# Sequences

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• Sep 29th 2007, 10:45 AM
alikation0
Sequences
1a) Define $(x_n)$ as being the sequence:

$x_1 = 3$

$x_{n+1} = \frac{1}{2}\cdot (x_n + \frac{3}{x_n})$

Prove $x_n$ converges and find the lim.

b) Let $b > 1$ and define $(x_n)$ as being the sequence:

$x_1 = b$

$x_(n+1) = \frac{1}{2}\cdot (x_n + \frac{b}{x_n})$

Prove $x_n$ converges and find the lim.

c) Suppose you're on an island with only a solar-powered very basic calculator. Use the result from part b to approximate $sqrt(17)$
• Sep 29th 2007, 01:41 PM
red_dog
1) a
We'll prove that the sequance is low bounded by $\sqrt{3}$
$x_1=3>\sqrt{3}$

$\displaystyle x_{n+1}=\frac{x_n+\frac{3}{x_n}}{2}\geq\sqrt{x_n\c dot\frac{3}{x_n}}=\sqrt{3}$ (I used AM-GM inequality).

Now, we'll prove that the sequence is descrescent.
$x_2=2\Rightarrow x_1>x_2$.
Suppose that $x_{n-1}>x_n$.
Then, $x_n-x_{n+1}=x_n-\frac{x_n^2+3}{2x_n}=\frac{x_n^2-3}{2x_n}>0\Rightarrow x_n>x_{n+1}$ (I used the fact that the sequence is bounded).

So, the sequence is convergent. Let $\lim_{n\to\infty}x_n=x$.
Applying the limit in the recurrency relation we have

$\displaystyle x=\frac{1}{2}\left(x+\frac{3}{x}\right)\Rightarrow x^2=3\Rightarrow x=\sqrt{3}$
• Sep 29th 2007, 03:56 PM
Soroban
Hello, alikation0!

I recognized the function . . . and "eyeballed" the answers.

Quote:

b) Let $b > 1$ and define $(x_n)$ as being the sequence:

$x_1 = b$

$x_(n+1) = \frac{1}{2}\cdot (x_n + \frac{b}{x_n})$

Prove $x_n$ converges and find the lim.

c) Suppose you're on an island with only a solar-powered very basic calculator.
Use the result from part b to approximate $\sqrt(17)$

I'm old enough to remember those "very basic calculators".
They were the size of a TI-30 (but an inch thick), weighed a pound,
. . used a 9-volt battery (used up quickly by those red LEDs)
.. cost about \$100 and did only basic arithmetic.

Back then, we learned many tricks for approximating more complex answers.
. . And one of them was square roots.

Suppose we want $\sqrt{N}$.

Make a first approximation, $a_1.$

If we're very very lucky, $a_1$ is exact.
. . Then the two factors of $N$ are equal: . $a_1 \,=\,\frac{N}{a_1}$

Most likely, the two factors are not equal.
. . Obviously, one is too small and the other too large.

Then a better approximation is the average of these two factors.

So we will calculate: . $a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \;=\;\frac{1}{2}\left(a_1 + \frac{N}{a_1}\right)$

Hence, $a_2$ is a better approximation of $\sqrt{N}.$

Then: . $a_3 \:=\:\frac{1}{2}\left(a_2 + \frac{N}{a_2}\right)$ is an even better approximation.

And that is the source of that recursive function: . $a_{n+1} \;=\;\frac{1}{2}\left(a_n + \frac{N}{a_n}\right)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Actually, I used this form: . $a_{n+1} \;=\;\frac{a_n^2 + N}{2a_n}$