = 1
00000010 = 2
00000100 = 4
00001000 = 8
00010000 = 16
00100000 = 32
01000000 = 64
10000000 = 128

Other words: To multiply by two, left shift one bit, to divide by 2, right shift one bit.

Both the irrational solution (2^1/2) and the binary/computer-esque solution (powers of 2 to the powers of powers of two) are valid answers with the information given. Without knowing the next in the sequence, we will never know which answer it really is.

--ETA The irrational solution would continue above approaching 1 as Mike_Mills posted above, the "binary" solution would result in 1's after next step (always 2^0)

Posted: 9/21/2004 12:04:35 PM EDT

[#20]

Not to be a smartass (better than being a dumbass, though

), but this has relevance in real life how?

Posted: 9/21/2004 12:09:24 PM EDT

[#21]

Well... if you are a programmer, you'd better know a couple of things about bitshift...

Dunno about 20iner's pattern problem though...

Posted: 9/21/2004 3:28:13 PM EDT

[#22]

After reading my reply, I notice that 32 and 128 are conspiculously missing from the example sequence.

Therefore it is a squares solution and the answer is 1.414-ish (sqrt(2))

Posted: 9/21/2004 3:38:47 PM EDT

[#23]

1 is the answer as has been explained by the others.

Would it help to know that the number that comes before 256 in that sequence is 65536 and before that would be 4294967296?

Posted: 9/21/2004 3:42:46 PM EDT

[#24]

Quoted:
Not to be a smartass (better than being a dumbass, though ), but this has relevance in real life how?

People who can spot patterns and know how to explain and predict the trends can get jobs working in air conditioned buildings. If you lived here in Arizona, you'd know exactly how relevant that was to real life.

Posted: 9/21/2004 3:57:21 PM EDT

[#25]

Yep, people who don't know math tend to work for those who do. It's not always true, but is more often true than not.

Want to earn big bucks as you help your nation lead the way in technology, weaponry and economy? LEARN MATH!

Posted: 9/21/2004 4:54:26 PM EDT

[#26]

Quoted:
1 is the answer as has been explained by the others.

Would it help to know that the number that comes before 256 in that sequence is 65536 and before that would be 4294967296?

Ok, let me get this straight:

4294967296 (2^32)
65536 (2^16)
256 (2^8)
16 (2^4)
4 (2^2)
2 (2^1)
1 (2^0)

So your proposed exponents (bit positions) are 32,16,8,4,2,1,0. Or, if you wish, bit shifts of 16, 8, 4, 2, 1, 1. Sorry, that doesn't compute. 0*2 != 1.

Posted: 9/21/2004 6:13:55 PM EDT

[#27]

That entire train of reasoning is wrong. The series is as I explained. The next in the series is sqrt(2).

Posted: 9/21/2004 7:06:36 PM EDT

[#28]

Quoted:
Well... if you are a programmer, you'd better know a couple of things about bitshift...

Dunno about 20iner's pattern problem though...

actually, it is ANY shift with ANY base

Shifting, by the base, multiplies or divides by the base.
Example:
in base 10: 1 * 10 = 10 which is 1 shift left (2 * 10 = 20, and so on....)
and in base2 = you see above examples

here is the digit format for base2
....,2^7,2^6,2^5,2^4,2^3,2^2,2^1,2^0,2^-1,2^-2,2^-3,2^-4,....