Htykuut

Consider the sequence $x_{k+1}=frac{1}{2}(x_k+frac{a}{x_k}), agt 0, xinmathbb{R}$. Assume the sequence converges, what does it converge to?

I'm having trouble seeing how to start,

Any help would be appreciated

Thanks

The number $x$ to which it converges should satisfy $x =frac12left(x+frac a xright)$. Solve that equation for $x$. (It may help to begin by multiplying both sides by $x$, thereby getting rid of the fraction.)

If $x_n to g$, then, if $g ne 0$, we have $frac{1}{2}(x_n + frac{a}{x_n}) to frac{1}{2}(g+frac{a}{g})$. We also have $x_{n+1} to g$, so since $x_{n+1} = frac{1}{2}(x_n + frac{a}{x_n})$, we have that $g = frac{1}{2}(g + frac{a}{g})$.

This sequence has a closed form.
First, if $x_0=sqrt{a}$, $x_n=sqrt{a}$ for all $n$, and the sequence is constant. We will thus assume $x_0nesqrt{a}$ in the following.

Now, given $x_nnesqrt{a}$ for some $n$, we have
$$x_{n+1}-sqrt{a}=frac{1}{2}left(x_n+ frac{a}{x_n} right) - sqrt{a}=frac{1}{2}left( sqrt{x_n} - frac{sqrt{a}}{sqrt{x_n}}right)^2 > 0$$
Hence $x_n>sqrt{a}$ for all $n>0$. To simplify a bit, we can safely assume that $x_0 > sqrt{a}$ too.

We can thus write $x_n = sqrt{a} coth{t_n}$. Then
$$x_{n+1} = frac{1}{2}left( sqrt{a} coth{t_n} + frac{a}{sqrt{a} coth{t_n}} right) = frac{1}{2}sqrt{a} left( coth{t_n} + mathrm{th} t_nright) = sqrt{a} coth 2 t_n$$
We have thus $t_{n+1}=2 t_n$, and $t_n= 2^n t_0$, then for $x_n$ :
$$x_n=sqrt{a} coth left( 2^n arg coth frac{x_0}{sqrt{a}}right)$$
And the value inside parentheses tends to infinity, so $lim(x_n)=sqrt{a}$.