Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a=\frac{1+i\sqrt 7}{2}$ and $u_n=\Re(a^n)$ show that $(|u_n|)\to +\infty$

I think basics method does not works here.

Any ideas ?

share|improve this question
add comment

5 Answers 5

up vote 17 down vote accepted
+50

This isn't really a calculus problem because it amounts to a subtle question about how close a multiple of $\tan^{-1} \sqrt 7$ can get to an odd multiple of $\pi/2$. Robert Israel already explained why one expects that $|u_n|$ should grow faster than $n^{-\mu} 2^{n/2}$ for some constant $\mu$, which is much more than enough to prove the desired $|u_n| \rightarrow \infty$; and achille lui applied the "heavy sledgehammer" of Baker and Wüstholz to prove this for some explicit $\mu$. Here we give a much simpler approach that yields a result that, though much weaker than the sledgehammer, is still sufficient to prove that $|u_n| \rightarrow \infty$.

It was already noted that $u_n = (a^n + b^n) / 2$ where $b = 1-a$ is the complex conjugate of $a$, and thus that the $u_n$ are half-integers satisfying the recurrence $u_{n+2} = u_{n+1} - 2 u_n$. (The odd integers $2u_n$ are OEIS sequence A002249.) Thus $|u_n| \rightarrow \infty$ if and only if for every integer $m$ there are only finitely many solutions $n$ of $2u_n = m$. We shall show that every $m$ occurs at most once except for $1 = u_1 = u_4$ and $-5 = u_3 = u_9$.

The technique is basically what's known as "Skolem's $p$-adic method", which immediately gives a uniform upper bound on the number of solutions of $2u_n = m$; after some experimentation I was able to reduce this bound to $1$ (with the exception of the two solutions for each of $m=1$ and $m=-5$).

We start with some elementary observations. First, $2u_n \bmod 7$ has period $3$ with remainders $1,4,2,1,4,2,\ldots$, so if $u_n = u_{n'}$ then $n' \equiv n \bmod 3$. (This happened for $\{n,n'\} = \{1,4\}$ and $\{3,9\}$.) Likewise, $2u_n \bmod 16$ becomes periodic of period $4$ after the first three terms: $1, -3, -5$, and then then $1, -5, -7, 3, 1, -5, -7, 3,\ldots$. Thus $2u_2=-3$ never recurs, and if $u_n=u_{n'}$ with neither $n$ or $n'$ equal $1$ or $3$ then $n' \equiv n \bmod 4$, whence $n' \equiv n \bmod 12$ because we already knew the congruence mod $3$.

Assume henceforth that $n,n' \geq 4$, so $n' \equiv n \bmod 12$. Then we claim $n' \equiv n \bmod 84$. Indeed for each of the primes $p=29$ and $p=43$ we have $2u_{n+84} \equiv 2u_n \bmod p$ for all $n$, because in each case $-7$ is a square mod $p$ (so $a,b \bmod p$ can be identified with elements of $({\bf Z}/p{\bf Z})^*$) and $p-1$ is a factor of $84$. So it is enough to check that none of the $12$ subsequences $$ \{ 2u_n,\, 2u_{n+12},\, 2u_{n+24},\, 2u_{n+36},\, 2u_{n+48},\, 2u_{n+60},\, 2u_{n+72} \} \bmod 29 \cdot 43 $$ with $1 \leq n \leq 12$ contains a repeat, and this turns out to be the case. (The odds were in our favor: for $12$ random septuplets mod $29 \cdot 43$ the probability of success is over $90\%$, and by using $n \bmod 12$ we removed the known coincidences $u_1=u_4$ and $u_3=u_9$.)

Now for $p=29$ or $p=43$ each of $a^{84} \bmod p$ and $b^{84} \bmod p$ can be written as $1+pc$, $1+pd$ for some algebraic integers $c,d$. It follows by induction on $k$ that $a^{84p^k}$ and $b^{84p^k}$ are congruent to $1 + p^{k+1} c$ and $1 + p^{k+1} d \bmod p^{k+2}$. We thus find, for each $n_0 = 1, 2, \ldots, 84$, some $e_{n_0} \bmod p$ such that $$2u_{n_0+84p^k t} \equiv 2u_{n_0} + p^{k+1} e_{n_0} t \bmod p^{k+2}$$ for all integers $t$. So if $e_{n_0}$ is not zero mod $p$ then there are no repeats in the sequence $\{u_n \mid n \equiv n_0 \bmod 84\}$: if $n \neq n'$ then $n-n'$ has some finite $p$-valuation $k$, and $2u_n$ is congruent to $2u_{n'} \bmod p^{k+1}$ but not mod $p^{k+2}$. If some $e_{n_0}$ does vanish mod $p$ then we can try congruences mod $p^{k+3}$, $p^{k+4}$, etc. But it turns out that this is not necessary: we have two choices of $p$, and while there are some $n_0 \leq 84$ for which $e_{n_0} \equiv 0 \bmod p$ for one of $p=29$ and $p=43$, there is none for which both $e_{n_0} \bmod p$ vanish. (Again the odds were in our favor: a random $84$-tuple mod $29\cdot 43$ has no zero entries with probability $>93\%$.)

This completes the proof that the sequence $\{u_n \mid n \geq 4\}$ has no repeats, and thus establishes our claim that $|u_n| \rightarrow \infty$ as $n \rightarrow \infty$, QED.

share|improve this answer
    
In fact I haven't understand the answer by Robert Israel. And when I saw achille hui's answer, I thought that Robert Israel's answer is just a explanation more than a proof. Thanks for your answer, I have to work on it to fully understand. –  Free X Mar 17 at 9:59
add comment

I'm sure there is a more elementary method than the heavy sledgehammer I used here.
Since I only know this one, here is the sledgehammer.

Let $\displaystyle c = \frac{1+i\sqrt{7}}{2}$, we are going to prove $\Re(c^k) \to \infty$ as $k \to \infty$.
Consider the sequence $(c_k)_{k\in\mathbb{N}}$ where $c_k = 2\Re(c^k) = \left(\frac{1+i\sqrt{7}}{2}\right)^k + \left(\frac{1-i\sqrt{7}}{2}\right)^k$.
It is easy to check it satisfies a linear recurrence relation of the form

$$c_{k+2} = c_{k+1} - 2c_{k}$$

Since $c_0 = 2$ and $c_1 = 1$, this means for all $k > 0$, $c_k$ is an odd integer and hence $\Re(c^k) \ne 0$.

Let $\mathbb{A}$ be the set of algebraic numbers, there is something general we can say about any $c \in \mathbb{A}$ such that $|c| > 1$ and $\Re(c^k) \ne 0$ for all $k > 0$. For any $a$ in $\mathbb{A}$, let $H(a)$ be its naive height. Let us split $c$ as

$$c = |c| a\quad\text{ with }\quad |c| \in \mathbb{A}\quad\text{ and }\quad a \in \mathbb{A}\setminus\mathbb{R}, |a| = 1$$

Now $\Re(c^k) \ne 0$ for all $k > 0$ implies $a^k \ne i$ or $-i$ for all $k > 0$. This implies for any $b_1, b_2 \in \mathbb{Z}$, not both zero, the linear form of logarithm:

$$L(b_1,b_2) = b_1 \log(a) + b_2 \log(i) \ne 0$$

In 1993, Baker and Wüstholz has proved following theorem$\color{blue}{^{[1],[2]}}$:

For any $a_1, \ldots, a_n$ in $\mathbb{A}$, let $L : \mathbb{Z}^n \to \mathbb{C}$ be the linear form of logarithm $$L(b_1,\ldots,b_n) = b_1 \log(a_1) + \cdots + b_n \log(a_n)$$ $L$ is either $0$ for some non trivial $(b_1,\ldots,b_n)$ or bounded away from $0$ by $$|L| > \exp\left[ -(16nd)^{2n+4} \log(A_1) \cdots \log(A_n) \log(B) \right]$$ where $d = [\mathbb{Q}(a_1,\ldots,a_n) : \mathbb{Q}]$, $A_i = \max( H(a_i), e )$ and $B = \max( |b_1|, \ldots, |b_n|, e )$.

If we apply this theorem to our linear form, we get $| b_1 \log(a) + b_2 \log(i) | > B^{-\mu}$ for some constant $\mu$. This implies for large $k$,

$$ | a^k \pm i | > ( \text{constant}\cdot k)^{-\mu} \quad\implies\quad| \Re( c^k ) | > ( \text{constant}\cdot k)^{-\mu} |c|^k $$ As a result, $| \Re( c^k ) | \to \infty$ as $k \to \infty$.

Notes

  • $\color{blue}{[1]}$ A. Baker and G. Wüstholz, Logarithmic forms and group varieties, J. Reine Angew. Math. 442 (1993) 19-62
  • $\color{blue}{[2]}$ This is theorem 2.15 in A. Baker and G. Wüstholz's book Logarithmic Forms and Diophantine Geometry.
share|improve this answer
    
"I'm sure there is a more elementary method than the heavy sledgehammer I used here." I wouldn't be so sure. In fact I highly doubt there is a more elementary method (see Robert Israel's answer). –  Goos Mar 18 at 23:09
    
@Goos, we know a lot about sequence like $2 u_n$. It is a special case of Lucas sequence of second type. We know after at most the first 30 terms, each $2 u_n$ will contain a primitive prime divisor. The first result along this line (for Frobenius sequence) is available in 1920s???. If we can prove such a strong result 100 years ago. It is sort of hard to believe there isn't an elementary method for this much weaker statement. –  achille hui Mar 19 at 1:29
add comment

Write $a = r e^{i\theta}$ where $r = \sqrt{2}$ and $\theta = \arctan(\sqrt{7})$. The question is whether for any $N$ there exist infinitely many $n$ such that $|2^{n/2} \cos(n \theta)| < N$.
This can be viewed as a question about whether $2\theta/\pi$ has extremely good rational approximations. Namely, if $$ \left| \dfrac{2\theta}{\pi} - \dfrac{m}{n} \right| < \dfrac{2 N}{\pi n} 2^{-n/2}$$ for some odd integer $m$, then we would have $$|\cos(n \theta)| < \left| n \theta - m \pi/2 \right| < N 2^{-n/2}$$ Now almost surely it doesn't, in fact for any $\epsilon > 0$ and almost every real $x$ there are only finitely many pairs of positive integers $(m,n)$ such that $|x - m/n| < 1/(n^2 \log(n)^{1+\epsilon}$. But there are some irrational numbers $x$ that do have extremely good rational approximations (in fact these contain a dense $G_\delta$ subset of $\mathbb R$). As far as I know there is no way to prove that $2 \arctan(\sqrt{7})/\pi$ is not one of those numbers.

share|improve this answer
    
So this exercise it's harder than it looks..Anyway, it's an interesting idea (+1). –  user119228 Mar 10 at 0:33
    
Unfortunately I don't understand what have you proved here ? Could you please explain a little more ? –  Free X Mar 17 at 9:51
    
This is the correct way to look at the problem, but with regards to "As far as I know there is no way to prove..." it seems that achillehui's answer does indeed prove this. –  Goos Mar 18 at 23:08
    
Correction: "As far as I knew when I wrote that, ..." –  Robert Israel Mar 19 at 3:26
add comment

No idea if this will help but could not resist posting these striking images. Try plotting the absolute value of the real part of $a^n/2^{n/2}$ for $0\le n\le1000$.

enter image description here
Addendum. Even better, change the plot style to "point" so that you don't have lines joining successive points.

enter image description here

share|improve this answer
add comment

Here are a couple of ideas - though I think the problem may be harder than it looks.

Let $a=r(\cos \theta+i\sin \theta)$ so that $a^n=r^n (\cos n\theta+i\sin n\theta)$ and $u_n=r^n \cos n\theta$.

You will find that $r=\sqrt 2$ and $\theta = \arctan \sqrt 7$. This shows that the general growth is like $(\sqrt 2)^n$ but you have to bound the trigonometric bit sufficiently away from zero to avoid "too many" values which are "too small".


Let $b=\bar a$ then $u_n=\Re (a^n) = \cfrac {a^n+b^n}{2}$

Now $a+b=1$ and $ab=2$ so that $a$ and $b$ are roots of $x^2-x+2=0$ and hence of $$f_n(x)=x^{n+2}-x^{n+1}+2x^n=0$$ whence $$\frac {f_n(a)+f_n(b)}{2}=u_{n+2}-u_{n+1}+2u_n=0$$ so that $u_{n+2}=u_{n+1}-2u_n$

This gives a convenient way of calculating $u_n$, but shows that if it is growing it will regularly change sign (as we expect from the trigonometric formulation) and the absolute value is not monotone increasing.

Perhaps someone knows a trick which makes this easier.

share|improve this answer
    
The limit formulation $$\lim_{n\to\infty}2^{n/2}\cos(n\tan^{-1}(\sqrt 7))$$ is quite interesting. I am almost certain there is a theorem about that, but it seems challenging from first principles. –  Ian Mateus Mar 9 at 22:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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