Consider the following recursion:
a[k] = (1/2)(a[k-1] + 2/a[k-1]), k = 1, 2, 3... ; a[.] is positive real
We need a seed; it's not too important, try a[1] = 10

Does the sequence converge? (hint: it does)
What's the limit?

Using your troglodyte brain, you might calculate a few
terms and guess the limit. yuk, crude

Or you might exercise your homo sapiens brain
and derive the limit, elegantly.

--
Rich

On 11/21/2021 6:39 PM, RichD wrote:
> Consider the following recursion:
> a[k] = (1/2)(a[k-1] + 2/a[k-1]), k = 1, 2, 3... ; a[.] is positive real
> We need a seed; it's not too important, try a[1] = 10
>
> Does the sequence converge? (hint: it does)
> What's the limit?
>
> Using your troglodyte brain, you might calculate a few
> terms and guess the limit. yuk, crude
>
> Or you might exercise your homo sapiens brain
> and derive the limit, elegantly.
>
> --
> Rich
>

https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/using-the-formula-for-geometric-series/

https://nypost.com/2021/11/16/kamala-harris-is-a-very-weird-person/

RichD <r_delaney2001@yahoo.com> wrote:
> Consider the following recursion:
> a[k] = (1/2)(a[k-1] + 2/a[k-1]), k = 1, 2, 3... ; a[.] is positive real

Call that equation 1.

> We need a seed; it's not too important, try a[1] = 10
>
> Does the sequence converge? (hint: it does)
> What's the limit?
>
> Using your troglodyte brain, you might calculate a few
> terms and guess the limit. yuk, crude
>
> Or you might exercise your homo sapiens brain
> and derive the limit, elegantly.

The easiest way to find the limit is to suppose it converges to some
number a, at which point a = (a + 2/a)/2, or 2a^2 = a^2 + 2, or a^2=2,
a = sqrt(2)*. For convergence, just show a[n] > a[n+1] > sqrt(a)
after a[2].

Seeing the sqrt(2) value is a reminder that equation 1 looks a lot
like Newton's method; and indeed it's a form shown in refs A, B, C.
Reference C give the details of a[n] > a[n+1] > sqrt(a), while A & B
have calculus-based quadratic convergence proofs.

*) a = sqrt(2) if a[1] is positive, else -sqrt(2).
A) <https://en.wikipedia.org/wiki/Newton's_method#Square_root>
B) <https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method>
C) <https://math.mit.edu/~stevenj/18.335/newton-sqrt.pdf>

On November 21, James Waldby wrote:
>> Consider the following recursion:
>> a[k] = (1/2)(a[k-1] + 2/a[k-1]), k = 1, 2, 3... ; a[.] is positive real
>
> Call that equation 1.
>
>> We need a seed; it's not too important, try a[1] = 10
>> Does the sequence converge? (hint: it does)
>> What's the limit?
>
> The easiest way to find the limit is to suppose it converges to some
> number a, at which point a = (a + 2/a)/2, or 2a^2 = a^2 + 2, or a^2=2,
> a = sqrt(2)*.

Yes, that's the forward method: start with a^2 = 2, do some algebra,
then the recursion follows easily.

However, the backward method - given the recursion, decode it - is
more obscure. I posted it here, to see if anyone recognized it.

> For convergence, just show a[n] > a[n+1] > sqrt(a)
> after a[2].

It's a difference equation, which has a rich theory. But it's non-linear,
and the books only deal with linear systems. Stability proofs become
dicey (though maybe not in this case).

> Seeing the sqrt(2) value is a reminder that equation 1 looks a lot
> like Newton's method; and indeed it's a form shown in refs A, B, C.
> Reference C give the details of a[n] > a[n+1] > sqrt(a), while A & B
> have calculus-based quadratic convergence proofs.
>
> A) <https://en.wikipedia.org/wiki/Newton's_method#Square_root>

--
Rich

On Sunday, November 21, 2021 at 4:40:01 PM UTC-8, RichD wrote:
> Consider the following recursion:
> a[k] = (1/2)(a[k-1] + 2/a[k-1]), k = 1, 2, 3... ; a[.] is positive real
> We need a seed; it's not too important, try a[1] = 10
>
> Does the sequence converge? (hint: it does)
> What's the limit?
>
> Using your troglodyte brain, you might calculate a few
> terms and guess the limit. yuk, crude
>
> Or you might exercise your homo sapiens brain
> and derive the limit, elegantly.
>
> --
> Rich

You got a monkey brain rich...