# Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

• Dec 16th 2011, 04:42 AM
buenogilabert
Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}
Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

Thanks a lot!

The whole problem is:

"Let $\displaystyle f(x)=(x^2+3)/(2x)$ for $\displaystyle x \not= 0$. Define a sequence of real numbers $\displaystyle x_n$ by

$\displaystyle x_{n+1} =f(x_n)$ for $\displaystyle n \geq 1$, $\displaystyle x_1=2$.

A) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle f(x)> \sqrt 3$.
B) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle x>f(x)$. (Hint: show $\displaystyle x-f(x)>0$.)
C) Conclude that the sequence $\displaystyle x_n$ converges."
• Dec 16th 2011, 05:25 AM
alexmahone
Re: Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}
Quote:

Originally Posted by buenogilabert
Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

Thanks a lot!

The whole problem is:

"Let $\displaystyle f(x)=(x^2+3)/(2x)$ for $\displaystyle x \not= 0$. Define a sequence of real numbers $\displaystyle x_n$ by

$\displaystyle x_{n+1} =f(x_n)$ for $\displaystyle n \geq 1$, $\displaystyle x_1=2$.

A) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle f(x)> \sqrt 3$.
B) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle x>f(x)$. (Hint: show $\displaystyle x-f(x)>0$.)
C) Conclude that the sequence $\displaystyle x_n$ converges."

A) $\displaystyle f(x)=\frac{x^2+3}{2x}$

$\displaystyle =\frac{x}{2}+\frac{3}{2x}$

$\displaystyle \ge 2\sqrt{\frac{x}{2}\cdot \frac{3}{2x}}$ (AM-GM inequality, which holds for $\displaystyle x>0$)

$\displaystyle =\sqrt{3}$

So, if $\displaystyle x>0$, then $\displaystyle f(x)\ge \sqrt{3}$.

Equality holds only when

$\displaystyle \frac{x^2+3}{2x}=\sqrt{3}$

$\displaystyle x^2+3=2\sqrt{3}x$

$\displaystyle x^2-2\sqrt{3}x+3=0$

$\displaystyle (x-\sqrt{3})^2=0$

$\displaystyle x=\sqrt{3}$

So, if $\displaystyle x>0$ and $\displaystyle x\neq\sqrt{3}$, then $\displaystyle f(x)>\sqrt{3}$.

In particular, if $\displaystyle x>\sqrt{3}$, then $\displaystyle f(x)>\sqrt{3}$.