1. ## Sequences

1a) Define $\displaystyle (x_n)$ as being the sequence:

$\displaystyle x_1 = 3$

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

Prove $\displaystyle x_n$ converges and find the lim.

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

$\displaystyle x_1 = b$

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

Prove $\displaystyle x_n$ 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 $\displaystyle sqrt(17)$

2. 1) a
We'll prove that the sequance is low bounded by $\displaystyle \sqrt{3}$
$\displaystyle x_1=3>\sqrt{3}$

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

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

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

$\displaystyle \displaystyle x=\frac{1}{2}\left(x+\frac{3}{x}\right)\Rightarrow x^2=3\Rightarrow x=\sqrt{3}$

3. Hello, alikation0!

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

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

$\displaystyle x_1 = b$

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

Prove $\displaystyle x_n$ 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 $\displaystyle \sqrt(17)$

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 $100 and did only basic arithmetic. Back then, we learned many tricks for approximating more complex answers. . . And one of them was square roots. Suppose we want$\displaystyle \sqrt{N}$. Make a first approximation,$\displaystyle a_1.$If we're very very lucky,$\displaystyle a_1$is exact. . . Then the two factors of$\displaystyle N$are equal: .$\displaystyle a_1 \,=\,\frac{N}{a_1}$Most likely, the two factors are not equal. . . Obviously, one is too small and the other too large. Then a better approximation is the average of these two factors. So we will calculate: .$\displaystyle a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \;=\;\frac{1}{2}\left(a_1 + \frac{N}{a_1}\right)$Hence,$\displaystyle a_2$is a better approximation of$\displaystyle \sqrt{N}.$Then: .$\displaystyle a_3 \:=\:\frac{1}{2}\left(a_2 + \frac{N}{a_2}\right)$is an even better approximation. And that is the source of that recursive function: .$\displaystyle a_{n+1} \;=\;\frac{1}{2}\left(a_n + \frac{N}{a_n}\right)$~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Actually, I used this form: .$\displaystyle a_{n+1} \;=\;\frac{a_n^2 + N}{2a_n} \$