7
$\begingroup$

Find the fraction with integers for the numerator and denominator, where the decimal expansion is $0.11235.....$

The numerator and denominator must be less than $100$.

Find the fraction.

I believe I can use generating functions here to get $1+x+2x^2+3x^3+5x^4+.....$, but I do not know how to apply it.

$\endgroup$
6
  • $\begingroup$ What's the next digit, $8$? $\endgroup$ Dec 20, 2014 at 20:21
  • 2
    $\begingroup$ Good rational approximations can be found using partial fractions. $\endgroup$ Dec 20, 2014 at 20:31
  • $\begingroup$ @CameronBuie It was not given. $\endgroup$ Dec 20, 2014 at 20:33
  • $\begingroup$ @barakmanos: Not sure. It was not given. $\endgroup$ Dec 20, 2014 at 20:34
  • 1
    $\begingroup$ Is the fractional term supposed to be the simple stringing together of the Fibonacci numbers, i.e., 0.112358132134...? $\endgroup$
    – Mico
    Dec 21, 2014 at 8:56

4 Answers 4

15
$\begingroup$

Like you hint, we can observe that the value $$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$ where $F_k$ is the $k$th Fibonacci number (with the convention that $F_0 = 0$, $F_1 = 1$), is simply the value of the series $$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \cdots$$ at $x = \frac{1}{10}$.

Hint Using the defining recurrence relation $$F_{k + 2} = F_{k + 1} + F_k$$ (and the above convention) gives that $F(x)$ satisfies $$F(x) = x + x F(x) + x^2 F(x).$$

Rearranging gives that on the open interval of convergence of the series, which turns out to be $(-1/\phi, 1/\phi)$---where $\phi$ is the Golden Ratio, and which in particular contains the value $\frac{1}{10}$ of interest)---we have $$F(x) = \frac{x}{1 - x - x^2}.$$ Thus, the series has value $$F\left(\tfrac{1}{10}\right) = \frac{\left(\frac{1}{10}\right)}{1 - \left(\frac{1}{10}\right) - \left(\frac{1}{10}\right)^2} = \color{#bf0000}{\boxed{\frac{10}{89}}} .$$ Since this is a rational number, its decimal expansion repeats: $$0.\overline{1123595505617977528089887640449438202247191}.$$

$\endgroup$
1
  • $\begingroup$ @FlybyNight yes, that behavior is obvious and expected -- 1, 1, 2, 3, 5, 8, 13 and then the 8 and 13 overlap to give a 9 in the decimal expansion in place of the 8. $\endgroup$
    – obataku
    Dec 20, 2014 at 21:22
6
$\begingroup$

Continued fractions: \begin{align} \frac{11235}{100000}&=\frac{2247}{20000}\\ &=\cfrac{1}{8+\cfrac{2024}{2247}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{223}{2024}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{2}{17}}}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{1}{8+\cfrac{1}{2}}}}}}\\ &=[0;8,1,9,13,2] \end{align} The first convergent is $1/8$; the second convergent is $$ \cfrac{1}{8+1}=\frac{1}{9}; $$ the third convergent is $$ \cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9}}}=\cfrac{1}{8+\cfrac{9}{10}}=\frac{10}{89}; $$ The fourth convergent can be computed to $131/1166$ (see this continued fraction calculator)

$\endgroup$
3
$\begingroup$

An exhaustive search (by computer) of all fractions with numerator and denominator $< 100$ shows there is only one whose decimal representation starts 0.11235: $$ \frac{10}{89} $$

The decimal expansion continues $\ldots 955056179775\ldots$ (I haven't bothered finding aprogram that can calculate enough that we can see the period), that's completely impossible to guess- I would call the question poorly worded.

$\endgroup$
4
  • $\begingroup$ To save anyone from having to reach for a calculator, the first many digits of the expansion goes $0.1123595505617977\dots$. The question of if there is a better method than brute force remains. We could clearly see that for it to work for some $\frac{p}{q}$, we would need $8.5p < q<9p$ and so $p\leq 11$, reducing cases to test to less than 18 $\endgroup$
    – JMoravitz
    Dec 20, 2014 at 20:37
  • $\begingroup$ @Henrik: I am guessing that the decimal representation would represent Fibonacci numbers, but I'm not sure.. $\endgroup$ Dec 20, 2014 at 20:39
  • $\begingroup$ The period is $44$ since $m = 44$ is the smallest positive integer such that $10^m \equiv 1 \pmod{89}$. WolframAlpha gives the decimal expansion to enough digits. $\endgroup$
    – JimmyK4542
    Dec 20, 2014 at 20:40
  • $\begingroup$ @MathisLife: I would be surprised if the decimal representation of the Fibonacci numbers were rational. $\endgroup$ Dec 20, 2014 at 20:40
0
$\begingroup$

These are fractions $f$ with $11235/100000\le f<11236/100000.$ A direct, if slow, method for counting these would be $$ \sum_{d=1}^{100}\left\lfloor\frac{100000d}{11235}\right\rfloor-\left\lfloor\frac{100000d}{11236}\right\rfloor $$

or

sum(d=1,100,100000*d\11235-100000*d\11236)
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .