You'd need to solve for the positive value k first.

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

Use the k obtained and substitute it into the x equation to find x.

Printable View

- Dec 18th 2011, 02:17 PMArchie MeadeRe: Continued fraction
You'd need to solve for the positive value k first.

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

Use the k obtained and substitute it into the x equation to find x. - Dec 18th 2011, 03:30 PMehpocRe: Continued fraction
I don't get it, maybe my best bet is to hope one like that does not show up on the exam HAHA

- Dec 18th 2011, 04:44 PMArchie MeadeRe: Continued fraction
The k equation will end up as a quadratic to solve as before.

Then just put the value you get for k into the x equation to find x.

The two sets of brackets show you how to find k.

Remember the song by Desmond Decker ? (No, you're probably too young)

"You can get it if you really want, but you must try, try and try, try and try, you'll succeed at last". - Dec 18th 2011, 05:23 PMehpocRe: Continued fraction
So do I solve for just what k is in the second equation?

k=1+1/[3+1/k]

k=(4k+1)/(3k+1)

3k^2+k=4k+1

3k^2-3k-1=0

k=√(8/3)/2 - Dec 19th 2011, 02:53 AMArchie MeadeRe: Continued fraction
Yes, but what is that technique you are using for the final quadratic?

You seem to be also putting "-b" inside the square root instead of

$\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}$ - Dec 19th 2011, 04:50 AMehpocRe: Continued fraction
I had the correct value for b but I forgot the plus/minus. I divided that polynomial by 3. It helps keep it simpler for me personally. If I stray into too much algebra I will surely stumble somewhere :P

k^2-k-1/3 a=1 b=-1 c=-1/3

k=[1+√(1+4/3)]/2 correct?

I also added the fraction wrong LOL

It should work out to [1+√(7/3)]/2 correct?

Now I solved for the repeating part in #4. Do I do anything with the rest of that where you said you relate x to k or whatever. Or do i just state that? - Dec 19th 2011, 05:07 AMArchie MeadeRe: Continued fraction
Your equation to find x contains a k, so you substitute in that k value to find x.

If you could not find k, you cannot write the value of x.

But you should know this from algebraic substitution.

At this time, you should probably review solving algebraic equations in 2 unknowns.

It was not necessary to introduce k, but solving for x alone without introducing something like "k"

could be even messier. - Dec 19th 2011, 06:13 AMehpocRe: Continued fraction
Did I solve for K correctly? If so, how exactly did you derive x from the fraction.

Before I worry about solving algebraic equations with two unknowns, I should be able to get x properly right? - Dec 19th 2011, 02:54 PMArchie MeadeRe: Continued fraction
Just put in the value of k.