Is the Sequence $\{x_n\}$ Defined by $7x_{n+1} = x_n^3 + 6$ Increasing?

[Dustinsfl]

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]

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...

https://www.physicsforums.com/attachments/398._xfImport

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

$\chi$ $\sigma$

 

[Dustinsfl]

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...

https://www.physicsforums.com/attachments/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 let's 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)|

 

[chisigma]

By Theorem 4.1, we can take an x let's 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 $\chi$ $\sigma$

 

[blue_raver22]

To prove that the sequence is increasing, we can use mathematical induction.
First, we can see that for $n=1$, $x_1 = \frac{1}{2}$ and $x_2 = \frac{1}{2}(\frac{1}{2})^3+6 = \frac{1}{8}+6 = \frac{49}{8}$.
So, $x_2 > x_1$ and the sequence is increasing for the base case.
Now, we assume that $x_n < x_{n+1}$ for some $n\geq 1$.
Then, we have $7x_{n+1} = x_n^3 + 6 > x_n^3$, since $x_n > 0$.
Dividing both sides by $7$, we get $x_{n+1} > \frac{1}{7}x_n^3$.
Since $x_n > 0$, we can take the cube root of both sides, giving us $x_{n+1} > \sqrt[3]{\frac{1}{7}x_n^3} = \frac{1}{\sqrt[3]{7}}x_n$.
Since $\frac{1}{\sqrt[3]{7}} < 1$, we have $x_{n+1} > x_n$, which proves that the sequence is increasing.
To find the limit of the sequence, we can take the limit of both sides of the given equation.
As $n$ approaches infinity, the left side of the equation becomes $7\lim_{n\to\infty}x_{n+1}$, and the right side becomes $\lim_{n\to\infty}x_n^3 + 6$.
Since we have proved that the sequence is increasing, we can use the Monotone Convergence Theorem to say that the limit of the sequence exists.
Let the limit be $L$. Then we have $7L = L^3 + 6$, which can be rearranged to give $L^3 - 7L + 6 = 0$.
This can be factored as $(L-1)(L-2)(L+3) = 0$, giving us three possible values for $L$: $1$, $2$, and $-3$.
To determine which of these values is