# Can I use an analogus function to show that the sequence is increasing or decreasing?

1. Sep 23, 2004

Folks,

This is the solution I have for a problem in my textbook regarding sequences. I just need to know whether I have the right idea in mind.

Thank you very much!

We can use an analogus function to show that the sequence given by

$$a_{n+1} = \sqrt{2+a_n} \quad a_1 = \sqrt{2}$$

is increasing. Here it goes

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

$$\frac{dy}{dx}=\frac{1}{2\sqrt{2+x}}>0 \Longrightarrow a_{n+1}>a_n$$

Last edited: Sep 23, 2004
2. Sep 23, 2004

### phoenixthoth

You'd have to prove not that the function itself is increasing but that the nth iterate of x is less than the n+1st iterate of x. IF f is increasing then x<f(x) will to the job, I think, because then you could take f of both sides n times to get f^n(x)<f^n+1(x).

a_n=f^n(x).

3. Sep 23, 2004

### vsage

Ok I gave it some thought and I really can't give you a definitive answer but I can ask you why you think that your analogous function is analogous? The question posed in the book will have a_infinity = 2 and in fact the function converges to the function y^2 - y - 2 = 0, whereas y = sqrt(2+x) goes well beyond 2. It just doesn't add up too well to me. I'll delete this if someone gives a better response (or maybe my incorrectness will compell someone to answer correctly)

4. Sep 23, 2004

Guys,

I've just found out it is better to use mathematical induction or graphical methods instead. Thanks for the help.

5. Sep 24, 2004

### Pyrrhus

I've used Differential Calculus to find out when does it increase or decrease a sequence with an anologous function, and then just check for n>= 1 values. I see no problem using this method as long as you understand what you're doing.

6. Sep 28, 2004