Two problems regarding convergence

**Utherr** · June 26th 2010, 09:28 AM

1.

I need to prove that $(x_n)_{n\ge1}$ converges. Normally i'd know how to prove this (with Weierstrass theorem) but this one's a bit different :-s.

2.

In this problem i need to prove that $(x_n)_{n\ge1}$ converges to 2. This translates to $f_n(x_n)=x^{n}_{n}+\ln{x_n}=2^n \to x^{n}_{n}+\ln{x_n}-2^n=0$

**Also sprach Zarathustra** · June 26th 2010, 09:50 AM

I think you should "do" Newton-Raphson to $f(x)$

**Utherr** · June 26th 2010, 10:02 AM

I don't know what that is... never seen it in high-school :-s (or either i don't know it by this name)

**Also sprach Zarathustra** · June 26th 2010, 10:31 AM

Here: Newton's method - Wikipedia, the free encyclopedia

**Utherr** · June 26th 2010, 10:47 AM

i did $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ and end up with $x_{n+1}=\frac{x^{2}_{n}}{\arctan{x_n}+x^{2}_{n}\ar ctan{x_n}}$ i don't really know what to do with it

And i can't use this because:

1. i can't use theorems or methods they don't teach at high-school for my exam
2. it's hard for me to understand without the guidance of a teacher.

**chisigma** · June 26th 2010, 02:41 PM

The first difference equation can be written in the equivalent form...

$\Delta_{n} = x_{n+1} - x_{n} = x_{n}\ (\tan^{-1} x_{n} -1)= \varphi(x_{n})$ (1)

The function $\varphi(*)$ is represented here...

It has an attractive fixed point in $x=0$ and a repulsive fixed point at $x=- \frac{\pi}{4}$ so that any value $- 1 < x_{0} < \frac{\pi}{4}$ will produce a sequence converging at $x=0$ ...

Kind regards

$\chi$ $\sigma$

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