Let a > 0, a1 > 0. Define a(n+1) = 1/2(an+a/an) then,

[daon]

Let a > 0, a₁ > 0.

Define \(\displaystyle a_{n+1}=\frac{1}{2}(a_n +\frac{a}{a_n})\)

I need to show that \(\displaystyle a_n > 0\) and \(\displaystyle a_{n+1}^2 - a > 0\) for all n.

The first part doesn't seem that bad..

Assume BWOC that \(\displaystyle a_n < 0\) (since it can't be zero by its definition) then \(\displaystyle \frac{1}{2}a_n + \frac{a}{2a_n} < 0 \,\, \Rightarrow \,\, a_n^2 + a < 0\) This is a contradiction since both \(\displaystyle a_{n}^2\) and \(\displaystyle a\) are positive.


However, on the second part, I usually end up with something odd that requires a to be greater than 2 or a_n to be greater than 1 or something of the like.

edit: The best I can get is \(\displaystyle a_{n+1}^2-\frac{1}{2}a > 0\).

Thanks,
Daon

 

[pka]

Unless there is more to the given, I have a counterexample.
If a=1 & a<SUB>1</SUB>=1 then the second condition is not the case.

 

[daon]

Unless there is more to the given, I have a counterexample.
If a=1 & a<SUB>1</SUB>=1 then the second condition is not the case.

I apologize. It should read greater-than or equal to zero. Also, it is stated as: " Let a>0. Let a1 be any positive number and define.." (from here is is the same.) Thanks.

 

[pka]

For all a & b, \(\displaystyle a^2 + b^2 \ge 2ab\).


Thus \(\displaystyle \L
\begin{array}{rcl}
a_{n + 1} & = & \frac{{\left( {a_n } \right)^2 + a}}{{2a_n }} \\
& \ge & \frac{{2a_n \sqrt a }}{{2a_n }} \\
\Rightarrow a_{n + 1} & \ge & \sqrt a \\
\Rightarrow \left( {a_{n + 1} } \right)^2 & \ge & a \\
\Rightarrow \left( {a_{n + 1} } \right)^2 - a & \ge & 0 \\
\\
\end{array}\)

 

[daon]

Ah, the AGMI

One more quick question. If I want to show this is non-increasing (for n>= 2), would this suffice:

Using the fact that a²_n+1 >= a for n>=1, we also know a²_n >= a for n>=2. So,

\(\displaystyle \L a^2_{n} \ge a \,\, \Leftrightarrow \,\, 1 \ge \frac{a}{a_{n}^2} \\
\,\, \Leftrightarrow \,\, 2 \ge 1 + \frac{a}{a_{n}^2} \\ \,\, \Leftrightarrow \,\,
2a_n^2 \ge a_n^2 + a \\ \,\, \Leftrightarrow \,\, a_n \ge \frac{a_n^2+a}{2a_n} \,\, = \,\, a_{n+1}\)

Also, I am supposed to show this is bounded, but dooesn't the fact that this is non-increasing and greater than zero imply that it is, doesn't it also imply that it is convergent? I found the limit to be \(\displaystyle \sqrt{a}\). Hopefully that is correct.

Thanks.
-Daon

 

[pka]

Also, I am supposed to show this is bounded, but doesn’t the fact that this is non-increasing and greater than zero imply that it is, doesn't it also imply that it is convergent? I found the limit to be \(\displaystyle \sqrt{a}\).

Yes that works. This is a standard way of approximating square roots.

 

[soroban]

Re: Let a > 0, a1 > 0. Define a(n+1) = 1/2(an+a/an) th

Hello, daon!

Let \(\displaystyle k\,>\,0,\;a_1\,>\, 0\)

Define \(\displaystyle a_{n+1}\:=\:\frac{1}{2}\left(a_n\,+\,\frac{k}{a_n}\right)\)

Show that: \(\displaystyle \;[1]\;a_n\,>\,0\;\;\;\;[2]\;\left(a_{n+1}\right)2\, -\,k\:>\:0\) for all \(\displaystyle n\).

If part [2] has \(\displaystyle \geq\), I have a proof . . .


We have: \(\displaystyle \:a_{n+1}\:=\:\frac{1}{2}\left(a_n\,+\,\frac{k}{a_n}\right)\;\;\Rightarrow\;\;2a_{n+1} \:=\:a_n\,+\,\frac{k}{a_n}\)

Square both sides: \(\displaystyle \:4(a_{n+1})^2 \;= \;\left(a_n\,+\,\frac{k}{a_n}\right)^2 \;= \;a_n^2\,+\,2k\,+\,\frac{k^2}{a_n^2}\)

Subtract \(\displaystyle 4k\) from both sides: \(\displaystyle \:4(a_{n+1})^2\,-\,4k\;=\;a_n^2\,+\,2k\,+\,\frac{k^2}{a_n^2}\,-\,4k\;=\;a_n^2\,-\,2k\,+\,\frac{k^2}{a_n^2}\)

And we have: \(\displaystyle \:4\left[(a_{n+1})^2 - k\right] \;= \;\left(a_n\,-\,\frac{k}{a_n}\right)^2 \;=\;\left(\frac{a_n^2\,-\,k}{a_n}\right)^2\)

Divide by 4: \(\displaystyle \a_{n+1})^2\,-\,k \:=\;\frac{1}{4}\left(\frac{a_n^2\,-\,k}{a_n}\right)^2\)

Hence: \(\displaystyle \a_{n+1})^2\,-\,k\;=\;\left(\frac{a_n^2\,-\,k}{2a_n}\right)^2\)


Since the square of a real number is nonnegative,

. . we have: \(\displaystyle \L\a_{n+1})^2\,-\,k\:\geq\:0\)

 

[daon]

Thank you soroban. I believe at first I was attempting to do something similar to what you just did, but I never got that far. I like seeing alternative methods.

 

[daon]

I've been thinking a lot about this problem, specifically why the 1/2 has any significance.

∙ Instead of 1/2 I tried 1/3 and found that the limit when a=18 gives an exact 3 (to 128-bit precision).

∙ With 1/4 and a=27, I get the limit as 3.

∙ With 1/5 and a=36 I also get 3.

∙ With 1/6 and a=45 I get 3 as well.

With this pattern I get that when I have:

\(\displaystyle lim \,\, \frac{1}{k}(a_n + \frac{a}{a_n}) = 3\) when \(\displaystyle a=(9)(k-1)\)

I just thought I'd share this for anyone who might be interested. Any thoughts, comments? I'll keep fiddleing with numbers

edit: Additionally, through some more runs through the code:

If \(\displaystyle lim \,\, \frac{1}{k}(a_n + \frac{a}{a_n}) = A\) then \(\displaystyle a=A^2(k-1)\)

So the square root aproximation would be a special case of this, namely when k=2.