6
$\begingroup$

enter image description here goes on till infinity.

I get two solutions by rewriting the term in the form of the equation $x = 3-(2/x)$, which are $1$ and $2$.

But in my opinion this term should have only one possible value. Then which one is wrong and why?

$\endgroup$
  • 1
    $\begingroup$ Maybe it doesn't converge, but instead alternates around two poles $1$ and $2$? $\endgroup$ – Colm Bhandal Aug 6 '15 at 16:19
  • $\begingroup$ Maybe, but I haven't ever seen such cases. Can you please give me another example, where the final value keeps "jumping" among certain fixed values? $\endgroup$ – Phoenix Aug 6 '15 at 16:26
  • $\begingroup$ Seems like the answers cleared that up :) $\endgroup$ – Colm Bhandal Aug 6 '15 at 16:41
7
$\begingroup$

If you continue adding numbers to the expression one at a time, then you have the sequence $$ 3,\; 3-2,\; 3-\frac{2}{3},\; 3-\frac{2}{3-2},\; 3-\frac{2}{3-\frac{2}{3}},\; 3-\frac{2}{3-\frac{2}{3-2}},\;3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}\ldots, $$ or $$ 3,\; 1,\; \frac{7}{3},\; 1,\; \frac{15}{7},\; 1,\;\frac{31}{15},\;\ldots, $$ which consists of two alternating subsequences: one is identically $1$, and the other converges to $2$. Depending on exactly how you define the value of the infinite fraction (i.e., what sequence you define it to be the limit of), it could be $1$, or $2$, or non-convergent.

$\endgroup$
  • $\begingroup$ You might be interested in some other aspects I mention in my answer. Regards, $\endgroup$ – Markus Scheuer Aug 7 '15 at 15:00
3
$\begingroup$

Look at it as two different series and you will understand why both 1 and 2 are possible solutions of this:

series 1: $\{3-2, 3-\frac{2}{3-2}, 3-\frac{2}{3-\frac{2}{3-2}}, ...\}$
series 2: $\{3-\frac{2}{3}, 3-\frac{2}{3-\frac{2}{3}}, 3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}, ...\}$.

series 1 converges to 1 whereas series 2 converges to 2.

$\endgroup$
0
$\begingroup$

Note: OP's expression is usually regarded as continued fraction. We will show it has the single solution $2$.

Before we analyse OPs expression, let's have a look at a continued fraction representation of $\sqrt{2}$

\begin{align*} \sqrt{2}=[1;2,2,2,\ldots]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots}}} \end{align*}

The convenient notation $[1;2,2,2,\ldots]$ shows the integer part $1$ of $\sqrt{2}$ in the first position, followed by the successive values $2$ in the denominators left from the '$+$' sign.

Since the numerator is always $1$ it's called a simple continued fraction. In general a simple continued fraction can be written as

\begin{align*} x=[a_0;a_1,a_2,a_3,\ldots]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ddots}}} \end{align*}

The approximation by finite continued fractions of $x$ is the sequence \begin{align*} [a_0],[a_0;a_1],[a_0;a_1,a_2],[a_0;a_1,a_2,a_3],\ldots \end{align*}

In case of $\sqrt{2}$ we obtain \begin{align*} [a_0]&=[1]=1\\ [a_0;a_1]&=[1;2]=1+\frac{1}{2}=\frac{3}{2}\\ [a_0;a_1,a_2]&=[1;2,2]=1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}\\ [a_0;a_1,a_2,a_3]&=[1;2,2,2]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}\\ &\ldots \end{align*}

In general we consider a continued fraction in the form

\begin{align*} x=a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3+\ddots}}}\tag{1} \end{align*}

and the approximation of $x$ is analogously

\begin{align*} x_0&=a_0\\ x_1&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.}=a_0+\frac{b_1}{a_1}\\ x_2&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2}}\\ x_3&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} +\frac{\left.b_3\right|}{\left|a_3\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3}}}\\ &\ldots \end{align*}

Now we are well prepared to take a look at OPs expression

\begin{align*} x=3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3+\ddots}}} \end{align*}

Here we put the '$-$' sign to the denominator in order to get the same representation as in (1). In a more compact way we can write

\begin{align*} x=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\cdots \end{align*}

The approximation of $x$ by its finite continued fractions is

\begin{align*} x_0&=3\\ x_1&=3+\frac{\left.-2\right|}{\left|3\right.}=3+\frac{-2}{3}=\frac{7}{3}\\ x_2&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3}=\frac{15}{7}\\ x_3&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3}}}=\frac{31}{15}\tag{2}\\ &\ldots \end{align*}

We see from (2) that $x$ converges to $\lim_{n\rightarrow \infty}\left(2+\frac{1}{n}\right)=2$ which is easily to prove.

Conclusion: The infinite continued fraction $x$ is defined as the limit of the corresponding approximating finite continued fractions in (2) and we so obtain $x=2$ as the only solution.

Epilog: Another nice gem is the continued fraction of $e$

\begin{align*} e=[2;1,2,1,1,4,1,1,6,1,1,8,...] \end{align*}

and here is a short proof of it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.