General rule for a sequence..

hi,

I am having trouble trying to find a general rule for the sequence defined as $\displaystyle a_1 =10$, $\displaystyle a_{k+1} = \frac{(a_k)^2}{a_k+1}$

I have the first n terms $\displaystyle a_1 = 10, a_2 = 4/3, a_3 = 9/4, a_4 = 16/5 .. $ but cant see a pattern..

Thanks,

Re: General rule for a sequence..

Quote:

Originally Posted by

**Oiler** hi,

I am having trouble trying to find a general rule for the sequence defined as $\displaystyle a_1 =10$, $\displaystyle a_{k+1} = \frac{(a_k)^2}{a_k+1}$

I have the first n terms $\displaystyle a_1 = 10, a_2 = 4/3, a_3 = 9/4, a_4 = 16/5 .. $ but cant see a pattern..

Thanks,

Assuming the correction I made to your LaTeX is correct those are not the terms of the sequence, but those of (other than the first which you were given):

$\displaystyle b_{k}=\frac{k^2}{k+1}$

Assuming the correction:

$\displaystyle a_2=\frac{a_1^2}{a_1+1}=\frac{100}{11}$

CB

Re: General rule for a sequence..

Quote:

Originally Posted by

**Oiler** hi,

I am having trouble trying to find a general rule for the sequence defined as $\displaystyle a_1 =10$, $\displaystyle a_{k+1} = \frac{(a_k)^2}{a_k+1}$

I have the first n terms $\displaystyle a_1 = 10, a_2 = 4/3, a_3 = 9/4, a_4 = 16/5 .. $ but cant see a pattern..

Thanks,

The difference equation...

$\displaystyle a_{n+1}= \frac{a_{n}^{2}}{1+a_{n}}\ ,\ a_{0}=10$ (1)

... is nonlinear and a direct solution is a difficult task. However, if You are interested to the 'asyntotic behaviour' of the $\displaystyle a_{n}$ then You can write (1) as...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= -\frac{a_{n}}{1+a_{n}}= f(a_{n})\ ,\ a_{0}=10$ (2)

... and follow the procedure illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html

In Your case is...

$\displaystyle f(x)= -\frac{x}{1-x}$ (3)

... and f(x) has only one 'attractive fixed point' in $\displaystyle x_{0}=0$. You can verify that for any $\displaystyle a_{0} \ne 0$ is...

$\displaystyle |f(x)|<|x_{0}-x|$ (4)

... so that the conditions of theorem 4.1 are satisfied and the $\displaystyle a_{n}$ do converge monotonically at 0...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$