# Thread: continued fraction

1. ## continued fraction

Let a = sqrt(3) -1. prove that

a = 1/(1 + 1/(2 + a) )

Use this to find the continued fraction expansion for a.

Deduce the continued fraction expansion for sqrt(3).

Check that your answer makes sense - that is , use the first 6 or 7 terms of the continued fraction expansion to give an approximation for sqrt(3) and make sure that this approximation is reasonable.

Can you teach me how to solve this question? Thank you very much.

2. Hello, beta12!

Now that LaTeX is back, I'll revise this post . . .

Let $\displaystyle a \:=\:\sqrt{3} -1$

Prove that: .$\displaystyle a\;=\;\frac{1}{1 + \frac{1}{2+a}}$

Simplify that awful equation: .$\displaystyle a\;=\;\frac{1}{1 + \frac{1}{2+a}} \;=\;\frac{1}{\frac{3+a}{2+a}} \;=\;\frac{2+a}{3+a}$

We have: .$\displaystyle a \:=\:\frac{2+a}{3+a}\quad\Rightarrow\quad a^2 + 2a - 2 \:=\:0$

Quadratic Formula: .$\displaystyle a \:=\:\frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} \:=\:-1 \pm \sqrt{3}$

Since $\displaystyle a$ is positive: .$\displaystyle a \:=\:\sqrt{3} - 1$

3. Originally Posted by beta12
Let a = sqrt(3) -1. prove that

a = 1/(1 + 1/(2 + a) )

Use this to find the continued fraction expansion for a.

Deduce the continued fraction expansion for sqrt(3).

Check that your answer makes sense - that is , use the first 6 or 7 terms of the continued fraction expansion to give an approximation for sqrt(3) and make sure that this approximation is reasonable.

Can you teach me how to solve this question? Thank you very much.
Look at Soroban's Post
We know that,
a=[0;1,2,a]
Intuitively you can substitute "a" in the brackets for "a":
a=[0;1,2,1,2,a]
Again and again,
a=[0;1,2,1,2,1,2,1,2...]
Periodical expansion.

You can quickly evaluate the convergents of this continued fraction using the following recusion relations (I am sure you know them):

p_k=a_k*p_{k-1}+p_{k-2}
q_k=a_k*q_{k-1}+q_{k-2}
Where,

p_0=0
q_0=1

p_1=1
q_1=1

Thus, use those equations above to get,

p_2=(2)(1)+0=2
q_2=(2)(1)+1=3

p_3=(1)(2)+1=3
q_3=(1)(3)+1=4

p_4=(2)(3)+2=8
q_4=(2)(4)+3=11

p_5=(1)(8)+3=11
q_5=(1)(11)+4=15

p_6=(2)(11)+8=30
q_6=(2)(15)+11=41

Thus, the fraction,
30/41 Is the best possible approximation of that number.
The error is less than 1/(q_6)^2

4. Hi Soroban,

Thank you very much. I got the first part.

5. Originally Posted by beta12
Hi Soroban,

Thank you very much. I got the first part.

But how should I do the second part " Deduce the continued fraction expansion for sqrt(3)? Could you please teach me? Thank you very much.
I showed you, look at my post.

Since,
a=[0:1,2,a]
Substitute the "a" into the "a" in the fraction,
a=[0,1,2,1,2,a]
Do that again,
a=[0,1,2,1,2,1,2,a]
See the pattern?

6. Hi Perfecthacker,

Ok.

======================
Since,
a=[0:1,2,a]
Substitute the "a" into the "a" in the fraction,
a=[0,1,2,1,2,a]
Do that again,
a=[0,1,2,1,2,1,2,a]
See the pattern?

Yes. I got this part.
This should be belong to my question " use this to find the continued fraction expansion for a"
=====================
You can quickly evaluate the convergents of this continued fraction using the following recusion relations (I am sure you know them):

p_k=a_k*p_{k-1}+p_{k-2}
q_k=a_k*q_{k-1}+q_{k-2}
Where,

p_0=0
q_0=1

p_1=1
q_1=1

Thus, use those equations above to get,

p_2=(2)(1)+0=2
q_2=(2)(1)+1=3

p_3=(1)(2)+1=3
q_3=(1)(3)+1=4

p_4=(2)(3)+2=8
q_4=(2)(4)+3=11

p_5=(1)(8)+3=11
q_5=(1)(11)+4=15

p_6=(2)(11)+8=30
q_6=(2)(15)+11=41

Thus, the fraction,
30/41 Is the best possible approximation of that number.

Is this part answering my question " Deduce the continued fraction expansion for sqrt(3)?

7. Hi Perfecthacker,

Thank you very much. I got it now.

8. Originally Posted by beta12
Hi Perfecthacker,

Is this part answering my question " Deduce the continued fraction expansion for sqrt(3)?
No it is answering the other question to find the 6th or 7th Convergent.

9. Yes. I noticed this point.

I used the relation between a and sqrt(3) to deduce the continued fraction expansion for sqrt(3).

Thank you very much for teaching me.