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Let $c>0$, $a_{1} = 1$, and $$a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)$$

I need to:

  1. Show that $a_{n}$ is defined for every $n\geq 1$
  2. Show that this sequence is convergent.
  3. Find its limit.

I proved the first part by showing by induction that this sequence is positive for every $n$. To show that this sequence is convergent I'm thinking of showing that this sequence is a Cauchy series, yet can't figure out how.

For the third part I'm clueless at the moment.

Martin Sleziak
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  • This is just heron's famous method of computing the square root, look here https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method –  Nov 27 '16 at 17:08
  • For the second part : try to show that the sequence is decreasing (at least for $n\geq 2$). – Arnaud D. Nov 27 '16 at 17:09
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    use $AM-GM$ to solve the problem – Dr. Sonnhard Graubner Nov 27 '16 at 17:09
  • Once you have proven that the sequence converges, you can then call that limit $a := \lim\limits_{n \rightarrow \infty}{a_n}$ and use the fact that $\lim\limits_{n \rightarrow \infty}{a_{n+1}} = \lim\limits_{n \rightarrow \infty}{a_n}$ to calculate what $a$ is. – Tom Nov 27 '16 at 17:20
  • https://math.stackexchange.com/questions/981819/demonstrating-the-convergence-of-x-n1-1-2x-n-c-x-n – Workaholic Nov 27 '16 at 19:04
  • See also: Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$. Other posts linked there might be of interst, too. – Martin Sleziak Dec 04 '16 at 01:46
  • You posted this problem twice. Why? – marty cohen Dec 04 '16 at 01:53
  • The first part, that $a_n$ is defined for every $n \geq 1$, is not by induction! Rather, it's by the principle of recursion (a theorem whose proof involves a not-too-trivial use of the principle of induction). It would be fallacious to argue along the following lines: "First, $a_1$ is defined because its value is given. Now let $n \geq 1$ and assume that $a_1, a_2, \dots, a_n$ have already been defined. Then the recurrence relation gives a value for $a_{n+1}$, too." Yes, a lot of introductory texts try to browbeat you into believing this argument; but it's bogus. – murray Aug 10 '17 at 22:08

3 Answers3

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For (2). You have to show that the sequence is bounded and decreasing, therefore convergent.

For (3). what you need to do is set $\lim{a_n} = L$ and solve. We have
$L = (1/2)L+\frac{c}{L}$, since $\lim{a_{n+1}} = \lim{a_n}$. Solving yields $L = \sqrt{c}$.

Retired account
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It might be interesting that the recursion

$$a(n+1) = \frac{1}{2} (a(n) + \frac{c}{a(n)})\tag{1}$$

can be solved explicitly.

The solution is

$$a(n) = \sqrt{c}\; \coth {(2^{n-1}\;\text{arccoth}( a(1)/\sqrt{c}))}\tag{2} $$

For large $n$ the argument of the $\text{coth}$ gets large as well so that the $\text{coth} \to 1$ and $a(n) \to \sqrt{c}$, as expected.

Derivation

First we normalize the problem letting

$$d(n) = a(n) \sqrt{c}$$

which leads to the recursion

$$d(n+1) = \frac{1}{2} (d(n) + \frac{1}{d(n)})\tag{3}$$

The sum of $d(n)$ and its inverse suggests the ansatz

$$d(n) = \coth(f(n)) = \frac { e^{f(n)} + e^{- f(n)} } { {e^{ f(n)} - e^{- f(n)} }}\tag{4}$$

Inserting this into the r.h.s of (3) gives

$$\frac{ e^{2 f(n)} + e^{-2 f(n)} } { {e^{2 f(n)} - e^{-2 f(n)} }}$$

which has the same form as $d(n+1)$ if we let $f(n)$ obey the recursion

$$f(n+1) = 2 f(n)\tag{5}$$

But this can be solved immediately

$$f(n) = 2^{n-1} f(1) $$

so that the solution to (3) is given by

$$d(n) = \coth {(2^{n-1}f(1))}$$

The constant $f(1)$ has to be adapted to the initial condition leading to

$$f(1) = \text{arccoth}( d(1)) $$

Hence the solution to (3) becomes

$$d(n) = \coth {(2^{n-1}\text{arccoth}( d(1)))} $$

And, reversing the nomalization, we find the solution (2).

Dr. Wolfgang Hintze
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1

Hints:

  1. One can prove that if $a_1>0$ and $n \geq 2$ then $a_n \geq \sqrt{c}$.
  2. One can prove that if $a_1>0$ and $n \geq 2$ then $a_{n+1} \leq a_n$. Knowing that and the bound in 1, convergence follows.
  3. Once you know that a recursive sequence is convergent, its limit can only be a fixed point of the recursion mapping, i.e. in your case a solution to $\frac{x}{2}+\frac{c}{2x}=x$.
Ian
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  • Could you show the way to prove that $a_{n} \geq \sqrt(c)$ for every $n\geq2$ ? –  Nov 27 '16 at 21:38
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    @S.Peter For $n \geq 2$, $a_n$ is the arithmetic mean of $a_{n-1}$ and $c/a_{n-1}$, so by the arithmetic-geometric mean inequality it is bounded below by $\sqrt{a_{n-1}(c/a_{n-1})}=\sqrt{c}$. – Ian Nov 27 '16 at 21:42
  • Could you also explain how the second hint can be proved? –  Nov 27 '16 at 22:33
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    @S.Peter It boils down to showing $\frac{c}{a_n}-a_n \leq 0$, since $a_{n+1}-a_n$ is just half of that. Can you do that using the first hint? – Ian Nov 27 '16 at 22:35