I name this number the Near Integer Ratio Rv

[AntonVrba]

Rv=
3.7320508075688772935274463415058723669428052538103806280558069794519330169088000370811461867572485756756261414154067030299699450949989524788116555120943736485280932319023055820679748201010846749232650153123432669033228866506722546689218379712270471316603678615880190499865373798593894676503475065760507566183481296061009476021871903250831458295239598329977898245082887144638329173472241639845878553976679580638183536661108431737808943783161020883055249016700235207111442886959909563657970871684980728994932964842830207864086039887386975375823173178313959929830078387028770539133695633121037072640192491067682311992883756411414220167427521023729942708310598984594759876642888977961478379583902288548529035760338528080643819723446610596897228728652641538226646984200211954841552784411812865345070351916500166892944154808460712771439997629268346295774383618951101271486387469765459824517885509753790138806649619119622229571105552429237231921977382625616314688420328537166829386496119170497388363954959381457576718533736331259108996554246248347871976052359977691923235702203053028403859154149710724295592067062025095201759631858727663599752836634310801506658537106473285386259222605822205104036802702975047987280794616581004170526819400190957334621759438936702493204226910343698124637201111852610842689102997203112021000635071763745824052038475551972799337976149061078949855442233260040188513036315611448868472815892881632451872650666453848775991625766428721112408420680167635171001029431807155151909616424609070394081292169035174929613640041396704310412536323270309225773279602923765977455370954691157421404242307819923276174019064245124548775168626961053336942162136053946042456541401285330078136334498567364067039773422298119610429255345016014059404795471545345484072717376562623665491666402330060132657440701078368584684523131604677544805004022406399119703622186029202388671507110171694002968687596635000408953162142334252279568340670134701859020283607167621477434934495635958080821304425864694685226……………


The above number is generated to 2000 decimal places. I name this number the Near Integer Ratio Rv, it is generated by a simple algorithm and is approximated by Rv=P[n]/P[n-1]. The number of iteration steps n gives the relevant number of decimal places or precision. Interesting the integer quantities P[n] and P[n-1] have no common divisors or factors.

The reason why I call it the near integer ratio is that Rv exhibits following property that when Rv raised to the power of 2m (m<n) or 2m+1 the result has more than m recurring nines after the decimal point, i.e. Rv^2m is a near integer quantity.

Rv^2 = 13.928..
Rv^3 = 51.9807..
Rv^4 = 193.9948..
Rv^5 = 723.99961..
Rv^6 = 2701.99962..
.
.
Rv^10 = 524173.999998092...
Rv^20 = 274758382273.99999999999636...
Rv^21 = 1025412242451.999999999999024...

Rv^41=281740608900066187095843.999999999999999999999996450...

OK I think I made my point.

Is Rv unique or do other such numbers exist?

Is Rv new or are there existing references to it?

Is Rv a irrational number even though the approximation is rational?

Can I say that Rv raised to infinity is a infinate large integer? [edit] I rephrase after Data's comment"Rv raised to infinity is a integer quantity?"

 

[James R]

Do you mean:

$$Rv = \lim_{n \rightarrow \infty} P(n)/P(n-1)$$

What's P(1)?

How do we get P(2) from P(1)?
How do we get P(n) from P(n-1)?

 

[Data]

Can I say that Rv raised to infinity is a infinate large integer?

what does it mean to "raise" a number to infinity? As well, I'm pretty sure that all integers are finite.

As to your uniqueness question, I think a pretty good example of another number exhibiting this property is 1 (or any other nonzero integer). 1^m has a rather nice string of infinitely many nines following the decimal regardless of what $m$ is. Hey, even square roots of integers work, if we're only raising things to even powers, and lost of those are irrational (you really should specify what you mean by 'unique' a little more clearly )!

 

[AntonVrba]

.

Hey, even square roots of integers work, if we're only raising things to even powers, and lost of those are irrational (you really should specify what you mean by 'unique' a little more clearly )!

No they don't, look at my example even and odd powers

sqrt(2) = 1.414...
sqrt(2)^2 = 2
sqrt(2)^3 = 2.828..

is it not unique that Rv^20 ...273.999999999996... multiplied by 3.732050807... = ...451.9999999999990...

Show me another non-integer number that can do that

 

[Data]

I said they did if you only raised things to even powers . I didn't notice you had changed it to include odd powers too, though~

 

[AntonVrba]

How do we get P(2) from P(1)?
How do we get P(n) from P(n-1)?

I was waiting for this but after giving up my battle with LATEX or TEX I remain with ASCII.

Using matrix representation and the best abilities of ASCII (Just join the vertical lines)



|R(2m)| |-1 2 2| |R(2m-1)|
|Q(2m)|=|-2 1 2|.|Q(2m-1)|
|P(2m)| |-2 2 3| |P(2m-1)|

|R(2m+1)| |1 -2 2| |R(2m)|
|Q(2m)+1|=|2 -1 2|.|Q(2m)|
|P(2m)+1| |2 -2 3| |P(2m)|



You can start with any number triplet R(1), Q(1) and P(1), I started with r,q,p=1,0,1. (other combinations could result in P(n) and P(n-1) having common factors or divisors)

Another interesting property

for all (n) R^2+Q^2-P^2 = constant

Rz is approached by the three ratios P(n)/P(n-1), R(n)/Q(n-1) and Q(n)/R(n-1)

and a further interesting property is the ratio

P(n)/P(n-1) - Q(n)/R(n-1)
----------------------------- = 2.99999999999999...
P(n)/P(n-1) - R(n)/Q(n-1)

the number of 9's again dependendant on (n)

 

[Hurkyl]

There's a closed-form solution for P(m), Q(m), and R(m). I'll work it out if nobody else has done it by the time I get back from work.

 

[snoble]

I'm pretty sure you're just looking at $$2+\sqrt{3}$$. I'm guessing if you look at its continued fraction the fact that you can generate it like this becomes obvious.

 

[snoble]

Yeah it is $$2+\sqrt{3}$$. Given your to Matrices call the first one A and the second B you just have to look at the dominant eigenvector of $$A\cdot B$$. Then just multiply that eigenvector by B and take the appropriate ratio.

 

[AntonVrba]

I'm pretty sure you're just looking at $$2+\sqrt{3}$$. I'm guessing if you look at its continued fraction the fact that you can generate it like this becomes obvious.

Hey SNoble you are a star!
why it is obvius I do not know but you are 100% correct.

Now the question remains why $$Rv=(2+\sqrt{3})^n$$ has the property of being a near integer that is atleast n/2 recuring 9's after the decimal point AND if other irrational numbers exist that have this property or if Rv is the only such number.

 

[Zurtex]

Hey SNoble you are a star!
why it is obvius I do not know but you are 100% correct.

Now the question remains why $$Rv=(2+\sqrt{3})^n$$ has the property of being a near integer that is atleast n/2 recuring 9's after the decimal point AND if other irrational numbers exist that have this property or if Rv is the only such number.

Numbers of the form:

$$\frac{a + \sqrt{b}}{c}$$

Become quite obvious their of this form when you look at their continued fraction. http://mathworld.wolfram.com/ContinuedFraction.html

As for your problem, if it helps:

$$\left(2 + \sqrt{3} \right)^n = \mathcal{A}_1(n) + \mathcal{A}_0(n) \sqrt{3}$$

Where:

$$\mathcal{A}_j(n)= 4\mathcal{A}_j(n-1) - \mathcal{A}_j(n-2)$$

And:

$$\mathcal{A}_j(0) = j \quad \mathcal{A}_j(1) = j + 1$$

 

[strid]

Numbers of the form:

$$\frac{a + \sqrt{b}}{c}$$

Become quite obvious their of this form when you look at their continued fraction. http://mathworld.wolfram.com/ContinuedFraction.html

As for your problem, if it helps:

$$\left(2 + \sqrt{3} \right)^n = \mathcal{A}_1(n) + \mathcal{A}_0(n) \sqrt{3}$$

Where:

$$\mathcal{A}_j(n)= 4\mathcal{A}_j(n-1) - \mathcal{A}_j(n-2)$$

And:

$$\mathcal{A}_j(0) = j \quad \mathcal{A}_j(1) = j + 1$$

didnt find anything corresponding on that link (might have missed it)...
quite a fasciniating number...impressive..

i don't get $$\frac{a + \sqrt{b}}{c}$$...
becase if i set a=4 b=5 c=6 I don't get that sort of cool number... actually, i didnt get any number like that by that method. what did you mean with that formula?

also-- I am interested on how the topic-creater managed to find this fascnianig number :)

coool

 

[shmoe]

You can find many more irrational numbers that behave like this. Take any integer a, then consider b's near a^2. Then $$a+\sqrt{b}$$ will be "close" to an integer when you raise it to a high power. For example a=13, b=167 then $$(13+\sqrt{167})^{50}$$ has 55 9's following the decimal.

Notice if $$(2+\sqrt3)^n=A(n)+B(n)\sqrt{3}$$ then it looks like we have (for m>1 and assuming I've got the parity correct):

$$A(2m-1)=2R(2m)-Q(2m)$$
$$B(2m-1)=P(2m)$$
$$A(2m)=2R(2m+1)-P(2m+1)$$
$$B(2m)=Q(2m+1)$$

So the p,q,r are really just encoding the powers of 2+sqrt(3). It might be simpler if you could relate the coeffecients of these powers with the continued fraction expansion if you wanted to prove how close to an integer you are. I'm not sure how easy this would be though, or if it will even work.

Another possibility would be to directly examine the nature of the B(n). It's a polynomial with integer coefficients in 2 and 3 (or a, b, in the more general case) that appears to be trying to converge to an integer times sqrt(3) (or sqrt(b) in the general case if a and sqrt (b) are "close"). I'm abusing the word "converge" here, really I mean $$|B(n)-[B(n)]|\rightarrow 0$$ as n grows and [x] denotes the closest integer to x. I don't know if there's a simple reason for this (or if it's even true).

 

[Zurtex]

strid I never said they were interesting, what you will find is that numbers of that form can be represented by the continued fraction:

$$\begin{multline*} \frac{a + \sqrt{b}}{c} = \\ a_0 + \left(\frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{\ddots + \frac{1}{a_n + \frac{1}{a_1 + \frac{1}{\ddots}}}}}}\right) \end{multline*}$$

Or in usual notation:

$$\frac{a + \sqrt{b}}{c} = [a_0; a_1, a_2, \ldots, a_n, a_1, a_2 \ldots] = [a_0; \overline{a_1, \ldots, a_n}]$$

Where of course $a_i \in \mathbb{N}$. You can actually say even more about the continued fraction, but what really matters is that when looking at the continued fraction of something you can easily see if it is of the above form or not.

Anyway, shmoe makes some good points.

 

[AntonVrba]

You can find many more irrational numbers that behave like this. Take any integer a, then consider b's near a^2. Then $$a+\sqrt{b}$$ will be "close" to an integer when you raise it to a high power. For example a=13, b=167 then $$(13+\sqrt{167})^{50}$$ has 55 9's following the decimal.

Shmoe five stars to you ! !

 

[Zurtex]

Haha, made a bit of a mistake in my last post, I actually put up the thing that represents, $\sqrt{d}$. However basically the only difference is that there are a few terms on the front that don't have to repeat. Took me ages to do so can't be bothered redoing lol.

 

[AntonVrba]

also-- I am interested on how the topic-creater managed to find this fascnianig number :)
coool

I was actually studying near equilateral triangles, two sides being equal, one side (the base) being either one less or one more than the two sides and having an area which is an integer quantity

( R, Q, P)
( 3, 4, 5)
( 15, 8, 17)
( 33, 56, 65)
( 209, 120, 241)
( 451, 780, 901)
( 2911, 1680, 3361)
( 6273, 10864, 12545)
( 40545, 23408, 46817)

The algorithm in the opening thread generates the series of near equilateral triangles P-P-2*min(R,Q) that is 5-5-6, 17-17-16, 65-65-66 and whose area are integer quantities

and then I looked at the ratio's ...

 

[Hurkyl]

Solving linear recurrences

Let A and B denote the two matrices in your linear recurrence, and x be the solution vector. That is,

$$A := \left( \begin{array}{ccc} -1 & 2 & 2 \\ -2 & 1 & 2 \\ -2 & 2 & 3 \end{array} \right)$$
$$B := \left( \begin{array}{ccc} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{array} \right)$$
$$x_n := \left( \begin{array}{c} R_n \\ Q_n \\ P_n \end{array} \right)$$

Then, your recurrence is merely:

$$x_{2n} = A x_{2n-1}$$
$$x_{2n+1} = B x_{2n}$$

The solution to this recurrence is easy to see:

$$x_{2n} = (AB)^n x_0$$
$$x_{2n+1} = B (AB)^n x_0$$

So the only trick is to figure out how to compute $(AB)^n$.

Well, usually what you want to do is diagonalize the matrix:

$$AB = \left( \begin{array}{ccc} 7 & -4 & 8 \\ 4 & -1 & 4 \\ 8 & -4 & 9 \end{array} \right)$$

Well, the eigenvalues are 7 - 4√3, 7 + 4 √3, and 1. Anyone want to find the eigenvectors for me? TI-89's only do it numerically, and I really don't want to do it by hand.

Anyways, once you can write AB as $PDP^{-1}$, you know that $(AB)^n = PD^nP^{-1}$, allowing you to get an explicit formula for $x_n$ in terms of $x_0$.

 

[Data]

The eigenvectors are (in the same order as you listed the eigenvalues, according to Maple 9.5)

$$\left(\begin{array}{c} -\sqrt{3} \\ 1 \\ 2 \end{array}\right), \ \left( \begin{array}{c} \sqrt{3} \\ 1 \\ 2 \end{array} \right), \ \left( \begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right)$$

 

[Hurkyl]

Okay then, so we have:

$$x_{2n} = \frac{1}{6} \left( \begin{array}{ccc} -\sqrt{3} & \sqrt{3} & 0 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{array} \right) \left( \begin{array}{ccc} (7 - 4 \sqrt{3})^n & 0 & 0 \\ 0 & (7 + 4 \sqrt{3})^n & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} -\sqrt{3} & -1 & 2 \\ \sqrt{3} & -1 & 2 \\ 0 & 4 & -2 \end{array} \right) x_0$$

Which you could multiply out, then, to get the explicit formula for $x_{2n}$.

 

[AntonVrba]

Okay then, so we have:

$$x_{2n} = \frac{1}{6} \left( \begin{array}{ccc} -\sqrt{3} & \sqrt{3} & 0 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{array} \right) \left( \begin{array}{ccc} (7 - 4 \sqrt{3})^n & 0 & 0 \\ 0 & (7 + 4 \sqrt{3})^n & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} -\sqrt{3} & -1 & 2 \\ \sqrt{3} & -1 & 2 \\ 0 & 4 & -2 \end{array} \right) x_0$$

Which you could multiply out, then, to get the explicit formula for $x_{2n}$.

and is
$$x_{2n} = \frac{1}{6} \left( \begin{array}{ccc} 3{{(7-4 {\sqrt{3}})}^n}+3 {{(7+4 {\sqrt{3}})}^n}& \sqrt{3}} {{(7-4 {\sqrt{3}})}^n}-{\sqrt{3}} {{(7+4 {\sqrt{3}})}^n}& -2 {\sqrt{3}} {{(7-4 {\sqrt{3}})}^n}+2 {\sqrt{3}} {{(7+4 {\sqrt{3}})}^n} \\ -{\sqrt{3}} {{(7-4 {\sqrt{3}})}^n}+{\sqrt{3}} {{(7+4 {\sqrt{3}})}^n}& 8-{{(7-4 {\sqrt{3}})}^n}-{{(7+4 {\sqrt{3}})}^n}& -4+2 {{(7-4 {\sqrt{3}})}^n}+2 {{(7+4 {\sqrt{3}})}^n} \\ -2 {\sqrt{3}} {{(7-4 {\sqrt{3}})}^n}+2 {\sqrt{3}} {{(7+4{\sqrt{3}})}^n}& 4-2 {{(7-4 {\sqrt{3}})}^n}-2 {{(7+4 {\sqrt{3}})}^n}& -2+4 {{(7-4 {\sqrt{3}})}^n}+4 {{(7+4 {\sqrt{3}})}^n} \\ \end{array} \right) x_0$$

 

[strid]

i just discovered a cool thing.. if we use this instead:

$$(1+\sqrt{2})^{n}$$

than we will get the cool thing that, at n= even numbers, we will get an answer just below an integer with 999999... as decimals...
but if we have n= odd number, the answer will be just above and integer with decimals as 0000000001...

any explanation?