[QUOTE=jinydu;85897]Mathematica seems to be behaving strangely. I'm getting:

ContinuedFraction[0.3] = {0, 3}

ContinuedFraction[3/10] = {0, 3, 3}

Even worse:

ContinuedFraction[0.2] = {0}[/QUOTE]

Use PARI/GP, then. :cool:

gp>contfrac(0.3)
%104 = [0, 3, 3]
gp>contfrac(3/10)
%105 = [0, 3, 3]
gp>contfrac(0.2)
%106 = [0, 5]

[QUOTE=jinydu;85897]Mathematica seems to be behaving strangely. I'm getting:

ContinuedFraction[0.3] = {0, 3}

ContinuedFraction[3/10] = {0, 3, 3}

Even worse:

ContinuedFraction[0.2] = {0}[/QUOTE]

The Mathematica strange behaviour continues by introducing a small error

ContinuedFraction[0.2-10^-10] returns {0,5}
and
ContinuedFraction[0.2+10^-10] returns {0,4,1}

This is on version 4.1

I'm using Version 4.0.2. Are you saying that this problem is fixed in versions newer than 4.1?

In any case, given that such errors occur, I'm not so sure that I can trust the answer that Mathematica gives for, say:

ContinuedFraction[[tex]\pi^3[/tex]]

[QUOTE=alpertron;85861]Yes, it agrees with the first line I wrote above: 0.3 = [3, 3]. It appears that your notation [0;3,3] is 0+[3,3] (or [3,3])

Continuing with the problem set by the original poster, if the number in question is 0.12345678910111213141516... (which is obviously irrational) we find its continued fraction expansion:

[ 8 , 9 , 1 , 149083 , 1 , 1 , 1 , 4 , 1 , 1 , 1 , 3 , 4 , 1 , 1 , 1 , 15 ,
4575401113910310764836466282429561185996039397104575\
5500066200439309026265925631493795320774712865626829\
1352511788034072479309738318986794132119711326023863\
081376170,
6 , 4 , 1 , 5 , 1 , 1 , ...]

If we convert the continued fraction to standard fractions just before the huge coefficients we get:

10 / 81
60499999499 / 490050000000

The first 189 digits of the last number agrees with the decimal expansion of our irrational number.[/QUOTE]Indeed, the presence of a particularly large term in the CF expansion of a number indicates that it is especially well represented by a rational approximation. There can be practical benefits of observing such things. For instance, it can have cryptographic significance. See the justly famous Blum, Blum & Shub paper for a concrete example. Yes, I am being deliberately obscure in the hope that you'll put just a little effort into tracking down the paper mentioned. The effort brings its own rewards.

Paul

[QUOTE=alpertron;85861]Yes, it agrees with the first line I wrote above: 0.3 = [3, 3]. It appears that your notation [0;3,3] is 0+[3,3] (or [3,3])

Continuing with the problem set by the original poster, if the number in question is 0.12345678910111213141516... (which is obviously irrational) we find its continued fraction expansion:

[ 8 , 9 , 1 , 149083 , 1 , 1 , 1 , 4 , 1 , 1 , 1 , 3 , 4 , 1 , 1 , 1 , 15 ,
4575401113910310764836466282429561185996039397104575\
5500066200439309026265925631493795320774712865626829\
1352511788034072479309738318986794132119711326023863\
081376170,
6 , 4 , 1 , 5 , 1 , 1 , ...]

If we convert the continued fraction to standard fractions just before the huge coefficients we get:

10 / 81
60499999499 / 490050000000

The first 189 digits of the last number agrees with the decimal expansion of our irrational number.[/QUOTE]Simple exercise: predict when the next huge coefficient occurs in the CF expansion of this number. Explain your reasoning.

Slightly harder exercise: find the next huge coefficient and see whether its position is in accordance with your prediction.

Further exercise: repeat the previous two exercises until you run out of computational resources.


Paul

[QUOTE=xilman;85913]Indeed, the presence of a particularly large term in the CF expansion of a number indicates that it is especially well represented by a rational approximation. There can be practical benefits of observing such things. For instance, it can have cryptographic significance. See the justly famous Blum, Blum & Shub paper for a concrete example. Yes, I am being deliberately obscure in the hope that you'll put just a little effort into tracking down the paper mentioned. The effort brings its own rewards.

Paul[/QUOTE]

Indeed. The presence of such a large term early on also strongly
suggests that the number is transcendental, because Thue's Theorem
limits how quickly the denominators can grow for ALGEBRAIC irrationals.

[QUOTE=xilman;85913]Indeed, the presence of a particularly large term in the CF expansion of a number indicates that it is especially well represented by a rational approximation. There can be practical benefits of observing such things. For instance, it can have cryptographic significance. See the justly famous Blum, Blum & Shub paper for a concrete example. Yes, I am being deliberately obscure in the hope that you'll put just a little effort into tracking down the paper mentioned. The effort brings its own rewards.

Paul[/QUOTE]

Indeed. The presence of such a large term early on also strongly
suggests that the number is transcendental, because Thue's Theorem
limits how quickly the denominators can grow for ALGEBRAIC irrationals.

[QUOTE=xilman;85914]Simple exercise: predict when the next huge coefficient occurs in the CF expansion of this number. Explain your reasoning.

Slightly harder exercise: find the next huge coefficient and see whether its position is in accordance with your prediction.

Further exercise: repeat the previous two exercises until you run out of computational resources.

Paul[/QUOTE]

This is a known constant. See: [URL="http://mathworld.wolfram.com/ChampernowneConstant.html"]http://mathworld.wolfram.com/ChampernowneConstant.html[/URL]

It is clear that this number must have very large coefficients in the continued fraction expansion:

From:

[tex]\large \frac{1}{\left(10^n-1\right)^2} = \frac{1}{10^{2n}} + \frac{2}{10^{3n}} + \frac{3}{10^{4n}} + ...[/tex]

Now we multiply by a power p of 10 such that the consecutive numbers of n digits are in the same location than the Champernowne constant C. For example when n=2, p=11. Let B be this product.

Now the product has 9n(10[sup]n-1[/sup]) - n - 1 digits equals to C starting from the digit G(n) (G(2) = 10, G(3) = 190, G(4) = 2890, ..., G(n) = G(n-1) + 9(n-1)(10[sup]n-2[/sup])). (Notice that G(n) is near (n-1)10[sup]n-1[/sup]). This means that the difference between C and the product is very near a fraction whose denominator is 10[sup]G(n)-1[/sup]. Call it D.

Now B - D is extremely near to C, with about n 10[sup]n[/sup] digits correct, while the denominator of D - B is lcm ((10[sup]n[/sup]-1)[sup]2[/sup], 10[sup]G(n)-1[/sup]) which has about (n-1)10[sup]n-1[/sup] digits.

This implies a very huge coefficient in the continued fraction expansion of the constant, which are larger when n increases.

[QUOTE=alpertron;85926]It is clear that this number must have very large coefficients in the continued fraction expansion:[/QUOTE]Thank you! :bow:

i'm always pleased when someone takes seriously my suggestions of worthwhile "exercises", which are sometimes trivial to answer and sometimes turn out to be research problems.


Paul

I would like to repeat the question I asked earlier, lest it be forgotten:

[QUOTE=jinydu;85911]I'm using Version 4.0.2. Are you saying that this problem is fixed in versions newer than 4.1?

In any case, given that such errors occur, I'm not so sure that I can trust the answer that Mathematica gives for, say:

ContinuedFraction[[tex]\pi^3[/tex]][/QUOTE]