1. ## Continued fraction

I am asked the following question..

9. Evaluate the continued fraction x = 1 + 1/ [ 2 + 1/ [ 3 + 1/ [ 2 + 1/ [ 3 + 1/ ... ] ] ] ]
Tabulate the first 6 convergents, and find the exact value of x.
I am not wondering the answer, but just what is being asked here.

What does "tabulate the first 6 convergents" mean exactly?

Does this mean just continue the pattern up to 6 terms?

2. ## Re: Continued fraction

Hello, ehpoc!

$x \;=\;1 + \dfrac{1}{2 + \dfrac{1}{3 + \dfrac{1}{2 + \dfrac{1}{3 + \hdots }}}}}}$

I am not wondering the answer, but just what is being asked here.

What does "tabulate the first 6 convergents" mean exactly?

Does this mean just continue the pattern up to 6 terms? . Yes!

Crank out the first six "terms".

$x_1 \;=\;1$

$x_2 \;=\;1 + \frac{1}{2} \;=\;\frac{3}{2}$

$x_3 \;=\;1 + \frac{1}{2 + \frac{1}{3}} \;=\;\frac{10}{7}$

$x_4 \;=\;1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{2}}} \;=\;\frac{23}{16}$

$x_5 \;=\;1 + \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3}}}} \;=\;\frac{79}{55}$

$x_6 \;=\;1 + \frac{1}{2+\frac{1}{3 +\frac{1}{2 + \frac{1}{3 + \frac{1}{2}}}}} \;=\;\frac{181}{126}$

But check my work . . . please!

3. ## Re: Continued fraction

Is there any trick to solving a really messy looking fraction like x6?

4. ## Re: Continued fraction

Yes. It is called "using a computer". I cannot for the life of me see why someone would want to do continued fractions by hand.

5. ## Re: Continued fraction

I cannot for the life of me see why someone would want to do continued fractions by hand.
What does "tabulate the first 6 convergents" mean exactly?
Well sorroban suggested that this means calculate the first 6 convergents.

He said..

But check my work . . . please!
So I assume he calculated it by hand.

EDIT:

I found the formula for taking the previous convergent to make solving the next convergent easier

Term A is a rational number pn/qn

Pn = AnP(n-1)+P(n-2)
Qn = AnQ(n-1)+Q(n-2)

6. ## Re: Continued fraction

Originally Posted by ehpoc
I am asked the following question..

I am not wondering the answer, but just what is being asked here.

What does "tabulate the first 6 convergents" mean exactly?

Does this mean just continue the pattern up to 6 terms?
It may be useful to observe that if You want to obtain 'directly' the limit of the 'continous fraction'...

$x=1 + \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+...}}} }$ (1)

... You can write...

$x=1 + \frac{1}{2 + \frac{1}{3 + x-1}}$ (2)

... and then solve (2) for x...

Kind regards

$\chi$ $\sigma$

7. ## Re: Continued fraction

I am working on some more continued fractions.

I am doing this one.

x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]

I would be so so SO happy if someone could confirm if my solutions are correct.

To tabulate the first 6 convergents I got

x1=2
x2=5/2
x3=12/5
x4=29/2
x5=70/29
x6=169/70

To solve for the value to which x converges I did...

x=2+(1/(2+1/x))

x=2+(x/(2x+1))

x=(5x+2)/(2x+1)

2x^2+x=5x+2

x^2-2x-1=0

x=(2√8)/2

x=√8

Like I said I would be VERY VERY appreciative if someone could confirm if that solution is correct or not. Thanks.

8. ## Re: Continued fraction

Originally Posted by ehpoc
I am asked the following question..

I am not wondering the answer, but just what is being asked here.

What does "tabulate the first 6 convergents" mean exactly?

Does this mean just continue the pattern up to 6 terms?
The first 6 convergents give you an idea of what x is.

To find x

$x-1=k=\frac{1}{2+\frac{1}{3+k}}}$

$\frac{1}{k}=2+\frac{1}{3+k}$

$\frac{1-2k}{k}=\frac{1}{3+k}$

$(3+k)(1-2k)=k\Rightarrow\ 3-6k+k-2k^2=k\Rightarrow\ 2k^2+6k-3=0$

For k>0, solve for k using the quadratic formula and deduce the value of x from x = 1+k.

9. ## Re: Continued fraction

The first 6 convergents give you an idea of what x is.

To find x

$x-1=k=\frac{1}{2+\frac{1}{3+k}}}$

$\frac{1}{k}=2+\frac{1}{3+k}$

$\frac{1-2k}{k}=\frac{1}{3+k}$

$(3+k)(1-2k)=k\Rightarrow\ 3-6k+k-2k^2=k\Rightarrow\ 2k^2+6k-3=0$

For k>0, solve for k using the quadratic formula and deduce the value of x from x = 1+k.
Thanks. But that was my old question. Could you check my work in question 7. I am really anxious to see if I solved this one right.

10. ## Re: Continued fraction

Originally Posted by ehpoc
I am working on some more continued fractions.

I am doing this one.

x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]

I would be so so SO happy if someone could confirm if my solutions are correct.

To tabulate the first 6 convergents I got

x1=2
x2=5/2
x3=12/5
x4=29/2
x5=70/29
x6=169/70

To solve for the value to which x converges I did...

x=2+(1/(2+1/x))

x=2+(x/(2x+1))

x=(5x+2)/(2x+1)

2x^2+x=5x+2

x^2-2x-1=0

x=(2√8)/2

x=√8

Like I said I would be VERY VERY appreciative if someone could confirm if that solution is correct or not. Thanks.
You have typos on $x_4$

$x_1=2$

$x_2=2+\frac{1}{2}=\frac{5}{2}$

$x_3=2+\frac{2}{5}=\frac{12}{5}$

$x_4=2+\frac{5}{12}=\frac{29}{12}$

$x_5=2+\frac{12}{29}=\frac{70}{29}$

$x_6=2+\frac{29}{70}=\frac{169}{70}$

A quicker way to solve for x (as it is simpler than your first example) is

$x=2+\frac{1}{x}=\frac{2x+1}{x}$

$x^2-2x-1=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+4}}{2}$

and as x is positive

$x=\frac{2+\sqrt{8}}{2}=1+\frac{\sqrt{8}}{\sqrt{4}} =1+\sqrt{2}$

11. ## Re: Continued fraction

see I think my biggest problem is recognizing exactly what I need to have as the equation that x equals.

x=2+(1/(2+1/x))

x=2+(1/x)

So I guess first before I worry about solving the equation I must first be able to recognize it properly.

I have a list of practice examples. Can someone confirm if I am recognizing the equation that x equals properly when finding what x converges to?

1. x = 1 + 1/[1 + 1/[1 + 1/[1 + ...]]]
x=1+1/x?

2. x = 2 + 1/[2 + 1/[2 + 1/[2 + ...]]]
x=2+1/x?

3. x = 1 + 1/[2 + 1/[3 + 1/[2 + 1/[3 + ...]]]]
this one confuses me because a 1 shows up in the first term and never again.

Is it...

x=1+1/[2+1/[3+1/x]]?

or

x=2+1/[3+1/x]?

4. x = 1 + 1/[2 + 1/[4 + 1/[1 + 1/[3 + 1/[1 + 1/[3 + ...]]]]]]
This one confuses me so much!!

is it?

x=2 + 1/[4 + 1/[1 + 1/[3 + 1/[1 + 1/[3 + 1/x]]]]?

Any help would be greatly appreciated as I need to nail this before I can move on to solving the fractions right?

12. ## Re: Continued fraction

Could you write out number 4 above again please?

$x=1+\frac{1}{\left(1+\frac{1}{1+......} \right)}$

Notice the expression in parentheses is also x, so this is

$x=1+\frac{1}{x}$

$x=2+\frac{1}{\left(2+\frac{1}{2+....}\right)}$

$x=2+\frac{1}{x}$

$x=1+\left(\frac{1}{2+\frac{1}{3+\left(\frac{1}{2+ \frac{1}{3+....}\right)}}}\right)$

The terms in brackets are recurring, so start by subtracting 1 from both sides.

$x=1+k$

$k=\frac{1}{2+\frac{1}{3+k}}$

Solve for k, then for x.

13. ## Re: Continued fraction

thanks SO much for the clarification!!!!

PS. WHy do you want me to write out 4 again? Nothing will change LOL.

14. ## Re: Continued fraction

Ok, it's just a bit deeper then, though you have written it twice and they differ.

If it's

$x=1+\frac{1}{2+\frac{1}{4+\frac{1}{\left(1+\frac{1 }{3+ \frac{1}{\left(1+\frac{1}{3+.....}\right)}}\right) }}}$

then you get

$\frac{1}{x-1}=2+\frac{1}{4+\frac{1}{k}}$

so use that to write the relationship between x and k,
and use the "nesting" to write

$k=1+\frac{1}{3+\frac{1}{k}}$

to solve for k

15. ## Re: Continued fraction

I am a little confused. Do I simplify for x first equation. Then simplify for k, then sub k into the other equation?

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