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I have the following sequence (derived from law of cosines):

$$d_{n+1}^2 = d_n^2 + r^2 - 2rd_n \cos \theta$$

such that $r, d_n>0$ and $\theta < \frac{\pi}{2}$.

I would like to show that the limit exists for this sequence. I believe the limit to be $\frac{r}{2\cos \theta}$.

Attempt: I originally tried to show that the sequence is bounded and monotonic. While it is bounded, it is not monotonic. So I thought maybe I could try to show that the sequence is Cauchy. But I am having trouble showing that as well.

Any hints/ideas? Thanks.

edit: If $\theta = 0$ the limit does not exist in general. So the restriction on $\theta$ is $0<\theta<\frac{\pi}{2}$.

edit 2: Example configuration to make $d_n - \frac{r}{2 \cos \theta}$ alternate +/-:

$d_0 = 1$, $r=4$, $\theta = \frac{\pi}{6}$

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    $\begingroup$ Just checking: is it $d_n^2\color{red}{+}r^2-..$ or $d_n^2-r^2-..$ on the RHS? $\endgroup$
    – Anurag A
    Apr 26 '20 at 22:24
  • $\begingroup$ @anuragA yes! Thanks for catching that. $\endgroup$
    – T. Fo
    Apr 26 '20 at 22:26
  • $\begingroup$ I don't think it matters, but what is $d_0$? $\endgroup$ Apr 26 '20 at 22:33
  • $\begingroup$ @VVejalla, $d_0 > 0$ only condition. I don't think it should matter. $\endgroup$
    – T. Fo
    Apr 26 '20 at 22:38
  • $\begingroup$ It seems that the seequence is alternately increasing and decreasing. That is, $d_{2n}$ and $d_{2n+1}$ are both monotonic. They approach their common limit in geometric progression. $\endgroup$
    – Somos
    Apr 27 '20 at 0:41
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Let us rewrite the problem as follows.

$d_{n+1}^2 = d_n^2 + r^2 - mrd_n,\space where \space m := 2 cos(\theta), \space r > 0, \space\forall n (d_n > 0)$

We can see that $m \in (0;2)$ as $cos(\theta) \in (0;1)$ for $\theta \in (0;\frac \pi 2)$.

It will be easier to consider the sequence $(a_n) := (\frac {d_n} r)$ as it only depends on $m$ and, possibly, $d_0$.

As we can see, $a_{n+1}^2 = a_n^2 + 1 - ma_n$

Now, we can observe that if the limit exists, it is equal to $\frac 1 m$.

If $\exists L (\lim_{n\to\infty} {a_n} = L) \space \Rightarrow \exists L (\lim_{n\to\infty} {a_n} = L \wedge \lim_{n\to\infty} {a_{n+1}^2} = \lim_{n\to\infty} {a_n^2 + 1 - ma_n} = L^2 + 1 - mL = L^2) \Rightarrow \exists L (\lim_{n\to\infty} {a_n} = L \wedge L = \frac 1m) \Rightarrow \lim_{n\to\infty} {a_n} = \frac 1m \space (*)$

It is also easy to derive the formula for the $n$-th term, by recursively applying the formula.

$a_n^2 = a_0^2 + n - m\sum_{i=0}^{n-1}{a_i}$

Now we will show that if an element of the sequence is below $\frac 1 m$, then every element after it is above the element in question.

$a_n < \frac 1m \Rightarrow ( k = n + 1 \Rightarrow a_k^2 = a_{n+1}^2 = a_n^2 + 1 - ma_n > a_n^2 )$

$a_n < \frac 1m \Rightarrow ( \forall t ( n < t < k \Rightarrow a_t > a_n) \Rightarrow a_k^2 - a_n^2 = (k - n) - m\sum_{i=n}^{k-1}{a_i} > (k - n) - m\sum_{i=n}^{k-1}{a_n} = (k - n) - m(k - n)a_n > (k - n) - m(k - n)\frac 1m > 0 \Rightarrow a_k > a_n )$

$\therefore a_n < \frac 1m \Rightarrow( k > n \Rightarrow a_k > a_n )$

Similarly, if an element is above $\frac 1 m$, then the sequence will always be below it.

$a_n > \frac 1m \Rightarrow ( k = n + 1 \Rightarrow a_k^2 = a_{n+1}^2 = a_n^2 + 1 - ma_n < a_n^2 )$

$a_n > \frac 1m \Rightarrow ( \forall t ( n < t < k \Rightarrow a_t < a_n) \Rightarrow a_k^2 - a_n^2 = (k - n) - m\sum_{i=n}^{k-1}{a_i} < (k - n) - m\sum_{i=n}^{k-1}{a_n} = (k - n) - m(k - n)a_n < (k - n) - m(k - n)\frac 1m < 0 \Rightarrow a_k < a_n )$

$\therefore a_n > \frac 1m \Rightarrow( k > n \Rightarrow a_k < a_n )$

We will partition the sequence $(a_n)$ into three other, depending on the relative positions of the elements and $\frac 1 m$.

$(b_n)$ - subsequence of $(a_n)$, such that $\forall n (b_n > \frac 1m)$

$(c_n)$ - subsequence of $(a_n)$, such that $\forall n (c_n < \frac 1m)$

$(w_n)$ - subsequence of $(a_n)$, such that $\forall n (w_n = \frac 1m)$

Observe that if $a_n = \frac 1m\Rightarrow a_{n+1} = \frac 1m$.

This implies, that once the sequence reaches $\frac 1 m$ it stays there.

$\therefore a_k\in(w_n) \Rightarrow \lim_{n\to\infty}{a_n} = a_k = \frac 1m\blacksquare$

We will, therefore, consider the other possibility, i.e. $(w_n)\equiv\emptyset$.

Thus every element is either strictly below or strictly above $\frac 1 m$.

$(b_n)\cup(c_n) \equiv (a_n)$

Now, if some subsequence is finite, the other one dominates on large indices. Additionally, each subsequence is monotone and bounded by $\frac 1 m$. Therefore, in this case, the infinite subsequence converges, implying the convergence of the sequence as a whole.

$|(b_n)| < \aleph_0 \Rightarrow \exists N \forall n > N (a_n < \frac 1m \wedge \forall i \forall j (n < i < j \Rightarrow a_i < a_j)) \Rightarrow \exists L (\lim_{n\to\infty}{a_n} = L)\blacksquare $

$|(c_n)| < \aleph_0 \Rightarrow \exists N \forall n > N (a_n > \frac 1m \wedge \forall i \forall j (n < i < j \Rightarrow a_i > a_j)) \Rightarrow \exists L (\lim_{n\to\infty}{a_n} = L)\blacksquare $

In these cases, we have only shown that the limit $L$ exists and $L\ge\frac 1m$ and $L\le\frac 1m$ respectively. The proof is completed by $(*)$.

We will now consider the case, when both subsequences are infinite, i.e. $|(b_n)| = |(c_n)| = \aleph_0$.

We will call their limits $G_1$ and $G_2$. The limits exist because both sequences are monotone and bounded.

$G_1 := \lim_{n\to\infty} {b_n}$

$G_2 := \lim_{n\to\infty} {c_n}$

Notice that $\nexists n (b_n = G_1)$ and $\nexists n (c_n = G_2)$, because the sequences are strictly monotone, and, therefore, cannot attain their respective limits.

$\therefore \forall n > 0 (c_0 < c_n < G_2 \le \frac 1m \le G_1 < b_n < b_0)$

If $G_1 = G_2 \Rightarrow \lim_{n\to\infty} {a_n} = G_1 = G_2 = \frac 1m\blacksquare$

Otherwise $G_1 > G_2$. We will consider this case next.

From the definition of a limit for a sequence of real numbers,

$\forall \rho > 0 \exists N \forall n > N ((G_1 < b_n < G_1 + \rho) \wedge (G_2 - \rho < c_n < G_2))$

We will now consider a function

$f(x) := x^2 + 1 - mx$

It is a polynomial and, therefore, it is continuous.

Also $\forall n (f(a_n) = a_{n+1}^2)$.

From the definition of a limit of a real-argument-real-valued function,

$\therefore \forall \epsilon > 0 \exists \delta \forall y (|x - y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon)$

If it is the case, that the sequence infinitely often switches from being close to $G_2$ several times in a row to being close to $G_1$, then it will have "big" jumps for "small" differences in the initial values, which would mean, that $f(x)$ is not continuous.

If $\forall N \exists n > N (a_n < G_2 \wedge a_{n+1} < G_2 \wedge a_{n+2} > G_1) \Rightarrow \forall \rho > 0 \exists N \exists n > N (|a_n - a_{n+1}| < \rho \wedge |a_{n+1} - a_{n+2}||a_{n+1} + a_{n+2}| > (G_1 - G_2) 2c_0 ) \Rightarrow \exists \epsilon > 0 \forall \rho > 0 \exists y (|G_2 - y| < \rho \wedge |f(G_2) - f(y)| > \epsilon)\Rightarrow\Leftarrow$

A similar argument applies if we interchange $G_1$ and $G_2$.

If $\forall N \exists n > N (a_n < G_1 \wedge a_{n+1} < G_1 \wedge a_{n+2} > G_2) \Rightarrow \forall \rho > 0 \exists N \exists n > N (|a_n - a_{n+1}| < \rho \wedge |a_{n+1} - a_{n+2}||a_{n+1} + a_{n+2}| > (G_1 - G_2) 2c_0 ) \Rightarrow \exists \epsilon > 0 \forall \rho > 0 \exists y (|G_1 - y| < \rho \wedge |f(G_1) - f(y)| > \epsilon)\Rightarrow\Leftarrow$

This means that after a certain point, the sequence has to change between closeness to $G_1$ and $G_2$ at each step.

$\therefore \exists N \forall n > N ((a_n < G_2 \Rightarrow a_{n+1} > G_1)\wedge (a_n > G_1 \Rightarrow a_{n+1} < G_2))$

Thus after some point $(b_n)$ and $(c_n)$ alternate with some constant difference in indices.

$\exists p \exists q \exists N \forall n > N (b_{n+1+p}^2 = c_{n+q}^2 + 1 - mc_{n+q} \wedge c_{n+1+q}^2 = b_{n+p}^2 + 1 - mb_{n+p})$

If we take the limits of the equations, we will be left with equations in terms of $G_1$ and $G_2$.

$\therefore \lim_{n\to\infty} {b_n^2} = \lim_{n\to\infty} {c_n^2 + 1 - mc_n} \wedge \lim_{n\to\infty} {c_n^2} = \lim_{n\to\infty} {b_n^2 + 1 - mb_n}$

$G_1^2 = G_2^2 + 1 - mG_2$

$G_2^2 = G_1^2 + 1 - mG_1$

We can solve this system as follows.

$G_1^2 + G_2^2 = (G_2^2 + 1 - mG_2) + (G_1^2 + 1 - mG_1)$

$0 = 2 - m (G_1 + G_2)$

Now we know the sum of $G_1$ and $G_2$.

$\therefore G_1 + G_2 = \frac 2m$

$G_1^2 - G_2^2 = (G_2^2 + 1 - mG_2) - (G_1^2 + 1 - mG_1)$

$G_1^2 - G_2^2 = G_2^2 - G_1^2 - mG_2 + mG_1$

$2(G_1^2 - G_2^2) = m(G_1 - G_2)$

$2(G_1 - G_2)(G_1 + G_2) = m(G_1 - G_2)$

We have already established that $G1 > G2$, therefore, $G_1 - G_2 \neq 0$ and as such

$2(G_1 + G_2) = m$

$G_1 + G_2 = \frac m2$

We already know another formula for the sum of $G_1$ and $G_2$. Substituting it, we get

$\frac m 2 = \frac 2 m$

This is only possible if $m$ is $2$ or $-2$.

$m = 2 \vee m = -2 \Rightarrow\Leftarrow$

Nevertheless, we know it is not possible, because $m \in (0;2)$. Therefore, our assertion that $G_1 > G_2$ was false.

At this point, we have exhausted the cases and have shown that each of them is either impossible or implies $\lim_{n\to\infty} {d_n} = \lim_{n\to\infty} {r a_n} = \frac r m = \frac r {2 cos(\theta)}$. $\blacksquare$

Visual aid

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  • $\begingroup$ This is a fantastically detailed answer. I admit it's taking me a while to parse exactly what is going on. In particular, it's taking me a while to understand your arguments for G1 and G2. I'm also not following how we know that the subsequences $b_n$ and $c_n$ are monotonic. I need to give this another read or two... $\endgroup$
    – T. Fo
    May 15 '20 at 4:57
  • $\begingroup$ @T. Fo We have proven that if $a_n > \frac 1m$ then every $a_k$ after it is smaller, therefore, if $i > j$ then $b_i > b_j$, note that $b_n > \frac 1m$. Simmilar argument applies for $(c_n)$. $\endgroup$
    – Fullfungo
    May 15 '20 at 22:56

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