Continued & Infinite Continued Fractions/Chapter 8

Below is Grace's work on fractions to continued fractions, and continued fractions to fractions in Chapter 8:

Can you make a continued fraction equal to  27/8 ?

Make a continued fraction, then find the simplest fraction equal to it.


6 May 2010 It's been a long time since I introduced a 9 year old to continued fractions, and in the process Sara did dividing fractions, reciprocals, reducing fractions, changing an mixed number to an improper fraction, drawing a picture of 2/3 and 1 2/3, dividing and multiplying fractions and using a TI-84 a TI-84- changing a fraction to a decimal and vice-versa and writing a mixed number as an improper fraction on the calculator.

I gave Sara the continued fraction below; after a little thought, she said 3 1/5.  It soon became clear to me that Sara didn't know how to divide fractions, which was the first thing  she needed to do: 1 divided by 3 1/5. So instead of getting the answer to the original question, I worked on the division 1 divided by 1/2. To set the stage I asked her how many 4's make 12, or 12 4, which she knew was 3. Then I had her do 1 = 2 = 2/1, 1 1/3= 3..Then I gave her 1 2/3 for which I drew the 1 cake and she showed me what 2/3 looked like and she figured out that  there are 1 1/2 = 3/2 of the 2/3 in 1, so 1 2/3 = 3/2. I told her the reciprocal of 2/3 is 3/2. 

Then Sara made up the problem 9 6/8 and she figured out this would equal 12. I showed her that for 9 6/8, 1 6/8 = 8/6, so 9 6/8= 9x8/6 = 72/6 = 12 ( 9/1x8/6 = 72/6= 12) and this agrees with her answer. Sara was really excited about dividing fractions and multiplying she knew. We didn't finish the problem, but she learned how to divide fractions!

The next week, after a hug for Don, she finished the continued fraction, getting 3 16/37 !

Fine job Sara!


I gave Maya the continued fraction below which was simpler than the one I gave Sara.

Maya knew that 2+1/7 = 15/7, then she did the rest, 1 15/7 = 7/15, and she finished the problem with 1+7/15 = 1 7/15.

Fine job Maya!


Solving a quadratic equation by getting an infinite continued fraction- method #2 (see also solving a quadratic equation 10 ways on the map):

We can solve the quadratic equation x2 - 5x + 6 = 0 to get x = .
We then take this function and in place of x we'll put , since this is equal to x, and we get . Then we'll put in for x again, and we get . If we keep doing that we get an infinite continued fraction. Jonathan, at age 7, did this at home with his Mom in CT, using Don's original book. He got all excited about this being infinite! (He came with his Dad for a week, in each of the next 2 summers).


If you put 1 in the first of these you get -1. If you put 1 in the second one you get 11. If you put 1 in for x in the third one you get 4.4545.. You get an infinite sequence which approaches 3, which is one of the two roots of the original quadratic equation!


If you start with -100, you would still get an infinite sequence approaching 3 as well. Notice, you only get the one root, 3, using this method #2 for solving the quadratic equation x2 - 5x + 6 = 0.

 

Solving the same quadratic equation x2 - 5x + 6 = 0 by getting an infinite continued fraction- method #3

Solve for x another way to get:  x = then iterate the right side to get and so on to get an infinite continued fraction. You can also put a number in for x, in , get the number out, put that number in for x, get a new number out, and continue that way to get an infinite sequence.

 

Solving the same quadratic equation x2 - 5x + 6 = 0 by getting an infinite continued fraction- method #4

Solve for x another way to get:  x = then iterate the right side to get

and so on to get an infinite continued fraction

Iterating this same function -6/(x-5) in Mathematica using NestList, starting with 5, doing 1000 iterations, showing 250 digits, and looking at the last one, we get an infinite sequence that approaches 2, one of the roots of the original equation.

h[x_]:= -6/(x-5)

Take[N[NestList[h,5,1000],250],-1]

Power::infy: Infinite expression \[NoBreak]1/0\[NoBreak] encountered.

{1.99999999999999999999999999999999999999999999999999999999999

9999999999999999999999999999999999999999999999999999999999999

99999999999999999999999999999999999999999999999999999998784283

8015208649994273601205071405411499043769607328752211168623914688150}

Notice, when it uses 5, it gives "infinite expression encountered, then continues to do 999 more. I wondered what Mathematica did, so I started with 5 again, but only did 6 iterations and looked at all of them:

N[NestList[h,5,6],10]

Power::infy: Infinite expression \[NoBreak]1/0\[NoBreak] encountered.

{5.000000000,ComplexInfinity, 0,1.200000000,1.578947368,1.753846154,1.848341232}

Looking at the 6 numbers above, Mathematica showed the complexInfinity when 5 is put in for x, then put 0 in for x, and then continued on as if nothing happened. Hmmm.

 


Solve  x2 - x -1 = 0 for an infinite continued fraction:

We got the infinite continued fraction for , the golden mean, from the equation  x2 - x -1 = 0 . We added x and 1 to both sides to get  x2   = x + 1. Then we divided both sides by x to get  

     . We then put            in 

 

for x on the right side, to get  

 

and continued this process. to get the

infinite continued fraction. On the way we graphed         ,                          and

                                              on the same axes above.

 

Notice above, we are iterating the function 

The straight line and parts of 2 hyperbolas all intersect at the two points (1.618, 1.618) and ( -0.618, -0.618), which are the 2 solutions of the quadratic equation x2 - x -1 = 0 and are (the Golden Mean), and 1- (see chapter 7). The infinite continued fraction for is:


An infinite continued fraction for 4/p, by Lord Brouncker, circa 1658--see Olds

 

An infinite continued fraction for e, see chapter 11), by Euler, circa 1737- see Olds:

 

 


Find the infinite continued fraction for (hint: start with x2 = 2, and x2 -1 = 1 and (x-1)*(x+1)=1 and solve for x on the left side). Try this yourself.