quote:

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The never-ending Fraction

Evaluate the following to three decimal places:

1/(1 + 1/(1 + 1/(1 + 1/( ...

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I've found that if you go down the row (stopping at each point instead of going on), there are a few patterns that show up. For example, 1/(1 + 1) = 1/2. 1/(1 + 1/(1 + 1)) = 2/3. 1/(1 + 1/(1 + 1/(1 + 1))) = 3/5 Each time, the new fraction that is created has the numerator of the previous fraction's denominator. The difference between the new fraction's numerator and denominator is the sum of the difference of the numerator and denominator of the previous two fractions. In other words, the difference between 1 and 2 is 1 and the difference between 2 and 3 is 1, so 1 + 1 = 2, the difference between 3/5's numerator and denominator is 2. You can go on and see this trend. The next fraction will have a numerator of 5 and a denominator of 5 + 1 + 2 or 8, so it will be 5/8.

I would appreciate any help I can get on this. The teacher has changed the directions a little, too. She wants one solution explanation for getting it to three decimal places and one for getting it exactly. I have a pretty good idea on how to get it to three decimal places (I believe it's going to be roughly 0.618). I can likely explain that one. It's the exact one that gives me grief since I know this somehow uses a series, but I can't figure out how to create such a series.

Again, thank you for helping with this.

[ 09-22-2002: Message edited by: »Waisztarroz« ]

quote:

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Sabratiz probably says this to all the girls:
Personnaly I get 1/8 but im stupid and well dont get it that well.

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It doesn't divide quite like that. It's 1/(1 + (1/1 + 1/ ...

In other words, it's 1 divided by 1 plus 1 divided by 1 plus 1 divided by 1 plus 1 divided by 1 plus 1 divided by...

It continues on forever.

EDIT: Better diagram:

code:

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  1
-----
1 + 1
    ------
     1 + 1
         ------
          1 + 1
              ------
                ...   

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[ 09-22-2002: Message edited by: »Waisztarroz« ]

[ 09-22-2002: Message edited by: •Delidgamond• ]

quote:

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Tegadil had this to say about Knight Rider:
Where do they want you to stop?

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Positive Infinity.

the fractions are continually getting closer and closer to 1, and eventualy you will get a number so close that the difference is irrelevant.

The numbers on the bottom keep getting smaller and smaller and smaller because you are dividing by a number greater than 1, so each time you divide the number 1 with by the demoninator, you will get a number smaller then 1 each time which keeps getting progressivly smaller. So EVENTUALLY (think of limits) you will be adding 0 (not 0, but close enough that you can consider it 0) to the dominator so by the time you get to the top, the final line will be

code:

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  1
-----
1 + 0

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which = 1

[ 09-22-2002: Message edited by: •Delidgamond• ]

I'm not sure what I should put as to three decimal places, either...or how to explain how I got there.

[ 09-22-2002: Message edited by: »Waisztarroz« ]

quote:

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Verily, »Waisztarroz« doth proclaim:
Well, shit, now I have even less of an idea of how to get to a final answer than I did before.

I'm not sure what I should put as to three decimal places, either...or how to explain how I got there.

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Think Hyperbolas.

If you go down for a little while (I assumed my pattern would hold up the whole time, so I'm using it), you get 10946/17711~0.61803399. Another pattern I found that works to get the next denominator is to add the previous fractions numerator and denominator, so the next number is 17711/(17711 + 10946), 17711/28657, ~0.61803399.

Dammit, I bet I'm just doing something wrong. Yay, Ford sees my answer coming, also.

[ 09-22-2002: Message edited by: »Waisztarroz« ]

code:

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1
---------
1+1
  -------
  1+1
    -----
    1+1
      ---
      1+1
etc...

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code:

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1
----------
1 + n

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thus, you get 1/(1+n)=n
therefore, 1=n(1+n)
therefore 1=n+n*n
therefore n^2 + n - 1 = 0
using the quadratic formula
____________________________
Ax^2 + Bx + C=0

x= (-B plusminus sqrt( B^2-4AC )) / 2A
____________________________
n= (sqrt(5)-1) /2
n= 0.618033988749894848204586834365638

delid, where did your math go?

I'm kicking myself for not seeing such an easy algebra method, too.

[ 09-22-2002: Message edited by: »Waisztarroz« ]

quote:

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Ford Prefect thought this was the Ricky Martin Fan Club Forum and wrote:
Bah. BigMac pegged it less than a minute before me

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meh, it was the type of thing where if you've seen something similar before, you'll know how to tackle it as well...

although i must have spent a good 5 minutes typing up that damn solution.

quote:

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A sleep deprived Big_Mac stammered:
although i must have spent a good 5 minutes typing up that damn solution.

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My math went out the window when I started to think in terms of limits =)

[ 09-22-2002: Message edited by: •Delidgamond• ]

You set n to equal everything after the

code:

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  1
-----
1 +(1
   -----
   1 + 1
etc...)

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even though it stretches to infinite, the entire thing goes to infinite + 1.

code:

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  1
----- not = n
1 + n

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[ 09-22-2002: Message edited by: •Delidgamond• ]

when i said that 1/(1+n)=n it was a transformation similar to 0.99999999... =1.

It is perfectly correct, and perfectly legal.

-Actually, I think it's one of those problems you take the limit of. You know, as the permutations head to infinite, the result heads to 0.

-if 1/(1+n) is how you want to look at it, n=1/(1+n) (and thus 1/(1 + 1/(1 + 1/(1+... It's a recursive function for any of you that are programmers. 1+n will be greater than 1 unless n = 0, and you can get n=0 from taking the limit of the problem.

It's a very strange problem actually. I suppose it's one of those math problems that can have more than one answer.

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Check out the big brain on Landale!
heh nm. I did like 10000 permutations, and came up with like 0.61 and some more numbers after it. Sorry dude, guess that prolly doesn't help a whole lot.

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I already solved it, so no, it doesn't.