[SOLVED]Show a sequence is increasing

dwsmith

Well-known member
A real sequence $\{x_n\}$ satisfies $7x_{n + 1} = x_n^3 + 6$ for $n\geq 1$. If $x_1 = \frac{1}{2}$, prove that the sequence increases and find its limit.

To be increasing, we must have $s_n\leq s_{n + 1}$. What next? My Analysis game is weak.

chisigma

Well-known member
A real sequence $\{x_n\}$ satisfies $7x_{n + 1} = x_n^3 + 6$ for $n\geq 1$. If $x_1 = \frac{1}{2}$, prove that the sequence increases and find its limit.

To be increasing, we must have $s_n\leq s_{n + 1}$. What next? My Analysis game is weak.
The difference equation can be written as...

$\displaystyle \Delta_{n}= x_{n-1}-x_{n}= \frac{x_{n}^{3}}{7} -x_{n}+ \frac{6}{7}= f(x_{n})$ (1)

The f(x) is represented here...

There is only one attractive fixed point in x=1 and any initial value $-3 < x_{0} < 2$ will generate a sequence monotonically convergent to 1, increasing for $x_{0}<1$, decreasing for $x_{0}>1$. The initial values $x_{0}<-3$ and $x_{0}>2$ will generate diverging sequences. The solving procedure is descrpted in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

Kind regards

$\chi$ $\sigma$

dwsmith

Well-known member
The difference equation can be written as...

$\displaystyle \Delta_{n}= x_{n-1}-x_{n}= \frac{x_{n}^{3}}{7} -x_{n}+ \frac{6}{7}= f(x_{n})$ (1)

The f(x) is represented here...

View attachment 398

There is only one attractive fixed point in x=1 and any initial value $-3 < x_{0} < 2$ will generate a sequence monotonically convergent to 1, increasing for $x_{0}<1$, decreasing for $x_{0}>1$. The initial values $x_{0}<-3$ and $x_{0}>2$ will generate diverging sequences. The solving procedure is descrpted in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

Kind regards

$\chi$ $\sigma$
By Theorem 4.1, we can take an x lets say equal to 1/2.
Now, 1/2 is in the prescribed range.
$$|f(x)| = |x^3 - 7x+6| < |x_0-x| = |1-x|$$
Plugging in 1/2, we have
$$\frac{21}{8}>\frac{1}{2}$$
which is not less than. So for it to be strictly increasing, we need it to be less than. How is it strictly increasing then based on that Theorem?

I believe Theorem 4.1 should say |f'(x)| not |f(x)|

Last edited:

chisigma

Well-known member
By Theorem 4.1, we can take an x lets say equal to 1/2.
Now, 1/2 is in the prescribed range.
$$|f(x)| = |x^3 - 7x+6| < |x_0-x| = |1-x|$$
Plugging in 1/2, we have
$$\frac{21}{8}>\frac{1}{2}$$
which is not less than. So for it to be strictly increasing, we need it to be less than. How is it strictly increasing then based on that Theorem?

I believe Theorem 4.1 should say |f'(x)| not |f(x)|
The 'true' f(*) is $\displaystyle \frac{x^{3}-7\ x +6}{7}$ and not $\displaystyle x^{3}-7\ x +6$...

Kind regards

$\chi$ $\sigma$